Question 2515 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\frac{\text{x}}{2}-\sin\frac{\pi\text{x}}{6}\text{ on }[-1,0]$
AnswerThe given function is $\text{f}(\text{x})=\frac{\text{x}}{2}-\sin\frac{\pi\text{x}}{6}$ Since $\sin\text{x}\ \&\ \frac{\text{x}}{2}$ are everywhere continuous and differentiable, f(x) is continuous on [-1, 0] and differentiable on (-1, 0). Also, f(-1) - f(0) = 0 Thus, f(x) satisfies all the conditions of Rolle's theorem. Now, we have to show that there must exist a point $\text{c}\in(-1,0)$ such that f'(c) = 0.We have
$\text{f}(\text{x})=\frac{\text{x}}{2}-\sin\frac{\pi\text{x}}{6}$ $\Rightarrow\text{f}'(\text{x})=\frac{1}{2}-\frac{\pi}{6}\cos\frac{\pi\text{x}}{6}$ $\therefore\ \text{f}'(\text{x})=0$ $\Rightarrow\frac{1}{2}-\frac{\text{x}}{6}\cos\frac{\pi\text{x}}{6}=0$ $\Rightarrow\cos\frac{\pi\text{x}}{6}=\frac{3}{\pi}$ $\Rightarrow\text{x}=\frac{-6}{\pi}\cos^{-1}\Big(\frac{3}{\pi}\Big)$ Thus, $\text{c}=\frac{-6}{\pi}\cos^{-1}\Big(\frac{3}{\pi}\Big)\in(-1,0)$ such that f'(c) = 0. Hence, Rolle's theorem is verified.
View full question & answer→Question 2525 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f(x)}=\begin{cases}\frac{\text{x}^\text{x}-1}{\log(1+2\text{x})}&,\text{ if x}\neq0&\\ \ &&\text{at x}=0\\7&,\text{ if x}=0\end{cases}$
AnswerIn this problem we need to check continuity at x = 0
Given function is,
$\text{f(x)}=\begin{cases}\frac{\text{x}^\text{x}-1}{\log(1+2\text{x})}&,\text{ if x}\neq0&\\ \ &&\text{at x}=0\\7&,\text{ if x}=0\end{cases}$
$\therefore$ We need to check L.H.L, R.H.L and value of function at x = 0
Idea of logarithmic limit and exponential limit,
$\lim\limits_{\text{x} \rightarrow 0}\frac{\log(1+\text{x})}{\text{x}}=1\ ...(\text{i})$
$\lim\limits_{\text{x} \rightarrow 0}\frac{\text{e}^{\text{x}-1}}{\text{x}}=1\ ...(\text{ii})$
You must have read such limits. You can verify these by expanding log(1 + x) and $e^x$ in its taylor form.
Numerator and denominator conditions also hold for this limit like sandwich theorem.
E.g: $\lim\limits_{\text{x} \rightarrow 0}\frac{\log(1+2\text{x})}{2\text{x}}=1$
But, $\lim\limits_{\text{x} \rightarrow 0}\frac{\log(1+2\text{x})}{\text{x}}\neq1$ as denominator does not have 2x
Now we are ready to solve the problem.
Given function is,
$\text{f(x)}=\begin{cases}\frac{\text{e}^\text{x}-1}{\log(1+2\text{x})}&,\text{ if x}\neq0&\\ \ \ \ \ &&\text{at x}=0\\7&,\text{ if x}=0\end{cases}\ ...(\text{iii})$
Clearly,
f(0) = 7 [from equation 2]
$\text{L.H.L}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h})=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h})$ [putting x = -h in equation iii]
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{e}^{-\text{h}}-1}{\log1+2(-\text{h})}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{e}^{-\text{h}}-1}{\log(1-2\text{h})}$
Using logarithmic and exponential limit as explained above, we have,
$\text{L.H.L}=\frac{1}{2}\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{(\text{e}^{-\text{h}}-1)}{-\text{h}}}{\frac{\log(1-2\text{h})}{-2\text{h}}}=\frac{1}{2}$
$\text{RHL}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h})=\lim\limits_{\text{h} \rightarrow 0}\text{f(h)}$ [putting x = h in equation iii]
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{e}^{\text{h}}-1}{\log1+2\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{e}^{\text{h}}-1}{\log(1+2\text{h})}$
Using logarithmic and exponential limit as explained above, we have,
$\text{R.H.L}=\frac{1}{2}\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{(\text{e}^\text{h}-1)}{\text{h}}}{\frac{\log(1+2\text{h})}{2\text{h}}}=\frac{1}{2}$
Thus, $\text{L.H.L}=\text{R.H.L}\neq\text{f(0)}$
$\therefore\ \text{f(x)}$ is discontinuous at x = 0
View full question & answer→Question 2535 Marks
Differentiate the following functions with respect to x:
$\big(\sin^{-1}\text{x}^4\big)^4$
AnswerConsider $\text{y}=\big(\sin^{-1}\text{x}^4\big)^4$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}^4\big)^4$
$=4\big(\sin^{-1}\text{x}^4\big)\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}^4\big)$
[Using chain rule]
$=4\big(\sin^{-1}\text{x}^4\big)^3\frac{1}{\sqrt{1-\big(\text{x}^4\big)^2}}\frac{\text{d}}{\text{dx}}\big(\text{x}^4\big)$
$=4\big(\sin^{-1}\text{x}^4\big)^3\frac{4\text{x}^3}{\sqrt{1-\text{x}^8}}$
$=\frac{16\text{x}^3\big(\sin^{-1}\text{x}^4\big)^3}{\sqrt{1-\text{x}^8}}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}^4\big)=\frac{16\text{x}^3\big(\sin^{-1}\text{x}^4\big)^3}{\sqrt{1-\text{x}^8}}$
View full question & answer→Question 2545 Marks
Differentiate the following functions with respect to x:
$\frac{2^\text{x}\cos\text{x}}{(\text{x}^2+3)^2}$
AnswerLet $\text{y}=\frac{\text{2}^\text{x}\cos\text{x}}{(\text{x}^2+3)^3}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\frac{\text{2}^\text{x}\cos\text{x}}{(\text{x}^2+3)^3}\Big]$
$=\bigg[\frac{(\text{x}^2+3)^2\frac{\text{d}}{\text{dx}}(2^\text{x}\cos\text{x})-(2^\text{x}\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}^2+3)^2}{\big[(\text{x}^2+3)\big]^2}\bigg]$
[Using quotient rule]
$=\Bigg[\frac{(\text{x}^2+3)^2\Big\{2^\text{x}\frac{\text{d}}{\text{dx}}\cos\text{x}+\cos\text{x}\frac{\text{d}}{\text{dx}}2^\text{x}\Big\}-(2^\text{x}+3)2(\text{x}^2+3)\frac{\text{d}}{\text{dx}}(\text{x}^2+3)}{(\text{x}^2+3)^4}\Bigg]$
[Using Product rule and chain rule]
$=\bigg[\frac{(\text{x}^2+3)^2\big\{-2^\text{x}\sin\text{x}+\cos\text{x}2^\text{x}\log_\text{e}2\big\}-2(2^\text{x}\cos\text{x})(\text{x}^2+3)(2\text{x})}{(\text{x}^2+3)^4}\bigg]$
$=\bigg[\frac{2^\text{x}(\text{x}^2+3)\big\{(\text{x}^2+3)(\cos\text{x}\log_\text{e}2-\sin\text{x})-4\text{x}\cos\text{x}\big\}}{(\text{x}^2+3)^4}\bigg]$
$=\frac{2^\text{x}}{(\text{x}^2+3)^2}\bigg[\cos\text{x}\log_\text{e}2-\sin\text{x}-\frac{4\text{x}\cos\text{x}}{(\text{x}^2+3)}\bigg]$
So,
$\frac{\text{d}}{\text{dx}}\Big[\frac{2^\text{x}\cos\text{x}}{(\text{x}^2+3)^2}\Big]=\frac{2^\text{x}}{(\text{x}^2+3)^2}\bigg[\cos\text{x}\log_\text{e}2-\sin\text{x}-\frac{4\text{x}\cos\text{x}}{(\text{x}^2+3)}\bigg]$
View full question & answer→Question 2555 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\frac{6\text{x}}{\pi}-4\sin^{2}\text{x}\text{ on }\Big[0,\frac{\pi}{6}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\frac{6\text{x}}{\pi}-4\sin^{2}\text{x}$ Since $\sin\text{x}\ \&\ \text{x}$ are everywhere continuous and differentiable, f(x) is continuous on $\Big[0,\frac{\pi}{6}\Big]$ and differentiable on $\Big(0,\frac{\pi}{6}\Big)$ Also, $\text{f}\Big(\frac{\pi}{6}\Big)=\text{f}(0)=0$ Thus, f(x) satisfies all the conditions of Rolle's theorem. Now, we have to show that there must exist a point $\text{c}\in\Big(0,\frac{\pi}{6}\Big)$ such that f'(c) = 0.We have
$\text{f}(\text{x})=\frac{6\text{x}}{\pi}-4\sin^{2}\text{x}$ $\Rightarrow\text{f}'(\text{x})=\frac{6}{\pi}-8\sin\text{x}\cos\text{x}$ $\therefore\ \text{f}'(\text{x})=0$ $\Rightarrow\frac{6}{\pi}-8\sin\text{x}\cos\text{x}=0$ $\Rightarrow\sin2\text{x}=\frac{3}{2\pi}$ $\Rightarrow\text{x}=\frac{1}{2}\sin^{-1}\Big(\frac{3}{2\pi}\Big)$ Thus, $\text{c}=\frac{1}{2}\sin^{-1}\Big(\frac{3}{2\pi}\Big)\in\Big(0,\frac{\pi}{6}\Big)$ such that f'(c) = 0. Hence, Rolle's theorem is verified.
View full question & answer→Question 2565 Marks
If $xy^2 = 1$, prove that $2\frac{\text{dy}}{\text{dx}}+\text{y}^3=0$
AnswerWe have $xy^2 = 1$ .....(i)
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}(\text{xy}^2)=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\text{x}\frac{\text{d}}{\text{dx}}(\text{y}^2)+\text{y}^2\frac{\text{d}}{\text{dx}}(\text{x})=0$
$\Rightarrow\text{x}(2\text{y})\frac{\text{d}}{\text{dx}}+\text{y}^2(1)=0$
$\Rightarrow2\text{xy}\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2}{2\text{xy}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{2\text{x}}$
Put $\text{x}=\frac{1}{\text{y}^2}$ from equation (i)
$ \Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{2\Big(\frac{1}{\text{y}^2}\Big)}$
$\Rightarrow2\frac{\text{dy}}{\text{dx}}=-\text{y}^3$
$\Rightarrow2\frac{\text{dy}}{\text{dx}}+\text{y}^3=0$
View full question & answer→Question 2575 Marks
Differentiate the functions given in Exercise:
$\text{x}^{\sin\text{x}}+(\sin\text{x})^{\cos\text{x}}$
AnswerLet $\text{y}=\text{x}^{\sin\text{x}}+(\sin\text{x})^{\cos\text{x}}$
Putting $\text{u}=\text{x}^{\sin\text{x}}\text{ and v }(\sin\text{x})^{\cos\text{x}},\text{we get }\text{ y}=\text{u}+\text{v}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ \dots\text{(i)}$
Now $\text{u}=\text{x}^{\sin\text{x}}\ \Rightarrow\ \log\text{u}=\log\text{x}^{\sin\text{x}}=\sin\text{x}\log\text{x}$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{u}=\frac{\text{d}}{\text{dx}}(\sin\text{x}\log\text{x})$ $\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\sin\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\sin\text{x}$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\sin\text{x}\frac{1}{\text{x}}+\log\text{x}(\cos\text{x})$ $\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\log\text{x}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}\Big(\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\log\text{x}\Big)$ $\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big(\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\log\text{x}\Big) \dots\text{(ii)}$
Again $\text{v}=(\sin\text{x})^{\cos\text{x}}\ \Rightarrow\ \log\text{v}=\log(\sin\text{x})^{\cos\text{x}}=\cos\text{x}\log\sin\text{x}$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{v}=\frac{\text{d}}{\text{dx}}[\cos\text{x}\log(\sin\text{x})]$ $\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}\cos\text{x}$
$\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\cos\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}\sin\text{x}+\log\sin\text{x}(-\sin\text{x})$
$\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\cot\text{x}.\cos\text{x}-\sin\text{x}\log\sin\text{x}$ $\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{v}(\cot\text{x}.\cos\text{x}-\sin\text{x}\log\sin\text{x})$
$ \Rightarrow\ \frac{\text{dv}}{\text{dx}}=(\sin\text{x})^{\cos\text{x}}(\cot\text{x}.\cos\text{x}-\sin\text{x}\log\sin\text{x})\ \dots\text{(iii)}$
Putting values from eq. (ii) and (iii) in eq. (i),
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big(\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\log\text{x}\Big)+(\sin\text{x})^{\cos\text{x}}(\cot\text{x}.\cos\text{x}-\sin\text{x}\log\sin\text{x})$
View full question & answer→Question 2585 Marks
If $\text{y}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big),$, find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere,
$\text{y}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
Differentiating it with respect to x using chain rule and quotinet rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\big(\frac{1-\text{x}}{1+\text{x}}\big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
$=\frac{(1+\text{x})^2}{(1+\text{x}^2+2\text{x}+1+\text{x}^2-2\text{x})}\bigg[\frac{(1+\text{x})\frac{\text{d}}{\text{dx}}(1-\text{x})-(1-\text{x})\frac{\text{d}}{\text{dx}}(1+\text{x})}{(1+\text{x})^2}\bigg]$
$=\frac{(1+\text{x})^2}{2\text{x}^2+2}\Big[\frac{(1+\text{x})(-1)-(1-\text{x})(1)}{(1+\text{x})^2}\Big]$
$=\frac{(1+\text{x})^2}{2(\text{x}^2+1)}\Big(\frac{-\text{x}-1-1+\text{x}}{(1+\text{x})^2}\Big)$
$=\frac{(1+\text{x})^2}{2(\text{x}^2+1)}\times\frac{-2}{(1+\text{x})^2}$
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\text{x}^2+1}$
View full question & answer→Question 2595 Marks
Verify mean value theorem for the function:
$\text{f(x)}=\frac{1}{4\text{x}-1}\text{ in }[1,4].$
AnswerWe have, $\text{f(x)}=\frac{1}{4\text{x}-1}\text{ in }[1,4]$
f(x) is continuous in [1, 4].
Also, at $\text{x}=\frac{1}{4}.$ f(x) is discontinuous.
Hence, f(x) is continuous in [1, 4].
$\text{f}'(\text{x})=-\frac{4}{(4\text{x}-1)^2},$ which exists in (1, 4).
Since, conditions of mean value theorem are satisfied.
Hence, there exists a real number $\text{c}\in[1,4]$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(1)}{4-1}$
$\Rightarrow\ \frac{-4}{(4\text{c}-1)^2}=\frac{\frac{1}{16-1}-\frac{1}{4-1}}{4-1}=\frac{\frac{1}{15}-\frac{1}{3}}{3}$
$\Rightarrow\ \frac{-4}{(4\text{c}-1)^2}=\frac{1-5}{45}=\frac{-4}{45}$
$\Rightarrow\ (4\text{c}-1)^2=45$
$\Rightarrow\ 4\text{c}-1=\pm3\sqrt{5}$
$\Rightarrow\ \text{c}=\frac{3\sqrt{5}+1}{4}\in(1,4)$ [neglecting (-ve) value]
Hence, mean value theorem has been verified.
View full question & answer→Question 2605 Marks
If $\text{y}=\text{x}^3\log\text{x},$ Prove that $\frac{\text{d}^4\text{y}}{\text{dx}^4}=\frac{6}{\text{x}}$
Answerhere,
$\text{y}=\text{x}^3\log\text{x},$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=3\text{x}^2\log{x}+\text{x}^3\times\frac{1}{\text{x}}$
$=3\text{x}^2\log{\text{x}}+\text{x}^2$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{x}\log\text{x}+3\text{x}^2\times\frac{1}{\text{x}}+2\text{x}$
$=6\text{x}\log\text{x}+5\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\log\text{x}+6\text{x}\times\frac{1}{\text{x}}+5=6\log\text{x}+11$
Differentiating w.r.t.x, we get
View full question & answer→Question 2615 Marks
Write the number of points where f(x) = |x| + |x − 1| is continuous but not differentiable.
AnswerGiven:
f(x) = |x| + |x - 1|
$\Rightarrow\text{f(x)}=\begin{cases}-\text{x}-(\text{x}-1),&\text{x}<0\\\text{x}-(\text{x}-1),&0\leq\text{x}<1\\\text{x}+(\text{x}-1),&\text{x}\geq1\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}-2\text{x}+1,&\text{x}<0\\1,&0\leq\text{x}<12\\\text{x}-1,&\text{x}\geq1\end{cases}$
When x < 0, we have:
f(x) = -2x + 1 which, being a polynomial function is continuous and differentiable.
When $0\leq\text{x}<1,$ we have:
f(x) = 1 which, being a polynimial function is continuous and differentiable on (0, 1).
When $\text{x}\leq1,$ we have:
f(x) = 2x - 1 which, being a polynimial function is continuous and differentiable on x > 2.
Thus, the possible points of non differentiability of f(x) are 0 and 1.
Now,
(LHL at x = 0)
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{2\text{x}+1-1}{\text{x}-0}\ [\because\text{f(x)}=-2\text{x}+1,\text{x}<0]$
$\lim\limits_{\text{x}\rightarrow0}\frac{-2\text{x}}{\text{x}}=-2$
(RHL at x = 0)
$=\lim\limits_{\text{x}\rightarrow0^{+}}\text{f(x)}-\text{f}(0)\text{x}-0$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{1-1}{\text{x}-1}$
$=0\ [\because\text{f(x)}=1,0\leq\text{x}<1]$
Thus, (LHL at x = 1) $\neq$ (RHL at x = 1)
Hence f(x) is not differentiable at x = 1.
Therefore, 0, 1 are the points where f(x) is continuous but not differentiable.
View full question & answer→Question 2625 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\bigg[\frac{\text{x}^\frac{1}{3}+\text{a}^{\frac{1}{3}}}{1-(\text{ax})^\frac{1}{3}}\bigg]$
AnswerLet $\text{y}=\tan^{-1}\bigg[\frac{\text{x}^\frac{1}{3}+\text{a}^{\frac{1}{3}}}{1-(\text{ax})^\frac{1}{3}}\bigg]$
$\Rightarrow\text{y}=\tan^{-1}\big(\text{x}^\frac{1}{3}\big)+\tan^{-1}\big(\text{a}^\frac{1}{3}\big)$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiate it with respect to x usign chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\big(\text{x}^\frac{1}{3}\big)^2}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^\frac{1}{2}\big)+0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\bigg(\frac{1}{2}\times\text{x}^{\frac{1}{3}-1}\bigg)}{1+\text{x}^\frac{2}{3}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{1}{3\text{x}^\frac{2}{3}\Big(1+\text{x}^\frac{2}{3}\Big)}$
View full question & answer→Question 2635 Marks
Differentiate the following functions with respect to x:
$\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}$
AnswerLet $\text{y}=\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2\frac{\text{d}}{\text{dx}}(\text{e}^\text{x}\log\text{x})-(\text{e}^\text{x}\log\text{x})\frac{\text{d}}{\text{dx}}\text{x}^2}{\big(\text{x}^2\big)^2}$
[Using quotient rule]
$=\frac{\text{x}^2\Big\{\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})\Big\}-\text{e}^\text{x}\log\text{x}\times2\text{x}}{\text{x}^4}$
[Using product rule]
$=\frac{\text{x}^2\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]-2\text{xe}^\text{x}\log\text{x}}{\text{x}^4}$
$=\frac{\frac{\text{x}^2\text{e}^\text{z}(1+\text{x}\log\text{x})}{\text{x}}-2\text{xe}^\text{z}\log\text{x}}{\text{x}^4}$
$=\frac{\text{xe}^\text{x}[1+\text{x}\log\text{x}-2\log\text{x}]}{\text{x}^4}$
$=\frac{\text{xe}^\text{x}}{\text{x}^3}\Big[\frac{1}{\text{x}}+\frac{\text{x}\log\text{x}}{\text{x}}-\frac{2\log\text{x}}{\text{x}}\Big]$
$=\text{e}^\text{x}\text{x}^{-2}\Big[\frac{1}{\text{x}}+\log\text{x}-\frac{2}{\text{x}}\log\text{x}\Big]$
So,
$\frac{\text{d}}{\text{dx}}\Big[\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}\Big]=\text{e}^\text{x}\text{x}^{-2}\Big[\frac{1}{\text{x}}+\log\text{x}-\frac{2}{\text{x}}\log\text{x}\Big]$
View full question & answer→Question 2645 Marks
Differentiate the following functions with respect to x:
$\sqrt{\frac{1-\text{x}^2}{1+\text{x}^2}}$
AnswerLet $\text{y}=\sqrt{\frac{1-\text{x}^2}{1+\text{x}^2}}$
$\text{y}=\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^\frac{1}{2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^\frac{1}{2}$
$=\frac{1}{2}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^{\frac{-1}{2}}\bigg[\frac{(1+\text{x}^2)\frac{\text{d}}{\text{dx}}(1-\text{x}^2)-(1-\text{x}^2)\frac{\text{d}}{\text{dx}}(1+\text{x}^2)}{\big(1+\text{x}^2\big)^2}\bigg]$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^{\frac{1}{2}}\bigg[\frac{(1+\text{x}^2)(-2\text{x})-(1-\text{x}^2)(2\text{x})}{\big(1+\text{x}^2\big)^2}\bigg]$
$=\frac{1}{2}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^{\frac{1}{2}}\bigg[\frac{-2\text{x}-2\text{x}^3-2\text{x}+2\text{x}^3}{\big(1+\text{x}^2\big)^2}\bigg]$
$=\frac{1}{2}\frac{-4\text{x}}{\sqrt{1-\text{x}^2}\big(1+\text{x}^2)^\frac{3}{2}}$
So,
$\frac{\text{d}}{\text{dx}}\bigg(\sqrt{\frac{1-\text{x}^2}{1+\text{x}^2}}\bigg)=\frac{-4\text{x}}{\sqrt{1-\text{x}^2}\big(1+\text{x}^2)^\frac{3}{2}}$
View full question & answer→Question 2655 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$y^3 - 3xy^2 = x^3 + 3x^2y$
AnswerHere, $y^3 - 3xy^2 = x^3 + 3x^2y$ Differentiating with respect to x,
$\Rightarrow\frac{\text{d}}{\text{dy}}(\text{y}^3)-\frac{\text{d}}{\text{dx}}(3\text{xy}^2)=\frac{\text{d}}{\text{dx}}(\text{x}^3)+\frac{\text{d}}{\text{dx}}(3\text{x}^2\text{y})$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-3\Big[\text{x}\frac{\text{d}}{\text{dx}}\text{y}^2\frac{\text{d}}{\text{dx}}(\text{x})\Big]=3\text{x}^2+3\Big[\text{x}^2\frac{\text{d}}{\text{dx}}(\text{y})+\text{y}\frac{\text{d}}{\text{dx}}(\text{x}^2)\Big]$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-3\Big[\text{x}(2\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big]=3\text{x}^2+3\Big[\text{x}^2\frac{\text{dy}}{\text{dx}}+\text{y}(2\text{x})\Big]$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-6\text{xy}\frac{\text{dy}}{\text{dx}}-3\text{y}^2+3\text{x}^2+3\text{x}^2\frac{\text{dy}}{\text{dx}}+6\text{xy}$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-6\text{xy}\frac{\text{dy}}{\text{dx}}-3\text{x}^2\frac{\text{dy}}{\text{dx}}=3\text{x}^2+6\text{xy}+3\text{y}^2$
$=3\frac{\text{dy}}{\text{dx}}(\text{y}^2-2\text{xy}-\text{x}^2)=3(\text{x}^2+2\text{xy}+\text{y}^2)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3(\text{x}+\text{y})^2}{3(\text{y}^2-2\text{xy}-\text{x}^2)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}+\text{y})^2}{\text{y}^2-2\text{xy}-\text{x}^2)}$
View full question & answer→Question 2665 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{a}+\text{bx}}{\text{b}-\text{ax}}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{a}+\text{bx}}{\text{b}-\text{ax}}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}+\text{bx}}{\text{b}}}{\frac{\text{b}-\text{ax}}{\text{b}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}}{\text{b}}+\frac{\text{bx}}{\text{b}}}{\frac{\text{b}}{\text{a}}-\frac{\text{ax}}{\text{b}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}}{\text{b}}+\text{x}}{1-\big(\frac{\text{a}}{\text{b}}\big)\text{x}}\bigg)$
$\text{y}=\tan^{-1}\big(\frac{\text{a}}{\text{b}}\big)+\tan^{-1}\text{x}$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+\frac{1}{1+\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$
View full question & answer→Question 2675 Marks
Differentiate the following functions with respect to x:
$\log(\text{x}+\sqrt{\text{x}^2+1})$
AnswerLet $\text{y}=\log(\text{x}+\sqrt{\text{x}^2+1})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dy}}\log\big(\text{x}+\sqrt{\text{x}^2+1}\big)$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\frac{\text{d}}{\text{dx}}\Big(\text{x}+\big(\text{x}^2+1\big)^\frac{1}{2}\Big)$
[Using chain rule]
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[1+\frac{1}{2}\big(\text{x}^2+1\big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+1\big)\Big]$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[1+\frac{1}{2\sqrt{\text{x}^2+1}}\times2\text{x}\Big]$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[\frac{\sqrt{\text{x}^2+1}+\text{x}}{\sqrt{\text{x}^2+1}}\Big]$
$=\frac{1}{\sqrt{\text{x}^2+1}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\log\big(\text{x}+\sqrt{\text{x}^2+1}\big)\Big)=\frac{1}{\sqrt{\text{x}^2+1}}$
View full question & answer→Question 2685 Marks
If $\text{y}=3\cos(\log\text{x})+4\sin(\log\text{x}),$ prove that $\text{x}^2\text{y}_2+\text{xy}_1+\text{y}=0$
AnswerHere,
$\text{y}=3\cos(\log\text{x})+4\sin(\log\text{x}),$
Differentiating w.r.t.x, we get
$\text{y}_1=-3\sin(\log\text{x})\times\frac{1}{\text{x}}+4\cos(\log\text{x})\times\frac{1}{\text{x}}$
$=\frac{-3\sin(\log\text{x})+4\cos(\log\text{x})}{\text{x}}$
Differentiating w.r.t.x, we get
$\text{y}_2=\frac{\Big(\frac{-3\cos(\log\text{x})}{\text{x}}-\frac{4\sin(\log\text{x})}{\text{x}}\Big)\times \text{ x}-\{-3\sin(\log\text{x})+4\cos(\log\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{y}_2=\frac{-3\cos(\log\text{x})-4\sin(\log\text{x})-\{-3\sin(\log\text{x})+4\cos(\log\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{y}_2=\frac{-3\cos(\log\text{x})-4\sin(\log\text{x})}{\text{x}^2}-\frac{\{-3\sin(\log\text{x})+4\cos(\log\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{y}_2=\frac{-3\cos(\log\text{x})+4\sin(\log\text{x})}{\text{x}^2}-\frac{\{-3\sin(\log\text{x})+4\cos(\log\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{y}_2=\frac{-\text{y}}{\text{x}^2}-\frac{\text{y}_1}{\text{x}}$
$\Rightarrow\text{x}^2\text{y}_2=-\text{y}-\text{xy}_1$
$\Rightarrow\text{x}^2\text{y}_2+\text{y}+\text{xy}_1=0$
Hence proved
View full question & answer→Question 2695 Marks
If $f(x) = Ax^2 + Bx + C$ is such that f(a) = f(b), then write the value of c in Rolle's theorem.
AnswerWe have,
$f(x) = Ax^2 + Bx + C$
Differentiating the given function with respect to x, we get
$f'(x) = 2Ax + B$
$\Rightarrow f'(c) = 2Ac + B$
$\therefore f'(c) = 0$
$\Rightarrow 2Ac + B = 0$
$\Rightarrow\text{c}=\frac{-\text{b}}{2\text{A}}\ ...1$
$\because f(a) = f(b)$
$\therefore Aa^2+ Ba + C = Ab^2+ bB + C$
$\Rightarrow Aa^2+ Ba = Ab^2+ bB$
$\Rightarrow A(a^2- b^2) + B(a - b) = 0$
$\Rightarrow A(a^- b)(a + b) + B(a - b) = 0$
$\Rightarrow (a^- b)[A(a + b) + B] = 0$
$\Rightarrow\text{a}=\text{b},\text{A}=\frac{-\text{B}}{\text{a}+\text{b}}$
$\Rightarrow(\text{a}+\text{b})=\frac{-\text{B}}{\text{A}}$
$(\because\ \text{a}\neq\text{b})$
From (1) we have,
$\text{c}=\frac{\text{a}+\text{b}}{2}$
View full question & answer→Question 2705 Marks
If $\text{y}=(\sin^{-1}\text{x})^2,$ prove that $(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0$
AnswerGiven,
$\text{y}=(\sin^{-1}\text{x})^2\dots\text{ eq.1}$
To prove: $(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0$
Let's find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, lets first find $\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})^2$
Using chain rule we will differentiate the above expression:
Let $\text{t}=\sin^{-1}\text{x}$
$\Rightarrow\frac{\text{dt}}{\text{dx}}=\frac{1}{\sqrt{(1-\text{x}^2)}}$ $[$using formula for derivative of $\sin^{-1}\text{x}]$
And $y = t^2$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\frac{\text{dt}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=2\text{t}\frac{1}{\sqrt{(1-\text{x}^2)}}=2\sin^{-1}\text{x}\frac{1}{\sqrt{(1-\text{x}^2)}}\dots\text{ eq. 2}$
Again differentiating with respect to x applying product rule:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)+\frac{2}{\sqrt{(1-\text{x}^2)}}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{2\sin^{-1}\text{x}}{2(1-\text{x}^2)\sqrt{1-\text{x}^2}}(-2\text{x})+\frac{2}{(1-\text{x}^2)}$ $\bigg[\text{using }\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})=\text{nx}^{\text{n}-1}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}=\frac{1}{\sqrt{(1-\text{x}^2)}}\bigg]$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}+\frac{2}{(1-\text{x}^2)}$
$(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=2+\frac{2\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
Using eq. 2:
$(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=2+\frac{\text{dy}}{\text{dx}}$
$\therefore(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0\dots\text{ proved.}$
View full question & answer→Question 2715 Marks
Find which of the function in Exercises 2 to 10 is continuous or discontinuos at the indicated point:
$\text{f(x)}=\begin{cases}\frac{\text{e}^{\frac{1}{\text{x}}}}{1+\text{e}^{\frac{1}{\text{x}}}},&\text{if x}\neq0\\0,&\text{if x }=\text{a}\end{cases}$
at x = 0
AnswerWe have, $\text{f(x)}=\begin{cases}\frac{\text{e}^{\frac{1}{\text{x}}}}{1+\text{e}^{\frac{1}{\text{x}}}},&\text{if x}\neq0\\0,&\text{if x }=\text{a}\end{cases}$ at x = 0.
At x = 0 $\text{L.H.L}=\lim\limits_{\text{h}\rightarrow0^-}\frac{\text{e}^{\frac{1}{\text{x}}}}{1+\text{e}^{\frac{1}{\text{x}}}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\frac{1}{0-\text{h}}}{1+\text{e}^{\frac{1}{0-\text{h}}}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\frac{1}{\text{h}}}}{1+\text{e}^{-\frac{1}{\text{h}}}}$ $=\frac{\text{e}^{-\infty}}{1+\text{e}^{-\infty}}=\frac{0}{1+0}=0$
$\text{R.H.L}=\lim\limits_{\text{h}\rightarrow0^+}\frac{\text{e}^{\frac{1}{\text{x}}}}{1+\text{e}^{\frac{1}{\text{x}}}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\frac{1}{0+\text{h}}}}{1+\text{e}^{\frac{1}{0+\text{h}}}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\frac{1}{\text{h}}}}{1+\text{e}^{\frac{1}{\text{x}}}}=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{e}^{-\frac{1}{\text{h}}}+1}$ $=\frac{1}{\text{e}^{-\infty}+1}=\frac{1}{0+1}=1$
Thus, L.H.L ≠ R.H.L at x = 0.
So, f(x) is discontinuous at x = 0.
View full question & answer→Question 2725 Marks
$\text{If}\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),\ \text{with}\cos\text{a}\neq\pm1,\ \text{prove that}\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
AnswerIt is given that, $\cos\text{y}=\text{x}\cos(\text{a}+\text{y})$ $\therefore\ \frac{\text{d}}{\text{dx}}[\cos\text{y}]=\frac{\text{d}}{\text{dx}}[\text{x}\cos(\text{a}+\text{y})]$ $\Rightarrow\ -\sin\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y}).\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}.\frac{\text{d}}{\text{dx}}[\cos(\text{a}+\text{y})]$ $\Rightarrow\ -\sin\text{y}\frac{\text{dy}}{\text{dt}}=\cos(\text{a}+\text{y})+\text{x}.[-\sin(\text{a}+\text{y})]\frac{\text{dy}}{\text{dx}}$ $\Rightarrow\ [\text{x}\sin(\text{a}+\text{y})-\sin\text{y}]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})\ \dots(1)$Since $\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),\text{x}=\frac{\cos\text{y}}{\cos(\text{a}+\text{y})}$
Then, equation (1) reduces to $\Big[\frac{\cos\text{y}}{\cos(\text{a}+\text{y})}.\sin(\text{a}+\text{y})-\sin\text{y}\Big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$ $\Rightarrow\ \ [\cos\text{y}.\sin(\text{a}+\text{y})-\sin\text{y}.\cos(\text{a}+\text{y})].\frac{\text{dy}}{\text{dx}}=\cos^2(\text{a}+\text{y})$ $\Rightarrow\ \sin(\text{a}+\text{y}-\text{y})\frac{\text{dy}}{\text{dx}}=\cos^2(\text{a}+\text{b})$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{b})}{\sin\text{a}}$ Hence, proved.
View full question & answer→Question 2735 Marks
Verify mean value theorem for the function:
$\text{f(x)}=\sqrt{25-\text{x}^2}\text{ in }[1,5].$
AnswerWe have, $\text{f(x)}=\sqrt{25-\text{x}^2}\text{ in }[1,5]$
Since $25 - x^2$ and square root function are continuous and differentiable in their domain, given function f(x) is also continuous and differentiable.
So, conditions of mean value thecorem are satisfied.
Hence, there exists atleast one $\text{c}\in(1,5)$ such that,
$\text{f}'(\text{c})=\frac{\text{f}(5)-\text{f}(1)}{5-1}$
$\Rightarrow\ \frac{-\text{c}}{\sqrt{25-\text{c}^2}}=\frac{0-\sqrt{24}}{4}$
$\Rightarrow\ 16\text{c}^2=24(25-\text{c}^2)$
$\Rightarrow\ 40\text{c}^2=600$
$\Rightarrow\ \text{c}^2=15$
$\Rightarrow\ \text{c}=\sqrt{15}\in(1,5)$
Hence, mean value theorem has been verified.
View full question & answer→Question 2745 Marks
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find0 $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\text{a}\Big(\cos\text{t}+\log\tan\frac{\text{t}}{2}\Big)\text{y}=\text{a}\sin\text{t}$
AnswerThe given equations are $\text{x}=\text{a}\Big(\cos\text{t}+\log\tan\frac{\text{t}}{2}\Big)\text{ and y}=\text{a}\sin\text{t}$
Then, $\frac{\text{dx}}{\text{dt}}= \text{a}.\Big[\frac{\text{d}}{\text{dt}}(\cos\text{t})+\frac{\text{d}}{\text{dt}}\Big(\log\tan\frac{\text{t}}{2}\Big)\Big]$
$=\text{a}\Bigg[-\sin\text{t}+\frac{1}{\tan\frac{\text{t}}{2}}.\frac{\text{d}}{\text{dt}}\Big(\tan\frac{\text{t}}{2}\Big)\Bigg]$
$=\text{a}\Big[-\sin\text{t}+\cot\frac{\text{t}}{2}.\sec^2\frac{\text{t}}{2}.\frac{\text{d}}{\text{dt}}\Big(\frac{\text{t}}{2}\Big)\Big]$
$=\text{a}\Bigg[-\sin\text{t}+\frac{\cos\frac{\text{t}}{2}}{\sin\frac{\text{t}}{2}}\times\frac{1}{\cos^2\frac{\text{t}}{2}}\times\frac{1}{2}\Bigg]$
$=\text{a}\Bigg[-\sin\text{t}+\frac{1}{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}\Bigg]$
$=\text{a}\Big(-\sin\text{t}+\frac{1}{\sin\text{t}}\Big)$
$=\text{a}\Big(\frac{-\sin^2\text{t}+1}{\sin\text{t}}\Big)$
$=\text{a}\frac{\cos^2\text{t}}{\sin\text{t}}$
$\frac{\text{dy}}{\text{dt}}=\text{a}\frac{\text{d}}{\text{dt}}(\sin\text{t})=\text{a}\cos\text{t}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{\text{a}\cos\text{t}}{\Big(\text{a}\frac{\cos^2\text{t}}{\sin\text{t}}\Big)}=\frac{\sin\text{t}}{\cos\text{t}}=\tan\text{t}$
View full question & answer→Question 2755 Marks
Differentiate the following functions with respect to x:
$\log_\text{x}3$
AnswerLet, $\text{y}=\log_\text{x}3$
$\Rightarrow\ \text{y}=\frac{\log3}{\log\text{x}}\ \Big[\because\ \log_\text{a}\text{b}=\frac{\log\text{b}}{\log\text{a}}\Big]$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\log3}{\log\text{x}}\Big)$
$=\log3\frac{\text{d}}{\text{dx}}(\log\text{x})^{-1}$
$=\log3\times\Big[-1(\log\text{x})^{-2}\Big]\frac{\text{d}}{\text{dx}}(\log\text{x})$
[Using chain rule]
$=-\frac{\log 3}{(\log\text{x})^2}\times\frac{1}{\text{x}}$
$=-\Big(\frac{\log 3}{\log\text{x}}\Big)^2\times\frac{1}{\text{x}}\times\frac{1}{\log3}$
$=-\frac{1}{\text{x}\log3(\log_3\text{x})^2} \Big[\because \frac{\log\text{b}}{\log\text{a}}=\log_\text{a}\text{b}\Big]$
So,
$\frac{\text{d}}{\text{dx}}(\log_\text{x}3)=-\frac{1}{\text{x}\log3(\log_3\text{x})^2}$
View full question & answer→Question 2765 Marks
If $xy = e^{x-y}$, find $\frac{\text{dy}}{\text{dx}}$
AnswerThe given function is $xy = e^{x-y}$
Taking logarithm on both the sides, we obtain
$\log(\text{xy})=\log\big(\text{e}^{\text{x}-\text{y}})$
$\Rightarrow\log\text{x}+\log\text{y}=(\text{x}-\text{y})\log\text{e}$
$\Rightarrow\log\text{x}=\log\text{y}=(\text{x}-\text{y})\times1$
$\Rightarrow\log\text{x}=\log\text{y}=\text{x}-\text{y}$
Differentiating both sides with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\log\text{x})+\frac{\text{d}}{\text{dx}}(\log\text{y})=\frac{\text{d}}{\text{dx}}(\text{x})-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}}+\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\big(1+\frac{1}{\text{y}}\big)\frac{\text{dy}}{\text{dx}}=1-\frac{1}{\text{x}}$
$\Rightarrow\big(\frac{\text{y}+1}{\text{y}}\big)\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{\text{x}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}-1)}{\text{x}(\text{y}+1)}$
View full question & answer→Question 2775 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$(x^2 + y^2)^2 = xy$
AnswerGiven, $(x^2 + y^2)^2 = xy$
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\Big(\big(\text{x}^2+\text{y}^2\big)^2\Big)=\frac{\text{d}}{\text{dx}}(\text{xy})$
$\Rightarrow2(\text{x}^2+\text{y}^2\big)\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2\big)=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}\big(\text{x}\big)$
[Using chain rule]
$\Rightarrow2(\text{x}^2+\text{y}^2\big)\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)$
$\Rightarrow4\text{x}(\text{x}^2+\text{y}^2\big)+4\text{y}\big(\text{x}^2+\text{y}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}$
$\Rightarrow4\text{y}\big(\text{x}^2+\text{y}^2\big)\frac{\text{dy}}{\text{dx}}-\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-4\text{x}(\text{x}^2+\text{y}^2\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[4\text{y}\text{x}^2+4\text{y}^3-\text{x}\big]=\text{y}-4\text{x}^3-4\text{x}\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{y}-4\text{x}^3-4\text{x}\text{y}^2}{4\text{y}\text{x}^2+4\text{y}^3-\text{x}}\Big)$
View full question & answer→Question 2785 Marks
If $\text{y}=\cos^{-1}\Big\{\frac{2\text{x}-3\sqrt{1-\text{x}^2}}{\sqrt{13}}\Big\},$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\cos^{-1}\Big\{\frac{2\text{x}-3\sqrt{1-\text{x}^2}}{\sqrt{13}}\Big\}$
Let $\text{x}=\cos\theta, \text{So},$
$\text{y}=\cos^{-1}\Big\{\frac{2\cos\theta-3\sqrt{1-\cos^2\theta}}{\sqrt{13}}\Big\}$
$=\cos^{-1}\Big\{\frac{2}{\sqrt{13}}\cos\theta-\frac{3}{13}\sin\theta\Big\}$
Let $\cos\phi=\frac{2}{\sqrt{13}}$
$\Rightarrow\sin\phi=\sqrt{1-\cos^2\phi}$
$=\sqrt{1-\Big(\frac{2}{\sqrt{13}}\Big)^2}$
$=\sqrt{\frac{13-4}{13}}$
$=\sqrt{\frac{9}{13}}$
$\sin\phi=\frac{3}{\sqrt{13}}$
So,
$\text{y}=\cos^{-1}\big\{\cos\phi\cos\theta-\sin\phi\sin\theta\big\}$
$=\cos^{-1}\big[\cos(\theta+\phi)\big]$
$\text{y}=\phi+\theta$
$\text{y}=\cos^{-1}\Big(\frac{2}{\sqrt{13}}\Big)+\cos^{-1}\text{x}$
$\Big[\text{Since, x}=\cos\theta,\cos\phi=\frac{2}{\sqrt{13}}\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+\Big(-\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 2795 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\sin3\text{x}\text{ on }[0,\pi]$
AnswerThe given function is $\text{f}(\text{x})=\sin3\text{x}$
Since $\sin3\text{x}$ everywhere continuous and differentiable,
$\sin3\text{x}$ is continuous on $[0,\pi]$ and differentiable on $(0,\pi).$
Also,
$\text{f}(\pi)=\text{f}(0)=0$
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists $\text{c}\in(0,\pi)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\sin3\text{x}$
$\Rightarrow \text{f}'(\text{x})=3\cos3\text{x}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow3\cos3\text{x}=0$
$\Rightarrow\cos3\text{x}=0$
$\Rightarrow3\text{x}=\frac{\pi}{2},\frac{3\pi}{2},....$
$\Rightarrow\text{x}=\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}$
Since, $\text{c}=\frac{\pi}{4}\in(0,\pi)$ such that f'(c) = 0
Hence, Rolle's theorem is verified.
View full question & answer→Question 2805 Marks
Verify Rolle's theorem for the following function on the indicated intervals$f(x) = x(x^- 4)^2$ on the interval $[0, 4]$
AnswerGiven function is $f(x) = x(x^- 4)^2$. Which can be rewritten as $f(x) = x^3 - 8x^2 + 16x.$
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function f(x) is continuous and derivable on$ [0, 4].$
Also,
$f(0) = f(4) = 0$
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists $\text{c}\in[0,4]$ such that $f'(c) = 0.$
We have
$f(x) = x^3- 8x^2 + 16x$
$\Rightarrow f'(x) = 3x^2- 16x + 16$
$\therefore f'(x) = 0$
$\Rightarrow 3x^2- 16x + 16 = 0$
$\Rightarrow 3x^2- 12x - 4x + 16 = 0$
$\Rightarrow 3x(x - 4) - 4(x - 4) = 0$
$\Rightarrow (x - 4)(3x - 4)$
$\Rightarrow\text{x}=4,\frac{4}{3}$
Thus, $\text{c}=\frac{4}{3}\in(0,4)$ such that $f'(c) = 0.$
Hence, Rolle's theorem is verified.
View full question & answer→Question 2815 Marks
Differentiate the following functions with respect to x:
$\frac{\text{x}^2(1-\text{x}^2)}{\cos2\text{x}}$
AnswerConsider $\text{y}=\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
Differentiate it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\cos2\text{x}\frac{\text{d}}{\text{dx}}\text{x}^2(1-\text{x}^2)^3-\text{x}^2(1-\text{x}^2)^3\frac{\text{d}}{\text{dx}}\cos2\text{x}}{\cos^2 2\text{x}}$
$=\frac{\cos2\text{x}\Big[\text{x}^2\frac{\text{d}}{\text{dx}}(1-\text{x}^2)^3+(1-\text{x}^2)^3\frac{\text{d}}{\text{dx}}\text{x}^2-\text{x}^2(1-\text{x}^2)^3(-2\sin2\text{x})\Big]}{\cos^2 2\text{x}}$
$=\frac{\cos2\text{x}\Big[-6\text{x}^2(1-\text{x}^2)^2+(1-\text{x}^2)^32\text{x}+2\text{x}^2(1-\text{x}^2)^3\sin2\text{x}\Big]}{\cos^2 2\text{x}}$
$=\frac{2\text{x}(1-\text{x}^2)^2}{\cos2\text{x}}-\frac{6\text{x}^3(1-\text{x}^2)^2}{\cos2\text{x}}+\frac{2\text{x}^2(1-\text{x}^2)^3\sin2\text{x}}{\cos^2 2\text{x}}$
$=2\text{x}(1-\text{x}^2)\sec2\text{x}\big\{1-4\text{x}^2+\text{x}(1-\text{x}^2)\tan2\text{x}\big\}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=2\text{x}(1-\text{x}^2)\sec2\text{x}\big\{1-4\text{x}^2+\text{x}(1-\text{x}^2)\tan2\text{x}\big\}$
View full question & answer→Question 2825 Marks
If $e^x + x^y = e^{x+y}$, prove that $\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{y}-\text{x}}=0$
AnswerHere,
$e^x + e^y = e^{x+y}$ ......(i)
Differentiating both the sides using chain rule,
$\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})+\frac{\text{d}}{\text{dx}}(\text{e}^\text{y})=\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}+\text{y}})$
$\text{e}^{\text{x}}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\text{e}^{\text{x}}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\Big[1+\frac{\text{d}}{\text{dx}}\Big]$
$\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}-\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}-\text{e}^\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}+\text{y}}-\text{e}^\text{x}}{\text{e}^\text{y}-\text{e}^{\text{x}+\text{y}}}$
$=\Big(\frac{\text{e}^\text{x}+\text{e}^\text{x}-\text{e}^\text{x}}{\text{e}^\text{y}-\text{e}^\text{x}-\text{e}^\text{y}}\Big)$
[Using equation (i)]
$\frac{\text{dy}}{\text{dx}}=-\text{e}^{\text{y}-\text{x}}$
$\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{y}-\text{x}}=0$
View full question & answer→Question 2835 Marks
Differentiate $(\log\text{x})^\text{x}$ with respect to x.
AnswerLet $\text{u}=(\log1+\text{x})^\text{x}$
Taking log on both sides,
$\log\text{u}=\log(\log\text{x})^\text{x}$
$\Rightarrow\log\text{u}=\text{x}\log(\log\text{x})$
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\big\{\log(\log\text{x})\big\}+\log(\log\text{x})\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\Big(\frac{1}{\log\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\log\text{x}(1)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\frac{\text{x}}{\log\text{x}}\big(\frac{1}{\text{x}}\big)+\log\log\text{x}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]\ .....(\text{i})$
Again, let $\text{v}=\log\text{x}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{\text{x}}\ .....(\text{ii})$
Dividing equation (i) by (ii), we get
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]}{\frac{1}{\text{x}}}$
$\Rightarrow\frac{\text{du}}{\text{dv}}=\frac{(\log\text{x})^\text{x}\Big[\frac{1+\log\text{x}(\log\log\text{x})}{\log\text{x}}\Big]}{\frac{1}{\text{x}}}$
$\Rightarrow\frac{\text{du}}{\text{dv}}=\text{x}(\log\text{x})^{\text{x}{-1}}(1+\log\text{x}\times\log\log\text{x})$
View full question & answer→Question 2845 Marks
Prove that $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$ is everywhere continuous.
AnswerWhen x < 0, we have
$\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$
We know that $\sin\text{x}$ as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\sin\text{x}}{\text{x}}$ is continuous at each x < 0
When x > 0, we have
f(x) = x + 1, which is a polynomial function.
Therefore, f(x) is continuous at each x > 0
Now,
Let us consider the point x = 0
Given, $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$
We have,
$(\text{LHL at x = 0})=\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(0-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin(-\text{h})}{-\text{h}}\Big)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin(\text{h})}{\text{h}}\Big)=1$
$(\text{RHL at x = 0})=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}\\=\lim\limits_{\text{h}\rightarrow0}\text{f}(0+\text{h})=\lim\limits_{\text{h}\rightarrow0}\text{f(h)}=\lim\limits_{\text{h}\rightarrow0}(\text{h}+1)=1$
Also, $\text{f}(0)=0+1=1$
$\therefore\ \lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}=\text{f}(0)$
Thus, f(x) is continuous at x = 0
Hence, f(x) is everywherefore continuous.
View full question & answer→Question 2855 Marks
Let $\text{f(x)}=\begin{cases}\frac{1-\sin^3\text{x}}{3\cos^2\text{x}},&\text{if }\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if }\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})}^2,&\text{x}>\frac{\pi}{2}\end{cases}$ if f(x) is continuous at $\text{x}=\frac{\pi}{2},$ find a and b.
AnswerIt is given that the function is continuous at $\text{x}=\frac{\pi}{2}$
$\therefore\ \text{LHL}=\text{RHL}=\text{f}\Big(\frac{\pi}{2}\Big)\ ...(\text{i})$
Now, $\text{f}\Big(\frac{\pi}{2}\Big)=\text{a}$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow\frac{\pi^-}{2}}\text{f(x)}=\lim_\limits{\text{h}\rightarrow 0}\text{f}\Big(\frac{\pi}{2}-\text{h}\Big)$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{1-\sin^3\Big(\frac{\pi}{2}-\text{h}\Big)}{3\cos^2\Big(\frac{\pi}{2}-\text{h}\Big)}=\lim_\limits{\text{h}\rightarrow 0}\frac{1-\cos^3\text{h}}{3\sin^2\text{h}}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{(1-\cos\text{h})(1+\cos^2\text{h}+\cos\text{h})}{3\sin^2\text{h}}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{2\sin^2\frac{\text{h}}{2}(1+\cos^2\text{h}+\cos\text{h})}{3\sin^2\text{h}}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{2\Bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\Bigg)^2\times\frac{\text{h}^2}{4}\times(1+\cos^2\text{h}+\cos\text{h})}{3\Big(\frac{\sin\text{h}}{\text{h}}\Big)^2\times\text{ h}^2}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{2\times\frac{1}{4}(1+\cos^2\text{h}+\cos\text{h})}{3}=\frac{1}{2}$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow\frac{\pi^-}{2}}\text{f(x)}=\lim_\limits{\text{h}\rightarrow 0}\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b}\Big(1-\sin\Big(\frac{\pi}{2}+\text{h}\Big)\Big)}{\Big(\pi-2\Big(\frac{\pi}{2}+\text{h}\Big)\Big)^2}=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b}(1-\cos\text{h})}{(\pi-\pi-2\text{h})^2}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b }\times\ 2\sin^2\frac{\text{h}}{2}}{(2\text{h})^2}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b}}{2}\Bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\Bigg)^2\times\frac{1}{4}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b}}{8}=\frac{\text{b}}{8}$
Thus, using (i) we get,
$\text{a}=\frac{1}{2}$
And $\frac{\text{b}}{8}=\frac{1}{2}$
$\Rightarrow\text{b}=4$
Thus, $\text{a}=\frac{1}{2}$ and $\text{b}=4$
View full question & answer→Question 2865 Marks
Differentiate $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ with respect to $\sec^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big),$ if:
$\text{x}\in\Big(0,\frac{1}{\sqrt{2}}\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\sin\theta\sqrt{1-\sin^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\sin\theta\cos\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
And,
Let $\text{v}=\sec^{-1}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\text{v}=\sec^{-1}\Big(\frac{1}{\sqrt{1-\sin^2\theta}}\Big)$
$\Rightarrow\text{v}=\sec^{-1}\Big(\frac{1}{\cos\theta}\Big)$
$\Rightarrow\text{v}=\sec^{-1}(\sec\theta)$
$\Rightarrow\text{v}=\cos^{-1}\bigg(\frac{1}{\frac{1}{\cos\theta}}\bigg)\ \Big[\text{Since},\sec^{-1}\text{x}=\cos^{-1}\big(\frac{1}{\text{x}}\big)\Big]$
$\Rightarrow\text{v}=\cos^{-1}(\cos\theta)\ .....(\text{ii})$
Here,
$\text{x}\in\Big(0,\frac{1}{\sqrt{2}}\Big)$
$\Rightarrow\sin\theta\in\Big(0,\frac{1}{\sqrt{2}}\Big)$
$\Rightarrow\theta\in\Big(0,\frac{\pi}{4}\Big)$
So, from equation (i),
$\text{u}=2\theta\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
Let, $\text{u}=2\sin^{-1}\text{x}\big[\text{Since},\text{x}=\sin\theta\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=2\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
And, from equation (ii),
$\text{v}=\theta\big[\text{Since},\cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi]\big]$
$\Rightarrow\text{v}=\sin^{-1}\text{x}\big[\text{Since},\text{x}=\sin\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{1}$
$\therefore\frac{\text{du}}{\text{dv}}=2$
View full question & answer→Question 2875 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}2,&\text{if }\text{ x}\leq3\\\text{ax}+\text{b},&\text{if }3<\text{ x}<5\\9,&\text{if }\text{ x}\geq5\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}2,&\text{if }\text{ x}\leq3\\\text{ax}+\text{b},&\text{if }3<\text{ x}<5\\9,&\text{if }\text{ x}\geq5\end{cases}$
If f(x) is continuous at x = 3 and 5, then
$\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}$ and $\lim_\limits{\text{x}\rightarrow5^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow5^+}\text{f(x)}$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\text{f}(3-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{f}(3+\text{h})$ and $\lim_\limits{\text{h}\rightarrow0}\text{f}(5-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{f}(5+\text{h})$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}(2)=\lim_\limits{\text{h}\rightarrow0}\big(\text{a}(3+\text{h})+\text{b}\big)$ and $\lim_\limits{\text{h}\rightarrow0}\big(\text{a}(5-\text{h})+\text{b}\big)=\lim_\limits{\text{h}\rightarrow0}(9)$
$\Rightarrow2=3\text{a}+\text{b}$ and $5\text{a}+\text{b}=9$
View full question & answer→Question 2885 Marks
Show that the function g(x) = x - [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.
AnswerThe given function is g(x) = x - [x]
It is evident that g is defined at all integral points.
Let n be an integer.
Then,
g(n) = n - [n] = n - n = 0
The left hand limit of f at x = n is,
$\lim\limits_{{\text{x}}\rightarrow\text{n}^-}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{n}^-}\big(\text{x}-[\text{x}]\big)=\lim\limits_{{\text{x}}\rightarrow\text{n}^-}(\text{x})-\lim\limits_{{\text{x}}\rightarrow\text{n}^-}[\text{x}]\\=\text{n}-(\text{n}-1)=1$
The right hand limit of f at x = n is,
$\lim\limits_{{\text{x}}\rightarrow\text{n}^+}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{n}^+}\big(\text{x}-[\text{x}]\big)\\=\lim\limits_{{\text{x}}\rightarrow\text{n}^+}(\text{x})-\lim\limits_{{\text{x}}\rightarrow\text{n}^+}[\text{x}]=\text{n}-\text{n}=0$
It is observed that the left and right hand limits of f at x = n do not coincide.
Therefore, f is not continuous at x = n
Hence, g is discontinuous at all integral points.
View full question & answer→Question 2895 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}}+\cos\text{x},&\text{if }\text{ x}\neq0\\5,&\text{if }\text{ x}=0\end{cases}$
AnswerWhen $\text{x}\neq0,$ then
$\text{f(x)}=\frac{\sin\text{x}}{\text{x}}+\cos\text{x}$
We know that $\sin\text{x}$ as well as the identity function x both are everwhere continuous.
So, the quotient function $\frac{\sin\text{x}}{\text{x}}$ is continuous at each $\text{x}\neq0$
Also, $\cos\text{x}$ is everwhere continuous.
Therefore, $\text{f(x)}=\frac{\sin\text{x}}{\text{x}}+\cos\text{x}$ is continuous at each $\text{x}\neq0$
Let us consider the point x = 0
Given, $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}}+\cos\text{x},&\text{if }\text{ x}\neq0\\5,&\text{if }\text{ x}=0\end{cases}$
We have
$(\text{LHL at x}=0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(-\text{h})}{-\text{h}}+\cos(-\text{h})\Big)$
$=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(-\text{h})}{-\text{h}}\Big)+\lim_\limits{\text{h}\rightarrow0}\cos(-\text{h})=1+1=2$
$(\text{RHL at x}=0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(\text{h})}{\text{h}}+\cos(\text{h})\Big)$
$=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(\text{h})}{\text{h}}\Big)+\lim_\limits{\text{h}\rightarrow0}\cos(\text{h})=1+1=2$
Also, f(0)=5
$\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}\neq\text{f}(0)$
Thus, f(x) is discontinuous at x = 0
Hence, the only point of discontinuity for f(x) is x = 0
View full question & answer→Question 2905 Marks
Prove that the function f defined by $\text{f(x)}=\begin{cases}\frac{\text{x}}{|\text{x}|+2\text{x}^2},&\text{if x}\neq0\\\text{k},&\text{ if x}=0\end{cases}$ remains discontinuous at x = 0, regardless the choice of k.
AnswerWe have, $\text{f(x)}=\begin{cases}\frac{\text{x}}{|\text{x}|+2\text{x}^2},&\text{if x}\neq0\\\text{k},&\text{ if x}=0\end{cases}$
At x = 0, $\text{L.H.L}=\lim\limits_{\text{x}\rightarrow0^-}\frac{\text{x}}{|\text{x}|+2\text{x}^2}=\lim\limits_{\text{h}\rightarrow0}\frac{(0-\text{h})}{|0-\text{h}|+2(0-\text{h})^2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}+2\text{h}^2}=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}(1+2\text{h})}=-1$
$\text{R.H.L}=\lim\limits_{\text{x}\rightarrow0^+}\frac{\text{x}}{|\text{x}|+2\text{x}^2}=\lim\limits_{\text{h}\rightarrow0}\frac{0+\text{h}}{|0+\text{h}|+2(0+\text{h})^2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}+2\text{h}^2}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}(1+2\text{h})}=1$
And f(0) = k
Since, L.H.L ≠ R.H.L for any value of k.
Hence, f(x) is discontinuous at x = 0 regardless the choice of k.
View full question & answer→Question 2915 Marks
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cos^{-1}\text{x},$ if
$\text{x}\in(0, 1)$
AnswerLet $\text{u}=\sin^{-1}\sqrt{1-\text{x}^2}$
Put $\text{x}=\cos\theta$
$\Rightarrow\text{u}=\sin^{-1}\sqrt{1-\cos^2\theta}$
$\Rightarrow\text{u}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
And, $\text{v}=\cos^{-1}\text{x}\ .....(\text{ii})$
Now, $\text{x}\in(0,1)$
$\Rightarrow\cos\theta\in(0,1)$
$\Rightarrow\theta\in\Big(0,\frac{\pi}{2}\Big)$
So, from equation (i),
$\text{u}=\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta\text{ if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\cos^{-1}\text{x}\big[\text{Since},\cos\theta=\text{x}\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\cos^{-1}\text{x}$
Differentaiting it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-1}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{-1}$
$\therefore\frac{\text{du}}{\text{dx}}=1$
View full question & answer→Question 2925 Marks
Differentiate the following w.r.t. x:
$\tan^{-1}\Big(\frac{\text{a}\cos\text{x}-\text{b}\sin\text{x}}{\text{b}\cos\text{x}+\text{a}\sin\text{x}}\Big),-\frac{\pi}{2}<\text{x}<\frac{\pi}{2}\text{ and }\frac{\text{a}}{\text{b}}\tan\text{x}>-1$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{a}\cos\text{x}-\text{b}\sin\text{x}}{\text{b}\cos\text{x}+\text{a}\sin\text{x}}\Big)$
$=\tan^{-1}\Bigg[\frac{\frac{\text{a}\cos\text{x}}{\text{b}\cos\text{x}}-\frac{\text{b}\sin\text{x}}{\text{b}\cos\text{x}}}{\frac{\text{b}\cos\text{x}}{\text{b}\cos\text{x}}+\frac{\text{a}\sin\text{x}}{\text{b}\cos\text{x}}}\Bigg]$
$=\tan^{-1}\Bigg[\frac{\frac{\text{a}}{\text{b}}-\tan\text{x}}{1+\frac{\text{a}}{\text{b}}\tan\text{x}}\Bigg]$
$=\tan^{-1}\frac{\text{a}}{\text{b}}-\tan^{-1}\tan\text{x}$ $\bigg[\because\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\bigg]$
$\therefore\ \text{y}=\tan^{-1}\frac{\text{a}}{\text{b}}-\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\tan^{-1}\frac{\text{a}}{\text{b}}\Big)-\frac{\text{d}}{\text{dx}}(\text{x})$
$=0-1\bigg[\because\ \frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}}{\text{b}}\Big)=0\bigg]$
$=-1$
View full question & answer→Question 2935 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\frac{{1}-\text{x}^\text{n}}{1-\text{x}}, & \text{x} \neq1\\\text{n}-1, & \text{ x} = 1\end{cases}\text{ n }\in\ \text{N at x}=1$
AnswerWe want, to check the continuity at x = 1
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 1^-}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0} \text{f}(1-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{1-(1-\text{h})^\text{n}}{1-(1-\text{h})}=\lim\limits_{\text{h} \rightarrow 0}\frac{1-\Big[1-\text{nh}+\frac{\text{n}(\text{n}-1)}{2}\text{h}^2+\dots\Big]}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{n}-\frac{\text{n(n-1)}}{2!}\text{h}+\dots$
$=\text{n}$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow 1^+}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0} \text{f}\text{(1+h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{1-(1+\text{h})^\text{n}}{1-(1+\text{h})}=\lim\limits_{\text{h} \rightarrow 0}\frac{1-\Big[1-\text{nh}+\frac{\text{n}(\text{n}-1)}{2}\text{h}^2+\dots\Big]}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{n}+\frac{\text{n}(\text{n}-1)}{2!}\text{h}+\dots$
$=\text{n}$
$\text{f}(1)=\text{n}-1$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}( 1)$
Hence, funtion is discontinuous at x = 1
This is removable discotinuity.
View full question & answer→Question 2945 Marks
If $\text{x}=\text{a}(\theta+\sin\theta)\ \text{and}\ \text{y}=\text{a}(1+\cos\theta)$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{a}}{\text{y}^2}.$
AnswerHere
$\text{x}=\text{a}(\theta+\sin\theta)\ \text{and}\ \text{y}=\text{a}(1+\cos\theta)$
Differentiating w.r.t.$\theta$, we get
$\frac{\text{dx}}{\text{d}\theta}=\text{a}+\text{a}\cos\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=-\text{a}\sin\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{-\text{a}\sin\theta}{\text{a}+\text{a}\cos\theta}=\frac{-\sin\theta}{1+\cos\theta}$
Differentiating w.r.t.$\theta$, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big\{\frac{(1+\cos\theta)\cos\theta+\sin^2\theta}{(1+\cos\theta)^2}\Big\}\frac{\text{d}\theta}{\text{dx}}$
$=\frac{-\cos\theta-\cos^2\theta-\sin^2\theta}{(1+\cos\theta)^2}\times\frac{1}{\text{a}+\text{a}\cos\theta}$
$=\frac{-(1+\cos\theta)}{\text{a}(1+\cos\theta)^3}$
$=\frac{-1}{\text{a}(1+\cos\theta)^2}$
$=\frac{-\text{a}}{\text{y}^2}\ [\because\text{y}=\text{a}(1+\cos\theta)]$
Hence proved
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Differentiate $\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$ with respect to $\sec^{-1}\text{x}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}-\frac{\text{x}}{2}\Big)\Big]$
$\Rightarrow\text{u}=\frac{\pi}{4}-\frac{\text{x}}{2}$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0-\big(\frac{1}{2}\big)$
$\frac{\text{du}}{\text{dx}}=-\frac{1}{2}\ .....(\text{i})$
Let, $\text{v}=\sec^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{\text{x}\sqrt{\text{x}^2-1}}\ .....\text{(ii)}$
Differentiating equation (i) by (ii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=-\frac{1}{2}\times\frac{\text{x}\sqrt{\text{x}^2-1}}{1}$
$\frac{\text{du}}{\text{dv}}=\frac{-\text{x}\sqrt{\text{x}^2-1}}{2}$
View full question & answer→Question 2965 Marks
Discuss the applicability of Rolle’s theorem on the function given by.
$\text{f(x)}=\begin{cases}\text{x}^2+1,&\text{if }0\leq\text{x}\leq1\\3-\text{x},&\text{if }1\leq\text{x}\leq2\end{cases}$
AnswerConsider, $\text{f(x)}=\begin{cases}\text{x}^2+1,&\text{if }0\leq\text{x}\leq1\\3-\text{x},&\text{if }1\leq\text{x}\leq2\end{cases}$
We know that, polynomial function is everywhere continuous and differentiability.
So, f(x) is continuous and differentiable at all points except possibly at x = 1.
Now, check the differentiability at x = 1,
At x = 1 $\text{L.D.H}=\lim\limits_{\text{x}\rightarrow1^-}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}^2+1)-(1+1)}{\text{x}-1}$ $[\because\ \text{f(x)}=\text{x}^2+1,\forall\ 0\leq\text{x}\leq1]$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2-1}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}+1)(\text{x}-1)}{\text{x}-1}$
and $\text{R.D.H}=\lim\limits_{\text{x}\rightarrow1^+}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow1}\frac{(3-\text{x})\text{f}(1+1)}{(\text{x}-1)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{3-\text{x}-2}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow1}\frac{-(\text{x}-1)}{\text{x}-1}=-1$
$\therefore$ L.H.D ≠ R.H.D
So, f(x) is not differentiable at x = 1.
Hence, polle’s theorem is not applicable on the interval [0, 2]
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Find a point on the parabola $y = (x - 4)^2$, where the tangent is parallel to the chord joining $(4, 0)$ and $(5, 1)$.
AnswerHere,
curve is $y = (x - 4)^2$
Since, it is a polynomial function so it is differentiable and continuous. So, it Lagrange's mean value theorem is applicable, so, there exist a point c such that,
$\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$
$\Rightarrow2(\text{c}-4)=\frac{\text{f}(5)-\text{f}(4)}{5-4}$
$\Rightarrow2\text{c}-8=\frac{1-0}{1}$
$\Rightarrow2\text{c}=9$
$\Rightarrow\text{c}=\frac{9}{2}$
$\Rightarrow\text{y}=\Big(\frac{9}{2}-4\Big)^2$
$\Rightarrow\text{y}=\frac{1}{4}$
Thus, $(\text{c},\text{y})=\Big(\frac{9}{2},\frac{1}{4}\Big)$ is a required point.
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If $\text{y}=(\cot^{-1}\text{x})^2$ prove that $\text{y}^2(\text{x}^2+1)^2+2\text{x}(\text{x}^2+1)\text{y}_1=2.$
Answer$\text{y}=(\cot^{-1}\text{x})^2$Differentiating w.r.t.x,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}_1=\frac{-2\cot^{-1}\text{x}}{1+\text{x}^2}$
$=\frac{-2\cot^{-1}\text{x}}{1+\text{x}^2}\ (\text{chain rule})$
$\Rightarrow(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}=-2\cot^{-1}\text{x}$
Differentiating w.r.t.x,
$\Rightarrow(1+\text{x}^2)\text{y}^2+2\text{xy}_1=+2\Big(\frac{+1}{1+\text{x}^2}\Big)$
(Multiplication rule on LHS)
$\Rightarrow(1+\text{x}^2)^2\text{y}_2+2\text{x}(1+\text{x}^2)\text{y}_1=2$
Hence proved
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If $\text{e}^{\text{x}}+\text{e}^{\text{y}}=\text{e}^{\text{x}+\text{y}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{e}^{\text{x}}(\text{e}^\text{y}-1)}{\text{e}^{\text{y}}(\text{e}^{\text{x}}-1)}$ or $\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{y}-\text{x}}=0$
Answer$\text{e}^\text{x}+\text{e}^\text{y}=\text{e}^{\text{x}+\text{y}}$
$\Rightarrow\text{e}^\text{x}+\text{e}^\text{y}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\text{e}^\text{x}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}+\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\text{e}^\text{y}\frac{\text{dy}}{\text{dx}}-\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}(\text{e}^\text{y}-\text{e}^{\text{x}+\text{y}})=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}}{\text{x}^\text{y}-\text{e}^{\text{x}+\text{y}}}$
$=\frac{\text{e}^\text{x}(\text{e}^\text{y}-1)}{\text{e}^\text{y}({1-\text{e}}^\text{x})}$
$=-\frac{\text{e}^\text{x}(\text{e}^\text{y}-1)}{\text{e}^\text{y}(\text{e}^\text{x}-1)}$
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Verify Rolle's theorem for the following function on the indicated intervals$\text{f}(\text{x})=\cos2\Big(\text{x}-\frac{\pi}{4}\Big)\text{ on }\Big[0,\frac{\pi}{2}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\cos2\Big(\text{x}-\frac{\pi}{4}\Big)$ $=\cos\Big(2\text{x}-\frac{\pi}{2}\Big)=\sin2\text{x}.$ Thus, we have to show that there exists $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.We have
$\text{f}(\text{x})=\sin2\text{x}$ $\Rightarrow\text{f}'(\text{x})=2\cos2\text{x}$ $\Rightarrow\text{f}'(\text{x})=0$ $\Rightarrow2\cos2\text{x}=0$ $\Rightarrow\cos2\text{x}=0$ $\Rightarrow\text{x}=\frac{\pi}{4}$ Thus, $\text{c}=\frac{\pi}{4}\in\Big(0,\frac{\pi}{2}\Big)$ such that $\text{f}'(\text{c})=0.$ Hence, Rolle's theorem is verified .
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