Q100 / 160 · AP EAMCET 2023 (17 May Shift 1) · Mathematics

Let $[t]$ represents the greatest integer not exceeding $t$ and $\mathrm{C}=1-2 \mathrm{e}^2$. If the function $f(x)=\left\{\begin{array}{cc} {\left[e^x\right],} & x < 0 \\ a e^x+[x-2], & 0 \leq x < 2 \\ {\left[e^{-x}\right]-C,} & x \geq 2 \end{array}\right.$ is continuous at $\mathrm{x}=2$, then $\mathrm{f}(\mathrm{x})$ is discontinuous at

  • Correct answer
    A
    $\mathrm{x}=1$ only
  • B
    $x=0$ and $x=1$
  • C
    $x=0$ only
  • D
    $x=0, x=1$ and $x=\frac{1}{2}$

Solution

$\because f(x)$ is continuous at $x=2$ $\begin{aligned} & \therefore \mathrm{LHL}=f(2) \\ & \Rightarrow a e^2+[2-2]=\left[e^{-2}\right]-\mathrm{C} \\ & \Rightarrow a e^2-1=0-\left(1-2 e^2\right) \Rightarrow a=2 \end{aligned}$

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

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