Maths 4 Marks

$\text{u}=\text{x}^{\text{x cos x}}\Rightarrow\log\text{u}=\text{x cos x}\cdot{\text{log x}}$

$\therefore\frac{1}{\text{u}}\cdot\frac{\text{du}}{\text{dx}}=\frac{\text{x cos x}}{\text{x}}+\text{cos x}\cdot{\text{log x - sin x log x}}$

$\Rightarrow$$\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x cos x}}(\text{cos x+cos x log x}\text{ - sin x log x})$

$\text{v}=\frac{\text{x}^{2}+1}{\text{x}^{2}-1},\frac{\text{dv}}{\text{dx}}=\frac{(\text{x}^{2}-1)\text{2x}-\text{(x}^{2}+1)\text{2x}}{\text{(x}^{2}-1)^{2}}=-\frac{\text{4x}}{\text{(x}^{2}-1)^{2}}$

$\therefore\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x cos x}}\text{(cos x + log x}\cdot\text{cos x - sin x log x})-\frac{\text{4x}}{(\text{x}^{2}-1)^{2}}$

$\text{Let x}^{\sin x} = \text{u, and} ( \sin \text{x)}^{\cos \text{x}} = \text{v} \therefore \text{y = u + v} \Rightarrow \frac{\text {dy}}{\text{dx}} = \frac{\text{du}}{\text{dx}} + \frac{\text{dv}}{\text{dx}}$

$\text{Getting} \frac{\text{du}}{\text{dx}} = \text{x}^{\sin{\text{x}}} \bigg[\frac{\sin\text{x}}{\text{x}} + \log \text{x} . \cos{\text{x}} \bigg]$

$\text{and} \frac{\text{dv}}{\text{dx}} = ( \sin \text{x})^{\cos\text{x}}[\cos\text{x} .\cot\text{x} - \sin\text{x} \log \sin \text{x}]$

$\therefore \frac{\text{dy}}{\text{dx}} = \text{x}^{\sin \text{x}} \bigg[ \frac{\sin \text{x}}{\text{x}} + \log\text{x}.\cos\text{x}\bigg] + (\sin\text{x}) ^{\cos{\text{x}}} [\cos\text{x}. \cot\text{x} -\sin\text{x} \log\sin\text{x}]$

$\text{Let y} = \bigg[\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x + \sqrt{ 1 - x}}}\bigg] = \tan^{-1} \bigg[\frac{1 - \tan \theta}{1 + \tan \theta}\bigg] = \tan^{-1} \tan \bigg(\frac{\pi}{4} - \theta\bigg)$

$= \frac{\pi}{4} - \theta = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} \text{x}$

$\therefore \frac{dy}{dx} = \frac{-1}{2} \bigg(\frac{-1}{\sqrt{1 - x^{2}}}\bigg) = \frac{1}{2\sqrt{1 - x^{2}}}$

$\lim\limits_{h \rightarrow 0} f (2 -h) = f(2) = \lim\limits_{h \rightarrow 0} f(2 + h)$

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 $f(2) = 5.$

$\therefore k = 5.$

$ \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}$

$y = \triangle y = \sin (2x +2\triangle x + 3)$

$\therefore \triangle y \sin (2x + 2\triangle x + 3) -\sin (2x + 3)$

$= 2\cos (2x + 3 + \triangle x) \sin \triangle x$

$\therefore \lim\limits_{\triangle x\rightarrow 0} \frac{\triangle y}{\triangle x} = \lim\limits_{\triangle x \rightarrow 0} 2\cos (2x +3 +\triangle x) \lim\limits_{\triangle x\rightarrow 0} \frac {\sin\triangle x}{\triangle x}$

OR

$\frac{dy}{dx} = \text2\cos (2x + 3) 1 = 2\cos (2x + 3)$

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

$\therefore f' (x) = (x - 1) 2 (x -2) + (x - 2)^{2}$

$= (x - 2)[2x - 2 + x- 2] = (x - 2) (3x - 4)$

$\therefore f' (c) = 0 \Rightarrow c = 2 \text{ and}\ {c}\frac{4}{3}$

$As 1<\frac{4}{3}<2, \text{the Roll's theorem is verfied}$

$\lim\limits_{X\rightarrow \frac{\pi}{6}} \Bigg[\frac{2\big(\cos \frac{\pi}{6}.\sin x-\sin\frac{\pi}{6}.\cos x\big)}{x-\frac{\pi}{6}}\Bigg]$

$2\lim\limits_{X\rightarrow\frac{\pi}{6}}\frac{\sin\big(x-\frac{\pi}{6}\big)}{\big(x - \frac{\pi}{6}\big)}$

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

$\frac{1}{2} \sqrt{\frac{x - 3)(x ^{2} + 4)}{3x^{2} + 4x +5}} \bigg[ \frac{1}{x - 3} + \frac{2x}{x^{2} + 4} - \frac{6x + 4}{3x^{2} + 4x + 5}\bigg]$

f(x) = x² - 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) = x² - 2x + 4

⇒ f'(x) = 2x - 2

⇒ f(5) = 25 - 10 + 4 = 19

⇒ 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.

$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}\text{ on }[1,3]$

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.

$\text{f}(\text{x})=\log\Big(\frac{\text{x}^2+2}{3}\Big)$

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.

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),

$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{1}{1+\text{x}^2}\times\frac{\sqrt{1-\text{x}^2}}{3}$

$\therefore\frac{\text{du}}{\text{dv}}=\frac{\sqrt{1-\text{x}^2}}{3(1+\text{x}^2)}$

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.

f(x) = 2x - x² 

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 - x² 

⇒ f'(x) = 2 - 2x,

⇒ f(1) = 2 - 1

⇒ f(1) = 1,

⇒ 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.

$\text{y}=\tan^{-1}\sqrt{\text{x}}+\tan^{-1}\sqrt{\text{a}}$

$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}\Big]$

Differentiating it with respect to x using chain rule,

$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\tan^{-1}\sqrt{\text{x}})+\frac{\text{d}}{\text{dx}}(\tan^{-1}\sqrt{\text{a}})$

$=\frac{1}{1+\big(\sqrt{\text{x}}\big)^2}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)+0$

$=\Big(\frac{1}{1+\text{x}}\Big)\Big(\frac{1}{2\sqrt{\text{x}}}\Big)$

$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}(1+\text{x})}$

f(x) = x(x + 4)²

⇒ f(x) = x(x² + 16 + 8x)

⇒ f(x) = x³ + 8x² + 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) = x³ + 8x² + 16x

⇒ f(x) = 3x² + 16x + 16,

⇒ f(4) = 64+ 128 + 64 = 256,

⇒ 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.

$\Rightarrow\ \frac{\text{d}}{\text{dx}}(\sin\text{xy})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}\Big)=\frac{\text{d}}{\text{dx}}\text{x}^2-\frac{\text{d}}{\text{dx}}\text{y}$

$\Rightarrow\ \cos\text{xy}\cdot\frac{\text{d}}{\text{dx}}(\text{xy})+\frac{\text{y}\frac{\text{d}}{\text{dx}}\text{x}-\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{y}}{\text{y}^2}=2\text{x}\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\ \cos\text{xy}\cdot\Big[\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{y}+\text{y}\cdot\frac{\text{d}}{\text{dx}}\cdot\text{x}\Big]+\frac{\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}}{\text{y}^2}=2\text{x}-\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\ \text{x}\cos\text{xy}\cdot\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{xy}+\frac{\text{y}}{\text{y}^2}\frac{\text{x}}{\text{y}^2}\frac{\text{dy}}{\text{dx}}=2\text{x}\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\ \text{x}\cos\text{xy}\cdot\frac{\text{dy}}{\text{dx}}-\frac{\text{x}}{\text{y}^2}\frac{\text{dy}}{\text{dx}}+\frac{\text{dy}}{\text{dx}}=2\text{x}-\text{y}\cos\text{xy}-\frac{1}{\text{y}}$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big[\text{x}\cos\text{xy}-\frac{\text{x}}{\text{y}^2}+1\Big]=2\text{x}-\text{y}\cos\text{xy}-\frac{1}{\text{y}}$

$\therefore\ \frac{\text{dy}}{\text{dx}}=\Big[\frac{2\text{xy}-\text{y}^2\cos\text{xy}-1}{\text{y}}\Big]\Big[\frac{\text{y}^2}{\text{xy}^2\cos\text{xy}-\text{x}+\text{y}^2}\Big]$

$=\frac{2\text{xy}^2-\text{y}^3\cos\text{xy}-\text{y}}{\text{xy}^2\cos\text{xy}-\text{x}+\text{y}^2}$

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.

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]

$\Rightarrow\text{y}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+1}\big)+\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+1}-\text{x}\big)$

$\Rightarrow\frac{\text{y}}{2\sqrt{\text{x}^2+1}}\times\frac{\text{d}}{\text{dx}}(\text{x}^2+1)+\sqrt{\text{x}^2+1}\frac{\text{d}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\times\Big[\frac{1}{2\sqrt{\text{x}^2+1}}\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-1\Big]$

$\Rightarrow\ \frac{2\text{xy}}{2\sqrt{\text{x}^2+1}}+\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\Big[\frac{2\text{x}}{2\sqrt{\text{x}^2+1}}-1\Big]$

$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\Big[\frac{1}{\sqrt{\text{x}^2+1}-\text{x}}\Big]\Big[\frac{\text{x}-\sqrt{\text{x}^2+1}}{\sqrt{\text{x}^2+1}}\Big]-\frac{\text{xy}}{\sqrt{\text{x}^2+1}}$

$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}}=\frac{-(1+\text{xy})}{\sqrt{\text{x}^2+1}}$

$\Rightarrow\big(\text{x}^2+1\big)\frac{\text{dy}}{\text{dx}}=-(1+\text{xy})$

$\Rightarrow(\text{x}^2+1)\frac{\text{dy}}{\text{dx}}+1+\text{xy}=0$

$\text{f}(\text{x})=\tan^{-1}\text{x}$

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.

Then, $\frac{\text{dx}}{\text{dt}}= \frac{\text{d}}{\text{dt}}\Big[\frac{\sin^3\text{t}}{\sqrt{\cos2\text{t}}}\Big]$

$=\frac{\sqrt{\cos2\text{t}}.\frac{\text{d}}{\text{dt}}(\sin^3\text{t})-\sin^3\text{t}.\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$

$=\frac{\sqrt{\cos2\text{t}}.3\sin^2\text{t}.\frac{\text{d}}{\text{dt}}(\sin\text{t})-\sin^3\text{t}\times\frac{1}{2\sqrt{\cos2\text{t}}}.\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$

$=\frac{3\sqrt{\cos2\text{t}}.\sin^2\text{t}\cos\text{t}-\frac{\sin^3\text{t}}{2\sqrt{\cos2\text{t}}}.(-2\sin2\text{t})}{\cos2\text{t}}$

$=\frac{3\cos2\text{t}\sin^2\text{t}\cos\text{t}+\sin^3\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$

$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}}\Big]$

$=\frac{\sqrt{\cos2\text{t}}.\frac{\text{d}}{\text{dt}}(\cos^3\text{t})-\cos^3\text{t}.\frac{\text{d}}{\text{dt}}(\sqrt{\cos2\text{t}})}{\cos2\text{t}}$

$=\frac{\sqrt{\cos2\text{t}}\cos^2\text{t}.\frac{\text{d}}{\text{dt}}(\cos\text{t})-\cos^3\text{t}.\frac{1}{2\sqrt{\cos2\text{t}}}.\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$

$=\frac{3\sqrt{\cos2\text{t}}.\cos^2\text{t}(-\sin\text{t})-\cos^3\text{t}.\frac{1}{2\sqrt{\cos2\text{t}}}.(-2\sin2\text{t})}{\cos2\text{t}}$

$=\frac{-3\cos2\text{t}\cdot\cos^2\text{t}\cdot\sin\text{t}+\cos^3\text{t}.\sin2\text{t}}{\cos2\text{t}\cdot\sqrt{\cos2\text{t}}}$

$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{-3\cos2\text{t}.\cos^2\text{t}.\sin\text{t}+\cos^3\text{t}\sin2\text{t}}{3\cos2\text{t}\sin^2\text{t}\cos\text{t}+\sin^3\text{t}\sin2\text{t}}$

$=\frac{-3\cos2\text{t}.\cos^2\text{t}.\sin\text{t}+\cos^3\text{t}(2\sin\text{t}\cos\text{t})}{3\cos2\text{t}\sin^2\text{t}\cos\text{t}+\sin^3\text{t}(2\sin\text{t}\cos\text{t})}$

$=\frac{\sin\text{t}\cos\text{t}[-3\cos2\text{t}.\cos\text{t}+2\cos^3\text{t}]}{\sin\text{t}\cos\text{t}[3\cos2\text{t}\sin\text{t}+2\sin^3\text{t}]}$

$=\frac{[-3(2\cos^2\text{t}-1)\cos\text{t}+2\cos^3\text{t}]}{[3(1-2\sin^2\text{t})\sin\text{t}+2\sin^3\text{t}]}$ $\begin{bmatrix}\cos2\text{t}=(2\cos^2\text{t}-1). \\\cos2\text{t}=(1-2\sin^2\text{t}) \end{bmatrix}$

$=\frac{-4\cos^3\text{t}+3\cos\text{t}}{3\sin\text{t}-4\sin^3\text{t}}$

$=\frac{-\cos3\text{t}}{\sin3\text{t}}$ $\begin{bmatrix}\cos3\text{t}=4\cos^3\text{t}-3\cos\text{t}. \\\sin3\text{t}=3\sin\text{t}-4\sin^3\text{t} \end{bmatrix}$

$=-\cot3\text{t}$

$\text{x}=\text{a}\sin\text{t}-\text{b}\cos\text{t},\text{y}=\text{a}\cos\text{t}+\text{b}\sin\text{t},$

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}$

$=\frac{-(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2-(-\text{a}\sin\text{t}+\text{b}\cos\text{t})^2}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})}$

$=\frac{-(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2-(\text{a}\sin\text{t}+\text{b}\cos\text{t})^2}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})}$

$=\frac{-\text{y}^2-\text{x}^2}{\text{y}^3}$

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]

$\Rightarrow\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]-\sin(\text{x}+\text{y})\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=0$

$\Rightarrow\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big]-\sin(\text{x}+\text{y})+\sin(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\text{x}\cos(\text{xy})\frac{\text{dy}}{\text{dx}}+\text{y}\cos(\text{xy})-\sin(\text{x}+\text{y})+\sin(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\big[\text{x}\cos(\text{xy})+\sin(\text{x}+\text{y})\big]\frac{\text{dy}}{\text{dx}} \\ =\big[\sin(\text{x}+\text{y})-\text{y}\cos(\text{xy})\big]$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{\sin(\text{x}+\text{y})-\text{y}\cos\text{xy}}{\text{x}\cos\text{xy}+\sin(\text{x}+\text{y})}\Big]$

f(x) = x³ - 5x² - 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) = x³ - 5x² - 3x

f'(x) = 3x² - 10x - 3

⇒ f(3) = -27

⇒ 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.

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

1. 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.

2. $\text{f(x)} = \text{cosec x}= \frac{1}{\sin \text x}$ 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.

3. $\text{f(x)} = \sec \text{x}= \frac{1}{\cos \text x}$ 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.

4. $\text{f(x)} = \cot \text{x}= \frac{1}{\tan \text x}$ 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.

x¹⁶y⁹ = (x + y)²⁰

Taking log on both the siede,

$\log(\text{x}^{16}\times\text{y}^{19})=\log(\text{x}^2+\text{y})^{17}$

$\big[\text{since}, \log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$

$16\log\text{x}+9\log\text{y}=17\log(\text{x}^2+\text{y})$

Differentiating it with respect to x using chain rule

$16\frac{\text{d}}{\text{dx}}(\log\text{x})+9\frac{\text{d}}{\text{dx}}(\log\text{y})=17\frac{\text{d}}{\text{dx}}\log(\text{x}^2+\text{y})$

$\frac{16}{\text{x}}+\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}=17\frac{1}{(\text{x}^2+\text{y})}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y})$

$\frac{16}{\text{x}}+\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{17}{\text{x}^2+\text{y}}\Big[2\text{x}+\frac{\text{dy}}{\text{dx}}\Big]$

$\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}-\frac{17}{(\text{x}^2+\text{y})}\frac{\text{dy}}{\text{dx}}=\Big(\frac{34\text{x}}{\text{x}^2+\text{y}}\Big)-\frac{16}{\text{x}}$

$\frac{\text{dy}}{\text{dx}}\Big[\frac{9}{\text{y}}-\frac{17}{(\text{x}^2+\text{y})}\Big]=\frac{34\text{x}^2-16\text{x}^2-16\text{y}}{\text{x}(\text{x}^2+\text{y})}$

$\frac{\text{dy}}{\text{dx}}\Big[\frac{9\text{x}^2+9\text{y}-17\text{y}}{\text{y}(\text{x}^2+\text{y})}\Big]=\frac{18\text{x}^2-16\text{y}}{\text{x}(\text{x}^2+\text{y})}$

$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\Big(\frac{2(9\text{x}^2-8\text{y})}{9\text{x}^2-8\text{y}}\Big)$

$\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}}$

$\text{x}\frac{\text{dy}}{\text{dx}}=2\text{y}$

f(x) = (x² - 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

⇒ 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) = (x² - 1)(x - 2)

f'(x) = 2x(x - 2) + (x² - 1)

= 2x² - 4 + x² - 1

f'(x) = 3x² - 5

Now,

f'(c) = 0

⇒ 3x² - 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.