MHTCET 25 Apr 2025 Shift 1 Paper

Concept:A differentiable function is strictly decreasing on an interval if its derivative is negative throughout that interval.Explanation:The given function is $f(x)=\frac{x}{2}+\frac{2}{x}$ with $x \neq 0$.Differentiating with respect to $x$, we get:$$f'(x)=\frac{1}{2}-\frac{2}{x^{2}}$$For strictly decreasing, we need $f'(x) < 0$.Therefore, $\frac{1}{2}-\frac{2}{x^{2}} < 0$.Rearranging gives $\frac{2}{x^{2}} > \frac{1}{2}$, which implies $x^{2} < 4$.Also, $x \neq 0$ because the original function is not defined at zero.Hence, $x \in (-2,0) \cup (0,2)$.Among the given options, the interval $(1,2)$ lies completely inside $(0,2)$.The interval $(-2,2)$ includes $x=0$, where the function is not defined, so it is not a valid interval of decrease.Therefore, the function is strictly decreasing in $(1,2)$.Answer:Option (D) $(1,2)$.