JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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 $M =$

Choose the correct answer:

Explanation

Solving

Step 1: Derivative aur Roots

f'(x) = 8x^3 - 36x + 8 = 4(x - 2)(2x^2 + 4x - 1)

Roots of $2x^2 + 4x - 1 = 0$ :

x = \frac{-4 \pm \sqrt{16 + 8}}{4} = -1 \pm \frac{\sqrt{6}}{2}

Local Maxima at $x = \frac{\sqrt{6} - 2}{2}$

Step 2: $f(x)$ ko simplify karna

Pehle $f(x)$ ko $f'(x)$ ke terms mein likhte hain (Long Division):

f(x) = \frac{1}{4}(x + 1)(8x^3 - 36x + 8) - 9x^2 + 15x + 10

Chunki $f'(x) = 0$ local maxima par:

M = -9\left(\frac{\sqrt{6}-2}{2}\right)^2 + 15\left(\frac{\sqrt{6}-2}{2}\right) + 10

Step 3: Final Calculation

M = -9\left(\frac{6 + 4 - 4\sqrt{6}}{4}\right) + \frac{15\sqrt{6} - 30}{2} + 10

M = -9\left(\frac{10 - 4\sqrt{6}}{4}\right) + \frac{15\sqrt{6} - 30}{2} + 10

M = \frac{-45 + 18\sqrt{6}}{2} + \frac{15\sqrt{6} - 30}{2} + \frac{20}{2}

M = \frac{18\sqrt{6} + 15\sqrt{6} - 45 - 30 + 20}{2}

M = 12\sqrt{6} - \frac{33}{2}

Correct Option: (3)

Choose the correct answer:

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