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, $\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}=\text{f}'(0)=0$.
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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.
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$\text{If y}=\text{e}^{\text{y}}(\text{x}+1)=1,\text{ show that }\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
AnswerHere $\text{e}^{\text{y}}(\text{x}+1)=1$
$\therefore\ \log[\text{e}^{\text{y}}(\text{x}+1)]=\log1$
$\therefore\ \log\text{e}^{\text{y}}(\text{x}+1)=0$
$\therefore\ \text{y}\log\text{e}=-\log(\text{x}+1)$
$\therefore\ \text{y}=-\log(\text{x}+1)\ \ [\because\text{loge}=1]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=-\frac{1}{\text{x}+1}\ \dots(1)$
$\therefore\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Bigg[\frac{(\text{x}+1).\frac{\text{d}}{\text{dx}}(1)-1.\frac{\text{d}}{\text{dx}}(\text{x}+1)}{(\text{x}+1)^2}\Bigg]$ $=-\Big[\frac{(\text{x}+1).0-1.1}{(\text{x}+1)^2}\Big]$
$=\frac{1}{(\text{x}+1)^2}=\Big(-\frac{1}{(\text{x}+1)^2}\Big)^2$
$\therefore\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\ \dots[\because\text{of }(1)]$
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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→