NIMCET 2010 — Mathematics PYQ

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Question

NIMCET 2010 — Mathematics PYQ

NIMCET | Mathematics | 2010

$f(x) = \begin{vmatrix} x & 8 & 8 \\ 2 & x & 8 \\ 2 & 2 & x \end{vmatrix}$ has local maximum at?

Choose the correct answer:

Explanation

Solving:

1. Expanding the Determinant:

$f(x) = x(x^2 - 16) - 8(2x - 16) + 8(4 - 2x)$

$f(x) = x^3 - 16x - 16x + 128 + 32 - 16x$

$f(x) = x^3 - 48x + 160$

2. Finding Critical Points:

Differentiate $f(x)$ with respect to $x$ :

$f'(x) = 3x^2 - 48$

Set $f'(x) = 0$ :

$3x^2 = 48$

$x^2 = 16 \implies x = \pm 4$

3. Applying Second Derivative Test:

$f''(x) = 6x$

At $x = 4$ :

$f''(4) = 6(4) = 24 > 0$

So, $x = 4$ is a point of local minimum.
At $x = -4$ :

$f''(-4) = 6(-4) = -24 < 0$

So, $x = -4$ is a point of local maximum.

Correct Option:

(b) -4