JEE MAIN - Mathematics (2025 - 7th April Morning Shift - No. 18)
Let $x=-1$ and $x=2$ be the critical points of the function $f(x)=x^3+a x^2+b \log _{\mathrm{e}}|x|+1, x \neq 0$.
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 $|\mathrm{M}+m|$ is equal to
$\left(\right.$ Take $\left.\log _{\mathrm{e}} 2=0.7\right):$
21.1
19.8
22.1
20.9
Explanation
_7th_April_Morning_Shift_en_18_1.png)
$\left.\begin{array}{c}3-2 a-b=0 \\ 12+4 a+\frac{b}{2}=0\end{array}\right\} \begin{aligned} & a=\frac{-9}{2} \\ & b=12\end{aligned}$
$$\begin{aligned} & \therefore f(x)=x^3-\frac{9}{2} x^2+12 \ln |x|+1 \\ & f(-1)=-1-\frac{9}{2}+1=-\frac{9}{2}=-4.5 \\ & f(-2)=-8-18+12 \ln 2+1 \\ & \quad=-25+12 \ln 2=-16.6 \\ & f\left(-\frac{1}{2}\right)=-\frac{1}{8}-\frac{9}{8}+12 \ln \left(\frac{1}{2}\right)+1=-8.5 \\ & |M+m|=|-16.6-4.5|=21.1 \end{aligned}$$
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