MHTCET 11 May 2024 Shift 2 Paper

Concept:To find the maximum value of a continuous function on a closed interval, check critical points and endpoints.Explanation:Given, $\mathrm{f}(x)=2x^3 - 15x^2 + 36x - 48$.The set is $A = \{x \mid x^2 + 20 \leq 9x\}$.Rewrite: $x^2 - 9x + 20 \leq 0$.Factor: $(x-4)(x-5) \leq 0$.So, $A = [4, 5]$.Now, differentiate: $\mathrm{f}'(x) = 6x^2 - 30x + 36$.Set $\mathrm{f}'(x) = 0$: $6(x-2)(x-3) = 0$.Thus, critical points are $x = 2$ and $x = 3$.But $2, 3 \notin A$, so they are rejected.Therefore, check endpoints of $[4, 5]$.$\mathrm{f}(4) = 2(4)^3 - 15(4)^2 + 36(4) - 48 = -16$.$\mathrm{f}(5) = 2(5)^3 - 15(5)^2 + 36(5) - 48 = 7$.Since $7 > -16$, the maximum value is $7$.Answer:The maximum value of $\mathrm{f}(x)$ on set $A$ is $7$, i.e. option D.