JEE Main 2025 Apr 7 Shift 1, Mathematics Q3: Maxima & Minima

JEE Main2025Apr 7, Shift 1Mathematics

Let $x = - 1$ and $x = 2$ be the critical points of the function $f (x) = x^{3} + a x^{2} + b lo g_{e} ∣ x ∣ + 1$ , $x \neq = 0$ . Let $m$ and $M$ respectively be the absolute minimum and the absolute maximum values of $f$ in the interval $[- 2, - \frac{1}{2}]$ . Then $∣ M + m ∣$ is equal to (Take $lo g_{e} 2 = 0.7$ ):

A
$21.1$
B
$19.8$
C
$22.1$
D
$20.9$

Solution

$f (x) = x^{3} + a x^{2} + b ln ∣ x ∣ + 1, x \neq = 0$

$f^{'} (x) = 3 x^{2} + 2 a x + \frac{b}{x}$

$f^{'} (- 1) = 3 - 2 a - b = 0$

$f^{'} (- 2) = 12 + 4 a - \frac{b}{2} = 0$

$\Rightarrow a = - \frac{9}{2}, b = 12$

$f'(x)=3x^2-9x+\dfrac{12}{x} =\dfrac{3(x+1)(x+2)^2}{x}$

Hence, the maximum occurs at $x = - 1$ .

$f (x) = x^{3} - \frac{9}{2} x^{2} + 12 ln ∣ x ∣ + 1$

$f (- 1) = - 1 - \frac{9}{2} + 1 = - \frac{9}{2}$

$∴ M = - 4.5$

The minimum occurs at $x = - 2$ .

$f (- 2) = - 8 - 18 + 12 ln 2 + 1$

$= - 25 + 12 ln 2 \approx - 16.6$

$∴ m = - 16.6$

$∣ M + m ∣ = ∣ - 4.5 + (- 16.6) ∣ \approx 21.1$

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JEE Main 2025 Apr 7 Shift 1, Mathematics Q3: Maxima & Minima

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