MHTCET 13 Apr 2026 Shift 1 Paper

Concept:For $f(x)=e^{\sqrt{5x-3-2x^2}}$ to be defined, the expression inside the square root must be non-negative.Explanation:The exponential part $e^y$ is defined for all real $y$.So we only need $\sqrt{5x-3-2x^2}$ to be real.Therefore, $5x-3-2x^2 \ge 0$.Rewrite it as $-2x^2+5x-3 \ge 0$.Multiply by $-1$ and reverse the inequality sign:$2x^2-5x+3 \le 0$.Factorising gives $(2x-3)(x-1) \le 0$.The roots are $x=1$ and $x=\frac{3}{2}$.Since the quadratic opens upwards, its value is non-positive between the roots.Hence, $x \in \left[1, \frac{3}{2}\right]$.Answer:The domain is $\left[1, \frac{3}{2}\right]$, which is Option C.