Question 515 Marks
If sin y = x sin (a + y), prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{sin}^{2}\text{(a+y)}}{\text{sin a}}$.
Answersin y = x sin (a + y) $\Rightarrow$ cos y $\frac{\text{dy}}{\text{dx}}$ = x cos (a + y) $\frac{\text{dy}}{\text{dx}}$ + sin (a + y)
$\therefore\frac{\text{dy}}{\text{dx}}= \frac{\text{sin (a + y)}}{\text{cos y - x cos (a + y)}}$
$\text{x} = \frac{\text{sin y}}{\text{sin (a+y)}}\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\text{sin (a+y)}}{\text{cos y -}\frac{\text{sin y}}{\text{sin (a+y)}}.\text{cos (a+y)}}$
$\therefore \frac{\text{dy}}{\text{dx}}= \frac{\text{sin}^{2}\text{(a+y)}}{\text{sin (a+y) cos y - cos (a + y) sin y}} = \frac{\text{sin}^{2}\text{(a + y)}}{\text{sin a}}$.
View full question & answer→Question 525 Marks
Using differentials, find the approximate value of $\sqrt{49.5}$
AnswerLet y = $\sqrt{x}\therefore\text{ y }+\Delta\text{y}=\sqrt{\text{x}+\Delta\text{x}}$
$ \Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}.\Delta\text{x}\simeq\sqrt{\text{x}+\Delta\text{x}}$
$ \Rightarrow\sqrt{\text{x}}+\frac{\text{1}}{\text{2}\sqrt{x}}.\Delta\text{x}\simeq\sqrt{\text{x}+\Delta\text{x}}$
Putting x = 49 and Δx = 0.5 we get
$ \Rightarrow\sqrt{\text{49}}+\frac{\text{1}}{\text{2}\sqrt{49}}(0.5)\simeq\sqrt{\text{49.5}}$
$ \Rightarrow\sqrt{\text{49}}=7+\frac{\text{1}}{\text{28}}=\text{7.0357}$.
View full question & answer→Question 535 Marks
$\text{If (cos x)}^{\text{y}}=\text{(cos y)}^{\text{x}},\text{find }\frac{\text{dy}}{\text{dx}}.$
Answer$\text{(cos x)}^{\text{y}}=\text{(cos y)}^{\text{x}}\Rightarrow$ y log cos x = x log cos y.$\therefore\text{y}.\frac{\text{(-sin x)}}{\text{cos x}}+\text{log cos x.}\frac{\text{dy}}{\text{dx}}=\text{x}.\frac{\text{(-sin y)}}{\text{cos y}}\frac{\text{dy}}{\text{dx}}+\text{log cos y.}$
(log cos x + x tan y) $\frac{\text{dy}}{\text{dx}}$ = log cos y + y tan x
$\therefore\frac{\text{dy}}{\text{dx}}= \frac{\text{log cos y + y tan x}}{\text{log cos x + x tan y}}$.
View full question & answer→Question 545 Marks
Differentiate $x^{x \cos x}$ + $\frac{\text{x}^{2}+1}{\text{x}^{2}-1}$ w.r.t.x.
Answer$\text{y}=\text{x}^{\text{x cos x}}+\frac{\text{x}^{2}+1}{\text{x}^{2}-1}=\text{u + v}$$\text{u}=\text{x}^{\text{x cos x}}\Rightarrow\log\text{u}=\text{x cos x}\cdot{\text{log x}}$
$\therefore\frac{1}{\text{u}}\cdot\frac{\text{du}}{\text{dx}}=\frac{\text{x cos x}}{\text{x}}+\text{cos x}\cdot{\text{log x - sin x log x}}$
$\Rightarrow$$\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x cos x}}(\text{cos x+cos x log x}\text{ - sin x log x})$
$\text{v}=\frac{\text{x}^{2}+1}{\text{x}^{2}-1},\frac{\text{dv}}{\text{dx}}=\frac{(\text{x}^{2}-1)\text{2x}-\text{(x}^{2}+1)\text{2x}}{\text{(x}^{2}-1)^{2}}=-\frac{\text{4x}}{\text{(x}^{2}-1)^{2}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x cos x}}\text{(cos x + log x}\cdot\text{cos x - sin x log x})-\frac{\text{4x}}{(\text{x}^{2}-1)^{2}}$
View full question & answer→Question 555 Marks
Find the value of 'a' for which the function f defined as
$ \begin{matrix} & \text{a sin}\frac{\pi}{2}\text{(x + 1)}, & x\leq0 \\ \text{f(x)} \\ & \frac{\text{tan x - sin x}}{\text{x}^{3}}, & x<0 \\ \end{matrix}$
is continuous at X = 0.
AnswerL.H.L = $\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{-}}$ f(x) = a
f(0) = a.sin$\pi/\text{r}$ = a
R.H.L. = $\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{-}}$ $\frac{\text{tan x}}{\text{x}}\cdot\frac{\text{(1 - cos x)}}{\text{x}^{2}}$
= $\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{-}}\frac{\tan \text{x}}{\text{x}}\cdot2\Bigg(\frac{\sin\text{x/2}}{2\text{x/2}}\Bigg)^{2}=\frac{1}{2}$
$\therefore\text{a}=\frac{1}{2}$
View full question & answer→Question 565 Marks
Show that the function f defined as follows, is continuous at x = 2, but not differentiable thereat:
f(x) = $\begin{matrix} \text{3x - 2}, && 02 \end{matrix}$.
Answer$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{ f(2 - h)}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}[2(2 - \text{h})^{2}-(2 - \text{h})]=6\ .......\text{(i)}$
$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{ f(2 + h)}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}[5(2 + \text{h})-4 ]=6\ ............\text{(ii)}$
f(2) = 8 – 2 = 6 ...........(iii)
From (i), (ii), and (iii), f(x) is continuous at x = 2
$\text{RHD}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\Bigg[\frac{\left\{5{(2+h)-4}-(6)\right\}}{h}\Bigg]\neq\text{LHD}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\Bigg[\frac{\left\{(2h - 3)(h - 2) - 6\right\}}{-h}\Bigg]\text{as 5}\neq7$
$\therefore$ f(x) is not differentiable there at.
View full question & answer→Question 575 Marks
$\text{Find} \frac{\text{dy}}{\text{dx}} \text{if (x}^{2} + \text{y}^2)^{2} = \text{xy.}$
Answer$\text{(x}^{2} + \text{y}^{2})^{2} = \text{xy} \Rightarrow 2 \text{(x}^{2} + \text{y}^{2}) \bigg(\text{2x + 2y} \frac{\text{dy}}{\text{dx}}\bigg) = \text{x} \frac{\text{dy}}{\text{dx}} + \text{y}$$\Rightarrow \text{4y} \frac{\text{dy}}{\text{dx}}\text{(x}^{2} + \text{y}^{2}) -\text{(x}^{2} + \text{y}^{2})- \text{x} \frac{\text{dy}}{\text{dx}} = \text{y - 4x} \text{(x}^{2} + \text{y}^{2})$
$\therefore \frac{\text{dy}}{\text{dx}} = \frac{\text{y - 4x}\text{(x}^{2} + \text{y}^{2})}{\text{4y}\text{(x}^{2} + \text{y}^{2}) - \text{x}}$
View full question & answer→Question 585 Marks
Differentiate the following function w.r.t. x:$X^{\sin x} + (\sin x)^{\cos x} $
Answer$\text{Let x}^{\sin x} = \text{u, and} ( \sin \text{x)}^{\cos \text{x}} = \text{v} \therefore \text{y = u + v} \Rightarrow \frac{\text {dy}}{\text{dx}} = \frac{\text{du}}{\text{dx}} + \frac{\text{dv}}{\text{dx}}$
$\text{Getting} \frac{\text{du}}{\text{dx}} = \text{x}^{\sin{\text{x}}} \bigg[\frac{\sin\text{x}}{\text{x}} + \log \text{x} . \cos{\text{x}} \bigg]$
$\text{and} \frac{\text{dv}}{\text{dx}} = ( \sin \text{x})^{\cos\text{x}}[\cos\text{x} .\cot\text{x} - \sin\text{x} \log \sin \text{x}]$
$\therefore \frac{\text{dy}}{\text{dx}} = \text{x}^{\sin \text{x}} \bigg[ \frac{\sin \text{x}}{\text{x}} + \log\text{x}.\cos\text{x}\bigg] + (\sin\text{x}) ^{\cos{\text{x}}} [\cos\text{x}. \cot\text{x} -\sin\text{x} \log\sin\text{x}]$
View full question & answer→Question 595 Marks
$\text{If y = 3} \cos (\log\text{x}) + 4\sin (\log \text{x}), \text{then show that x}^{2} .\frac{\text{d}^{2}{\text{y}}}{\text{dx}^{2}} + \text{y} = 0$
Answer$\text{y} = 3\cos(\log\text{x}) + 4\sin (\log\text{x}) \Rightarrow \frac{\text{dy}}{\text{dx}} = - \frac{3\sin(\log\text{x})}{\text{x}} +\frac{4\cos(\log\text{x})}{\text{x}}$$\text{x} \frac{\text{dy}}{\text{dx}} = -3\sin (\log\text{x}) + 4\cos(\log\text{x})$
$\Rightarrow \text{x} \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = - \frac{3\cos(\log\text{x})}{\text{x}} -\frac{4\sin(\log\text{x})}{\text{x}}$
$\Rightarrow \text{x} \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} + \text{x} \frac{\text{dy}}{\text{dx}} = -[3\cos (\log\text{x}) + 4\sin (\log{\text{x}})] = \text{-y}$
$\text{or x}^{2} \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} +\text{y = o} $
View full question & answer→Question 605 Marks
Differentiate the following with respect to x: $\tan^{-1} \bigg(\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x + \sqrt{ 1 - x}}}\bigg)$
Answer$\text{Let x} = \cos 2 \theta , \sqrt{1 + x} = \sqrt{2} \cos\theta, \sqrt{1 - x} = \sqrt{2} \sin\theta$$\text{Let y} = \bigg[\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x + \sqrt{ 1 - x}}}\bigg] = \tan^{-1} \bigg[\frac{1 - \tan \theta}{1 + \tan \theta}\bigg] = \tan^{-1} \tan \bigg(\frac{\pi}{4} - \theta\bigg)$
$= \frac{\pi}{4} - \theta = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} \text{x}$
$\therefore \frac{dy}{dx} = \frac{-1}{2} \bigg(\frac{-1}{\sqrt{1 - x^{2}}}\bigg) = \frac{1}{2\sqrt{1 - x^{2}}}$
View full question & answer→Question 615 Marks
For what value of $ k$ is the following function continuous at $x = 2$$f(x) = \begin{matrix} 2x + 1 & ; & x< 2 \\ k & ; & x = 2 \\ 3x - 1 & ; & x> 2 \end{matrix} $
AnswerFor continuity of the function at $x = 2$$\lim\limits_{h \rightarrow 0} f (2 -h) = f(2) = \lim\limits_{h \rightarrow 0} f(2 + h)$
Now, $f (2 - h) = 2 (2 - h) + 1 = 5 - 2h$
$\therefore \lim\limits_{h \rightarrow 0} f(2- h) = 5$
Also, $f(2 + h) = 3(2 + h) -1 = 5 + 3h$
$ \lim\limits_{h \rightarrow 0} f(2 + h) = 5$
So, for continuity $f(2) = 5.$
$\therefore k = 5.$
View full question & answer→Question 625 Marks
Find the value of k if the function: $f (x) = \begin{matrix} kx^{2}& x\geq 1 & \\ 4 & x< 1 & \\ \end{matrix} \text{is continuous at x = 1} $
Answer$\lim\limits_{ x \rightarrow 1^{-}} f (x) = \lim\limits_{ h \rightarrow{0}} k ( 1 + h)^{2} = k \dots\dots\dots\text{(i)}$$f(1) = k \dots\dots\dots\dots\dots\text{(ii)}$
$\lim\limits_{ x \rightarrow 1^{-}} = 4 \dots\dots\dots\dots\text{(iii)}$
$\text{As f (x) is continuous at x} = 1 , \therefore k = 4$
View full question & answer→Question 635 Marks
$\text{Evaluate} \lim\limits_{x \rightarrow\frac{\pi}{4}} \bigg( \frac{ \sin x - \cos x}{x- \frac{\pi}{4}} \bigg)$
Answer$\lim\limits_{ x \rightarrow \frac{\pi}{4}} \Bigg[\frac {\sin x - \cos x} { x - \frac{\pi}{4}}\Bigg] = \lim\limits_{ x \rightarrow \frac{\pi}{4}} \sqrt{2} \Bigg[ \frac{\sin x.{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}.\cos x}}{ x- \frac{\pi}{4}}\Bigg]$$ \lim\limits_{ x \rightarrow \frac{\pi}{4}} \sqrt{2} \Bigg[ \frac{\sin x.\cos {\frac{\pi}{{4}} - \cos x . \sin \frac{\pi}{4}}}{ x- \frac{\pi}{4}}\Bigg]$
$ \lim\limits_{ x \rightarrow \frac{\pi}{4}} \sqrt{2} \Bigg[ \frac{\sin\bigg( x - \frac{\pi}{4}\bigg){}}{ x- \frac{\pi}{4}}\Bigg] = \sqrt{2}. 1 = \sqrt{2}$
View full question & answer→Question 645 Marks
Verify Rolle's theorem for the function $f ( x) = x^{2} - 5x + 4\text{ on} [ 1, 4].$
Answer$\text{f (x)} = x^{2} - 5x + 4 \text {( a polynomial function)}$$\text{(i) the function is continuous on [1,4] }$
$\text{(ii) It is differentiable on (1,4)}$
$\text{(iii) f (1)} = \text{f (4)} = 0$
$\therefore$ All the conditions of Rolles' Theoremare satisfied.
$\therefore \text{f' (c)} = 0 \Rightarrow \text{2 c - 5} = 0 \Rightarrow \text{c} = \frac{5}{2}$
$\text{As c} = \frac{5}{2} \in (1,4),$ the Rolle's theorem is verified.
View full question & answer→Question 655 Marks
Differentiate $\sin(x^{2} + 1)$with respect to $ x$from first principle.
Answer$\frac{\triangle y}{\triangle x} = \frac {\sin \bigg[( {x + \triangle x)^{2}}+ 1\bigg] -\sin ( x^{2} + 1)}{\triangle x} $$\therefore \frac{dy}{dx} = \lim\limits_{\triangle x \rightarrow 0} \frac{2\cos\bigg[ \frac{( x + \triangle x)^{2} + x^{2}}{2}\bigg] .\sin \bigg[\frac{\triangle x( 2x + \triangle x}{2} \bigg]} {\triangle x}$
$= 2.\cos ( x^{2} + 1) . \lim\limits_{\triangle x \rightarrow 0} \frac{\sin\bigg[\frac{(\triangle x+{2 x} +\triangle{x)}}{2}\bigg] \bigg[\frac{(2x +\triangle x)}{2}\bigg]}{\triangle x.\bigg(\frac{2x + \triangle x}{2}\bigg)}$
$= \text{2 x} .\cos ( x^{2} + 1)$
View full question & answer→Question 665 Marks
$\text{If y} = \sin (\log x) , \text{prove that}$$x^{2} \frac{d^{2}y}{dx^{2}} + x \frac{dy}{dx} + y =0$
Answer$y = \sin(\log x)$$\therefore \frac{dy}{dx} = \cos (\log x) . \frac{1}{x} \Rightarrow x \frac{dy}{dx} = \cos (\log x)$
$\therefore x \frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} = \frac{-1}{x} \sin (\log x) = -\frac{y}{x}$
$\Rightarrow x^{2} \frac{d^{2}y}{dx^{2}} + x \frac{dy}{dx} + y = 0$
View full question & answer→Question 675 Marks
Verify Rolle's Theorem for the following function:$f (x) = (x - 1) (x - 2)^{2}, [1,2]$
Answer$f (x) = (x - 1) (x - 2)^{2}, [1,2]$The function f(x) is differentiable on [1, 2] and so it is continuous on [1, 2]
$\therefore$ Al conditions of Rolles Theorem are satisfied
$\therefore f' (x) = (x - 1) 2 (x -2) + (x - 2)^{2}$
$= (x - 2)[2x - 2 + x- 2] = (x - 2) (3x - 4)$
$\therefore f' (c) = 0 \Rightarrow c = 2 \text{ and}\ {c}\frac{4}{3}$
$As 1<\frac{4}{3}<2, \text{the Roll's theorem is verfied}$
View full question & answer→Question 685 Marks
Evaluate:$\lim\limits_{X\rightarrow \frac{\pi}{6}} \bigg[ \frac{\sqrt{3}\sin x - \cos x}{x- \frac{\pi}{6}}\bigg]$
Answer$\lim\limits_{X\rightarrow \frac{\pi}{6}} \Bigg[ \frac{\sqrt{3}\sin x - \cos x}{x- \frac{\pi}{6}}\Bigg] \lim\limits_{X\rightarrow \frac{\pi}{6}} \Bigg[2\frac{\Big( \frac{\sqrt{3}}{2}.\sin x-\frac{1}{2}\cos x\Big)}{x-\frac{\pi}{6}}\Bigg]$$\lim\limits_{X\rightarrow \frac{\pi}{6}} \Bigg[\frac{2\big(\cos \frac{\pi}{6}.\sin x-\sin\frac{\pi}{6}.\cos x\big)}{x-\frac{\pi}{6}}\Bigg]$
$2\lim\limits_{X\rightarrow\frac{\pi}{6}}\frac{\sin\big(x-\frac{\pi}{6}\big)}{\big(x - \frac{\pi}{6}\big)}$
View full question & answer→Question 695 Marks
$\text{if y } = \sqrt{\frac{(x - 3)(x^{2} + 4)}{3x^{2} + 4x + 5}}, \text{find} \frac{dy}{dx}$
Answer$y = \sqrt{\frac{(x - 3)(x^{2} + 4)}{3x^{2} + 4x + 5}} \dots\dots\dots \text{(i})$Talking log of both sides of (i), we get
$\text{\log y} = \frac{1}{2} \bigg[\log (x - 3) + \log (x^{2} + 4) - \log (3x^{2} + 4x +5)\bigg]$
$\therefore \frac{1}{y} \frac{dy}{dx} = \frac{1}{2}\bigg[ \frac{1}{x-3} + \frac{2x}{x^{2} + 4} - \frac{6x + 4}{3x^{2} + 4x + 5}\bigg]$
$\therefore \frac{dy}{dx} = \frac{y}{2}\bigg[ \frac{1}{x-3} + \frac{2x}{x^{2} + 4} - \frac{6x + 4}{3x^{2} + 4x + 5}\bigg]$
OR
$\frac{1}{2} \sqrt{\frac{x - 3)(x ^{2} + 4)}{3x^{2} + 4x +5}} \bigg[ \frac{1}{x - 3} + \frac{2x}{x^{2} + 4} - \frac{6x + 4}{3x^{2} + 4x + 5}\bigg]$
View full question & answer→Question 705 Marks
Differentiate sin (2x + 3) w.r.t. x from first principle.
Answer$y = \sin(2x + 3)$$y = \triangle y = \sin (2x +2\triangle x + 3)$
$\therefore \triangle y \sin (2x + 2\triangle x + 3) -\sin (2x + 3)$
$= 2\cos (2x + 3 + \triangle x) \sin \triangle x$
$\therefore \lim\limits_{\triangle x\rightarrow 0} \frac{\triangle y}{\triangle x} = \lim\limits_{\triangle x \rightarrow 0} 2\cos (2x +3 +\triangle x) \lim\limits_{\triangle x\rightarrow 0} \frac {\sin\triangle x}{\triangle x}$
OR
$\frac{dy}{dx} = \text2\cos (2x + 3) 1 = 2\cos (2x + 3)$
View full question & answer→Question 715 Marks
If $\log(\text{x}^2+\text{y}^2)=2\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big),$ show that $=\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+\text{y}}{\text{x}-\text{y}}.$
AnswerWe have, $\log(\text{x}^2+\text{y}^2)=2\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\frac{1}{2}.\frac{\text{d}}{\text{dx}}\log(\text{x}^2+\text{y}^2)=\frac{\text{d}}{\text{dx}}\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\frac{1}{2}\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{y}^2\big)=\frac{1}{1+\Big(\frac{\text{y}}{\text{x}}\Big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\frac{1}{2}\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\Big[2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big]=\frac{\text{x}^2}{(\text{x}^2+\text{y}^2)}\bigg[\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}\bigg]$
$\Rightarrow\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\Big(\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{x}^2}{(\text{x}^2+\text{y}^2)}\bigg[\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}\bigg]$
$\Rightarrow\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\Big(\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{x}^2}{(\text{x}^2+\text{y}^2)}\bigg[\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}(1)}{\text{x}^2}\bigg]$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}\frac{\text{dy}}{\text{dx}}=-\text{y}-\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\text{y}-\text{x})=-(\text{y}+\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-(\text{y}+\text{x})}{\text{y}-\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+\text{y}}{\text{x}-\text{y}}$
View full question & answer→Question 725 Marks
If $\text{x}^\text{y}-\text{y}^\text{x}=\text{a}^\text{b},$ find $=\frac{\text{dy}}{\text{dx}}.$
AnswerThe given function is $\text{x}^\text{y}-\text{y}^x=\text{a}^\text{b}$
Let $\text{x}^\text{y}=\text{u}$ and $\text{y}^\text{x}=\text{v}$
Then, the function becomes $\text{u}-\text{v}=\text{a}^\text{b}$
$\frac{\text{du}}{\text{dx}}-\frac{\text{dv}}{\text{dx}}=0\ \dots(1)$
$\text{u}=\text{x}^\text{y}$
$\Rightarrow\log\text{u}=\log(\text{x}^\text{y})$
$\Rightarrow\log\text{u}=\text{y}\log\text{x}$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{u}}.\frac{\text{du}}{\text{dx}}=\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}.\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}.\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^\text{y}\Big(\log\text{x}\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}\Big)\ \dots(2)$
$\text{v}=\text{y}^\text{x}$
$\Rightarrow\log\text{v}=\log(\text{y}^\text{x})$
$\Rightarrow\log\text{v}=\text{x}\log\text{y}$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}.\frac{\text{dv}}{\text{dx}}=\log\text{y}.\frac{\text{dy}}{\text{dx}}(\text{x})+\text{x}.\frac{\text{d}}{\text{dx}}(\log\text{y})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big(\log\text{y}.1+\text{x}.\frac{1}{\text{y}}.\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{y}^\text{x}\Big(\log\text{y}+\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}\Big)\ \dots(3)$
From (1), (2) and (3), we obtain
$\text{x}^\text{y}\Big(\log\text{x}\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}\Big)-\text{y}^\text{x}\Big(\log\text{y}+\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow\text{x}^\text{y}\log\text{x}\frac{\text{dy}}{\text{dx}}-\text{x}\text{y}^{\text{x}-1}\frac{\text{dy}}{\text{dx}}+\text{x}^{\text{y}-1}\text{y}-\text{y}^\text{x}\log\text{y}=0$
$\Rightarrow(\text{x}^\text{y}\log\text{x}-\text{xy}^{\text{x}-1})\frac{\text{dy}}{\text{dx}}=\text{y}^\text{x}\log\text{y}-\text{x}^{\text{y}-1}\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^\text{x}\log\text{y}-\text{x}^{\text{y}-1}\text{y}}{(\text{x}^\text{y}\log\text{x}-\text{xy}^{\text{x}-1})}$
View full question & answer→Question 735 Marks
Differentiate $\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$ with respect to $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big),$ if $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
Put $\text{x}=\sin\theta$
$\Rightarrow\theta=\sin^{-1}\text{x}$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{\sin\theta}{\sqrt{1-\sin^{2}\theta}}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{\sin\theta}{\cos\theta}\Big)$
$\Rightarrow\text{u}=\tan^{-1}(\tan\theta)\ .....(\text{i})$
And
Let, $\text{v}=\sin^{-1}(2\text{x}\sqrt{1-\text{x}^2})$
$\text{v}=\sin^{-1}(2\sin\theta\sqrt{1-\sin^{2}\theta})$
$\text{v}-\sin^{-1}(2\sin\theta\cos\theta)$
$\text{v}=\sin^{-1}(\sin2\theta)\ .....(\text{ii})$
Here,
$-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i)
$\text{u}=\theta\Big[\text{since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\sin^{-1}\text{x}$
Differentiatiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=2\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{v}=2\sin^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)\Big(\frac{\sqrt{1-\text{x}^2}}{2}\Big)$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{1}{2}$
View full question & answer→Question 745 Marks
Differentiate $\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$ with respect to $\sqrt{1-\text{x}^2},$ if -1 < x < 1.
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
Put $\text{x}=\tan\theta$
$\Rightarrow\theta=\tan^{-1}\text{x}$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{1-\tan\theta}{1+\tan\theta}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}-\theta\Big)\Big]\ .....(\text{i})$
Here,
$-1<\text{x}<1$
$\Rightarrow-1<\tan\theta<1$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
$\Rightarrow\frac{\pi}{4}>-\theta>\frac{\pi}{4}$
$\Rightarrow-\frac{\pi}{4}<-\theta<\frac{\pi}{4}$
$\Rightarrow0<\frac{\pi}{4}-\theta<\frac{\pi}{2}$
So, from equation (i)
$\text{u}=\frac{\pi}{4}-\theta$
$\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\frac{\pi}{4}-\tan^{-1}\text{x}$
$\frac{\text{du}}{\text{dx}}=0-\Big(\frac{1}{1+\text{x}}\Big)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=-\frac{1}{1+\text{x}^2}\ .....(\text{ii})$
And
Let, $\text{v}=\sqrt{1-\text{x}^2}$
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}\times\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}(-2\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-\text{x}}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
Dividing equation (ii) by (iii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=-\frac{1}{1+\text{x}^2}\times\frac{\sqrt{1-\text{x} ^2}}{-\text{x}}$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{\sqrt{1-\text{x}^2}}{\text{x}(1+\text{x}^2)}$
View full question & answer→Question 755 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}|\text{x}-3|,&\text{if }\text{ x}\geq1\\\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4},&\text{if }\text{ x}<1\end{cases}$
AnswerWhen x >, then
f(x) = |x - 3|
Since modulus function is a continuous function, f(x) is continuous for each x > 1
When x < 1, then
$\text{f(x)}=\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4}$
Since, $x^2$ & $3x$ are continuous being polynomial functions, $x^2$ & $3x$ will also be continuous.
Also, $\frac{13}{4}$ is continuous being a polymomial function.
$\Rightarrow\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4}$ is continuous for eqch x < 1
⇒ f(x) is continuous for each x < 1
At x = 1, we have
$(\text{LHL at x}=1)=\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\Big[\frac{(1-\text{h}^2)}{4}-\frac{3(1-\text{h})}{2}+\frac{13}{4}\Big]=\frac{1}{4}-\frac{3}{2}+\frac{13}{4}=2$
$(\text{RHL at x}= 1)=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|1+\text{h}-3|\big]=|-2|=2$
Also, $\text{f}(1)=|1-3|=|-2|=2$
Thus, $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\text{f}=(1)$
Hence, f(x) is continuous at x = 1
Thus, the given function is now where discontinuous.
View full question & answer→Question 765 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\sqrt{\cot\text{x}}}$
AnswerLet, $\text{y}=\text{e}^\sqrt{{\cot\text{x}}}$
$\Rightarrow\ \text{y}=\text{e}^{(\cot\text{x})^\frac{1}{2}}$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\text{e}^{(\cot\text{x})^\frac{1}{2}}\Big)$
$=\text{e}^{(\cot\text{x})^\frac{1}{2}}\frac{\text{d}}{\text{dx}}(\cot\text{x})^\frac{1}{2}$
[Using chain rule]
$=\text{e}^\sqrt{\cot\text{x}}\times\frac{1}{2}(\cot\text{x})^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}(\cot\text{x})$
$=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\text{e}^\sqrt{\cot\text{x}}\Big)=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}$
View full question & answer→Question 775 Marks
Differentiate $\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$ with respect to $\sqrt{1-4\text{x}^2},$ if:
$\text{x}\in\Big(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$
Put $2\text{x}=\cos\theta \text{ So},$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\times\cos\theta\sqrt{1-\cos^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\cos\theta\sin\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let, $\text{v}=\sqrt{1-4\text{x}^2}\ .....(\text{ii})$
Here,
$\text{x}\in\Big(\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\Rightarrow2\text{x}\in\Big(-1,-\frac{1}{\sqrt{2}},\Big)$
$\Rightarrow\theta\in\Big(\frac{3\pi}{4},\pi\Big)$
So, from equation (i),
$\text{u}=\pi-2\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\pi-\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{3\pi}{2}\Big]\Big]$
$\Rightarrow\text{u}=\pi-2\cos^{-1}(2\text{x})\big[\text{Since},2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{du}}{\text{dx}}=0-2\bigg(\frac{-1}{\sqrt{1-(2\text{x})^2}}\bigg)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{-2}{\sqrt{1-4\text{x}^2}}(2)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\ .....(\text{vi})$
From equation (iv)
$\frac{\text{dv}}{\text{dx}}=\frac{4}{\sqrt{1-4\text{x}^2}}$
But, $\text{x}\in\Big(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\frac{\text{dv}}{\text{dx}}=\frac{-4(-\text{x})}{\sqrt{1-4(-\text{x})^2}}$
$\frac{\text{dv}}{\text{dx}}=\frac{4\text{x}}{\sqrt{1-4\text{x}^2}}\ .....(\text{vii})$
Dividing equation (vi) by (vii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\times\frac{\sqrt{1-4\text{x}^2}}{4\text{x}}$
$\frac{\text{du}}{\text{dv}}=-\frac{1}{\text{x}}$
View full question & answer→Question 785 Marks
At what points on the following curves, is the tangent parallel to x-axis?
$y = 12(x + 1)(x - 2)$ on $[-1, 2]$
AnswerLet $f(x) = 12(x + 1)(x - 2) ...(1)$
$\Rightarrow f(x) = 12(x^2 - x - 2)$
$\Rightarrow f(x) = 12x^2 - 12x - 24$
Since f(x) is a polynomial function, f(x) is continuous on [-1, 2] and differentiable on (1, 2).
Also,
f(2) = f(-1) = 0
Thus, all the conditions of Rolle's theorem are satisfied.
Consequently, there exists at least one point $\text{c}\in(-1,2)$ for which f'(c) = 0.
But $\text{f}'(\text{c})=0$
$\Rightarrow24\text{c}-12=0$
$\Rightarrow\text{c}=\frac{1}{2}$
$\therefore\ \text{f}(\text{c})=\text{f}\Big(\frac{1}{2}\Big)=-12\Big(\frac{3}{2}\Big)\Big(\frac{3}{2}\Big)=-27$
By the geometrical interepretetion of Rolle's theorem, $\Big(\frac{1}{2}\Big),-27$ is the point on y = 12(x + 1)(x - 2) where the tangent is parallel to the x-axis.
View full question & answer→Question 795 Marks
If $\text{x}=\cot\text{t and y}=\sin\text{t},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{3}}\text{ at t}=\frac{2\pi}{3}$
AnswerWe have, $\text{x}=\cos\text{t}$ and $\text{y}=\sin\text{t}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\cos\text{t}) $ and $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\sin\text{t}) $
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\sin\text{t}$ and $\frac{\text{dy}}{\text{dt}}=\cos\text{t}$
$\therefore\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\cos\text{t}}{-\sin\text{t}}=-\cot{\text{t}}$
Now, $\big(\frac{\text{dy}}{\text{dx}}\big)_{\text{t}=\frac{2\pi}{3}}=-\cot\big(\frac{2\pi}{3}\big)=\frac{1}{\sqrt{3}}$
View full question & answer→Question 805 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{x}}{1+6\text{x}^3}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{x}}{1+6\text{x}^3}\Big)$
$\Rightarrow\ \text{y}=\tan^{-1}\Big(\frac{3\text{x}-2\text{x}}{1+(3\text{x})(2\text{x})}\Big)$
$\Rightarrow\text{y}=\tan^{-1}3\text{x}-\tan^{-1}2\text{x}$
$\Big[\text{Since}, \tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\Big]$
Differentiate it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+(3\text{x})^2}\frac{\text{d}}{\text{dx}}(3\text{x})-\frac{1}{1+(2\text{x})^3}\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+9\text{x}^2}(3)-\frac{1}{1+4\text{x}^2}(2)$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{3}{1+9\text{x}^2}-\frac{2}{1+4\text{x}^2}$
View full question & answer→Question 815 Marks
If u, v and w are functions of x, then show that
$\frac{\text{d}}{\text{dx}}(\text{u. v. w)}=\frac{\text{du}}{\text{dx}}\text{ v. w}+\text{u }\frac{\text{dv}}{\text{dx}}\text{ w}+\text{u.v}\frac{\text{dw}}{\text{dx}}$
in two ways - first by repeated application of product rule, second by logarithmic
differentiation.
AnswerGiven: u, v and w are functions of x.
To prove: $\frac{\text{d}}{\text{dx}}(\text{u. v. w)}=\frac{\text{du}}{\text{dx}}\text{ y.w}+\text{u }\frac{\text{dv}}{\text{dx}}\text{ w}+\text{u.v}\frac{\text{dw}}{\text{dx}}$
- By repeated application of product rule:
$\text{L.H.S } \frac{\text{d}}{\text{dx}}(\text{u.v.w)}=\frac{\text{d}}{\text{dx}}[(\text{uv).w}]=\text{uv}\frac{\text{d}}{\text{dx}}\text{w}+\text{w}\frac{\text{d}}{\text{dx}}\text{(uv)}$
$=\text{uv}\frac{\text{dw}}{\text{dx}}+\text{w}\Big[\text{u}\frac{\text{d}}{\text{dx}}\text{v}+\text{v}\frac{\text{d}}{\text{dx}}\text{u}\Big]=\text{uv}\frac{\text{dw}}{\text{dx}}+\text{uw}\frac{\text{dv}}{\text{dx}}+\text{vw}\frac{\text{du}}{\text{dx}}$
$=\frac{\text{du}}{\text{dx}}\text{.v.w}+\text{u}\frac{\text{dv}}{\text{dx}}\text{.w}+\text{u.v.}\frac{\text{dw}}{\text{dx}}=\text{R.H.S}\ \ \ \text{Hence proved}.$
- By Logarithmic differentiation:
Let y = uvw $\ \Rightarrow\ \log\text{y}=\log(\text{u.v.w)}$
$\Rightarrow\ \log\text{y}=\log\text{u}+\log\text{v}+\log\text{w}\ \Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{y}=\frac{\text{d}}{\text{dx}}\log\text{u}+\frac{\text{d}}{\text{dx}}\log\text{v}+\frac{\text{d}}{\text{dx}}\log\text{w}$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}+\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}+\frac{1}{\text{w}}\frac{\text{dw}}{\text{dx}}\ \Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}+\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}+\frac{1}{\text{w}}\frac{\text{dw}}{\text{dx}}\Big]$
Putting y = uvw, we get
$\frac{\text{d}}{\text{dx}}(\text{u.v.w)}=\text{uvw}\Big[\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}+\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}+\frac{1}{\text{w}}\frac{\text{dw}}{\text{dx}}\Big]$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\text{(u.v.w)}=\frac{\text{du}}{\text{dx}}\text{.v.w}+\text{u.}\frac{\text{dv}}{\text{dx}}\text{.w}+\text{u.v.}\frac{\text{dw}}{\text{dx}}\ \ \ \text{Hence proved}.$ View full question & answer→Question 825 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\text{x}^4-16}{\text{x}-2},&\text{if }\text{ x}\neq2\\16,&\text{if }\text{ x}=2\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\text{x}^4-16}{\text{x}-2},&\text{if }\text{ x}\neq2\\16,&\text{if }\text{ x}=2\end{cases}$
When $\text{x}\neq2,$ then
$\text{f(x)}=\frac{\text{x}^4-16}{\text{x}-2}=\frac{\text{x}^4-2^4}{\text{x}-2}=\frac{(\text{x}^2-4)(\text{x}-2)(\text{x}+2)}{\text{x}-2}=(\text{x}^2+4)(\text{x}+2)$
We know that a polynomial function is everywhere continuous.
Therefore, the functions $(x^2 + 4)$ and $(x + 2)$ are everywhere continuous.
So, the product function $x^2 + 4x + 2$ is everywhere continuous.
Thus, f(x) is continuous at every $\text{x}\neq2$
At x = 2, we have
$(\text{LHL at x}=2)=\lim_\limits{\text{x}\rightarrow2^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[(2-\text{h})^2+4\big](2-\text{h}+2)=8(4)=32$
$(\text{RHL at x}= 2)=\lim_\limits{\text{x}\rightarrow2^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[(2+\text{h})^2+4\big](2+\text{h}+2)=8(4)=32$
Also, f(2) = 16
$\therefore\ \lim_\limits{\text{x}\rightarrow2^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow2^+}\text{f(x)}\neq\text{f}(2)$
Thus, f(x) is discontinuous at x = 2
Hence, the only point of discontinuity for f(x) is x = 2
View full question & answer→Question 835 Marks
Differentiate the functions given in Exercise:
$\text{x}^\text{x}-2^{\sin\text{x}}$
AnswerLet $\text{y}=\text{x}^\text{x}-2^{\sin\text{x}}$
Putting $\text{u}=\text{x}^\text{x}\text{ and v }=2^{\sin\text{x}}$
$\Rightarrow\ \text{y}=\text{u}-\text{v}\ \Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}-\frac{\text{dv}}{\text{dx}}\ \dots\text{(i)}$
Now, $u = x^x$ $\ \Rightarrow\ \log\text{u}=\log\text{x}^\text{x}=\text{x}\log\text{x}$
$\therefore\ \frac{\text{d}}{\text{dx}}\log\text{u}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$ $\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{1}{\text{x}}+\log\text{x}.1$ $\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=1+\log\text{x}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}(1+\log\text{x})$ $=\text{x}^\text{x}(1+\log\text{x})\ \dots\text{(ii)}$
Again, $\text{v}=2^{\sin\text{x}}\ \Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}2^{\sin\text{x}}$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=2^{\sin \text{x}}\log2\frac{\text{d}}{\text{dx}}\sin\text{x}=\ \Big[\because\frac{\text{d}}{\text{dx}}\text{a}^{\text{f(x)}}=\text{a}^{\text{f(x)}}\log\text{a}\frac{\text{d}}{\text{dx}}\text{f(x)}\Big]$
$\frac{\text{dv}}{\text{dx}}=2^{\sin\text{x}}(\log2).\cos\text{x}=\cos\text{x}.2^{\sin\text{x}}\log2\ \dots\text{(iii)}$
Putting the values from eq. (ii) and (iii) in eq. (i),
$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})-\cos\text{x}.2^{\sin\text{x}}\log2$
View full question & answer→Question 845 Marks
If $\text{y}\log(1+\cos\text{x}),$ prove that $\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^\text{y}}{\text{dx}^2}.\frac{\text{d}\text{y}}{\text{dx}}=0$
Answer$\text{y}\log(1+\cos\text{x}),$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\cos\text{x}}\times-\sin\text{x}=\frac{-\sin\text{x}}{1+\cos\text{x}}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big[\frac{(1+\cos\text{x})\cos\text{x}-\sin(-\sin\text{x})}{(1+\cos\text{x})^2}\Big]$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big[\frac{\cos\text{x}+\cos^2\text{x}+\sin^2\text{x}}{(1+\cos\text{x})^2}\Big]=-\Big[\frac{1+\cos\text{x}}{(1+\cos\text{x})^2}\Big]=\frac{-1}{1+\cos\text{x}}$
Differentiating w.r.t.x
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^3}=-\Big(\frac{+1}{(1+\cos\text{x})^2}\times+\sin\text{x}\Big)\\=-\Big(\frac{-\sin\text{x}}{1+\cos\text{x}}\Big)\times\Big(\frac{-1}{1+\cos\text{x}}\Big)=-\frac{\text{dy}}{\text{dx}}.\frac{\text{d}^2\text{y}}{\text{dx}^2}$
$\Rightarrow\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^2\text{y}}{\text{dx}^2}.\frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 855 Marks
If $\sec\Big(\frac{\text{x}+\text{y}}{\text{x}-\text{y}}\Big)=\text{a}$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
AnswerWe have, $\sec\Big(\frac{\text{x}+\text{y}}{\text{x}-\text{y}}\Big)=\text{a}$
$\Rightarrow\frac{\text{x}+\text{y}}{\text{x}-\text{y}}=\sec^{-1}({\text{a}})$
Differentiate with respect to x, we get,
$\Rightarrow\bigg[\frac{(\text{x}-\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})-(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}-\text{y})}{(\text{x}-\text{y}}\bigg]=0$
$\Rightarrow(\text{x}-\text{y})\Big(1+\frac{\text{d}}{\text{dx}}\Big)-(\text{x}+\text{y})\Big(1-\frac{\text{d}}{\text{dx}}\Big)=0$
$\Rightarrow(\text{x}-\text{y})+(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}-(\text{x}+\text{y})+(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}[\text{x}-\text{y}+\text{x}+\text{y}]=\text{x}+\text{y}-\text{x}+\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{x})=2\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
View full question & answer→Question 865 Marks
Determine if f defined by:
$\text{f(x)}=\begin{cases}\text{x}^{2} \sin\frac{1}{\text{x}}, \text{if} \ \text{x}\neq0\\0, \ \ \ \ \ \ \ \ \ \ \ \text{if}\ \text{x} = 0\end{cases}$
AnswerIt is given that $\text{f(x)}=\begin{cases}\text{x}^{2} \sin\frac{1}{\text{x}}, \text{if} \ \text{x}\neq0\\0, \ \ \ \ \ \ \ \ \ \ \ \text{if}\ \text{x} = 0\end{cases}$
We know that f is defined at all point of the real line.
Let k be a real number.
Case I: $\text{k} \neq 0,$
Then $\text{f(k)} =\text{k}^{2} \sin\frac{1}{\text{k}}$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\Big(\text{x}^{2}\sin\frac{1}{\text{x}}\Big) = \text{k}^{2}\sin \frac{\text{1}}{\text{k}}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$
Thus, f is continuous at all points x that is $\text{x}\neq0.$
Case II: k = 0
Then f(k) = f(0) = 0
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\Big(\text{x}^{2}\sin\frac{1}{\text{x}}\Big) =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\Big( \text{x}^{2}\sin \frac{\text{1}}{\text{x}}\Big)$
We know that $-1 \leq\sin\frac{1}{\text {x}}\leq1, \text{x}\neq0$
$\rightarrow \text{x}^{2}\leq \text{x}^{2}\sin\frac{1}{\text {x}}\leq0$
$\Rightarrow^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}}\Big(\text{x}^{2}\sin\frac{1}{\text{x}}\Big) = 0$
$\Rightarrow^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = 0$
similarly,
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\Big(\text{x}^{2}\sin\frac{1}{\text{x}}\Big) =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}}\Big( \text{x}^{2}\sin \frac{\text{1}}{\text{x}}\Big) = 0$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = \text{f(0)}= ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)}$
Therefore , f is continuous at x = 0.
Therefore, f is has no point of discontinuity.
View full question & answer→Question 875 Marks
If $\text{y}=\sqrt{\text{x}^2+\text{a}^2},$ prvoe that $\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}=0$
AnswerHere, $\text{y}=\sqrt{\text{x}^2+\text{a}^2}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+\text{a}^2}\big)$
$=\frac{1}{2\sqrt{\text{x}^2+\text{a}^2}}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{a}^2\big)$
[Using chain rule]
$=\frac{1}{2\sqrt{\text{x}^2+\text{a}^2}}\times(2\text{x})$
$=\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\ \Big[\text{Since},\sqrt{\text{x}^2+\text{a}^2}=\text{y}\Big]$
$\Rightarrow \text{y}\frac{\text{dy}}{\text{dx}}=\text{x}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}=0$
View full question & answer→Question 885 Marks
If $\text{y}=\text{x}\sin(\text{a}+\text{y}),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
AnswerHere,
$\text{y}=\text{x}\sin(\text{a}+\text{y})$
Differentiating with respect to x using the chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\sin(\text{a}+\text{y})+\sin(\text{a}+\text{y})\frac{\text{dx}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}+\sin(\text{a}+\text{y})$
$(1-\text{x}\cos(\text{a}+\text{y}))\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{(1-\text{x}\cos(\text{a}+\text{y}))}$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{\Big(1-\frac{\text{y}}{\sin(\text{a}+\text{y})}\cos(\text{a}+\text{y})\Big)}\ \Big[\text{Since}\frac{\text{y}}{\sin(\text{a}+\text{y})}=\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
View full question & answer→Question 895 Marks
If $\text{y}=\text{e}^{\text{x}}+\text{e}^{-\text{x}},$ prvoe that $\frac{\text{dy}}{\text{dx}}=\sqrt{\text{y}^2-4}$
AnswerDifferentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)$
$=\frac{\text{d}}{\text{dx}}\text{e}^{\text{x}}+\frac{\text{d}}{\text{dx}}{\text{e}}^{-\text{x}}$
$=\text{e}^{\text{x}}+\text{e}^{-\text{x}}\frac{\text{d}}{\text{dx}}\big(-\text{x}\big)$
[Using chain rule]
$=\text{e}^{\text{x}}+\text{e}^{-\text{x}}(-1)$
$=\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)$
$=\sqrt{\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)^2-4\text{e}^{\text{x}}\times\text{e}^{-\text{x}}}$
$\Big[\text{Since},(\text{a}-\text{b}=\sqrt{(\text{a}+\text{b})^2-4\text{ab}}\Big]$
$=\sqrt{\text{y}^2-4}$
$\big[\text{Since e}^\text{x}+\text{e}^{-\text{x}}=\text{y}\big]$
Hence, the solution is, $\frac{\text{dy}}{\text{dx}}=\sqrt{\text{y}^2-4}$
View full question & answer→Question 905 Marks
Prove that the function $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$ 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 $\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$ 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,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(-\text{h})}{-\text{h}}\Big)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(\text{h})}{\text{h}}\Big)=1$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}(\text{h}+1)=1$
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 everywhere continuous.
View full question & answer→Question 915 Marks
If $\text{x}=\sin\text{t}$ and $\text{y}=\sin\text{pt},$ prove that $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}+\text{p}^2\text{ y}=0.$
AnswerWe have, $\text{x}=\sin\text{t}$ and $\text{y}=\sin\text{pt}$
$\therefore\ \frac{\text{dx}}{\text{dt}}=\cos\text{t}$ and $\frac{\text{dy}}{\text{dt}}=\text{p}\cos\text{pt}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{p}\cdot\cos\text{pt}}{\cos\text{t}}$
$\Rightarrow\ \text{y}'=\frac{\text{p}\cdot\cos\text{pt}}{\cos\text{t}}$
$\Rightarrow\ \text{y}'^2\cos^2\text{t}=\text{p}^2\cos^2\text{pt}$
$\Rightarrow\ \text{y}'^{2}(1-\sin^2\text{t})=\text{p}^2(1-\sin^2\text{pt})$
$\Rightarrow\ \text{y}'^2(1-\text{x}^2)=\text{p}^2(1-\text{y}^2)$
Differentiating above w.r.t. x, we get
$2\text{y}'\text{y}''(1-\text{x}^2)-2\text{xy}'^2=\text{p}^2(-2\text{yy}')$
$\Rightarrow\ \text{y}''(1-\text{x}^2)-\text{xy}'=-\text{p}^2\text{y}$
$\Rightarrow\ \text{y}''(1-\text{x}^2)-\text{xy}'+\text{p}^2\text{y}=0$
View full question & answer→Question 925 Marks
If $\text{y}=(\sin\text{x})^{(\sin\text{x})^{(\sin\text{x})^{....\infty}}},$ prove that $\frac{\text{y}^2\cot\text{x}}{(1-\text{y}\log\sin\text{x})}$
AnswerHere,
$\text{y}=(\sin\text{x})^{(\sin\text{x})^{(\sin\text{x})^{....\infty}}}$
$\Rightarrow\text{y}=(\sin\text{x})^\text{y}$
Taking log on both sides,
$\log\text{y}=\log(\sin\text{x})^{\text{y}}$
$\log\text{y}=\text{y}(\log\sin\text{x})$
Differentiating it with respect to x, using product rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\sin\text{x}\Big)=\frac{\text{y}}{\sin\text{x}}(\cot\text{x})$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1-\text{y}\log\sin\text{x}}{\text{y}}\Big)=\text{y}\cot\text{x}$
$\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{y}^2\cot\text{x}}{1-\text{y}\log\sin\text{x}}\Big)$
View full question & answer→Question 935 Marks
$\text{If (x}-\text{a})^2+(\text{y}-\text{b})^2=\text{c}^2,$ for some c > 0 , prove that
$\frac{\Big[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]^{\frac{3}{2}}}{\frac{\text{d}^2\text{y}}{\text{dx}^2}}$
is a constant independent of a and b.
AnswerIt is given that, $\text{(x}-\text{a})^2+(\text{y}-\text{b})^2=\text{c}^2$
Differentiating both sides with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}[(\text{x}-\text{a})^2]+\frac{\text{d}}{\text{dx}}[(\text{y}-\text{b})^2]=\frac{\text{d}}{\text{dx}}(\text{c}^2)$
$\Rightarrow\ 2(\text{x}-\text{a}).\frac{\text{d}}{\text{dx}}(\text{x}-\text{a})+2(\text{y}-\text{b}).\frac{\text{d}}{\text{dx}}(\text{y}-\text{b})=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-(\text{x}-\text{a})}{\text{y}-\text{b}}\ \dots(1)$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big[\frac{-(\text{x}-\text{a})}{\text{y}-\text{b}}\Big]$
$=-\Bigg[\frac{(\text{y}-\text{b}).\frac{\text{d}}{\text{dx}}(\text{x}-\text{a})-(\text{x}-\text{a}).\frac{\text{d}}{\text{dx}}(\text{y}-\text{b})}{(\text{y}-\text{b})^2}\Bigg]$
$=-\Bigg[\frac{(\text{y}-\text{b})-(\text{x}-\text{a}).\frac{\text{dy}}{\text{dx}}}{(\text{y}-\text{b})^2}\Bigg]$
$=-\Bigg[\frac{(\text{y}-\text{b})-(\text{x}-\text{a}).\Big\{\frac{-(\text{x}-\text{a})}{\text{y}-\text{b}}\Big\}}{(\text{y}-\text{b})^2}\Bigg]\ \ [\text{using (1)}]$
$=-\Big[\frac{(\text{y}-\text{b})^2+(\text{x}-\text{a})^2}{(\text{y}-\text{b})^3}\Big]$
$\therefore\ \Bigg[\frac{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}{\frac{\text{d}^2\text{y}}{\text{dx}^2}}\Bigg]^{\frac{3}{2}}$ $=\frac{\Bigg[1+\frac{(\text{x}-\text{a})^2}{(\text{y}-\text{b})^2}\Bigg]^{\frac{3}{2}}}{-\Bigg[\frac{(\text{y}-\text{b})^2+(\text{x}-\text{a})^2}{(\text{y}-\text{b})^3}\Bigg]}$ $=\frac{\Bigg[\frac{(\text{y})-\text{b})^2+(\text{x}-\text{a})^2}{(\text{y}-\text{b})^2}\Bigg]^{\frac{3}{2}}}{-\Bigg[\frac{(\text{y}-\text{b})^2+(\text{x}-\text{a})^2}{(\text{y}-\text{b})^3}\Bigg]}$
$=\frac{\Bigg[\frac{\text{c}^2}{(\text{y}-\text{b})^2}\Bigg]^{\frac{3}{2}}}{-\frac{\text{c}^2}{(\text{y}-\text{b})^3}}=\frac{\frac{\text{c}^3}{(\text{y}-\text{b})^3}}{-\frac{\text{c}^2}{(\text{y}-\text{b})^3}}$
= -c, which is constant and is independent of a and b
Hence, proved.
View full question & answer→Question 945 Marks
Find all point of discontinuity of the function $\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2},$ where $\text{t}=\frac{1}{\text{x}-1}$
Answer$\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2},$ where $\text{t}=\frac{1}{\text{x}-1}$
Clearly $\text{t}=\frac{1}{\text{x}-1}$ is discontinuous at x = 1
For $\text{x}\neq1,$ we get
$\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2}=\frac{1}{(\text{t}-2)(\text{t}-1)}$
This is discontinuous at t = -2 and t = 1
For t = -2, $\text{t}=\frac{1}{\text{x}-1}$
$\Rightarrow\text{x}=\frac{1}{2}$
For t = 1, $\text{t}=\frac{1}{\text{x}-1}$
⇒ x = 2
Hence, f is discontinuous at $\text{x}=\frac{1}{2},$ x = 1 and x = 2
View full question & answer→Question 955 Marks
Differentiate the following functions with respect to x:
$(\tan\text{x})^\frac{1}{\text{x}}$
AnswerLet $\text{y}=(\tan\text{x})^\frac{1}{\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\tan\text{x})^\frac{1}{\text{x}}$
$\log\text{y}=\frac{1}{\text{x}}\log(\tan\text{x})\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule and chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}\log(\tan\text{x})+\log(\tan\text{x})\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}}\Big)$
$=\frac{1}{\text{x}}\times\frac{1}{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log(\tan\text{x})\Big(-\frac{1}{\text{x}^2}\Big)$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}\tan\text{x}}(\sec^2\text{x})-\frac{\log(\tan\text{x})}{\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\sec^2\text{x}}{\text{x}\tan\text{x}}-\frac{\log(\tan\text{x})}{\text{x}^2}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^\frac{1}{\text{x}}\Big[\frac{\sec^2\text{x}}{\text{x}\tan\text{x}}-\frac{\log(\tan\text{x})}{\text{x}^2}\Big]$
[Using equation (i)]
View full question & answer→Question 965 Marks
Differentiate the following functions with respect to x:
$\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
AnswerLet $\text{y}=\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dy}}\Big(\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Big)$
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
[Using chain rule]
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Bigg[\frac{(1-\text{x}^2)\frac{\text{d}}{\text{dx}}(1+\text{x}^2)-(1+\text{x}^2)\frac{\text{d}}{\text{dx}}(1-\text{x}^2)}{(1-\text{x})^2}\Bigg]$
[Using chain rule]
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\bigg[\frac{(1-\text{x}^2)(2\text{x})-(1+\text{x}^2)(-2\text{x})}{(1-\text{x}^2)^2}\bigg]$
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Big[\frac{2\text{x}-2\text{x}^3+2\text{x}+2\text{x}^3}{(1-\text{x}^2)^2}\Big]$
$=\frac{4\text{x}}{\big(1-\text{x}^2\big)^2}\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
So,
$\frac{\text{d}}{\text{dx}}\Big(\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Big)=\frac{4\text{x}}{\big(1-\text{x}^2\big)^2}\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
View full question & answer→Question 975 Marks
If $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{a}(\text{x}-\text{y}),$ prove that $\frac{\text{dy}}{\text{dx}}=\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}}.$
AnswerWe have, $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{a}(\text{x}-\text{y})$
On putting $\text{x}=\sin\alpha$ and $\text{y}=\sin\beta,$ we get
$\sqrt{1-\sin^2\alpha}+\sqrt{1-\sin\beta}=\text{a}(\sin\alpha-\sin\beta)$
$\Rightarrow\ \cos\alpha+\cos\beta=\text{a}(\sin\alpha-\sin\beta)$
$\Rightarrow\ 2\cos\frac{\alpha+\beta}{2}\cdot\cos\frac{\alpha-\beta}{2}=2\text{a}\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$
$\Rightarrow\ \cos\frac{\alpha-\beta}{2}=\text{a}\sin\frac{\alpha-\beta}{2}$
$\Rightarrow\ \cot\frac{\alpha-\beta}{2}=\text{a}$
$\Rightarrow\ \frac{\alpha-\beta}{2}=\cot^{-1}\text{a}$
$\Rightarrow\ \alpha-\beta=2\cot^{-1}\text{a}$
$\Rightarrow\ \sin^{-1}\text{x}-\sin^{-1}\text{y}=2\cot^{-1}\text{a}$ $[\because\text{x}=\sin\alpha\text{ and y}=\sin\beta]$
Differentiating both sides w.r.t. x, we get
$\frac{1}{\sqrt{1-\text{x}^2}}-\frac{1}{\sqrt{1-\text{y}^2}}\frac{\text{dy}}{\text{dx}}=0$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\sqrt{1-\text{y}^2}}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 985 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}^2}}\Big)$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}^2}}\Big)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}^2}}\Big)\Big\}$
$=\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)$
[Using chain rule and quotient rule]
$=\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)^2}}\times\Bigg[\frac{(\text{x}^2+\text{a}^2)^\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x})-\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{a}^2)^\frac{1}{2}}{\Big[(\text{x}^2+\text{a}^2)^\frac{1}{2}\Big]^2}\Bigg]$
$=\frac{\sqrt{\text{x}^2+\text{a}^2}}{\sqrt{\text{x}^2+\text{a}^2-\text{x}^2}}\times\Bigg[\frac{\sqrt{\text{x}^2+\text{a}^2}-\frac{\text{x}}{2\sqrt{\text{x}^2+\text{a}^2}}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{a}^2)}{(\text{x}^2+\text{a}^2)}\Bigg]$
$=\frac{\sqrt{\text{x}^2+\text{a}^2}}{\text{a}(\text{x}^2+\text{a}^2)}\times\Big[\sqrt{\text{x}^2+\text{a}^2}-\frac{\text{x}}{2\sqrt{\text{x}^2+\text{a}^2}}\times2\text{x}\Big]$
$=\frac{\sqrt{\text{x}^2+\text{a}^2}}{\text{a}(\text{x}^2+\text{a}^2)}\times\Big[\frac{\text{x}^2+\text{a}^2-\text{x}^2}{\sqrt{\text{x}^2+\text{a}^2}}\Big]$
$=\frac{\text{a}^2}{\text{a}(\text{x}^2+\text{a}^2)}$
$=\frac{\text{a}}{(\text{x}^2+\text{a}^2)}$
So,
$\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)\Big\}=\frac{\text{a}^2}{\text{a}(\text{x}^2+\text{a}^2)}$
View full question & answer→Question 995 Marks
If $(\text{x}-\text{y})\text{e}^{\frac{\text{x}}{\text{x}-\text{y}}}=\text{a},$ prove that $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=2\text{y}$
AnswerConsider the given function, $(\text{x}-\text{y})\text{e}^{\frac{\text{x}}{\text{x}-\text{y}}}=\text{a}.$
We need to prove that $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=2\text{y}.$
Differentiating the given equation w.r.t 'x' we get
$(\text{x}-\text{y})\Bigg[\text{e}^{\frac{\text{x}}{\text{x}-\text{y}}}\Bigg(\frac{(\text{x}-\text{y})-\text{x}\big(1-\frac{\text{dy}}{\text{dx}}\big)}{(\text{x}-\text{y})^2}\Bigg)\Bigg]+\text{e}^\frac{\text{x}}{\text{x}-\text{y}}\Big(1-\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow\frac{(\text{x}-\text{y})-\text{x}\Big(1-\frac{\text{dy}}{\text{dx}}\Big)}{(\text{x}-\text{y})}+\Big(1-\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big(1-\frac{\text{x}}{\text{x}-\text{y}}\Big)+1=0$
$\Rightarrow\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big(\frac{-\text{y}}{\text{x}-\text{y}}\Big)+1=0$
$\Rightarrow-\text{y}+\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}-\text{y}=0$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=2\text{y}$
View full question & answer→Question 1005 Marks
Show that $\text{f}\text{(x)}=\begin{cases}\frac{\sin 3\text{x}}{\tan2\text{x}},&\text{if } \text{x}<0\\\frac{3}{2},&\text{if }\text{x} = 0\\\frac{\log(1+3\text{x})}{\text{e}^{2\text{x}}},&\text{if}\text{ x}>0\end{cases}$ is discontinuous at x = 0.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{\sin 3\text{x}}{\tan2\text{x}},&\text{if } \text{x}<0\\\frac{3}{2},&\text{if }\text{x} = 0\\\frac{\log(1+3\text{x})}{\text{e}^{2\text{x}}},&\text{if}\text{ x}>0\end{cases}$
We observe
$\text{(LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\\=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sin3(-\text{h})}{\tan2(-\text{h})}\Big)=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sin3\text{h}}{\tan2\text{h}}\Big)$
$=\lim\limits_{\text{h} \rightarrow 0}\Bigg(\frac{\frac{3\sin3\text{h}}{3\text{h}}}{\frac{2\tan2\text{h}}{2\text{h}}}\Bigg)$
$=\frac{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{3\sin3\text{h}}{3\text{h}}\Big)}{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{2\tan2\text{h}}{2\text{h}}\Big)}$
$=\frac{3\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sin3\text{h}}{3\text{h}}\Big)}{2\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\tan2\text{h}}{2\text{h}}\Big)}$
$=\frac{3\times1}{2\times1}=\frac{3}{2}$
$\text{(RHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}\\=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\log(1+3\text{h})}{\text{e}^{2\text{h}}-1}\Big)$
$=\lim\limits_{\text{h} \rightarrow 0}\Bigg(\frac{3\text{h}\frac{\log(1+3\text{h})}{3\text{h}}}{2\text{h}\frac{(\text{e}^{2\text{h}-1})}{2\text{h}}}\Bigg)$
$\frac{3}{2}\lim\limits_{\text{h} \rightarrow 0}\Bigg(\frac{\frac{\log(1+3\text{h})}{3\text{h}}}{\frac{(\text{e}^{2\text{h}}-1)}{2\text{h}}}\Bigg)$
$=\frac{3}{2}\frac{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\log(1+3\text{h})}{3\text{h}}\Big)}{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{(\text{e}^{2\text{h}-1})}{2\text{h}}\Big)}$
$=\frac{3\times1}{2\times1}=\frac{3}{2}$
And, $\text{f}(0)=\frac{3}{2}$
$\therefore\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0^+}\text{f}\text{(x)}=\text{f}(0)$
Thus, f(x) is continuous at x = 0.
View full question & answer→