JEE MAIN - Mathematics (2022 - 28th July Evening Shift - No. 6)
The function $$f(x)=x \mathrm{e}^{x(1-x)}, x \in \mathbb{R}$$, is :
increasing in $$\left(-\frac{1}{2}, 1\right)$$
decreasing in $$\left(\frac{1}{2}, 2\right)$$
increasing in $$\left(-1,-\frac{1}{2}\right)$$
decreasing in $$\left(-\frac{1}{2}, \frac{1}{2}\right)$$
Explanation
$$f(x) = x{e^{x(1 - x)}},\,x \in R$$
$$f'(x) = x{e^{x(1 - x)}}\,.\,(1 - 2x) + {e^{x(1 - x)}}$$
$$ = {e^{x(1 - x)}}[x - 2{x^2} + 1]$$
$$ = - {e^{x(1 - x)}}[2{x^2} - x - 1]$$
$$ = - {e^{x(1 - x)}}(2x + 1)(x - 1)$$
$$\therefore$$ $$f(x)$$ is increasing in $$\left( { - {1 \over 2},1} \right)$$ and decreasing in $$\left( { - \infty ,\, - {1 \over 2}} \right) \cup \left( {1,\infty } \right)$$
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