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

For $$x \neq 0$$ the function is $$f(x)=x^{3}+ax^{2}+b\,\log_e|x|+1$$.

Critical points satisfy $$f'(x)=0$$.
Differentiate: $$f'(x)=3x^{2}+2ax+\dfrac{b}{x}$$.

The points $$x=-1$$ and $$x=2$$ are given to be critical, so

At $$x=-1$$: $$3(-1)^{2}+2a(-1)+\dfrac{b}{-1}=0 \;\;\Longrightarrow\;\; 3-2a-b=0$$ $$\Rightarrow\; 2a+b=3 \;-(1)$$

At $$x=2$$: $$3(2)^{2}+2a(2)+\dfrac{b}{2}=0 \;\;\Longrightarrow\;\; 12+4a+\dfrac{b}{2}=0$$ Multiply by $$2$$: $$24+8a+b=0 \;-(2)$$

From $$(2)$$ we get $$b=-24-8a$$.
Put this in $$(1)$$:

$$2a+(-24-8a)=3 \;\;\Longrightarrow\;\; -6a=27 \;\;\Longrightarrow\;\; a=-\dfrac{9}{2}$$

Hence $$b=-24-8\left(-\dfrac{9}{2}\right)=12$$.

The required interval is $$\left[-2,\,-\dfrac{1}{2}\right]$$. Inside it, possible locations of absolute extrema are:

• the critical point $$x=-1$$ (it lies inside the interval)
• the endpoints $$x=-2$$ and $$x=-\dfrac{1}{2}$$.

Substitute $$a=-\dfrac{9}{2},\; b=12$$ in $$f(x)$$:

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

Case 1:

$$x=-1$$:
$$f(-1)=(-1)^{3}-\dfrac{9}{2}(1)+12\log_e 1+1$$ $$f(-1)=-1-\dfrac{9}{2}+0+1=-\dfrac{9}{2}=-4.5$$

Case 2:

$$x=-2$$:
$$f(-2)=(-8)-\dfrac{9}{2}(4)+12\log_e 2+1$$ $$f(-2)=-8-18+12\log_e 2+1=-25+12\log_e 2$$

Case 3:

$$x=-\dfrac{1}{2}$$:
$$f\!\left(-\dfrac{1}{2}\right)=\left(-\dfrac{1}{8}\right)-\dfrac{9}{2}\!\left(\dfrac{1}{4}\right)+12\log_e\!\left(\dfrac{1}{2}\right)+1$$ $$= -\dfrac{1}{8}-\dfrac{9}{8}-12\log_e 2+1$$ $$= -\dfrac{10}{8}-12\log_e 2+1=-\dfrac{5}{4}-12\log_e 2+1$$ $$= -\dfrac{1}{4}-12\log_e 2$$ (numerically about $$-8.57$$)

Compare the three values:

$$f(-2)=-25+12\log_e 2 \approx -16.68$$ (smallest)
$$f\!\left(-\dfrac{1}{2}\right)\approx -8.57$$
$$f(-1)=-4.5$$ (largest)

Therefore
Absolute minimum $$m=f(-2)=-25+12\log_e 2$$,
Absolute maximum $$M=f(-1)=-\dfrac{9}{2}$$.

Compute $$|M+m|$$:

$$M+m=-\dfrac{9}{2}+\bigl(-25+12\log_e 2\bigr)=-\dfrac{59}{2}+12\log_e 2$$ Since this value is negative, $$|M+m|=\dfrac{59}{2}-12\log_e 2$$.

Numerically: $$\dfrac{59}{2}=29.5,\; 12\log_e 2\approx 8.3178$$ $$|M+m|\approx 29.5-8.3178\approx 21.18$$.

This matches Option A (21.1). Hence

$$|M+m| \approx 21.1$$ (Option A).