Mathematics · Application of Derivatives

JEE Main 2025 — 2 April, Morning Shift — Question 37

If the function $f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1$ , where a $>0$ , attains its local maximum and local

minimum values at $p$ and $q$ , respectively, such that $p^{2}=q$ , then $f(3)$ is equal to

Answer: B

$f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1, a>0$

\begin{aligned} f^{\prime}(x) & =6 x^{2}-18 a x+12 a^{2}=0 \\& =6\left(x^{2}-3 a x+2 a^{2}\right) \\& =6(x-a)(x-2 a)=0 \end{aligned}

$x=a, 2 a$

$\therefore \quad x=a$ is point of maxima $x=2 a$ is point of minima

$\therefore \quad p=a, q=2 a$

$p^{2}=q$ (Given)

$a^{2}=2 a$

$\Rightarrow \quad a=2$

$f(x)=2 x^{3}-18 x^{2}+48 x+1$

$f(3)=37$

Answer key and solution verified before publishing.

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