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
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 545 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 555 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 565 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 575 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 585 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}}$$\text{x} \frac{\text{dy}}{\text{dx}} = -3\sin (\log\text{x}) + 4\cos(\log\text{x})$
$\Rightarrow \text{x} \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = - \frac{3\cos(\log\text{x})}{\text{x}} -\frac{4\sin(\log\text{x})}{\text{x}}$
$\Rightarrow \text{x} \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} + \text{x} \frac{\text{dy}}{\text{dx}} = -[3\cos (\log\text{x}) + 4\sin (\log{\text{x}})] = \text{-y}$
$\text{or x}^{2} \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} +\text{y = o} $
View full question & answer→Question 595 Marks
Differentiate the following with respect to x: $\tan^{-1} \bigg(\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x + \sqrt{ 1 - x}}}\bigg)$
Answer$\text{Let x} = \cos 2 \theta , \sqrt{1 + x} = \sqrt{2} \cos\theta, \sqrt{1 - x} = \sqrt{2} \sin\theta$$\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 605 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 615 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)}$$f(1) = k \dots\dots\dots\dots\dots\text{(ii)}$
$\lim\limits_{ x \rightarrow 1^{-}} = 4 \dots\dots\dots\dots\text{(iii)}$
$\text{As f (x) is continuous at x} = 1 , \therefore k = 4$
View full question & answer→Question 625 Marks
$\text{Evaluate} \lim\limits_{x \rightarrow\frac{\pi}{4}} \bigg( \frac{ \sin x - \cos x}{x- \frac{\pi}{4}} \bigg)$
Answer$\lim\limits_{ x \rightarrow \frac{\pi}{4}} \Bigg[\frac {\sin x - \cos x} { x - \frac{\pi}{4}}\Bigg] = \lim\limits_{ x \rightarrow \frac{\pi}{4}} \sqrt{2} \Bigg[ \frac{\sin x.{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}.\cos x}}{ x- \frac{\pi}{4}}\Bigg]$$ \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 635 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)}$$\text{(i) the function is continuous on [1,4] }$
$\text{(ii) It is differentiable on (1,4)}$
$\text{(iii) f (1)} = \text{f (4)} = 0$
$\therefore$ All the conditions of Rolles' Theoremare satisfied.
$\therefore \text{f' (c)} = 0 \Rightarrow \text{2 c - 5} = 0 \Rightarrow \text{c} = \frac{5}{2}$
$\text{As c} = \frac{5}{2} \in (1,4),$ the Rolle's theorem is verified.
View full question & answer→Question 645 Marks
Differentiate $\sin(x^{2} + 1)$with respect to $ x$from first principle.
Answer$\frac{\triangle y}{\triangle x} = \frac {\sin \bigg[( {x + \triangle x)^{2}}+ 1\bigg] -\sin ( x^{2} + 1)}{\triangle x} $$\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 655 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 665 Marks
Differentiate sin (2x + 3) w.r.t. x from first principle.
Answer$y = \sin(2x + 3)$$y = \triangle y = \sin (2x +2\triangle x + 3)$
$\therefore \triangle y \sin (2x + 2\triangle x + 3) -\sin (2x + 3)$
$= 2\cos (2x + 3 + \triangle x) \sin \triangle x$
$\therefore \lim\limits_{\triangle x\rightarrow 0} \frac{\triangle y}{\triangle x} = \lim\limits_{\triangle x \rightarrow 0} 2\cos (2x +3 +\triangle x) \lim\limits_{\triangle x\rightarrow 0} \frac {\sin\triangle x}{\triangle x}$
OR
$\frac{dy}{dx} = \text2\cos (2x + 3) 1 = 2\cos (2x + 3)$
View full question & answer→Question 675 Marks
Verify Rolle's Theorem for the following function:$f (x) = (x - 1) (x - 2)^{2}, [1,2]$
Answer$f (x) = (x - 1) (x - 2)^{2}, [1,2]$The function f(x) is differentiable on [1, 2] and so it is continuous on [1, 2]
$\therefore$ Al conditions of Rolles Theorem are satisfied
$\therefore f' (x) = (x - 1) 2 (x -2) + (x - 2)^{2}$
$= (x - 2)[2x - 2 + x- 2] = (x - 2) (3x - 4)$
$\therefore f' (c) = 0 \Rightarrow c = 2 \text{ and}\ {c}\frac{4}{3}$
$As 1<\frac{4}{3}<2, \text{the Roll's theorem is verfied}$
View full question & answer→Question 685 Marks
Evaluate:$\lim\limits_{X\rightarrow \frac{\pi}{6}} \bigg[ \frac{\sqrt{3}\sin x - \cos x}{x- \frac{\pi}{6}}\bigg]$
Answer$\lim\limits_{X\rightarrow \frac{\pi}{6}} \Bigg[ \frac{\sqrt{3}\sin x - \cos x}{x- \frac{\pi}{6}}\Bigg] \lim\limits_{X\rightarrow \frac{\pi}{6}} \Bigg[2\frac{\Big( \frac{\sqrt{3}}{2}.\sin x-\frac{1}{2}\cos x\Big)}{x-\frac{\pi}{6}}\Bigg]$$\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
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 715 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 725 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 735 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 745 Marks
$\text{If y}=3\cos(\log \text{x})+4\sin(\log\text{x}),\text{ show that }\text{x}^2\text{y}_2+\text{xy}_1+\text{y}=0$
AnswerGiven: $\text{y}=3\cos(\log\text{x})+4\sin(\log\text{x})\ \dots\text{(i)}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{y}_1=-3\sin(\log\text{x})\frac{\text{d}}{\text{dx}}\log\text{x}+4\cos(\log\text{x})\frac{\text{d}}{\text{dx}}\log\text{x}$
$\Rightarrow\ \text{y}_1=-3\sin(\log\text{x})\frac{1}{\text{x}}+4\cos(\log\text{x})\frac{1}{\text{x}}$ $=\frac{1}{\text{x}}[-3\sin(\log\text{x})+4\cos(\log\text{x})]$
$\Rightarrow\ \text{xy}_1=-3\sin(\log\text{x})+4\cos(\log\text{x})$
Now $\frac{\text{d}}{\text{dx}}(\text{xy}_1)=-3\cos(\log\text{x})\frac{\text{d}}{\text{dx}}\log\text{x}-4\sin(\log\text{x})\frac{\text{d}}{\text{dx}}\log\text{x}$
$\Rightarrow\ \text{x}\frac{\text{d}}{\text{dx}}(\text{y}_1)+\text{y}_1\frac{\text{d}}{\text{dx}}\text{x}$ $=-3\cos(\log\text{x})\frac{1}{\text{x}}-4\sin(\log\text{x})\frac{1}{\text{x}}$
$\Rightarrow\ \text{xy}_2+\text{y}_1=-\frac{[3\cos(\log\text{x})+4\sin(\log\text{x})]}{\text{x}}$
$\Rightarrow\ \text{x}(\text{xy}_2+\text{y}_1)=-[3\cos(\log\text{x})+4\sin(\log\text{x})]$
$\Rightarrow\ \text{x}(\text{xy}_2+\text{y}_1)=-\text{y}\ \ [\text{From eq.(i)}]$
$\Rightarrow\ \text{x}^2\text{y}_2+\text{xy}_1+\text{y}=0 \ \text{ Hence proved}.$
View full question & answer→Question 755 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big\},-\frac{3\pi}{4}<\text{x}<\frac{\pi}{4}$
AnswerLet $\text{y}=\sin^{-1}\Big\{\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big\}$
$=\sin^{-1}\bigg\{\sin\text{x}\Big(\frac{1}{\sqrt{2}}\Big)+\cos\text{x}\Big(\frac{1}{\sqrt{2}}\Big)\bigg\}$
$=\sin^{-1}\Big\{\sin{\text{x}}\cos\frac{\pi}{4}+\cos\text{x}\times\sin\frac{\pi}{4}\Big\}$
$\text{y}=\sin^{-1}\Big\{\sin\Big(\text{x}+\frac{\pi}{4}\Big)\Big\}$
Here, $\frac{-3\pi}{4}<\text{x}<\frac{\pi}{4}$
$\Rightarrow\Big(\frac{-3\pi}{4}+\frac{\pi}{4}\Big)$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta, \text{ if }\theta\in\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=1+0$
$\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 765 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\tan^{-1}\text{x}}$
AnswerLet $\text{y}=\text{x}^{\tan^{-1}\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\tan^{-1}\text{x}}$
$\log\text{y}=\tan^{-1}\text{x}\log\text{x}\ \big[\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}\Big(\frac{1}{1+\text{x}^2}\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\tan^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{1+\text{x}^2}\Big]$
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\tan^{-1}\text{x}}\Big[\frac{\tan^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{1+\text{x}^2}\Big]$
[Using equation (i)]
View full question & answer→Question 775 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{a}+\text{bx}}{\text{b}-\text{ax}}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{a}+\text{bx}}{\text{b}-\text{ax}}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}+\text{bx}}{\text{b}}}{\frac{\text{b}-\text{ax}}{\text{b}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}}{\text{b}}+\frac{\text{bx}}{\text{b}}}{\frac{\text{b}}{\text{a}}-\frac{\text{ax}}{\text{b}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}}{\text{b}}+\text{x}}{1-\big(\frac{\text{a}}{\text{b}}\big)\text{x}}\bigg)$
$\text{y}=\tan^{-1}\big(\frac{\text{a}}{\text{b}}\big)+\tan^{-1}\text{x}$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+\frac{1}{1+\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$
View full question & answer→Question 785 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point $'c\ '$ in the indicated interval as stated by the Lagrange's mean value theorem. $f(x) = x^2 - 2x + 4$ on $[1, 5]$
AnswerWe have,$f(x) = x^2 - 2x + 4$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, $f(x)$ is continuous on $[1, 5]$ and differentiable on $(1, 5).$
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $\text{c}\in(1,5)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(5)-\text{f}(-1)}{5-1}=\frac{\text{f}(5)-\text{f}(-1)}{4}$
Now,$ f(x) = x^2 - 2x + 4$
$\Rightarrow f'(x) = 2x - 2$
$\Rightarrow f(5) = 25 - 10 + 4 = 19$
$\Rightarrow f(1) = 1 - 2 + 4 = 3$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(5)-\text{f}(-1)}{4}$
$\Rightarrow2\text{x}-2=\frac{19-3}{4}$
$\Rightarrow2\text{x}-2-4=0$
$\Rightarrow\text{x}=\frac{6}{2}=3$
Thus, $\text{c}=3\in(1,5)$ such that $\text{f}'(\text{c})=\frac{\text{f}(5)-\text{f}(-1)}{5-1}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 795 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}\text{ on }[1,3]$
AnswerHere,$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}\text{ on }[1,3]$
f(x) attains a unique value for each $\text{x}\in[1,3],$ so it is continuous
$\text{f}'(\text{x})=1-\frac{1}{\text{x}^2}$ is defined for each $\text{x}\in(1,3)$
⇒ f(x) is differentiable in (1,3), so Lagrange's mean value theorem is applicable, so there exist a point $\text{c}\in(1,3)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
$\Rightarrow1-\frac{1}{\text{c}^2}=\frac{\Big(3+\frac{1}{3}-(1+1)\Big)}{2}$
$\Rightarrow1-\frac{1}{\text{c}^2}=\frac{\frac{10}{3}-2}{2}$
$\Rightarrow1-\frac{1}{\text{c}^2}=\frac{4}{3\times2}$
$\Rightarrow1-\frac{2}{3}=\frac{1}{\text{c}^2}$
$\Rightarrow\text{c}^2=3$
$\text{c}=\sqrt{3}\in(1,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
So, Lagrange's mean value theorem is verified.
View full question & answer→Question 805 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\log(\text{x}^2+2)-\log3\text{ on }[-1,1]$
AnswerThe given function is $\text{f}(\text{x})=\log(\text{x}^2+2)-\log3,$ which can be rewritten as$\text{f}(\text{x})=\log\Big(\frac{\text{x}^2+2}{3}\Big)$
We know that, logarithmic function is differentiable and so continuous in its domain $\text{f}(\text{x})=\log\Big(\frac{\text{x}^2+2}{3}\Big)$ is continuous is $[-1,1]$ and differentiable is $(-1,1).$
Also,
f(1) = f(-1) = 0
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have show that there must exists $\text{c}\in(-1,1)$ such that f'(c) = 0.
We have,
$\text{f}(\text{x})=\log\Big(\frac{\text{x}^2+2}{3}\Big)$
$\text{f}'(\text{x})=\frac{3(2\text{x})}{\text{x}^2+2}=\frac{6\text{x}}{\text{x}^2+2}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow\frac{6\text{x}}{\text{x}^2+2}=0$
$\Rightarrow\text{x}=0$
Thus, $\text{c}=0\in(-1,1)$ such that f'(c) = 0.
Hence, Rolle's theorem is verified.
View full question & answer→Question 815 Marks
Find all points of discontinuity of f, where f is defined by:
$\text f(\text x)=\begin{cases}2\text{x}+3, \text {if}\ \ \text x\leq2 \\2\text{x} - 3, \text {if}\ \ \text x > 2\end{cases}$
AnswerHere $\text{f(x)} = \begin{cases}2\text{x} +3, \text{if}\ \text{x}\leq2\\ 2\text{x}-3, \ \text{if}\ \text{x} > 2\end{cases}$Function f is defined for all points of thr real line.
Let c be any real number.
Three cases arise:
Case I: c < 2
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}(2\text{x}+ 3) = 2\text{c} + 3$
Also f(c) = 2c + 3
$\therefore$ f is continuous at all points x < 2.
Case II: c > 2
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}(2\text{x}- 3) = 2\text{c} - 3$
Also f(c) = 2c - 3
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)}= \text{f(c)}$
$\therefore$ f is continuous at all points x >2.
Case III: c = 2
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}{-}}(2\text{x}+ 3) = 4 + 3 = 7$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}{=}}(2\text{x}- 3) = 4 - 3 = 1$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}^{-}}\text{f(x)}\neq \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}^{+}}\text{f(x)}$
$\therefore$ f is not continuous at x = 2
$\therefore$ x = 2 is the only point of discontinuity of f.
View full question & answer→Question 825 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point $'c\ '$ in the indicated interval as stated by the Lagrange's mean value theorem. $f(x) = 2x - x^{2}$ on $[0, 1]$
AnswerWe have$,f(x) = 2x - x^{2 }$
Since a polynomial function is everywhere continuous and differentiable.
Therefore$, f(x)$ is continuous on $0, 1$ and differentiable on $0,1$
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $\text{c}\in0,1$ such that
$\text{f}\ '(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}=\frac{\text{f}(1)-\text{f}(0)}{1}$
Now,
$f(x) = 2x - x^{2 }$
$\Rightarrow f'(x) = 2 - 2x,$
$\Rightarrow f(1) = 2 - 1$
$\Rightarrow f(1) = 1,$
$\Rightarrow f(0) = 0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow2-2\text{x}=\frac{1-0}{1}$
$\Rightarrow-2\text{x}=1-2$
$\Rightarrow\text{x}=\frac{1}{2}$
Thus, $\text{c}=\frac{1}{2}\in(1,0)$ such that $\text{f}\ '(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 835 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{1-\sqrt{\text{xa}}}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{1-\sqrt{\text{xa}}}\Big)$$\text{y}=\tan^{-1}\sqrt{\text{x}}+\tan^{-1}\sqrt{\text{a}}$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}\Big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\tan^{-1}\sqrt{\text{x}})+\frac{\text{d}}{\text{dx}}(\tan^{-1}\sqrt{\text{a}})$
$=\frac{1}{1+\big(\sqrt{\text{x}}\big)^2}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)+0$
$=\Big(\frac{1}{1+\text{x}}\Big)\Big(\frac{1}{2\sqrt{\text{x}}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}(1+\text{x})}$
View full question & answer→Question 845 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point $'c\ '$ in the indicated interval as stated by the Lagrange's mean value theorem. $f(x) = x(x + 4)^2$ on $[0,4]$
AnswerHere$,f(x) = x(x + 4)^2$
$\Rightarrow f(x) = x(x^2+ 16 + 8x)$
$\Rightarrow f(x) = x^3 + 8x^2+ 16x$
Since $f(x)$ is a polynomial function which is everywhere continuous and differentiable.
Therefore$, f(x)$ is continuous on $[0, 4]$ and derivable on $(0, 4)$
Thus, both the conditions of Lagrange's theorem is satisfied.
Consequently, there exists some $\text{c}\in(0,4)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}=\frac{\text{f}(4)-\text{f}(0)}{4}$
Now$, f(x) = x^3 + 8x^2+ 16x$
$\Rightarrow f(x) = 3x^2+ 16x + 16,$
$\Rightarrow f(4) = 64+ 128 + 64 = 256,$
$\Rightarrow f(0) = 0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
$\Rightarrow3\text{x}^2+16\text{x}+16=\frac{256}{4}$
$\Rightarrow3\text{x}^2+16\text{x}-48=0$
$\Rightarrow\text{x}=-\frac{4}{3}\big(2+\sqrt{13}\big),\frac{4}{3}\big(\sqrt{13}-2\big)$
Thus, $\text{c}=\frac{-8+4\sqrt{13}}{3}\in(0,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 855 Marks
Find all points of discontinuity of f, where f is defined by:
$\text f(\text x)=\begin{cases}\left|\text x\right|+3, \text{if x}\leq-3\\-2 \text{x},\text{if}-3 < \text x > 3\\6\text{x}+2,\text{if}\ \text{x}\geq3\end{cases}$
AnswerHere $\text f(\text x)=\begin{cases}\left|\text x\right|+3, \text{if x}\leq-3\\-2 \text{x},\text{if}-3 < \text x > 3\\6\text{x}+2,\text{if}\ \text{x}\geq3\end{cases}$Function f is defined for all points of thr real line.
Let c be any real number.
Five cases arise:
Case I: c < -3
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}(|\text{x}|+ 3) = |\text{c}| + 3$
Also f(c) = |c| + 3
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at all point x < -3.
Case II: c = -3
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}{-}}(|\text{x}|+ 3) = |-3| + 3 = 3 +3 =6$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}{ +}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}{ +}}\text{(-2x)} = 6$
Also f(c) = f(-3) = |-3|+3 = 3 + 3 = 6
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{+}}\text{f(x)} =\text{f(c)}= 6$
$\therefore$ f is continuous at x= -3
Case III: -3 < c < 3
f(x) = -2 x is a continuous function as it is a polynomial.
Case IV: c = 3
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{-}}\text{(-2x)} = -6$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{+}}\text{(6x + 2)} = 18 + 2 = 20 $
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{-}}\text{f(x)}\neq ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{+}}\text{f(x)}$
$\therefore$ f is discontinuous at x = 3.
Case V: c > 3
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{(6x+2)}= \text{6c}+ 2$
Also f(c) = 6c +2
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\ \text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at all points x > 3.
View full question & answer→Question 865 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point $'c\ '$ in the indicated interval as stated by the Lagrange's mean value theorem. $f(x) = x^{3 }- 5x^2 - 3x$ on $[1, 3]$
AnswerWe have$, f(x) = x^{3 }- 5x^2 - 3x$
Since, polynomial function is everywhere continuous and differentiable.
Therefore$, f(x)$ is continuous on $1, 3$ and differentiable on $1, 3$
Thus, both the conditions of Lagrange's theorem is satisfied.
Concequently, there exist some $\text{c}\in1,3$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}=\frac{\text{f}(3)-\text{f}(1)}{2}$
Now$, f(x) = x^{3 }- 5x^2 - 3x$
$f'(x) = 3x^2 - 10x - 3$
$\Rightarrow f(3) = -27$
$\Rightarrow f(1) = -7$
$\therefore\ \text{f}\ '(\text{x})=\frac{\text{f}(3)-\text{f}(1)}{2}$
$\Rightarrow3\text{x}^2-10\text{x}-3=\frac{-20}{2}$
$\Rightarrow3\text{x}^2-10\text{x}+7=0$
$\Rightarrow\text{x}=1,\frac{7}{3}$
Thus, $\text{c}=\frac{7}{3}\in(1,3)$ such that $\text{f}\ '(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
Hence, Lagrange's theorem is verified.
View full question & answer→Question 875 Marks
Find $\frac{\text{dy}}{\text{dx}}$ when x and y are connected by the relation:
$\sin(\text{xy})+\frac{\text{x}}{\text{y}}=\text{x}^2-\text{y}$
AnswerConsider, $\sin(\text{xy})+\frac{\text{x}}{\text{y}}=\text{x}^2-\text{y}$$\Rightarrow\ \frac{\text{d}}{\text{dx}}(\sin\text{xy})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}\Big)=\frac{\text{d}}{\text{dx}}\text{x}^2-\frac{\text{d}}{\text{dx}}\text{y}$
$\Rightarrow\ \cos\text{xy}\cdot\frac{\text{d}}{\text{dx}}(\text{xy})+\frac{\text{y}\frac{\text{d}}{\text{dx}}\text{x}-\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{y}}{\text{y}^2}=2\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \cos\text{xy}\cdot\Big[\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{y}+\text{y}\cdot\frac{\text{d}}{\text{dx}}\cdot\text{x}\Big]+\frac{\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}}{\text{y}^2}=2\text{x}-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \text{x}\cos\text{xy}\cdot\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{xy}+\frac{\text{y}}{\text{y}^2}\frac{\text{x}}{\text{y}^2}\frac{\text{dy}}{\text{dx}}=2\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \text{x}\cos\text{xy}\cdot\frac{\text{dy}}{\text{dx}}-\frac{\text{x}}{\text{y}^2}\frac{\text{dy}}{\text{dx}}+\frac{\text{dy}}{\text{dx}}=2\text{x}-\text{y}\cos\text{xy}-\frac{1}{\text{y}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big[\text{x}\cos\text{xy}-\frac{\text{x}}{\text{y}^2}+1\Big]=2\text{x}-\text{y}\cos\text{xy}-\frac{1}{\text{y}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\Big[\frac{2\text{xy}-\text{y}^2\cos\text{xy}-1}{\text{y}}\Big]\Big[\frac{\text{y}^2}{\text{xy}^2\cos\text{xy}-\text{x}+\text{y}^2}\Big]$
$=\frac{2\text{xy}^2-\text{y}^3\cos\text{xy}-\text{y}}{\text{xy}^2\cos\text{xy}-\text{x}+\text{y}^2}$
View full question & answer→Question 885 Marks
Is the function f defined by
$\text{f(x)} = \begin{cases}\text{x}, \text{if}\ \text{x}\leq1\\5, \text{if}\ \text{x} > 1\end{cases}$
continuous at x = 0? At x = 1? At x = 2?
AnswerHere $\text{f(x)} = \begin{cases}\text{x}, \text{if}\ \text{x}\leq1\\5, \text{if}\ \text{x} > 1\end{cases}$At x = 0
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}}\text{(x)}\ [ \text{Put}\ \text{x} = 0 - \text{h}, \text{h}>0\ \text{so that}\ \text{h}\rightarrow 0\ \text{as}\ \text{x}\rightarrow 0^-]$
$=\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}(0-\text{h})$
$=\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\text{(-h)} = (0) = 0$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{(x)}\ [ \text{Put}\ \text{x} = 0 + \text{h}, \text{h}>0\ \text{so that}\ \text{h}\rightarrow 0\ \text{as}\ \text{x}\rightarrow 0^+]$
$=\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}(0 + \text{h})$
= 0 + 0 = 0
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = 0$
$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}}\text{f(x)} = 0$
Also f(0) = value of x at x = 0
= 0
$\therefore$ f is continous at x = 0.
At x = 1
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{x}\ [ \text{Put}\ \text{x} = 1 - \text{h}, \text{h}>0,\ \text{so that}\ \text{h}\rightarrow 0\ \text{on}\ \text{x}\rightarrow 1^-]$
$^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}(1 - \text{h}) = 1 - 0 = 1$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}(5) = 5$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)}\neq\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)}$
$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)}$ does not exist
$\therefore$ f is discontinuous at x =1.
At x = 2
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}}(5) = 5$
Also f(2) = 5
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}}\text{f(x)} = \text{f(2)} = 5$
$\therefore$ f is continous at x = 2.
View full question & answer→Question 895 Marks
If $\text{y}\sqrt{\text{x}^2+1}=\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big),$ prove that $\big(\text{x}^2+1\big)\frac{\text{dx}}{\text{dx}}+\text{xy}+1=0$
AnswerWe have, $\text{y}\sqrt{\text{x}^2+1}=\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big)$Differentiating with respect to x, we get,
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{y}\sqrt{\text{x}^2+1}\Big)=\frac{\text{d}}{\text{dx}}\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big)$
[Using Product rule and chain rule]
$\Rightarrow\text{y}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+1}\big)+\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+1}-\text{x}\big)$
$\Rightarrow\frac{\text{y}}{2\sqrt{\text{x}^2+1}}\times\frac{\text{d}}{\text{dx}}(\text{x}^2+1)+\sqrt{\text{x}^2+1}\frac{\text{d}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\times\Big[\frac{1}{2\sqrt{\text{x}^2+1}}\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-1\Big]$
$\Rightarrow\ \frac{2\text{xy}}{2\sqrt{\text{x}^2+1}}+\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\Big[\frac{2\text{x}}{2\sqrt{\text{x}^2+1}}-1\Big]$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\Big[\frac{1}{\sqrt{\text{x}^2+1}-\text{x}}\Big]\Big[\frac{\text{x}-\sqrt{\text{x}^2+1}}{\sqrt{\text{x}^2+1}}\Big]-\frac{\text{xy}}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}}=\frac{-(1+\text{xy})}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\big(\text{x}^2+1\big)\frac{\text{dy}}{\text{dx}}=-(1+\text{xy})$
$\Rightarrow(\text{x}^2+1)\frac{\text{dy}}{\text{dx}}+1+\text{xy}=0$
View full question & answer→Question 905 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$\text{f}(\text{x})=\tan^{-1}\text{x}\text{ on }[0,1]$
AnswerWe have,$\text{f}(\text{x})=\tan^{-1}\text{x}$
Clearly, f(x) is continuous on 0, 1 and derivable on 0, 1
Thus, both the conditions of Lagrange's theorem are satisfied.
Concequently, there exist some $\text{c}\in-3,4$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}=\frac{\text{f}(1)-\text{f}(0)}{1}$
Now,
$\text{f}(\text{x})=\tan^{-1}\text{x}$
$\text{f}'(\text{x})=\frac{1}{1+\text{x}^2},\text{f}(1)=\frac{\pi}{4},\text{f}(0)=0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow\frac{1}{1+\text{x}^2}=\frac{\pi}{4}-0$
$\Rightarrow49\Big(\frac{\pi}{4}-1\Big)=\text{x}^2$
$\Rightarrow\text{x}=\pm\sqrt{\frac{4-\pi}{\pi}}$
Thus, $\text{c}=\sqrt{\frac{4-\pi}{\pi}}\in(0,1)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 915 Marks
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos2\text{t}}},\text{y}\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}}$
AnswerThe given equations are $\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos2\text{t}}}\text{ and y}=\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}}$Then, $\frac{\text{dx}}{\text{dt}}= \frac{\text{d}}{\text{dt}}\Big[\frac{\sin^3\text{t}}{\sqrt{\cos2\text{t}}}\Big]$
$=\frac{\sqrt{\cos2\text{t}}.\frac{\text{d}}{\text{dt}}(\sin^3\text{t})-\sin^3\text{t}.\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
$=\frac{\sqrt{\cos2\text{t}}.3\sin^2\text{t}.\frac{\text{d}}{\text{dt}}(\sin\text{t})-\sin^3\text{t}\times\frac{1}{2\sqrt{\cos2\text{t}}}.\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$=\frac{3\sqrt{\cos2\text{t}}.\sin^2\text{t}\cos\text{t}-\frac{\sin^3\text{t}}{2\sqrt{\cos2\text{t}}}.(-2\sin2\text{t})}{\cos2\text{t}}$
$=\frac{3\cos2\text{t}\sin^2\text{t}\cos\text{t}+\sin^3\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}}\Big]$
$=\frac{\sqrt{\cos2\text{t}}.\frac{\text{d}}{\text{dt}}(\cos^3\text{t})-\cos^3\text{t}.\frac{\text{d}}{\text{dt}}(\sqrt{\cos2\text{t}})}{\cos2\text{t}}$
$=\frac{\sqrt{\cos2\text{t}}\cos^2\text{t}.\frac{\text{d}}{\text{dt}}(\cos\text{t})-\cos^3\text{t}.\frac{1}{2\sqrt{\cos2\text{t}}}.\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$=\frac{3\sqrt{\cos2\text{t}}.\cos^2\text{t}(-\sin\text{t})-\cos^3\text{t}.\frac{1}{2\sqrt{\cos2\text{t}}}.(-2\sin2\text{t})}{\cos2\text{t}}$
$=\frac{-3\cos2\text{t}\cdot\cos^2\text{t}\cdot\sin\text{t}+\cos^3\text{t}.\sin2\text{t}}{\cos2\text{t}\cdot\sqrt{\cos2\text{t}}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{-3\cos2\text{t}.\cos^2\text{t}.\sin\text{t}+\cos^3\text{t}\sin2\text{t}}{3\cos2\text{t}\sin^2\text{t}\cos\text{t}+\sin^3\text{t}\sin2\text{t}}$
$=\frac{-3\cos2\text{t}.\cos^2\text{t}.\sin\text{t}+\cos^3\text{t}(2\sin\text{t}\cos\text{t})}{3\cos2\text{t}\sin^2\text{t}\cos\text{t}+\sin^3\text{t}(2\sin\text{t}\cos\text{t})}$
$=\frac{\sin\text{t}\cos\text{t}[-3\cos2\text{t}.\cos\text{t}+2\cos^3\text{t}]}{\sin\text{t}\cos\text{t}[3\cos2\text{t}\sin\text{t}+2\sin^3\text{t}]}$
$=\frac{[-3(2\cos^2\text{t}-1)\cos\text{t}+2\cos^3\text{t}]}{[3(1-2\sin^2\text{t})\sin\text{t}+2\sin^3\text{t}]}$ $\begin{bmatrix}\cos2\text{t}=(2\cos^2\text{t}-1). \\\cos2\text{t}=(1-2\sin^2\text{t}) \end{bmatrix}$
$=\frac{-4\cos^3\text{t}+3\cos\text{t}}{3\sin\text{t}-4\sin^3\text{t}}$
$=\frac{-\cos3\text{t}}{\sin3\text{t}}$ $\begin{bmatrix}\cos3\text{t}=4\cos^3\text{t}-3\cos\text{t}. \\\sin3\text{t}=3\sin\text{t}-4\sin^3\text{t} \end{bmatrix}$
$=-\cot3\text{t}$
View full question & answer→Question 925 Marks
If $\text{x}=\text{a}\sin\text{t}-\text{b}\cos\text{t},\text{y}=\text{a}\cos\text{t}+\text{b}\sin\text{t},$ Prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{x}^2+\text{y}^2}{\text{y}^2}$
AnswerWe have$\text{x}=\text{a}\sin\text{t}-\text{b}\cos\text{t},\text{y}=\text{a}\cos\text{t}+\text{b}\sin\text{t},$
On differentiating with respect to t, we get
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\sin\text{t}-\text{b}\cos\text{t})=\text{a}\cos\text{t}+\text{b}\sin\text{t}$
and
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\cos\text{t}+\text{b}\sin\text{t})=-\text{a}\sin\text{t}+\text{b}\cos\text{t}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-\text{a}\sin\text{t}+\text{b}\cos\text{t}}{\text{a}\cos+\text{b}\sin\text{t}}$
Therefore
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}\Big(\frac{-\text{a}\sin\text{t}+\text{b}\cos\text{t}}{\text{a}\cos\text{t}+\text{b}\sin\text{t}}\Big)$
$=\frac{\text{d}}{\text{dt}}\Big(\frac{-\text{a}\sin\text{t}+\text{b}\cos\text{t}}{\text{a}\sin\text{t}+\text{b}\sin\text{t}}\Big)\times\frac{\text{dt}}{\text{dx}}$
$=\frac{(\text{a}\cos\text{t}+\text{b}\sin\text{t})\frac{\text{d}}{\text{dt}}(-\text{a}\sin\text{t}+\text{b}\cos\text{t})-(-\text{a}\sin\text{t}+\text{b}\cos\text{t})\frac{\text{d}}{\text{dt}}(\text{a}\cos\text{t}+\text{b}\sin\text{t})}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2}$
$$$=\frac{(\text{a}\cos\text{t}+\text{b}\sin\text{t})(\text{a}\cos\text{t}+\text{b}\sin\text{t})(-\text{a}\sin\text{t}+\text{b}\cos\text{t})-(-\text{a}\sin\text{t}+\text{b}\cos\text{t})}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2}$
$=\frac{-(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2-(-\text{a}\sin\text{t}+\text{b}\cos\text{t})^2}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})}$
$=\frac{-(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2-(\text{a}\sin\text{t}+\text{b}\cos\text{t})^2}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})}$
$=\frac{-\text{y}^2-\text{x}^2}{\text{y}^3}$
View full question & answer→Question 935 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\sin\text{ xy}+\cos(\text{x}+\text{y})=1$
AnswerWe have, $\sin\text{ xy}+\cos(\text{x}+\text{y})=1$Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\sin\text{xy}+\frac{\text{d}}{\text{dx}}\cos(\text{x}+\text{y})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\cos \text{xy}\frac{\text{d}}{\text{dx}}(\text{xy})-\sin(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=0$
[Using chain rule]
$\Rightarrow\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]-\sin(\text{x}+\text{y})\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=0$
$\Rightarrow\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big]-\sin(\text{x}+\text{y})+\sin(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}\cos(\text{xy})\frac{\text{dy}}{\text{dx}}+\text{y}\cos(\text{xy})-\sin(\text{x}+\text{y})+\sin(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\big[\text{x}\cos(\text{xy})+\sin(\text{x}+\text{y})\big]\frac{\text{dy}}{\text{dx}} \\ =\big[\sin(\text{x}+\text{y})-\text{y}\cos(\text{xy})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{\sin(\text{x}+\text{y})-\text{y}\cos\text{xy}}{\text{x}\cos\text{xy}+\sin(\text{x}+\text{y})}\Big]$
View full question & answer→Question 945 Marks
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Answer
- Let a be an arbitrary real number then $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}^{+}}\text{f(x)}\Rightarrow^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}^{+}}\cos\text{x} \Rightarrow^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\cos\text{(a + h)}$
$\Rightarrow\ ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}(\cos\text{a}\cos\text{ h} - \sin\text{a} \sin \text{h})= \cos \text{a}\ ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\ \cos\text{h}- \sin\text{a}\ ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\sin\text{h}$
$= \cos\text{a} \times 1 - \sin\text{a}\times0 = \cos \text{a} = \text{f(a)}$
$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}}\text{f(x)} = \text{f(a)}$ for all $\text{a}\in\text{R}$
Therefore, f(x0 is continuous at x = a.
Since, a is an arbitrary real number, therefore $\cos\text{x}$ is continuous.
- $\text{f(x)} = \text{cosec x}= \frac{1}{\sin \text x}$ and domain $\text{x} = \text{R} - (\text{x}\pi), \text{x}\in \text{I}$
$\Rightarrow ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}}\frac{1}{{\sin}\text{x}} = \frac{1}{^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\sin(\text{a + h})}=\frac{1}{{^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}(\sin}\text{a}\cos\text{h} + \cos\text{a}\sin\text{h})}$$=\frac{1}{{\sin}\text{a}\cos{0} + \cos\text{a}\sin0}$
$= \frac{1}{{\sin}\text{a}(1)+ \cos\text{a}(0)} = \frac{1}{\sin\text{a}} = \text{f(a)}$
Therefore, f(x) is continuous at x = a.
Since, a is an arbitrary real number, therefore, f(x) = cosec x is continuous.
- $\text{f(x)} = \sec \text{x}= \frac{1}{\cos \text x}$ and domain $\text{x} = \text{R} -(2\text{x} + 1) \frac\pi{2}, \text{x}\in \text{I}$
$\Rightarrow ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}}\frac{1}{{\cos}\text{x}} = \frac{1}{^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\cos(\text{a + h})}=\frac{1}{{^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}(\cos}\text{a}\cos\text{h} - \sin\text{a}\sin\text{h})}$$=\frac{1}{{\cos}\text{a}\cos{0} - \sin\text{a}\sin0}$
$= \frac{1}{{\cos}\text{a}(1)- \sin\text{a}(0)} = \frac{1}{\cos\text{a}} = \text{f(a)}$
Therefore, f(x) is continuous at x = a.
Since, a is an arbitrary real number, therefore, f(x) = sec x is continuous.
- $\text{f(x)} = \cot \text{x}= \frac{1}{\tan \text x}$ and domain $\text{x} = \text{R} - (\text{x}\pi), \text{x}\in \text{I}$
$\Rightarrow^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}}\frac{1}{{\tan}\text{x}}= \frac{1}{\lim\limits_{\text{h}\rightarrow0}\tan(\text{a + h})}=\frac{1}{\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan\text{a}+\tan\text{h}}{1-\tan\text{a}\tan\text{h}}\Big)}$$=\frac{1}{\frac{\tan\text{a}+0}{1-\tan\text{a}\tan0}}$
$\frac{1 - 0}{\tan\text{a}} = \frac{1}{\tan \text{a}} = \text{f(a)}$
Therefore, f9x) is continuous at x = a.
Since, a is an arbitrary real number, therefore, $\text{f(x)} = \cot \text{x}$ is continuous. View full question & answer→Question 955 Marks
Differentiate $\tan^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$ with respect to $\sin^{-1}\big(3\text{x}-4\text{x}^3\big),$ if $-\frac{1}{2}<\text{x}<\frac{1}{2}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$Put $\text{x}=\tan\theta$
$\Rightarrow \text{u}=\tan^{-1}\Big(\frac{\tan\theta-1}{\tan\theta+1}\Big)$
$\Rightarrow \text{u}=\tan^{-1}\bigg(\frac{\tan\theta-\tan\frac{\pi}{4}}{1+\tan\theta\tan\frac{\pi}{4}}\bigg)$
$\Rightarrow\text{u}\tan^{-1}\Big[\tan\big(\theta-\frac{\pi}{4}\big)\Big]\ .....(\text{i})$
Here, $-\frac{1}{2}<\text{x}<\frac{1}{2}$
$\Rightarrow-\frac{1}2{}<\tan\theta<\frac{1}{2}$
$\Rightarrow-\tan^{-1}\big(\frac{1}{2}\big)<\theta<\tan^{-1}\big(\frac{1}{2}\big)$
So, from equation (i)
$\text{u}=\theta-\frac{\pi}{4}$
$\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\tan^{-1}\text{x}-\frac{\pi}{4} \ [\text{Since,x}= \tan\theta]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{1}{1+\text{x}^2}-0$
$\Rightarrow \frac{\text{du}}{\text{dx}}=\frac{1}{1+\text{x}^2}\ ..... \text{(ii)}$
And,
Let, $\text{v}=\sin^{-1}(3\text{x}-4\text{x}^3)$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{v}=\sin^{-1}(3\sin\theta-4\sin^3\theta)$
$\Rightarrow\text{v}=\sin^{-1}(\sin3\theta)\ .....\text{(iii)}$
Now, $-\frac{1}{2}<\text{x}<\frac{1}{2}$
$\Rightarrow-\frac{1}{2}<\sin\theta<\frac{1}{2}$
$\Rightarrow-\frac{1}{6}<\theta<\frac{\pi}{6}$
So, from equation (iii),
$\text{v}=3\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{v}=3\sin^{-1}\text{x}[\text{Since, x}=\sin\theta]$
Dirrerentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{3}{\sqrt{1-\text{x}^2}}\ .....\text{(iv)}$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{1}{1+\text{x}^2}\times\frac{\sqrt{1-\text{x}^2}}{3}$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{\sqrt{1-\text{x}^2}}{3(1+\text{x}^2)}$
View full question & answer→Question 965 Marks
If $\text{y}=(\cot^{-1}\text{x})^2$ prove that $\text{y}^2(\text{x}^2+1)^2+2\text{x}(\text{x}^2+1)\text{y}_1=2.$
Answer$\text{y}=(\cot^{-1}\text{x})^2$Differentiating w.r.t.x,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}_1=\frac{-2\cot^{-1}\text{x}}{1+\text{x}^2}$
$=\frac{-2\cot^{-1}\text{x}}{1+\text{x}^2}\ (\text{chain rule})$
$\Rightarrow(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}=-2\cot^{-1}\text{x}$
Differentiating w.r.t.x,
$\Rightarrow(1+\text{x}^2)\text{y}^2+2\text{xy}_1=+2\Big(\frac{+1}{1+\text{x}^2}\Big)$
(Multiplication rule on LHS)
$\Rightarrow(1+\text{x}^2)^2\text{y}_2+2\text{x}(1+\text{x}^2)\text{y}_1=2$
Hence proved
View full question & answer→Question 975 Marks
If $x^{16}y^9 = (x + y)^{17},$ prove that $\text{x}\frac{\text{dy}}{\text{dx}}=2\text{y}$
AnswerHere$,x^{16}y^9 = (x + y)^{20}$
Taking $\log$ on both the siede,
$\log(\text{x}^{16}\times\text{y}^{19})=\log(\text{x}^2+\text{y})^{17}$
$\big[$since $, \log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$16\log\text{x}+9\log\text{y}=17\log(\text{x}^2+\text{y})$
Differentiating it with respect to $x$ using chain rule
$16\frac{\text{d}}{\text{dx}}(\log\text{x})+9\frac{\text{d}}{\text{dx}}(\log\text{y})=17\frac{\text{d}}{\text{dx}}\log(\text{x}^2+\text{y})$
$\frac{16}{\text{x}}+\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}=17\frac{1}{(\text{x}^2+\text{y})}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y})$
$\frac{16}{\text{x}}+\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{17}{\text{x}^2+\text{y}}\Big[2\text{x}+\frac{\text{dy}}{\text{dx}}\Big]$
$\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}-\frac{17}{(\text{x}^2+\text{y})}\frac{\text{dy}}{\text{dx}}=\Big(\frac{34\text{x}}{\text{x}^2+\text{y}}\Big)-\frac{16}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{9}{\text{y}}-\frac{17}{(\text{x}^2+\text{y})}\Big]=\frac{34\text{x}^2-16\text{x}^2-16\text{y}}{\text{x}(\text{x}^2+\text{y})}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{9\text{x}^2+9\text{y}-17\text{y}}{\text{y}(\text{x}^2+\text{y})}\Big]=\frac{18\text{x}^2-16\text{y}}{\text{x}(\text{x}^2+\text{y})}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\Big(\frac{2(9\text{x}^2-8\text{y})}{9\text{x}^2-8\text{y}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}}$
$\text{x}\frac{\text{dy}}{\text{dx}}=2\text{y}$
View full question & answer→Question 985 Marks
If $\text{x}=\cos\theta,\text{y}=\sin^3$ prove that $\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}^2}\Big)=3\sin^2\theta(5\cos^2\theta-1)$
AnswerHere$\text{x}=\cos\theta,\text{y}=\sin^3$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{d}\theta}=-\sin\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=3\sin^2\theta\cos\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{3\sin^2\theta\cos\theta}{-\sin\theta}=-3\sin\theta\cos\theta$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=(-3\cos^2\theta+3\sin^2\theta)\frac{\text{d}\theta}{\text{dx}}\frac{)-3\cos^2\theta+3\sin^2\theta)}{-\sin\theta}$
Now,
$\text{LHS}=\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$=\sin^3\theta\times\frac{(-3\cos^2\theta+3\sin^2\theta)}{\sin\theta}+(-3\sin\theta\cos\theta)^2$
$=3\sin^2\theta\cos^2\theta-3\sin^4\theta+9\sin^2\theta\cos^2\theta$
$=12\sin^2\theta\cos^2\theta-3\sin^4\theta$
$=3\sin^2\theta(4\cos^2\theta-\sin^2\theta)$
$=3\sin^2\theta(5\cos^2\theta-1)$
$=\text{RHS}$
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Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point $'c\ '$ in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = x^3 - 2x^2 - x + 3 on [0, 1]$
AnswerHere,$f(x) = x^3 - 2x^2 - x + 3$
Since $f(x)$ is polynomial function. So, $f(x)$ is continuous in $[0, 1]$ and differentiable in $(0, 1).$
Thus, both conditions of Lagrange's mean value theorem is appplicable.
Thus, there exist a point $\text{c}\in(0,1)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow3\text{c}^2-4\text{c}-1=\frac{[(1)^3-2(1)^2-(1)+3]-3}{1}$
$\Rightarrow 3c^2 - 4c - 1 = 1^{-2}$
$\Rightarrow 3c^2 - 4c + 1 = 0$
$\Rightarrow 3c^2- 3c - c + 1 = 0$
$\Rightarrow 3c(c - 1) - 1(c - 1) = 0$
$\Rightarrow (3c - 1)(c - 1) = 0$
$\Rightarrow\text{c}=\frac{1}{3}\in(0,1)$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 1005 Marks
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = (x^2- 1)(x - 2)$ on $[-1, 2]$
AnswerHere,$f(x) = (x^2- 1)(x - 2)$ on $[-1, 2]$
$f(x)$ is continuous is $[-1, 2]$ and differentiable in $(-1, 2)$ as it is a polynomial functions.
Now,
$f(-1) = (1-1)(-1-2) = 0$
$f(2) = (4-1)(2-2) = 0$
$\Rightarrow f(-1) = f(2)$
So, Rolle's theorem is applicable on $f(x)$ is $[-1, 2]$
therefore, we have to show that there exist a $\text{c}\in(-1,2)$ such that $f'(c) = 0$
Now,
$f(x) = (x^2- 1)(x - 2)$
$f'(x) = 2x(x - 2) + (x^2 - 1)$
$= 2x^2 - 4 + x^2 - 1$
$f'(x) = 3x^2 - 5$
Now,
$f'(c) = 0$
$\Rightarrow 3x^2 - 5 = 0$
$\Rightarrow\text{x}=-\sqrt{\frac{5}{3}}$ or $\text{x}=\sqrt{\frac{5}{3}}\in(-1,2)$
Thus, Rolle's theorem is verified.
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