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JEE Advanced 2021 Paper 2

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Let

g_i : \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \rightarrow \mathbb{R}, i=1,2

, and

f : \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \rightarrow \mathbb{R}

be functions such that

g_1(x)=1, g_2(x)=|4x-\pi|

and

f(x)=\sin^2 x

, for all

x \in \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right]

Define

S_i = \int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_i(x) \, dx, i=1,2

Section: Mathematics

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Question : 49 of 57

Marks: +1, -0

The value of

\;\frac{16 S_{1}}{\pi}

[JEE Adv 2021 P2]

Your Answer:

Solution:

S_{i}=\int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) dx \equiv \int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^{2} x dx

......(i)

=\int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^{2}\left(\;\frac{\pi}{8}+\;\frac{3\pi}{8}-x\right) dx=\int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \cos^{2} x dx

.....(ii)

Adding (i) and (ii) we get

2 S_{1}=\int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} (\sin^{2} x+\cos^{2} x) dx = \;\frac{3\pi}{8}-\;\frac{\pi}{8} = \;\frac{\pi}{4} \Rightarrow \;\frac{16 S_{1}}{\pi}=2

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