%%

The function $$f(x)=x \mathrm{e}^{x(1-x)}, x \in \mathbb{R}$$, is :

JEE · Math · previous-year question

  1. A.increasing in $$\left(-\frac{1}{2}, 1\right)$$correct
  2. B.decreasing in $$\left(\frac{1}{2}, 2\right)$$
  3. C.increasing in $$\left(-1,-\frac{1}{2}\right)$$
  4. D.decreasing in $$\left(-\frac{1}{2}, \frac{1}{2}\right)$$

Answer

A. increasing in $$\left(-\frac{1}{2}, 1\right)$$

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 $$\left( { - \infty ,\, - {1 \over 2}} \right) \cup \left( {1,\infty } \right)$$

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions