Let $x=-1$ and $x=2$ be the critical points of the function $\mathrm{f}(\mathrm{x})=\mathrm{x}^3+\mathrm{ax}^2+\mathrm{b} \log _{\mathrm{c}}|\mathrm{x}|+1, \mathrm{x}
eq 0$. Let $m$ and $M$ respectively be the absolute minimum and the absolute maximum values of $f$ in the interval $\left[-2…

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{f}(\mathrm{x})=\mathrm{x}^2+\mathrm{ax}^2+\mathrm{b} \ell \mathrm{n}|\mathrm{x}|+1, \quad \mathrm{x} 
eq 0 \ & \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^2+2 \mathrm{ax}+\frac{\mathrm{b}}{\mathrm{x}} \ & \mathrm{f}^{\prime}(-1)=3-2 \mathrm{a}-\mathrm{b}=0 \ & \mathrm{f}^{\prime}(-2)=12+4 \mathrm{a}-\frac{\mathrm{b}}{2}=0 \ & \mathrm{a}=\frac{-9}{2}, \mathrm{~b}=12 \ & \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^2-9 \mathrm{x}+\frac{12}{\mathrm{x}}=\frac{3(\mathrm{x}+1)(\mathrm{x}+2)^2}{\mathrm{x}} \end{aligned}$ Max. at $\mathrm{n}=-1$ $f(x)=x^2-\frac{9}{2} x^2+12 \ln |x|+1$ $\begin{aligned} & \mathrm{f}(-1)=-1-\frac{9}{2}+1=-\frac{9}{2} \ & \mathrm{M}=-4.5 \end{aligned}$ Min. value at $\mathrm{x}=-2$ $\begin{aligned} & \mathrm{f}(-2)=-8-18+12 \ell \mathrm{n} 2+1 \ & \mathrm{~m}=-25+12 \ell \mathrm{n} 2=-16.6 \ & |\mathrm{M}+\mathrm{m}|=21.1 \end{aligned}$

Let $x=-1$ and $x=2$ be the critical points of the function…

Solution