Q48 / 100 · BITSAT 2024 (Memory Based Paper 3) · Mathematics

Let $f$ be the function defined by $f(x)=\left\{\begin{array}{cc} \frac{x^2-1}{x^2-2|x-1|-1}, & x \neq 1 \\ \frac{1}{2}, & x=1 \end{array}\right.$

  • A
    The function is continuous for all values of $x$
  • B
    The function is continuous only for $x\gt1$
  • C
    The function is continuous at $x=1$
  • Correct answer
    D
    The function is not continuous at $x=1$

Solution

$f(x)=\left\{\begin{array}{cc}\frac{x^2-1}{x^2-2 x+1} ; & x\gt1 \\ \frac{1}{2} \quad ; & x=1 \\ \frac{x^2-1}{x^2+2 x-3} ; & x \lt 1\end{array}\right.$ $\because \quad x^2+3 x-x-3$ $=x(x+3)-1(x+3)$ Lets Find LHL, $\lim _{x \rightarrow 1^{-}} \frac{x^2-1}{x^2-2 x+1}$ and RHL, $\lim _{x \rightarrow 1^{+}} \frac{x^2-1}{x^2+2 x-3}$ $\begin{aligned} & \Rightarrow \lim _{x \rightarrow 1^{-}} \frac{(x-1)(x+1)}{(x-1)^2} \Rightarrow \lim _{x \rightarrow 1^{+}} \frac{(x-1)(x+1)}{(x+3)(x-1)} \\ & \Rightarrow \lim _{x \rightarrow 1^{-}} \frac{x+1}{x-1}=\infty \Rightarrow \lim _{x \rightarrow 1^{+}} \frac{x+1}{x+3}=\frac{2}{4}=\frac{1}{2} \end{aligned}$ $\therefore$ LHL $\neq f(1) \Rightarrow f(x)$ is not continuous at $x=1$

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

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