MHTCET 22 Apr 2025 Shift 1 Paper

Concept:The sign of $f'(x)$ determines whether $f(x)$ is increasing or decreasing.Explanation:Given $f(x)=x e^{x(1-x)}$.Using the product rule:$f'(x)=e^{x(1-x)} + x e^{x(1-x)}(1-2x)$Simplify:$f'(x)=e^{x(1-x)}(1+x-2x^2)=e^{x(1-x)}(-2x^2+x+1)$Since $e^{x(1-x)}>0$ for all $x$, the sign of $f'(x)$ depends only on $-2x^2+x+1$.For $f(x)$ to be increasing, $f'(x) \ge 0$$-2x^2+x+1 \ge 0$Multiplying by $-1$ and reversing the inequality:$2x^2-x-1 \le 0$Factorizing:$(2x+1)(x-1) \le 0$Thus, $x \in \left[-\frac{1}{2},1\right]$.So $f(x)$ is increasing on $\left(-\frac{1}{2},1\right)$.For $f(x)$ to be decreasing, $f'(x) \le 0$:$(2x+1)(x-1) \ge 0$Hence, $x \in \left(-\infty,-\frac{1}{2}\right] \cup [1,\infty)$.Therefore, the correct option is B.Answer:B. increasing in $\left(-\frac{1}{2},1\right)$