#### Please solve RD Sharma class 12 chapter Continuity exercise 8.2 question 4 subquestion (vi) maths textbook solution

$p=-1/2$

Hint:

Put LHL = RHL at x = 0

Given:

$f(x)= \begin{cases}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} & -1 \leq x<0 \\ \frac{2 x+1}{x-2} & 0 \leq x \leq 1\end{cases}$

Explanation:

At x = 0

L.H.L

\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} \\ &=\lim _{x \rightarrow 0} \frac{p}{\sqrt{1+p x}}+\frac{p}{2 \sqrt{1+p x}} \quad\left[0 / 0_{\text {from }}\right] \\ &=\lim _{x \rightarrow 0} \frac{2 p}{\sqrt{1+p x}} \times 2 \end{aligned}

\begin{aligned} &=\frac{2 p}{2 \times \sqrt{1+p \times 0}}=\frac{2 p}{2}=p \\ &R . H \cdot L=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{2 h+1}{h-2}=\frac{-1}{2} \end{aligned}

As f(x) is continuous at x = 0

Hence, p = -1/2
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