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,
$\frac{\text{d}}{\text{dx}}=\big(\text{e}^{\sin^\sqrt{\text{x}}}\big)=\frac{1}{2\sqrt{\text{x}}}\cos\sqrt{\text{x}}\times\text{e}^{\sin\sqrt{\text{x}}}$
View full question & answer→Question 1993 Marks
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, $\text{f}'(0)=\text{m}$
View full question & answer→Question 2003 Marks
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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