Question 13 Marks
If $\text{y}=\sqrt{\cos\text{x}+\sqrt{\cos\text{x}+\sqrt{\cos\text{x}+\ .... \text{to }\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{x}}{1-2\text{y}}$
AnswerHere,
$\text{y}=\sqrt{\cos\text{x}+\sqrt{\cos\text{x}+\sqrt{\cos\text{x}+\ .... \text{to }\infty}}}$
$\text{y}=\sqrt{\cos\text{x}+\text{y}}$
Squaring both the sides,
$\text{y}^2=\cos\text{x}+\text{y}$
Differentiating it with respect to x,
$2\text{y}\frac{\text{dy}}{\text{dx}}=-\sin\text{x}+\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}(2\text{y}-1)=-\sin\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{-\sin\text{x}}{(2\text{y}-1)}$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{x}}{1-2\text{y}}$
View full question & answer→Question 23 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$
AnswerGiven,
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$
Differentiating with resepct to x,
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}\Big)=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{\text{a}^2}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}^2}{\text{b}^2}\Big)=0$
$\Rightarrow\frac{1}{\text{a}^2}(2{\text{x}})+\frac{1}{\text{b}^2}(2\text{y})\frac{\text{d}}{\text{dx}}=0$
$\Rightarrow\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=-\frac{2{\text{x}}}{\text{a}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\Big(\frac{2{\text{x}}}{\text{a}^2}\Big)\Big(\frac{\text{b}^2}{2\text{y}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{{\text{b}^2\text{x}}}{\text{a}^2\text{y}}$
View full question & answer→Question 33 Marks
If $\text{x}=\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$ and $\frac{\text{d}^2\text{x}}{\text{dt}^2}=\lambda\text{x}$ then find the value of $\lambda.$
Answerwe have $\text{x}=\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=-\text{a}\sin(\text{nt})\times\text{n}-\text{bn}\cos)\text{nt}$$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\text{an}^2\cos(\text{nt})+\text{bn}^2\sin(\text{nt})$
since, $\frac{\text{d}^2\text{y}}{\text{dt}^2}=\lambda\text{x}$
$\Rightarrow-\text{an}^2\cos(\text{nt})+\text{bn}^2\sin(\text{nt})=\lambda(\text{a}\cos\text{nt}-\text{b}\sin\text{nt})$
$\Rightarrow\lambda=\text{n}^2$
View full question & answer→Question 43 Marks
Differentiate the following functions with respect to x:
$3^{\text{x}^2+2\text{x}}$
AnswerConsider $\text{y}=3^{\text{x}^2+2\text{x}}$
Differentiate it with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(3^{\text{x}^2+2\text{x}}\Big)$
$=3^{\text{x}^2+2\text{x}}\times\log3\frac{\text{d}}{\text{dx}}(\text{x}^2+2\text{x})$
[Using chain rule]
$=(2\text{x}+2)\log3\times3^{\text{x}^2+2\text{x}}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(3\text{x}^2+2\text{x}\big)=(2\text{x}+2)\log3\times3^{\text{x}^2+2\text{x}}$
View full question & answer→Question 53 Marks
If f(x) is an even function, then write whether f'(x) is even of odd.
AnswerHere,
f(x) is even function, so
f(-x) = f(x)
Differentiating it with respect to x,
$\frac{\text{d}}{\text{dx}}(\text{f}(-\text{x}))=\frac{\text{d}}{\text{dx}}(\text{f}(\text{x}))$
$\text{f}'(-\text{x})\frac{\text{d}}{\text{dx}}(-\text{x})=\text{f}'\text{(x)}$
$\text{f}'(-\text{x})\times(-1)=\text{f}'(\text{x})$
$-\text{f}'(-\text{x})=\text{f}'(\text{x})$
$\text{f}'(-\text{x})=-\text{f}'(\text{x})$
So,
f'(x) is odd function.
View full question & answer→Question 63 Marks
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\text{a}(\cos\theta+\theta\sin\theta),\text{y}=\text{a}(\sin\theta-\theta\cos\theta)$
AnswerThe given equations are $\text{x}=\text{a}(\cos\theta+\theta\sin\theta)\text{ and y}=\text{a}(\sin\theta-\theta\cos\theta)$
Then, $\frac{\text{dx}}{\text{d}\theta}= \text{a}\Big[\frac{\text{d}}{\text{d}\theta}\cos\theta+\frac{\text{d}}{\text{d}\theta}(\theta\sin\theta)\Big]$ $=\text{a}\Big[-\sin\theta+\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta)+\sin\theta\frac{\text{d}}{\text{d}\theta}(\theta)\Big]$
$=\text{a}[-\sin\theta+\theta\cos\theta+\sin\theta]=\text{a}\theta\cos\theta$
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\frac{\text{d}}{\text{d}\theta}(\sin\theta)-\frac{\text{d}}{\text{d}\theta}(\theta\cos\theta)\Big]$ $=\text{a}\Big[\cos\theta-\Big\{\theta\frac{\text{d}}{\text{d}\theta}(\cos\theta)+\cos\theta.\frac{\text{d}}{\text{d}\theta}(\theta)\Big\}\Big]$
$=\text{a}[\cos\theta+\theta\sin\theta-\cos\theta]$
$=\text{a}\theta\sin\theta$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{d}\theta}\Big)}{\Big(\frac{\text{dx}}{\text{d}\theta}\Big)}=\frac{\text{a}\theta\sin\theta}{\text{a}\theta\cos\theta}=\tan\theta$
View full question & answer→Question 73 Marks
If $\text{y}=\text{x}+\tan\text{x},$ show that $\cos^2\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{y}+2\text{x}=0$
Answer$\text{y}=\text{x}+\tan\text{x},$
differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1+\sec^2\text{x}$
differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=0+2\sec^2\times\tan\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\sin\text{x}}{\cos^3\text{x}}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\tan\text{x}+2\text{x}-2\text{x}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}=2(\text{x}+\tan\text{x})-2\text{x}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{y}-2\text{x}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{y}+2\text{x}=0$
View full question & answer→Question 83 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = 2x^2 - 3x + 1 on [1, 3]$
AnswerHere,$f(x) = 2x^2 - 3x + 1$ on $[1, 3]$
We know that a polynomial function is continuous and differentiable.
So, f(x) is continuous in [1, 3] and f(x) differentiable in (1, 3).
So, Lagrange's mean value theorem is applicable.
So, there must exist at least one real number $\text{c}\in(1,3)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(-1)}{3-1}$
$\Rightarrow4\text{c}-3=\frac{(2(3)^2-3(3)+1)-(2-3+1)}{3-1}$
$\Rightarrow4\text{c}-3=\frac{10}{2}$
$\Rightarrow4\text{c}=5+3$
$\Rightarrow4\text{c}=8$
$\Rightarrow\text{c}=2\in(1,3)$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 93 Marks
If $\text{y}=\text{x}\sin\text{y},$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\sin^2\text{y}}{(1-\text{x}\cos\text{y})}$
AnswerHere,
$\text{y}=\text{x}\sin\text{y}$
Differentiate with respect to x,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{y})+\sin\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
[Using product rule]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\cos\frac{\text{dy}}{\text{dx}}+\sin\text{y}(1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(1-\text{x}\cos\text{y})=\sin\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{y}}{1-\text{x}\cos\text{y}}$
View full question & answer→Question 103 Marks
Find the value of k in this question, so that the function f is continuous at the indicated point:
$\text{f(x)}=\begin{cases}3\text{x}-8,&\text{if x}\leq5\\2\text{k},&\text{if x}>5\end{cases}$ at x = 5.
AnswerWe have, $\text{f(x)}=\begin{cases}3\text{x}-8,&\text{if x}\leq5\\2\text{k},&\text{if x}>5\end{cases}$ at x = 5.
Since, f(x) is continuous at x = 5.
$\therefore$ L.H.L = R.H.L = f(5)
Now, $\text{L.H.L}=\lim\limits_{\text{h}\rightarrow5^-}(3\text{x}-8)=\lim\limits_{\text{h}\rightarrow0}[(5-\text{h})-8]$
$=\lim\limits_{\text{h}\rightarrow0}\ [15-3\text{h}-8]=7$
$\text{R.H.L}=\lim\limits_{\text{h}\rightarrow5^+}2\text{k}=\lim\limits_{\text{h}\rightarrow0}2\text{k}=2\text{x}=7$ $[\because\ \text{L.H.L}=\text{R.H.L}]$
And f(5) = 3 × 5 - 8 = 7
$2\text{k}=7\Rightarrow\ \text{k}=\frac{7}{2}$
View full question & answer→Question 113 Marks
Differentiate the following w.r.t. x:
$(\sin\text{x})^{\cos\text{x}}$
AnswerLet $\text{y}=(\sin\text{x})^{\cos\text{x}}$
$\Rightarrow\ \log\text{y}=\log(\sin\text{x})^{\cos\text{x}}=\cos\text{x}\log(\sin\text{x})$
Differentiate both sides w.r.t.x, we get
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{d}}{\text{dx}}=\cos\text{x}\cdot\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x }\frac{\text{d}}{\text{dx}}(\cos\text{x})$
$=\cos\text{x}\cdot\frac{1}{\sin\text{x}}\cdot\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\cdot(-\sin\text{x})$
$=\cos\text{x}\cdot\cos\text{x}-\log(\sin\text{x})\cdot\sin\text{x}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{y}\big[\cot\text{x}\cos\text{x}-\sin\text{x}\cdot\log(\sin\text{x})\big]$
$=(\sin\text{x})^{\cos\text{x}}\big[\cot\text{x}\cos\text{x}-\sin\text{x}\cdot\log(\sin\text{x})\big]$
View full question & answer→Question 123 Marks
If $e^{x+y} - x = 0$, prove that $\frac{\text{dy}}{\text{dx}}=\frac{1-\text{x}}{\text{x}}$
AnswerHere,
$e^{x+y} - x = 0$
$e^{x+y} = x$ .....(i)
Differentiating it with respect to x using chain rule,
$\frac{\text{d}}{\text{dx}}\big(\text{x}^{\text{x}+\text{y}}\big)=\frac{\text{d}}{\text{dx}}(\text{x})$
$\text{e}^{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=1$
$\text{x}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=1$
[Using euqation (i)]
$1+\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}-1$
$\frac{\text{dy}}{\text{dx}}=\frac{1-\text{x}}{\text{x}}$
View full question & answer→Question 133 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big),\text{x}\in\text{R}$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
$\Rightarrow\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
$\Big[\text{Since},\sec^{-1}\text{x}=\cos^{-1}\big(\frac{1}{\text{x}}\big)\Big]$
$\Rightarrow \text{y}=\frac{\pi}{2}\Big[\text{Since}, \sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
Differentiate it with respect to x,
$\therefore \frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 143 Marks
If $y = x^x$, find $\frac{\text{dy}}{\text{dx}}\text{at x}=\text{e}$
AnswerWe have, $y = x^x$ .....(i)
Taking log on both sides,
$\log\text{y}=\log\text{x}^\text{x}$
$\Rightarrow\log\text{y}=\text{x}\log\text{x}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1+\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}(1+\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})$
[Using equation (i)]
Putting x = e, we get,
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{e}(1+\log_\text{e}\text{e})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^\text{e}(1+1)\big[\because\log_\text{e}\text{e}=1\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\text{e}^\text{e}$
View full question & answer→Question 153 Marks
If $\text{y}=\cos^{-1}\text{x},$ Find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ in terms of y alone.
AnswerHere,
$\text{y}=\cos^{-1}\text{x},$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-2\text{x}}{2\sqrt{1-\text{x}^2}^\frac{3}{2}}=\frac{-\text{x}}{(1-\text{x}^2)}$
Now,
$\text{y}=\cos^{-1}\text{x}$
$\Rightarrow\text{x}=\cos\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\cos\text{y}}{(1-\cos^2\text{y})^\frac{3}{2}}=-\frac{\cos\text{y}}{(\sin^2\text{y})^\frac{3}{2}}=-\cot\text{y}\ \text{cosec}^2\text{y}$
View full question & answer→Question 163 Marks
$\text{If y}=(\tan ^1 \text{x})^2,\text{show that }(\text{x}^2+1)^2 \ \text{y}_2+ 2\text{x}(\text{x}^2+1)_\text{y}=2$
Answer$\text{y}=(\tan^{-1}\text{x})^2\ \dots(1)$ $\therefore\ \frac{\text{dy}}{\text{dx}}=2 \tan ^{-1}\text{x}.\frac{1}{1+\text{x}^2}$ $\Rightarrow \ (1+ \text{x}^2) \frac{\text{dy}}{\text{dx}} =2 \tan^{-1} \text{x}$ $\Rightarrow\ (1+\text{x}^2)^2\bigg(\frac{\text{dy}}{\text{dx}}\bigg)^2=4 (\tan ^{-1}\text{x})^2$ $\Rightarrow\ (1+\text{x}^2)^2 \bigg(\frac{\text{dy}}{\text{dx}}\bigg)^2= 4\text{y}\ \ [\because\text{of } (1)]$ Differenitiating both sides w.r.t.x, we get,$(1 +\text{x}^2)^2.2 \frac{\text{dy}}{\text{dx}}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\bigg(\frac{\text{dy}}{\text{dx}}\bigg)^2.2(1+\text{x}^2).2 \text{x}=4 \frac{\text{dy}}{\text{dx}}$
Divide both sides by $2 \frac{\text{dy}}{\text{dx}},$ we get, $(1+\text{x}^2)^2.\frac{\text{d}^2\text{y}}{\text{dx}^2}+2 \text{x}(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}=2$ $\text{Or}\ \ (\text{x}{^2}+1)\text{y}_2+2 \text{x}(\text{x}^2+1)\text{y}_1=2$
View full question & answer→Question 173 Marks
Find $\frac{\text{dy}}{\text{dx}}$ when x and y are connected by the relation:
$\big(\text{x}^2+\text{y}^2\big)^2=\text{xy}$
AnswerWe have, $\big(\text{x}^2+\text{y}^2\big)^2=\text{xy}$
On differentiating both sides w.r.t. x, we get
$\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{y}^2\big)^2=\frac{\text{d}}{\text{dx}}(\text{xy})$
$\Rightarrow\ 2\big(\text{x}^2+\text{y}^2\big)^2\cdot\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2)=\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{y}+\text{y}\cdot\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ 2(\text{x}^2+\text{y}^2)\cdot\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dy}}\Big)=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}$
$\Rightarrow\ 2\text{x}^2\cdot2\text{x}+2\text{x}^2\cdot2\text{y}\frac{\text{dy}}{\text{dx}}+2\text{y}^2\cdot2\text{x}+2\text{y}^2\cdot2\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\big[4\text{x}^2\text{y}+4\text{y}^3-\text{x}\big]=\text{y}-4\text{x}^3-4\text{xy}^2$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{(\text{y}-4\text{x}^3-4\text{xy}^2)}{(4\text{x}^2\text{y}+4\text{y}^3-\text{x})}$
View full question & answer→Question 183 Marks
If $\text{x}=\text{a}(\cos2\text{t}+2\text{t}\sin2\text{t})\ \text{and}\ \text{y}=\text{a}(\sin2\text{t}-2\text{t}\cos2\text{t}),$ then find $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$
Answer$\text{x}=\text{a}(\cos2\text{t}+2\text{t}\sin2\text{t})$
$\frac{\text{dx}}{\text{dt}}=-2\text{a}\sin2\text{t}+2\text{a}\sin2\text{t}+4\text{at}\cos2\text{t}=4\text{at}\cos2\text{t}$
$\text{y}=\text{a}(\sin2\text{t}-2\text{t}\cos2\text{t})$
$\frac{\text{dy}}{\text{dt}}=2\text{a}\cos2\text{t}-2\text{a}\cos2\text{t}+4\text{at}\sin2\text{t}=4\text{at}\sin2\text{t}$
$\frac{\text{dy}}{\text{dx}}=\tan2\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}(\tan2\text{t})$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sec^22\text{t}\frac{\text{d}}{\text{dx}}(\text{t})$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sec^22\text{t}\times\frac{1}{4\text{at}\cos2\text{t}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{2\text{a}}\sec^32\text{t}$
View full question & answer→Question 193 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = x^2 - 3x + 2$ on $[-1, 2]$
AnswerWe have $f(x) = x^2 - 3x + 2$ Since a polynomial function is everywhere continuous and differentiable.
Therefore, $f(x)$ is continuous on $-1, 2$ and differentiable on $-1, 2.$
Thus, both conditions of Lagrange's mean value theorem are satisfied. So, there must exist at least one real number $\text{c}\in-1,2$
such that $\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(-1)}{2+1}$
$\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(-1)}{3}$
Now, $f(x) = x^2 - 3x + 2$
$\Rightarrow f'(x) = 2x - 3$
$\Rightarrow f(2) = 0$
$\Rightarrow f(-1) = (-1)^2 - 3(-1) + 2$
$\Rightarrow f(-1) = 6$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(2)-\text{f}(-1)}{3}$
$\Rightarrow2\text{x}-3=-2$
$\Rightarrow2\text{x}-1=0$
$\Rightarrow\text{x}=\frac{1}{2}$
Thus, $\text{c}=\frac{1}{2}\in(1,2)$
such that $\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(-1)}{2-(-1)}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 203 Marks
Determine the value of the constant k so that the function $\text{f(x)}=\begin{cases}\text{kx}^2,&\text{if }\text{ x}\leq2\\3,&\text{if }\text{ x}>2\end{cases}$ is continuous at x = 2.
AnswerGiven, $\text{f(x)}=\begin{cases}\text{kx}^2,&\text{if }\text{ x}\leq2\\3,&\text{if }\text{ x}>2\end{cases}$ If f(x) is continuous at x = 2, then $\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}=\text{f}(2)\ ...(\text{i})$ Now, $\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\text{k}(2-\text{h})^2=4\text{k}$ And, f(2) = 3 From (i) we have, $4\text{k}=3$$\Rightarrow\text{k}=\frac{3}{4}$
View full question & answer→Question 213 Marks
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\sin^{2}\text{y}+\cos\text{xy}=\pi$
AnswerThe given relationship is $\sin^{2}\text{y}+\cos\text{xy}=\pi$
differenting this relationship with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\sin^{2}\text{y}+\cos\text{xy})=\frac{\text{d}}{\text{dx}}(\pi)$
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin^{2}\text{y})+\frac{\text{d}}{\text{dx}}(\cos\text{xy})=0\ ...(\text{i})$
Using chain rule, we obtain
$\frac{\text{d}}{\text{dx}}(\sin^{2}\text{y})= 2\sin\text{y}\frac{\text{d}}{\text{dx}}(\sin\text{y})=2\sin\text{y}\cos\text{y}\frac{\text{dy}}{\text{dx}} ...\text{(ii)}$
$\frac{\text{d}}{\text{dx}}(\cos\text{xy})=-\sin\text{xy}\frac{\text{d}}{\text{dx}}(\text{xy})=-\sin\text{xy}\Big[\text{y}\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{dy}}{\text{dx}}\Big]$
$=-\sin\text{xy}\Big[\text{y}.1+\text{x}\frac{\text{dy}}{\text{dx}}\Big]= -\text{y}\sin\text{xy}-\text{x}\sin\text{xy}\frac{\text{dy}}{\text{dy}} ...(\text{iii})$
From (1), (2) and (3), we obtain
$2\sin\text{y}\cos\text{y}\frac{\text{dy}}{\text{dx}}-\text{y}\sin\text{xy}-\text{x}\sin\text{xy}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow(2\sin\text{y}\cos\text{y}-\text{x}\sin\text{xy})\frac{\text{dy}}{\text{dx}}=\text{y}\sin\text{xy}$
View full question & answer→Question 223 Marks
Differentiate the following w.r.t. x:
$\sin^{-1}\Big(\frac{1}{\sqrt{\text{x}+1}}\Big)$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{1}{\sqrt{\text{x}+1}}\Big)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin^{-1}\Big(\frac{1}{\sqrt{\text{x}+1}}\Big)$
$=\frac{1}{\sqrt{-1\Big(\frac{1}{\sqrt{\text{x+1}}}\Big)^2}}\cdot\frac{\text{d}}{\text{dx}}\frac{1}{(\text{x}+1)^{\frac{1}{2}}}$ $\Big[\because\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})=\frac{1}{\sqrt{1-\text{x}^2}}\Big]$
$=\frac{1}{\sqrt{\frac{\text{x}+1-1}{\text{x}+1}}}\cdot\frac{\text{d}}{\text{dx}}(\text{x+1})^{\frac{-1}{2}}$
$=\sqrt{\frac{\text{x}+1}{\text{x}}}\cdot\frac{-1}{2}(\text{x}+1)^{\frac{1}{2}-1}\cdot\frac{\text{d}}{\text{dx}}(\text{x+1})$
$=\frac{(\text{x}+1)^{\frac{1}{2}}}{\text{x}^{\frac{1}{2}}}\cdot\Big(-\frac{1}{2}\Big)(\text{x}+1)^{-\frac{3}{2}}$
$=\frac{-1}{2\sqrt{\text{x}}}\cdot\Big(\frac{1}{\text{x}+1}\Big)$
View full question & answer→Question 233 Marks
Differentiate the following functions with respect to x:
$\cos(\log\text{ x})^2$
AnswerConsider $\text{y}=\cos(\log\text{ x})^2$
Differentiate it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\cos(\log\text{ x})^2$
$=-\sin(\log\text{x})^2\frac{\text{d}}{\text{dx}}(\log\text{ x})^2$
$=-\sin(\log\text{x})^2\frac{2\log\text{x}}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{-2\log\text{x}\sin(\log\text{x})^2}{\text{x}}$
So, The solution is $\frac{\text{dy}}{\text{dx}}=\frac{-2\log\text{x}\sin(\log\text{x})^2}{\text{x}}$
View full question & answer→Question 243 Marks
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{x}^{3}+\text{x}^{2}\text{y} +\text{x} \text{y}^{2}+\text{y}^{3} = 81$
AnswerThe given relationship is $\text{x}^{3}+\text{x}^{2}\text{y} +\text{x} \text{y}^{2}+\text{y}^{3} = 81$ differenting this relationship with respect to x, we obtain $\frac{\text{d}}{\text{dx}}(\text{x}^{3}+\text{x}^{2}\text{y} +\text{x} \text{y}^{2}+\text{y}^{3}) =\frac{\text{d}}{\text{dx}}(81)$ $\Rightarrow\frac{\text{d}}{\text{dx}}(\text{x}^{3})+\frac{\text{d}}{\text{dx}}(\text{x}^{2}\text{y})+\frac{\text{d}}{\text{dx}}(\text{x}\text{y}^{2})+\frac{\text{d}}{\text{dx}}\text{(y}^{3}) =0$ $\Rightarrow 3\text{x}^{2}+\Big[\text{y}\frac{\text{d}}{\text{dx}}(\text{x}^{2})+\text{x}^{2}\frac{\text{dy}}{\text{dx}}\Big] +\Big[\text{y}^{2}\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{dy}}{\text{dx}}(\text{y}^{2})\Big]+3\text{y}^{2}\frac{\text{dy}}{\text{dx}}=0$$\Rightarrow3\text{x}^{2}+\Big[\text{y}.2\text{x}+\text{x}^{2}\frac{\text{dy}}{\text{dx}}\Big]+\Big[\text{y}^{2}.1+\text{x}.2\text{y}.\frac{\text{dy}}{\text{dx}}\Big]+3\text{y}^{2}\frac{\text{dy}}{\text{dx}}= 0$
$\Rightarrow(\text{x}^{2} + 2\text{xy}+3\text{y}^{2})\frac{\text{dy}}{\text{dx}}+(3\text{x}^{2}+2\text{xy}+\text{y}^{2})= 0$$\therefore\frac{\text{dy}}{\text{dx}}= \frac{-(3\text{x}^{2}+2\text{xy}+\text{y}^{2})}{(\text{x}^{2}+2\text{xy}+3\text{y}^{2})}$
View full question & answer→Question 253 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\text{kx}+5,&\text{if }\text{ x}\leq2\\\text{x}-1,&\text{if }\text{ x}>2\end{cases}$
Answer$\text{f(x)}=\begin{cases}\text{kx}+5,&\text{if }\text{ x}\leq2\\\text{x}-1,&\text{if }\text{ x}>2\end{cases}$
It is given that the function is continuous
$\therefore\ \text{LHL}=\text{RHL}=\text{f}(2)\ ...(\text{i})$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow2^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{k}(2-\text{h})+5=2\text{k}+5$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow2^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})=\lim_\limits{\text{h}\rightarrow0}(2+\text{h})-1=1$
Thus, using (i) we get
$2\text{k}+5=1$
$\text{k}=-2$
View full question & answer→Question 263 Marks
Find the second order derivatives of the following functions:
$\text{y}=\log(\log\text{x})$
AnswerWe have,
$\text{y}=\log(\log\text{x})$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\log\text{x}}\times\frac{1}{\text{x}}=\frac{1}{\text{x}\log\text{x}}$
Differentiating w.r.t.x, we get
$ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{0-(\log\text{x}+1)}{(\text{x}\log\text{x})^2}=-\frac{(1+\log\text{x})}{(\text{x}\log\text{x})^2}$
View full question & answer→Question 273 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{kx}}{\text{x}^2},&\text{if}\text{ x}\neq0\\8,&\text{if}\text{ x}=0\end{cases}\text{at x}=0$
AnswerGiven,
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{kx}}{\text{x}^2},&\text{if}\text{ x}\neq0\\8,&\text{if}\text{ x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow 0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{1-\cos2\text{kx}}{\text{x}^2}=8$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{2\text{k}^2\sin^2\text{kx}}{\text{k}^2\text{x}^2}=8$
$\Rightarrow2\text{k}^2\lim_\limits{\text{x}\rightarrow 0}\Big(\frac{\sin\text{kx}}{\text{kx}}\Big)^2=8$
$\Rightarrow2\text{k}^2\times1=8$
$\Rightarrow\text{k}^2=4$
$\Rightarrow\text{k}=\pm2$
View full question & answer→Question 283 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big)$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big)$
Put $\text{x}=\cot\theta$
$\text{y}=\sin^{-1}\Big(\frac{1}{\sqrt{1+\cot^2\theta}}\Big)$
$=\sin^{-1}\Big(\frac{1}{\sqrt{\text{cosec}^2\theta}}\Big)$
$=\sin^{-1}(\sin\theta)$
$=\theta$
$\text{y}=\cot^{-1}\text{x}\ [\text{Since}, \cot\theta=\text{x}]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{(1+\text{x}^2)}$
View full question & answer→Question 293 Marks
Differentiate the following functions with respect to x:
$10^{(10^\text{x})}$
AnswerLet $\text{y}=10^{(10^\text{x})}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log_\text{e}10^{(10^\text{x})}$
$\log\text{y}=10^{\text{x}}\log_\text{e}10$
Differentiating with respect to x,
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log_\text{e}10\times10^\text{x}\log_\text{e}10$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=10^\text{x}\times(\log_\text{e}10)^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[10^{\text{x}}\times(\log_\text{e}10)^2\big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=10^{(10\text{x})}\times10^\text{x}\times(\log_\text{e}10)^2$
[Using equation (i)]
View full question & answer→Question 303 Marks
Find which of the function:
$\text{f(x)}=|\text{x}|+|\text{x}-1|\text{ at x}=1$
AnswerThe function f will be continuous at x = a, if $=\lim\limits_{\text{h}\rightarrow\text{a}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow\text{a}^+}\text{f(x)}=\text{f(a)}.$
Consider, $\text{f(x)}=|\text{x}|+|\text{x}-1|\text{ at x}=1$
At x = 1, $\text{L.H.L}=\lim\limits_{\text{h}\rightarrow1^-}\big[|\text{x}|+|\text{x}-1|\big]$
$=\lim\limits_{\text{h}\rightarrow0}\big[|1-\text{h}|+|1-\text{h}-1|\big]=1+0=1$
At x = 1, $\text{R.H.L}=\lim\limits_{\text{h}\rightarrow1^-}\big[|\text{x}|+|\text{x}-1|\big]$
$=\lim\limits_{\text{h}\rightarrow0}\big[|1+\text{h}|+|1+\text{h}-1|\big]=1+0=1$
$\text{f}(1)=|1|+|0|=1$
Since, L.H.L = R.H.L = f(1)
Hence, f(x) is continuous at x = 1.
View full question & answer→Question 313 Marks
If $f(0) = f(1) = 0, f'(1) = 1$ and $y = f(e^x) e^{f(x)}$, write the value of $\frac{\text{dy}}{\text{dx}}\text{ at x} = 0.$
AnswerHere,
$f(0) = f(1) = 0, f'(1) = 2$
And, $y = f(e^x)d^{f(x)}$
Differentiating ti with respect to x using product rule, chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{f}(\text{e}^\text{x})\times\text{e}^{\text{f(x)}}\big]$
$=\text{f}(\text{e}^\text{x})\frac{\text{d}}{\text{dx}}\text{e}^{\text{f(x)}}+\text{e}^{\text{f(x)}}\frac{\text{d}}{\text{dx}}\text{f}(\text{e}^\text{x})$
$=\text{f}(\text{e}^\text{x})\text{e}^{\text{f(x)}}\frac{\text{d}}{\text{dx}}\text{f(x)}+\text{e}^{\text{f(x)}}\times\text{f}'(\text{e}^\text{x})\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})$
$=\text{f}(\text{e}^\text{x})\times\text{e}^{\text{f(x)}}\times\text{f}'\text{(x)}+\text{e}^{\text{f(x)}}+\text{f}'(\text{e}^\text{x})\times\text{e}^\text{x}$
Put x = 0
$=\text{f}(\text{e}^0)\text{e}^{\text{f}(0)}\text{f}'(0)+\text{e}^{\text{f}(0)}\text{f}^{1}(\text{e}^0)\times\text{e}^0$
$=\text{f}(1)\text{e}^{\text{f}(0)}\times\text{f}'(0)+\text{e}^{\text{f}(0)}\times\text{f}'(1)\times1$
$=0\times\text{e}^0\times\text{f}'(0)+\text{e}^02\times1$
$\big[\text{Since},\text{f}(0)=\text{f}(1)=0,\text{f}'(1)=2\big]$
$=0+1\times2\times1$
$=2$
So,
$\frac{\text{dy}}{\text{dx}}=2$
View full question & answer→Question 323 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}\text{x}^2,&\text{x}\geq1\\4,&\text{x}<1\end{cases}\text{at x} =1$
AnswerGiven,
$\text{f(x)}=\begin{cases}\text{k}\text{x}^2,&\text{x}\geq1\\4,&\text{x}<1\end{cases}$
We have,
$(\text{LHL at x}= 1)=\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})$
$\lim_\limits{\text{h}\rightarrow0}4=4$
$(\text{RHL at x}= 1)=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$\lim_\limits{\text{h}\rightarrow0}\text{k}(1+\text{h})^2=\text{k}$
If f(x) is continuous at x = 1, then
$\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}$
$\Rightarrow\text{k}=4$
View full question & answer→Question 333 Marks
If $\text{y}=\log_\text{a}\text{x},$, find $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\log_\text{a}\text{x},$
$\Rightarrow\text{y}=\frac{\log\text{x}}{\log\text{a}} \Big[\because\log_\text{a}\text{b}=\frac{\log\text{b}}{\log\text{a}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\log\text{a}}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\log\text{a}}\Big(\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}\log\text{a}}$
View full question & answer→Question 343 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$\text{f}(\text{x})=\sqrt{\text{x}^2-4}\text{ on }[2,4]$
AnswerWe have,$\text{f}(\text{x})=\sqrt{\text{x}^2-4}$
Here, f(x) will exist,
if
$\text{x}^2-4\geq0$
$\Rightarrow\text{x}\leq-2\text{ or }\text{x}\geq2$
Since, for each $\text{x}\in2,4,$ the function f(x) attains a unique definite value.
So, f(x) is continuous on 2, 4
Also,
$\text{f}'(\text{x})=\frac{1}{2\sqrt{\text{x}^2-4}}(2\text{x})=\frac{\text{x}}{\sqrt{\text{x}^2-4}}$
Exists for all $\text{x}\in2,4$
So, f(x) is differentiable on 2, 4.
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists some $\text{c}\in2,4$ such that
$\text{f}'(\text{x})=\frac{1}{2\sqrt{\text{x}^2-4}}(2\text{x})=\frac{\text{x}}{\sqrt{\text{x}^2-4}}$
Now,
$\text{f}(\text{x})=\sqrt{\text{x}^2-4}$
$\text{f}'(\text{x})=\frac{1}{\sqrt{\text{x}^2-4}},\text{f}(4)=2\sqrt3,\text{f}(2)=0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(2)}{4-2}$
$\Rightarrow\frac{\text{x}}{\sqrt{\text{x}^2-4}}=\frac{2\sqrt3}{2}$
$\Rightarrow\frac{\text{x}}{\sqrt{\text{x}^2-4}}=\sqrt3$
$\Rightarrow\frac{\text{x}^2}{\text{x}^2-4}=3$
$\Rightarrow\text{x}^2=3\text{x}^2-12$
$\Rightarrow\text{x}^2=6$
$\Rightarrow\text{x}=\pm\sqrt6$
Thus, $\text{c}=\sqrt6\in(2,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(2)}{4-2}$
Hence, Lagrange's theorem is verified.
View full question & answer→Question 353 Marks
If $\text{y}=\log|3\text{x}|,\text{x}\neq0,$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\log|3\text{x}|$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\log|3\text{x}|)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{3\text{x}}\frac{\text{d}}{\text{dx}}(3\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{3\text{x}}(3)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
View full question & answer→Question 363 Marks
Find which of the function:
$\text{f(x)}=\begin{cases}\frac{|\text{x}-4|}{2(\text{x}-4)},&\text{if x}\neq4\\0,&\text{if x}=4\end{cases}$
at x = 4
AnswerThe condition for function f to be a continuous at x = a is given by $=\lim\limits_{\text{x}\rightarrow\text{a}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\text{a}^+}\text{f(x)}=\text{f(a)}$
Consider, $\text{f(x)}=\begin{cases}\frac{|\text{x}-4|}{2(\text{x}-4)},&\text{if x}\neq4\\0,&\text{if x}=4\end{cases}$ at x = 4.
At x = 4, $\text{L.H.L}=\lim\limits_{\text{x}\rightarrow4^-}\frac{|\text{x}-4|}{2(\text{x}-4)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{|4-\text{h}-4|}{2\big[(4-\text{h})-4\big]}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{|-\text{h}|}{-2\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{-2\text{h}}=\frac{-1}{2}\text{ and f}(4)=0\neq\text{L.H.L}$
So, f(x) is discontinuous at x = 4.
View full question & answer→Question 373 Marks
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{y}=\tan^{-1}\Bigg(\frac{3\text{x}-\text{x}^{3}}{1-3\text{x}^{2}}\Bigg), -\frac{1}{\sqrt{3}}<\text{x}<\frac{1}{\sqrt{3}}$
AnswerThe given relationship is $\text{y}=\tan^{-1}\Bigg(\frac{3\text{x}-\text{x}^{3}}{1-3\text{x}^{2}}\Bigg)$
$\Rightarrow\tan\text{y}=\frac{3\text{x}-\text{x}^{3}}{1-3\text{x}^{2}} ...\text{(i)}$
$\text{y}=\tan^{-1}\Bigg(\frac{3\text{x}-\text{x}^{3}}{1-3\text{x}^{2}}\Bigg)$
It is known that, $\tan\text{y}=\frac{3\tan\frac{\text{y}}{3}-\tan^{3}\frac{\text{y}}{3}}{1-3\tan^{2}\frac{\text{y}}{3}} ...\text{(ii)}$
Comparing equations (1) and (2), we obtain
$\text{x} =\tan\frac{\text{y}}{3}$
Differenting this relationship with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\text{x})=\frac{\text{d}}{\text{dx}}\Bigg(\tan\frac{\text{y}}{3}\Bigg)$
$\Rightarrow1 =\sec^{2}\frac{\text{y}}{3}.\frac{\text{d}}{\text{dx}}\Bigg(\frac{\text{y}}{3}\Bigg)$
$\Rightarrow1 =\sec^{2}\frac{\text{y}}{3}.\frac{\text{1}}{\text{3}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}} = \frac{3}{\sec^{2}\frac{\text{y}}{3}} =\frac{3}{1 +\tan^{2}\frac{\text{y}}{3}}$
$\therefore \frac{\text{dy}}{\text{dx}}= \frac{3}{1+ \text{x}^{2}}$
View full question & answer→Question 383 Marks
Find $\frac{\text{dy}}{\text{dx}},\ \text{if y}=12(1-\cos \text{t}),\ \text{x}=10(\text{t}-\sin\text{t}),\ -\frac{\pi}{2}<\text{t}<\frac{\pi}{2}$
AnswerIt is given that, $\text{y}=12(1-\cos\text{t}),\text{x}=10(\text{t}-\sin\text{t})$
$\therefore\ \frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[10(\text{t}-\sin\text{t})]$ $=10.\frac{\text{d}}{\text{dt}}(\text{t}-\sin\text{t})=10(1-\cos\text{t})$
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[12(1-\cos\text{t})$ $=12.\frac{\text{d}}{\text{dt}}(1-\cos\text{t})=12.[0-(-\sin\text{t})]=12\sin\text{t}$
$\therefore\ \frac{\text{dx}}{\text{dx}}\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{12\sin\text{t}}{10(1-\cos\text{t})}$ $=\frac{12.2\sin\frac{\text{t}}{2}.\cos\frac{\text{t}}{2}}{10.2\sin^2\frac{\text{t}}{2}}=\frac{6}{5}\cot\frac{\text{t}}{2}$
View full question & answer→Question 393 Marks
Discuss the continuity of the following functions:
$\text{f(x)} = \sin \text{x} . \cos \text{x}$
AnswerLet a be an arbitrary real number then $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}^{+}}\text{f(x)} \Rightarrow^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\text{f(a + h)} $
$\Rightarrow\ ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\sin\text{(a + h)} . \cos (\text{a} + \text{h})$
$\Rightarrow\ ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}(\sin\text{a}\cos\text{ h} + \cos\text{a} \sin \text{h})(\cos \text{a}\cos\text{h}-\sin\text{a}\sin\text{h})$
$= (\sin \text{a}\cos0+\cos\text{a}\sin0) (\cos\text{a}\cos0 - \sin\text{a}\sin0)$
$=( \sin \text{a} + 0) ( \cos\text{a}-0)$
$= \sin \text{a} . \cos\text{a}= \text{f(a)}$
Similarly, we have $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}^{-}}\text{f(x)} = \text{f(a)}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}^{-}}\text{f(x)}= \text{f(a)}= ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}^{+}}\text{f(x)}$
Therefore, f(x) is continuous at x = a.
Since, a is an arbitrary real number, therefore, $\text{f(x)}= \sin\text{x} . \cos\text{x}$ is continuous.
View full question & answer→Question 403 Marks
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\text{a}(\theta-\sin\theta),\text{y}=\text{a}(1+\cos\theta)$
AnswerThe given equations are $\text{x}=\text{a}(\theta-\sin\theta)\text{ and y}=\text{a}(1+\cos\theta)$
Then, $\frac{\text{dx}}{\text{d}\theta}= \text{a}\Big[\frac{\text{d}}{\text{d}\theta}(\theta)-\frac{\text{d}}{\text{d}\theta}(\sin\theta)\Big]=\text{a}(1-\cos\theta)$
$\frac{\text{dy}}{\text{d}\theta}= \text{a}\Big[\frac{\text{d}}{\text{d}\theta}(1)+\frac{\text{d}}{\text{d}\theta}(\cos\theta)\Big]=\text{a}[0+(-\sin\theta)]=-\text{a}\sin\theta$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{d}\theta}\Big)}{\Big(\frac{\text{dx}}{\text{d}\theta}\Big)}=\frac{-\text{a}\sin\theta}{\text{a}(1-\cos\theta)}=\frac{-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}=\frac{-\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}=-\cot\frac{\theta}{2}$
View full question & answer→Question 413 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big),\pi<\text{x}<\pi$
AnswerLet $\text{f(x)}=\tan^{-1}\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
This function is defined for all real numbers where $\cos\text{x}\neq1$
$\text{f(x)}=\tan^{-1}\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
$\Rightarrow\ \text{f(x)}=\tan^{-1}\Bigg[\frac{2\sin\big(\frac{\text{x}}{2}\big)\cos\big(\frac{\text{x}}{2}\big)}{2\cos^2\big(\frac{\text{x}}{2}\big)}\Bigg]$
$\Rightarrow\ \text{f(x)}=\tan^{-1}\big[\tan\big(\frac{\text{x}}{2}\big)\big]=\frac{\text{x}}{2}$
Thus, $\text{f'(x)}=\frac{\text{d}}{\text{dx}}\big(\frac{\text{x}}{2}\big)=\frac{1}{2}$
View full question & answer→Question 423 Marks
If $\text{y}=\sin(\log\text{x})$ prove that $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
AnswerHere,
$\text{y}=\sin(\log\text{x})$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\cos(\log\text{x})}{\text{x}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\sin(\log\text{x})-\cos(\log\text{x})}{\text{x}^2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\sin(\log\text{x})}{\text{x}^2}-\frac{\cos(\log\text{x})}{\text{x}^2}{}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{y}}{\text{x}^2}-\frac{1}{\text{x}}\times\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
View full question & answer→Question 433 Marks
Differentiate the following w.r.t. x:
$\tan^{-1}\Big(\frac{3\text{a}^2\text{x}-\text{x}^3}{\text{a}^3-3\text{ax}^2}\Big),\frac{-1}{\sqrt{3}}<\frac{\text{x}}{\text{a}}<\frac{1}{\sqrt{3}}$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{3\text{a}^2\text{x}-\text{x}^3}{\text{a}^3-3\text{ax}^2}\Big)$
Put $\text{x}=\text{a}\tan\theta\Rightarrow\ \theta=\tan^{-1}\frac{\text{x}}{\text{a}}$
$\therefore\ \text{y}=\tan^{-1}\bigg[\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\bigg]$ $\bigg[\because\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\bigg]$
$=\tan^{-1}(\tan3\theta)=3\theta$
$=3\tan^{-1}\frac{\text{x}}{\text{a}}\Big[\because\theta=\tan^{-1}\frac{\text{x}}{\text{a}}\Big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=3\cdot\frac{\text{d}}{\text{dx}}\tan^{-1}\frac{\text{x}}{\text{a}}$ $=3\cdot\Bigg[\frac{1}{1+\frac{\text{x}^2}{\text{a}^2}}\Bigg]\cdot\frac{\text{d}}{\text{dx}}\cdot\Big(\frac{\text{x}}{\text{a}}\Big)$
$=3\cdot\frac{\text{a}^2}{\text{a}^2+\text{x}^2}.=\frac{1}{\text{a}}=\frac{3\text{a}}{\text{a}^2+\text{x}^2}$
View full question & answer→Question 443 Marks
Write the derivative of $f(x) = |x|^3$ at $x = 0$.
AnswerGiven: $\text{f(x)}=|\text{x}^3|=\begin{cases}\text{x}^3,&\text{x}\geq0\\-\text{x}^3,&\text{x}<0\end{cases}$
(LHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{\text{x}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{h}^3}{-\text{h}}$
$=0$
(RHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{\text{x}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{h}^3-0}{-\text{h}}$
$=0$
And f(0) = 0.
Thus, (LHL at x = 0) = (RHL at x = 0) = f(0)
Hence, $\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}=\text{f}'(0)=0$.
View full question & answer→Question 453 Marks
Show that the Lagrange's mean value theorem is not applicable to the function
$\text{f}(\text{x})=\frac{1}{\text{x}}\text{ on }[-1,1]$
AnswerGiven,
$\text{f}(\text{x})=\frac{1}{\text{x}}$
Clearly, f(x) is does not exist for x = 0
Thus, the given function is discontinuous on [-1, 1]
Hence, Lagrange's mean value theorem is not applicable for the given function on [-1, 1].
View full question & answer→Question 463 Marks
$\text{If x}\sqrt{1+\text{y}}+\text{y}\sqrt{1+\text{x}}=0,$ for, < x < 1, prove that
AnswerIt is given that,
$\text{x}\sqrt{1+\text{y}}+\text{y}\sqrt{1+\text{x}}=0$
$\Rightarrow\ \text{x}\sqrt{1+\text{y}}=-\text{y}\sqrt{1+\text{x}}$
Squaring both sides, we obtain
$\text{x}^2(1+\text{y})=\text{y}^2(1+\text{x})$
$\Rightarrow\ \text{x}^2+\text{x}^2\text{y}=\text{y}^2+\text{xy}^2$
$\Rightarrow\ \text{x}^2-\text{y}^2=\text{xy}^2-\text{x}^2\text{y}$
$\Rightarrow\ \text{x}^2-\text{y}^2=\text{xy}(\text{y}-\text{x})$
$\Rightarrow\ (\text{x}+\text{y})(\text{x}-\text{y})=\text{xy}(\text{y}-\text{x})$
$\therefore\ \text{x}+\text{y}=-\text{xy}$
$\Rightarrow\ (1+\text{x})\text{y}=-\text{x}$
$\Rightarrow\ \text{y}=\frac{-\text{x}}{(1+\text{x})}$
Differentiating both sides with respect to x, we obtain
$ \text{y}=\frac{-\text{x}}{(1+\text{x})}$
$\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x})\frac{\text{d}}{\text{dx}}(\text{x)}-\text{x}\frac{\text{d}}{\text{dx}}(1+\text{x})}{(1+\text{x})^2}$ $=-\frac{(1+\text{x})-\text{x}}{(1+\text{x})^2}=-\frac{1}{(1+\text{x})^2}$
Hence, proved.
View full question & answer→Question 473 Marks
$\text{If y}=\text{e}^{\text{y}}(\text{x}+1)=1,\text{ show that }\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
AnswerHere $\text{e}^{\text{y}}(\text{x}+1)=1$
$\therefore\ \log[\text{e}^{\text{y}}(\text{x}+1)]=\log1$
$\therefore\ \log\text{e}^{\text{y}}(\text{x}+1)=0$
$\therefore\ \text{y}\log\text{e}=-\log(\text{x}+1)$
$\therefore\ \text{y}=-\log(\text{x}+1)\ \ [\because\text{loge}=1]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=-\frac{1}{\text{x}+1}\ \dots(1)$
$\therefore\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Bigg[\frac{(\text{x}+1).\frac{\text{d}}{\text{dx}}(1)-1.\frac{\text{d}}{\text{dx}}(\text{x}+1)}{(\text{x}+1)^2}\Bigg]$ $=-\Big[\frac{(\text{x}+1).0-1.1}{(\text{x}+1)^2}\Big]$
$=\frac{1}{(\text{x}+1)^2}=\Big(-\frac{1}{(\text{x}+1)^2}\Big)^2$
$\therefore\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\ \dots[\because\text{of }(1)]$
View full question & answer→Question 483 Marks
State Lagrange's mean value theorem.
AnswerLagrange's Mean Value Theorem:
Let f(x) be a function defined on [a, b] such that
- It is continuous on [a, b] and
- It is differentiable on (a, b).
Then, there exists a real number $\text{c}\in(\text{a},\text{b})$ such that $\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}.$ View full question & answer→Question 493 Marks
Find the value of k for which the function $\text{f(x)}=\begin{cases}\frac{\text{x}^{2} + 3\text{x} - 10}{\text{x} - 2},&\text{x}\neq2\\\text{k},&\text{x} = {2}\end{cases}$ is continues at x = 2.
AnswerGiven
$\text{f(x)} = \frac{\text{x}^{2} + 3\text{x} - 10}{\text{x} - 2}$
Continuity
$\text{x} = 2$
$\lim\limits_{\text{x} \rightarrow 2} \frac{\text{x}^{2} + 3\text{x} - 10}{\text{x} -2} = \text{k}$
$\lim\limits_{\text{x}\rightarrow 2} \frac{\text{x}^{2} + 5\text{x} - 2\text{x} - 10}{\text{x} - 2 } = \text{k}$
$\lim\limits_{\text{x} \rightarrow 2}\frac{\text{x} (\text{x} + 5) - 2 (\text{x} + 5)}{\text{x} - 2} = \text{k}$
$\lim\limits_{\text{x} \rightarrow 2} \frac{(\text{x} - 2) (\text{x} + 5)}{\text{(x} - 2)} = \text{k}$
When x = 2
x + 5 = k
k = 5 + 2 = 7
k = 7
View full question & answer→Question 503 Marks
AnswerRolle's theorem: Let f(x) be a real value function defined on the closed interval [a, b] such that
- It is continuous on [a, b]
- It is differentiable on (a, b)
- f(a) = f(b)
Then, there exists a real number $\text{c}\in(\text{a},\text{b})$ such that f'(c) = 0. View full question & answer→Question 513 Marks
Using Rolle’s theorem, find the point on the curve $\text{y}=\text{x}(\text{x}-4),\text{x}\in[0,4].$ where the tangent is parallel to x-axis.
AnswerWe have, $\text{y}=\text{x}(\text{x}-4),\text{x}\in[0,4]$
Since given function is polynomial it is continuous and differentiable
Also y(0) = y(4) = 0
So, conditions of Rolle's theorem are satisfied.
Hence there exists a point $\text{c}\in(0,4)$ such that
f'(c) = 0
⇒ 2c - 4 = 0
⇒ c = 2
⇒ x = 2 and y(2) = 2(2 - 4) = -4
Therefore, the required point on the curve, where the tangent drawn is parallel to the x-axix is (2, -4).
View full question & answer→Question 523 Marks
Examine the differentialiblilty of the function f defined by $\text{f(x)}=\begin{cases}2\text{x}+3 & \text{if}-3\leq\text{x}\leq-2\\\text{x}+1 & \text{if} -2\leq\text{x}\leq0\\\text{x}+2&\text{if}\ 0\leq\text{x}\leq1\end{cases}$
Answer$\text{f(x)}=\begin{cases}2\text{x}+3 & \text{if}-3\leq\text{x}\leq-2\\\text{x}+1 & \text{if} -2\leq\text{x}\leq0\\\text{x}+2&\text{if}\ 0\leq\text{x}\leq1\end{cases}$
$\text{f}'(\text{x})=\begin{cases}2 & \text{if}-3\leq\text{x}\leq-2,\\1 & \text{if} -2\leq\text{x}\leq0\\1&\text{if}\ 0\leq\text{x}\leq1\end{cases}$
Now,
$\text{LHL}=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{-}}2=2$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{+}}1=1$
Since, at $\text{x}=-2,\text{LHL}\neq\text{RHL}$
Hence, f(x) is not differentiable at x = -2
Again,
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow0^{-}}1=1$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow0^{+}}1=1$
Since, at $\text{x}=0,$
$\text{LHL}=\text{RHL}$
Hence, f(x) is differentiable at x = 0
View full question & answer→Question 533 Marks
Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=\text{x}^{\frac{2}{3}}\text{ on }[-1,1]$
AnswerHere, $\text{f}(\text{x})=\text{x}^{\frac{2}{3}}\text{ on }[-1,1]$
$\text{f}'(\text{x})=\frac{2}{3\text{x}^{\frac{1}{3}}}$
$\text{f}'(0)=\frac{2}{3(0)^{\frac{1}{3}}}$
$\text{f}'(0)=\infty$
So, f'(x) does not exist at $\text{x}=0\in(-1,1)$
⇒ f(x) is not differentialble in $\text{x}\in(-1,1)$
So, Rolle's theorem is not applicable on f(x) in [-1, 1].
View full question & answer→Question 543 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\text{e}^{\text{x}-\text{y}}=\log\Big(\frac{\text{x}}{\text{y}}\Big)$
AnswerWe have, $\tan^{-1}\big(\text{x}^2+\text{y}^2\big)=\text{a}$
Differentiating with respect to x, we get
$\frac{\text{d}}{\text{dx}}\big[\tan^{-1}\big(\text{x}^2+\text{y}^2\big)\big]=\frac{\text{d}}{\text{dx}}(\text{a})$
$\Rightarrow\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{y}^2\big)=0$
$\Rightarrow\Big[\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\Big]\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}}$
View full question & answer→Question 553 Marks
Test the continuity of the function on f(x) at the origin:
$\text{f}\ (\text{x})=\begin{cases}\frac{\text{x}}{\text{|x|}},& \text{x}\neq0\\1, & \text{x} = 0\end{cases}$
AnswerGiven,
$\text{f}\ (\text{x})=\text{x},\text{ x}\neq0$
$\text{f}\ (\text{x})=1,\text{ x}=0$
We observe
$\text{(LHL at x}= 0)$
$\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f} \ (0-\text{h})$
$\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (-h)}=\lim\limits_{\text{h} \rightarrow 0} \frac{\text{-h}}{\text{h}}$
$\lim\limits_{\text{h} \rightarrow 0}-1=-1$
$\text{(RHL at x}=0)$
$\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\ \text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h})$
$\lim\limits_{\text{h} \rightarrow 0}\text{f}\ \text{(h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}}{\text{h}}$
$\lim\limits_{\text{h} \rightarrow 0}1=1$
Hence, f(x) is discontinuous at the origin.
View full question & answer→Question 563 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2+3\text{x}),&\text{if }\text{ x}<0\\\cos2\text{x},&\text{if }\text{ x}\geq0\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\text{k}(\text{x}^2+3\text{x}),&\text{if }\text{ x}<0\\\cos2\text{x},&\text{if }\text{ x}\geq0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\text{f}{(-\text{h)}}=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\Big(\text{k}\big((-\text{h})^2-3\text{h}\big)\Big)=\lim_\limits{\text{h}\rightarrow0}(\cos2\text{h})$
$\Rightarrow0=1$ [It is not possible]
Hence, there does not exist any value of k, which can make the given function continuous.
View full question & answer→Question 573 Marks
For what value of k is the following function continuous at x = 2?
$\text{f(x)}=\begin{cases}2\text{x}+1,&\text{if }\text{ x}<2\\\text{k},&\text{x}=2\\3\text{x}-1,&\text{x}>2\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}2\text{x}+1,&\text{if }\text{ x}<2\\\text{k},&\text{x}=2\\3\text{x}-1,&\text{x}>2\end{cases}$
We have,
$(\text{LHL at x}= 2)=\lim_\limits{\text{x}\rightarrow2^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(2(2-\text{h})+1)=5$
$(\text{RHL at x}= 2)=\lim_\limits{\text{x}\rightarrow2^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})=\lim_\limits{\text{h}\rightarrow0}3(2+\text{h})-1=5$
Also, $\text{f}(2)=\text{k}$
If f(x) is continuous at x = 2, then
$=\lim_\limits{\text{x}\rightarrow2^-}\text{f}(\text{x})=\lim_\limits{\text{x}\rightarrow2^+}\text{f}(\text{x})=\text{f}(2)$
$\Rightarrow5=5=\text{k}$
Hence, for k = 5, (fx) is continuous at x = 2
View full question & answer→Question 583 Marks
Find f(x) is continuse at x = 0, then $\text{f(x)}=\frac{\text{x}}{1-\sqrt{1-\text{x}}}$ becomes continuous at x = 0.
AnswerIf f(x) is continuous at x = 0, then $\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})=\text{f}(0)\ ....(\text{i})$
Given, $\text{f(x)}=\frac{\text{x}}{1-\sqrt{1-\text{x}}}$
$\Rightarrow\text{f(x)}=\frac{\text{x}\big(1+\sqrt{1-\text{x}}\big)}{\big(1-\sqrt{1-\text{x}}\big)\big(1+\sqrt{1-\text{x}}\big)}$
$\Rightarrow\text{f(x)}=\frac{\text{x}\big(1+\sqrt{1-\text{x}}\big)}{1-(1-\text{x})}$
$\Rightarrow\text{f(x)}=\big(1+\sqrt{1-\text{x}}\big)$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}=\big(1+\sqrt{1-\text{x}}\big)=\text{f}(0)$ [From eq. (i)]
$\Rightarrow\text{f}(0)=2$
So, for f(0) = 2, the function f(x) becomes continuous x = 0
View full question & answer→Question 593 Marks
Differentiate the following functions with respect to x:
$\tan^2\text{x}$
AnswerLet,
$\text{y}=\tan^2\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=2\tan\text{x }\frac{\text{d}}{\text{dx}}(\tan\text{x})\ \big[\text{using chain rule}\big]$
$=2\tan\text{x}\times\sec^2\text{x}$
So,
$\frac{\text{d}}{\text{dx}}=\big(\tan^2\text{x}\big)=2\tan\text{x }\sec^2\text{x}.$
View full question & answer→Question 603 Marks
Using mathematical induction prove that $\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})=\text{nx}^{\text{n-1}}$ for all positive.
AnswerTo prove: $\text{P(n)}:\frac{\text{d}}{\text{dx}}\text{(x}^\text{n})=\text{nx}^{\text{n}-1}$ for all positive integers n
For n = 1,
$\text{P}(1):\frac{\text{d}}{\text{dx}}\text{(x)}=1=1.\text{x}^{1-1}$
$\therefore\ $P(n) is true for n = 1
Let P(k) is true for some positive integer k.
That is, $\text{P(k)}:\frac{\text{d}}{\text{dx}}\text{(x}^\text{k})=\text{kx}^{\text{k}-1}$
It has to be proved that P(k + 1) is also true.
Consider $\frac{\text{d}}{\text{dx}}(\text{x}^{\text{k}+1})=\frac{\text{d}}{\text{dx}}(\text{x}.\text{x}^\text{k})$
$=\text{x}^\text{k}.\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}.\frac{\text{d}}{\text{dx}}(\text{x}^\text{k})$ $[\text{By applying product rule}]$
$=\text{x}^\text{k}.1+\text{x}.\text{k}.\text{x}^{\text{k}-1}$
$=\text{x}^\text{k}+\text{kx}^\text{k}$
$=(\text{k}+1).\text{x}^\text{k}$
$=(\text{k}+1).\text{x}^{\text{(k}+1)-1}$
Thus, P(k + 1) is true whenever P(k) is true.
Therefore, by the principle of mathematical lnduction, the statement P(n) is true for every positive integer n.
Hence, proved.
View full question & answer→Question 613 Marks
Find $\frac{\text{dy}}{\text{dx}},\ \text{if y}=\sin^{-1}\text{x}+\sin^{-1} \sqrt{1-\text{x}^2},\ -1\leq\text{x}\leq1$
AnswerIt is given that, $\text{y}=\sin^{-1}\text{x}+\sin^{-1}\sqrt{1-\text{x}^2}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\sin^{-1}\text{x}+\sin^{-1}\sqrt{1-\text{x}^2}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})+\frac{\text{d}}{\text{dx}}(\sin^{-1}\sqrt{1-\text{x}^2})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+\frac{1}{\sqrt{1-(\sqrt{1-\text{x}^2})^2}}.\frac{\text{d}}{\text{dx}}(\sqrt{1-\text{x}^2})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+\frac{1}{\text{x}}.\frac{1}{2\sqrt{1-\text{x}^2}}.\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+\frac{1}{2\text{x}\sqrt{1-\text{x}^2}}(-2\text{x})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}-\frac{1}{\sqrt{1-\text{x}^2}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 623 Marks
Differentiate the following functions with respect to x:
$\log(\tan^{-1}\text{x})$
AnswerConsider $\text{y}=\log\big(\tan^{-1}\text{x}\big)$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\big(\tan^{-1}\text{x}\big)$
$=\frac{1}{\tan^{-1}\text{x}}\times\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)$
[Using chain rule]
$=\frac{1}{\big(1+\text{x}^2\big)\tan^{-1}\text{x}}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\log\tan^{-1}\text{x}\big)=\frac{1}{\big(1+\text{x}^2\big)\tan^{-1}\text{x}}$
View full question & answer→Question 633 Marks
At what points on the following curves, is the tangent parallel to x-axis?
$\text{y}=\text{x}^2\text{ on }[-2,2]$
AnswerLet $f(x) = x^2$
Since f(x) is a polynomial function, it is continuous on [-2, 2] and differentiable on (-2, 2)
Also, $f(2) = f(-2) = 4$
Thus, all the conditions of Rolle's theorem are satisfied.
Concequently, there exists at least one point $\text{c}\in(-2,2)$ for which f'(c) = 0.
But $\text{f}'(\text{c})=0$
$\Rightarrow2\text{c}=0$
$\Rightarrow\text{c}=0$
$\therefore\text{f}_\text{c}=\text{f}_0=0$
By the geometrical interpretetion of Rolle's theorem, (0, 0) is the point on $y = x^2$, where the tangent is parallel to the x-axis.
View full question & answer→Question 643 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq5\\3\text{x}-5,&\text{if}\text{ x}>5\end{cases}\text{at x} =5$
Answer$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq5\\3\text{x}-5,&\text{if}\text{ x}>5\end{cases}\text{at x} =5$We have given that function is continuous at x = 5
$\therefore\ \text{LHL}=\text{RHL}=\text{f}(5)\ ....(\text{i})$
$\text{f}(5)=5\text{k}+1$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow5^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}(5+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}3(5+\text{h})-5=10$
Thus, using (i) we get,
$5\text{k}+1=10$
$5\text{k}=9$
$\text{k}=\frac{9}{5}$
View full question & answer→Question 653 Marks
Write the value of b which $\text{f(x)}=\begin{cases}5\text{x}-4,&0<\text{x}\leq1\\4\text{x}^2+3\text{bx},&1<\text{x}<2\end{cases}$ is continuous at x = 1.
AnswerGiven, $\text{f(x)}=\begin{cases}5\text{x}-4,&0<\text{x}\leq1\\4\text{x}^2+3\text{bx},&1<\text{x}<2\end{cases}$
If f(x) is continuous at x = 1, then
$\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x})=\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x})=\text{f}(1)\ ...(\text{i})$
Now,
$\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x})=\lim\limits_{{\text{h}}\rightarrow0}\text{f}(1-\text{h})\\=\lim\limits_{{\text{h}}\rightarrow0}5(1-\text{h})-4=5-4=1$
$\lim\limits_{{\text{x}}\rightarrow1^+}\text{f(x})=\lim\limits_{{\text{h}}\rightarrow0}\text{f}(1+\text{h})\\=\lim\limits_{{\text{h}}\rightarrow0}4(1+\text{h})^2+3\text{b}(1+\text{h})=4+3\text{b}$
Also,
$\text{f}(1)=5(1)-4=1$
$=\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow1^+}\text{f(x)}=\text{f}(1)$ [From eq. (i)]
$\Rightarrow1=4+3\text{b}=1$
$\Rightarrow1=4+3\text{b}$
$\Rightarrow-3=3\text{b}$
$\Rightarrow\text{b}=-1$
Thus, for b = -1, the function f(x) is continuous at x = 1.
View full question & answer→Question 663 Marks
Differentiate the following w.r.t. x:
$\frac{8^\text{x}}{\text{x}^8}$
AnswerLet $\text{y}=\frac{8^\text{x}}{\text{x}^8}$
$\Rightarrow\ \log\text{y}=\log\frac{8^\text{x}}{\text{x}^8}$
$\Rightarrow\ \frac{\text{d}}{\text{dy}}\log\text{y}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\log8^\text{x}-\log\text{x}^8\big]$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\big[\text{x}.\log8-8.\log\text{x}]$
On differentiating w.r.t. x, we get
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\log8.1-8.\frac{1}{\text{x}}$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\log8-\frac{8}{\text{x}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big(\log8-\frac{8}{\text{x}}\Big)=\frac{8^\text{x}}{\text{x}^8}\Big(\log8-\frac{8}{\text{x}}\Big)$
View full question & answer→Question 673 Marks
For what value of k is the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin2\text{x}}{\text{x}}, & \text{x} \neq 0\\\text{k}, &\text{x} = 0\end{cases}$ continuous at x = 0.
AnswerGiven, $\text{f}\text{(x)}=\begin{cases}\frac{\sin2\text{x}}{\text{x}}, & \text{x} \neq 0\\\text{k}, &\text{x} = 0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{\sin2\text{x}}{\text{x}}=\text{k}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin2\text{x}}{2\text{x}}=\text{k}$
$\Rightarrow2\lim\limits_{\text{x} \rightarrow 0}\frac{\sin2\text{x}}{2\text{x}}=\text{k}$
$\Rightarrow2\times1=\text{k}$
$\Rightarrow\text{k} =2$
View full question & answer→Question 683 Marks
Given the funcation $\text{f(x)}=\frac{1}{\text{x}+2}.$ Find the points of discontinuity of the function f(f(x)).
Answer$\text{f}\big[\text{f(x)}\big]=\frac{1}{\frac{1}{\text{x}+2}+2}=\frac{\text{x}+2}{2\text{x}+5}$
So, f[f(x)] is not defind at x + 2 = 0 and 2x + 5 = 0
If x + 2, then x = - 2
If 2x + 5 = 0, then $\text{x}=-\frac{5}{2}$
Hence, the function is dicontinuous at $\text{x}=-\frac{5}{2}$ and -2
View full question & answer→Question 693 Marks
Find the values of k so that the function f is continuous at the indicated point:
$\text{f(x)}\begin{cases}\frac{\text{k}\cos\text{x}}{\pi -2\text{x}}\ \text{if}\ \text{x}\neq \frac{\pi}{2}\\3, \ \ \ \ \ \ \ \ \text{if}\ \text{x} =\frac{\pi}{2}\end{cases}$
$\text{at} \text{x} = \frac{\pi}{2}$
AnswerHere $\text{f(x)}\begin{cases}\frac{\text{k}\cos\text{x}}{\pi -2\text{x}},\ \text{if}\ \text{x}\neq \frac{\pi}{2}\\3, \ \ \ \ \ \ \ \ \text{if}\ \text{x} =\frac{\pi}{2}\end{cases}$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\frac{\pi}{2}}\text{f(x)}= ^{\ \ \text{Lt}}_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\text{k}\cos\text{x}}{\pi - 2\text{x}}$ $\left[ \text{Put}\ \text{x} = \frac{\pi}{2} + \text{h},\text{h} > 0 \ \text{so that h} \rightarrow0\ \text{as x} \rightarrow\frac{\pi}{2}\right]$
$ ^{\ \ \text{Lt}}_{\text{h}\rightarrow0}\frac{\text{k}\cos\Big(\frac{\pi}{2}+\text{h}\Big)}{{\pi}- {2}\Big(\frac{\pi}{2}-\text{h}\Big)}=^{\ \ \text{Lt}}_{\text{h}\rightarrow0}\frac{-\text{k}\sin\text{h}}{-2\text{h}}$
$\frac{\text{k}}{2}\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}} = \frac{\text{k}}{2}\times1 = \frac{\text{k}}{2}$
Also $\text{f}(\frac{\pi}{2})= 3$
Since f is continuous at $\text{x}= \frac{\pi}{2}$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow{\frac{\pi}{2}}}\text {f(x}) = \text{f}\Big(\frac{\pi}{2}\Big)\Rightarrow \frac{\text{k}}{2} = 3 \Rightarrow\text{k} =6$
View full question & answer→Question 703 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\frac{\sin2\text{x}}{5\text{x}},&\text{if }\text{ x}\neq0\\3\text{k},&\text{if }\text{ x}=0\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\sin2\text{x}}{5\text{x}},&\text{if }\text{ x}\neq0\\3\text{k},&\text{if }\text{ x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\Rightarrow\lim_\limits{\text{x}\rightarrow0}\frac{\sin2\text{x}}{5\text{x}}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow0}\frac{2\sin2\text{x}}{2\times5\text{x}}=\text{f}(0)$
$\Rightarrow\frac{2}{5}\lim_\limits{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}=\text{f}(0)$
$\Rightarrow\frac{2}{5}=3\text{k}$
$\Rightarrow\text{k}=\frac{2}{15}$
View full question & answer→Question 713 Marks
Find $\frac{\text{dy}}{\text{dx}}$ of the functions given in Exercise:
$\text{y}^\text{x}=\text{x}^\text{y}$
AnswerGiven: $\text{y}^\text{x}=\text{x}^\text{y}\ \Rightarrow\ \text{x}^\text{y}=\text{y}^\text{x}$
$\Rightarrow\ \log\text{x}^\text{y}=\log\text{y}^\text{x}\ \Rightarrow\ \text{y}\log\text{x}=\text{x}\log\text{y}$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}(\text{y}\log\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})\ \Rightarrow\ \text{y}.\frac{1}{\text{x}}+\log\text{x}.\frac{\text{dy}}{\text{dx}}=\text{x}.\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}.1$
$\Rightarrow\ \Big(\log\text{x}-\frac{\text{x}}{\text{y}}\Big) \frac{\text{dy}}{\text{dx}}=\log\text{y}-\frac{\text{y}}{\text{x}}\ \Rightarrow\ \Big(\frac{\text{y}\log\text{x}-\text{x}}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\text{x}\log\text{y}-\text{y}}{\text{x}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}\log\text{y}-\text{y})}{\text{x}(\text{y}\log\text{x}-\text{x})}$
View full question & answer→Question 723 Marks
For what value of k is the following function continuous at x = 1
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-1}{\text{x}-1}, & \text{x} \neq 1\\\text{k}, & \text{x}= 1\end{cases}$
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-1}{\text{x}-1}, & \text{x} \neq 1\\\text{k}, & \text{x}= 1\end{cases}$
If f(x) is continuous at x = 1, then
$\lim\limits_{\text{x} \rightarrow 1}\text{f}\text{(x)}=\text{f}(1)$
$\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^2-1}{\text{x}-1}=\text{k}$
$\lim\limits_{\text{x} \rightarrow 1}\frac{\text{(x}-1)(\text{x}+1)}{\text{x}-1}=\text{k}$
$\lim\limits_{\text{x} \rightarrow 1}(\text{x}+1)=\text{k}$
$\text{k}=2$
View full question & answer→Question 733 Marks
Determine the value of the constant k so that the function
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-3\text{x}+2}{\text{x}-1}, &\text{if}\text{ x}\neq1\\\text{k}, &\text{if}\text{ x}=1\end{cases}$ is continuous at x = 1
AnswerWe have that the function is continuous at x = 1
$\therefore$ LHL = RHL = f(1) ....(1)
Now,
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(1-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{(1-\text{h})^2-3(1-\text{h})+2}{(1-\text{h})-1}$
$\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+\text{h}}{-\text{h}}=\lim\limits_{\text{h} \rightarrow 0}-\text{h}-1=-1$
$\text{f}(1)=\text{k}$
from (1), We get,
k = -1
View full question & answer→Question 743 Marks
Find all points of discontinuity of the function $\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2},$ where $\text{t}=\frac{1}{\text{x}-1}.$
AnswerWe have, $\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2},$ where $\text{t}=\frac{1}{\text{x}-1}$
$\therefore\ \text{f(x)}=\frac{1}{\Big(\frac{1}{\text{x}-1}\Big)^2+\frac{1}{\text{x}-1}-2}$
$=\frac{(\text{x}-1)^2}{1+(\text{x}-1)-2(\text{x}-1)^2}=\frac{(\text{x}-1)^2}{-(2\text{x}^2-5\text{x}+2)}=\frac{(\text{x}-1)^2}{(2\text{x}-1)(2-\text{x})}$
So, f(t) is discontinuous at 2x - 1 = 0
$\Rightarrow\ \text{x}=\frac{1}{2}$
and 2 - x = 0 ⇒ x = 2
Also f(t) is discontinuous at x = 1, where $\text{t}=\frac{1}{\text{x}-1}$ is discontinuous.
View full question & answer→Question 753 Marks
For what value of k is the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin5\text{x}}{3\text{x}}, &\text{if}\text{ x}\neq0\\\text{k}, &\text{if}\text{ x}=0\end{cases}$ is continuous at x = 0?
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{\sin5\text{x}}{3\text{x}}, &\text{if}\text{ x}\neq0\\\text{k}, &\text{if}\text{ x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{\sin5\text{x}}{3\text{x}}=\text{k}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{5\sin5\text{x}}{3\times5\text{x}}=\text{k}$
$\Rightarrow\frac{5}{3}\lim\limits_{\text{x} \rightarrow 0}\frac{\sin5\text{x}}{5\text{x}}=\text{k}$
$\Rightarrow\frac{5}{3}\times1=\text{k}$
$\Rightarrow\text{k}=\frac{5}{3}$
View full question & answer→Question 763 Marks
Differentiate the following functions with respect to x:
$\log_7(2\text{x}-3)$
AnswerLet, $\text{y}=\log_7(2\text{x}-3)$
$\Rightarrow\ \text{y}=\frac{\log(2\text{x}-3)}{\log_7}\ \Big[\text{Since}, \log^\text{b}_\text{a}=\frac{\log\text{b}}{\log\text{a}}\Big]$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\log7}\frac{\text{d}}{\text{dx}}\big(\log(2\text{x}-3)\big)$
$=\frac{1}{\log7}\times\frac{1}{(2\text{x}-3)}\frac{\text{d}}{\text{dx}}(2\text{x}-3)$
[Using chain rule]
$=\frac{2}{(2\text{x}-3)\log7}$
Hence, $\frac{\text{d}}{\text{dx}}(\log_7(2\text{x}-3))=\frac{2}{(2\text{x}-3)\log7}$
View full question & answer→Question 773 Marks
Find which of the function:
$\begin{cases}\frac{1-\cos2\text{x}}{\text{x}^2},&\text{ if x}\neq0\\5,&\text{if x}=0\end{cases}$
at x = 0
View full question & answer→Question 783 Marks
Find the second order derivatives of the following functions:
$\text{y}=\text{x}^3\log\text{x}$
AnswerWe have
$\text{y}=\text{x}^3\log\text{x}$
differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dy}}=3\text{x}^2\log\text{x}+\text{x}^3\times\frac{1}{\text{x}}$
$=3\text{x}^2\log\text{x}+\text{x}^2$
differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{x}\log\text{x}+3\text{x}^2\times\frac{1}{\text{x}}+2\text{x}$
$=6\text{x}\log\text{x}+5\text{x}$
View full question & answer→Question 793 Marks
If f(x) = |x - 2| write whether f(2) exists or not.
AnswerGiven: $\text{f(x)}=|\text{x}-2|=\begin{cases}\text{x}-2, & \text{x}> 2\\-\text{x}+2, & \text{x}\leq 2\end{cases}$
Now,
(LHL at x = 2)
$\lim_\limits{\text{x}\rightarrow2^{-}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(2-\text{h})-\text{f}(2)}{2-\text{h}-2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(-2+\text{h}+2)-0}{-\text{h}}$
$=-1$
(RHL at x = 2)
$\lim_\limits{\text{x}\rightarrow2^{+}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(2+\text{h})-\text{f}(2)}{2+\text{h}-2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2+\text{h}+2-0}{\text{h}}$
$=1$
Thus, (LHL at x = 2) $\neq$ (RHL at x = 2)
Hence, $\lim_\limits{\text{x}\rightarrow2}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}=\text{f'}(2)$ does not exist.
View full question & answer→Question 803 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2-2\text{x}),&\text{if}\text{ x}<0\\\cos\text{x},&\text{if}\text{ x}\geq0\end{cases}\text{at x} = 0$
AnswerGiven,
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2-2\text{x}),&\text{if}\text{ x}<0\\\cos\text{x},&\text{if}\text{ x}\geq0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{k}(\text{h}^2+2\text{h})=0$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}\cos\text{h}=1$
$\therefore\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}\neq\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}$
Thus, no value of k exists for which f(x) is continuous at x = 0.
View full question & answer→Question 813 Marks
If $\text{f(x)}=\frac{2\text{x}+3\sin\text{x}}{3\text{x}+2\sin\text{x}},\text{ x}\neq0$ is continuous at x = 0, then find f(0).
AnswerIt is given that the function is continuous at x = 0
$\therefore\text{LHL}=\text{RHL}=\text{f}(0)\ ....(\text{i})$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow 0}=\lim_\limits{\text{h}\rightarrow 0}(0-\text{h})=\lim_\limits{\text{h}\rightarrow 0}\frac{2(-\text{h})+3\sin(-\text{h})}{3(-\text{h})+2\sin(-\text{h})}$
$=\lim_\limits{\text{x}\rightarrow 0}\frac{-2\text{h}-3\sin\text{h}}{-3\text{h}-2\sin\text{h}}$
$=\lim_\limits{\text{x}\rightarrow 0}\frac{\frac{2\text{h}+3\sin\text{h}}{\text{h}}}{\frac{3\text{h}+2\sin\text{h}}{\text{h}}}$
$=\lim_\limits{\text{x}\rightarrow 0}\frac{2+3\frac{\sin\text{h}}{\text{h}}}{3+2\frac{\sin\text{h}}{\text{h}}}$
$=\frac{2+3}{3+2}=1$ $\Big[\because=\lim_\limits{\theta\rightarrow 0}\frac{\sin\theta}{\theta}=1\Big]$
Using (i) we get
$=\text{f}(0)=1$
View full question & answer→Question 823 Marks
Discuss the continuity of the function f(x) at the point $\text{x}=\frac{1}{2}$ where
$\text{f}\text{(x)}=\begin{cases}\text{x}, & 0\leq\text{x} < \frac{1}{2}\\\frac{1}{2},&\text{x}=\frac{1}{2}\\1-\text{x}, &\frac{1}{2}< \text{x}\leq 1\end{cases}$
AnswerWe want to discuss the continuity of the function at $\text{x}=\frac{1}{2}.$
$\text{LHL}=\lim\limits_{\text{x} \rightarrow \frac{1}{2}^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{f}\Big(\frac{1}{2}-\text{h}\Big)=\lim\limits_{\text{x} \rightarrow 0}\frac{1}{2}-\text{h}=\frac{1}{2}$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow \Big(\frac{1}{2}^-\Big)}=\lim\limits_{\text{x} \rightarrow 0}\text{f}\Big(\frac{1}{2}+\text{h}\Big)=\lim\limits_{\text{x} \rightarrow 0}1-\Big(\frac{1}{2}+\text{h}\Big)=\frac{1}{2}$
$\text{f}\Big(\frac{1}{2}\Big)=\frac{1}{2}$
Thus, $\text{LHL}=\text{RHL}=\text{f}\Big(\frac{1}{2}\Big)=\frac{1}{2}$
Hence, the function is continuous at $\text{x}=\frac{1}{2}.$
View full question & answer→Question 833 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\text{|x|}\cos\Big(\frac{1}{\text{x}}\Big), & \text{ x}\neq 0\\0 &\text{ x} = 0\end{cases}\text{at x}=0$
AnswerGiven,
$\text{f}\text{(x)}=\text{x}\cos\Big(\frac{1}{\text{x}}\Big),\text{x}\neq0$
$\text{f}\text{(x)}=0,\ \text{x}=0$
We observe
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{x}\cos\Big(\frac{1}{\text{x}}\Big)$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{x}\lim\limits_{\text{x} \rightarrow 0}\cos\Big(\frac{1}{\text{x}}\Big)$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=0\times\lim\limits_{\text{x} \rightarrow 0}\cos\Big(\frac{1}{\text{x}}\Big)$
$=0$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
Hence, f(x) is continuous at x = 0.
View full question & answer→Question 843 Marks
Determine that value of the constant 'k' so that function $\text{f(x)}=\begin{cases}\frac{\text{kx}}{|\text{x}|},&\text{if }\text{ x}<0\\3,&\text{if }\text{ x}\geq0\end{cases}$ is continuous at x = 0.
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\text{kx}}{|\text{x}|},&\text{if }\text{ x}<0\\3,&\text{if }\text{ x}\geq0\end{cases}$
Since, the function is continuous at x = 0, therefore,
$\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x})=\lim\limits_{{\text{x}}\rightarrow0^+}\text{f(x})=\text{f}(0)$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\frac{-\text{kx}}{\text{x}}=\lim\limits_{{\text{x}}\rightarrow0}3=3$
$\Rightarrow-\text{k}=3$
$\Rightarrow\text{k}=-3$
View full question & answer→Question 853 Marks
Show that the function $\text{f(x)}=\big|\sin\text{x}+\cos\text{x}|$ is continuous at $\text{x}=\pi.$
AnswerConsider, $\text{f(x)}=\big|\sin\text{x}+\cos\text{x}\big|\text{ at x}=\pi$
Let $\text{g(x)}=\sin\text{x}+\cos\text{x}$
And $\text{h(x)}=|\text{x}|$
$\therefore\ \text{hog (x)}=\text{h}[\text{g (x)}]$
$=\text{h }(\sin\text{x}+\cos\text{x})$
$=|\sin\text{x}+\cos\text{x}|$
Since, g(x) and h(x) are continuous functions and f(x) is a composite functions.
We know that composite functions of two continuous functions is also a continuous function.
Hence, $\text{f(x)}=|\sin\text{x}+\cos\text{x}|$ is a continuous function everywhere,
So, f(x) is continuous at $\text{x}=\pi.$
View full question & answer→Question 863 Marks
Examine the following functions for continuity.
$\text f(\text x)=\begin{vmatrix}\text x-5\end{vmatrix}$
AnswerHere f(x) = | x - 5|
Function f is defined for all real numbers.
Let c be any real number.
$\therefore \text{f}(\text{c}) = \left| \text{c} - 5\right|$
Also $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\left|\text{x}- 5\right| = \left|\text{c}-5\right|$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at x = c,
But c is any real number
$\therefore$ f is continuous at every real number.
View full question & answer→Question 873 Marks
Examine the following functions for continuity.
$\text f(\text X)=\frac{\text X^{2} - 25}{\text {X} + 5}$
Answer$\text f(\text X)=\frac{\text X^{2} - 25}{\text {X} + 5}$
For f to be defined,
x + 5 $\neq$ 0 i. e. x $\neq$ -5
$\therefore$ $D_f=$ Set of all real numbers except -5 = R - { -5}
Letv $\text{c} \neq -5$ be any real number.
$\therefore$ f(c) = $\frac {\text{c}^2 - 25}{\text{c} + 25} = \frac{(\text{c}-5)(\text{c} + 5)}{\text{c}+5} = \text{c}-5$
Also $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\frac{\text{x}^2-25}{\text{x}+5} =\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\frac{({\text{x} -5)(\text{x} +5)}}{\text{x}+ 5}$
$= ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{(x}- 5) = \text{c} - 5$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x}) = \text {f(c)}$
$\therefore$ f is continuous at x = c.
But $\text{c} \neq -5$ is any real number.
$\therefore$ f is continuous at every real number $\text{c} \in \text{D}_\text{f}.$
$\therefore$ f is continuous function.
View full question & answer→Question 883 Marks
$\text{If x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t})\text{ and y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t}),\ \text{find}\dfrac{\text{d}^2\text{y}}{\text{dx}^2}$
AnswerIt is given that, $\text{x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t})\text{ and y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t})$
$\therefore\ \frac{\text{dx}}{\text{dt}}=\text{a}.\frac{\text{d}}{\text{dt}}(\cos\text{t}+\text{t}\sin\text{t})$
$=\text{a}\Big[-\sin\text{t}+\sin\text{t}.\frac{\text{d}}{\text{dx}}(\text{t})+\text{t}.\frac{\text{d}}{\text{dt}}(\sin\text{t})\Big]$
$=\text{a}[-\sin\text{t}+\sin\text{t}+\text{t}\cos\text{t}]=\text{at}\cos\text{t}$
$\frac{\text{dy}}{\text{dt}}=\text{a}.\frac{\text{d}}{\text{dt}}(\sin\text{t}-\text{t}\cos\text{t})$
$=\text{a}\Big[\cos\text{t}-\Big\{\cos\text{t}.\frac{\text{d}}{\text{dt}}(\text{t})+\text{t}.\frac{\text{d}}{\text{dt}}(\cos\text{t})\Big\}\Big]$
$=\text{a}[\cos\text{t}-\{\cos\text{t}-\text{t}\sin\text{t}\}]=\text{at}\sin\text{t}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{\text{at}\sin\text{t}}{\text{at}\cos\text{t}}=\tan\text{t}$
Then, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan\text{t})=\sec^2\text{t}.\frac{\text{dt}}{\text{dx}}$
$=\sec^2\text{t}.\frac{1}{\text{at}\cos\text{t}}\ \Big[\frac{\text{dx}}{\text{dt}}=\text{at}\cos\text{t}\Rightarrow\ \frac{\text{dt}}{\text{dx}}=\frac{1}{\text{at}\cos\text{t}}\Big]$
$=\frac{\sec^3\text{t}}{\text{at}}.0<\text{t}<\frac{\pi}{2}$
View full question & answer→Question 893 Marks
Differentiate the following functions with respect to x:
$3^{\text{x}\log\text{x}}$
AnswerLet $\text{y}=3^{\text{x}\log\text{x}}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(3^{\text{x}\log\text{x}}\big)$
$=3^{\text{x}\log\text{x}}\times\log_\text{e}3\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$
[Using chain rule]
$=3^{\text{x}\log\text{x}}\times\log_\text{e}3\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=3^{\text{x}\log\text{x}}\times\log_\text{e}3\Big[\frac{\text{x}}{\text{x}}+\log\text{x}\Big]$
$=3^{\text{x}\log\text{x}}\big[1+\log\text{x}\big]\times\log_\text{e}3$
So,
$\frac{\text{d}}{\text{dx}}\big(3^{\text{x}\log\text{x}}\big)=3^{\text{x}\log\text{x}}\big[1+\log\text{x}\big]\log_\text{e}3$
View full question & answer→Question 903 Marks
If $\text{y}=\sqrt{\text{x}+\sqrt{\text{x}+\sqrt{\text{x}+\ .... \text{to }\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}-1}$
AnswerWe have, $\text{y}=\sqrt{\text{x}+\sqrt{\text{x}+\sqrt{\text{x}+\ .... \text{to }\infty}}}$
Squaring both sides, we get,
$y^2 = x + y$
$\Rightarrow 2\text{y}\frac{\text{dy}}{\text{dx}}=1+\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{y}-1)=1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}-1}$
View full question & answer→Question 913 Marks
If `$\text{y}=(\cos\text{x})^{(\cos\text{x})^{(\cos\text{x})\dots\infty}},$ show that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2\tan\text{x}}{\text{y}\log\cos\text{x}-1}.$
AnswerWe have, $\text{y}=(\cos\text{x})^{(\cos\text{x})^{(\cos\text{x})\dots\infty}}$
$\Rightarrow\ \text{y}=(\cos\text{x})^\text{y}$
$\therefore\ \log\text{y}=\log(\cos\text{x})^\text{y}$
Differentiating w.r.t. x, we get
$\frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\text{y}\cdot\frac{\text{d}}{\text{dx}}(\log\cos\text{x})+\log\cos\text{x}\cdot\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\cos\text{x}}\cdot\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\cdot\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big[\frac{1}{\text{y}}-\log\cos\text{x}\Big]$ $=\frac{-\text{y}\sin\text{x}}{\cos\text{x}}=-\text{y}\tan\text{x}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2\tan\text{x}}{(1-\text{y}\log\cos\text{x})}=\frac{\text{y}^2\tan\text{x}}{\text{y}\log\cos\text{x}-1}$
View full question & answer→Question 923 Marks
Differentiate w.r.t. x the function in Exercise:
$(\log\text{x})^{\log\text{x}},\text{x}>1$
AnswerLet $\text{y}=(\log\text{x})^{\log\text{x}}$
Tanking logarithm on both the sides, we obtain
$\log\text{y}=\log\text{x}.\log(\log\text{x})$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}[\log\text{x}.\log(\log\text{x})]$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\log\text{x)}.\frac{\text{d}}{\text{dx}}(\log\text{x)}+\log\text{x}.\frac{\text{d}}{\text{dx}}[\log(\log\text{x})]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log(\log\text{x)}.\frac{1}{\text{x}}+\log\text{x}.\frac{1}{\log\text{x}}.\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{1}{\text{x}}\log(\log\text{x)}+\frac{1}{\text{x}}\Big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=(\log\text{x)}^{\log\text{x}}\Big[\frac{1}{\text{x}}+\frac{\log(\log\text{x})}{\text{x}}\Big]$
View full question & answer→Question 933 Marks
Show that the function defined by $g (x) = x – [x]$ is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.
Answerg(x) = x - [x]
Let a be any integer.
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{-}}\text{g(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{-}}\left\{\text{x-[x]}\right\}$
[Put x - a - h, h > 0 so that $h\rightarrow 0$ as $x\rightarrow a^-$
$ = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}}\left\{\text{(a}-\text{h})-(\text{a}-\text{h})\right\} = ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\left\{\text{(a}-\text{h})-(\text{a}-\text{1})\right\}$
$ = ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\text{(a}-\text{h}-\text{a}+1) = \text{a}- 0-\text{a} + 1 = 1$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{+}}\text{g(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{+}}\left\{\text{x - [x]}\right\}$
[Put x = a + h, h > 0 so that $h\rightarrow 0$ as $x\rightarrow a^+$]
$= ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\left\{\text{(a}+\text{h})-(\text{a}+\text{h})\right\}= ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\left\{\text{(a}+\text{h})-\text{a}\right\} =^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\text{(h)}$
= 0
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{-}}\text{f(x)}\neq^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{+}}\text{f(x)}$
$\therefore$ f is discontinuous at x = a
But a is any integral point.
$\therefore$ f is discontinuous at all integgral points.
View full question & answer→Question 943 Marks
Find $\frac{\text{dy}}{\text{dx}}$ of the functions expressed in parametric:
$\text{x}=\text{t}+\frac{1}{\text{t}},\text{ y}=\text{t}-\frac{1}{\text{t}}$
AnswerConsider, $\text{x}=\text{t}+\frac{1}{\text{t}}$ and $\text{y}=\text{t}\frac{1}{\text{t}}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=1+\Big(-\frac{1}{\text{t}^2}\Big)$ and $\frac{\text{dy}}{\text{dt}}=1-\Big(-\frac{1}{\text{t}^2}\Big)$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=1-\frac{1}{\text{t}^2}$ and $\frac{\text{dy}}{\text{dt}}=1+\frac{1}{\text{t}^2}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\text{t}^2-1}{\text{t}^2}$ and $\frac{\text{dy}}{\text{dt}}=\frac{\text{t}^2+1}{\text{t}^2}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$
$=\frac{\text{t}^2+\frac{1}{\text{t}^2}}{\text{t}^2-\frac{1}{\text{t}^2}}$
$=\frac{\text{t}^2+1}{\text{t}^2-1}$
View full question & answer→Question 953 Marks
Differentiate the following functions with respect to x:
$2^{\text{x}^3}$
AnswerConsider $\text{y}=2^{\text{x}^3}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(2^{\text{x}^3}\Big)$
$=2^{\text{x}^3}\times\log_2\frac{\text{d}}{\text{dx}}(\text{x}^3)$
[Using chain rule]
$=3\text{x}^2\times2^{\text{x}^3}\times\log_2$
Hence, the solution is $\frac{\text{d}}{\text{dx}}\big(2^{\text{x}^3}\big)=3\text{x}^2\times2^{\text{x}^3}\log_2$
View full question & answer→Question 963 Marks
Prove that the greatest integer function defined by
$\text{f(x)} = [\text{x} ], 0 < \text{x} < 3$
is not differentiable at x = 1 and x = 2.
AnswerGiven: $\text{f(x)} = [\text{x} ], 0 < \text{x} < 3$
$\text{R}\text{f}{'}(1) = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{\text{f(1 + h)}-\text{f}(1)}{\text{h}} = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|1 + \text{h} |-1}{\text{h}}$$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{1-1}{\text{h}}= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{0}{\text{h}}=0$
$\text{And}\ \text{L}\text{f}{'}(1) = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{\text{f(1 - h)}-\text{f}(1)}{-\text{h}} = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|1 - \text{h}| - 1}{-\text{h}}$$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|1-\text{h}|-1}{\text{h}}= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{0-1}{-\text{h}}=\infty$
Since $\text{R }\text{f}{'}(1)\neq \text{L}{\text{f}}{'}(1)$
Therefore, f(x) = [x] is not differentiable at x =1.
Similarly f(x) = [x] is not differentiable at x = 2.
View full question & answer→Question 973 Marks
$\text{If y}=5\cos\text{x}-3\sin\text{x},\text{ prove that }\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0$
AnswerLet $\text{y}=5\cos\text{x}-3\cos\text{x}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=-5\sin\text{x}-3\cos\text{x}$
$\Rightarrow\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-5\cos\text{x}+3\sin\text{x}=-(5\cos\text{x}-3\sin\text{x})=-\text{y}$
$\Rightarrow\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{y}\ \Rightarrow\ \frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0$
View full question & answer→Question 983 Marks
Differentiate the following w.r.t. x:
$\log\big[\log(\log\text{x}^5)\big]$
AnswerLet $\text{y}=\log\big[\log(\log\text{x}^5)\big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\big[\log(\log\text{x}^5)\big]$
$=\frac{1}{\log\log\text{x}^5}\cdot\frac{\text{d}}{\text{dx}}\big(\log\cdot\log\text{x}^5\big)$
$=\frac{1}{\log\log\text{x}^5}\cdot\frac{1}{\log\text{x}^5}\cdot\frac{\text{d}}{\text{dx}}\log\text{x}^5$
$=\frac{1}{\log\log\text{x}^5}\cdot\frac{1}{\log\text{x}^5}\cdot\frac{\text{d}}{\text{dx}}(5\log\text{x})$
$=\frac{5}{\text{x}\cdot\log(\log\text{x}^5)\cdot\log(\text{x}^5)}$
View full question & answer→Question 993 Marks
If $\text{y}=\log(\sin\text{x})$ Prove that $\frac{\text{d}^3\text{y}}{\text{dx}^3}=2\cos\text{x}\ \text{cosec}^3\text{x}$
AnswerHere,
$\text{y}=\log(\sin\text{x})$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sin\text{x}}\times\cos\text{x}=\cot\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{cosec}^2\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^3\text{y}}{\text{dx}^3}=-2\text{cosec}\ \text{x}\times(-\text{cosec}\ \text{x}\cot\text{x})$
$=2\cot\ \text{x}\ \text{cosec}^2\text{x}=2\cos\ \text{x}\ \text{cosec}^3\text{x}$
View full question & answer→Question 1003 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\tan3\text{x}}$
AnswerLet, $\text{y}=\text{e}^{\tan3\text{x}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan3\text{x}}\big)$
$=\text{e}^{\tan3\text{x}}\frac{\text{d}}{\text{dx}}(\tan3\text{x})$
[Using chain rule]
$\text{e}^{\tan3\text{x}}\times\sec^23\text{x}\times\frac{\text{d}}{\text{dx}}(3\text{x})$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan3\text{x}}\big)=3\text{e}^{\tan3\text{x}}\times\sec^2 3\text{x}$
View full question & answer→Question 1013 Marks
$\text{If y}=\text{Ae}^{\text{mx}}+\text{Be}^{\text{nx}},\text{ show that }\frac{\text{d}^2\text{y}}{\text{dx}^2}-(\text{m}+\text{n})\frac{\text{dy}}{\text{dx}}+\text{mny}=0$
Answer$\text{y}=\text{Ae}^\text{mx}+\text{Be}^{\text{nx}}\dots(1)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{Ame}^{\text{mx}}+\text{Bne}^{\text{nx}}\dots(2)$
and $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{Am}^2\text{e}^{\text{mx}}+\text{Bn}^2\text{e}^{\text{nx}}\dots(3)$
$\text{L.H.S.}=\frac{\text{d}^2\text{y}}{\text{dx}^2}-(\text{m}+\text{n})\frac{\text{dy}}{\text{dx}}+\text{mny} $
$=(\text{Am}^2\text{e}^{\text{mx}}+\text{Bn}^2\text{e}^{\text{nx}})-(\text{m}+\text{n})(\text{Ame}^{\text{mx}}+\text{Bne}^{\text{nx}})+\text{mn}(\text{Ae}^{\text{mx}}+\text{Be}^{\text{nx}})$
$[\because \text{of }(1),(2),(3)]$$$
$=\text{Am}^2\text{e}^{\text{mx}}+\text{Bn}^2\text{e}^{\text{nx}}-\text{Am}^2\text{e}^{\text{mx}}-\text{Bmne}^{\text{nx}}-\text{Amne}^{\text{mx}}$
$-\text{Bn}^2\text{e}^{\text{nx}}+\text{Amne}^{\text{mx}}+\text{Bmne}^{\text{nx}}$
$=0$
$=\text{R.H.S.}$
View full question & answer→Question 1023 Marks
Differentiate the functions given in Exercise:
$(\text{x}+3)^2.(\text{x}+4)^3.(\text{x}+5)^4$
AnswerLet $\text{y}=(\text{x}+3)^2.(\text{x}+4)^3.(\text{x}+5)^4\ \dots\text{(i)}$
Taking logs on both sides, we have
$\log\text{y}=2\log(\text{x}+3)+3\log(\text{x}+4)+4\log(\text{x}+5)^4$
$\therefore\ \frac{\text{d}}{\text{dx}}\log\text{y}=2\frac{\text{d}}{\text{dx}}\log(\text{x}+3)+3\frac{\text{d}}{\text{dx}}\log(\text{x}+4)+4\frac{\text{d}}{\text{dx}}\log(\text{x}+5)$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=2\frac{1}{\text{x}+3}\frac{\text{d}}{\text{dx}}(\text{x}+3)+3\frac{1}{\text{x}+4}\frac{\text{d}}{\text{dx}}(\text{x}+4)+4\frac{1}{\text{x}+5}\frac{\text{d}}{\text{dx}}(\text{x}+5)$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{2}{\text{x}+3}+\frac{3}{\text{x}+4}+\frac{4}{\text{x}+5}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big(\frac{2}{\text{x}+3}+\frac{3}{\text{x}+4}+\frac{4}{\text{x}+5}\Big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(\text{x}+3)^2(\text{x}+4)^3(\text{x}+5)^4\Big(\frac{2}{\text{x}+3}+\frac{3}{\text{x}+4}+\frac{4}{\text{x}+5}\Big)\ \text{[From eq.(i)]}$
View full question & answer→Question 1033 Marks
If $\text{y}=\sqrt{\tan\text{x}+\sqrt{\tan\text{x}+\sqrt{\tan\text{x}+\ .... \text{to }\infty}}}$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sec^2\text{x}}{2\text{y}-1}$
AnswerWe have, $\text{y}=\sqrt{\tan\text{x}+\sqrt{\tan\text{x}+\sqrt{\tan\text{x}+\ .... \text{to }\infty}}}$
$\Rightarrow\text{y}=\sqrt{\tan\text{x}+\text{y}}$
Squaring both sides, we get,
$\text{y}^2=\tan\text{x}+\text{y}$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=\sec^2\text{x}+\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{y}-1)=\sec^2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sec^2\text{x}}{2\text{y}-1}$
View full question & answer→Question 1043 Marks
If $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}, 0 <\text{x}<\frac{1}{2},$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}$
Put $2\text{x}=\cos\theta$
$\therefore\ \text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}\sqrt{1-\cos^2\theta}$
$\Rightarrow \text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}(\sin\theta)$
$\therefore\ \text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}\Big[\cos\big(\frac{\pi}{2}\big)-\theta\Big]\ .....(\text{i})$
Here, $0<\text{x}<\frac{1}{2}$
$\Rightarrow 0<2\text{x}<1$
$\Rightarrow 0<\cos\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{2}$
And
$\Rightarrow 0> -\theta>-\frac{\pi}{2}$
$\Rightarrow\ \frac{\pi}{2}>\big(\frac{\pi}{2}-\theta\big)>0$
View full question & answer→Question 1053 Marks
If $\text{y}=\text{e}^{\text{a}\cos^{-1}}\text{x}$ prove that $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{a}^2\text{y}=0$
AnswerHere,
$\text{y}=\text{e}^{\text{a}\cos^{-1}}\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=-\text{e}^{\text{a}\cos^{-1}}\text{x}\ \times\frac{\text{a}}{\sqrt{1-\text{x}^2}}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{e}^{\text{a}\cos^{-1}}\text{x}\ \times\frac{\text{a}^2}{1-\text{x}^2}+\frac{\text{xa }\text{e}^{\text{a}\cos^{-1}}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{y}\times\frac{\text{a}^2}{1-\text{x}^2}-\frac{\text{x}\frac{\text{dy}}{\text{dx}}}{(1-\text{x}^2)}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{a}^2\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}-\text{a}^2\text{y}=0$
View full question & answer→Question 1063 Marks
Verify Mean Value Theorem, if $f(x) = x^3 – 5x^2 – 3x$ in the interval $[a, b]$, where $a = 1$ and $b = 3$. Find all $\text{c}\in(1,\ 3)$ for which $f′(c) = 0$.
AnswerFunction is continuous in [1, 3] as it is a polynomial function and polynomial function is always continuous.
$f'(x) = 3x^2 -10x$, f'(x) exists in [1, 3], hence derivable. Conditions of MVT theorem are satisfied, hence there exists, at least one $\text{c}\in(1,\ 3)$ such that
$\frac{\text{f}(3)-\text{f}(1)}{3-1}=\text{f}'\text{(c)}\ \Rightarrow\ \frac{-21-(-7)}{2}=3\text{c}^2-10\text{c}$
$\Rightarrow\ -7 = 3\text{c}^2 - 10\text{c}$ $\ \Rightarrow\ 3\text{c}^2 - 10\text{c} + 7 = 0$
$\Rightarrow\ 3\text{c}^2-7\text{c}-3\text{c}+7=0$ $\ \Rightarrow\ \text{c}(3\text{c} -7) -1(3\text{c}-7)=0$
$\Rightarrow\ (3\text{c}-7)(\text{c} -1)=0$ $\ \Rightarrow\ (3\text{c}-7)=0\text{ or } (\text{c}-1)=0$
$\Rightarrow\ 3\text{c}=7 \text{ or}\text{ c}=1 $ $\ \Rightarrow\ \text{c}=\frac{7}{3}\text{or c}=1$
$\therefore\ \text{c}=\frac{7}{3}\in(1, 3)$ and other value $\in(1,\ 3)$
Since $\text{f}(1)\neq\text{f}(3),$ therefore the value of 'c' does not exist such that f(c) = 0.
View full question & answer→Question 1073 Marks
Find the second order derivatives of the following functions:$\sin(\log\text{x})$
AnswerLet $\text{y}=\sin(\log\text{x})$
Then
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}[\sin(\log\text{x})]=\cos(\log\text{x}).\frac{\text{d}}{\text{dx}}(\log\text{x})=\frac{\cos(\log\text{x})}{\text{x}}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big[\frac{\cos(\log\text{x})}{\text{x}}\Big]$
$=\frac{\text{x}.\frac{\text{d}}{\text{dx}}[\cos(\log\text{x})]-\cos(\log\text{x}).\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}$
$=\frac{\text{x}.\Big[-\sin(\log\text{x}).\frac{\text{d}}{\text{dx}(\log\text{x})}\Big]-\cos(\log\text{x}.1}{\text{x}^2}$
$\frac{-\text{x}\sin(\log\text{x}).\frac{1}{\text{x}}-\cos(\log\text{x})}{\text{x}^2}$
$=\frac{[-\sin(\log\text{x})+\cos(\log\text{x})]}{\text{x}^2}$
View full question & answer→Question 1083 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\frac{\text{|x}^2-1|}{\text{x}-1},\text{for} & \text{x} \neq1\\2, &\text{for} \text{ x} = 1\end{cases} \text{at x}=1$
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{|x}^2-1|}{\text{x}-1}, & \text{x} \neq1\\2, &\text{ x} = 1\end{cases}$
$\Rightarrow\text{f}\text{(x)}=\begin{cases}\text{x}+1, & \text{x}< -1\\-\text{x}-1, & -1\leq \text{x} <0\\\text{x}+1,& \text{x}>1\\2,&\text{x}=1\end{cases}$
We obseve
$(\text{LHL at x}=1)=\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(1-h)}$
$=\lim\limits_{\text{h} \rightarrow 0}-\text{(1-h)}-1=\lim\limits_{\text{h} \rightarrow 0}-2+\text{h}=-2$
And, $\text{f}(1)=2$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}\neq\text{f}(1)$
Hence, f(x) is discontinuous at x = 1
View full question & answer→Question 1093 Marks
Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=[\text{x}]\text{ for }-1\leq\text{x}\leq1,$ where [x] denotes the greatest integer not exceeding x.
AnswerHere, $\text{f}(\text{x})=[\text{x}]$ and $\text{x}\in[-1,1],$ at n = 1 $\text{LHL}=\lim\limits_{\text{x}\rightarrow(1-\text{h})}[\text{x}]$ $=\lim\limits_{\text{h}\rightarrow0}[1-\text{h}]$ $=0$ $\text{RHL}=\lim\limits_{\text{x}\rightarrow(1+\text{h})}[\text{x}]$ $=\lim\limits_{\text{h}\rightarrow0}[1+\text{h}]$ $=1$ $\text{LHL}\neq\text{RHL}$So, f(x) is not continuos at $1\in[-1,1]$
Hence, Rolle's theorem is not applicable on f(x) in [-1, 1].
View full question & answer→Question 1103 Marks
If $\log\text{y}=\tan^{-1}$ show that $(1+\text{x}^2)\text{y}_2+(2\text{x}-1)\text{y}_1=0$
AnswerHere$\log\text{y}=\tan^{-1}$
Differentiating w.r.t.x, we get
$\frac{1}{\text{y}}\times\text{y}_1=\frac{1}{1+\text{x}^2}$ $\Rightarrow(1+\text{x}^2)\text{y}_1=\text{y}$ $\Rightarrow(1+\text{x}^2)\text{y}_2+2\text{xy}_1=\text{y}_1$ $\Rightarrow(1+\text{x}^2)\text{y}_2+2\text{xy}_1-\text{y}_1=0$$\Rightarrow(1+\text{x}^2)\text{y}_2+(25\text{x}-1)\text{y}_1=0$
hence proved
View full question & answer→Question 1113 Marks
Differentiate of the following w.r.t. x:
$\sin^\text{m}\text{x}\cdot\cos^\text{n}\text{x}$
AnswerLet $\text{y}=\sin^\text{m}\text{x}\cdot\cos^\text{n}\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[(\sin\text{x})^\text{m}\cdot(\cos\text{x})^\text{n}\big]$
$=(\sin\text{x})^\text{m}\cdot\frac{\text{d}}{\text{dx}}(\cos\text{x})^\text{n}+(\cos\text{x})^\text{n}\cdot\frac{\text{d}}{\text{dx}}(\sin\text{x})^\text{m}$
$=(\sin\text{x})^\text{m}\cdot\text{n}(\cos\text{x})^{\text{n}-1}\cdot\frac{\text{d}}{\text{dx}}\cos\text{x}+(\cos\text{x})^\text{n}\text{m}(\sin\text{x})^{\text{m}-1}\cdot\frac{\text{d}}{\text{dx}}\sin\text{x}$
$=(\sin\text{x})^\text{m}\cdot\text{n}(\cos\text{x})^{\text{n}-1}(-\sin\text{x})+(\cos\text{x})^\text{n}\cdot\text{m}(\sin\text{x})^{\text{m}-1}\cos\text{x}$
$=-\text{n}\cdot\sin^\text{m}\text{x}\cdot\sin\text{x}\cdot\cos^\text{n}\text{x}\cdot\frac{1}{\cos\text{x}}+\text{m}\cdot\sin^\text{m}\text{x}\cdot\frac{1}{\sin\text{x}}\cdot\cos^\text{n}\text{x}\cdot\cos\text{x}$
$=-\text{n}\sin^\text{m}\text{x}\cdot\cos^\text{n}\text{x}\cdot\tan\text{x}+\text{m}\sin^\text{m}\text{x}\cdot\cos^\text{n}\text{x}\cdot\cot\text{x}$
$=\sin^\text{m}\text{x}\cdot\cos^\text{n}\text{x}\big[\text{m}\cot\text{x}-\text{n}\tan\text{x}\big]$
View full question & answer→Question 1123 Marks
Discuss the continuity of the function $\text{f(x)}=\begin{cases}\frac{\text{x}}{|\text{x}|},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$
AnswerWhen $\text{x}\neq0,$
$\text{f(x)}=\frac{\text{x}}{|\text{x}|}=\begin{cases}\frac{-\text{x}}{\text{x}}=-1;&\text{x}<0\\\frac{\text{x}}{|\text{x}|}=1;&\text{x}>0\end{cases}$
So, f(x) is a constant function when $\text{x}\neq0,$
Hence, is continuous for all x < 0 and x > 0
Now, Consider the point x = 0
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})=\lim_\limits{\text{h}\rightarrow0}\frac{-\text{h}}{|-\text{h}|}=-1$
$\text{RHL }=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}}{|\text{h}|}=1$
So, $\text{LHL}\neq\text{RHL}$
Hence, function is discontinuous at x = 0
View full question & answer→Question 1133 Marks
Differentiate the following functions with respect to x:
$\log(\cos\text{x}^2)$
AnswerConsider $\text{y}=\log(\cos\text{x}^2)$
Differentiate it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log(\cos\text{x}^2)$
$=\frac{-2\text{x}\sin\text{x}^2}{\cos\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=-2\text{x}\tan\text{x}^2$
Therefore,
$\frac{\text{dy}}{\text{dx}}=-2\text{x}\tan\text{x}^2$
View full question & answer→Question 1143 Marks
Differentiate the functions with respect to x.
$\cos\text{x}^{3}. \sin^{2}(\text{x}^{5})$
Answer$\text{Let y} =\cos\text{x}^{3}. \sin^{2}(\text{x}^{5})$
$\therefore \frac{\text{dy}}{\text{dx}} =\cos\text{x}^{3} \frac{\text{d}}{\text{dx}}\sin^{2}(\text{x}^{5})+\sin^{2}(\text{x}^{5})\frac{\text{d}}{\text{dx}}\cos\text{x}^{3}$
$=\cos\text{x}^{3}.2\sin(\text{x}^{5}) \frac{\text{d}}{\text{dx}}\sin(\text{x}^{5})+\sin^{2}(\text{x}^{5})(-\sin\text{x}^{3})\frac{\text{d}}{\text{dx}}\text{x}^{3}$
$=\cos\text{x}^{3}.2\sin\text{x}^{5} .\cos\text{x}^{5}\frac{\text{d}}{\text{dx}}\text{x}^5+\sin^{2}(\text{x}^{5})(-\sin\text{x}^{3})3\text{x}^{2}$
$=\cos\text{x}^{3}.2\sin(\text{x}^{5}) \cos(\text{x}^{5})(5\text{x}^{4})-\sin^{2}(\text{x}^{5})\sin\text{x}^{3}.3\text{x}^{2}$
$=10\text{x}^{4}\cos\text{x}^{3} \sin(\text{x}^{5})\cos(\text{x}^{5})-3\text{x}^{2}\sin^{2}(\text{x}^{5})\sin\text{x}^{3}$
View full question & answer→Question 1153 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$xy = c^2$
AnswerWe have, $xy = c^2$
Differentiating with respect to x, we get
$\frac{\text{d}}{\text{dx}}(\text{xy})=\frac{\text{d}}{\text{dx}}(\text{c}^2)$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})=0$
[Using product rule]
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=-\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$
View full question & answer→Question 1163 Marks
If $\text{x}=3\sin\text{t}-\sin3\text{t},\text{y}=3\cos3\text{t}-\cos3\text{t}$ find $\frac{\text{dy}}{\text{dx}}\text{ at t}=\frac{\pi}{3}$
Answer$\text{x}=3\sin\text{t}-\sin3\text{t and } \text{y}=3\cos3\text{t}-\cos3\text{t}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=3\cos\text{t}-3\cos3\text{t}\text{ and} \\ \frac{\text{dy}}{\text{dt}}=-3\sin\text{t}-3\sin3\text{t}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-3\sin\text{t}+3\sin3\text{t}}{3\cos\text{t}-3\cos3\text{t}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}=\frac{\pi}{3}}=\frac{-3\sin\frac{\pi}{3}+3\sin\pi}{3\cos\frac{\pi}{3}-3\cos\pi}$
$=\frac{3\times\frac{\sqrt{3}}{2}+0}{3\times\frac{1}{2}+3}$
$=\frac{\frac{-3\sqrt{3}}{2}}{\frac{9}{2}}$
$=-\frac{1}{\sqrt{3}}$
View full question & answer→Question 1173 Marks
Write the derivative of $\sin\text{x}$ with respect to $\cos\text{x}$.
AnswerLet $\text{u}=\sin\text{x}\text{ and v}=\cos\text{x}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\cos\text{x and }\frac{\text{dv}}{\text{dx}}=-\sin\text{x}$
$\therefore\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{\cos\text{x}}{-\sin\text{x}}$
$\Rightarrow\frac{\text{du}}{\text{dv}}=-\cot\text{x}$
View full question & answer→Question 1183 Marks
Differentiate the function given in Exercise:
$\cos\text{x}.\cos 2\text{x}.\cos 3\text{x}$
AnswerLet $\text{y}=\cos\text{x}\cos2\text{x}\cos3\text{x}\ \dots\text{(i})$
Taking logs on both sides, we have
$\log\text{y}=\log(\cos\text{x}\cos2\text{x}\cos3\text{x})$ $=\log\cos\text{x}+\log\cos2\text{x}+\log\cos3\text{x}$
$\therefore\ \frac{\text{d}}{\text{dx}}\log\text{y}=\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\frac{\text{d}}{\text{dx}}\log\cos2\text{x}+\frac{\text{d}}{\text{dx}}\log\cos3\text{x}$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{d}}{\text{dx}}=\frac{1}{\cos\text{x}}\frac{\text{d}}{\text{dx}}\cos\text{x}+\frac{1}{\cos2\text{x}}\frac{\text{d}}{\text{dx}}\cos2\text{x}+\frac{1}{\cos3\text{x}}\frac{\text{d}}{\text{dx}}\cos3\text{x}$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{d}}{\text{dx}}=\frac{1}{\cos\text{x}}(-\sin\text{x})+\frac{1}{\cos2\text{x}}(-\sin2\text{x})\frac{\text{d}}{\text{dx}}2\text{x}+\frac{1}{\cos3\text{x}}(-\sin3\text{x})\frac{\text{d}}{\text{dx}}3\text{x}$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=-\tan\text{x}-(\tan2\text{x})2-\tan3\text{x}(3)$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}=-\text{y}(\tan\text{x}+2\tan2\text{x}+3\tan3\text{x})$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}=-\cos\text{x}\cos2\text{x}\cos3\text{x}(\tan\text{x}+2\tan2\text{x}+3\tan3\text{x})\ \ [\text{From eq. (i)}]$
View full question & answer→Question 1193 Marks
Find the values of a and b such that the function defined by
$\text{f(x)}=\begin{cases}5,&\text{if}\ \text{x}\leq{2}\\\text{ax} + \text{b},& \text{if}\ 2<\text{x}<10\\21,&\text{if}\ \text{x}\geq10\end{cases}$
is a continuous function.
Answer$\text{f(x)}=\begin{cases}5,&\text{if}\ \text{x}\leq{2}\\\text{ax} + \text{b},& \text{if}\ 2<\text{x}<10\\21,&\text{if}\ \text{x}\geq10\end{cases}$
$\therefore$ f is continuous function
$\therefore$ f is continuous at x = 2 and x = 10
$\therefore$ f is right continuous at x = 2 and left continuous at x = 10.
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}^{+}}\text{f(x)} = \text{f}(2)\Rightarrow 2\text{a} + \text{b} = 5\ ...{(\text i)}$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{10}^{-}}\text{f(x)} = \text{f}(10)\Rightarrow 10\text{a} + \text{b} = 21\ ...{(\text {ii})}$
Subtracting (1) from(2), we get,
8a = 16 or a = 2
$\therefore$ from (1), 4 + 6 = 5 ⇒ b = 1
$\therefore$ we have a = 2, b = 1
View full question & answer→Question 1203 Marks
Discuss the continuity of $\text{f}\text{(x)}=\begin{cases}2\text{x}-1, & \text{x} < 0\\2\text{x}+1, & \text{x} \geq 0\end{cases}\text{at}\text{ x}=0$
Answer$\text{f}\text{(x)}=\begin{cases}2\text{x}-1, & \text{x} < 0\\2\text{x}+1, & \text{x} \geq 0\end{cases}\text{at}\text{ x}=0$
$\text{(LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=2(0)-1=-1$
$\text{(RHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}=2(0)+1=1$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$
Hence, f(x) is discontinuous at x = 0.
View full question & answer→Question 1213 Marks
Find $\frac{\text{dy}}{\text{dx}}$ when x and y are connected by the relation:
$\sec(\text{x}+\text{y})=\text{xy}$
AnswerConsider, $\sec(\text{x}+\text{y})=\text{xy}$
$\Rightarrow\ \sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})\cdot\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{y}+\text{y}\cdot\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ \sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})\cdot\Big(1+\frac{\text{dy}}{\text{dx}}\Big)=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}$
$\Rightarrow\ \sec(\text{x}+\text{y})\tan(\text{x}+\text{y})+\sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})\cdot\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}$
$\Rightarrow\ \sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})\cdot\frac{\text{dy}}{\text{dx}}-\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\sec(\text{x}+\text{y})\tan(\text{x}+\text{y})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\big[\sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})-\text{x}\big]=\text{y}-\sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})}{\sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})-\text{x}}$
View full question & answer→Question 1223 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\text{e}^{\text{x}}}{\log_\text{e}(1+2\text{x})},&\text{if }\text{ x}\neq0\\7,&\text{if }\text{ x}=0\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\text{e}^{\text{x}}}{\log_\text{e}(1+2\text{x})},&\text{if }\text{ x}\neq0\\7,&\text{if }\text{ x}=0\end{cases}$
We have,
$\lim_\limits{\text{x}\rightarrow0}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0}\frac{\text{e}^{\text{x}}}{\log_\text{e}(1+2\text{x})}=\lim_\limits{\text{x}\rightarrow0}\frac{\big(\frac{\text{e}^{\text{x}}-1}{\text{x}}\Big)}{\Big(\frac{2\log_\text{e}(1+2\text{x})}{2\text{x}}\Big)}$
$=\frac{1}{2}\times\frac{\big(\frac{\text{e}^{\text{x}}-1}{\text{x}}\Big)}{\Big(\frac{2\log_\text{e}(1+2\text{x})}{2\text{x}}\Big)}=\frac{1}{2}$
It is given that f(0) = 7
$\lim_\limits{\text{x}\rightarrow0}\text{f(x)}\neq\text{f}(0)$
Hence, the given function is discontinuous at x = 0 and continuous elsewhere.
View full question & answer→Question 1233 Marks
If $\text{y}=\sec^{-1}\Big(\frac{\text{x}+1}{\text{x}-1}\Big)+\sin^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big),\text{x}>0.$ Find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\sec^{-1}\Big(\frac{\text{x}+1}{\text{x}-1}\Big)+\sin^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$
$\text{y}=\cos^{-1}\Big(\frac{\text{x}+1}{\text{x}-1}\Big)+\sin^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$
$\Big[\text{Since, } \sec^{-1}(\text{x})=\cos^{-1}\Big(\frac{1}{\text{x}}\Big)\Big]$
$\text{y}=\frac{\pi}{2} \Big[\text{Since}, \cos^{-1}\text{x}+\sin^{-1}=\frac{\pi}{2}\Big]$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 1243 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},& \text{when}\text{ x}\neq0 \\1,&\text{when} \text{ x}=0\end{cases}$ Find whether f(x) is continuous at x = 0.
AnswerGiven,
$\text{f}\text{(x)}=\frac{\text{x}^2-1}{\text{x}-1},\text{ if}\text{ x}\neq1$
$\text{f}\text{(x)}=2,\text{ if}\text{ x}=1$
We observe
$\text{(LHL at x = 1)}$
$\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{ (x)}=\lim\limits_{\text{x} \rightarrow 0}(1-\text{h})$
$\lim\limits_{\text{x} \rightarrow 0}(1-\text{h})=\lim\limits_{\text{x} \rightarrow 0}\frac{(1-\text{h})^2-1}{(1-\text{h})^2-1}$
$\lim\limits_{\text{x} \rightarrow 0}\frac{1-\text{h}^2-2\text{h}-1}{1-\text{h}-1}$
$\lim\limits_{\text{x} \rightarrow 0}\frac{\text{h}^2-2\text{h}}{-\text{h}}$
$\lim\limits_{\text{x} \rightarrow 0}2-\text{h}$
$= 2$
$(\text{RHL at x}=1)$
$\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}(1+\text{h)}$
$\lim\limits_{\text{h} \rightarrow 0}(1-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{(1+\text{h})^2-1}{(1+\text{h})-1}$
$\lim\limits_{\text{h} \rightarrow 0}\frac{1+\text{h}^2+2\text{h}-1}{1+\text{h}-1}$
$\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+2\text{h}}{\text{h}}$
$\lim\limits_{\text{h} \rightarrow 0}\text{h}+2$
$=2$
Also f(x) = 2
$\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}=\text{f}(1)$
Hence f(x) is continuous at x = 1.
View full question & answer→Question 1253 Marks
Determine the value of the constant k so that the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin2\text{x}}{5\text{x}}, &\text{if}\text{ x}\neq0\\\text{k}, &\text{if}\text{ x}=0\end{cases}$ is continuous at x = 0.
AnswerWe have given that the funtion is continuous at x = 0
So, LHL = RHL = f(0) ....(i)
Now,
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin2(-\text{h})}{5(-\text{h})}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\sin2\text{h}}{-5\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin2\text{h}}{2\text{h}}\times\frac{2\text{h}}{5\text{h}}=\frac{2}{5}$
$\text{f}(0)=\text{k}$
Using(i), $\text{k}=\frac{2}{5}$
View full question & answer→Question 1263 Marks
Differentiate following w.r.t. x:
$\sin^\text{n}\big(\text{ax}^2+\text{bx}+\text{c}\big)$
AnswerLet $\text{y}=\sin^\text{n}\big(\text{ax}^2+\text{bx}+\text{c}\big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\sin\big(\text{ax}^2+\text{bx}+\text{c}\big)\Big]^\text{n}$
$=\text{n}\cdot\Big[\sin\big(\text{ax}^2+\text{bx}+\text{c}\big)\Big]^{\text{n}-1}\cdot\frac{\text{d}}{\text{dx}}\sin\big(\text{ax}^2+\text{bx}+\text{c}\big)$
$=\text{n}\cdot\sin^{\text{n}-1}\big(\text{ax}^2+\text{bx}+\text{c}\big)\cdot\cos\big(\text{ax}^2+\text{bx}+\text{c}\big)\cdot\frac{\text{d}}{\text{dx}}\big(\text{ax}^2+\text{bx}+\text{c}\big)$
$=\text{n}\cdot\sin^{\text{n}-1}\big(\text{ax}^2+\text{bx}+\text{c}\big)\cdot\cos\big(\text{ax}^2+\text{bx}+\text{c}\big)\cdot(2\text{ax + b})$
$=\text{n}\cdot(2\text{ax + b})\cdot\sin^{\text{n}-1}\big(\text{ax}^2+\text{bx}+\text{c}\big)\cdot\cos\big(\text{ax}^2+\text{bx}+\text{c}\big)$
View full question & answer→Question 1273 Marks
Differentiate w.r.t. x the function in Exercise:
$(5\text{x})^{3\cos2\text{x}}$
AnswerLet $\text{y}=(5\text{x})^{3\cos2\text{x}}$ Taking logaritthm on both the sides, we obtain $\log\text{y}=3\cos2\text{x}\log5\text{x}$ Differentiating both sides with respect to x, we obtain$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3\Big[\log5\text{x}.\frac{\text{d}}{\text{dx}}(\cos2\text{x})+\cos2\text{x}.\frac{\text{d}}{\text{dx}}(\log5\text{x})\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=3\text{x}\Big[\log5\text{x}(-\sin2\text{x}).\frac{\text{d}}{\text{dx}}(2\text{x})+\cos2\text{x}.\frac{1}{5\text{x}}.\frac{\text{d}}{\text{dx}}(5\text{x})\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=3\text{y}\Big[-2\sin2\text{x}\log5\text{x}+\frac{\cos2\text{x}}{\text{x}}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=3\text{y}\Big[\frac{3\cos2\text{x}}{\text{x}}-6\sin2\text{x}\log5\text{x}\Big]$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(5\text{x})^{\text{x}\cos2\text{x}}\Big[\frac{3\cos2\text{x}}{\text{x}}-6\sin2\text{x}\log5\text{x}\Big]$
View full question & answer→Question 1283 Marks
Find the values of k so that the function f is continuous at the indicated point:
$\text{f(x)}= \begin{cases}\text{k}\text{x}+1,\ \text{if}\ \text{x}\leq{\pi}\\ \cos\text{x}, \ \ \ \ \text{if}\ \text{x} >{\pi}\end{cases}$
$\text{at}\ \text{x} = {\pi}$
AnswerHere $\text{f(x)}= \begin{cases}\text{k}\text{x}+1,\ \text{if}\ \text{x}\leq{\pi}\\ \cos\text{x}, \ \ \ \ \text{if}\ \text{x} >{\pi}\end{cases}$ $^{\ \ \text{Lt}}_{\text{x}\rightarrow{\pi}^{-}}\text{f(x)}= ^{\ \ \text{Lt}}_{\text{x}\rightarrow{\pi}^{-}}(\text{k}\text{x}+1)$ $\left[\text{Put}\ \text{x} = \pi -\text{h}, \text{h}>0\ \text{so that}\ \text{h}\rightarrow0\ \text{as}\ \text{x}\rightarrow\pi^{-}\right]$ $= ^{\ \ \text{Lt}}_{\text{x}\rightarrow0}\left\{\text{k}(\pi - \text{h})+1\right\}={\text{k}}(\pi - 0) = 1 = \text{k}\pi + 1$ $^{\ \ \text{Lt}}_{\text{x}\rightarrow\pi^{+}}\text{f(x)} =^{\ \ \text{Lt}}_{\text{x}\rightarrow\pi^{+}}\cos\pi$ $\left[\text{Put}\ \text{x} = \pi +\text{h}, \text{h}>0\ \text{so that}\ \text{h}\rightarrow0\ \text{as}\ \text{x}\rightarrow\pi^{+}\right]$ $^{\ \ \text{Lt}}_{\text{h}\rightarrow0}\cos(\pi + \text{h}) = ^{\ \ \text{Lt}}_{\text{h}\rightarrow0}(\cos\pi\cos\text{h} - \sin\pi\sin\text{h})$ $\cos \pi. 1 - \sin \pi. 0 = \cos\pi = -1$Since f(x) is continuous at $\text{x} = \pi$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow{\pi}^{-}}\text {f(x}) =^{\ \ \text{Lt}}_{\text{x}\rightarrow{\pi}^{+}}\text{f(x)}$ $\therefore \text{k}\pi = -1 \Rightarrow \text{k} = -\frac{1}{\pi}$
View full question & answer→Question 1293 Marks
If $\text{x}=\frac{1+\log\text{t}}{\text{t}^2},\text{y}=\frac{3+2\log\text{t}}{\text{t}},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$\text{x}=\frac{1+\log\text{t}}{\text{t}^2},\text{y}=\frac{3+2\log\text{t}}{\text{t}}$
$\frac{\text{dy}}{\text{dt}}=\frac{\text{t}^2\big(\frac{1}{\text{t}}\big)-(1+\log\text{t})(2\text{t})}{\text{t}^4} \\ =\frac{\text{t}-2\text{t}-2\text{t}\log\text{t}}{\text{t}^4}=\frac{-2\log\text{t}-1}{\text{t}^3}$
$\frac{\text{dy}}{\text{dt}}=\frac{\text{t}\big(\frac{2}{\text{t}}\big)-(3+2\log\text{t})(1)}{\text{t}^2} \\ =\frac{2-3-2\log\text{t}}{\text{t}^2}=\frac{-2\log\text{t}-1}{\text{t}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\frac{-2\log\text{t}-1}{\text{t}^2}}{\frac{-2\log\text{t}-1}{\text{t}^3}}=\text{t}$
View full question & answer→Question 1303 Marks
Differentiate the following w.r.t. x:
$\cos\big(\tan\sqrt{\text{x}+1}\big)$
AnswerLet $\text{y}=\cos\big(\tan\sqrt{\text{x}+1}\big)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\cos\big(\tan\sqrt{\text{x}+1}\big)$ $=-\sin\big(\tan\sqrt{\text{x}+1}\big)\cdot\frac{\text{d}}{\text{dx}}\big(\tan\sqrt{\text{x}+1}\big)$
$=-\sin\big(\tan\sqrt{\text{x}+1}\big).\sec^2\sqrt{\text{x}+1}.\frac{\text{d}}{\text{dx}}(\text{x}+1)^{\frac{1}{2}}$ $\Big[\because\frac{\text{d}}{\text{dx}}(\tan\text{x})=\sec^2\text{x}\Big]$
$=-\sin\big(\tan\sqrt{\text{x}+1}\big).\big(\sec\sqrt{\text{x}+1}\big)^2.\frac{1}{2}(\text{x}+1)^{\frac{1}{2}}.\frac{\text{d}}{\text{dx}}(\text{x}+1)$
$=\frac{-1}{2\sqrt{\text{x}+1}}.\sin\big(\tan\sqrt{\text{x}+1}\big).\sec^2\big(\sqrt{\text{x}+1}\big)$
View full question & answer→Question 1313 Marks
Examine the continuity of the function $f(x) = x^3 + 2x^2 - 1 at x = 1.$
AnswerWe know that, function f will be continuous at x = a, if $\lim\limits_{\text{x}\rightarrow\text{a}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\text{a}^+}\text{f(x)}=\text{f(a)}.$
Consider, $f(x) = x^3 + 2x^2 - 1 $at $x = 1.$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1+\text{h})^3+2(1+\text{h})^2-1=2$
and
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1-\text{h})^3+2(1-\text{h})^2-1=2$
$\because\ \lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}$
And $f(1) = 1 + 2 - 1 = 2$
Thus, f(x) is continuous at $x = 1.$
View full question & answer→Question 1323 Marks
Differentiate the functions given in Exercise:
$(\log\text{x})^{\cos\text{x}}$
AnswerLet $\text{y}=(\log\text{x})^{\cos\text{x}}\ \dots\text{(i)}$
Taking logs on both sides, we have
$\log\text{y}=\log(\log\text{x})^{\cos\text{x}}=\cos\text{x}\log(\log\text{x})$
$\therefore\ \frac{\text{d}}{\text{dx}}\log\text{y}=\frac{\text{d}}{\text{dx}}\ [\cos\text{x}\log(\log\text{x)}]$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}\log(\log\text{x})+\log(\log\text{x})\frac{\text{d}}{\text{dx}}\cos\text{x}\ \ \text{[By product rule}]$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log(\log\text{x})(-\sin\text{x})$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{\log\text{x}}.\frac{1}{\log\text{x}}-\sin\text{x}\log(\log\text{x})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\cos\text{x}}{\log\text{x}}-\sin\text{x}\log(\log\text{x})\Big]$ $=(\log\text{x})^{\cos\text{x}}\Big[\frac{\cos\text{x}}{\log\text{x}}-\sin\text{x}\log(\log\text{x})\Big]$
View full question & answer→Question 1333 Marks
Examine the continuity of f, where f is defined by
$\text{f(x)}=\begin{cases} \sin{\text{x}- \cos\text{x}}, \text{if} \ \text{x}\neq0\\-1, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\ \text{x} = 0\end{cases}$
Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.
AnswerIt is given that $\text{f(x)}=\begin{cases} \sin{\text{x}- \cos\text{x}}, \text{if} \ \text{x}\neq0\\-1, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\ \text{x} = 0\end{cases}$
We know that f is defined at all point of the real line.
Let k be a real number.
Case I: $\text{k} \neq 0,$
Then $\text{f(k)} =\sin\text{k}- \cos\text{k}$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(\sin\text{x}- \cos\text{x}) = \sin\text{k} - \cos\text{k}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$
Thus, f is continuous at all points x that is $\text{x}\neq0.$
Case II: $\text{k} = 0$
Then f(k) = f(0) = 0
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}(\sin\text{x} - \cos\text{x}) = \sin0 - \cos0 = 0 - 1 = - 1$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}(\sin\text{x} - \cos\text{x}) = \sin0 - \cos0 = 0 - 1 = - 1$
$\therefore \ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = \text{f(0)}$
Therefore , f is continuous at x = 0.
Therefore, f is has no point of discontinuity.
View full question & answer→Question 1343 Marks
Find the value of k in this question, so that the function f is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{2^{\text{x}+2}-16}{4^\text{x}-16},&\text{if x}\neq2\\\text{k},&\text{if x}=2\end{cases}$ at x = 2.
AnswerConsider, $\text{f(x)}=\begin{cases}\frac{2^{\text{x}+2}-16}{4^\text{x}-16},&\text{if x}\neq2\\\text{k},&\text{if x}=2\end{cases}$ at x = 2Since, f(x) is continuous at x = 2.
$\therefore$ L.H.L = R.H.L = f(2)
At x = 2, $=\lim\limits_{\text{h}\rightarrow2}\frac{2^\text{x}\cdot2^2-2^4}{4^\text{x}-4^2}=\lim\limits_{\text{h}\rightarrow2}\frac{4\cdot(2^\text{x}-4)}{(2^\text{x})^2-(4)^2}$
$=\lim\limits_{\text{h}\rightarrow2}\frac{4\cdot(2^\text{x}-4)}{(2^\text{x}-4)-(2^\text{x}+4)}$
$=\lim\limits_{\text{h}\rightarrow2}\frac{4}{2^\text{x}+4}=\frac{8}{4}=\frac{1}{2}$
But f(2) = k
$\therefore\ \text{k}=\frac{1}{2}$
View full question & answer→Question 1353 Marks
Find $\frac{\text{dy}}{\text{dx}}$ of the functions expressed in parametric:
$\sin\text{x}=\frac{2\text{t}}{1+\text{t}},\ \tan\text{y}=\frac{2\text{t}}{1-\text{t}^2}.$
AnswerWe have, $\sin\text{x}=\frac{2\text{t}}{1+\text{t}}$ and $\tan\text{y}=\frac{2\text{t}}{1-\text{t}^2}$
Let $\text{t}=\tan\text{z}$
$\therefore\ \sin\text{x}=\frac{2\tan\text{z}}{1+\tan^2\text{z}}=\sin2\text{z}$
$\therefore\ \text{x}=2\text{z}$
Also $\tan\text{y}=\frac{2\tan\text{z}}{1-\tan^2\text{z}}=\tan2\text{z}$
$\therefore\ \text{y}=2\text{z}$
$\therefore\ \text{y}=\text{x}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 1363 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\tan^{-1}\big(\text{x}^2+\text{y}^2\big)=\text{a}$
AnswerWe have, $\tan^{-1}\big(\text{x}^2+\text{y}^2\big)=\text{a}$Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\big[\tan^{-1}\big(\text{x}^2+\text{y}^2\big)\big]=\frac{\text{d}}{\text{dx}}(\text{a})$
$\Rightarrow\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{y}^2\big)=0$
$\Rightarrow\Big[\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\Big]\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}}$
View full question & answer→Question 1373 Marks
If $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{2},&\text{if }0\leq\text{ x}\leq1\\2\text{x}^2-3\text{x}+\frac{3}{2},&\text{if }1<\text{x}\leq2\end{cases}$ Show that f is continuous at x = 1.
AnswerWe want to discuss the continuity of the function at x = 1
We need to prove that
$\text{LHL}=\text{RHL}=\text{f}(1)$
$\text{f}(1)=\frac{1^2}{2}=\frac{1}{2}$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})=\lim_\limits{\text{h}\rightarrow0}\frac{(1-\text{h})^2}{2}=\frac{1}{2}$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}(1+\text{h})=\lim_\limits{\text{h}\rightarrow0}2(1+\text{h}^2)-3(1+\text{h})+\frac{3}{2}$
$=2-3+\frac{3}{2}=\frac{1}{2}$
Thus, $\text{LHL}=\text{RHL}=\text{f}(1)=\frac{1}{2}$
Hence, function is continuous at x = 1
View full question & answer→Question 1383 Marks
Discuss the continuity of the function f(x) at the point x = 0, where$\text{f}\text{(x)}=\begin{cases}\text{x}, & \text{x} > 0\\1,&\text{x}=0\\\text{-x}, & \text{x} > 0\end{cases}$
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\text{x}, & \text{x} > 0\\1,&\text{x}=0\\\text{-x}, & \text{x} > 0\end{cases}$
$\text{(LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(-h)}=\lim\limits_{\text{h} \rightarrow 0}-(-\text{h)}=0$
$\text{(RHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}$
$\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(h)}=0$
And, $\text{f}(0)=1$
$\therefore\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}\neq\text{f}(0).$
Hence, f(x) is discontinuous at x = 0.
View full question & answer→Question 1393 Marks
Let g(x) be the inverse of an invertible function f(x) which is derivable at $x = 3$. If $f(3) = 3$ and $f(3) = 9$, write the value of $g'(9)$.
AnswerWe have, $f(3) = 9, f'(3) = 9$
and $g(x) = f^{-1} (x)$
$\Rightarrow(\text{gof})(\text{x})=\text{x}$
$\Rightarrow\text{g}\{\text{f(x)}\}=\text{x}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big[\text{g}\{\text{f}\text{(x)}\}\big] = 1$
$\Rightarrow\text{g}'\big\{\text{f}\text{(x)}\big\}\frac{\text{d}}{\text{dx}}\big\{\text{f}(\text{x})\big\}=1$
$\Rightarrow\text{g}'\big\{\text{f}\text{(x)}\big\}\times\text{f}'\text{(x)}=1$
Putting x = 3, we get,
$\text{g}'\big\{\text{f}(3)\big\}\times\text{f}'(3)=1$
$\Rightarrow\text{g}'(9)\times9=1\big[\because\text{f}(3)=9,\text{f}'(3)=9\big]$
$\Rightarrow\text{g}'(9)=\frac{1}{9}$
View full question & answer→Question 1403 Marks
Find which of the function:
$\text{f(x)}=\begin{cases}3\text{x}+5,&\text{if x}\geq2\\\text{x}^2,&\text{if x}<2\end{cases}$
at x = 2
AnswerWe have, $\text{f(x)}=\begin{cases}3\text{x}+5,&\text{if x}\geq2\\\text{x}^2,&\text{if x}<2\end{cases}$
At x = 2, $\text{L.H.L}=\lim\limits_{\text{x}\rightarrow2^-}(\text{x})^2$
$=\lim\limits_{\text{h}\rightarrow0}(2-\text{h}^2)=\lim\limits_{\text{h}\rightarrow0}(4+\text{h}^2-4\text{h})=4$
And $\text{R.H.L}=\lim\limits_{\text{x}\rightarrow2^+}(3\text{x}+5)$
$=\lim\limits_{\text{h}\rightarrow0}\big[3(2+\text{h})+5\big]=11$
Since, L.H.L ≠ R.H.L at x = 2
So, f(x) is discontinuous at x = 2.
View full question & answer→Question 1413 Marks
Show that $\text{f}\text{(x)}=\begin{cases}1+\text{x}^2,&\text{if } 0\leq\text{x}\leq 1\\2-\text{x},&\text{if }\text{x} > 1\end{cases}$ is discontinuous at x = 1.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}1+\text{x}^2,&\text{if } 0\leq\text{x}\leq 1\\2-\text{x},&\text{if }\text{x} > 1\end{cases}$
We observe
$\text{(LHL at x}=1)\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(1-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}(1+1-\text{h)}^2=\lim\limits_{\text{h} \rightarrow 0}(2+\text{h}^2-\text{2h)}=2$
$\text{(RHL at x}=1)\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(1+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}(2-(1+\text{h))}=\lim\limits_{\text{h} \rightarrow 0}(1-\text{h)}=1$
$\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}$
Thus, f(x) is discontinuous at x = 1.
View full question & answer→Question 1423 Marks
Using the fact that $\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$and the differentiation, obtain the sum formula for cosines.
Answer$\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$
Differentiating both sides with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}[\sin(\text{A}+\text{B})]=\frac{\text{d}}{\text{dx}}(\sin\text{A}\cos\text{B})+\frac{\text{d}}{\text{dx}}(\cos\text{A}\sin\text{B})$
$\Rightarrow\ \cos(\text{A}+\text{B}).\frac{\text{d}}{\text{dx}}(\text{A}+\text{B})$
$=\cos\text{B}.\frac{\text{d}}{\text{dx}}(\sin\text{A})+\sin\text{A}.\frac{\text{d}}{\text{dx}}(\cos\text{B})+\sin\text{B}.\frac{\text{d}}{\text{dx}}(\cos\text{A})+\cos\text{A}.\frac{\text{d}}{\text{dx}}(\sin\text{B})$
$\Rightarrow\ \cos(\text{A}+\text{B}).\frac{\text{d}}{\text{dx}}(\text{A}+\text{B})$
$=\cos\text{B}.\cos\text{A}\frac{\text{dA}}{\text{dx}}+\sin\text{A}(-\sin\text{B})\frac{\text{dB}}{\text{dx}}+\sin\text{B}(-\sin\text{A})\frac{\text{dA}}{\text{dx}}+\cos\text{A}\cos\text{B}\frac{\text{dB}}{\text{dx}}$
$\Rightarrow\ \cos(\text{A}+\text{B}).\Big[\frac{\text{dA}}{\text{dx}}+\frac{\text{dB}}{\text{dx}}\Big]$
$=(\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}).\Big[\frac{\text{dA}}{\text{dx}}+\frac{\text{dB}}{\text{dx}}\Big]$
$\therefore\ \cos(\text{A}+\text{B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$
View full question & answer→Question 1433 Marks
If f(x) is an odd function, then write whether f'(x) is even of odd.
AnswerWe have, f(x) is an odd function.
$\Rightarrow\text{f}(-\text{x})=-\text{f}(\text{x})$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big\{\text{f}(-\text{x})\big\}=-\frac{\text{d}}{\text{dx}}\big\{\text{f}(\text{x})\big\}$
$\Rightarrow\text{f}'(-\text{x})\frac{\text{d}}{\text{dx}}(-\text{x})=-\text{f}'\text{(x)}$
$\Rightarrow\text{f}'(-\text{x})\times(-1)=-\text{f}'\text{(x)}$
$\Rightarrow\text{f}'(-\text{x})=-\text{f}'\text{(x)}$
$\Rightarrow\text{f}'(-\text{x})=\text{f}'\text{(x)}$
Thus, f'(x) is an even function.
View full question & answer→Question 1443 Marks
Differentiate the function given in Exercise:
$\sqrt{\frac{(\text{x}-1)(\text{x}-2)}{(\text{x}-3)(\text{x}-4)(\text{x}-5)}}$
AnswerLet $\text{y}=\sqrt{\frac{(\text{x}-1)(\text{x}-2)}{(\text{x}-3)(\text{x}-4)(\text{x}-5)}}=\Big(\frac{(\text{x}-1)(\text{x}-2)}{(\text{x}-3)(\text{x}-4)(\text{x}-5)}\Big)^{\frac{1}{2}}\ \dots\text{(i)}$
Taking logs on both sides, we have
$\log\text{y}=\frac{1}{2}[\log(\text{x}-1)+\log(\text{x}-2)-\log(\text{x}-3)-\log(\text{x}-4)-\log(\text{x}-5)]$
$\therefore\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big[\frac{1}{\text{x}-1}\frac{\text{d}}{\text{dx}}(\text{x}-1)+\frac{1}{\text{x}-2}\frac{\text{d}}{\text{dx}}(\text{x}-2)-\frac{1}{\text{x}-3}\frac{\text{d}}{\text{dx}}(\text{x}-3)-\frac{1}{\text{x}-4}\frac{\text{d}}{\text{dx}}(\text{x}-4)-\frac{1}{\text{x}-5}\frac{\text{d}}{\text{dx}}(\text{x}-5)\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{y}\Big[\frac{1}{\text{x}-1}+\frac{1}{\text{x}-2}-\frac{1}{\text{x}-3}-\frac{1}{\text{x}-4}-\frac{1}{\text{x}-5}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}\sqrt{\frac{(\text{x}-1)(\text{x}-2)}{(\text{x}-3)(\text{x}-4)(\text{x}-5)}}\Big[\frac{1}{\text{x}-1}+\frac{1}{\text{x}-2}-\frac{1}{\text{x}-3}-\frac{1}{\text{x}-4}-\frac{1}{\text{x}-5}\Big]\ \text{[From eq.(i)}]$
View full question & answer→Question 1453 Marks
If $\text{y}=\frac{\log\text{x}}{\text{x}},$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\log\text{x}-3}{\text{x}^3}.$
AnswerHere,
$\text{y}=\frac{\log\text{x}}{\text{x}},$
Differentiating w.r.t.x, we get
$\frac{\text{d}\text{y}}{\text{dx}}=\frac{1-\log\text{x}}{\text{x}^2}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{x}-2\text{x}(1-\log\text{x})}{\text{x}^4}$
$=\frac{-\text{x}-2\text{x}+2\text{x}\log\text{x}}{\text{x}^4}$
$=\frac{-3+2\log\text{x}}{\text{x}^3}$
$=\frac{2\log\text{x}-3}{\text{x}^3}$
Hence proved
View full question & answer→Question 1463 Marks
Differentiate the following functions with respect to x:
$\sin(\log\text{x})$
AnswerConsider $\text{y}=\sin(\log\text{x})$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin(\log\text{x})$
$=\cos(\log\text{x})\frac{\text{d}}{\text{dx}}(\log\text{x})$
[Using chain rule]
$=\frac{1}{\text{x}}\cos(\log\text{x})$
Hence, the solution is $\frac{\text{d}}{\text{dx}}=(\sin(\log\text{x}))=\frac{1}{\text{x}}\cos(\log\text{x})$
View full question & answer→Question 1473 Marks
If $\text{y}=\log\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{2\text{x}(\text{x}+1)}$
AnswerWe have, $\text{y}=\log\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
$=\frac{1}{\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}}\frac{\text{d}}{\text{dx}}\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
$=\frac{\sqrt{\text{x}}}{\text{x}+1}\Big(\frac{1}{2\sqrt{\text{x}}}-\frac{1}{2\text{x}\sqrt{\text{x}}}\Big)$
$=\frac{1}{2}\frac{\sqrt{\text{x}}}{\text{x}+1}\Big(\frac{\text{x}-1}{\text{x}\sqrt{\text{x}}}\Big)$
$=\frac{\text{x}-1}{2\text{x}(\text{x}+1)}$
So,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{2\text{x}(\text{x}+1)}$
View full question & answer→Question 1483 Marks
If $\text{y}=\text{ax}^{\text{n+1}}+\text{bx}^{-\text{n}}$ and $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\lambda\text{y}$ then write the value of $\lambda$
Answer$\text{y}=\text{ax}^{\text{n}+1}+\text{b}\text{x}^{-\text{-n}}$
and $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\lambda\text{y}$
Now,
$\frac{\text{dy}}{\text{dx}}=\text{a}(\text{n}+1)\text{x}^{\text{n}}-\text{bn x}^{-\text{n-1}}$
and $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{an}(\text{n}+1)\text{x}^{\text{n}+1}-\text{bn}(-\text{n}-1)\text{x}^{-\text{n}-2}$
Now, $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\lambda\text{y}[\text{given}]$
$\Rightarrow\text{x}^2[\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}]=\lambda(\text{ax}^{\text{n+1}}+\text{b x}^{-\text{n}})$
$\Rightarrow\text{an}(\text{n}+1)\text{x}^{\text{n}+1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}}=\lambda\text{ax}^{\text{n+1}}+\text{b x}^{-\text{n}}$
$\Rightarrow\text{n}(\text{n}+1)\text{ax}^{\text{n}+1}+\text{bx}^{-n}=\lambda\text{ax}^{\text{n}+1}+\text{dx}^{\text{-n}}$
$\Rightarrow\lambda=\text{n}(\text{n}+1)$
View full question & answer→Question 1493 Marks
Differentiate the following w.r.t.x: $\sin(\tan^{-1}\text{e}^{-\text{x}})$
Answer$\text{Let}\ \text{y}=\sin(\tan^{-1}\text{e}^{-\text{x}})$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\cos(\tan^{-1}\text{e}^{-\text{x}}).\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{e}^{-\text{x}})= \bigg[\because\frac{\text{d}}{\text{dx}}\sin\text{f(x)}=\cos\text{f(x)}\frac{\text{d}}{\text{dx}}\text{f(x)}\bigg]$
$=\cos(\tan^{-1}\text{e}^{-\text{x}}).\frac{1}{1+(\text{e}^{-\text{x}})^{2}}\frac{\text{d}}{\text{dx}}\text{e}^{-\text{x}}= \bigg[\because\frac{\text{d}}{\text{dx}}\ \tan^{-1} \text{f(x)}=\frac{1}{(\text{f(x)})^{2}}\frac{\text{d}}{\text{dx}}\text{f(x)}\bigg]$
$=\cos(\tan^{-1}\text{e}^{-\text{x}}).\frac{1}{1+\text{e}^{-\text{2x}}}\text{e}^{\text{-x}}\frac{\text{d}}{\text{dx}}({-\text{x}})$
$=\frac{-\text{e}^{-\text{x}}.\cos(\tan^{-1}\text{e}^{-\text{x}})}{1+\text{e}^{-2\text{x}}}$
View full question & answer→Question 1503 Marks
Differentiate the following w.r.t. x:
$2^{\cos^2}\text{x}$
AnswerLet $\text{y}=2^{\cos^2}\text{x}$
Taking logarithm on both sides, we get
$\log\text{y}=\log2\cos^2\text{x}$
$\Rightarrow\ \frac{\text{d}}{\text{dy}}(\log\text{y})\cdot\frac{\text{dy}}{\text{dx}}=\log2\frac{\text{d}}{\text{dx}}(\cos^2\text{x})$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\log2(2\cos\text{x})\frac{\text{d}}{\text{dx}}\cos\text{x}$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\log2\cdot2\cos\text{x}\cdot(-\sin\text{x})$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\log2\cdot[-(\sin2\text{x})]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-\text{y}\cdot\log2(\sin2\text{x})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-2^{\cos^2}\text{x}\cdot\log2(\sin2\text{x})$ $\big[\because\text{y}=2^{\cos^2}\text{x}\big]$
View full question & answer→Question 1513 Marks
Find the value of k for which $\text{f(x)}=\begin{cases}\frac{1-\cos4\text{x}}{8\text{x}^2},&\text{when x}\neq0\\\text{k},&\text{when x}=0\end{cases}$ is continous at x = 0.
AnswerGiven,
$\text{f(x)}=\begin{cases}\frac{1-\cos4\text{x}}{8\text{x}^2},&\text{when x}\neq0\\\text{k},&\text{when x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow 0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{1-\cos4\text{x}}{8\text{x}^2}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{2\sin^22\text{x}}{8\text{x}^2}=\text{f}(0)$
$\Rightarrow\frac{2}{2}\lim_\limits{\text{x}\rightarrow 0}\frac{\sin^22\text{x}}{4\text{x}^2}=\text{f}(0)$
$\Rightarrow\frac{2}{2}\lim_\limits{\text{x}\rightarrow 0}\Big(\frac{\sin2\text{x}}{2\text{x}^2}\Big)^2=\text{f}(0)$
$\Rightarrow1\times1=\text{f}(0)$
$\Rightarrow\text{k}=1$ $(\because\text{f}(0)=\text{k})$
View full question & answer→Question 1523 Marks
If $\text{y}=\text{e}^\text{x}(\sin\text{x}+\cos\text{x})$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
Answer$\text{y}=\text{e}^\text{x}(\sin\text{x}+\cos\text{x})$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}(\cos\text{x}-\sin\text{x})+(\sin\text{x}+\text{cos}\text{x})\text{e}^\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}+\text{e}^\text{x}(\cos\text{x}-\sin\text{x})$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+\text{e}^\text{x}(-\sin\text{x}-\cos\text{x})+(\cos\text{x}-\sin\text{x})\text{e}^\text{x}$
$=\frac{\text{dy}}{\text{dx}}-\text{y}+(\cos\text{x}-\sin\text{x})\text{e}^\text{x}$
Adding and substracting y on RHS
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+\text{e}^\text{x}(-\sin\text{x}-\cos\text{x})+(\cos\text{x}-\sin\text{x})\text{e}^\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
View full question & answer→Question 1533 Marks
If $\text{y}=(\tan^{-1}\text{x})^2$ then prove that $(1+\text{x}^2)\text{y}_2+2\text{x}(1+\text{x}^2)\text{y}_1=2$
AnswerHere,
$\text{y}=(\tan^{-1}\text{x})^2$
Differentiating w.r.t.x, we get
$\text{y}_1=\frac{2\tan^{-1}\text{x}}{1+\text{x}^2}$
Differentiating w.r.t.x, we get
$\text{y}_2=\frac{2-4\text{x}\tan^{-1}\text{x}}{(1+\text{x}^2)^2}$
$\Rightarrow\text{y}_2=\frac{2}{(1+\text{x}^2)^2}-\frac{2\tan^{-1}\text{x}\times2\text{x}}{(1+\text{x}^2)^2}$
$\Rightarrow\text{y}_2=\frac{2}{(1+\text{x}^2)^2}-\frac{2\text{xy}_1}{(1+\text{x}^2)^2}$
$\Rightarrow(1+\text{x}^2)^2\text{y}_2=2-2\text{x}(1+\text{x}^2)\text{y}_1$
$\Rightarrow(1+\text{x}^2)^2\text{y}_2+2\text{x}(1+\text{x}^2)\text{y}_2=2$
Hence proved
View full question & answer→Question 1543 Marks
Differentiate the following w.r.t. x:
$\tan^{-1}(\sec\text{x}+\tan\text{x}),\frac{-\pi}{2}<\text{x}<\frac{\pi}{2}$
AnswerLet $\text{y}=\tan^{-1}(\sec\text{x}+\tan\text{x})$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\tan^{-1}(\sec\text{x}+\tan\text{x})$
$=\frac{1}{1+(\sec\text{x}+\tan\text{x})^2}\cdot\frac{\text{d}}{\text{dx}}(\sec\text{x}+\tan\text{x})$
$=\frac{1}{1+\sec^2\text{x}+\tan^2\text{x}+2\sec\text{x}\cdot\tan\text{x}}\big[\sec\text{x}\cdot\tan\text{x}+\sec^2\text{x}\big]$
$=\frac{1}{\big(\sec^2\text{x}+\sec^2\text{x}+2\sec\text{x}\cdot\tan\text{x}\big)}\cdot\sec\text{x}\cdot(\sec\text{x}+\tan\text{x})$
$=\frac{1}{2\sec\text{x}(\tan\text{x}+\sec\text{x})}\cdot\sec\text{x}(\sec\text{x}+\tan\text{x})=\frac{1}{2}$
View full question & answer→Question 1553 Marks
If $\text{y}=\text{e}^{-\text{x}}\cos\text{x},$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{e}^{-\text{x}}\sin\text{x}.$
AnswerHere,
$\text{y}=\text{e}^{-\text{x}}\cos\text{x},$
differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=-\text{e}^{-\text{x}}\cos\text{x}$
$=-\text{e}-\text{x}\sin\text{x}+\text{e}-\text{x}\cos\text{x}$
differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{e}^{-\text{x}}\cos\text{x}-\text{e}^{-\text{x}}\sin\text{x}-\text{e}^{\text{-x}}\cos\text{x}$
$=2\text{e}^{-\text{x}}\sin\text{x}$
View full question & answer→Question 1563 Marks
$\text{If y}=\cos^{-1}\text{x},\text{ Find }\frac{\text{d}^2\text{y}}{\text{dx}^2}\text{ in terms of y alone}.$
Answer$\text{y}=\cos^{-1}\text{x}\ \ \dots(1)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}=-(1-\text{x}^2)^{-\frac{1}{2}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}(1-\text{x}^2)^{\frac{-3}{2}}(-2\text{x})=-\frac{\text{x}}{(1-\text{x}^2)^{\frac{3}{2}}}=-\frac{\cos\text{y}}{(1-\cos^2\text{y})^{\frac{3}{2}}}\ \ [\because\text{of }(1)]$
$=-\frac{\cos\text{y}}{(1-\cos^2\text{y})^{\frac{3}{2}}}=-\frac{\cos\text{y}}{\sin^3\text{y}}=-\frac{\cos\text{y}}{\sin\text{y}}.\frac{1}{\sin^2\text{y}}=-\cot\text{y cosec}^2\text{y}.$
View full question & answer→Question 1573 Marks
Write the value of the derivative of f(x) = |x − 1| + |x − 3| at x = 2.
AnswerGiven: f(x) = |x - 1| + |x - 3|
$\Rightarrow\text{f(x)}=\begin{cases}-(\text{x}-1)-(\text{x}-3), & \text{x}<1\\ \text{x}-1-(\text{x}-3),& 1\leq\text{x}<3\$\text{x}-1)+(\text{x}-3),&\text{x}\geq3\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}-2\text{x}+4, & \text{x}<1\\ 2,& 1\leq\text{x}<3\\2\text{x}-4,&\text{x}\geq3\end{cases}$
Wecheck differentiable at x = 2
(LHL at x = 2)
$\lim\limits_{\text{x}\rightarrow2^{-}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(2-\text{h})-\text{f}(2)}{2-\text{h}-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2-2}{-\text{h}}$
$=0$
View full question & answer→Question 1583 Marks
Prove that the function f given by:
$\text{f(x)} = |\text{x} - 1|, \text{x} \in \text{R}$
is not differentiable at x = 1.
AnswerGiven: $\text{f(x)} = |\text{x} - 1|\ \therefore \ \text{f(1)} = |1 - 1| = 0$
$\text{R}\text{f}{'}(1) = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{\text{f(1 + h)}-\text{f}(1)}{\text{h}} = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|1 + \text{h} - 1|-0}{\text{h}}$$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|\text{h}|}{\text{h}}= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{\text{h}}{\text{h}}=1$
$\text{And}\ \text{L}\text{f}{'}(1) = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{\text{f(1 - h)}-\text{f}(1)}{-\text{h}} = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|1 - \text{h} - 1|-0}{-\text{h}}$$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|-\text{h}|}{\text{h}}= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{-\text{h}}{\text{h}}=-1$
Since $\text{R }\text{f}{'}(1)\neq \text{L}{\text{f}}{'}(1)$
Therefore, f(x) is not differentiable at x =1.
View full question & answer→Question 1593 Marks
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\text{a}\sec\theta,\text{y}=\text{b}\tan\theta$
AnswerThe given equations are $\text{x}=\text{a}\sec\theta\text{ and y}=\text{b}\tan\theta$
Then, $\frac{\text{dx}}{\text{d}\theta}= \text{a}.\frac{\text{d}}{\text{d}\theta}(\sec\theta)=\text{a}\sec\theta\tan\theta$
$\frac{\text{dy}}{\text{d}\theta}=\text{b}\frac{\text{d}}{\text{d}\theta}(\tan\theta)=\text{b}\sec^2\theta$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{d}\theta}\Big)}{\Big(\frac{\text{dx}}{\text{d}\theta}\Big)}=\frac{\text{b}\sec^2\theta}{\text{a}\sec\theta\tan\theta}=\frac{\text{b}}{\text{a}}\sec\theta\cot\theta$ $=\frac{\text{b}\cos\theta}{\text{a}\cos\theta\sin\theta}=\frac{\text{b}}{\text{a}}\times\frac{1}{\sin\theta}=\frac{\text{b}}{\text{a}}\ \text{cose}\theta$
View full question & answer→Question 1603 Marks
Differentiate $\log(1+\text{x}^2)$ with respect to $\tan^{-1}\text{x}$
AnswerLet $\text{u}=\log(1+\text{x}^2)$
Differentiating it with respect to x using chain rule,
$\frac{\text{du}}{\text{dx}}=\frac{1}{(1+\text{x}^2)}\frac{\text{d}}{\text{dx}}(1+\text{x}^2)$
$=\frac{1}{(1+\text{x}^2)}(2\text{x})$
$\frac{\text{du}}{\text{dx}}=\frac{2\text{x}}{(1+\text{x}^2)}\ .....(\text{i})$
Let $\text{v}=\tan^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{1+\text{x}^2}\ .....(\text{ii})$
Dividing equation (i) by (ii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2\text{x}}{(1+\text{x}^2)}\times\frac{(1+\text{x}^2)}{1}$
$\frac{\text{du}}{\text{dx}}=2\text{x}$
View full question & answer→Question 1613 Marks
Differentiate the following functions with respect to x:
$\log\Big\{\cot\Big(\frac{\pi}{4}+\frac{\pi}{2}\Big)\Big\}$
Answer$\frac{\text{d}}{\text{dx}}\Big[\log\Big\{\cot\Big(\frac{\Pi}{4}+\frac{\pi}{2}\Big)\Big\}\Big]$
$\frac{1}{\cot\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)}\times\Big(-\text{cosec}^2\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big)\times\frac{1}{2}$
$\frac{-1}{2\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)}=-\frac{1}{\sin2\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)}$
$=-\frac{1}{\sin2\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)}=-\frac{1}{\cos\text{x}}=-\sec\text{x}$
View full question & answer→Question 1623 Marks
Differentiate the following functions with respect to x:
$\tan(\text{e}^{\sin\text{x}})$
AnswerConsider $\text{y}=\tan(\text{e}^{\sin\text{x}})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\tan\text{e}^{\sin\text{x}}\big]$
$=\sec^2\big(\text{e}^{\sin\text{e}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin\text{x}}\big)$
[Using chain rule]
$=\sec^2\big(\text{e}^{\sin\text{x}}\big)\times\text{e}^{\sin\text{x}}\times\frac{\text{d}}{\text{dx}}(\sin\text{ x})$
View full question & answer→Question 1633 Marks
Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.
AnswerHere f(x) = 5x - 3
- $\ \ \ \text{Lt}\ \ \ \ \ \ \text{f(x)}\\ \text{x} \rightarrow 0$ $= \ \ \ \text{Lt}\ \ \ \ \ \ (5\text{x}-3) = 5(0) - 3 = 0 - 3 = -3\\ \ \ \ \ \text{x} \rightarrow 0$
Now f is defined at x = 0
and f(0) = 5(0) - 3 = 0 - 3 = -3
$\therefore \ \ \text{Lt}\ \ \ \ \ \ \ \ \ \ \ \text{f(x)} = \text{f}(0) = -3\\ \ \ \ \text{x}\rightarrow0$
$\therefore$ f is continous at x = 0
- $\ \ \ \text{Lt}\ \ \ \ \ \ \text{f(x)}\\ \text{x} \rightarrow -3$$= \ \ \ \text{Lt}\ \ \ \ \ \ (5\text{x}-3) = 5(-3) - 3 = -15- 3 = -18\\ \ \ \ \ \text{x} \rightarrow -3$
Now f is defined at x = -3
and f(-3) = 5(-3) - 3 = -15 - 3 = -18
$\therefore \ \ \text{Lt}\ \ \ \ \ \ \ \ \ \ \ \text{f(x)} = \text{f}(-3) = -18\\ \ \ \ \text{x}\rightarrow-3$
$\therefore$ f is continous at x = -3
- $\ \ \ \text{Lt}\ \ \ \ \ \ \text{f(x)}\\ \text{x} \rightarrow 5$$= \ \ \ \text{Lt}\ \ \ \ \ \ (5\text{x}-3) = 5(5) - 3 = 25- 3 = 22\\ \ \ \ \ \text{x} \rightarrow 5$
Now f is defined at x = 5
and f(5) = 5(5) - 3 = 25 - 3 = 22
$\therefore \ \ \text{Lt}\ \ \ \ \ \ \ \ \ \ \ \text{f(x)} = \text{f}(5) = 22\\ \ \ \ \text{x}\rightarrow5$
$\therefore$ f is continous at x = 5 View full question & answer→Question 1643 Marks
A function f(x) is defined as,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-\text{x}-6}{\text{x}-3}&; &\text{if} \text{x}\neq3\\5 &;&\text{if}\text{ x}=3\end{cases}$
show that f(x) is continuous that x = 3.
AnswerWe have, to check the continuity at x = 3.
$\text{L.H.L}=\lim\limits_{\text{x} \rightarrow 3^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(3-\text{h})=\lim\limits_{\text{h} \rightarrow 0}\frac{(3-\text{h})^2-(3-\text{h})-6}{(3-\text{h})-3}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2-5\text{h}}{-\text{h}}=\lim\limits_{\text{h} \rightarrow 0}-\text{h}+5=5$
$\text{R.H.L}=\lim\limits_{\text{x} \rightarrow 3^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(3\text{+h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{(3+\text{h})^2-(3+\text{h})-6}{(3+\text{h})-3}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+5\text{h}}{\text{h}}=\lim\limits_{\text{h}\rightarrow 0}\text{h}+5=5$
$\text{f}(3)=5$
Thus, We have, LHL = RHL = f(3) = 5
So,The function is continus at x = 3
View full question & answer→Question 1653 Marks
If $\text{x}=\text{e}^{\frac{\text{x}}{\text{y}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}.$
AnswerWe have, $\text{x}=\text{e}^{\frac{\text{x}}{\text{y}}}$
Differentiating both sides w.r.t. x, we get
$\therefore\ 1=\text{e}^{\frac{\text{x}}{\text{y}}}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}\Big)$
$\Rightarrow\ 1=\text{e}^{\frac{\text{x}}{\text{y}}}\bigg[\frac{\text{y}\cdot1-\text{x}\frac{\text{dy}}{\text{dx}}}{\text{y}^2}\bigg]$
$\Rightarrow\ \text{y}^2=\text{y}\cdot\text{e}^{\frac{\text{x}}{\text{y}}}-\text{x}\cdot\frac{\text{dy}}{\text{dx}}\cdot\text{e}^{\frac{\text{x}}{\text{y}}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}\Big(\text{e}^{\frac{\text{x}}{\text{y}}}-\text{y}\Big)}{\text{x}\cdot\text{e}^{\frac{\text{x}}{\text{y}}}}$
$=\frac{\text{e}^{\frac{\text{x}}{\text{y}}}-\text{y}}{\frac{\frac{\text{x}}{\text{y}}\text{e}^{\frac{\text{x}}{\text{y}}}}{}}$
$=\frac{\text{x}-\text{y}}{\text{x}\cdot\log\text{x}}$ $\Big[\because\ \text{x}=\text{e}^{\frac{\text{x}}{\text{y}}}\Rightarrow\log\text{x}=\frac{\text{x}}{\text{y}}\Big]$
Hence proved
View full question & answer→Question 1663 Marks
Find the second order derivatives of the following functions:
$\text{y}=\text{x}.\cos\text{x}$
Answerlet $\text{y}=\text{x}.\cos\text{x}$
Then,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}.\cos\text{x})=\cos\text{x}.\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})$
$=\cos.1+\text{x}(-\sin\text{x})=\cos\text{x}-\text{x}\sin\text{x}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}[\cos\text{x}-\text{x}\sin\text{x}]=\frac{\text{d}}{\text{dx}}(\cos\text{x})-\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{x})$
$=-\sin\text{x}-\Big[\sin\text{x}.\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}.\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]$
$=-\sin\text{x}=(\sin\text{x}+\text{x}\cos\text{x})$
View full question & answer→Question 1673 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}(\text{e}^{\text{x}})$
AnswerConsider $\text{y}=\tan^{-1}(\text{e}^{\text{x}})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{e}^{\text{x}}\big)$
$=\frac{1}{1+\big(\text{e}^{2\text{x}}\big)^2}=\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\big)$
[using chain rule]
$=\frac{1}{1+\text{e}^{2\text{x}}}\times\text{e}^\text{x}$
$=\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{e}^\text{x}\big)=\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}$
View full question & answer→Question 1683 Marks
Find $\frac{\text{dy}}{\text{dx}}$ of the functions given in Exercise:
$\text{xy}=\text{e}^{(\text{x}-\text{y})}$
AnswerGiven: $\text{xy}=\text{e}^{\text{x}-\text{y}}\ \Rightarrow\ \log\text{xy}\ \log=\text{e}^{\text{x}-\text{y}}$
$\Rightarrow\ \log\text{x}+\log\text{y}=(\text{x}-\text{y})\log\text{e}\ \Rightarrow\ \log\text{x}+\log\text{y}=(\text{x}-\text{y})\ \ [\because\log\text{e}=1]$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{x}+\frac{\text{d}}{\text{dx}}\log\text{y}=\frac{\text{d}}{\text{dx}}(\text{x}-\text{y})\ \Rightarrow\ \frac{1}{\text{x}}+\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}+\frac{\text{dy}}{\text{dx}}=1-\frac{1}{\text{x}}\ \Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}+1\Big)=\frac{\text{x}-1}{\text{x}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big(\frac{1+\text{y}}{\text{y}}\Big)=\frac{\text{x}-1}{\text{x}}\ \Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}-1)}{\text{x}(1+\text{y})}$
View full question & answer→Question 1693 Marks
Differentiate:$\tan(\text{x}^\circ+45^\circ)$
AnswerLet, $\text{y}=\tan(\text{x}^\circ+45^\circ)$
$\Rightarrow \text{y}=\tan\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\tan\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}$
$=\sec^2\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}\times\frac{\text{d}}{\text{dx}}(\text{x}+45)\frac{\pi}{180}$
[Using chain rule]
$=\frac{\pi}{180}\sec^2(\text{x}^\circ+45^\circ)$
So,
$\frac{\text{d}}{\text{dx}}\big\{\tan(\text{x}^\circ+45^\circ)\big\}=\frac{\pi}{180}\sec^2(\text{x}^\circ+45^\circ)$
View full question & answer→Question 1703 Marks
If $\text{f(x)}=\begin{cases}2\text{x}^2+\text{k},&\text{if }\text{ x}\geq0\\-2\text{x}^2+\text{k},&\text{if }\text{ x}<0\end{cases},$ then what should be the value of k so that f(x) is continuous at x = 0.
AnswerIt is given that function is continous at x = 0 then,
$\text{LHL}=\text{RHL}=\text{f}(0)\ ....(\text{i})$
Now, $\text{f}(0)=2\times0+\text{k}=\text{k}$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}\lim_\limits{\text{h}\rightarrow0}=\lim_\limits{\text{h}\rightarrow0}-2(-\text{h})^2+\text{k}=\text{k}$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})=\lim_\limits{\text{h}\rightarrow0}2(\text{h}^2)+\text{k}=\text{k}$
Thus, the function will be continuous for any $\text{k}\in\text{R}$
View full question & answer→Question 1713 Marks
Differentiate $(x^2– 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
by using product rule
AnswerLet $y = (x^2- 5x + 8)(x^3 + 7x + 9)$ ....(i)
$\frac{\text{dy}}{\text{dx}}=(\text{x}^2-5\text{x}+8)\frac{\text{d}}{\text{dx}}(\text{x}^3+7\text{x}+9)+(\text{x}^3+7\text{x}+9)\frac{\text{d}}{\text{dx}}(\text{x}^2-5\text{x}+8)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(\text{x}^2-5\text{x}+8)(3\text{x}^2+7)+(\text{x}^3+7\text{x}+9)(2\text{x}-5)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=3\text{x}^4+7\text{x}^2-15\text{x}^3-35\text{x}+24\text{x}^2+56+2\text{x}^4-5\text{x}^3+14\text{x}^2-35\text{x}+18\text{x}-45$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=5\text{x}^4-20\text{x}^3+45\text{x}^2+11\ \dots\text{(ii)}$
View full question & answer→Question 1723 Marks
Differentiate the following functions with respect to x:
$\tan(\text{x}^\circ+45^\circ)$
AnswerLet, $\text{y}=\tan(\text{x}^\circ+45^\circ)$
$\Rightarrow\text{y}=\tan\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}$
Differentiating it with respect to x we get,
$\frac{\text{dx}}{\text{dy}}=\frac{\text{d}}{\text{dx}}\tan\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}$
$=\sec^2\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}\times\frac{\text{d}}{\text{dx}}(\text{x}+45)\frac{\pi}{180}$
[Using chain rule]
$=\frac{\pi}{180}\sec^2(\text{x}^\circ+45^\circ)$
So,
$=\frac{\text{d}}{\text{dx}}\Big\{\tan(\text{x}^\circ+45^\circ)\Big\}=\frac{\pi}{180}\sec^2(\text{x}^\circ+45^\circ)$
View full question & answer→Question 1733 Marks
Differentiate the following functions with respect to x:
$\sin^2(2\text{x}+1)$
AnswerCobnsider $\text{y}=\sin^2(2\text{x}+1)$
Differentiate it with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\sin^2(2\text{x}+1)\big]$
$=2\sin(2\text{x}+1)\frac{\text{d}}{\text{dx}}\sin(2\text{x}+1)$
[Using chain rule]
$=2\sin(2\text{x}+1)\cos(2\text{x}+1)\frac{\text{d}}{\text{dx}}\sin(2\text{x}+1)$
[Using chain rule]
$=4\sin(2\text{x}+1)\cos(2\text{x}+1)$
$=2\sin(2\text{x}+1)$
$\Big[\text{Since}, \sin^2\text{A}=2\sin\text{A}\cos\text{A}\Big]$
$2\sin(4\text{x}+2)$
Hence, the solution is $\frac{\text{d}}{\text{dx}}\big(\sin^2(2\text{x}+1)\big)=2\sin(4\text{x}+2)$
View full question & answer→Question 1743 Marks
Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=\begin{cases}-4\text{x}+5,&0\leq\text{x}\leq1\\2\text{x}-3,&1<\text{x}\leq2\end{cases}$
View full question & answer→Question 1753 Marks
Using Rolle's theorem, find points on the curve $\text{y}=16-\text{x}^2,\text{x}\in[-1,1],$ where tagent is parallel to x-axis.
AnswerThe equation of the curve is,
$\text{y}=16-\text{x}^2\ ....(1)$
Let $P(x_1,y_1)$ be a point on it where the tangent is parallel to x-axis.
Then,
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=0\ ....(2)$
Differentiating (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=-2\text{x}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=-2\text{x}_1$
$\Rightarrow-2\text{x}_1=0$ (from(2))
$\Rightarrow\text{x}_1=0$
$P(x_1, y_1)$ lies on the curve $y = 16 - x^2$
$\therefore\text{y}_1=16-\text{x}_1^2$
When $x_1 = 0$,
$y_1 = 16$
Hence, $(0, 16)$ is the required point.
View full question & answer→Question 1763 Marks
If $\text{y}=\log\sqrt{\tan\text{x}},$ write $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\log\sqrt{\tan\text{x}}$
$\Rightarrow\text{y}=\log(\tan\text{x})^\frac{1}{2}$
$\Rightarrow\text{y}=\frac{1}{2}\log(\tan\text{x})\big[\because\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\frac{1}{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\frac{1}{\tan\text{x}}(\sec^2\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\frac{\sin\text{x}}{\cos\text{x}}}\times\cos^2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sin\text{x}\cos\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\sin2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{cosec }2\text{x}$
View full question & answer→Question 1773 Marks
Differentiate w.r.t. x the function in Exercise:
$\frac{\cos^{-1}\frac{\text{x}}{2}}{\sqrt{2\text{x}+7}},-2<\text{x}<2$
AnswerLet $\text{y}=\frac{\cos^{-1}\frac{\text{x}}{2}}{\sqrt{2\text{x}+7}}$By quotient rule, we obtain
$\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{2\text{x}+7}\frac{\text{d}}{\text{dx}}\Big(\cos^{-1}\frac{\text{x}}{2}\Big)-\Big(\cos^{-1}\frac{\text{x}}{2}\Big)\frac{\text{d}}{\text{dx}}(\sqrt{2\text{x}+7)}}{(\sqrt{2\text{x}+7})^2}$
$=\frac{\sqrt{2\text{x}+7}\Bigg[\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{2}\Big)^2}}.\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{2}\Big)\Bigg]-\Big(\cos^{-1}\frac{\text{x}}{2}\Big)\frac{1}{2\sqrt{2\text{x}+7}}.\frac{\text{d}}{\text{dx}}(2\text{x}+7)}{2\text{x}+7}$
$=\frac{\sqrt{2\text{x}+7}\frac{-1}{\sqrt{4-\text{x}^2}}-\Big(\cos^{-1}\frac{\text{x}}{2}\Big)\frac{2}{2\sqrt{2\text{x}+7}}}{2\text{x}+7}$
$=\frac{-\sqrt{2\text{x}+7}}{\sqrt{4-\text{x}^2}\times(2\text{x}+7)}-\frac{\cos^{-1}\frac{\text{x}}{2}}{(\sqrt{2\text{x}+7})(2\text{x}+7)}$
$=-\Bigg[\frac{1}{\sqrt{4-\text{x}^2}\sqrt{2\text{x}+7}}+\frac{\cos^{-1}\frac{\text{x}}{2}}{(2\text{x}+7)^{\frac{3}{2}}}\Bigg]$
View full question & answer→Question 1783 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.$\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}-\text{x}\text{ on }[0,\pi]$
AnswerWe have,$\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}-\text{x}$
Since, $\sin\text{x},\sin2\text{x}\ \&\ \text{x}$ are everywhere continuous and differentiable.
Therefore, f(x) is continuous on $[0,\pi]$ and differentiable on $(0,\pi)$
Concequently, there exist some $\text{c}\in(0,\pi)$such that
$\text{f}'(\text{c})=\frac{\text{f}(\pi)-\text{f}(0)}{\pi-0}=\frac{\text{f}(\pi)-\text{f}(0)}{\pi}$
Now, $\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}-\text{x}$
$\text{f}'(\text{x})=\cos\text{x}-2\cos2\text{x}-1,\text{f}(\pi)=-\pi,\text{f}(0)=0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(\pi)-\text{f}(0)}{\pi-0}$
$\Rightarrow\cos\text{x}-2\cos2\text{x}-1=-1$
$\Rightarrow\cos\text{x}-2\cos2\text{x}=0$
$\Rightarrow\cos\text{x}-4\cos^2\text{x}=-2$
$\Rightarrow4\cos^2\text{x}-\cos\text{x}-2=0$
$\Rightarrow\cos\text{x}=\frac{1}{8}\big(1\pm\sqrt{33}\big)$
$\Rightarrow\text{x}=\cos^{-1}\Big[\frac{1}{8}\big(1\pm\sqrt{33}\big)\Big]$
Thus, $\text{c}=\cos^{-1}\Big(\frac{1\pm\sqrt{33}}{8}\Big)\in(0,\pi)$ such that $\text{f}'(\text{c})=\frac{\text{f}(\pi)-\text{f}(0)}{\pi-0}.$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 1793 Marks
Differentiate the following w.r.t. x:
$\sin\sqrt{\text{x}}+\cos^2\sqrt{\text{x}}$
AnswerLet $\text{y}=\sin\sqrt{\text{x}}+\big(\cos^2\sqrt{\text{x}}\big)^2$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin\Big(\text{x}^{\frac{1}{2}}\Big)+\frac{\text{d}}{\text{dx}}\Big[\cos\Big(\text{x}^{\frac{1}{2}}\Big)\Big]^2$
$=\cos\text{x}^{\frac{1}{2}}\cdot\frac{\text{d}}{\text{dx}}\text{x}^{\frac{1}{2}}+\cos\Big(\text{x}^{\frac{1}{2}}\Big)\frac{\text{d}}{\text{dx}}\Big[\cos\Big(\text{x}^{\frac{1}{2}}\Big)\Big]$
$=\cos\sqrt{\text{x}}\cdot\frac{1}{2\sqrt{\text{x}}}+2\cos\sqrt{\text{x}}\Big[-\sin\sqrt{\text{x}}\cdot\frac{1}{2\sqrt{\text{x}}}\Big]$
$=\frac{1}{2\sqrt{\text{x}}}\Big[\cos\big(\sqrt{\text{x}}\big)-\sin\big(2\sqrt{\text{x}}\big)\Big]$
View full question & answer→Question 1803 Marks
Differentiate the following functions with respect to x:
$\tan5\text{x}^\circ$
AnswerLet, $\text{y}=\tan5\text{x}^\circ$
$\Rightarrow\ \text{y}=\tan\Big(5\text{x}\times\frac{\pi}{180}\Big)$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\tan\Big(5\text{x}\times\frac{\pi}{180}\Big)$
$=\sec^2\Big(5\text{x}\times\frac{\pi}{180}\Big)\frac{\text{d}}{\text{dx}}\Big(5\text{x}\times\frac{\pi}{108}\Big) $
[Using chain rule]
$=\Big(\frac{5\text{x}}{180}\Big)\sec^2\Big(5\text{x}\times\frac{\pi}{180}\Big)$
$=\frac{5\pi}{180}\sec^2(5\text{x}^\circ)$
Hence, $\frac{\text{d}}{\text{dx}}(\tan5\text{x}^\circ)=\frac{5\pi}{180}\sec^2(5\text{x}^\circ)$
View full question & answer→Question 1813 Marks
Differentiate the following functions from first principles:
$e^{3x}$.
AnswerLet $f(x) = e^{3x}$
$\Rightarrow f(x + h) = e^{3(x + h)}$
$\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3(\text{x}+\text{h})}-\text{e}^{3\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3\text{x}}\text{e}^{3\text{h}}-\text{e}^{3\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{3\text{x}}\left\{\frac{(\text{e}^{3\text{h}}-1)}{3\text{h}}\right\}\times3$
$=3\text{e}^{3\text{x}}\Big[\text{Since, }\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
Hence,
$\frac{\text{d}}{\text{dx}}(\text{e}^{3\text{x}})=3\text{e}^{3\text{x}}$
View full question & answer→Question 1823 Marks
If $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big),$ write the value of $\frac{\text{dy}}{\text{dx}}\text{ for x}>1.$
AnswerWe have, $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
Putting $\text{x}=\tan\theta$
$\Rightarrow 1 <\tan\theta<\infty$
$\Rightarrow\frac{\pi}{4}<\theta<\frac{\pi}{2}$
$\frac{\pi}{2}<2\theta<\pi$
$\therefore\text{y}=\sin^{-1}(\sin2\theta)$
$\Rightarrow\text{y}=\sin^{-1}\big\{\sin(\pi-2\theta)\big\}$
$\Rightarrow\text{y}=\pi-2\theta$
$\Rightarrow\text{y}=\pi-2\tan^{-1}\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=0-\frac{2}{1+\text{x}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{2}{1+\text{x}^2}$
View full question & answer→Question 1833 Marks
If $\text{y}=\text{e}^\text{x}\cos\text{x},$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{e}^\text{x}\cos(\text{x}+\frac{\pi}{2}).$
AnswerHere
$\text{y}=\text{e}^\text{x}\cos\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}\cos\text{x}-\text{e}^\text{x}\sin\text{x}=\text{e}^\text{x}(\cos\text{x}-\sin\text{x})$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=\text{e}^\text{x}(\cos\text{x}-\sin\text{x})+\text{e}^\text{x}(-\sin\text{x}-\cos\text{x})$
$=\text{e}^\text{x}\cos\text{x}-\text{e}^\text{x}\sin\text{x}-\text{e}^\text{x}\sin\text{x}-\text{e}^\text{x}\cos\text{x}$
$=-2\text{e}^\text{x}\sin\text{x}$
$=2\text{e}^\text{x}\cos(\text{x}+\frac{\pi}{2})$
View full question & answer→Question 1843 Marks
If $\text{x}=\text{a}(\theta-\sin\theta)\text{ and},\text{y}=\text{a}(1+\cos\theta),$ find $\frac{\text{dy}}{\text{dx}}\text{ at }\theta=\frac{\pi}{3}$
AnswerHere,
$\text{x}=\text{a}(\theta-\sin\theta)\text{ and y}=\text{a}(1+\cos\theta)$
Then,
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\big[\text{a}(\theta-\sin\theta)\big]=\text{a}(1-\cos\theta)$
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\big[\text{a}(1+\sin\theta)\big]=\text{a}(1-\sin\theta)$
$\therefore\frac{\text{dy}}{\text{dx}}=\Bigg[\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1-\cos\theta)}\Bigg]_{\theta=\frac{\pi}{3}}$
$=-\frac{\sin\frac{\pi}{2}}{1-\cos\frac{\pi}{3}}=\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}=-\sqrt{3}$
View full question & answer→Question 1853 Marks
If $\text{y}=\sqrt{\log\text{x}+\sqrt{\log\text{x}+\sqrt{\log\text{x}+\ .... \text{to }\infty}}},$ prove that $(2\text{y}-1)\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
AnswerWe have, $\text{y}=\sqrt{\log\text{x}+\sqrt{\log\text{x}+\sqrt{\log\text{x}+\ .... \text{to }\infty}}}$
$\Rightarrow\text{y}=\sqrt{\log\text{x}+\text{y}}$
Squaring both sides, we get,
$\text{y}^2=\log\text{x}+\text{y}$
$=2\text{y}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}+\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{y}-1)=\frac{1}{\text{x}}$
View full question & answer→Question 1863 Marks
Give an example of a function which is continuos but not differentiable at at a point.
AnswerConsider a function, $\text{f(x)}=\begin{cases}\text{x}, & \text{x}> 0\\-\text{x}, & \text{x}\leq 0\end{cases}$
This mod function is continuous at x = 0 but not differentiable at x = 0.
Continuity at x - 0, We have:
(LHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}$
$=\lim_\limits{\text{x}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{x}\rightarrow0}-(0-\text{h})$
$=0$
(RHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}$
$=\lim_\limits{\text{x}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{x}\rightarrow0}(0+\text{h})$
$=0$
and f(0) = 0
Thus, $\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}=\text{f}(0).$
Hence, f(x) is continuous at x = 0.
Now, we will check the differentiability at x = 0, we have:
(LHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0-\text{h}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-(0-\text{h})-0}{-\text{h}}=-1$
(RHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{0+\text{h}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(0+\text{h})-0}{-\text{h}}=1$
Thus, $\lim_\limits{\text{h}\rightarrow0^{-}}\text{f(x)}\neq\lim_\limits{\text{h}\rightarrow0^{+}}\text{f(x)}$
Hence f(x) is not differentiable at x = 0.
View full question & answer→Question 1873 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{\text{x}^2-25}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}\text{at x} =5$
AnswerGiven,
$\text{f(x)}=\begin{cases}\frac{\text{x}^2-25}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}\frac{(\text{x}-5)(\text{x}+5)}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}$
If f(x) is continuous at x = 5, then,
$\lim_\limits{\text{x}\rightarrow5}\text{f(x)}=\text{f}(5)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow5}\text{(x}+5)=\text{k}$
$\Rightarrow\text{k}=5+5=10$
View full question & answer→Question 1883 Marks
If $\text{y}=\tan^{-1}$ show that $(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\text{x}\frac{\text{dy}}{\text{dx}}=0$
Answer$\text{y}=\tan^{-1}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$
$\Rightarrow(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}=1$
Differentiating w.r.t.x, we get
$\Rightarrow(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\text{x}\frac{\text{dy}}{\text{dx}}=0$
Hence proved
View full question & answer→Question 1893 Marks
Differentiate the following functions with respect to x:
$(\log\sin\text{x})^2$
AnswerLet $\text{y}=(\log\sin\text{x})^2$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\log\sin\text{x})^2$
$=2(\log\sin\text{x})\frac{\text{d}}{\text{dx}}(\log\sin\text{x})$
$=2(\log\sin\text{x})\times\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})$
$=2(\log\sin\text{x})\times\frac{1}{\sin\text{x}}\cos\text{x}$
$=2(\log\sin\text{x})\cot\text{x}$
So,
$\frac{\text{d}}{\text{dx}}(\log\sin\text{x})^2=2(\log\sin\text{x})\cot\text{x}$
View full question & answer→Question 1903 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}(\text{x}-\text{a}){\sin}\Big(\frac{1}{\text{x}-\text{a}}\Big) & \text{x} \neq \text{a}\\\ 0, & \text{ x} = \text{a}\end{cases}\text{at x}=\text{a}$
AnswerWe want, to check the continuity at x = 0.
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{ h)}=\lim\limits_{\text{x} \rightarrow 0}(-\text{h)}^2$
$\sin\Big(\frac{1}{\text{-h}}\Big)=0$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}=\lim\limits_{\text{x} \rightarrow 0}\text{h}^2$
$\sin\Big(\frac{1}{\text{h}}\Big)=0$
$\text{f}(0)=0$
Thus, LHL = RHL = f(0) = 0
Hence, the function is continuous at x = 0.
View full question & answer→Question 1913 Marks
Differentiate the following w.r.t. x:
$\sec^{-1}\Big(\frac{1}{4\text{x}^3-3\text{x}}\Big),0<\text{x}<\frac{1}{\sqrt{2}}$
AnswerLet $\text{y}=\sec^{-1}\Big(\frac{1}{4\text{x}^3-3\text{x}}\Big)$
On putting $\text{x}=\cos\theta,$ we get
$\text{y}=\sec^{-1}\frac{1}{4\cos^3\theta-3\cos\theta}$
$=\sec^{-1}\frac{1}{\cos3\theta}$
$=\sec^{-1}(\sec3\theta)$
$=3\theta$
$=3\cos^{-1}\text{x}$ $\big[\because\theta=\cos^{-1}\text{x}\big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(3\cos^{-1}\text{x})$
$=\frac{-3}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1923 Marks
Differentiate the following w.r.t. x:
$\log\Big(\text{x}+\sqrt{\text{x}^2+\text{a}}\Big)$
AnswerLet $\text{y}=\log\Big(\text{x}+\sqrt{\text{x}^2+\text{a}}\Big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\big(\text{x}+\sqrt{\text{x}^2+\text{a}}\big)}\cdot\frac{\text{d}}{\text{dx}}\Big[\text{x}+\sqrt{\text{x}^2+\text{a}}\Big]$
$=\frac{1}{\big(\text{x}+\sqrt{\text{x}^2+\text{a}}\big)}\Big[1+\frac{1}{2}(\text{x}^2+\text{a})^{\frac{-1}{2}}\cdot2\text{x}\Big]$
$=\frac{1}{\big(\text{x}+\sqrt{\text{x}^2+\text{a}}\big)}\cdot\Big(1+\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}}}\Big)$
$-\frac{\big(\sqrt{\text{x}^2+\text{a}}+\text{x}\big)}{\big(\text{x}+\sqrt{\text{x}^2+\text{a}}\big)\big(\sqrt{\text{x}^2+\text{a}}\big)}$
$=\frac{1}{\big(\sqrt{\text{x}^2+\text{a}}\big)}$
View full question & answer→Question 1933 Marks
Differentiate the following functions with respect to x:
$\sin(\log\sin\text{x})$
AnswerConsider $\text{y}=\sin(\log\sin\text{x})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin(\log\sin\text{x})$
$=\cos(\log\sin\text{x})\frac{\text{d}}{\text{dx}}(\log\sin\text{x})$
[Using chain rule]
$=\cos(\log\sin\text{x})\times\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}0\sin\text{x}$
$=\cos(\log\sin\text{x})\frac{\cos\text{x}}{\sin\text{x}}$
$=\cos(\log\sin\text{x})\times\cot\text{x}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}(\sin(\log\sin\text{x}))=\cos(\log\sin\text{x})\text{x}\cot\text{x}$
View full question & answer→Question 1943 Marks
Differentiate the following w.r.t. x:
$\sin\text{x}^2+\sin^2\text{x}+\sin^2(\text{x}^2)$
AnswerLet $\text{y}=\sin\text{x}^2+\sin^2\text{x}+\sin^2(\text{x}^2)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin\big(\text{x}^2\big)+\frac{\text{d}}{\text{dx}}\big(\sin\text{x}\big)^2+\frac{\text{d}}{\text{dx}}(\sin\text{x}^2)^2$
$=\cos\big(\text{x}^2\big)\frac{\text{d}}{\text{dx}}\big(\text{x}^2\big)+2\sin\text{x}\cdot\frac{\text{d}}{\text{dx}}\sin\text{x}+2\sin^2\cdot\frac{\text{d}}{\text{dx}}\sin\text{x}^2$
$=2\text{x}\cos\text{x}^2+2\cdot\sin\text{x}\cdot\cos\text{x}+2\sin\text{x}^2\cos\text{x}^2\cdot\frac{\text{d}}{\text{dx}}\text{x}^2$
$=2\text{x}\cos\text{x}^2+2\cdot\sin\text{x}\cdot\cos\text{x}+2\sin\text{x}^2\cos\text{x}^2\cdot2\text{x}$
$=2\text{x}\cos\text{x}^2+\sin2\text{x}+\sin\big(2\text{x}^2\big)\cdot2\text{x}$
$=2\text{x}\cos\text{x}^2+2\text{x}\cdot\sin2\big(\text{x}^2\big)+\sin2\text{x}$
View full question & answer→Question 1953 Marks
If $\text{y}=3\text{e}^{2\text{x}}+2\text{e}^{3\text{x}}$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}-5\frac{\text{dy}}{\text{dx}}+6\text{y}=0$
Answer$\text{y}=3\text{e}^{2\text{x}}+2\text{e}^{3\text{x}}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=6\text{e}^{2\text{x}}+6\text{e}^{3\text{x}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=12\text{e}^{2\text{x}}+18\text{e}^{3\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=5(6\text{e}^{2\text{x}}+6\text{e}^{3\text{x}})-6(3\text{e}^{2\text{x}}+2\text{e}^{3\text{x}})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=5\Big(\frac{\text{dy}}{\text{dx}}\Big)-6\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-5\Big(\frac{\text{dy}}{\text{dx}}\Big)+6\text{y}=0$
View full question & answer→Question 1963 Marks
If $\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}},$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
AnswerHere,
$\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=2\text{a}\text{e}^{2\text{x}}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=4\text{a}\text{e}^{2\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{a}\text{e}^{2\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
Hence proved.
View full question & answer→Question 1973 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{a}+\text{x}}{1-\text{ax}}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{a}+\text{x}}{1-\text{ax}}\Big)$
$\text{y}=\tan^{-1}\text{a}+\tan^{-1}\text{x}$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{a})+\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})$
$=0+\frac{1}{1+\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$
View full question & answer→Question 1983 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\sin\sqrt{\text{x}}}$
AnswerLet, $\text{y}=\text{e}^{\sin\sqrt{\text{x}}}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin\sqrt{\text{x}}}\big)$
$=\text{e}^{\sin\sqrt{\text{x}}}\frac{\text{d}}{\text{dx}}\big(\sin\sqrt{\text{x}}\big)$
[Using chain rule]
$=\text{e}^{\sin\sqrt{\text{x}}}\times\cos\sqrt{\text{x}}\frac{\text{d}}{\text{dx}}\sqrt{\text{x}}$
[Using chain rule]
$=\text{e}^{\sin\sqrt{\text{x}}}\times\cos\sqrt{\text{x}}\times\frac{1}{2\sqrt{\text{x}}}$
$=\frac{1}{2\sqrt{\text{x}}}\times\cos\sqrt{\text{x}}\times\text{e}^{\sin\sqrt{\text{x}}}$
So,
$\frac{\text{d}}{\text{dx}}=\big(\text{e}^{\sin^\sqrt{\text{x}}}\big)=\frac{1}{2\sqrt{\text{x}}}\cos\sqrt{\text{x}}\times\text{e}^{\sin\sqrt{\text{x}}}$
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Find the derivative of the function f defined by f(x) = mx + c at x = 0.
AnswerGiven: f(x) = mx + c
Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of f at x is given by:
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h}-\text{f(x)})}{\text{h}}$
$\Rightarrow\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{m}(\text{x}+\text{h})+\text{c}-\text{mx}-\text{c}}{\text{h}}$
$\Rightarrow\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{mx}+\text{mh}+\text{c}-\text{mx}-\text{c}}{\text{h}}$
$\Rightarrow\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{mh}}{\text{h}}$
$\Rightarrow\text{f}'(\text{x})=\text{m}$
Thus, $\text{f}'(0)=\text{m}$
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If $f(x)$ is defined by $f(x) x^2$. find $f(2)$.
AnswerGiven: $f(x) = x^2$.
We know a polynomial function is everywhere differentiable. Therefore f(x) is differentiable at x = 2.
$\text{f}'(2)=\lim_\limits{\text{k}\rightarrow0}\text{f}\frac{(2+\text{h})-\text{f}(2)}{\text{h}}$
$\Rightarrow\text{f}'(2)=\lim_\limits{\text{k}\rightarrow0}\text{f}\frac{(2+\text{h})2-22}{\text{h}}$
$\Rightarrow\text{f}'(2)=\lim_\limits{\text{k}\rightarrow0}\text{f}\frac{(4+\text{h}2-4\text{h})-4}{\text{h}}$
$\Rightarrow\text{f}'(2)=\lim_\limits{\text{k}\rightarrow0}\text{f}\frac{\text{h}(\text{h}+4)}{\text{h}}$
$\Rightarrow\text{f}'(2)=4$
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Differentiate the functions with respect to x.
$\frac{\sin(\text{ax + b)}}{\cos(\text{cx + d})}$
Answer$\text{Let y} = \frac{\sin(\text{ax + b)}}{\cos(\text{cx + d})}$
Using quotient rule,
$\therefore \frac{\text{dy}}{\text{dx}} = \frac{\cos(\text{cx + d})\frac{\text{d}}{\text{dx}}\sin\text(\text{ax + b})-\sin(\text{ax + b)}\frac{\text{d}}{\text{dx}}\cos(\text{cx} + \text{d})}{\cos^2(\text{cx} + \text{d})}$
$= \frac{\cos(\text{cx + d})\cos\text(\text{ax + b})\frac{\text{d}}{\text{dx}}(\text{ax + b)}-\sin(\text{ax + b)}\left\{-\sin(\text{cx + d)}\right\}\frac{\text{d}}{\text{dx}}(\text{cx} + \text{d})}{\cos^2(\text{cx} + \text{d})}$
$= \frac{\cos(\text{cx + d})\cos\text(\text{ax + b})(\text{a)}+\sin(\text{ax + b})\sin(\text{cx + d)}(\text{c})}{\cos^2(\text{cx} + \text{d})}$
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show that $\text{f}\text{(x)}=\begin{cases}\frac{\text{x}-|\text{x}|}{2}, & \text{when} \text{ x}\neq 0\\2, & \text{when}\text{ x} = 0\end{cases}$ is discontinuous at x = 0.
AnswerWe want, to check the continuty of the function at x = 0.
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{x)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\text{h}-|-\text{h|}}{2}=\lim\limits_{\text{h} \rightarrow 0}\frac{-\text{h}-\text{h}}{2}=0$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}-\text{(|h|)}}{2}=0$
$\text{f}(0)=2$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}(0)$
Hence,The function is discotinuous at x = 0
This is rem ovable discontinuty.
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