Let x = –1 and x = 2 be the critical points of the function $f (x) = x^{3} + a x^{2} + b \log_{e} | x | + 1, x \neq 0$ . 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 $\log_{e} 2 = 0.7$ ): [2025]

Question

Let x = –1 and x = 2 be the critical points of the function $f (x) = x^{3} + a x^{2} + b \log_{e} | x | + 1, x \neq 0$ . 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 $\log_{e} 2 = 0.7$ ): [2025]

Smart Achievers · Accepted Answer

(3)

We have, $f (x) = x^{3} + a x^{2} + b \log_{e} | x | + 1, x \neq 0$

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

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

$f' (2) = 12 + 4 a + \frac{b}{2} = 0 \Rightarrow a = \frac{- 9}{2}, b = 12$

$∴ f (x) = x^{3} - \frac{9}{2} x^{2} + 12 \log_{e} | x | + 1$

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

M = – 4.5

Min. value at x = – 2

$f (- 2) = - 8 - 18 + 12 \log_{e} 2 + 1$

m = – 25 + 12(0.7) = – 16.6

$∴$ |M + m| = 21.1

Solution

1 19.8

2 22.1

3 21.1

4 20.9

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