Skip to content
dronacharya.ai
← JEE Advanced 2021 · Paper 2
MathematicsNumerical+2Question 49 of 57

JEE Advanced 2021 · Paper 2, Mathematics question 11

Passage

Question Stem for Question Nos. 11 and 12

Let gi:[π8,3π8]→Rg_i: \left[\frac\pi8, \frac{3\pi}8\right] \to \mathbb R, i=1,2i = 1, 2, and f:[π8,3π8]→Rf: \left[\frac\pi8, \frac{3\pi}8\right] \to \mathbb R be functions such that
g1(x)=1,g2(x)=∣4x−π∣andf(x)=sin⁡2x,for all x∈[π8,3π8]g_1(x) = 1,\quad g_2(x) = |4x - \pi| \quad\text{and}\quad f(x) = \sin^2 x,\quad\text{for all } x \in \left[\frac\pi8, \frac{3\pi}8\right]
Define
Si=∫π/83π/8f(x)⋅gi(x) dx,i=1,2S_i = \int_{\pi/8}^{3\pi/8} f(x)\cdot g_i(x)\,dx,\qquad i = 1, 2

The value of 16S1π\dfrac{16S_1}{\pi} is ____.

Answer and worked solution

Official answer: 1.99

Compute S1S_1. Using sin⁡2x=1−cos⁡2x2\sin^2x=\frac{1-\cos2x}2,
S1=∫π/83π/8sin⁡2x dx=12[x−sin⁡2x2]π/83π/8=12⋅π4=π8.S_1=\int_{\pi/8}^{3\pi/8}\sin^2x\,dx=\frac12\left[x-\frac{\sin2x}2\right]_{\pi/8}^{3\pi/8}=\frac12\cdot\frac\pi4=\frac\pi8.
(The sine terms cancel, since sin⁡3π4=sin⁡π4\sin\frac{3\pi}4=\sin\frac\pi4.)

Value. 16S1π=2\dfrac{16S_1}\pi=2.