Q242 / 270 · NDA 2021 (Phase 1) · Mathematics

If a differentiable function $f(x)$ satisfies $\lim _{x \rightarrow-1} \frac{f(x)+1}{x^2-1}=-\frac{3}{2}$ then what is $\lim _{x \rightarrow-1} f(x)$ equal to?

  • Correct answer
    A
    $-\frac{3}{2}$
  • B
    $-1$
  • C
    $0$
  • D
    $1$

Solution

$\begin{aligned} & \lim _{x \rightarrow-1} \frac{f(x)+1}{x^2-1}=\frac{-3}{2} \\ & {\left[\frac{f^{\prime}(x)}{2 x}\right]=\frac{-3}{2}}\end{aligned}$ On integrating both sides, $\begin{aligned} & f(x)=\frac{-3}{2} x^2+c \quad \text { where } c=\text { Constant } \\ & \lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow-1}\left[\frac{-3}{2} x^2+c\right]=\frac{-3}{2}\end{aligned}$

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

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