Q103 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics

If $f(x)=\left\{\begin{array}{cc}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & -1 \leq x < 0 \\ \frac{2 x+1}{x-2}, & 0 \leq x \leq 1\end{array}\right.$ is continuous on $[-1,1]$, then $p=$

  • Correct answer
    A
    $-\frac{1}{2}$
  • B
    $-\frac{1}{4}$
  • C
    $\frac{1}{2}$
  • D
    2

Solution

$f(x)=\left\{\begin{array}{c}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x},-1 \leq x < 0 \\ \frac{2 x+1}{x-2}, 0 \leq x \leq 1\end{array}\right.$ Continues in $[-1,1]$, so it is continues at $x=0$ also So, $\quad \lim _{x \rightarrow 0^{\circ}} f(x)=\lim _{x \rightarrow 0^{\oplus}} f(x)=f(0)$ $$ \begin{aligned} \lim _{x \rightarrow 0^{\circ}} \frac{\sqrt{1-P x}-\sqrt{1+p x}}{-x} & =\frac{2(0)+1}{0-2} \\ \Rightarrow \lim _{x \rightarrow 0} \frac{1-\frac{1}{2} p x-1-\frac{1}{2} p x+\ldots}{-x} & =-\frac{1}{2} \\ \Rightarrow \quad \frac{-p x}{-x} & =-\frac{1}{2} \Rightarrow p=-\frac{1}{2} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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