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Applications of Derivatives (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Applications of Derivatives (p-1)1 Mark
Question
The function $f(x)=x \cdot e^{x(1-x)}$ is increasing on $\left(\frac{-1}{2}, 1\right)$.
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Answer

TRUE
$
\begin{gathered}
f(x)=x e^{x(1-x)}=x e^{x-x^2} \\
\therefore f^{\prime}(x)=x \cdot \frac{d}{d x}\left(e^{x-x^2}\right)+e^{x-x^2} \cdot \frac{d}{d x}(x)
\end{gathered}
$
$
\begin{aligned}
& =x \cdot e^{x-x^2} \cdot \frac{d}{d x}\left(x-x^2\right)+e^{x-x^2} \times 1 \\
& =x \cdot e^{x-x^2} \times(1-2 x)+e^{x-x^2} \\
& =e^{x-x^2}\left(x-2 x^2+1\right) \\
& =-2 e^{x-x^2}\left(x^2-\frac{1}{2} x-\frac{1}{2}\right) \\
& =-2 e^{x-x^2}\left[\left(x^2-\frac{1}{2} x+\frac{1}{16}\right)-\frac{1}{2}-\frac{1}{16}\right] \\
\therefore f^{\prime}(x) & =-2 e^{x-x^2}\left[\left(x-\frac{1}{4}\right)^2-\frac{9}{16}\right]
\end{aligned}
$
Now, $x \in\left(-\frac{1}{2}, 1\right) \quad \therefore-\frac{1}{2}<x<1$
$
\begin{aligned}
& \therefore-\frac{1}{2}-\frac{1}{4}<x-\frac{1}{4}<1-\frac{1}{4}=\frac{3}{4} \\
& \therefore 0<\left(x-\frac{1}{4}\right)^2<\frac{9}{16} \\
& \therefore\left(x-\frac{1}{4}\right)^2-\frac{9}{16}<0
\end{aligned}
$
Also, $2 e^{x-x^2}>0$ for $x \in\left(-\frac{1}{2}, 1\right)$
$
\therefore-2 e^{x-x^2}<0
$
From (2) and (3),
$
\begin{aligned}
& -2 e^{x-x^2}\left[\left(x-\frac{1}{4}\right)^2-\frac{9}{16}\right]>0 \\
& \therefore f^{\prime}(x)>0
\end{aligned}
$
Hence, function $f(x)$ is increasing on $\left(\frac{-1}{2}, 1\right)$.

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