Let x = –1 and x = 2 be the critical points of the function f(x) = x^3 + ax^2 + b loge |x| + 1, x ≠ 0.

Question

Let x = -1 and 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 \(\log _{\mathrm{e}} 2=0.7\)):

(1) 21.1

(2) 19.8

(3) 22.1

(4) 20.9

Answer 1

Correct option is: (1) 21.1  

\(f(x)=x^{2}+a x^{2}+b \ell n |x|+1, \quad x \neq 0\)

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

\(\mathrm{f}^{\prime}(-1)=3-2 \mathrm{a}-\mathrm{b}=0\)

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

\(\mathrm{a}=\frac{-9}{2}, \mathrm{~b}=12\)

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

Max. at \(\mathrm{n}=-1\)

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

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

\(\mathrm{M}=-4.5\)

Min. value at x = -2

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

\(m=-25+12\ \ell \mathrm{n} 2=-16.6\)

\(|\mathrm{M}+\mathrm{m}|=21.1\)