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COMEDK 2025 Evening Solved Paper

Question : 131 of 180

Marks: +1, -0

Let

f(x)=x \sqrt{4 a x-x^2}, a>0

f^{\prime}(x)

x=2 a

is :

Solution:

Concept:

The derivative at a point indicates whether the function is increasing or decreasing there.

Explanation:

Given

f(x)=x\sqrt{4ax-x^{2}}

with

a>0

Domain:

0\le x\le4a

Differentiate using product rule:

f'(x)=\sqrt{4ax-x^{2}}+x\cdot\frac{4a-2x}{2\sqrt{4ax-x^{2}}}

x=2a

\sqrt{4a(2a)-(2a)^{2}}=\sqrt{8a^{2}-4a^{2}}=2a

Second term:

2a\cdot\frac{4a-4a}{2\cdot2a}=0

Thus

f'(2a)=2a

Since

a>0

f'(2a)>0

, so the derivative exists and is positive.

Hence

f

is increasing at

x=2a

Answer:

Option D (Increasing).

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