1. Given function and critical points

We have $f(x)=x^3+ax^2+b\ln|x|+1, \qquad x\ne 0$ and the critical points are $x=-1$ and $x=2$.

So, $f'(x)=3x^2+2ax+\frac{b}{x}$ for $x\ne 0$.

Since $x=-1$ and $x=2$ are critical points, $f'(-1)=0, \qquad f'(2)=0.$

2. Use the critical point conditions to find $a,b$

From $f'(-1)=0$, $3(-1)^2+2a(-1)+\frac{b}{-1}=0$ $3-2a-b=0$ $2a+b=3 \quad ...(1)$

From $f'(2)=0$, $3(2)^2+2a(2)+\frac{b}{2}=0$ $12+4a+\frac{b}{2}=0$ Multiplying by $2$, $24+8a+b=0 \quad ...(2)$

Now subtract (1) from (2): $(24+8a+b)-(2a+b)=0-3$ $24+6a=-3$ $6a=-27$ $a=-\frac{9}{2}$

Then from (1), $2\left(-\frac{9}{2}\right)+b=3$ $-9+b=3$ $b=12$

So the function is $f(x)=x^3-\frac{9}{2}x^2+12\ln|x|+1.$

3. Find absolute maximum and minimum on $\left[-2,-\frac12\right]$

Since the interval is closed and lies away from $0$, absolute extrema occur at critical points inside the interval or at endpoints.

The only given critical points are $x=-1$ and $x=2$. In the interval $\left[-2,-\frac12\right]$, only $x=-1$ lies inside.

So we evaluate $f(x)$ at:

• $x=-2$
• $x=-1$
• $x=-\frac12$

4. Compute function values

At $x=-2$

$f(-2)=(-2)^3-\frac{9}{2}(-2)^2+12\ln 2+1$ $=-8-\frac{9}{2}\cdot 4+12\ln 2+1$ $=-8-18+12\ln 2+1$ $=-25+12\ln 2$ Using $\ln 2=0.7$, $f(-2)=-25+8.4=-16.6$

At $x=-1$

$f(-1)=(-1)^3-\frac{9}{2}(1)+12\ln 1+1$ Since $\ln 1=0$, $f(-1)=-1-\frac{9}{2}+0+1=-\frac{9}{2}=-4.5$

At $x=-\frac12$

$f\left(-\frac12\right)=\left(-\frac12\right)^3-\frac{9}{2}\left(\frac14\right)+12\ln\left(\frac12\right)+1$ $=-\frac18-\frac98+12(-\ln 2)+1$ $=-\frac{10}{8}+1-12\ln 2$ $=-\frac54+1-12\ln 2$ $=-\frac14-12\ln 2$ Using $\ln 2=0.7$, $f\left(-\frac12\right)=-0.25-8.4=-8.65$

5. Identify absolute maximum and minimum

From the values: $f(-2)=-16.6, \qquad f(-1)=-4.5, \qquad f\left(-\frac12\right)=-8.65$

Thus,

• Absolute maximum: $M=-4.5$
• Absolute minimum: $m=-16.6$

6. Compute $|M+m|$

$M+m=-4.5+(-16.6)=-21.1$ Hence, $|M+m|=21.1$

7. Match with options

$\boxed{21.1}$ So the correct option is A.