Q53 / 150 · MHT CET 2023 (13 May Shift 1) · Mathematics

The function $\mathrm{f}(\mathrm{t})=\frac{1}{\mathrm{t}^2+\mathrm{t}-2}$ where $\mathrm{t}=\frac{1}{x-1}$ is discontinuous at

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

Solution

$$ f(t)=\frac{1}{t^2+t-2}=\frac{1}{(t+2)(t-1)} $$ $\mathrm{f}(\mathrm{t})$ is not defined at $\mathrm{t}=-2$ and $\mathrm{t}=1$. $$ \begin{aligned} & \mathrm{t}=-2 \\ & \Rightarrow \frac{1}{x-1}=-2 \\ & \Rightarrow x=\frac{1}{2} \\ & \mathrm{t}=1 \\ & \Rightarrow \frac{1}{x-1}=1 \\ & \Rightarrow x=2 \end{aligned} $$ $\therefore \quad$ The function $\mathrm{f}(\mathrm{t})$ is discontinuous at $x=\frac{1}{2}$ and $x=2$.

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

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