#### Explain solution RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 20 maths

The correct option is (b)

Hint:

(i) A function f(x) is said to be continuous at a point x = a of its domain, if

$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$

Given:

$f(x)=\left\{\begin{array}{c} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x},-1 \leq x<0 \\ \frac{2 x+1}{x-2} \quad, 0 \leq x \leq 1 \end{array}\right.$

Step 1 : Rationalize the function

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)$

$\lim _{h \rightarrow 0} \frac{\sqrt{1-p h}-\sqrt{1+p h}}{-h}$

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{(\sqrt{1-p h}-\sqrt{1+p h}) \times(\sqrt{1-p h}+\sqrt{1+p h})}{(-h)(\sqrt{1-p h}+\sqrt{1+p h})} \\ &=\lim _{h \rightarrow 0} \frac{((-p h)-(1+p h)}{(-h) \sqrt{1-p h}+\sqrt{1+p^{n}}} \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-2 p h}{-n(\sqrt{1-p h}+\sqrt{1+p h})} \\ &=\lim _{h \rightarrow 0} \frac{+2 p}{\sqrt{1-p h}+\sqrt{1+p h}} \end{aligned}

\begin{aligned} &=\frac{2 p}{\sqrt{1}+\sqrt{1}} \\ &=\frac{2 p}{1+1} \\ &=\frac{2 p}{2} \\ &=p \end{aligned}

\begin{aligned} &\text { And } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h) \\ &=\lim _{n \rightarrow 0} \frac{2 h+1}{h-2} \\ &=\frac{2 \times 0+1}{0-2} \\ &=\frac{1}{-2} \end{aligned}

\begin{aligned} &\text { And } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h) \\ &=\lim _{n \rightarrow 0} \frac{2 h+1}{h-2} \\ &=\frac{2 \times 0+1}{0-2} \\ &=\frac{1}{-2} \end{aligned}

\begin{aligned} &=\frac{-1}{2} \end{aligned}

Also,

\begin{aligned} &f(x)=\frac{2 \cdot 0+1}{0-2} \\ &=\frac{-1}{2} \end{aligned}

From (i)

\begin{aligned} &p=-\frac{1}{2}=-\frac{1}{2} \\ &\Rightarrow p=\frac{-1}{2} \end{aligned}

So, the correct option is (b)