JEE Main 2025 Apr 7 Shift 2, Mathematics Q8: Maxima & Minima

JEE Main2025Apr 7, Shift 2Mathematics

Let f : $R \to R$ be a polynomial function of degree four having extreme values at $x = 4$ and $x = 5$ .

If $x \to 0 lim \frac{f ( x )}{x ^{2}} = 5$ , then $f (2)$ is equal to :

A
12
B
10
C
8
D
14

Solution

Write $f (x) = a x^{4} + b x^{3} + c x^{2} + d x + e$ . For $lim_{x \to 0} f (x) / x^{2}$ to be a finite value 5, the constant and linear terms must vanish ( $e = 0, d = 0$ ) and the coefficient of $x^{2}$ must equal 5, i.e. $c = 5$ . So $f (x) = a x^{4} + b x^{3} + 5 x^{2}$ .

$f^{'} (x) = 4 a x^{3} + 3 b x^{2} + 10 x = x (4 a x^{2} + 3 b x + 10)$ . Extreme values at $x = 4$ and $x = 5$ require $f^{'} (4) = 0$ and $f^{'} (5) = 0$ :

$64 a + 12 b + 10 = 0$ and $100 a + 15 b + 10 = 0$ .

Solving these simultaneously gives $a = \frac{1}{8}$ and $b = - \frac{3}{2}$ .

$f (2) = \frac{1}{8} (16) - \frac{3}{2} (8) + 5 (4) = 2 - 12 + 20 = 10$ .

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JEE Main 2025 Apr 7 Shift 2, Mathematics Q8: Maxima & Minima

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