Question 13 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 23 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 33 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 43 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 53 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 63 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 73 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 83 Marks
If $\text{y}=\log(\sin\text{x})$ Prove that $\frac{\text{d}^3\text{y}}{\text{dx}^3}=2\cos\text{x}\ \text{cosec}^3\text{x}$
AnswerHere,
$\text{y}=\log(\sin\text{x})$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sin\text{x}}\times\cos\text{x}=\cot\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{cosec}^2\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^3\text{y}}{\text{dx}^3}=-2\text{cosec}\ \text{x}\times(-\text{cosec}\ \text{x}\cot\text{x})$
$=2\cot\ \text{x}\ \text{cosec}^2\text{x}=2\cos\ \text{x}\ \text{cosec}^3\text{x}$
View full question & answer→Question 93 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 103 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 113 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 123 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 133 Marks
Differentiate the following functions with respect to x:
$3^{\text{x}\log\text{x}}$
AnswerLet $\text{y}=3^{\text{x}\log\text{x}}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(3^{\text{x}\log\text{x}}\big)$
$=3^{\text{x}\log\text{x}}\times\log_\text{e}3\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$
[Using chain rule]
$=3^{\text{x}\log\text{x}}\times\log_\text{e}3\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=3^{\text{x}\log\text{x}}\times\log_\text{e}3\Big[\frac{\text{x}}{\text{x}}+\log\text{x}\Big]$
$=3^{\text{x}\log\text{x}}\big[1+\log\text{x}\big]\times\log_\text{e}3$
So,
$\frac{\text{d}}{\text{dx}}\big(3^{\text{x}\log\text{x}}\big)=3^{\text{x}\log\text{x}}\big[1+\log\text{x}\big]\log_\text{e}3$
View full question & answer→Question 143 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 153 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 163 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 173 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 183 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 193 Marks
If f(x) = |x - 2| write whether f(2) exists or not.
AnswerGiven: $\text{f(x)}=|\text{x}-2|=\begin{cases}\text{x}-2, & \text{x}> 2\\-\text{x}+2, & \text{x}\leq 2\end{cases}$
Now,
(LHL at x = 2)
$\lim_\limits{\text{x}\rightarrow2^{-}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(2-\text{h})-\text{f}(2)}{2-\text{h}-2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(-2+\text{h}+2)-0}{-\text{h}}$
$=-1$
(RHL at x = 2)
$\lim_\limits{\text{x}\rightarrow2^{+}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(2+\text{h})-\text{f}(2)}{2+\text{h}-2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2+\text{h}+2-0}{\text{h}}$
$=1$
Thus, (LHL at x = 2) $\neq$ (RHL at x = 2)
Hence, $\lim_\limits{\text{x}\rightarrow2}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}=\text{f'}(2)$ does not exist.
View full question & answer→Question 203 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 213 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 223 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 233 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 243 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 253 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 263 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 273 Marks
Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=\begin{cases}-4\text{x}+5,&0\leq\text{x}\leq1\\2\text{x}-3,&1<\text{x}\leq2\end{cases}$
AnswerThe given function is
$\text{f}(\text{x})=\begin{cases}-4\text{x}+5,&0\leq\text{x}\leq1\\2\text{x}-3,&1<\text{x}\leq2\end{cases}$
At x = 0, we have
$\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})=\lim\limits_{\text{h}\rightarrow0}\text{f}(1-\text{h})=\lim\limits_{\text{h}\rightarrow0}[-4(1-\text{h}+5)]=1$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})=\lim\limits_{\text{h}\rightarrow0}\text{f}(1+\text{h})=\lim\limits_{\text{h}\rightarrow0}[2(1+\text{h}-3)]=-1$
$\therefore\ \lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})\neq\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})$
Thus, f(x) is discontinuous at x = 1.
Hence, Rolle's theorem is not applicable for the given function.
View full question & answer→Question 283 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 293 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 303 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 313 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 323 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 333 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 343 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 353 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 363 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 373 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 383 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 393 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 403 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 413 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 423 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 433 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 443 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 453 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 463 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 $x = 2.$
$\text{f}'(2)=\lim_\limits{\text{k}\rightarrow0}\text{f}\frac{(2+\text{h})-\text{f}(2)}{\text{h}}$
$\Rightarrow\text{f}'(2)=\lim_\limits{\text{k}\rightarrow0}\text{f}\frac{(2+\text{h})2-22}{\text{h}}$
$\Rightarrow\text{f}'(2)=\lim_\limits{\text{k}\rightarrow0}\text{f}\frac{(4+\text{h}2-4\text{h})-4}{\text{h}}$
$\Rightarrow\text{f}'(2)=\lim_\limits{\text{k}\rightarrow0}\text{f}\frac{\text{h}(\text{h}+4)}{\text{h}}$
$\Rightarrow\text{f}'(2)=4$
View full question & answer→Question 473 Marks
Differentiate the functions with respect to x.
$\frac{\sin(\text{ax + b)}}{\cos(\text{cx + d})}$
Answer$\text{Let y} = \frac{\sin(\text{ax + b)}}{\cos(\text{cx + d})}$
Using quotient rule,
$\therefore \frac{\text{dy}}{\text{dx}} = \frac{\cos(\text{cx + d})\frac{\text{d}}{\text{dx}}\sin\text(\text{ax + b})-\sin(\text{ax + b)}\frac{\text{d}}{\text{dx}}\cos(\text{cx} + \text{d})}{\cos^2(\text{cx} + \text{d})}$
$= \frac{\cos(\text{cx + d})\cos\text(\text{ax + b})\frac{\text{d}}{\text{dx}}(\text{ax + b)}-\sin(\text{ax + b)}\left\{-\sin(\text{cx + d)}\right\}\frac{\text{d}}{\text{dx}}(\text{cx} + \text{d})}{\cos^2(\text{cx} + \text{d})}$
$= \frac{\cos(\text{cx + d})\cos\text(\text{ax + b})(\text{a)}+\sin(\text{ax + b})\sin(\text{cx + d)}(\text{c})}{\cos^2(\text{cx} + \text{d})}$
View full question & answer→Question 483 Marks
show that $\text{f}\text{(x)}=\begin{cases}\frac{\text{x}-|\text{x}|}{2}, & \text{when} \text{ x}\neq 0\\2, & \text{when}\text{ x} = 0\end{cases}$ is discontinuous at x = 0.
AnswerWe want, to check the continuty of the function at x = 0.
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{x)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\text{h}-|-\text{h|}}{2}=\lim\limits_{\text{h} \rightarrow 0}\frac{-\text{h}-\text{h}}{2}=0$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}-\text{(|h|)}}{2}=0$
$\text{f}(0)=2$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}(0)$
Hence,The function is discotinuous at x = 0
This is rem ovable discontinuty.
View full question & answer→Question 493 Marks
Differentiate w.r.t. x the function in Exercise:
$(\log\text{x})^{\log\text{x}},\text{x}>1$
AnswerLet $\text{y}=(\log\text{x})^{\log\text{x}}$
Tanking logarithm on both the sides, we obtain
$\log\text{y}=\log\text{x}.\log(\log\text{x})$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}[\log\text{x}.\log(\log\text{x})]$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\log\text{x)}.\frac{\text{d}}{\text{dx}}(\log\text{x)}+\log\text{x}.\frac{\text{d}}{\text{dx}}[\log(\log\text{x})]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log(\log\text{x)}.\frac{1}{\text{x}}+\log\text{x}.\frac{1}{\log\text{x}}.\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{1}{\text{x}}\log(\log\text{x)}+\frac{1}{\text{x}}\Big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=(\log\text{x)}^{\log\text{x}}\Big[\frac{1}{\text{x}}+\frac{\log(\log\text{x})}{\text{x}}\Big]$
View full question & answer→Question 503 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.
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