MHT CET · Maths · Application of Derivatives
The function \(f(x)=(x+2) e^{-x}\) is
- A decreasing in \((-\infty,-1)\) and increasing in \((-1, \infty)\)
- B decreasing for all \(x\)
- C increasing in \((-\infty,-1)\) and decreasing in \((-1, \infty)\)
- D increasing for all \(x\)
Answer & Solution
Correct Answer
(C) increasing in \((-\infty,-1)\) and decreasing in \((-1, \infty)\)
Step-by-step Solution
Detailed explanation
(C)
Given
\(\begin{array}{l}
f(x)=(x+2) e^{-x} \\
\begin{aligned}
\therefore f^{\prime}(x) &=(x+2)\left(e^{-x}\right)(-1)+e^{-x}(1) \\
&=e^{-x}[1-(x+2)]=e^{-x}(-x-1) \\
&=-e^{-x}(x+1)
\end{aligned}
\end{array}\)
Here \(e^{-x}\) is always positive.
\(\begin{array}{l}
\text { Now }-(x+1)>0 \Rightarrow-x>1 \Rightarrow x < -1 \\
\text { Also }-(x+1) < 0 \Rightarrow-x < 1 \Rightarrow x>-1
\end{array}\)
Thus \(\mathrm{f}(\mathrm{x})\) increases in \((-\infty,-1)\) and decreases in \((-1, \infty)\)
Given
\(\begin{array}{l}
f(x)=(x+2) e^{-x} \\
\begin{aligned}
\therefore f^{\prime}(x) &=(x+2)\left(e^{-x}\right)(-1)+e^{-x}(1) \\
&=e^{-x}[1-(x+2)]=e^{-x}(-x-1) \\
&=-e^{-x}(x+1)
\end{aligned}
\end{array}\)
Here \(e^{-x}\) is always positive.
\(\begin{array}{l}
\text { Now }-(x+1)>0 \Rightarrow-x>1 \Rightarrow x < -1 \\
\text { Also }-(x+1) < 0 \Rightarrow-x < 1 \Rightarrow x>-1
\end{array}\)
Thus \(\mathrm{f}(\mathrm{x})\) increases in \((-\infty,-1)\) and decreases in \((-1, \infty)\)
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