Q78 / 150 · MHT CET 2023 (11 May Shift 1) · Mathematics

If $\mathrm{f}(x)=\left\{\begin{array}{ll}\frac{\sqrt{1+\mathrm{m} x}-\sqrt{1-\mathrm{m} x}}{x} & ,-1 \leq x < 0 \\ \frac{2 x+1}{x-2} & , 0 \leq x \leq 1\end{array}\right.$ is continuous in the interval $[-1,1]$, then $\mathrm{m}$ is equal to

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

Solution

Since $\mathrm{f}(x)$ is continuous in $[-1,1]$, it is continuous at $x=0$. $$ \begin{aligned} \therefore \quad & \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x) \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{1+\mathrm{m} x}-\sqrt{1-\mathrm{m} x}}{x}=\lim _{x \rightarrow 0} \frac{2 x+1}{x-2} \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{(1+\mathrm{m} x-1+\mathrm{m} x)}{x(\sqrt{1+\mathrm{m} x}+\sqrt{1-\mathrm{m} x})}=\frac{2(0)+1}{0-2} \\ & \Rightarrow \frac{2 \mathrm{~m}}{1+1}=\frac{1}{-2} \\ & \Rightarrow \mathrm{m}=\frac{-1}{2} \end{aligned} $$

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

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