JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Hard

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Analyzing +4 | -1

Let

I(a)=\int_0^\pi\left(\frac{x}{a}+a \sin x\right)^2 d x

, where '

a

' is positive real. The value of '

a

' for which

I(a)

attains its minimum value, is

View answer and explanation

Correct option: A

\sqrt{\pi \sqrt{\frac{2}{3}}}

Option A (

\sqrt{{\pi \sqrt{{\frac{{2}}{{3}}}}}}

) is correct because the minimum value of I(a) is attained when the derivative with respect to 'a' is zero, leading to this value.

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