JEE Main 2025 Apr 4 Shift 2, Mathematics Q1: Maxima & Minima

JEE Main2025Apr 4, Shift 2Mathematics

Let $a > 0$ . If the function $f (x) = 6 x^{3} - 45 a x^{2} + 108 a^{2} x + 1$ attains its local maximum and minimum values at the points $x_{1}$ and $x_{2}$ respectively such that $x_{1} x_{2} = 54$ , then $a + x_{1} + x_{2}$ is equal to :-

A
15
B
18
C
24
D
13

Solution

Differentiate and set to zero to locate the critical points.

$f^{'} (x) = 18 x^{2} - 90 a x + 108 a^{2} = 18 (x^{2} - 5 a x + 6 a^{2}) = 18 (x - 2 a) (x - 3 a)$ .

So the critical points are $x = 2 a$ and $x = 3 a$ . Since the leading coefficient of $f$ is positive, $f$ has a local maximum at the smaller root ( $x_{1} = 2 a$ ) and a local minimum at the larger ( $x_{2} = 3 a$ ).

Using $x_{1} x_{2} = 54$ : $2 a \cdot 3 a = 6 a^{2} = 54 \Rightarrow a^{2} = 9 \Rightarrow a = 3$ (taking $a > 0$ ).

$∴ a + x_{1} + x_{2} = 3 + 6 + 9 = 18$ .

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JEE Main 2025 Apr 4 Shift 2, Mathematics Q1: Maxima & Minima

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