Question 513 Marks
Using Rolle’s theorem, find the point on the curve $\text{y}=\text{x}(\text{x}-4),\text{x}\in[0,4].$ where the tangent is parallel to x-axis.
AnswerWe have, $\text{y}=\text{x}(\text{x}-4),\text{x}\in[0,4]$
Since given function is polynomial it is continuous and differentiable
Also y(0) = y(4) = 0
So, conditions of Rolle's theorem are satisfied.
Hence there exists a point $\text{c}\in(0,4)$ such that
f'(c) = 0
⇒ 2c - 4 = 0
⇒ c = 2
⇒ x = 2 and y(2) = 2(2 - 4) = -4
Therefore, the required point on the curve, where the tangent drawn is parallel to the x-axix is (2, -4).
View full question & answer→Question 523 Marks
Examine the differentialiblilty of the function f defined by $\text{f(x)}=\begin{cases}2\text{x}+3 & \text{if}-3\leq\text{x}\leq-2\\\text{x}+1 & \text{if} -2\leq\text{x}\leq0\\\text{x}+2&\text{if}\ 0\leq\text{x}\leq1\end{cases}$
Answer$\text{f(x)}=\begin{cases}2\text{x}+3 & \text{if}-3\leq\text{x}\leq-2\\\text{x}+1 & \text{if} -2\leq\text{x}\leq0\\\text{x}+2&\text{if}\ 0\leq\text{x}\leq1\end{cases}$
$\text{f}'(\text{x})=\begin{cases}2 & \text{if}-3\leq\text{x}\leq-2,\\1 & \text{if} -2\leq\text{x}\leq0\\1&\text{if}\ 0\leq\text{x}\leq1\end{cases}$
Now,
$\text{LHL}=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{-}}2=2$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{+}}1=1$
Since, at $\text{x}=-2,\text{LHL}\neq\text{RHL}$
Hence, f(x) is not differentiable at x = -2
Again,
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow0^{-}}1=1$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow0^{+}}1=1$
Since, at $\text{x}=0,$
$\text{LHL}=\text{RHL}$
Hence, f(x) is differentiable at x = 0
View full question & answer→Question 533 Marks
Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=\text{x}^{\frac{2}{3}}\text{ on }[-1,1]$
AnswerHere, $\text{f}(\text{x})=\text{x}^{\frac{2}{3}}\text{ on }[-1,1]$
$\text{f}'(\text{x})=\frac{2}{3\text{x}^{\frac{1}{3}}}$
$\text{f}'(0)=\frac{2}{3(0)^{\frac{1}{3}}}$
$\text{f}'(0)=\infty$
So, f'(x) does not exist at $\text{x}=0\in(-1,1)$
⇒ f(x) is not differentialble in $\text{x}\in(-1,1)$
So, Rolle's theorem is not applicable on f(x) in [-1, 1].
View full question & answer→Question 543 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\text{e}^{\text{x}-\text{y}}=\log\Big(\frac{\text{x}}{\text{y}}\Big)$
AnswerWe have, $\tan^{-1}\big(\text{x}^2+\text{y}^2\big)=\text{a}$
Differentiating with respect to x, we get
$\frac{\text{d}}{\text{dx}}\big[\tan^{-1}\big(\text{x}^2+\text{y}^2\big)\big]=\frac{\text{d}}{\text{dx}}(\text{a})$
$\Rightarrow\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{y}^2\big)=0$
$\Rightarrow\Big[\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\Big]\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}}$
View full question & answer→Question 553 Marks
Test the continuity of the function on f(x) at the origin:
$\text{f}\ (\text{x})=\begin{cases}\frac{\text{x}}{\text{|x|}},& \text{x}\neq0\\1, & \text{x} = 0\end{cases}$
AnswerGiven,
$\text{f}\ (\text{x})=\text{x},\text{ x}\neq0$
$\text{f}\ (\text{x})=1,\text{ x}=0$
We observe
$\text{(LHL at x}= 0)$
$\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f} \ (0-\text{h})$
$\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (-h)}=\lim\limits_{\text{h} \rightarrow 0} \frac{\text{-h}}{\text{h}}$
$\lim\limits_{\text{h} \rightarrow 0}-1=-1$
$\text{(RHL at x}=0)$
$\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\ \text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h})$
$\lim\limits_{\text{h} \rightarrow 0}\text{f}\ \text{(h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}}{\text{h}}$
$\lim\limits_{\text{h} \rightarrow 0}1=1$
Hence, f(x) is discontinuous at the origin.
View full question & answer→Question 563 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2+3\text{x}),&\text{if }\text{ x}<0\\\cos2\text{x},&\text{if }\text{ x}\geq0\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\text{k}(\text{x}^2+3\text{x}),&\text{if }\text{ x}<0\\\cos2\text{x},&\text{if }\text{ x}\geq0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\text{f}{(-\text{h)}}=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\Big(\text{k}\big((-\text{h})^2-3\text{h}\big)\Big)=\lim_\limits{\text{h}\rightarrow0}(\cos2\text{h})$
$\Rightarrow0=1$ [It is not possible]
Hence, there does not exist any value of k, which can make the given function continuous.
View full question & answer→Question 573 Marks
For what value of k is the following function continuous at x = 2?
$\text{f(x)}=\begin{cases}2\text{x}+1,&\text{if }\text{ x}<2\\\text{k},&\text{x}=2\\3\text{x}-1,&\text{x}>2\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}2\text{x}+1,&\text{if }\text{ x}<2\\\text{k},&\text{x}=2\\3\text{x}-1,&\text{x}>2\end{cases}$
We have,
$(\text{LHL at x}= 2)=\lim_\limits{\text{x}\rightarrow2^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(2(2-\text{h})+1)=5$
$(\text{RHL at x}= 2)=\lim_\limits{\text{x}\rightarrow2^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})=\lim_\limits{\text{h}\rightarrow0}3(2+\text{h})-1=5$
Also, $\text{f}(2)=\text{k}$
If f(x) is continuous at x = 2, then
$=\lim_\limits{\text{x}\rightarrow2^-}\text{f}(\text{x})=\lim_\limits{\text{x}\rightarrow2^+}\text{f}(\text{x})=\text{f}(2)$
$\Rightarrow5=5=\text{k}$
Hence, for k = 5, (fx) is continuous at x = 2
View full question & answer→Question 583 Marks
Find f(x) is continuse at x = 0, then $\text{f(x)}=\frac{\text{x}}{1-\sqrt{1-\text{x}}}$ becomes continuous at x = 0.
AnswerIf f(x) is continuous at x = 0, then $\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})=\text{f}(0)\ ....(\text{i})$
Given, $\text{f(x)}=\frac{\text{x}}{1-\sqrt{1-\text{x}}}$
$\Rightarrow\text{f(x)}=\frac{\text{x}\big(1+\sqrt{1-\text{x}}\big)}{\big(1-\sqrt{1-\text{x}}\big)\big(1+\sqrt{1-\text{x}}\big)}$
$\Rightarrow\text{f(x)}=\frac{\text{x}\big(1+\sqrt{1-\text{x}}\big)}{1-(1-\text{x})}$
$\Rightarrow\text{f(x)}=\big(1+\sqrt{1-\text{x}}\big)$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}=\big(1+\sqrt{1-\text{x}}\big)=\text{f}(0)$ [From eq. (i)]
$\Rightarrow\text{f}(0)=2$
So, for f(0) = 2, the function f(x) becomes continuous x = 0
View full question & answer→Question 593 Marks
Differentiate the following functions with respect to x:
$\tan^2\text{x}$
AnswerLet,
$\text{y}=\tan^2\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=2\tan\text{x }\frac{\text{d}}{\text{dx}}(\tan\text{x})\ \big[\text{using chain rule}\big]$
$=2\tan\text{x}\times\sec^2\text{x}$
So,
$\frac{\text{d}}{\text{dx}}=\big(\tan^2\text{x}\big)=2\tan\text{x }\sec^2\text{x}.$
View full question & answer→Question 603 Marks
Using mathematical induction prove that $\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})=\text{nx}^{\text{n-1}}$ for all positive.
AnswerTo prove: $\text{P(n)}:\frac{\text{d}}{\text{dx}}\text{(x}^\text{n})=\text{nx}^{\text{n}-1}$ for all positive integers n
For n = 1,
$\text{P}(1):\frac{\text{d}}{\text{dx}}\text{(x)}=1=1.\text{x}^{1-1}$
$\therefore\ $P(n) is true for n = 1
Let P(k) is true for some positive integer k.
That is, $\text{P(k)}:\frac{\text{d}}{\text{dx}}\text{(x}^\text{k})=\text{kx}^{\text{k}-1}$
It has to be proved that P(k + 1) is also true.
Consider $\frac{\text{d}}{\text{dx}}(\text{x}^{\text{k}+1})=\frac{\text{d}}{\text{dx}}(\text{x}.\text{x}^\text{k})$
$=\text{x}^\text{k}.\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}.\frac{\text{d}}{\text{dx}}(\text{x}^\text{k})$ $[\text{By applying product rule}]$
$=\text{x}^\text{k}.1+\text{x}.\text{k}.\text{x}^{\text{k}-1}$
$=\text{x}^\text{k}+\text{kx}^\text{k}$
$=(\text{k}+1).\text{x}^\text{k}$
$=(\text{k}+1).\text{x}^{\text{(k}+1)-1}$
Thus, P(k + 1) is true whenever P(k) is true.
Therefore, by the principle of mathematical lnduction, the statement P(n) is true for every positive integer n.
Hence, proved.
View full question & answer→Question 613 Marks
Find $\frac{\text{dy}}{\text{dx}},\ \text{if y}=\sin^{-1}\text{x}+\sin^{-1} \sqrt{1-\text{x}^2},\ -1\leq\text{x}\leq1$
AnswerIt is given that, $\text{y}=\sin^{-1}\text{x}+\sin^{-1}\sqrt{1-\text{x}^2}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\sin^{-1}\text{x}+\sin^{-1}\sqrt{1-\text{x}^2}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})+\frac{\text{d}}{\text{dx}}(\sin^{-1}\sqrt{1-\text{x}^2})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+\frac{1}{\sqrt{1-(\sqrt{1-\text{x}^2})^2}}.\frac{\text{d}}{\text{dx}}(\sqrt{1-\text{x}^2})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+\frac{1}{\text{x}}.\frac{1}{2\sqrt{1-\text{x}^2}}.\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+\frac{1}{2\text{x}\sqrt{1-\text{x}^2}}(-2\text{x})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}-\frac{1}{\sqrt{1-\text{x}^2}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 623 Marks
Differentiate the following functions with respect to x:
$\log(\tan^{-1}\text{x})$
AnswerConsider $\text{y}=\log\big(\tan^{-1}\text{x}\big)$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\big(\tan^{-1}\text{x}\big)$
$=\frac{1}{\tan^{-1}\text{x}}\times\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)$
[Using chain rule]
$=\frac{1}{\big(1+\text{x}^2\big)\tan^{-1}\text{x}}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\log\tan^{-1}\text{x}\big)=\frac{1}{\big(1+\text{x}^2\big)\tan^{-1}\text{x}}$
View full question & answer→Question 633 Marks
At what points on the following curves, is the tangent parallel to x-axis?
$\text{y}=\text{x}^2\text{ on }[-2,2]$
AnswerLet $f(x) = x^2$
Since f(x) is a polynomial function, it is continuous on [-2, 2] and differentiable on (-2, 2)
Also, $f(2) = f(-2) = 4$
Thus, all the conditions of Rolle's theorem are satisfied.
Concequently, there exists at least one point $\text{c}\in(-2,2)$ for which f'(c) = 0.
But $\text{f}'(\text{c})=0$
$\Rightarrow2\text{c}=0$
$\Rightarrow\text{c}=0$
$\therefore\text{f}_\text{c}=\text{f}_0=0$
By the geometrical interpretetion of Rolle's theorem, (0, 0) is the point on $y = x^2$, where the tangent is parallel to the x-axis.
View full question & answer→Question 643 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq5\\3\text{x}-5,&\text{if}\text{ x}>5\end{cases}\text{at x} =5$
Answer$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq5\\3\text{x}-5,&\text{if}\text{ x}>5\end{cases}\text{at x} =5$We have given that function is continuous at x = 5
$\therefore\ \text{LHL}=\text{RHL}=\text{f}(5)\ ....(\text{i})$
$\text{f}(5)=5\text{k}+1$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow5^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}(5+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}3(5+\text{h})-5=10$
Thus, using (i) we get,
$5\text{k}+1=10$
$5\text{k}=9$
$\text{k}=\frac{9}{5}$
View full question & answer→Question 653 Marks
Write the value of b which $\text{f(x)}=\begin{cases}5\text{x}-4,&0<\text{x}\leq1\\4\text{x}^2+3\text{bx},&1<\text{x}<2\end{cases}$ is continuous at x = 1.
AnswerGiven, $\text{f(x)}=\begin{cases}5\text{x}-4,&0<\text{x}\leq1\\4\text{x}^2+3\text{bx},&1<\text{x}<2\end{cases}$
If f(x) is continuous at x = 1, then
$\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x})=\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x})=\text{f}(1)\ ...(\text{i})$
Now,
$\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x})=\lim\limits_{{\text{h}}\rightarrow0}\text{f}(1-\text{h})\\=\lim\limits_{{\text{h}}\rightarrow0}5(1-\text{h})-4=5-4=1$
$\lim\limits_{{\text{x}}\rightarrow1^+}\text{f(x})=\lim\limits_{{\text{h}}\rightarrow0}\text{f}(1+\text{h})\\=\lim\limits_{{\text{h}}\rightarrow0}4(1+\text{h})^2+3\text{b}(1+\text{h})=4+3\text{b}$
Also,
$\text{f}(1)=5(1)-4=1$
$=\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow1^+}\text{f(x)}=\text{f}(1)$ [From eq. (i)]
$\Rightarrow1=4+3\text{b}=1$
$\Rightarrow1=4+3\text{b}$
$\Rightarrow-3=3\text{b}$
$\Rightarrow\text{b}=-1$
Thus, for b = -1, the function f(x) is continuous at x = 1.
View full question & answer→Question 663 Marks
Differentiate the following w.r.t. x:
$\frac{8^\text{x}}{\text{x}^8}$
AnswerLet $\text{y}=\frac{8^\text{x}}{\text{x}^8}$
$\Rightarrow\ \log\text{y}=\log\frac{8^\text{x}}{\text{x}^8}$
$\Rightarrow\ \frac{\text{d}}{\text{dy}}\log\text{y}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\log8^\text{x}-\log\text{x}^8\big]$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\big[\text{x}.\log8-8.\log\text{x}]$
On differentiating w.r.t. x, we get
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\log8.1-8.\frac{1}{\text{x}}$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\log8-\frac{8}{\text{x}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big(\log8-\frac{8}{\text{x}}\Big)=\frac{8^\text{x}}{\text{x}^8}\Big(\log8-\frac{8}{\text{x}}\Big)$
View full question & answer→Question 673 Marks
For what value of k is the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin2\text{x}}{\text{x}}, & \text{x} \neq 0\\\text{k}, &\text{x} = 0\end{cases}$ continuous at x = 0.
AnswerGiven, $\text{f}\text{(x)}=\begin{cases}\frac{\sin2\text{x}}{\text{x}}, & \text{x} \neq 0\\\text{k}, &\text{x} = 0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{\sin2\text{x}}{\text{x}}=\text{k}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin2\text{x}}{2\text{x}}=\text{k}$
$\Rightarrow2\lim\limits_{\text{x} \rightarrow 0}\frac{\sin2\text{x}}{2\text{x}}=\text{k}$
$\Rightarrow2\times1=\text{k}$
$\Rightarrow\text{k} =2$
View full question & answer→Question 683 Marks
Given the funcation $\text{f(x)}=\frac{1}{\text{x}+2}.$ Find the points of discontinuity of the function f(f(x)).
Answer$\text{f}\big[\text{f(x)}\big]=\frac{1}{\frac{1}{\text{x}+2}+2}=\frac{\text{x}+2}{2\text{x}+5}$
So, f[f(x)] is not defind at x + 2 = 0 and 2x + 5 = 0
If x + 2, then x = - 2
If 2x + 5 = 0, then $\text{x}=-\frac{5}{2}$
Hence, the function is dicontinuous at $\text{x}=-\frac{5}{2}$ and -2
View full question & answer→Question 693 Marks
Find the values of k so that the function f is continuous at the indicated point:
$\text{f(x)}\begin{cases}\frac{\text{k}\cos\text{x}}{\pi -2\text{x}}\ \text{if}\ \text{x}\neq \frac{\pi}{2}\\3, \ \ \ \ \ \ \ \ \text{if}\ \text{x} =\frac{\pi}{2}\end{cases}$
$\text{at} \text{x} = \frac{\pi}{2}$
AnswerHere $\text{f(x)}\begin{cases}\frac{\text{k}\cos\text{x}}{\pi -2\text{x}},\ \text{if}\ \text{x}\neq \frac{\pi}{2}\\3, \ \ \ \ \ \ \ \ \text{if}\ \text{x} =\frac{\pi}{2}\end{cases}$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\frac{\pi}{2}}\text{f(x)}= ^{\ \ \text{Lt}}_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\text{k}\cos\text{x}}{\pi - 2\text{x}}$ $\left[ \text{Put}\ \text{x} = \frac{\pi}{2} + \text{h},\text{h} > 0 \ \text{so that h} \rightarrow0\ \text{as x} \rightarrow\frac{\pi}{2}\right]$
$ ^{\ \ \text{Lt}}_{\text{h}\rightarrow0}\frac{\text{k}\cos\Big(\frac{\pi}{2}+\text{h}\Big)}{{\pi}- {2}\Big(\frac{\pi}{2}-\text{h}\Big)}=^{\ \ \text{Lt}}_{\text{h}\rightarrow0}\frac{-\text{k}\sin\text{h}}{-2\text{h}}$
$\frac{\text{k}}{2}\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}} = \frac{\text{k}}{2}\times1 = \frac{\text{k}}{2}$
Also $\text{f}(\frac{\pi}{2})= 3$
Since f is continuous at $\text{x}= \frac{\pi}{2}$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow{\frac{\pi}{2}}}\text {f(x}) = \text{f}\Big(\frac{\pi}{2}\Big)\Rightarrow \frac{\text{k}}{2} = 3 \Rightarrow\text{k} =6$
View full question & answer→Question 703 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\frac{\sin2\text{x}}{5\text{x}},&\text{if }\text{ x}\neq0\\3\text{k},&\text{if }\text{ x}=0\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\sin2\text{x}}{5\text{x}},&\text{if }\text{ x}\neq0\\3\text{k},&\text{if }\text{ x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\Rightarrow\lim_\limits{\text{x}\rightarrow0}\frac{\sin2\text{x}}{5\text{x}}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow0}\frac{2\sin2\text{x}}{2\times5\text{x}}=\text{f}(0)$
$\Rightarrow\frac{2}{5}\lim_\limits{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}=\text{f}(0)$
$\Rightarrow\frac{2}{5}=3\text{k}$
$\Rightarrow\text{k}=\frac{2}{15}$
View full question & answer→Question 713 Marks
Find $\frac{\text{dy}}{\text{dx}}$ of the functions given in Exercise:
$\text{y}^\text{x}=\text{x}^\text{y}$
AnswerGiven: $\text{y}^\text{x}=\text{x}^\text{y}\ \Rightarrow\ \text{x}^\text{y}=\text{y}^\text{x}$
$\Rightarrow\ \log\text{x}^\text{y}=\log\text{y}^\text{x}\ \Rightarrow\ \text{y}\log\text{x}=\text{x}\log\text{y}$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}(\text{y}\log\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})\ \Rightarrow\ \text{y}.\frac{1}{\text{x}}+\log\text{x}.\frac{\text{dy}}{\text{dx}}=\text{x}.\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}.1$
$\Rightarrow\ \Big(\log\text{x}-\frac{\text{x}}{\text{y}}\Big) \frac{\text{dy}}{\text{dx}}=\log\text{y}-\frac{\text{y}}{\text{x}}\ \Rightarrow\ \Big(\frac{\text{y}\log\text{x}-\text{x}}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\text{x}\log\text{y}-\text{y}}{\text{x}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}\log\text{y}-\text{y})}{\text{x}(\text{y}\log\text{x}-\text{x})}$
View full question & answer→Question 723 Marks
For what value of k is the following function continuous at x = 1
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-1}{\text{x}-1}, & \text{x} \neq 1\\\text{k}, & \text{x}= 1\end{cases}$
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-1}{\text{x}-1}, & \text{x} \neq 1\\\text{k}, & \text{x}= 1\end{cases}$
If f(x) is continuous at x = 1, then
$\lim\limits_{\text{x} \rightarrow 1}\text{f}\text{(x)}=\text{f}(1)$
$\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^2-1}{\text{x}-1}=\text{k}$
$\lim\limits_{\text{x} \rightarrow 1}\frac{\text{(x}-1)(\text{x}+1)}{\text{x}-1}=\text{k}$
$\lim\limits_{\text{x} \rightarrow 1}(\text{x}+1)=\text{k}$
$\text{k}=2$
View full question & answer→Question 733 Marks
Determine the value of the constant k so that the function
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-3\text{x}+2}{\text{x}-1}, &\text{if}\text{ x}\neq1\\\text{k}, &\text{if}\text{ x}=1\end{cases}$ is continuous at x = 1
AnswerWe have that the function is continuous at x = 1
$\therefore$ LHL = RHL = f(1) ....(1)
Now,
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(1-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{(1-\text{h})^2-3(1-\text{h})+2}{(1-\text{h})-1}$
$\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+\text{h}}{-\text{h}}=\lim\limits_{\text{h} \rightarrow 0}-\text{h}-1=-1$
$\text{f}(1)=\text{k}$
from (1), We get,
k = -1
View full question & answer→Question 743 Marks
Find all points of discontinuity of the function $\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2},$ where $\text{t}=\frac{1}{\text{x}-1}.$
AnswerWe have, $\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2},$ where $\text{t}=\frac{1}{\text{x}-1}$
$\therefore\ \text{f(x)}=\frac{1}{\Big(\frac{1}{\text{x}-1}\Big)^2+\frac{1}{\text{x}-1}-2}$
$=\frac{(\text{x}-1)^2}{1+(\text{x}-1)-2(\text{x}-1)^2}=\frac{(\text{x}-1)^2}{-(2\text{x}^2-5\text{x}+2)}=\frac{(\text{x}-1)^2}{(2\text{x}-1)(2-\text{x})}$
So, f(t) is discontinuous at 2x - 1 = 0
$\Rightarrow\ \text{x}=\frac{1}{2}$
and 2 - x = 0 ⇒ x = 2
Also f(t) is discontinuous at x = 1, where $\text{t}=\frac{1}{\text{x}-1}$ is discontinuous.
View full question & answer→Question 753 Marks
For what value of k is the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin5\text{x}}{3\text{x}}, &\text{if}\text{ x}\neq0\\\text{k}, &\text{if}\text{ x}=0\end{cases}$ is continuous at x = 0?
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{\sin5\text{x}}{3\text{x}}, &\text{if}\text{ x}\neq0\\\text{k}, &\text{if}\text{ x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{\sin5\text{x}}{3\text{x}}=\text{k}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{5\sin5\text{x}}{3\times5\text{x}}=\text{k}$
$\Rightarrow\frac{5}{3}\lim\limits_{\text{x} \rightarrow 0}\frac{\sin5\text{x}}{5\text{x}}=\text{k}$
$\Rightarrow\frac{5}{3}\times1=\text{k}$
$\Rightarrow\text{k}=\frac{5}{3}$
View full question & answer→Question 763 Marks
Differentiate the following functions with respect to x:
$\log_7(2\text{x}-3)$
AnswerLet, $\text{y}=\log_7(2\text{x}-3)$
$\Rightarrow\ \text{y}=\frac{\log(2\text{x}-3)}{\log_7}\ \Big[\text{Since}, \log^\text{b}_\text{a}=\frac{\log\text{b}}{\log\text{a}}\Big]$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\log7}\frac{\text{d}}{\text{dx}}\big(\log(2\text{x}-3)\big)$
$=\frac{1}{\log7}\times\frac{1}{(2\text{x}-3)}\frac{\text{d}}{\text{dx}}(2\text{x}-3)$
[Using chain rule]
$=\frac{2}{(2\text{x}-3)\log7}$
Hence, $\frac{\text{d}}{\text{dx}}(\log_7(2\text{x}-3))=\frac{2}{(2\text{x}-3)\log7}$
View full question & answer→Question 773 Marks
Find which of the function:
$\begin{cases}\frac{1-\cos2\text{x}}{\text{x}^2},&\text{ if x}\neq0\\5,&\text{if x}=0\end{cases}$
at x = 0
View full question & answer→Question 783 Marks
Find the second order derivatives of the following functions:
$\text{y}=\text{x}^3\log\text{x}$
AnswerWe have
$\text{y}=\text{x}^3\log\text{x}$
differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dy}}=3\text{x}^2\log\text{x}+\text{x}^3\times\frac{1}{\text{x}}$
$=3\text{x}^2\log\text{x}+\text{x}^2$
differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{x}\log\text{x}+3\text{x}^2\times\frac{1}{\text{x}}+2\text{x}$
$=6\text{x}\log\text{x}+5\text{x}$
View full question & answer→Question 793 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 803 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2-2\text{x}),&\text{if}\text{ x}<0\\\cos\text{x},&\text{if}\text{ x}\geq0\end{cases}\text{at x} = 0$
AnswerGiven,
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2-2\text{x}),&\text{if}\text{ x}<0\\\cos\text{x},&\text{if}\text{ x}\geq0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{k}(\text{h}^2+2\text{h})=0$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}\cos\text{h}=1$
$\therefore\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}\neq\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}$
Thus, no value of k exists for which f(x) is continuous at x = 0.
View full question & answer→Question 813 Marks
If $\text{f(x)}=\frac{2\text{x}+3\sin\text{x}}{3\text{x}+2\sin\text{x}},\text{ x}\neq0$ is continuous at x = 0, then find f(0).
AnswerIt is given that the function is continuous at x = 0
$\therefore\text{LHL}=\text{RHL}=\text{f}(0)\ ....(\text{i})$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow 0}=\lim_\limits{\text{h}\rightarrow 0}(0-\text{h})=\lim_\limits{\text{h}\rightarrow 0}\frac{2(-\text{h})+3\sin(-\text{h})}{3(-\text{h})+2\sin(-\text{h})}$
$=\lim_\limits{\text{x}\rightarrow 0}\frac{-2\text{h}-3\sin\text{h}}{-3\text{h}-2\sin\text{h}}$
$=\lim_\limits{\text{x}\rightarrow 0}\frac{\frac{2\text{h}+3\sin\text{h}}{\text{h}}}{\frac{3\text{h}+2\sin\text{h}}{\text{h}}}$
$=\lim_\limits{\text{x}\rightarrow 0}\frac{2+3\frac{\sin\text{h}}{\text{h}}}{3+2\frac{\sin\text{h}}{\text{h}}}$
$=\frac{2+3}{3+2}=1$ $\Big[\because=\lim_\limits{\theta\rightarrow 0}\frac{\sin\theta}{\theta}=1\Big]$
Using (i) we get
$=\text{f}(0)=1$
View full question & answer→Question 823 Marks
Discuss the continuity of the function f(x) at the point $\text{x}=\frac{1}{2}$ where
$\text{f}\text{(x)}=\begin{cases}\text{x}, & 0\leq\text{x} < \frac{1}{2}\\\frac{1}{2},&\text{x}=\frac{1}{2}\\1-\text{x}, &\frac{1}{2}< \text{x}\leq 1\end{cases}$
AnswerWe want to discuss the continuity of the function at $\text{x}=\frac{1}{2}.$
$\text{LHL}=\lim\limits_{\text{x} \rightarrow \frac{1}{2}^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{f}\Big(\frac{1}{2}-\text{h}\Big)=\lim\limits_{\text{x} \rightarrow 0}\frac{1}{2}-\text{h}=\frac{1}{2}$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow \Big(\frac{1}{2}^-\Big)}=\lim\limits_{\text{x} \rightarrow 0}\text{f}\Big(\frac{1}{2}+\text{h}\Big)=\lim\limits_{\text{x} \rightarrow 0}1-\Big(\frac{1}{2}+\text{h}\Big)=\frac{1}{2}$
$\text{f}\Big(\frac{1}{2}\Big)=\frac{1}{2}$
Thus, $\text{LHL}=\text{RHL}=\text{f}\Big(\frac{1}{2}\Big)=\frac{1}{2}$
Hence, the function is continuous at $\text{x}=\frac{1}{2}.$
View full question & answer→Question 833 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\text{|x|}\cos\Big(\frac{1}{\text{x}}\Big), & \text{ x}\neq 0\\0 &\text{ x} = 0\end{cases}\text{at x}=0$
AnswerGiven,
$\text{f}\text{(x)}=\text{x}\cos\Big(\frac{1}{\text{x}}\Big),\text{x}\neq0$
$\text{f}\text{(x)}=0,\ \text{x}=0$
We observe
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{x}\cos\Big(\frac{1}{\text{x}}\Big)$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{x}\lim\limits_{\text{x} \rightarrow 0}\cos\Big(\frac{1}{\text{x}}\Big)$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=0\times\lim\limits_{\text{x} \rightarrow 0}\cos\Big(\frac{1}{\text{x}}\Big)$
$=0$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
Hence, f(x) is continuous at x = 0.
View full question & answer→Question 843 Marks
Determine that value of the constant 'k' so that function $\text{f(x)}=\begin{cases}\frac{\text{kx}}{|\text{x}|},&\text{if }\text{ x}<0\\3,&\text{if }\text{ x}\geq0\end{cases}$ is continuous at x = 0.
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\text{kx}}{|\text{x}|},&\text{if }\text{ x}<0\\3,&\text{if }\text{ x}\geq0\end{cases}$
Since, the function is continuous at x = 0, therefore,
$\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x})=\lim\limits_{{\text{x}}\rightarrow0^+}\text{f(x})=\text{f}(0)$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\frac{-\text{kx}}{\text{x}}=\lim\limits_{{\text{x}}\rightarrow0}3=3$
$\Rightarrow-\text{k}=3$
$\Rightarrow\text{k}=-3$
View full question & answer→Question 853 Marks
Show that the function $\text{f(x)}=\big|\sin\text{x}+\cos\text{x}|$ is continuous at $\text{x}=\pi.$
AnswerConsider, $\text{f(x)}=\big|\sin\text{x}+\cos\text{x}\big|\text{ at x}=\pi$
Let $\text{g(x)}=\sin\text{x}+\cos\text{x}$
And $\text{h(x)}=|\text{x}|$
$\therefore\ \text{hog (x)}=\text{h}[\text{g (x)}]$
$=\text{h }(\sin\text{x}+\cos\text{x})$
$=|\sin\text{x}+\cos\text{x}|$
Since, g(x) and h(x) are continuous functions and f(x) is a composite functions.
We know that composite functions of two continuous functions is also a continuous function.
Hence, $\text{f(x)}=|\sin\text{x}+\cos\text{x}|$ is a continuous function everywhere,
So, f(x) is continuous at $\text{x}=\pi.$
View full question & answer→Question 863 Marks
Examine the following functions for continuity.
$\text f(\text x)=\begin{vmatrix}\text x-5\end{vmatrix}$
AnswerHere f(x) = | x - 5|
Function f is defined for all real numbers.
Let c be any real number.
$\therefore \text{f}(\text{c}) = \left| \text{c} - 5\right|$
Also $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\left|\text{x}- 5\right| = \left|\text{c}-5\right|$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at x = c,
But c is any real number
$\therefore$ f is continuous at every real number.
View full question & answer→Question 873 Marks
Examine the following functions for continuity.
$\text f(\text X)=\frac{\text X^{2} - 25}{\text {X} + 5}$
Answer$\text f(\text X)=\frac{\text X^{2} - 25}{\text {X} + 5}$
For f to be defined,
x + 5 $\neq$ 0 i. e. x $\neq$ -5
$\therefore$ $D_f=$ Set of all real numbers except -5 = R - { -5}
Letv $\text{c} \neq -5$ be any real number.
$\therefore$ f(c) = $\frac {\text{c}^2 - 25}{\text{c} + 25} = \frac{(\text{c}-5)(\text{c} + 5)}{\text{c}+5} = \text{c}-5$
Also $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\frac{\text{x}^2-25}{\text{x}+5} =\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\frac{({\text{x} -5)(\text{x} +5)}}{\text{x}+ 5}$
$= ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{(x}- 5) = \text{c} - 5$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x}) = \text {f(c)}$
$\therefore$ f is continuous at x = c.
But $\text{c} \neq -5$ is any real number.
$\therefore$ f is continuous at every real number $\text{c} \in \text{D}_\text{f}.$
$\therefore$ f is continuous function.
View full question & answer→Question 883 Marks
$\text{If x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t})\text{ and y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t}),\ \text{find}\dfrac{\text{d}^2\text{y}}{\text{dx}^2}$
AnswerIt is given that, $\text{x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t})\text{ and y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t})$
$\therefore\ \frac{\text{dx}}{\text{dt}}=\text{a}.\frac{\text{d}}{\text{dt}}(\cos\text{t}+\text{t}\sin\text{t})$
$=\text{a}\Big[-\sin\text{t}+\sin\text{t}.\frac{\text{d}}{\text{dx}}(\text{t})+\text{t}.\frac{\text{d}}{\text{dt}}(\sin\text{t})\Big]$
$=\text{a}[-\sin\text{t}+\sin\text{t}+\text{t}\cos\text{t}]=\text{at}\cos\text{t}$
$\frac{\text{dy}}{\text{dt}}=\text{a}.\frac{\text{d}}{\text{dt}}(\sin\text{t}-\text{t}\cos\text{t})$
$=\text{a}\Big[\cos\text{t}-\Big\{\cos\text{t}.\frac{\text{d}}{\text{dt}}(\text{t})+\text{t}.\frac{\text{d}}{\text{dt}}(\cos\text{t})\Big\}\Big]$
$=\text{a}[\cos\text{t}-\{\cos\text{t}-\text{t}\sin\text{t}\}]=\text{at}\sin\text{t}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{\text{at}\sin\text{t}}{\text{at}\cos\text{t}}=\tan\text{t}$
Then, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan\text{t})=\sec^2\text{t}.\frac{\text{dt}}{\text{dx}}$
$=\sec^2\text{t}.\frac{1}{\text{at}\cos\text{t}}\ \Big[\frac{\text{dx}}{\text{dt}}=\text{at}\cos\text{t}\Rightarrow\ \frac{\text{dt}}{\text{dx}}=\frac{1}{\text{at}\cos\text{t}}\Big]$
$=\frac{\sec^3\text{t}}{\text{at}}.0<\text{t}<\frac{\pi}{2}$
View full question & answer→Question 893 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 903 Marks
If $\text{y}=\sqrt{\text{x}+\sqrt{\text{x}+\sqrt{\text{x}+\ .... \text{to }\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}-1}$
AnswerWe have, $\text{y}=\sqrt{\text{x}+\sqrt{\text{x}+\sqrt{\text{x}+\ .... \text{to }\infty}}}$
Squaring both sides, we get,
$y^2 = x + y$
$\Rightarrow 2\text{y}\frac{\text{dy}}{\text{dx}}=1+\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{y}-1)=1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}-1}$
View full question & answer→Question 913 Marks
If `$\text{y}=(\cos\text{x})^{(\cos\text{x})^{(\cos\text{x})\dots\infty}},$ show that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2\tan\text{x}}{\text{y}\log\cos\text{x}-1}.$
AnswerWe have, $\text{y}=(\cos\text{x})^{(\cos\text{x})^{(\cos\text{x})\dots\infty}}$
$\Rightarrow\ \text{y}=(\cos\text{x})^\text{y}$
$\therefore\ \log\text{y}=\log(\cos\text{x})^\text{y}$
Differentiating w.r.t. x, we get
$\frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\text{y}\cdot\frac{\text{d}}{\text{dx}}(\log\cos\text{x})+\log\cos\text{x}\cdot\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\cos\text{x}}\cdot\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\cdot\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big[\frac{1}{\text{y}}-\log\cos\text{x}\Big]$ $=\frac{-\text{y}\sin\text{x}}{\cos\text{x}}=-\text{y}\tan\text{x}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2\tan\text{x}}{(1-\text{y}\log\cos\text{x})}=\frac{\text{y}^2\tan\text{x}}{\text{y}\log\cos\text{x}-1}$
View full question & answer→Question 923 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 933 Marks
Show that the function defined by $g (x) = x – [x]$ is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.
Answerg(x) = x - [x]
Let a be any integer.
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{-}}\text{g(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{-}}\left\{\text{x-[x]}\right\}$
[Put x - a - h, h > 0 so that $h\rightarrow 0$ as $x\rightarrow a^-$
$ = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}}\left\{\text{(a}-\text{h})-(\text{a}-\text{h})\right\} = ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\left\{\text{(a}-\text{h})-(\text{a}-\text{1})\right\}$
$ = ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\text{(a}-\text{h}-\text{a}+1) = \text{a}- 0-\text{a} + 1 = 1$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{+}}\text{g(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{+}}\left\{\text{x - [x]}\right\}$
[Put x = a + h, h > 0 so that $h\rightarrow 0$ as $x\rightarrow a^+$]
$= ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\left\{\text{(a}+\text{h})-(\text{a}+\text{h})\right\}= ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\left\{\text{(a}+\text{h})-\text{a}\right\} =^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\text{(h)}$
= 0
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{-}}\text{f(x)}\neq^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{+}}\text{f(x)}$
$\therefore$ f is discontinuous at x = a
But a is any integral point.
$\therefore$ f is discontinuous at all integgral points.
View full question & answer→Question 943 Marks
Find $\frac{\text{dy}}{\text{dx}}$ of the functions expressed in parametric:
$\text{x}=\text{t}+\frac{1}{\text{t}},\text{ y}=\text{t}-\frac{1}{\text{t}}$
AnswerConsider, $\text{x}=\text{t}+\frac{1}{\text{t}}$ and $\text{y}=\text{t}\frac{1}{\text{t}}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=1+\Big(-\frac{1}{\text{t}^2}\Big)$ and $\frac{\text{dy}}{\text{dt}}=1-\Big(-\frac{1}{\text{t}^2}\Big)$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=1-\frac{1}{\text{t}^2}$ and $\frac{\text{dy}}{\text{dt}}=1+\frac{1}{\text{t}^2}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\text{t}^2-1}{\text{t}^2}$ and $\frac{\text{dy}}{\text{dt}}=\frac{\text{t}^2+1}{\text{t}^2}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$
$=\frac{\text{t}^2+\frac{1}{\text{t}^2}}{\text{t}^2-\frac{1}{\text{t}^2}}$
$=\frac{\text{t}^2+1}{\text{t}^2-1}$
View full question & answer→Question 953 Marks
Differentiate the following functions with respect to x:
$2^{\text{x}^3}$
AnswerConsider $\text{y}=2^{\text{x}^3}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(2^{\text{x}^3}\Big)$
$=2^{\text{x}^3}\times\log_2\frac{\text{d}}{\text{dx}}(\text{x}^3)$
[Using chain rule]
$=3\text{x}^2\times2^{\text{x}^3}\times\log_2$
Hence, the solution is $\frac{\text{d}}{\text{dx}}\big(2^{\text{x}^3}\big)=3\text{x}^2\times2^{\text{x}^3}\log_2$
View full question & answer→Question 963 Marks
Prove that the greatest integer function defined by
$\text{f(x)} = [\text{x} ], 0 < \text{x} < 3$
is not differentiable at x = 1 and x = 2.
AnswerGiven: $\text{f(x)} = [\text{x} ], 0 < \text{x} < 3$
$\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}{\text{h}}$$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{1-1}{\text{h}}= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{0}{\text{h}}=0$
$\text{And}\ \text{L}\text{f}{'}(1) = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{\text{f(1 - h)}-\text{f}(1)}{-\text{h}} = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|1 - \text{h}| - 1}{-\text{h}}$$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|1-\text{h}|-1}{\text{h}}= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{0-1}{-\text{h}}=\infty$
Since $\text{R }\text{f}{'}(1)\neq \text{L}{\text{f}}{'}(1)$
Therefore, f(x) = [x] is not differentiable at x =1.
Similarly f(x) = [x] is not differentiable at x = 2.
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$\text{If y}=5\cos\text{x}-3\sin\text{x},\text{ prove that }\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0$
AnswerLet $\text{y}=5\cos\text{x}-3\cos\text{x}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=-5\sin\text{x}-3\cos\text{x}$
$\Rightarrow\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-5\cos\text{x}+3\sin\text{x}=-(5\cos\text{x}-3\sin\text{x})=-\text{y}$
$\Rightarrow\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{y}\ \Rightarrow\ \frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0$
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Differentiate the following w.r.t. x:
$\log\big[\log(\log\text{x}^5)\big]$
AnswerLet $\text{y}=\log\big[\log(\log\text{x}^5)\big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\big[\log(\log\text{x}^5)\big]$
$=\frac{1}{\log\log\text{x}^5}\cdot\frac{\text{d}}{\text{dx}}\big(\log\cdot\log\text{x}^5\big)$
$=\frac{1}{\log\log\text{x}^5}\cdot\frac{1}{\log\text{x}^5}\cdot\frac{\text{d}}{\text{dx}}\log\text{x}^5$
$=\frac{1}{\log\log\text{x}^5}\cdot\frac{1}{\log\text{x}^5}\cdot\frac{\text{d}}{\text{dx}}(5\log\text{x})$
$=\frac{5}{\text{x}\cdot\log(\log\text{x}^5)\cdot\log(\text{x}^5)}$
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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
$\frac{\text{d}^3\text{y}}{\text{dx}^3}=-2\text{cosec}\ \text{x}\times(-\text{cosec}\ \text{x}\cot\text{x})$
$=2\cot\ \text{x}\ \text{cosec}^2\text{x}=2\cos\ \text{x}\ \text{cosec}^3\text{x}$
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Differentiate the following functions with respect to x:
$\text{e}^{\tan3\text{x}}$
AnswerLet, $\text{y}=\text{e}^{\tan3\text{x}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan3\text{x}}\big)$
$=\text{e}^{\tan3\text{x}}\frac{\text{d}}{\text{dx}}(\tan3\text{x})$
[Using chain rule]
$\text{e}^{\tan3\text{x}}\times\sec^23\text{x}\times\frac{\text{d}}{\text{dx}}(3\text{x})$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan3\text{x}}\big)=3\text{e}^{\tan3\text{x}}\times\sec^2 3\text{x}$
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