JEE Main 2025 Apr 2 Shift 1, Mathematics Q9: Maxima & Minima

JEE Main2025Apr 2, Shift 1Mathematics

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:

A
55
B
10
C
23
D
37

Solution

Differentiate:

$f^{'} (x) = 6 x^{2} - 18 a x + 12 a^{2} = 6 (x - a) (x - 2 a)$

The critical points are $x = a$ and $x = 2 a$ . Since $a > 0$ and $f^{''} (x) = 12 x - 18 a$ :

$f^{''} (a) = 12 a - 18 a = - 6 a < 0$ , so $x = a$ is the local maximum, i.e. $p = a$ .

$f^{''} (2 a) = 24 a - 18 a = 6 a > 0$ , so $x = 2 a$ is the local minimum, i.e. $q = 2 a$ .

The condition $p^{2} = q$ gives $a^{2} = 2 a$ , so $a = 2$ (since $a > 0$ ).

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

$f (3) = 54 - 162 + 144 + 1 = 37$

$∴$ Option (4) is correct.

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JEE Main 2025 Apr 2 Shift 1, Mathematics Q9: Maxima & Minima

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