Maths M.C.Q (1 Marks)

3. a = -1, b = 1

Solution:

Given, f(x) is continuous for $0\leq\text{x}<\infty.$

This means that f(x) is continuous for $\text{x}=1,\sqrt{2.}$

Now,

If(x) is continuons at x = 1, then

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

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(1-\text{h})=\text{a}$

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

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

$\Rightarrow\text{a}^2=1$

$\Rightarrow\text{a}=\pm1$

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

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

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(\sqrt{2}-\text{h})=\frac{2\text{b}^2-4\text{b}}{2}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{a=b}^2-2\text{b}$

$\Rightarrow\text{a = b}^2-2\text{b}$

$\Rightarrow\text{b}^2-2\text{b - a}=0$

$\therefore$ For a = -1, We have

$\text{b}^2-2\text{b}+1=0$

$\Rightarrow(\text{b}-1)^2=0$

$\Rightarrow\text{b}=1$

Thus, a = -1 and b=1

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 $\text{a}=1$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Solution:

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\frac{2\times-\sin^{-1}\text{x}}{2\times+\tan^{-1}\text{x}}$

$\text{f}\text{(0)}=\lim\limits_{\text{x}\rightarrow0}\frac{2\times-\frac{\sin^{-1}\text{x}}{\text{x}}}{2+\frac{\tan^{-1}\text{x}}{\text{x}}}$

$\text{f}(0)=\frac{2-1}{2+1}=\frac{1}{3}$

1. $\frac{1}{2}$ 

Solution:

Given, $\text{f(x)}=\begin{cases}\text{x}\sin\frac{\pi}{2}(\text{x}+1),&\text{x}\leq0\\\frac{\tan\text{x}-\sin\text{x}}{\text{x}^3},&\text{x}>0\end{cases}$

We have,

$(\text{LHL at x}=0)=\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(0-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim\limits_{\text{h}\rightarrow0}\text{a}\sin\Big(\frac{\pi}{2}(-\text{h+1})\Big)=\text{a}\sin\Big(\frac{\pi}{2}\Big)=\text{a}$

$(\text{RHL at x}=0)=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}\\=\lim\limits_{\text{h}\rightarrow0}\text{f}(0+\text{h})=\lim\limits_{\text{h}\rightarrow0}\text{f(h)}=\lim\limits\frac{\tan\text{h}-\sin\text{h}}{\text{h}^3}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\sin\text{h}}{\cos\text{h} }-\sin\text{h}}{\text{h}^3}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\sin\text{h}}{\cos\text{h}}(1-\cos\text{h})}{\text{h}^3}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{(1-\cos\text{h})\tan\text{h}}{\text{h}^3}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{2\sin^2\frac{\text{h}}{2}\tan\text{h}}{4\times\frac{\text{h}^2}{4}\times\text{h}}$

$=\frac{2}{4}\lim\limits_{\text{h}\rightarrow0}\frac{\sin^2\frac{\text{h}}{2}\tan\text{h}}{\frac{\text{h}^2}{4}\times\text{h}}$

$=\frac{1}{2}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\bigg)^2\times\lim\limits_{\text{h}\rightarrow0}\frac{\tan\text{h}}{\text{h}}$

$=\frac{1}{2}\times1\times1$

$=\frac{1}{2}$

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

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

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

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

Solution:

Given, $\text{f(x)}=\frac{\sqrt{\text{a}^2+\text{ax+x}^2}-\sqrt{\text{a}^2+\text{ax+x}^2}{}}{\sqrt{\text{a+x}}-\sqrt{\text{a-x}}}$

$\Rightarrow\text{f(x)}=\frac{\big(\sqrt{\text{a}^2-\text{ax+x}^2}-\sqrt{\text{a}^2+\text{ax+x}^2}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}{\big(\sqrt{\text{a+x}}-\sqrt{\text{a-x}}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}$

$\Rightarrow\text{f(x)}=\frac{\big(\text{a}^2-\text{ax+x}^2\big)-\big(\text{a}^2+\text{ax+x}^2\big)}{\big(\sqrt{\text{a+x}}-\sqrt{\text{a-x}}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}$

$\Rightarrow\text{f(x)}=\frac{(-2\text{ax})\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{\big(\sqrt{\text{a+x}}-\sqrt{\text{a-x}}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}$

$\Rightarrow\text{f(x)}=\frac{(-2\text{ax})\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{\big(\text{a+x-a+x}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2}+\text{ax+x}^2\big)}$

$\Rightarrow\text{f(x)}=\frac{(-2\text{ax})\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{(2\text{x})\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}$

$\Rightarrow\text{f(x)}=\frac{-\text{a}\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}$

So, if f(x) is continuous at x = 0, then

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

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Bigg[\frac{-\text{a}\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{\big(\sqrt{\text{a}^2-\text{ax+x}}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)} \Bigg]$

$\Rightarrow\bigg[\frac{-2\text{a}(\sqrt{\text{a}})}{(\sqrt{a}^2+\sqrt{\text{a}^2})}\bigg]=\text{f(0)}$

$\Rightarrow\begin{bmatrix}\frac{-2\text{a}(\sqrt{a})}{(\text{a+a})} \end{bmatrix}=\text{f(0)}$

$\Rightarrow\text{f}(0)$

$=-\sqrt{a}$

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

Solution:

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

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

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

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

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

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

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

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

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

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

4. none of these.

Solution:

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\frac{2-(256-7\text{x})^{\frac{1}{8}}}{(5\text{x}+32)^\frac{1}{5}-2}$

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\frac{-\bigg[(256-7\text{x})^\frac{1}{8}-256^\frac{1}{8}\bigg]}{(5\text{x+32})^\frac{1}{5}-32\frac{1}{5}}$

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{-\bigg[(256-7\text{x})^{\frac{1}{8}}-256\bigg]}{-7\text{x}}}{\frac{(5\text{x}+32)^{\frac{1}{5}}-32^\frac{1}{5}}{5\text{x}}}$

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{-\bigg[(256-7\text{x})^\frac{1}{8}-256^\frac{1}{8}\bigg]}{-7\text{x}}\text{x}-7}{\frac{(5\text{x}+32)^\frac{1}{5}-32^\frac{1}{5}}{5\text{x}}\times5}$

$\text{f}(0)=\frac{7}{5}\times\frac{\frac{1}{8}\times256^\frac{-7}{8}}{\frac{1}{5}\times32^\frac{-4}{5}}$

$\text{f}(0)=\frac{7}{5}\times\frac{5\times2^4}{8\times2^7}=\frac{7}{8}\times\frac{1}{8}=\frac{7}{64}$

4. -1

Solution:

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

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

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

$\text{a}=-1$

4. x except at x = 0 and x = 1

Solution:

Given function $\text{f(x)}=\frac{|\text{x}^2-\text{x}|}{\text{x}^2-\text{x}}$

Consider,

$\text{f}(0^+)=\lim\limits_{\text{x}\rightarrow0}\frac{|\text{x}^2-\text{x}|}{\text{x}^2-\text{x}}=\lim\limits_{\text{x}\rightarrow0}\frac{|\text{x(x}-1)|}{\text{x(x}-1)}=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x(x}-1)}{\text{x(x}-1)}=1$

$\text{f}(0^-)=\lim\limits_{\text{x}\rightarrow0}\frac{|\text{x}^2-\text{x}|}{\text{x}^2-\text{x}}=\lim\limits_{\text{x}\rightarrow0}\frac{|\text{x}(\text{x}-1)|}{\text{x}(\text{x}-1)}=\lim\limits_{\text{x}\rightarrow0}\frac{-\text{x}(\text{x}-1)}{\text{x}(\text{x}-1)}=-1$

Also, for f(1⁺) and f(1^-) you can check.

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$]$

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

4. (5, 2)

Solution:

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

$\lim\limits_{\text{x}\rightarrow1}\text{ax}^2+\text{b}=4$

a + b = 4

We have possible values as (2, 2), (3, 1), (4, 0)

But can not be (5, 2).

Function is can not be continuous at (5, 2).

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. 1

Solutuion:

$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}(\cos\text{x})^{\frac{1}{\text{x}}}$

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

$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\Big(1-2\sin^2\frac{\text{x}}{2}\Big)^\frac{1}{\text{x}}$

$\text{f(0)}=\lim\limits\Big(1-2\sin^2\frac{\text{x}}{1}\Big)^{\frac{1}{-2\sin^2\frac{\text{x}}{2}}\times\frac{-2\sin^2\frac{\text{x}}{2}}{\text{x}}}$

$\text{f(0)}= \lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin^2\frac{\text{x}}{2}}{\text{x}}}$

$\text{f(0)}=\lim \limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin\frac{\text{x}}{2}}{\text{x}}\times\sin\frac{\text{x}}{2}}$

$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin\frac{\text {x}}{2}}{\frac{\text{x}}{2}}\times\frac{1}{2}\sin\frac{\text{x}}{2}}$

$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\text{e}^{1\times\sin\frac{1}{2}}=\text{e}^0=1$

$\Rightarrow\text{k}=1$

1. -1

Solution:

Given, f(x) is continuous at every point of its domain. So, it is continuous at x = 1.

$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{+}}\text{f}\text{(x)}=\text{f}(1)$

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

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Big(4(1+\text{h})^2+3\text{b}(1+\text{h})\Big)=5(1)-4$

$\Rightarrow4+3\text{b}=1$

$\Rightarrow-3=3\text{b}$

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

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

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

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

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

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

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

2. 50

Solution:

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

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

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

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

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

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

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

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

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

$\Rightarrow\text{a}=50$

2. 5

Solution:

Given, $\text{f(x)}=\begin{cases}\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$

If f(x) is continuous at x = 0, then $\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f(0)}$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}}\Big)=\text{k}$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Big(\frac{3\log(1+3\text{x})}{3\text{x}}-\frac{2\log(1-2\text{x})}{2\text{x}}\Big)=\text{k}$

$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})}{3\text{x}}\Big)-2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1-2\text{x})}{2\text{x}}\Big)=\text{k}$

$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})}{3\text{x}}\Big)+2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1-2\text{x})}{-2\text{x}}\Big)=\text{k}$

$\Rightarrow3\times1+2\times1=\text{k}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}=1\Big]$

$\Rightarrow\text{k}=3+2$

$\Rightarrow\text{k}=5$

1. f(x) is continuous and $\text{f}'(1^+)=\log_{10}\text{e}$

4. f(x) is continuous and $(\text{f}'(1^-)=-\log_{10}\text{e})$

Solution:

Given,

$\text{f(x)}=|\log_{10}\text{x}|=\bigg|\frac{\log_{\text{e}}\text{x}}{\log_{\text{e}}10}\bigg|$

$=|(\log_{\text{e}}\text{x})\times(\log_{10}\text{e})|$

$=(\log_{10}\text{e})|\log_{10}\text{x}|$

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

$=\lim\limits_{\text{h}\rightarrow0}\frac{(\log_{{10}}\text{e})|\log_{\text{e}}(1+\text{h})|-(\log_{10}\text{e})|\log_{\text{e}}1|}{\text{h}}$

$=(\log_{10}\text{e})\lim\limits_{\text{h}\rightarrow0}\frac{|\log_{\text{e}}(1+\text{h})|}{\text{h}}$

$=(\log_{10}\text{e})\times1$

$=(\log_{10}\text{e})$

Also,

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

$=\lim\limits_{\text{h}\rightarrow0}\frac{(\log_{10}\text{e})|\log_{\text{e}}(1-\text{h})|-(\log_{10}\text{e})|\log_{\text{e}}1}{\text{h}}$

$=-(\log_{10}\text{e})\lim\limits_{\text{h}\rightarrow0}\frac{|\log_{\text{e}}(1+\text{h})|}{\text{h}}$

$=(\log_{10}\text{e})\times1=-(\log_{10}\text{e})$

1. 0

Solution:

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\times\sin\frac{1}{\text{x}}$

$-1\leq\sin\frac{1}{\text{x}}\leq1$

$\text{x}\times(-1)\leq\text{x}\sin\frac{1}{\text{x}}\leq\text{x}$

$\lim\limits_{\text{x}\rightarrow0}-\text{x}\leq\lim\limits_{\text{x}\rightarrow0}\text{x}\sin\frac{1}{\text{x}}\leq\lim\limits_{\text{x}\rightarrow0}\text{x}$

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

3. $\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\Big(\frac{2}{3}\Big),\text{ c}=1$

Solution:

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}(1+\text{ax})^{\frac{1}{\text{x}}}$

$\text{b}=\lim\limits_{\text{x}\rightarrow{\text{a}}}(1+\text{ax})^{\frac{1}{\text{ax}}\times\text{a}}$

$\text{b}=\text{e}^{\text{a}}$

$\text{a}=\log_{\text{e}}\text{b}$

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow\text{a}^+}\frac{(\text{x}+\text{c})^{\frac{1}{3}}-1}{(\text{x}+1)^{\frac{1}{2}}-1}$

Here, $\text{c}=1$

$\text{x}+1=\text{y}$

$\text{x}\rightarrow0\Rightarrow\text{y}\rightarrow1$

$\text{f}(0)=\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^{\frac{1}{3}}-1}{\text{y}^{\frac{1}{2}}-1}$

$\text{b}=\lim\limits_{\text{y}\rightarrow1}\frac{\frac{\text{y}^{\frac{1}{3}}-1}{\text{y}-1}}{\frac{\text{y}^{\frac{1}{2}}-1}{\text{y}-1}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$

$\text{a}=\log\text{b}=\log\frac{2}{3}$

2. is not continuous at x = 0

Solution:

Given, $\text{f(x)=}\begin{cases}\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1},\text{x}\neq0\\0,\text{x}=0\end{cases}$

We have

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1}\Bigg)$

if $\text{e}^\frac{1}{\text{x}}=\text{t},$ then 

$\text{x}\rightarrow0, \text{t}\rightarrow\infty$

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{t}\rightarrow\infty}\Big(\frac{\text{t}-1}{\text{t}+1}\Big)$

$=\lim\limits_{\text{t}\rightarrow\infty}\Bigg(\frac{1-\frac{1}{\text{t}}}{1+\frac{1}{\text{t}}}\Bigg)=\frac{1-0}{1+0}=1$

Also, f(0) = 0

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

Hence, f(x) is discontinuons at x = 0.

3. Is discontinuous only at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and x = 0

Solution:

Given function is

$\text{f{x}}\begin{cases}1,&|\text{x}|\geq1\\\frac{1}{\text{n}^2},&\frac{1}{\text{n}}<|\text{x}|<\frac{1}{\text{n}-1},\text{n}=2,3,...\end{cases}$

Consider,

$\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}^-}\text{f(x)}=\frac{1}{\text{n}^2}$

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

Hence, fnuction is discontinuous at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and x = 0

2. 6

Solution:

Given, $\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{x}\ne0\\\frac{\text{k}}{2},&\text{x}=0\end{cases}$

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

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

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{\text{x}}=\text{f}(0)$

$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}=\frac{\text{k}}{2}$

$\Rightarrow3\times1=\frac{\text{k}}{2}$

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

$\Rightarrow\text{k}=6$

2. a + b

Solution:

$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})-\log(1-\text{bx})}{\text{x}}=\text{k}$

$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})}{\text{x}}-\frac{\log(1-\text{bx})}{\text{x}}=\text{k}$

$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})}{\text{ax}}\times\text{a} -\frac{\log(1-\text{bx})}{\text{-bx}}\times(-\text{b})=\text{k}$

$\text{a}+\text{b}=\text{k}$

4. None of these.

Solution:

$\lim\limits_{\text{x}\rightarrow0}\frac{(4^\text{x}-1)^3}{\sin\Big(\frac{\text{x}}{\text{a}}\Big)\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}=12(\log4)^3$

$\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(4^\text{x}-1)^2}{\text{x}^3}}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\text{x}}}\times\frac{1}{\frac{\log\Big\{\Big(1+\frac{\text{x}}{3}\Big)\Big\}}{\text{x}^2}}=12(\log4)^3$

$\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(4^\text{x}-1)^3}{\text{x}^3}}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\text{x}}}\times\frac{1}{\frac{\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}{\text{x}^2}}=12(\log4)^3$

$\lim\limits_{\text{x}\rightarrow0}\frac{\Big(\frac{(4^\text{x}-1}{\text{x}^3}\Big)}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\frac{\text{x}}{\text{a}}}}\text{a}\text{x}\ {\times}\frac{\frac{1}{\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}}{\frac{\text{x}^3}{3}}\text{x}^3=12(\log4)^3$

$3(\log4)^3=12(\log4)^3$

$3\text{a}=12$

$\text{a}=12$

Note: The question is incorrect, so it has been modified.

1. 0

Solution:

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

$\text{x}\rightarrow\frac{\pi}{2}+\text{h}$

$\text{x}\rightarrow\frac{\pi}{2}\Rightarrow\text{h}\rightarrow0$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\Big(\cos\big(\frac{\pi}{2}+\text{h}\big)\Big)-\cos\big(\frac{\pi}{2}+\text{h}\big)}{\Big(\pi-2\big(\frac{\pi}{2}+\text{h}\big)\Big)^2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin(-\sin\text{h})+\sin\text{h}}{4\text{h}^2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}-\sin(\sin\text{h})}{4\text{h}^2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}-\sin(\sin\text{h})}{4\text{h}^2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)\sin\big(\frac{\text{h}-\sin\text{h}}{2}\big)}{4\text{h}^2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times\frac{\sin\big(\frac{\text{h}-\sin\text{h}}{2}\big)}{\frac{\text{h}-\sin\text{h}}{2}}\times\frac{\text{h}-\sin\text{h}}{2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times\frac{\text{h}-\sin\text{h}}{2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times(\text{h}-\sin\text{h})$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{1}{4}\frac{\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{\text{h}^2}\times(\text{h}-\sin\text{h})=0$

$\text{k}=0$

2. $\text{f}'(\text{a}^-)=-\phi(\text{a})$

Solution:

Given that $\text{f(x)}=|\text{x}-\text{a}|\ \phi\ (\text{x}),$ where $\phi(\text{x})$ continuous function.

$|\text{x}-\text{a}|\Rightarrow\text{x}-\text{a}$ if $\text{x}-\text{a}>0$

$|\text{x}-\text{a}|\Rightarrow-(\text{x}-\text{a})$ if $\text{x}-\text{a}<0$

By definition of continuity,

$\text{f}'(\text{a})= \lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{a}+\text{h})-\text{f(a)}}{\text{h}}$

Hence, $\text{f}(\text{a}^+)=\phi(\text{x})$ and $\text{f}'(\text{a}^-)=-\phi(\text{x})$

3. 2

Solution:

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

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

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

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

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

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

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

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

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

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

$=2$

4. R - {1, 2}

Solution:

Given:

$\text{f(x)=}\begin{cases}\frac{\text{x}^4-5\text{x}^2+4}{|(\text{x}-1)(\text{x-2})|},\text{x}\neq1,2\\6, \text{x}=1\\12,\text{x}=2\end{cases}$

Now,

$\Rightarrow\text{x}^4-5\text{x}^2+4=\text{x}^4-\text{x}^2-4\text{x}^2+4\\=\text{x}^2(\text{x}^2-1)-4(\text{x}^2-1)$

$=(\text{x}^2-1)(\text{x}^2-4)=(\text{x}-1)(\text{x}+1)(\text{x}-2)(\text{x}+2)$

$\Rightarrow\text{f(x)}=\begin{cases}\frac{(\text{x}-1)(\text{x}+1)(\text{x}-2)(\text{x}+2)}{|(\text{x}-2)(\text{x}-1)|},&\text{x}\neq1,2\\6,& \text{x}=1\\12,&\text{x}=2\end{cases}$

$\Rightarrow\text{f(x)}=\begin{cases}(\text{x}+1)(\text{x}+2),&\text{x}<1\\-(\text{x+1})(\text{x}+2),&1<\text{x}<2\$\text{x+1})(\text{x}+2),&\text{x}>2\\6,&\text{x=1}\\12,&\text{x}=2\end{cases}$

So,

$\Rightarrow\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(1-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(1-\text{h}+1)(1-\text{h}+2)\\=2\times3=6$

$\Rightarrow\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1+\text{h})\\=-\lim\limits_{\text{h}\rightarrow0}(1+\text{h}+1)(1+\text{h}+2)\\=-2\times3=-6$

Also,

$\Rightarrow\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h})\\=-\lim\limits_{\text{h}\rightarrow0}(2-\text{h}+1)(2-\text{h}+2)=-12$

$\Rightarrow\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2+\text{h}+1)(2+\text{h}+2)=12$

Thus,

$\Rightarrow\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}\neq\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}\neq\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}$

$\therefore$ The only point of discontinuities of the function f(x) are x = 1 and x = 2. Hence, the given function is continuous on the set R - {1, 2}.

2. {0, 1}

Solution:

Given, $\text{f}\text{(x)}=\frac{1}{1-\text{x}}$

Clearly, $\text{f}:\text{R}-\big\{1\big\}\rightarrow\text{R}$

Now,

$\text{f}\big(\text{f}\text{(x)}\big)=\text{f}\Big(\frac{1}{1-\text{x}}\Big)=\Bigg(\frac{1}{1-\big(\frac{1}{1-\text{x}}\big)}\Bigg)$

$=\Big(\frac{1-\text{x}}{-\text{x}}\Big)=\Big(\frac{\text{x}-1}{\text{x}}\Big)$

$\therefore\text{f}\text{o}\text{f}:\text{R}-\big\{0,1\big\}\rightarrow\text{R}$

Now,

$\text{f}\big(\text{f}\text{(x)}\big)=\text{f}\Big(\frac{1}{1-\text{x}}\Big)=\Bigg(\frac{1}{1-\big(\frac{1}{1-\text{x}}\big)}\Bigg)=\text{x}$

$\therefore\ \text{f}\text{o}\text{f}:\text{R}-\big\{0,1\big\}\rightarrow\text{R}$

Thus, f(f(f(x))) is not defind at x = 0, 1

Hence, f(f(f(x))) is discontinuous at {0, 1}

2. $\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$

Solution:

Given, $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x }>\frac{\pi}{2}\end{cases}$

We have

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

$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{1-\sin^2\big(\frac{\pi}{2}-\text{h}\big)}{3\cos^2\big(\frac{\pi}{2}-\text{h}\big)}\Bigg)$

$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{1-\cos^2\text{h}}{3\sin^2\text{h}}\Big)$

$=\frac{1}{3}\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin^2\text{h}}{\sin^2\text{h}}\Big)$

$=\frac{1}{3}$

$\Big(\text{RHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)$

$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\text{b}\big[1-\sin\big(\frac{\pi}{2}+\text{h})\big]}{\big[\text{x}-2\big(\frac{\pi}{2}+\text{h}\big)\big]^2}\Bigg)$

$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\text{b}(1-\cos\text{h})}{[-2\text{h}]^2}\bigg)$

$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^{2}\frac{\text{h}}{2}}{4\text{h}^2}\bigg)$

$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^2\frac{\text{h}}{2}}{16\frac{\text{h}^2}{4}}\bigg)$

$=\frac{\text{b}}{8}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\bigg)^2$

$=\frac{\text{b}}{8}\times1$

$=\frac{\text{b}}{8}$

Also, $\text{f}\big(\frac{\pi}{2}\big)=\text{a}$

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

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

$\Rightarrow\frac{1}{3}=\frac{\text{b}}{8}=\text{a}$

$\Rightarrow\text{a}=\frac{1}{3}\text{ and }\text{b}=\frac{8}{3}$

2. $\text{x}=\frac{5}{2}$

Solution:

$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}2\sqrt{\text{x}}=2$

$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}4-2\text{x}=2$

Function is continuous at x = 1

$\lim\limits_{\text{x}\rightarrow\frac{5}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{5}{2}}4-2\text{x}=-1$

$\lim\limits_{\text{x}\rightarrow\frac{5}{2}^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{5}{2}}2\text{x}-7=-2$

Function is discontinuous at $\text{x}=\frac{5}{2}$

1. 0

Solution:

Here,

x³ + x² - 16x + 20

= x³ - 2x² + 3x² - 6x - 10x + 20

= x²(x - 2) + 3x(x - 2) - 10(x - 2)

= (x - 2)(x² + 3x - 10)

= (x - 2)(x - 2) (x - 5)

= (x - 2)²(x + 5)

So, the given function can be rewritten as 

$\text{f(x)}=\frac{(\text{x}-2)^2(\text{x}+5)}{\text{x}-2}$

$\Rightarrow\text{f(x)}=(\text{x}-2)(\text{x}+5)$

If f(x) is continuous at x = 2, then

$\lim\limits_{\text{x}\rightarrow2}\text{f(x)}=\text{f}(2)$

$\Rightarrow\lim\limits_{\text{x}\rightarrow2}(\text{x}-2)(\text{x}+5)=\text{f}(2)$

$\Rightarrow\text{f}(2)=0$

Hence, in order to make f(x) continuous at x = 2, f(2) should be defined as 0.

2. x = 3

Solution:

$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}\frac{1}{5}(2\text{x}^2+3)=1$

$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}6-5\text{x}=1$

Function is continuou at x = 1

$\lim\limits_{\text{x}\rightarrow3^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}6-5\text{x}=-9$

$\lim\limits_{\text{x}\rightarrow3^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}\text{x}-3=0$

Function is discontinuous at x = 3

3. Discontinuous exactly at three points.

Solution:

Given,

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

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

$\Rightarrow\text{f(x)}=\frac{1}{\text{x}},\text{x}\neq0 $ and  $4-\text{x}^2\neq0 $  or $ \text{x}\neq0,\pm2$

Clearly, f(x) is defined and continuous for all real numbers except $\left\{0,\pm2\right\}$

Therefore, f(x) is discontinuous exactly at three points.

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 $\text{b}\neq0$

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

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

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

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

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

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

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

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

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

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

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