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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 : 50 of 57

Marks: +1, -0

The value of

\;\frac{48 S_{2}}{\pi^{2}}

is_____________

[JEE Adv 2021 P2]

Your Answer:

Solution:

S_{2} = \int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) g_{2}(x) \, dx

= \int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^{2} x \left| 4x - \pi \right| \, dx

.........(i)

= \int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^{2} \left( \frac{\pi}{8} + \frac{3\pi}{8} - x \right) \left| 4\left( \frac{\pi}{8} + \frac{3\pi}{8} - x \right) - \pi \right| dx

= \int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} (\cos^{2} x) \left| \pi - 4\pi \right| dx

= \int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \cos^{2} x \left| 4\pi - \pi \right| dx

.......(ii)

Adding equations (i) and (ii) we get

2 S_{2} = \int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \left| 4\pi - \pi \right| (\sin^{2} x + \cos^{2} x) dx = \int\limits_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \left| 4\pi - \pi \right| dx

= 2 \times \;\frac{1}{2} \times \;\frac{\pi}{8} \times \;\frac{\pi}{2} = \;\frac{\pi^{2}}{16} \Rightarrow \;\frac{48 S_{2}}{\pi^{2}} = \;\frac{3}{2} = 1.5

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