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Observe the statements given below :
Assertion (A) : $f(x)=x e^{-x}$ has the maximum at $x=1$
Reason (R) : $f^{\prime}(1)=0$ and $f^{\prime \prime}(1) < 0$
Which of the following is correct?
Options:
Assertion (A) : $f(x)=x e^{-x}$ has the maximum at $x=1$
Reason (R) : $f^{\prime}(1)=0$ and $f^{\prime \prime}(1) < 0$
Which of the following is correct?
Solution:
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Verified Answer
The correct answer is:
Both $(A)$ and (R) are true and (R) is the correct reason for (A)
Given, $f(x)=x e^{-x}$
$\begin{aligned} f^{\prime}(x) & =e^{-x}-x e^{-x} \\ f^{\prime \prime}(x) & =-e^{-x}-e^{-x}+x e^{-x} \\ & =-2 e^{-x}+x e^{-x}\end{aligned}$
For maximum, put
$f^{\prime}(1)=0 \Rightarrow x=1$ and $f^{\prime \prime}(1)=-1 < 0$
$\therefore$ Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is the correct reason for $\mathrm{A}$.
$\begin{aligned} f^{\prime}(x) & =e^{-x}-x e^{-x} \\ f^{\prime \prime}(x) & =-e^{-x}-e^{-x}+x e^{-x} \\ & =-2 e^{-x}+x e^{-x}\end{aligned}$
For maximum, put
$f^{\prime}(1)=0 \Rightarrow x=1$ and $f^{\prime \prime}(1)=-1 < 0$
$\therefore$ Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is the correct reason for $\mathrm{A}$.
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