JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(x=2\) be a local minima of the function \(f(x)=2 x^4-18 x^2+8 x+12, x \in(-4,4)\). If \(M\) is local maximum value of the function \(f\) in \((-4,4)\), then \(M =\)
- A \(12 \sqrt{6}-\frac{33}{2}\)
- B \(12 \sqrt{6}-\frac{31}{2}\)
- C \(18 \sqrt{6}-\frac{33}{2}\)
- D \(18 \sqrt{6}-\frac{31}{2}\)
Answer & Solution
Correct Answer
(A) \(12 \sqrt{6}-\frac{33}{2}\)
Step-by-step Solution
Detailed explanation
\(f^{\prime}(x)=8 x^3-36 x+8=4\left(2 x^3-9 x+2\right)\) \(f^{\prime}(x)=0\) \(\therefore x=\frac{\sqrt{6}-2}{2}\) Now \(f(x)=\left(x^2-2 x-\frac{9}{2}\right)\left(2 x^2+4 x-1\right)+24 x+7.5\) \(\therefore f\left(\frac{\sqrt{6}-2}{2}\right)=M=12 \sqrt{6}-\frac{33}{2}\)
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