Let \(\mathrm{x}=-1\) and \(\mathrm{x}=2\) be the critical points of the function \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{ax}^{2}+\mathrm{b} \log _{\mathrm{e}}|\mathrm{x}|+1, \mathrm{x} \neq 0\). Let \(m\) and M respectively be the absolute minimum and the absolute maximum values of \(f\) in the interval \(\left[-2,-\frac{1}{2}\right]\). Then \(|\mathrm{M}+m|\) is equal to (Take \(\log _{\mathrm{e}} 2=0.7\) ):

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Application of Derivatives

Let

x = - 1

and

x = 2

be the critical points of the function

f (x) = x^{3} + ax^{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

Answer: (A)

For

f (x) = x^{3} + a x^{2} + b lo g_{e} ∣ x ∣ + 1

the derivative is

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

. The conditions

f^{'} (- 1) = 0

and

f^{'} (2) = 0

give

3 - 2 a - b = 0

and

12 + 4 a + \frac{b}{2} = 0

, from which

a = - \frac{9}{2}

and

b = 12

, so

f^{'} (x) = \frac{3 ( x + 1 ) ( x - 2 ) ^{2}}{x}

. On

[- 2, - \frac{1}{2}]

the factor

(x - 2)^{2}

is positive and

x

is negative, so

f^{'}

carries the sign of

- (x + 1)

: positive on

(- 2, - 1)

and negative on

(- 1, - \frac{1}{2})

, making

x = - 1

the maximum. Thus

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

, while comparing the ends gives

f (- 2) = - 8 - 18 + 12 lo g_{e} 2 + 1 = - 25 + 12 (0.7) = - 16.6

and

f (- \frac{1}{2}) = - 8.65

, so the minimum is

m = - 16.6

. Therefore

∣ M + m ∣ = ∣ - 4.5 - 16.6∣ = 21.1

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option