Limits, Continuity and Differentiability

we have

$ \begin{aligned} & \lim _{x \rightarrow 1}\left(\frac{x+x^2+x^3+\ldots+x^n-n}{x-1}\right) \\ = & \lim _{x \rightarrow 1}\left[\frac{x+x^2+x^3+\ldots+x^n-n}{(x-1)}\right] \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 1}\left[\begin{array}{c} x+x^2+x^3+\ldots+x^n \\ -(1+1+\ldots+1 \ldots n \text { times }) \\ (x-1) \end{array}\right] \\ & =\lim _{x \rightarrow 1}\left[\frac{(x-1)+\left(x^2-1\right)+\left(x^3-1\right)+\ldots+\left(x^n-1\right)}{(x-1)}\right] \end{aligned} $

$ \begin{aligned} \ &\left.=\lim _{x \rightarrow 1}\left[\frac{(x-1)+\left(x^2-1^2\right)+\left(x^3-1^3\right)+\ldots+(x^n-1^n}{\left(x-1\right)}\right] \right. \end{aligned} $

So, divide each term is square bracket by $(x-1)$ as given in denominator and applying the limit, we get

$\mathop {\lim }\limits_{x \to 1} \left[ {{{(x - 1) + ({x^2} - {1^2}) + ({x^3} - {1^3}) + ... + ({x^n} - {1^n})} \over {(x - 1)}}} \right]$

$ \begin{aligned} & =\frac{1(1)^0+2(1)^1+3(l)^2+4(1)^3+\ldots+n(1)^{n-1}}{} \\ & {\left[\text { Formula used, } \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right]} \end{aligned} $

$ \begin{aligned} & =1+2+3+4+\ldots+n \\ & =\frac{n(n+1)}{2} \end{aligned} $

Given that the functions is continuous

$ \begin{aligned} & \Rightarrow \lim _{x \Rightarrow 0} f(x)=f(0) \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}=\frac{\sqrt{1+0}-1}{0}=\frac{0}{0} \\ & {\left[\text { which is } \frac{0}{0} \text { from }\right]} \end{aligned} $

So, we apply L-Hospital rule

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x}(\sqrt{1+x}-1)}{\frac{d}{d x} x}=\lim _{x \rightarrow 0} \frac{\frac{1}{2} \sqrt{1+x}}{1} \\ & =\lim _{x \rightarrow 0} \frac{1}{2 \sqrt{1+x}} \end{aligned} $

On applying the limit, we get

$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{1}{2 \sqrt{1+h}}=\frac{1}{2} \\ \therefore &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ f(0)=\frac{1}{2} \end{aligned} $

$ \begin{aligned} & \text { If } f(x)=\frac{5 x \cdot \operatorname{cosec}(\sqrt{x})-1}{(x-2) \operatorname{cosec}(\sqrt{x})} \\ & \text { then, } \lim _{x \rightarrow \infty} f\left(x^2\right)=\text { ? } \\ & f\left(x^2\right)=\frac{5\left(x^2\right) \operatorname{cosec}\left(\sqrt{x^2}\right)-1}{\left(x^2-2\right) \operatorname{cosec}\left(\sqrt{x^2}\right)} \\ & \lim _{x \rightarrow \infty} f\left(x^2\right)=\lim _{x \rightarrow \infty} \frac{5 x^2 \operatorname{cosec} x-1}{\left(x^2-2\right) \operatorname{cosec} x} \\ & =\lim _{x \rightarrow \infty} \frac{x^2 \operatorname{cosec} x\left(5-\sin x / x^2\right)}{x^2\left(1-\frac{2}{x^2}\right) \operatorname{cosec} x} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow \infty} \frac{\left[5-\frac{\sin x}{x^2}\right]}{\left(1-\frac{2}{x^2}\right)} \\ & \text { Here, } \frac{\sin x}{x^2} \rightarrow 0 \text { as } x \rightarrow \infty \\ & \Rightarrow \lim _{x \rightarrow \infty} \frac{5}{1-\frac{2}{x^2}}-\lim _{x \rightarrow \infty} \frac{\frac{\sin x}{x^2}}{1-\frac{2}{x^2}} \\ & \qquad \sin x=\frac{1}{\operatorname{cosec} x} \\ & \Rightarrow \lim _{x \rightarrow \infty} \frac{5}{1-2 / x^2}-\lim _{x \rightarrow \infty} \frac{\frac{1}{x^2 \operatorname{cosec} x}}{1-2 / x^2} \\ & \Rightarrow 5-0=5 \end{aligned} $

We have,

$ \lim \limits_{x \rightarrow 2} \frac{\sqrt{1+4 x}-\sqrt{3+3 x}}{x^3-8}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\frac{0}{0}) $

Using L,Hospital's rule,

$ \begin{aligned} & \lim _{x \rightarrow 2} \frac{\frac{1}{2} \frac{4}{\sqrt{1+4 x}}-\frac{1}{2} \frac{3}{\sqrt{3+3 x}}}{3 x^2} \\ & =\lim _{x \rightarrow 2} \frac{4 \sqrt{3+3 x}-3 \sqrt{1+4 x}}{6 x^2 \sqrt{3+3 x} \sqrt{1+4 x}} \\ & =\frac{4 \times 3-3 \times 3}{6 \times 4 \times 3 \times 3}=\frac{3}{72 \times 3}=\frac{1}{72} \end{aligned} $

Therefore, the limit is $1 / 72$.

$ \begin{aligned} & \quad[\sqrt{2 x+1}+\sqrt{2 x-1})^8 \\ & \lim _{x \rightarrow \infty} \frac{\left.+(\sqrt{2 x+1}-\sqrt{2 x-1})^8\right]\left(p x^4-18\right.}{\left(x+\sqrt{x^2-2}\right)^8+\left(x-\sqrt{x^2-2}\right)^8} =1 \end{aligned} $

Take the ' $x$ ' from numerator and denominator.

$ (x)^4\left(\sqrt{2+\frac{1}{x}}+\sqrt{2-\frac{1}{x}}\right)^8 $

$ \begin{aligned} & \lim _{x \rightarrow \infty} \frac{+\left(\sqrt{2+\frac{1}{x}}-\sqrt{2-\frac{1}{x}}\right)^8 \cdot x^4\left(p-\frac{16}{x}\right)}{x^8\left[\left(1+\sqrt{2+\frac{1}{x}}\right)^8+\left(1-\sqrt{2-\frac{1}{x}}\right)\right.} =1 \end{aligned} $

Canceled $x^8$ from numerator and denominator and put the value of $ x \rightarrow \infty $

$ \begin{aligned} & \frac{\left\{[\sqrt{(2+0)}+\sqrt{2-0}]^8+[\sqrt{2}-\sqrt{2}]^8\right\}\{P-0}{(1+\sqrt{1})^8+(1-\sqrt{1})^8} =1 \\ & \frac{\left[(\sqrt{2}+\sqrt{2})^8+(0)^8\right]\{P\}}{(2)^8+(0)^8}=1 \\ & \frac{(2 \sqrt{2})^8 P}{(2)^8}=1 \end{aligned} $

$ \begin{aligned} P=\frac{(2)^8}{(2 \sqrt{2})^8}=\frac{2^8}{2^8 \cdot(\sqrt{2})^8} \\ P=\frac{1}{2^4}=\frac{1}{16} \Rightarrow P=\frac{1}{16} \end{aligned} $

$ \begin{aligned} \lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x} \end{aligned} $

On putting the limit $x \rightarrow \frac{\pi}{4}$, we get $\frac{0}{0}$

which is an indeterminate form, using

$\begin{aligned} & \lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x} \\ & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{-5(\cos x+\sin x)^4 \cdot(-\sin x+\cos x)}{-\cos 2 x \cdot 2} \\ & =\frac{5}{2} \lim _{x \rightarrow \frac{\pi}{4}} \frac{(\cos x+\sin x)^4(\cos x-\sin x)}{\cos 2 x} \\ & =\frac{5}{2} \lim _{x \rightarrow \frac{\pi}{4}} \frac{(\cos x+\sin x)^4(\cos x-\sin x)}{\cos ^2 x-\sin ^2 x} \\ & =\frac{5}{2} \lim _{x \rightarrow \frac{\pi}{4}}(\cos x+\sin x)^3 \\ & =\frac{5}{2}\left(\cos \frac{\pi}{4}+\sin \frac{\pi}{4}\right)^3 \\ & =\frac{5}{2} \times 2 \sqrt{2}=5 \sqrt{2}\end{aligned}$

$\lim \limits_{x \rightarrow 0} \frac{e^x-a-\log (1+x)}{\sin x}=0$

$ \text { LHS }=\lim \limits_{x \rightarrow 0} \frac{\frac{\left\{e^x-a-\log (1+x)\right\}}{x}}{\frac{\sin x}{x}} $

Dividing by $x$ to both numerator and denominator

$\begin{aligned} & =\frac{\lim \limits_{x \rightarrow 0} \frac{e^x-a}{x}-\frac{\log (1+x)}{x}}{\lim \limits_{x \rightarrow 0} \frac{\sin x}{x}} \\ & =\frac{\lim \limits_{x \rightarrow 0} \frac{e^x-a}{x}-\lim \limits_{x \rightarrow 0} \frac{\log (1+x)}{x}}{\lim \limits_{x \rightarrow 0} \frac{\sin x}{x}} \\ & =\frac{\lim \limits_{x \rightarrow 0} \frac{e^x-a}{x}-1}{1}\end{aligned}$

$ \left[\because \lim \limits_{x \rightarrow 0} \frac{\log (1+x)}{x}=1, \lim \limits_{x \rightarrow 0} \frac{\sin x}{x}=1\right] $

$ =0=\text { RHS } $

$ \Rightarrow \quad \lim _{x \rightarrow 0} \frac{e^x-a}{x}-1=0 $

$ \Rightarrow \quad \lim \limits_{x \rightarrow 0} \frac{e^x-a}{x}=1 $

$ \therefore \quad a=1 \quad\left(\because \lim \limits_{x \rightarrow 0} \frac{e^x-1}{x}=1\right) $

Given,

$ f(x)=\left\{\begin{array}{cc}(1+|\sin x|)^{\frac{a}{|\sin x|}}, & -\frac{\pi}{6} < x < 0 \\ b, & x=0 \\ e^{\frac{\tan 2 x}{\tan 3 x}}, & 0 < x < \frac{\pi}{6}\end{array}\right. $

is continuous at $x=0$

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

Now,

$ \begin{gathered} \begin{aligned} \lim _{x \rightarrow 0^{-}}(1+|\sin x|)^{\frac{a}{|\sin x|}} & =e^{\lim _{x \rightarrow 0}|\sin x| \cdot \frac{a}{|\sin x|}} \\ & =e^a \\ \lim _{x \rightarrow 0^{+}} e^{\frac{\tan 2 x}{\tan 3 x}} & =\lim _{x \rightarrow 0} e^{\left(\frac{\tan 2 x}{2 x} \times \frac{3 x}{\tan 3 x} \times \frac{2 x}{3 x}\right)} \\ & =\lim _{x \rightarrow 0} e^{2 / 3}=e^{2 / 3} \end{aligned} \end{gathered} $

Since, $f$ is continuous at $x=0$

$ e^a=b=e^{2 / 3} \Rightarrow a=\frac{2}{3} $

and $b=e^{2 / 3}$

$f(x)=\left\{\begin{aligned} 2 x+3, & x \leq 1 \\ a x^2+b x, & x>1\end{aligned}\right.$

Given $f$ is differentiable for all $x \in R$.

$\therefore f$ is continuous everywhere.

At $x=1$

LHL $=$ RHL

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

$\Rightarrow \quad \lim \limits_{x \rightarrow 1^{-}} 2 x+3=\lim \limits_{x \rightarrow 1^{+}} a x^2+b x$

$\Rightarrow \quad 2(1)+3=a(1)^2+b(1)$

$\Rightarrow \quad 5=a+b\quad \ldots \text { (i) }$

Also, LHD = RHD

$ \begin{array}{ll} & \lim \limits_{x \rightarrow 1^{-}} f^{\prime}(x)=\lim \limits_{x \rightarrow 1^{+}} f^{\prime}(x) \\ \Rightarrow \quad & 2=\lim \limits_{x \rightarrow 1^{+}} 2 a x+b \\ \Rightarrow \quad & 2=2 a+b \quad \ldots \text { (i) }\end{array} $

From Eqs. (i) and (ii), we get

$\begin{array}{ll} & a=-3, b=8 \\ \therefore & f(x)=\left\{\begin{array}{cc}2 x+3, & x \leq 1 \\ -3 x^2+8 x, & x>1\end{array}\right.\end{array}$

Now, $f^{\prime}(x)=\left\{\begin{array}{cc}2, & x \leq 1 \\ -6 x+8, & x>1\end{array}\right.$

So, $\quad f^{\prime}(2)=-6(2)+8=-12+8=-4$

Given $f(x)=|x-1|+|x-2|$

$ \begin{aligned} f(x) & = \begin{cases}(1-x)+(2-x), & 0 \leq x<1 \\ (x-1)+(2-x), & 1 \leq x<2 \\ (x-1)+(x-2), & 2 \leq x \leq 3\end{cases} \\ & =\left\{\begin{array}{cc} 3-2 x, & 0 \leq x<1 \\ 1, & 1 \leq x<2 \\ 2 x-3, & 2 \leq x \leq 3 \end{array}\right. \end{aligned} $

At $x=1$

$ \begin{aligned} & \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 3-2 x=3-2(1)=1 \\ & \qquad f(1)=1, \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} 1=1 \\ & \therefore \lim _{x \rightarrow 1^{-}} f(x)=f(1)=\lim _{x \rightarrow 1^{+}} f(x) \end{aligned} $

Now, $\lim _{x \rightarrow 1^{-}} f^{\prime}(x)=\lim _{x \rightarrow 1^{-}}(3-2 x)^{\prime}$

$ \begin{gathered} =\lim _{x \rightarrow 1^{-}}-2=-2 \\ \lim _{x \rightarrow 1^{+}} f^{\prime}(x)=0 \\ \therefore \lim _{x \rightarrow 1^{-}} f^{\prime}(x) \neq \lim _{x \rightarrow 1^{+}} f^{\prime}(x) \end{gathered} $

So, $f(x)$ is continuous but not differentiable at $x=1$.

$ \text { At, } x=2 $

$ \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} 1=1 $

$ f(2)=2(2)-3=1 $

$ \lim \limits_{x \rightarrow 2^{+}} f(x)=\lim \limits_{x \rightarrow 2^{+}}(2 x-3)=4-3=1 $

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

$ \lim \limits_{x \rightarrow 2^{-}} f^{\prime}(x)=0, \lim \limits_{x \rightarrow 2^{+}} f^{\prime}(x)=\lim \limits_{x \rightarrow 2^{+}} 2 $

$ \therefore \lim \limits_{x \rightarrow 2^{-}} f^{\prime}(x) \neq \lim \limits_{x \rightarrow 2^{+}} f^{\prime}(x) $

So, $f(x)$ is continuous but not differentiable at $x=2$

$ \begin{aligned} & \text { } \lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2+x^5+x^6}}{x^4} \\ & =\lim _{x \rightarrow 0} \frac{\left[\sqrt{1+\sqrt{1+x^4}}-\sqrt{2+x^5+x^6}\right]\left[\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6}\right]}{x^4\left[\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6}\right]} \\ & =\lim _{x \rightarrow 0} \frac{1+\sqrt{1+x^4}-2-x^5-x^6}{x^4} \times-\lim _{x \rightarrow 0} \frac{1}{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6}} \\ & =\lim _{x \rightarrow 0}\left[\frac{\sqrt{1+x^4}-1}{x^4}-\frac{x^5(1+x)}{x^4}\right] \times \frac{1}{2 \sqrt{x}} \\ & =\frac{1}{2 \sqrt{2}} \lim _{x \rightarrow 0} \frac{\left(\sqrt{1+x^4}-1\right)\left(\sqrt{1+x^4}+1\right)}{x^4\left(\sqrt{1+x^4}+1\right)}=\frac{1}{2 \sqrt{2}} \times \frac{1}{2}=\frac{1}{4 \sqrt{2}} \end{aligned} $

$ =\lim \limits_{x \rightarrow 1} \frac{\frac{1}{2 \sqrt{x}}}{2 \cos ^{-1} x \times\left(\frac{-1}{\sqrt{1-x^2}}\right)}=\lim \limits_{x \rightarrow 1} \frac{-\sqrt{1-x^2}}{4 \sqrt{x} \cos ^{-1} x}\left(\frac{0}{0}\right) $

$ \begin{aligned} & =\frac{-1}{4} \lim _{x \rightarrow 1} \frac{\frac{1 \times(-2 x)}{2 \sqrt{1-x^2}}}{\sqrt{x} \times\left(-\frac{1}{\sqrt{1-x^2}}\right)+\cos ^{-1} \times \times \frac{1}{2 \sqrt{x}}} \\ & =-\frac{1}{4} \lim _{x \rightarrow 1} \frac{\frac{-x}{\sqrt{1-x^2}}}{\frac{-\sqrt{x}}{\sqrt{1-x^2}}+\frac{\cos ^{-1} x}{2 \sqrt{x}}}=\frac{-1}{4} \lim _{x \rightarrow 1} \frac{-x 2 \sqrt{x}}{-2 x+\left(\cos ^{-1} x\right) \sqrt{1-x^2}} \\ & =-\frac{1}{4} \end{aligned} $

To ensure the function $ f(x) $ is continuous at $ x = 0 $, the left-hand limit (LHL), right-hand limit (RHL), and the value of the function at zero must all be equal:

$ \text{LHL} = \text{RHL} = f(0) $

Calculating the Left-Hand Limit (LHL):

For $ x < 0 $, we have:

$ f(x) = \frac{\sin 3x - \tan 3x}{x^3} $

The LHL is evaluated as follows:

$ \begin{aligned} \lim_{x \to 0^{-}} \frac{\sin 3x - \tan 3x}{x^3} &= \lim_{x \to 0^{-}} \frac{\sin 3x(\cos 3x - 1)}{\cos 3x \times x^3} \\ &= \lim_{x \to 0^{-}} \frac{\sin 3x}{x} \times \lim_{x \to 0^{-}} \frac{1}{\cos 3x} \times \lim_{x \to 0^{-}} \frac{\cos 3x - 1}{x^2} \\ &= 3 \times 1 \times \lim_{x \to 0^{-}} (-2) \frac{\sin^2 \frac{3x}{2}}{\frac{9x^2}{4}} \times \frac{9}{4} \\ &= 3 \times (-2) \times \frac{9}{4} = -\frac{27}{2} \end{aligned} $

Calculating the Right-Hand Limit (RHL):

For $ x > 0 $, the function is:

$ f(x) = \frac{\tan((\alpha + 1)x) + \tan(2x)}{x} $

The RHL is computed as:

$ \begin{aligned} \lim_{x \to 0^{+}} \frac{\tan((\alpha + 1)x) + \tan(2x)}{x} &= \lim_{x \to 0^{+}} \frac{\tan((\alpha + 1)x)}{x} + \lim_{x \to 0^{+}} \frac{\tan(2x)}{x} \\ &= \alpha + 1 + 2 = \alpha + 3 \end{aligned} $

Value at $ x = 0 $:

$ f(0) = \beta $

Equating Limits for Continuity:

Given that LHL = RHL = $ f(0) $, we have:

$ \begin{aligned} -\frac{27}{2} &= \alpha + 3 = \beta \\ \alpha &= -\frac{27}{2} - 3 = -\frac{33}{2} \\ \beta &= -\frac{27}{2} \end{aligned} $

Calculate $ |\alpha| + |\beta| $:

$ |\alpha| = \frac{33}{2}, \quad |\beta| = \frac{27}{2} $

$ |\alpha| + |\beta| = \frac{33}{2} + \frac{27}{2} = \frac{60}{2} = 30 $

$ \begin{aligned} &\text {We have, } \lim _{x \rightarrow 3} \frac{x^3-27}{x^2-9}\\ &\begin{aligned} & =\lim _{x \rightarrow 3} \frac{(x-3)\left(x^2+9+3 x\right)}{(x-3)(x+3)} \\ & =\lim _{x \rightarrow 3} \frac{x^2+9+3 x}{x+3} \\ & =\frac{9+9+9}{6}=\frac{27}{6}=\frac{9}{2} \end{aligned} \end{aligned} $

Given the function $ f(x) = \begin{cases} 3ax - 2b, & x > 1 \\ ax + b + 1, & x < 1 \end{cases} $, we need to determine the condition for the existence of the limit $ \lim \limits_{x \rightarrow 1} f(x) $.

For the limit to exist at $ x = 1 $, the left-hand limit (LHL) and the right-hand limit (RHL) must be equal:

$ \lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x) $

Calculating the limits from both sides:

Left-hand limit (x approaches 1 from the left):

$ LHL = a(1) + b + 1 = a + b + 1 $

Right-hand limit (x approaches 1 from the right):

$ RHL = 3a(1) - 2b = 3a - 2b $

Setting the LHL equal to the RHL:

$ a + b + 1 = 3a - 2b $

Solving for the relationship between $ a $ and $ b $:

$ a + b + 1 = 3a - 2b $

Rearranging terms yields:

$ 2a - 3b = 1 $

Thus, the relation between $ a $ and $ b $ when the limit exists is $ 2a - 3b = 1 $.

We have, $f(x)= \begin{cases}\frac{2}{5-x}, & x<3 \\ 5-x, & x \geq 3\end{cases}$

$ \begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}} \frac{2}{5-x}=1 \\ & \mathrm{RHL}=\lim _{x \rightarrow 3^{+}} f(x)=5-3=2 \end{aligned} $

Hence, the function is left discontinuous at $x=3$.

We have, $f(x)=\left\{\begin{array}{cc}x^\alpha \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{array}\right.$

$ \begin{aligned} & \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{x^a\left(\sin \frac{1}{x}\right)}{\left(\frac{1}{x}\right)} \times \frac{1}{x} \\ & =\lim _{x \rightarrow 0} x^{\alpha-1}\left[\because \lim _{x \rightarrow 0} \frac{\sin \left(\frac{1}{x}\right)}{\frac{1}{x}}=1\right]=0 \end{aligned} $

$\therefore f(x)$ is continuous for all value except -1 .

Now, $f(x)$ is differentiable at $x=0$, if $\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$ exists.

$\Rightarrow \quad \lim _{x \rightarrow 0} x^{\alpha-1} \sin \frac{1}{x}$ exists finitely.

$ \Rightarrow \quad \alpha-1>0 $

$ \Rightarrow \quad \alpha>1 $

Hence, $f(x)$ is continuous and differentiable for $\alpha>1$.

The graph of $f(x)=\min \left\{x, x^2\right\}$ is shown alongside.

From the diagram, it is clear that $f(x)$ is continuous everywhere but $f(x)$ is not differentiable but $x=0$ and $x=1$.

$ f(x)= \begin{cases}x, & \text { if } x<0 \\ x^2, & \text { if } 01\end{cases} $

For $x>1, f^{\prime}(x)=1$

Given,

$ \begin{gathered} f(x)=\left\{\begin{array}{l} \frac{2 x e^{\frac{1}{2 x}}-3 x e^{-\frac{1}{2 x}}}{e^{\frac{1}{2 x}}+4 e^{-\frac{1}{2 x}}} \text { If } x \neq 0 \\ 0 \quad \text { If } x=0 \end{array}\right. \\ \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 x e^{1 / 2 x}-3 x e^{-1 / 2 x}}{e^{1 / 2 x}+4 e^{-1 / 2 x}} \\ =\lim _{x \rightarrow 0} \frac{x\left(2 e^{1 / x}-3\right)}{e^{1 / x}+4}=0 \end{gathered} $

$\therefore f(x)$ is continuous of $x=0$

$ \begin{aligned} & \text { LHD } \begin{aligned} f^{\prime}\left(0^{-}\right) & =\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{h\left(-2 e^{-1 / h}-3\right)}{-h\left(e^{-1 / h}+4\right)} \\ & =\lim _{h \rightarrow 0} \frac{2 e^{-1 / h}-3}{e^{-1 / h}+4}=\frac{-3}{4} \end{aligned} \end{aligned} $

$\begin{aligned} & f\left(0^{-}\right)=\frac{-3}{4} \\ & \begin{aligned} \text { RHD }=f^{\prime}\left(0^{+}\right) & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{h\left(2 e^{1 / h}-3\right)}{h\left(e^{1 / h}+4\right)} \\ & =\lim _{h \rightarrow 0} \frac{e^{1 / h}\left(2-\frac{3}{e^{1 / h}}\right)}{e^{1 / h}\left(1+\frac{4}{e^{1 / h}}\right)}=2\end{aligned} \\ & \text { RHD }=f^{\prime}\left(0^{-}\right)=2\end{aligned}$ LHD $\neq$ RHD $\quad \begin{aligned} & \therefore f \text { is not different of } x=0\end{aligned}$

$\begin{aligned} & \text { Here, } \lim _\limits{x \rightarrow-\infty} \log _e(\cosh x)+x \\ & =\lim _{x \rightarrow-\infty} \log \left(e^x+e^{-x}\right)+\ln 2 \\ & \Rightarrow \quad \lim _{x \rightarrow-\infty} \log \left(1+e^{2 x}\right)+\ln 2 \\ & =-\ln 2 \\ \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow \infty} \frac{(b-c) x^2+(c-a) x+(a-b)}{(a-b) x^2+(b-c) x+(c-a)}=\frac{1}{a} \\ & =\lim _{x \rightarrow \infty} \frac{x^2}{x^2}\left[\frac{(b-c)+(c-a) \frac{1}{x}+\frac{(a-b)}{x^2}}{(a-b)+(b-c) \frac{1}{x}+\frac{(c-a)}{x^2}}\right]=\frac{1}{2} \end{aligned}$

$\begin{aligned} & =\frac{b-c}{a-b}=\frac{1}{2} \\ \Rightarrow & 2 b-2 c=a-b \\ \Rightarrow & 3 b=a+2 c \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow-\infty} \frac{3|x|-x}{|x|-2 x}-\lim _{x \rightarrow 0} \frac{\log \left(1+x^3\right)}{\sin ^3 x} \\ & \lim _{x \rightarrow-\infty} \frac{-3 x-x}{-x-2 x}-\lim _{x \rightarrow 0} \frac{\log \left(1+x^3\right)}{x^3 \frac{\left(\sin x^3 x\right)}{x^3}} \\ \Rightarrow & \lim _{x \rightarrow-\infty} \frac{-4 x}{-3 x}-\lim _{x \rightarrow 0} \frac{1}{\left(\frac{\sin ^3 x}{x^3}\right)} \\ & \lim _{x \rightarrow-\infty} \frac{4}{3}-1=\frac{1}{3} \end{aligned}$

$\begin{aligned} & \lim _\limits{x \rightarrow \frac{-3}{5}} \frac{1}{x}\left[-\frac{1}{x}\right] \\ & =\frac{1}{\frac{-3}{5}}\left[-\frac{1}{\left(-\frac{3}{5}\right)}\right]=\frac{-5}{3}[1.66] \\ & =\frac{-5}{3} \cdot(1)=\frac{-5}{3} \end{aligned}$

If $ l $ and $ m $ (where $ l < m $) are the roots of the quadratic equation $ ax^2 + bx + c = 0 $, then the following limit expression is considered:

$ \lim\limits_{x \to \alpha} \frac{|ax^2 + bx + c|}{ax^2 + bx + c} $

Given that $ \alpha \in (l, m) $, by applying L'Hôpital's rule, we analyze the limit:

$ \lim\limits_{x \to \alpha} \frac{|ax^2 + bx + c|}{ax^2 + bx + c} $

This results in:

$ \frac{|a|}{a}, \text{ when } \alpha \in (l, m) $

Hence, the above contributes to the solution for the considered limit.

$\begin{aligned} \text { Let } f(x) & =\left\{\begin{array}{cc} \frac{1}{|x|}, & |x|>1 \\ a x^2+b, & |x| \leq 1 \end{array}\right. \\ & =\left\{\begin{array}{cc} \frac{1}{x}, & x>1 \\ a x^2+b & x \leq 1 \end{array}\right\} \end{aligned}$

Given, that $\lim _\limits{x \rightarrow 1^{+}} f(x)$ and $\lim _\limits{x \rightarrow 1^{+}} f(x)$ exist.

Thus, $\lim _\limits{x \rightarrow 1^{+}}(x)=\lim _\limits{x \rightarrow 1^{+}} \frac{1}{x}=1$

$\lim _\limits{x \rightarrow 1^{-}} f(x)=\lim _\limits{x \rightarrow 1^{-}}\left(a x^2+b\right)=a+b$

According to question, $a+b=1$ if take $a=\frac{3}{2}$ and $b=-\frac{1}{2}$.

Then, option (c) is only true all other option are incorrect.

$\begin{aligned} & \frac{d}{d x}\left(\lim _{y \rightarrow 2} \frac{1}{y-2}\left(\frac{1}{x}-\frac{1}{x+y-2}\right)\right) \\ & =\frac{d}{d x} \lim _{y \rightarrow 2}\left(\frac{\left(\frac{1}{x}-\frac{1}{x+y-2}\right)}{(y-2)}\right) \end{aligned}$

Applying L Hospital's rule,

$\begin{aligned} & \frac{d}{d x} \lim _{y \rightarrow 2} \frac{0+\frac{1}{(x+y-2)^2}}{1} \\ & =\frac{d}{d x}\left(\frac{1}{x^2}\right)=-\frac{2}{x^3} \end{aligned}$

Given, $\left\{\begin{array}{cc}\frac{x^2 \log (\cos x)}{\log (1+x)} & , x \neq 0 \\ 0 & , x=0\end{array}\right.$

Now, $\lim _\limits{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$

$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{x^2 \log (\cos x)}{x \log (1+x)}=\lim _{x \rightarrow 0} \frac{x \cdot \log (\cos x)}{\log (1+x)} \\ & =\lim _{x \rightarrow 0} \frac{x \log [1-(1-\cos x)]}{1-\cos x} \cdot \frac{1-\cos x}{\log (1+x)} \\ & =\lim _{x \rightarrow 0} \frac{\log [1-(1-\cos x)]}{1-\cos x} \cdot \frac{2 \sin ^2\left(\frac{x}{2}\right)}{4\left(\frac{x}{2}\right)^2}, x^2 \frac{x}{\log (1+x)}=0 \end{aligned}$

So, $f$ is differentiable hence it is also continuous.

$\begin{aligned} f(x \cdot f(y)) & =f(x y)+x \quad \text{... (i)}\\ x & \leftrightarrow y \end{aligned}$

$ \begin{aligned} & \Rightarrow \quad f(y \cdot f(x))=f(x y)+y \quad \text{... (ii)}\\ & x \leftrightarrow f(x) \text { in Eq. (i), } \\ \end{aligned}$

$ \begin{aligned} & f(f(x) \cdot f(y))=f(y \cdot f(x))+f(x) \quad \text{... (iii)}\\ \Rightarrow \quad & f(f x) \cdot f(y))=f(x y)+y+f(x) \\ & \text { [from Eq. (ii)]} \quad \text{.... (iv)} \end{aligned}$

$\begin{aligned} & \text { Again } x \leftrightarrow y \text { in Eq. (iv), } \\ & f(f(y) \cdot f(x))=f(y x)+x+f(y) \quad \text{... (v)}\\ & \text { Eq. (iv) = Eq. (v), } \\ & f(x y)+y+f(x)=f(x y)+x+f(y) \end{aligned}$

$\begin{aligned} & \Rightarrow f(x)-x=f(y)-y \quad \text{.... (vi)}\\ & \Rightarrow f(x)-x=\lambda=f(y)-y \end{aligned}$

Substituting $f(x)=\lambda+x$ in Eq. (i), we have

$\begin{aligned} & x \cdot f(y)+\lambda=(x y+\lambda)+x \\ & \Rightarrow \quad x f(y)=x y+x \quad [\because f(y)=\lambda+y]\\ & \therefore x(y+\lambda)=x y+x \\ & \Rightarrow \lambda x=x \\ & \Rightarrow \lambda=1 \quad [\because x>0]\\ & \therefore \quad f(x)=x+\lambda=x+1 \end{aligned}$

$\text { Now, } \lim _\limits{x \rightarrow 0} \frac{(f(x))^{\frac{1}{3}}-1}{(f(x))^{\frac{1}{2}}-1}=\lim _\limits{x \rightarrow 0} \frac{(1+x)^{\frac{1}{3}}-1}{(1+x)^{\frac{1}{2}}-1}$

$=\lim _\limits{x \rightarrow 0}\left(\frac{(1+x)^{\frac{1}{3}}-1}{1+x-1}\right)\left(\frac{1+x-1}{(1+x)^{\frac{1}{2}}-1}\right)$

$=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$

$\begin{aligned} &\begin{aligned} k & =\lim _{x \rightarrow 0} \frac{|x|}{\sqrt{x^4+4 x^2+5}} \\ & =\frac{|0|}{\sqrt{5}}=0 \\ l & =\lim _{x \rightarrow 0} x^4 \sin \left(\frac{1}{3 \sqrt{x}}\right)=(0)^4 \sin \left(\frac{1}{3 \sqrt{0}}\right) \\ \end{aligned}\\ &\begin{aligned} & =0 \times(\text { a value which belongs to }[-1,1]) \\ &=0\\ \therefore k+l & =0+0=0 \end{aligned} \end{aligned}$

$\lim _\limits{n \rightarrow \infty} x^n \log _e x=0$

It is possible only when $x \in(0,1)$ i.e. $x>0$ and $x<1$.

In this case,

$\begin{gathered} x^n \text { or } x^{\infty}=0 \\ \Rightarrow \quad x^{\infty} \log _c x=0 \end{gathered}$

Now, $\log _x 12$ :

We know, for $\log _a b$ : If $a \in(0,1)$, then $\log _a b=$ Negative

$\therefore \log _x 12=$ Negative \quad $[\because x \in(0,1)]$

$f(x)=\operatorname{Max}\{3-x, 3+x, 6\}$

Plotting the graph :

We know that a function is not differentiable at the points where it has sharp corner point.

In the above graph, we can see that only two sharp corner points are present.

Therefore, the above function is not differentiable at two points which are

$\begin{array}{r} x=-3 \text { and } x=3 \\ a=-3 \text { and } b=3 \\ \therefore|a|+|b|=3+3=6 \end{array}$

$\begin{aligned} & \lim _\limits{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots .\right. \\ & \left.=\quad+\frac{n^4}{n^5+n^5}\right) \\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^4}{r^5+n^5} \\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{(r / n)^4}{(r / n)^5+1} \cdot\left(\frac{1}{n}\right) \end{aligned}$

On putting $\frac{r}{n} \rightarrow x, \frac{1}{n} \rightarrow d x$ and $\Sigma \rightarrow \int$

$\begin{aligned} & =\int_\limits0^1 \frac{x^4}{1+x^5} \cdot d x=\frac{1}{5} \int_\limits0^1 \frac{5 x^4}{1+x^5} \cdot d x \\ & =\frac{1}{5}\left[\log \left(1+x^5\right)\right]_0^1=\frac{1}{5}[\log 2-\log 1] \\ & =\frac{1}{5}[\log 2-0]=\frac{1}{5} \log 2 \\ & =\log \sqrt[5]{2} \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow 0}\left(\frac{11 x^3-3 x+4}{13 x^3-5 x^2-7}\right) \\ & =\frac{11(0)-3(0)+4}{13(0)-5(0)-7}=\frac{+4}{-7}=\frac{a}{b} \\ & \Rightarrow \quad a=4 \alpha \text { and } b=-7 \alpha \\ & \Rightarrow a+b=-3 \alpha \text { or } a+b=3 \alpha \\ \end{aligned}$

Then, any multiple of 3 will be equal to $a+b$.

Among options 24 is multiple of 3.

$\therefore 24$ is required value of $a+b$.

$\begin{aligned} & \lim _{x \rightarrow 1} \frac{(1-x)\left(1-x^2\right) \ldots\left(1-x^{2 n}\right)}{\left\{(1-x)\left(1-x^2\right) \ldots\left(1-x^n\right)\right\}^2} \\ & =\lim _{x \rightarrow 1} \frac{(x-1)\left(x^2-1\right) \ldots\left(x^{2 n}-1\right)}{\left\{(x-1)\left(x^2-1\right) \ldots\left(x^n-1\right)\right\}^2} \\ & =\lim _{x \rightarrow 0} \frac{(x-1)\left(x^2-1\right) \ldots\left(x^{2 n}-1\right)}{\left\{\frac{(x-1)}{(x-1)} \cdot \frac{\left(x^2-1\right)}{(x-1)} \ldots \frac{\left(x^n-1\right)}{(x-1)}\right\}^2 \cdot(x-1)^{2 n}} \\ & =\lim _{x \rightarrow 1} \frac{\frac{(x-1)}{(x-1)} \frac{\left(x^2-1\right)}{(x-1)} \ldots \frac{\left(x^{2 n}-1\right)}{(x-1)}}{\left\{\frac{(x-1)}{(x-1)} \cdot \frac{\left(x^2-1\right)}{(x-1)} \cdots \frac{\left(x^n-1\right)}{(x-1)}\right\}^2} \\ & =\frac{1 \cdot 2 \cdot 3 \cdot \ldots 2 n}{\{1 \cdot 2 \ldots n\}^2} \\ & =\frac{(2 n) !}{(n !)^2}=\frac{(2 n) !}{n ! n !}={ }^{2 n} C_n\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a} n a^{n-1}\right]\end{aligned}$

$f(x)=\frac{\log _e\left(1+x^2 \tan x\right)}{\sin x^3} \cdot x \neq 0$

$f$ is continuous at $x=0$, then

$f(0)=\mathrm{LHL}=\mathrm{RHL}$

$\mathrm{RHL}=\lim _\limits{h \rightarrow 0} \frac{\log _e\left(1+h^2 \tan h\right)}{\sin h^3}$

$\begin{array}{rlr} & =\lim _{h \rightarrow 0} \frac{\log _e\left(1+h^2 \tan h\right)}{h^3 \cdot\left(\frac{\sin h^3}{h^3}\right)} & \\ & =\lim _{h \rightarrow 0} \frac{\log _e\left(1+h^2 \tan h\right)}{h^3} & {\left[\because \lim _{h \rightarrow 0} \frac{\sin h}{h}=1\right]} \\ & =\lim _{h \rightarrow 0} \frac{\log _e\left(1+h^3 \frac{\tan h}{h}\right)}{h^3} \\ & =\lim _{h \rightarrow 0} \frac{\log _e\left(1+h^3\right)}{h^3} & \left(\because \lim _{h \rightarrow 0} \frac{\tan h}{h}=1\right) \\ & =\lim _{h \rightarrow 0} \frac{\frac{1}{\left(1+h^3\right)}\left(3 h^2\right)}{3 h^2} \\ & =\lim _{h \rightarrow 0} \frac{1}{1+h^3}=1 \\ \therefore f(0) & =1 \end{array}$

[L-Hospital]

$\mathop {\lim }\limits_{n \to \infty } {{n{{(2n + 1)}^2}} \over {(n + 2)({n^2} + 3n - 1)}}$

$=\mathrm{\frac{Coefficient \,of \,n^3 \,in \,numerator}{Coefficient \,of \,n^3 \,in \,denominator}}=\frac{4}{1}$

$\lim _\limits{x \rightarrow 2^{-}} f(x)=f(2)$

$\because f(x)$ is continuous on $[0,8]$.

$\Rightarrow f(x)$ is continuous at 2 and 4 as well

$ \begin{aligned} 4+2 a+b & =8 \\ \Rightarrow 2 a+b & =4 \quad .....\text{(i)}\\ \lim _{x \rightarrow 4^{+}} f(x) & =f(4) \\ 8 a+5 b & =14\quad .....\text{(ii)} \end{aligned}$

From Eqs. (i) and (ii), we get

$a=3, b=-2$

$\mathop {\lim }\limits_{x \to {4^ - }} f(x) = f(4) = \mathop {\lim }\limits_{x \to {4^ - }} f(x)$

$ - 1 + a = a + b = 1 + b$

$ \Rightarrow b = - 1,a = 1$

$f$ is continuous $\Rightarrow f(x)$ will be continuous at $x=0$

\begin{array}{ll} \Rightarrow & \lim _\limits{x \rightarrow 0} f(x)=f(0) \\ \Rightarrow & \lim _\limits{x \rightarrow 0} \frac{e^{\alpha x}-e^x-x}{x^2}=\frac{3}{2} \\ \Rightarrow & \lim _\limits{x \rightarrow 0} \frac{\alpha e^{\alpha x}-e^x-1}{2 x}=\frac{3}{2}\end{array}$

[By $L$ hospital's rule]

This is solvable when

$\alpha e^0-e^0-1=0 \Rightarrow \alpha-2=0$

or $\alpha=2$

$\begin{aligned} & \text { } f(0)=\lim _{x \rightarrow 0} \frac{\left(e^x-1\right)^4}{\sin \left(\frac{x^2}{k^2}\right) \log \left\{1+\frac{x^2}{2}\right\}}=8 \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{2 k^2\left(\frac{e^x-1}{x}\right)^4}{\frac{\sin \left(\frac{x^2}{k^2}\right)}{\left(\frac{x^2}{k^2}\right)} \cdot \frac{\log \left\{1+\frac{x^2}{2}\right\}}{\frac{x^2}{2}}}=8 \\ & \end{aligned}$

$\begin{aligned} & \left\{\begin{array}{l}\because \lim _\limits{x \rightarrow} \frac{e^x-1}{x}=1 \Rightarrow \lim _\limits{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \lim _\limits{x \rightarrow 0} \frac{\log (1+x)}{x}=1\end{array}\right\} \\ & \Rightarrow \quad 2 k^2=8 \\ & \Rightarrow \quad k= \pm 2 \\ & \end{aligned}$

$\begin{aligned} & \text { } \lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^2} \quad \text{[0/0 form]}\\ & =\lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)-6 f^{\prime}(2 x)+4 f^{\prime}(4 x)}{2 x} \quad \text{[0/0 form]}\\ & =\lim _{x \rightarrow 0} \frac{2 f^{\prime \prime}(x)-12 f^{\prime \prime}(2 x)+16 f^{\prime \prime}(4 x)}{2} \\ & =\frac{2 f^{\prime \prime}(0)-12 f^{\prime \prime}(0)+16 f^{\prime \prime}(0)}{2} \\ & =3 f^{\prime \prime}(0)=3 \cdot 4=12 \\ & {\left[\because f^{\prime \prime}(0)=4\right]} \\ \end{aligned}$

$\lim _\limits{z \rightarrow 1} \frac{z^{1 / 3}-1}{z^{1 / 6}-1}$ $\left[\frac{0}{0} \text { form }\right]$

$\begin{aligned} & =\lim _{z \rightarrow 1} \frac{\frac{1}{3} z^{-2 / 3}}{\frac{1}{6} z^{-5 / 6}} \text {[L' Hospital rule]}\\ & =\frac{6}{3} \frac{(1)^{5 / 6}}{(1)^{2 / 3}}=\frac{6}{3}=2 \end{aligned}$

$f$ is continuous at $x=0$

$\begin{aligned} & \therefore \lim_\limits{x \rightarrow 0} f(x)=f(0)\\ & \Rightarrow \lim _\limits{x \rightarrow 0} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=k \log 2 \log 3 \\ & \Rightarrow k \log 2 \cdot \log 3 \\ & =\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(8^x-1\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} \end{aligned}$

$\begin{aligned} & =\lim _{x \rightarrow 0}\left[\begin{array}{c}\frac{\left(9^x-1\right)}{x} \cdot\left(\frac{8^x-1}{x}\right) \cdot \frac{x^2}{1-\cos x} \\ \cdot(\sqrt{2}+\sqrt{1+\cos x})\end{array}\right] \\ & =\log 9 \cdot \log 8 \cdot 2(\sqrt{2}+\sqrt{2}) \\ & =2 \log 3 \cdot 3 \log 2 \cdot 4 \sqrt{2} \\ & \Rightarrow k \log 3 \cdot \log 2=24 \sqrt{2} \log 3 \cdot \log 2 \Rightarrow k=24 \sqrt{2} \\ & \end{aligned}$

$\because f$ is continuous in $(0, \pi) . \Rightarrow f$ is continuous at $x=\frac{\pi}{4}, \frac{\pi}{2}$

$\therefore$ LHL of $f(x)\left(\right.$ at $\left.x=\frac{\pi}{4}\right)=$ RHL of $f(x)\left(\right.$ at $\left.x=\frac{\pi}{4}\right) =f\left(\frac{\pi}{4}\right)$

$\begin{array}{r} \Rightarrow \quad \lim _\limits{x \rightarrow \frac{\pi^{-}}{4}} f(x)=\lim _\limits{x \rightarrow \frac{\pi^{+}}{4}} f(x)=2\left(\frac{\pi}{4}\right) \cot \frac{\pi}{4}+b \\ \Rightarrow \quad \lim _\limits{x \rightarrow \frac{\pi^{-}}{4}}(x+a \sqrt{2} \sin x)=\lim _\limits{x \rightarrow \frac{\pi^{+}}{4}}(2 x \cot x+b) \\ =\frac{\pi}{2}+b\end{array}$

$\Rightarrow \frac{\pi}{4}+a \sqrt{2} \sin \frac{\pi}{4}=2 \cdot \frac{\pi}{4} \cot \frac{\pi}{4}+b$

$\Rightarrow a-b=\frac{\pi}{4}$ ..... (i)

Again, LHL of $f(x)\left(\right.$ at $\left.x=\frac{\pi}{2}\right)=$ RHL of $f(x)\left(\right.$ at $\left.x=\frac{\pi}{2}\right) \quad\left\{\because f(x)\right.$ is continuous at $\left.x=\frac{\pi}{2}\right\}$ $ \begin{array}{ll} \Rightarrow \quad & \lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}} f(x) \\ \Rightarrow \quad & \lim _\limits{x \rightarrow \frac{\pi^{-}}{2}}(2 x \cot x+b) \\ & =\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}}(a \cos 2 x-b \sin x) \\ \Rightarrow \quad & 2 \cdot \frac{\pi}{2} \cot \frac{\pi}{2}+b=a \cos \pi-b \sin \frac{\pi}{2} \end{array}$

$\Rightarrow a=-2b$ ..... (ii)

On solving Eqs. (i) and (ii), we get

$a=\frac{\pi}{6},b=-\frac{\pi}{12}$