JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(x=-1\) and \(x=2\) be the critical points of the function \(\mathrm{f}(\mathrm{x})=\mathrm{x}^3+\mathrm{ax}^2+\mathrm{b} \log _{\mathrm{c}}|\mathrm{x}|+1, \mathrm{x} \neq 0\). 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 \(\log _{\mathrm{c}} 2=0.7\) ):
- A 21.1
- B 19.8
- C 22.1
- D 20.9
Answer & Solution
Correct Answer
(A) 21.1
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{f}(\mathrm{x})=\mathrm{x}^2+\mathrm{ax}^2+\mathrm{b} \ell \mathrm{n}|\mathrm{x}|+1, \quad \mathrm{x} \neq 0 \\ & \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^2+2 \mathrm{ax}+\frac{\mathrm{b}}{\mathrm{x}} \\ & \mathrm{f}^{\prime}(-1)=3-2…
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