The curve 4 y=3 x^4-2 x² attains ----------- at the points x=-{1}{{3}} and x={1}{{3}}

Correct Answer is : both minimum values

a maximum value and a minimum value respectively

a minimum value and a maximum value respectively

Correct Answer is : both minimum values

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COMEDK 2025 Afternoon Solved Paper

Question : 129 of 180

Marks: +1, -0

The curve

4 y=3 x^4-2 x^2

x=-\frac{1}{\sqrt{3}}

x=\frac{1}{\sqrt{3}}

Solution:

Concept:

To determine the nature of the curve at the given points, we use the second derivative test on the function

y = \frac{3x^4 - 2x^2}{4}

Explanation:

Rewrite the equation as

y = \frac{3x^4 - 2x^2}{4}

First derivative:

y' = \frac{12x^3 - 4x}{4} = 3x^3 - x = x(3x^2 - 1)

Set

y' = 0

to get critical points:

x = 0

x = \pm \frac{1}{\sqrt{3}}

Second derivative:

y'' = 9x^2 - 1

x = \frac{1}{\sqrt{3}}

y'' = 9\left(\frac{1}{3}\right) - 1 = 3 - 1 = 2 > 0

, so local minimum.

x = -\frac{1}{\sqrt{3}}

y'' = 9\left(\frac{1}{3}\right) - 1 = 2 > 0

, also local minimum.

Thus both points are minima of the curve.

Answer:

Option A: both minimum values.

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