Maths M.C.Q (1 Marks)

4. 1

Solution:

$\text{U}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\text{ and V}=\tan^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$

Put, $\text{x}=\tan\theta$

$\text{U}=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)\text{ and V}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)$

$\text{U}=\sin^{-1}(\sin2\theta)\text{ and V}=\tan^{-1}(\tan2\theta)$

$\text{U}=2\theta\text{ and V}=2\theta$

$\text{U}=2\tan^{-1}\text{ x and V}=2\tan^{-1}\text{x}$

$\frac{\text{dU}}{\text{dx}}=\frac{\text{dV}}{\text{dx}}=\frac{2}{1+\text{x}^2}$

$\frac{\text{dU}}{\text{dV}}=\frac{\frac{\text{dU}}{\text{dx}}}{\frac{\text{dU}}{\text{dx}}}=1$

3. $\text{n}=\frac{\text{m}\pi}{2}$

Solution:

Here,

$\text{f}\Big(\frac{\pi}{2}\Big)=\frac{\text{m}\pi}{2}+1$

$\Big(\text{LHL}\text{ x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{2}-\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\text{m}\Big(\frac{\pi}{2}-\text{h}\Big)+1=\frac{\text{m}\pi}{2}+1$

$\Big(\text{RHL}\text{ x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)} = \lim\limits_{\text{h}\rightarrow0}\text{f}(\frac{\pi}{2}+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\Big[\sin\Big(\frac{\pi}{2}+\text{h}\Big)+\text{n}\Big]=\text{n}+1$

Thus,

if f(x) is continuons at $\text{x}= \frac{\pi}{2},$ then

$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}$

$\Rightarrow\frac{\text{m}\pi}{2}+1=\text{n}+1$

$\Rightarrow\frac{\text{m}\pi}{2}=\text{n}$

3. $\{(2\text{n}+1)\frac{\pi}{2}:\text{n}\in\text{z}\}$

Solution:

When $\tan(2\text{n}+1)\frac{\pi}{2}=\tan\Big(\text{n}\pi+\frac{\pi}{2}\Big)=-\cot\text{n}\pi,$ it is not defined at the integral points.

$[\text{n}\in\text{z}]$

Hence, f(x) is discontinuous on the set $\{(2\text{n}+1)\frac{\pi}{2}:\text{n}\in\text{z}\}$

3. $\text{a}=-\frac{3}{2},\text{b}\in\text{R}-\{0\},\text{c}=\frac{1}{2}$

Solution:

$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}}=\text{c}$

$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}}\times\frac{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c }$

$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x+bx}^2-\text{x}}{\text{bx}\sqrt{\text{x}}}\times\frac{1}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c}$

$\lim\limits_{\text{x}\rightarrow0}\frac{\text{bx}^2}{\text{bx}\sqrt{\text{x}}}\times\frac{1}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c}$

$\lim\limits_{\text{x}\rightarrow0}\sqrt{\text{x}}\times\frac{1}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c}$

$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x}}}{\sqrt{\text{x}\Big(1+\text{bx}^\frac{3}{2}\Big)}+\sqrt{\text{x}}}=\text{c}$

$\lim\limits_{\text{x}\rightarrow0}{\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}\Bigg[\sqrt{\Big(1+\text{b}^\frac{3}{2}}\Big)+1\Bigg]}}=\text{c}$

$\lim\limits_{\text{x}\rightarrow0}\frac{1}{\Bigg[\sqrt{\Big(1+\text{bx}^{\frac{3}{2}}\Big)}+1\Bigg]}=\text{c}$

$\text{c}=\frac{1}{2}$

$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a+1})\text{x}+\sin\text{x}}{\text{x}}=\text{c}$

$\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\bigg[\frac{(\text{a+1)x+x}}{2}\bigg]\cos\bigg(\frac{\text{ax}}{2}\bigg)}{\text{x}}=\frac{1}{2}$

$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\Big[\frac{(\text{a+2)x}}{2}\Big]}{\frac{(\text{a+2)x}}{2}}\times\frac{(\text{a}+2)}{2}\cos\Big(\frac{\text{ax}}{2}\Big)=\frac{1}{4}$

$1\times\frac{(\text{a+2)}}{2}\times1=\frac{1}{4}$

$\text{a}+2=\frac{1}{2}$

$\text{a}=\frac{-3}{2}$

$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}}$ exist if $\text{b}\neq0$

$\text{b }\in\text{ R}-(0)$

1. $-\frac{\text{y}}{\text{x}}$

Solution:

We have, $3\sin(\text{xy})+4\cos(\text{xy})=5$

$\Rightarrow3\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big]-4\sin(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big]=0$

$\Rightarrow\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big]\big[3\cos(\text{xy})-4\sin(\text{xy})\big]=0$

$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$

$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=-\text{y}$

$\therefore\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$

4. $\frac{3\pi}{4}$

Solution:

$\text{f}(\text{x})=\text{e}^{\text{x}}\sin\text{x}$

$\text{f}'(\text{x})=\text{e}^{\text{x}}\cos\text{x}+\text{e}^{\text{x}}\sin\text{x}$

$\text{f}'(\text{c})=0$

$\text{e}^\text{c}(\cos\text{c}+\sin\text{c})=0$

$\cos\text{c}+\sin\text{c}=0$

$\cos\text{c}=-\sin\text{c}$

$\tan\text{c}=-1$

$\text{c}=\frac{3\pi}{4}\in(0,\pi)$

3. 2

Solution:

For f(x) to be continuous at x = 0, we must have

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f}(0)$

$\Rightarrow\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\frac{(27-2\text{x})^\frac{1}{3}-3}{9-3(243+5\text{x})^\frac{1}{5}}$

$\Rightarrow\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{3\Big(243^{\frac{1}{5}}-(243+5\text{x})^\frac{1}{5}\Big)}$

$=\frac{1}{3}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{\text{x}}}{\frac{\bigg(243^\frac{1}{5}-(243+5\text{x})^\frac{1}{5}\bigg)}{\text{x}}}$

$=\frac{-1}{3}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{\text{x}}}{\frac{\bigg((243+5\text{x})^{\frac{1}{5}}-243\frac{1}{5}\bigg)}{\text{x}}}$

$=\frac{2}{15}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{-2\text{x}}}{\frac{\bigg((243+5\text{x})^\frac{1}{5}-243^\frac{1}{5}\bigg)}{5\text{x}}}$

$=\frac{2}{15}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{27-2\text{x}-27}}{\frac{\bigg((243+5\text{x})^\frac{1}{5}{-243^{\frac{1}{5}}\bigg)}}{243+5\text{x}-243}}$

$=\frac{2}{15}\times\frac{\frac{1}{3}\times27^{\frac{-2}{3}}}{\frac{1}{5}\times243^\frac{-4}{5}}$

$=\frac{2}{15}\times\frac{\frac{1}{3}\times\frac{1}{27^{\frac{-2}{3}}}}{\frac{1}{5}\times\frac{1}{243^\frac{-4}{5}}}$

$=2$

2. $\frac{\cos\text{x}}{2\text{y}-1}$

Solution:

$\text{y}=\sqrt{\sin\text{x}+\text{y}}$

$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\times\Big(\cos\text{x}+\frac{\text{dy}}{\text{dx}}\Big)$

$\frac{\text{dy}}{\text{dx}}=\frac{\cos}{2\sqrt{\sin\text{x}+\text{y}}}+\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\frac{\text{dy}}{\text{dx}}$

$\frac{\text{dy}}{\text{dx}}-\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\frac{\text{dy}}{\text{dx}}=\frac{\cos}{2\sqrt{\sin\text{x}+\text{y}}}$

$\Big(1-\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\sqrt{\sin\text{x}+\text{y}}}$

$\Big(1-\frac{1}{2\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}}$

$\Big(\frac{2\text{y}-1}{2\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}}$

$\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}-1}$

2. Not continuous at x = -2

Solution:

$\lim\limits_{\text{x}\rightarrow-2}\frac{|\text{x}+2|}{\tan^{-1}(\text{x}+2)}$

Let, x = -2 + h

x → -2 ⇒ h → 0

$\lim\limits_{\text{h}\rightarrow0}\frac{|-2+\text{h}+2|}{\tan^{-1}(-2+\text{h}+2)}$

$\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\tan^{-1}\text{h}}=1$

$\lim\limits_{\text{h}\rightarrow0}\frac{|\text{x}+2|}{\tan^{-1}(\text{x}+2)}\neq\text{f}(-2)$

Function is not continuous at x = -2.

4. Neither differentiable nor continuous at x = 3

Solution:

Given function can be writter as

$\text{f(x)}=-\text{x}+9\ \text{ x}<3$

$=\text{x}+4\ \text{ x}>3$

$\lim\limits_{\text{x}\rightarrow3^{+}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}-\text{x}+9=6$

$\lim\limits_{\text{x}\rightarrow3^{-}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}\text{x}+4=7$

Function is not continuous at x = 3

⇒ Function is not diffentiable at x = 3.

4. a = 1, b = -1

Solution:

Given,

$\text{f(x)=}\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if }\text{ x} < 4\\\text{a}+\text{b},&\text{if }\text{ x} =4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if }\text{ x} > 4\end{cases}$

We have

 $(\text{LHL at x = 4})=\lim\limits_{\text{x}\rightarrow4^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(4-\text{h})$

$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{4- \text{h}-4}{|4-\text{h}-4|}+\text{a}\Big)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{-\text{h}}{|-\text{h}|}+\text{a}\Big)=\text{a}-1$

$(\text{RHL at x = 4})=\lim\limits_{\text{x}\rightarrow4^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f(4+h)}$

$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{4+\text{h}-4}{|4+\text{h}-4|}+\text{b}\Big)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{h}}{|\text{h}|}+\text{b}\Big)=\text{b}+1$

Also,

$\text{f}(4)=\text{a+b}$

if(x) is continuous at x = 4, then

$\lim\limits_{\text{x}\rightarrow4^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow4^+}\text{f(x)}=\text{f}(4)$

$\Rightarrow\text{a}-1=\text{b}+1=\text{a + b}$

$\Rightarrow\text{a}-\text{1}=\text{a + b}$ and $\text{b}+1=\text{a + b} $

$\Rightarrow\text{b}=-1$ and $\text{a}=1$

2. $\text{a}=0,\text{b}=0;\text{c}\in\text{R}$

We have,

$\text{f(x)}=\text{a}|\sin\text{x}|+\text{be}^{|\text{x}|}+\text{c|x|}^3$

$=\begin{cases}\text{a}\sin\text{x}+\text{bx}^\text{x}+\text{cx}^3 & 0<\text{x}<\frac{\pi}{2}\\-\text{a}\sin\text{x}+\text{be}^{-\text{x}}-\text{cx}^3 & -\frac{\pi}{2}<\text{x}<0\end{cases}$

Here, f(x) is differentiable at x = 0

Therefore, (LHL at x = 0) = (RHL at x = 0)

$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{-\text{a}\sin\text{x}+\text{be}^{-\text{x}}-\text{cx}^3-\text{b}}{\text{x}}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{a}\sin\text{x}+\text{be}^{\text{x}}-\text{cx}^3-\text{b}}{\text{x}}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{-\text{a}\sin(0-\text{h})+\text{be}^{-(0-\text{h)}}-\text{c}(0-\text{h})^3-\text{b}}{0-\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\sin(0+\text{h})+\text{be}^{(0+\text{h)}}+\text{c}(0+\text{h})^3-\text{b}}{0+\text{h}}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\sin\text{h}+\text{be}^{\text{h}}+\text{ch}^3-\text{b}}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\sin\text{h}+\text{be}^{\text{h}}+\text{ch}^3-\text{b}}{\text{h}} $

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\cos\text{h}+\text{be}^{\text{h}}+3\text{ch}^2}{-1}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\cos\text{h}+\text{be}^{\text{h}}+3\text{ch}^2}{1} $ (By L'Hospital rule)

$\Rightarrow-(\text{a}+\text{b})=\text{a}+\text{b}$

$\Rightarrow-2(\text{a}+\text{b})=0$

$\Rightarrow\text{a}+\text{b}=0$

This is true for all value of c

$\therefore\text{c}\in\text{R}$

In the given option (b) satisfies a + b = 0 and $\text{c}\in\text{R.}$

2. $\sqrt3$

Solution:

We have

$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}=\frac{\text{x}^2+1}{\text{x}}$

Clearly, f(x) is continuous on [1, 3] and derivable on (1, 3).

Thus, both the conditions of Lagrange's theorem is satisfied.

Concequently there exists $\text{c}\in(1,3)$ such that

$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}=\frac{\text{f}(3)-\text{f}(1)}{2}$

Now, $\text{f}(\text{x})=\frac{\text{x}^2+1}{\text{x}}$

$\text{f}'(\text{x})=\frac{\text{x}^2-1}{\text{x}^2},\text{f}(1)=2,\text{f}(3)=\frac{10}{3}$

$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(1)}{2}$

$\Rightarrow\frac{\text{x}^2-1}{\text{x}^2}=\frac{4}{6}$

$\Rightarrow\frac{\text{x}^2-1}{\text{x}^2}=\frac{2}{3}$

$\Rightarrow3\text{x}^2-3=2\text{x}^2$

$\Rightarrow\text{x}=\pm\sqrt3$

Thus, $\text{c}=\sqrt3\in(1,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}.$

1. R - {3}

Solution:

(LHL at x = 3) $=\lim\limits_{\text{x}\rightarrow3^{-}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$

(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3-\text{h})-\text{f}(3)}{3-\text{h}-3}$

(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3-\text{h})-\text{f}(3)}{-\text{h}}$

(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{|3-\text{h}-3|\cos(3-\text{h})-\text{f}(3)}{-\text{h}}$

(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}\cos(3-\text{h})-0}{-\text{h}}=-\cos3$

(RHL at x = 3) $=\lim\limits_{\text{x}\rightarrow3^{+}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$

(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3+\text{h})-\text{f}(3)}{3+\text{h}-3}$

(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3+\text{h})-\text{f}(3)}{\text{h}}$

(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{|3+\text{h}-3|\cos(3+\text{h})-\text{f}(3)}{\text{h}}$

(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}\cos(3+\text{h})-0}{\text{h}}=\cos3$

So, f(x) is not diffrentiable at x = 3.

Also,f(x) is diffrentiable at all other points because both modulus and cosine function are differentiable and the product of two differentiable function is differentiable.

2. $\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{y}_1$

Solution:
$\text{y}=\text{a}+\text{bx}^2$

$\Rightarrow\text{y}_1=2\text{bx}$

$\Rightarrow\text{y}_2=2\text{b}$

Multiply by x on both sides we get

$\text{xy}_2=2\text{bx}=\text{y}_1$

$\Rightarrow\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{y}_1$

2. $\frac{1}{2}$

Solution:

$\text{f}\big(\frac{\pi}{2}\big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\tan\Big(\frac{\pi}{4}-\text{x}\Big)}{\cot2\text{x}}$

$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\frac{1-\tan\text{x}}{1+\tan\text{x}}}{\cot2\text{x}}$

$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\tan2\text{x}\Big(\frac{1-\tan\text{x}}{1+\tan\text{x}}\Big)$

$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2\tan\text{x}}{1-\tan^2\text{x}}\Big(\frac{1-\tan\text{x}}{1+\tan\text{x}}\Big)$

$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2\tan\text{x}}{(1-\tan\text{x})(1+\tan\text{x})}\Big(\frac{1-\tan\text{x}}{1+\tan\text{x}}\Big)$ $\begin{pmatrix}\because\tan\frac{\pi}{4}\rightarrow1\\1-\tan\frac{\pi}{4}\neq0 \end{pmatrix}$

$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2\tan\text{x}}{(1+\tan\text{x})^2}=\frac{2}{4}=\frac{1}{2}$

1. -(n - 1)²y

Solution:

Here,

$\text{y}=\text{x}^{\text{n}-1}\log\text{x}$

$\Rightarrow\text{y}_1=(\text{n}-1)\text{x}^{\text{n-2}}\log\text{x}+\frac{\text{x}^{\text{n}-1}}{\text{x}}$

$\Rightarrow\text{y}_1=\frac{(\text{n}-1)\text{x}^{\text{n}-1}\log\text{x}+\text{x}^{\text{n}-1}}{\text {x}}$

$\Rightarrow\text{xy}_1=(\text{n}-1)\text{y}+\text{x}^{\text{n-1}}$

$\Rightarrow\text{xy}_2+\text{y}_1=(\text{n}-1)\text{y}_1+(\text{n}-1)\text{x}^{\text{n-2}}$

$\Rightarrow\text{xy}_2+\text{y}_1=(\text{n}-1)\text{y}_1+\frac{(\text{n}-1)\text{x}^{\text{n}-1}}{\text{x}}$

$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)\text{x}^{\text{n}-1}$

$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)\{\text{xy}_1-(\text{n}-1)\text{y}\}$

$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)\text{xy}_1-(\text{n}-1)^2\text{y}$

$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=2\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)^2\text{y}$

$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1(1-2\text{n}+2)=-(\text{n}-1)^2\text{y}$

$\Rightarrow\text{x}^2\text{y}_2+(3-2\text{n})\text{xy}_1=-(\text{n}-1)^2\text{y}$

3. e

Solution:

Given, $\text{f(x)}=(\text{x+1})^{\cot\text{x}}$

$\log\text{f(x)}=(\cot\text{x})(\log(\text{x+1}))$ [Taking log on both sides]

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\log\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}(\cot\text{x})(\log(\text{x+1}))$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\log\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(\text{x+1})}{\tan\text{x}}\Big)$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\log\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\frac{\Big(\frac{\log(\text{x+1})}{\text{x}}\Big)}{\Big(\frac{\tan\text{x}}{\text{x}}\Big)}$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\log\text{f(x)}=\frac{\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(\text{x+1})}{\text{x}}\Big)}{\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\tan\text{x}}{\text{x}}\Big)}$

$\Rightarrow\log\Big(\lim\limits_{\text{x}\rightarrow0}\text{f(x)}\Big)=\frac{\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(\text{x+1})}{\text{x}}\Big)}{\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\tan\text{x}}{\text{x}}\Big)}$ $[\because$ f(x) is continuous at x =0$]$

$\Rightarrow\log\Big(\lim\limits_{\text{x}\rightarrow0}\text{f(x)}\Big)=1$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{e}$

$\Rightarrow\text{f(0)}=\text{e} $ $[\because$f(x) is continuous at x = 0$]$

1. $\text{e}^{\frac{1}{1}-\text{e}}$

Solution:

Given:

$\text{y}=\text{f}(\text{x})=\text{x}\log\text{x}$

Differentiating the given function with respect to x, we get

$\text{f}'(\text{x})=1+\log\text{x}$

⇒ Slope of the tangent to the curve $=1+\log\text{x}$

Also,

Slope of the chord joining the points (1, 0) and (e, e), $(\text{m})=\frac{\text{e}}{\text{e}-1}$

The tangent to the curve is parallel to chord joining the points (1, 0) and (e, e).

4. none of these

Solution:

If f(x) is continuous at x = 0, then

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f}(0)$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Big(\sin\frac{1}{\text{x}}\Big)=\text{k}$

But $\lim\limits_{\text{x}\rightarrow0}\Big(\sin\frac{1}{\text{x}}\Big)$ does not exist. Thus, there dose not exist any k thet makes f(x) a continuous function.

2. $2^{20}(\cos2\text{x}+2^{20}\cos4\text{x})$

Solution:

$\text{y}=2\cos\text{x}\cos3\text{x}=\cos(3\text{x}-\text{x})+\cos(3\text{x}+\text{x})=\cos2\text{x}+\cos4\text{x}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=-2\sin2\text{x}-4\sin4\text{x}=-2(\sin2\text{x}+2\sin4\text{x})$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4\cos2\text{x}-16\cos4\text{x}=-2^2(\cos2\text{x}+2^2\cos4\text{x})$

$\Rightarrow\frac{\text{d}^3\text{y}}{\text{dx}^3}=2^3(\sin2\text{x}+2^3\sin4\text{x})$

$\Rightarrow\frac{\text{d}^4\text{y}}{\text{dx}^4}=2^3(2\cos2\text{x}+4\times2^3\cos4\text{x})=2^4(\cos2\text{x}+2^4\cos4\text{x})$

$\therefore\frac{\text{d}^{20}(\cos2\text{x}+\cos4\text{x})}{\text{dx}^{20}}=2^{20}(\cos2\text{x}+2^{20}\cos4\text{x})$

1. xy₁ + 2

Solution:

Here,

$\text{y}=(\sin^{-1}\text{x})\frac{1}{\sqrt{1-\text{x}^2}}$

$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)^\frac{3}{2}}$

$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$

$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{\text{xy}_1}{(1-\text{x}^2)}$

$\Rightarrow\text{y}_2(1-\text{x}^2)=2+\text{xy}_1$

3. $16\sqrt{2}\text{ in }6\text{ in }3$

Solution:

$\text{k}=\lim\limits_{\text{x}\rightarrow0}\frac{36^{\text{x}}-9^{\text{x}}-4^{ \text{x}}+1}{\sqrt2-\sqrt{1+\cos\text{x}}}$

consider,

$=\lim\limits_{\text{x}\rightarrow0}\frac{36^{\text{x}}-9^{\text{x}}-4^{\text{x}}+1}{\sqrt{2}-\sqrt{1+\cos\text{x}}}$

$=\lim\limits_{\text{x}\rightarrow0}\frac{(4\times9)^{\text{x}}-9^{\text{x}}-4^{\text{x}}+1}{2-(1+\cos\text{x})}\times\frac{\sqrt{2}+\sqrt{1+\cos\text{x}}}{1}$

$=\lim\limits_{\text{x}\rightarrow0}\frac{(4^{\text{x}}\times9^{\text{x}})-9^{\text{x}}-4^{\text{x}}+1}{2-(1+\cos\text{x})}\times\frac{\sqrt{2}+\sqrt{1+\cos\text{x}}}{1}$

$=\lim\limits_{\text{x}\rightarrow0}\frac{(9^{\text{x}}-1)(4^{\text{x}}-1)(\sqrt{2}+\sqrt{1+\cos\text{x}})}{1-\cos\text{x}}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}$

$=\lim\limits_{\text{x}\rightarrow0}\frac{(9^{\text{x}}-1)(4^{\text{x}}-1)(\sqrt{2}+\sqrt{1+\cos\text{x}})(1+\cos\text{x})}{1-\cos^2\text{x}}$

$=\lim\limits_{\text{x}\rightarrow0}\frac{(9^{\text{x}}-1)(4^{\text{x}}-1)(\sqrt{2}+\sqrt{1+\cos\text{x}})(1+\cos\text{x})}{\sin^2\text{x}}$

dividing by x²

$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{9^{\text{x}}-1}{\text{x}}{\times\frac{4^{\text{x}}-1}{\text{x}}\times\big(\sqrt{2}+\sqrt{1+\cos\text{x}}\big)(1+\cos\text{x})}}{\frac{\sin^2\text{x}}{\text{x}^2}}$

$=(\log9)(\log4)\big(\sqrt{2}+\sqrt{1+1}\big)(1+1)$

$=4\sqrt{2}(\log9)(\log4)$

$=4\sqrt{2}(2\log3)(2\log2)$

$=16\sqrt{2}(\log3)(\log2)$

1. $\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}+1}$

Solution:

Let $\text{y}=\Big(1+\frac{1}{\text{x}}\Big)^\text{x}$

Taking log on both sides,

$\log\text{y}=\text{x}\log\Big(1+\frac{1}{\text{x}}\Big)$

$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log\Big(1+\frac{1}{\text{x}}\Big)+\log\Big(1+\frac{1}{\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\text{x})$

$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\bigg(\frac{1}{1+\frac{1}{\text{x}}}\bigg)\frac{\text{d}}{\text{dx}}\Big(1+\frac{1}{\text{x}}\Big)+\log\Big(1+\frac{1}{\text{x}}\Big)$

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

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

$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{-1}{\text{x}+1}+\log\Big(1+\frac{1}{\text{x}}\Big)$

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

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

2. 1

Solution:

We have, $\text{y}=\log\sqrt{\tan\text{x}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{\tan\text{x}}}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\tan\text{x}}\big)$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{\tan\text{x}}}\times\frac{1}{2\sqrt{\tan\text{x}}}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\tan\text{x}}\big)$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sec^2\text{x}}{2\tan\text{x}}$

Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{\Big[\sec\big(\frac{\pi}{4}\big)\Big]^2}{2\tan\big(\frac{\pi}{4}\big)}=\frac{2}{2\times1}=1$

2. Differentible at x = -1

Solution:

$\text{f(x)}=\begin{cases}1, & \text{x}\leq-1\\|\text{x}|, & -1 <\text{x} <1\\0,&\text{x}\geq1\end{cases}$

$\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(-1)}{\text{x}+1}=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{-\text{x}+1}{\text{x}+1}=0$

Similarly,

$\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{f(x)}-\text{f}(-1)}{\text{x}+1}=\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{x}+1}{\text{x}+1}=0$

Function is diffrentiable at x = -1.

2. -n²x

Solution:

Here

$\text{x}=\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$

Differentiating w.r.t.t, we get

$\frac{\text{dx}}{\text{dt}}=-\text{an}\sin\text{nt}-\text{bn}\cos\text{nt}$

Differentiating w.r.t.t, we get

$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{an}^2\cos\text{nt}+\text{bn}^2\sin\text{nt}$

$=-\text{n}^2\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$

$=-\text{n}^2\text{x}$

1. $\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$

Solution:

We have, $\sin\text{y}=\text{x}\cos(\text{a}+\text{y})$

$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin\text{y})=\frac{\text{d}}{\text{dx}}\big[\text{x}\cos(\text{a}+\text{y})\big]$

$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=1\times\cos(\text{a}+\text{y})-\text{x}\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{a}+\text{y})$

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

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

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

$\Rightarrow\Big[\cos\text{y}+\frac{\sin\text{y}}{\cos(\text{a}+\text{y})}\times\sin(\text{a}+\text{y})\Big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$

$\begin{bmatrix}\because\sin\text{y}=\text{x}\cos(\text{a}+\text{y}) \\ \because\text{x}=\frac{\sin\text{y}}{\cos(\text{a}+\text{y})} \end{bmatrix}$

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

$\Rightarrow\frac{\cos(\text{a}+\text{y}-\text{y})}{\cos(\text{a}+\text{y})}\times\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$

3. Discontinuous at exactly three points.

Solution:

We have, $\text{f(x)}=\frac{4-\text{x}^2}{4\text{x}-\text{x}^3}=\frac{(4-\text{x}^2)}{\text{x}(4-\text{x}^2)}$

$=\frac{(4-\text{x}^2)}{\text{x}(2^2-\text{x}^2)}=\frac{4-\text{x}^2}{\text{x}(2+\text{x})(2-\text{x})}$

Clearly, f(x) is discontinuous at exactly three points x = 0, x = -2 and x = 2.

3. Continuous at non-integer points only.

Solution:

f(x) = x - x

Consider n be an integer.

$\text{f(x)}=\text{x}-[\text{x}]=\begin{cases}\text{x}-(\text{n}-1)&\text{n}-1\leq\text{x}<\text{n}\\0&\text{x}=\text{n}\\\text{x}-\text{n}&\text{n}\leq\text{x}<\text{n}+1\end{cases}$

Now,

LHL at x = n

$=\lim\limits_{\text{x}\rightarrow\text{n}^{-}}\text{f(x)}=\text{x}-\text{n}-1=\text{x}-\text{n}+1$

RHL at x = n 

$=\lim\limits_{\text{x}\rightarrow\text{n}^{+}}\text{f(x)}=\text{x}-\text{n}=\text{x}-\text{nAs},$

$\text{LHL}\neq\text{RHL}$ at x = n

i.e., given function is not continuous at n.

Now, n is any integer. Therefore, given function is not continuous at integers.

Therefore, given points are continuous at non-integer points only.

1. $\big\{\text{x}=\text{n}\pi:\text{n}\in\text{Z}\big\}$

Solution:

Consider, $\text{f(x)}=\cos\text{x}=\frac{\cos\text{x}}{\sin\text{x}}$

We know that, $\big[\sin\text{x}=0\text{ at }\text{x}=\text{n}\pi,\text{n}\in\text{Z}\big]$

Hence, $\text{f(x)}=\cot\text{x}$ is discontinuous on the set $\big\{\text{x}=\text{n}\pi:\text{n}\in\text{Z}\big\}.$ 

4. $-\frac{1}{2\text{at}^3}$

Solution:

$\text{x}=\text{at}^2,\text{y}=2\text{at}$

$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2\text{a}}{2\text{at}}$

$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{d}}{\text{dt}}\Big(\frac{\text{dy}}{\text{dx}}\Big)}{\frac{\text{dx}}{\text{dt}}}=\frac{\frac{-1}{\text{t}^2}}{2\text{at}}=\frac{-1}{2\text{at}^3}$

3. f(x) is everywhere continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$

Solution:

$\text{f}(\text{x)} = |\cos\text{x}|$

Given function is trigonometric function.

⇒ Hence, it is continuous.

Function is not differentiable at odd multiples of $\frac{\pi}{2}.$

⇒ f(x) is not differentiable at $\text{x} = (2\text{n} + 1) \frac{\pi}{2}$

2. Everywhere continuous but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$

Solution:

As cos x is even function it is continuous everywhere but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$

$\cos\Big[(2\text{n}+1)\frac{\pi}{2}=\cos\Big(\text{n}\pi+\frac{\pi}{2}\Big)=-\sin\text{n}\pi$

For n as an integer $\Rightarrow\sin\text{n}\pi=0$

For n as rational $\Rightarrow\sin\text{n}\pi=-1$

2. f(x) is everywhere continuous but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}$

Solution:

$\text{f(x)}=|\sin\text{x}|$

Given function is continuous and differentiable on $(2\text{n}\pi,(2\text{n}+1)\pi)$

But not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}.$

As $\sin\text{n}\pi=0$ for $\text{n}\in\text{Z}.$

4. $\frac{3}{2}$

Solution:

We have

 f(x) = x(x - 2)

It can be rewritten as f(x) = x² - 2x

We know that a polynomial function is everywhere continuous and differentiable.

Since, f(x) is polynomial, it is continuous on [1, 2] and differentiable on [1, 2].

So, there must exist at least one real number $\text{c}\in(1,2)$ such that

$\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$

$=\frac{\text{f}(2)-\text{f}(1)}{1}$

Now, f(x) = x² - 2x

⇒ f'(x) = 2x - 2

and f(1) = -1, f(2) = 0

$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$

$\Rightarrow\text{f}(\text{x})=\frac{0+1}{1}$

$\Rightarrow2\text{x}-2=1$

$\Rightarrow\text{x}=\frac{3}{2}$

$\therefore\ \text{c}=\frac{3}{2}\in(1,2)$

2. Any interval $[0,\pi]$

Solution:

$\phi(\text{x})$ is continuous and differentiable function then using statement of Roll's theorem f(a) = f(b). Hence, here $\sin0=0$ also $\sin\pi=0.$

2. $\text{a}=-\frac{1}{2},\text{b}=\frac{3}{2}$

Solution:

Given function is continuous at x = 1.

$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{+}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^{-}}\text{f(x)}$

$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{1}{\text{x}}=\lim\limits_{\text{x}\rightarrow1^{-}}\text{ax}^2+\text{b}$

$\Rightarrow1=\text{a}+\text{b}\ \dots(1)$

Function is derivable at x = 1.

$\Rightarrow\lim\limits_{\text{h}\rightarrow0^{+}}\frac{\text{f}(1+\text{h}-\text{f}(1))}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{f}(0+\text{h}-\text{f}(1))}{\text{h}}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0^{+}}\frac{\frac{1}{1+\text{h}}+1}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{a}(1+\text{h})^2-\text{a}}{\text{h}}$

$\Rightarrow-1=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{h}(2\text{a}+\text{h})}{\text{h}}$

$\Rightarrow2\text{a}=-1$

$\Rightarrow\text{a}=\frac{-1}{2}$

$\text{a}+\text{b}=1\ (\text{From(1)})$

$\frac{-1}{2}+\text{a}=1$

$\Rightarrow\text{b}=\frac{3}{2}$

1. $1$

Solution:

Consider, $\text{f(x)}=\text{x}^3-3\text{x}$

$\Rightarrow\ \text{f(x)}=3\text{x}^2-3$

$\Rightarrow\ \text{f}'(\text{c})=3\text{c}^2-3$

Now, f(c) = 0

$\Rightarrow\ 3\text{c}^2-3=0$

$\Rightarrow\ \text{c}^2=\frac{3}{3}=1$

$\Rightarrow\ \text{c}=\pm1$ where $1\in\big(0,\sqrt{3}\big)$

3. 0

Solution:

$\text{y}=\frac{1}{1+\text{x}^{\text{a}-\text{b}}+\text{x}^{\text{c}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}+\text{x}^{\text{a}-\text{c}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}+\text{x}^{\text{c}-\text{a}}}$

$\text{y}=\frac{1}{1+\frac{\text{x}^\text{a}}{\text{x}^\text{b}}+\frac{\text{x}^\text{c}}{\text{x}^\text{b}}}+\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{c}}+\frac{\text{x}^\text{a}}{\text{x}^\text{c}}}+\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{a}}+\frac{\text{x}^\text{c}}{\text{x}^\text{a}}}$

$\text{y}=\frac{\text{x}^\text{b}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}+\frac{\text{x}^\text{c}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}+\frac{\text{x}^\text{a}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}$

$\text{y}=\frac{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}$

$\text{y}=1$

$\frac{\text{dy}}{\text{dx}}=0$

2. $-\frac{1}{2}$

Solution:

Given, $\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{px}}-\sqrt{1-\text{px}}}{\text{x}},&\text{if }0\leq\text{x}<0\\\frac{2\text{x}+1}{\text{x}-2},&\text{if }0\leq\text{x}\leq1\end{cases}$

If f(x) is continuous at x = 0,then

$\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\text{f(0)}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim\limits_{\text{h}\rightarrow0}\text{f(h)}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}}{-\text{h}}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}{-\text{h}\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{(1-\text{ph}-1-\text{ph})}{-\text{h}\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{(-2\text{ph})}{-\text{h}\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{(2\text{p})}{\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$

$\Rightarrow\Big(\frac{(2\text{p})}{(2)}\Big)=\Big(\frac{1}{-2}\Big)$

$\Rightarrow\text{p}=\frac{-1}{2}$

3. -2x + 9, if 4 < x < 5

Solution:

We have, f(x) = |x² - 9x + 20|

$\text{f}(\text{x})=\begin{Bmatrix} \text{x}^2-9\text{x}+20, & -\infty<\text{x}\leq4 \\ -\big(\text{x}^2-9\text{x}+20\big), & 4<\text{x}<5 \\ \text{x}^2-9\text{x}+20, & 5\leq\text{x}<\infty \end{Bmatrix}$

$\Rightarrow\text{f}'(\text{x})=\begin{Bmatrix} 2\text{x}-9\text{x}, & -\infty<\text{x}\leq4 \\ 2\text{x}-9, & 4<\text{x}<5 \\ 2\text{x}-9, & 5\leq\text{x}<\infty \end{Bmatrix}$

$\therefore$ f'(x) = -2x + 9 for 4 < x < 5

2. 50

Solution:

If f(x) is continuous at x = 0, then

$\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\text{f}(0)$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})=\text{f}(0)$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{(1-\cos(-10\text{h}))}{(-\text{h})^2}=\text{f}(0)$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{(1-\cos(10\text{h}))}{\text{h}^2}=\text{f}(0)$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{(2\sin^2(5\text{h}))}{\text{h}^2}=\text{a}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{2\times25(\sin^2(5\text{h}))}{25\text{h}^2}=\text{a}$

$\Rightarrow50\lim\limits_{\text{h}\rightarrow0}\frac{(\sin^2(5\text{h}))}{(5\text{h})^2}=\text{a}$

$\Rightarrow50\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin(5\text{h})}{5\text{h}}\Big)^2=\text{a}$

$\Rightarrow\text{a}=50$

3. $-\frac{1}{64}$

Solution:

if f(x) is continuous at $\text{x}=\frac{\pi}{2},$ then

$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\text{f(x)}=\text{f}(\frac{\pi}{2})$

$\text{f }\frac{\pi}{2}-\text{x = t},$ then

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\text{f}\big(\frac{\pi}{2}-\text{t}\big)=\text{f}(\frac{\pi}{2})$

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\Bigg(\frac{1-\sin(\frac\pi{2}-\text{t}{})}{4\text{t}^2}\times\frac{\log\sin(\frac{\pi}{2})}{\log\big(1+\pi^2-4\pi\big(\frac{\pi}{2}-\text{t}\big)+4\big(\frac{\pi}{2}-\text{t}\big)^2\big)}\Bigg)=\text{k}$

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\Bigg(\frac{(1-\cos\text{t})}{4\text{t}^2}\times\frac{\log\cos\text{t}}{\log\big(1+\pi^2-2\pi^2+4\pi\text{t}+4\big(\frac{\pi^2}{4}+\text{t}^2-\pi\text{t}\big)}\Bigg)=\text{k}$

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{(1-\cos\text{t})}{4\text{t}^2}\times\frac{\log\cos\text{t}}{\log(1-\pi^2+4\pi\text{t})+(\pi^2+4\text{t}^2-4\pi\text{t}}\bigg)=\text{t}$

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{(1-\cos\text{t})}{4\text{t}^2}\times\frac{\log \cos\text{t}}{\log(1+4\text{t}^2)}\bigg)=\text{k}$

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{2\sin^2\frac{\text{t}}{2}}{16\times\frac{\text{t}^2}{4}}\times\frac{\log\cos\text{t}}{\log(1+4\text{t}^2)}\bigg)=\text{k}$

$\Rightarrow\frac{2}{16}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{\bigg(\frac{\text{t}^2}{4}\bigg)}\times\frac{\log\cos\text{t}}{\bigg(\frac{4\text{t}^2\log(1+4\text{t}^2)}{4\text{t}^2}\bigg)} \end{pmatrix}=\text{k}$

$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{(\frac{\text{t}}{2})^2}\times\frac{\frac{\log\cos\text{t}}{4\text{t}^2}}{\bigg({\frac{\log(1+4\text{t}^2)}{4\text{t}}\bigg)}} \end{pmatrix}=\text{k}$

$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{(\frac{\text{t}}{2})}\times\frac{\frac{\log\sqrt{1-\sin^2\text{t}}}{4\text{t}^2}}{\bigg({\frac{\log(1+4\text{t}^2)}{4\text{t}^2}\bigg)}} \end{pmatrix}=\text{k}$

$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{\big(\frac{\text{t}}{2}\big)^2}\times\frac{\bigg(\frac{\log(1-\sin^2\text{t})}{(8\text{t}^2)}\bigg)}{\bigg({\frac{\log(1+4^2\text{t})}{4\text{t}^2}}\bigg)} \end{pmatrix}=\text{k}$

$\Rightarrow\frac{1}{64}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{\big(\frac{\text{t}}{2}\big)}\times\frac{\bigg(\frac{\log(1-\sin^2\text{t})}{\text{t}^2}\bigg)}{\bigg(\frac{\log(1+4\text{t}^2)}{4\text{t}^2}\bigg)} \end{pmatrix}=\text{k}$

$\Rightarrow\frac{1}{64}\begin{pmatrix}\lim\limits_{\text{t}\rightarrow0}\Bigg(\frac{\sin\frac{\text{t}}{2}}{\big({\frac{\text{t}}{2}\big)}}\Bigg)^2\times\frac{\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{\log(1-\sin^2\text{t})}{\text{t}^2}\bigg)}{\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{\log(1+4\text{t}^2)}{4\text{t}^2}\bigg)} \end{pmatrix}=\text{k}$

$\Rightarrow\frac{1}{64}\bigg(1\times\lim\limits_{\text{t}\rightarrow0}\frac{(-\sin^2\text{t})\log(1-\sin^2\text{t})}{\text{t}^2(-\sin^2\text{t})}\bigg)=\text{k}$

$\Rightarrow\frac{-1}{64}\bigg(\lim\limits_{\text{t}\rightarrow0}\frac{(\sin^2\text{t})\log(1-\sin^2\text{t})}{\text{t}^2(-\sin^2\text{t})}\bigg)=\text{k}$

$\Rightarrow\frac{-1}{64}\Bigg(\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{\sin\text{t}}{\text{t}}\bigg)^2\lim\limits_{\text{t}\rightarrow0}\frac{\log(1-\sin^2\text{t})}{(-\sin^2\text{t})}\Bigg)=\text{k}$

$\Rightarrow\frac{-1}{64}\Bigg(\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{\sin\text{t}}{\text{t}}\bigg)^2\lim\limits_{\text{t}\rightarrow0}\frac{\log(1-\sin^2\text{t})}{(-\sin^2\text{t})}\Bigg)=\text{k}$

$\Rightarrow\text{k}=\frac{-1}{64}$ $\bigg[\because\lim\limits_{\text{x}\rightarrow0}\frac{\log(1-\text{x})}{\text{x}}=1\bigg]$

1. $\Big(\frac{2}{3}\Big)^\frac{1}{2}$

Solution:

We have, $\text{y}=\cot^{-1}\Big(\sqrt{\cos2\text{x}}\Big)$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{1+\cot2\text{x}}\frac{\text{d}}{\text{dx}}\sqrt{\cos2\text{x}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\cos^2\text{x}}\times\frac{1}{2\sqrt{\cos2\text{x}}}\frac{\text{d}}{\text{dx}}(\cos2\text{x})$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\cos^2\text{x}}\times\frac{1}{2\sqrt{\cos2\text{x}}}\times-2\sin2\text{x}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin2\text{x}}{\cos^2\text{x}\times2\sqrt{\cos2\text{x}}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\sin\text{x}\cos\text{x}}{\cos^2\text{x}\times2\sqrt{\cos2\text{x}}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\tan\text{x}}{\sqrt{\cos2\text{x}}}$

So, at $\text{x}=\frac{\pi}{6},$ we get

$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{6}}=\frac{\tan\big(\frac{\pi}{6}\big)}{\sqrt{\cos2\big(\frac{\pi}{6}\big)}}=\frac{\big(\frac{1}{\sqrt{3}}\big)}{\sqrt{\frac{1}{2}}}=\Big(\frac{2}{3}\Big)^\frac{1}{2}$

4. -1

Solution:

$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}$

$\lim\limits_{\text{x}\rightarrow1}5\text{x}-4=\lim\limits_{\text{x}\rightarrow1}4\text{x}^2+3\text{ax}$

$1=4+3\text{a}$

$\text{a}=-1$

1. $\frac{1}{8}$

Solution:

If f(x) is continuous at $\text{x}=\frac{\pi}{2},$ then

$\lim\limits_{\text{x}\rightarrow\frac{\text{x}}{2}}\text{f}\text{(x)}=\text{f}\Big(\frac{\pi}{2}\Big)$

$\lim\frac{1-\sin\text{x}}{\text{x}\rightarrow\frac{\pi}{2}(\pi-2\text{x})^2}=\text{f}\Big(\frac{\pi}{2}\Big) \ ...(\text{i})$$$

suppose $\Big(\frac{\pi}{2}-\text{x}\Big)=\text{t},$ then

$\lim\limits_{\text{t}\rightarrow0}\begin{bmatrix}\frac{1-\sin\Big(\frac{\pi}{2}-\text{t}\Big)}{(2\text{t})^2} \end{bmatrix}=\text{f}\Big(\frac{\pi}{2}\Big)$ [From eq.(i)]

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\begin{bmatrix}\frac{1-\cos\text{t}}{(2\text{t})^2} \end{bmatrix}=\text{f}\Big(\frac{\pi}{2}\Big)$

$\Rightarrow\frac{1}{4}\lim\limits_{t\rightarrow0}\begin{bmatrix}\frac{2\sin^2\Big(\frac{\text{t}}{2}\Big)}{\text{t}^2} \end{bmatrix}=\text{f}\Big(\frac{\pi}{2}\Big)$

$\Rightarrow \frac{1}{4}\lim\limits_{\text{t} \rightarrow0} \begin{bmatrix}\frac{\frac{2}{4}\sin^{2}\big(\frac{\text{t}}{2}\big)}{\frac{\text{t}^{2}}{4}} \end{bmatrix} = \text{f} \Big(\frac{\pi}{2}\Big)$

$\Rightarrow \frac{1}{8}\lim\limits_{\text{t} \rightarrow0} \begin{bmatrix}\frac{\frac{2}{4}\sin^{2}\big(\frac{\text{t}}{2}\big)}{\frac{\text{t}^{2}}{4}} \end{bmatrix} = \text{f} \Big(\frac{\pi}{2}\Big)$

$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{bmatrix}\frac{\sin\Big(\frac{\text{t}}{2}\Big)}{\frac{\text{t}}{2}} \end{bmatrix}^2=\text{f}\Big(\frac{\pi}{2}\Big)$

$\Rightarrow\text{f}\Big(\frac{\pi}{2}\Big)=\lambda=\frac{1}{8}$

3. $(\text{x}\log\text{x})^{-1}$

Solution:

$\text{f(x)}=\log\text{x}$

$\Rightarrow\text{f}(\log\text{x})=\log(\log\text{x})$

$\Rightarrow\text{f}'(\log\text{x})=\frac{1}{\text{x}\log\text{x}}=(\text{x}\log\text{x})^{-1}$