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

$f(x) \begin{cases}=\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} & \text {, if } 1 \leq x < 0 \\ =\frac{2 x+1}{x-2} & \text {, if } 0 \leq x \leq 1\end{cases}$ is continuous in the interval $[-1,1]$, then $\mathrm{p}=$
  1. 1
  2. -1
  3. $\frac{-1}{2}$
  4. $\frac{1}{2}$

Solution

$\begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} \\ & =\lim _{x \rightarrow 0^{-1}} \frac{[(\sqrt{1+p x})-(\sqrt{1-p x})][(\sqrt{1+p x})+(\sqrt{1-p x})]}{x[(\sqrt{1+p x})+(\sqrt{1-p x})]} \\ & =\lim _{x \rightarrow 0^{-}} \frac{[(1+p x)-(1-p x)]}{x[\sqrt{1+p x}+\sqrt{1-p x}]}=\lim _{x \rightarrow 0^{-}} \frac{2 p}{\sqrt{1+p x}+\sqrt{1-p x}} \\ & =\frac{2 p}{2}=p \\ & \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{2 x+1}{x-2}=\frac{1}{-2} \end{aligned}$ Since $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$, we get $\mathrm{p}=\frac{-1}{2}$

Asked in: MHT CET 2021 (21 Sep Shift 2)

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