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
$\log\text{y}=\log2\cos^2\text{x}$
$\Rightarrow\ \frac{\text{d}}{\text{dy}}(\log\text{y})\cdot\frac{\text{dy}}{\text{dx}}=\log2\frac{\text{d}}{\text{dx}}(\cos^2\text{x})$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\log2(2\cos\text{x})\frac{\text{d}}{\text{dx}}\cos\text{x}$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\log2\cdot2\cos\text{x}\cdot(-\sin\text{x})$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\log2\cdot[-(\sin2\text{x})]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-\text{y}\cdot\log2(\sin2\text{x})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-2^{\cos^2}\text{x}\cdot\log2(\sin2\text{x})$ $\big[\because\text{y}=2^{\cos^2}\text{x}\big]$
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