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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
Options:
$f(x)=\left\{\begin{array}{cc}
{\left[e^x\right],} & x < 0 \\
a e^x+[x-2], & 0 \leq x < 2 \\
{\left[e^{-x}\right]-C,} & x \geq 2
\end{array}\right.$
is continuous at $\mathrm{x}=2$, then $\mathrm{f}(\mathrm{x})$ is discontinuous at
Solution:
1302 Upvotes
Verified Answer
The correct answer is:
$\mathrm{x}=1$ only
$\because f(x)$ is continuous at $x=2$
$\begin{aligned}
& \therefore \mathrm{LHL}=f(2) \\
& \Rightarrow a e^2+[2-2]=\left[e^{-2}\right]-\mathrm{C} \\
& \Rightarrow a e^2-1=0-\left(1-2 e^2\right) \Rightarrow a=2
\end{aligned}$
$\begin{aligned}
& \therefore \mathrm{LHL}=f(2) \\
& \Rightarrow a e^2+[2-2]=\left[e^{-2}\right]-\mathrm{C} \\
& \Rightarrow a e^2-1=0-\left(1-2 e^2\right) \Rightarrow a=2
\end{aligned}$
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