NIMCET 2013 — Mathematics PYQ

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Question

NIMCET 2013 — Mathematics PYQ

NIMCET | Mathematics | 2013

The minimum value of the function $y = 2x^3 - 21x^2 + 36x - 20$ is:

Choose the correct answer:

Explanation

Solution

1. Find the first derivative ( $y'$ ):

y = 2x^3 - 21x^2 + 36x - 20

\frac{dy}{dx} = 6x^2 - 42x + 36

2. Find critical points by setting $\frac{dy}{dx} = 0$ :

6x^2 - 42x + 36 = 0

Dividing by $6$ :

x^2 - 7x + 6 = 0

(x - 6)(x - 1) = 0

Points are $x = 1$ and $x = 6$ .

3. Use the second derivative test to find the minimum:

\frac{d^2y}{dx^2} = 12x - 42

At $x = 1$ : $\frac{d^2y}{dx^2} = 12(1) - 42 = -30 < 0$ (Local Maximum)
At $x = 6$ : $\frac{d^2y}{dx^2} = 12(6) - 42 = 72 - 42 = 30 > 0$ (Local Minimum)

4. Calculate the minimum value at $x = 6$ :

Substitute $x = 6$ into the original function:

y = 2(6)^3 - 21(6)^2 + 36(6) - 20

y = 2(216) - 21(36) + 216 - 20

y = 432 - 756 + 216 - 20

y = 648 - 776

y = -128

Choose the correct answer:

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