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JEE Advanced 2015Paper 2
Exam
JEE Advanced
Year
2015
Paper
Paper 2
Subject
Mathematics
Chapter
Definite Integration

JEE Advanced 2015 · Paper 2 · Mathematics · Definite Integration

The option(s) with the value of a and L that satisfy the following...

The option(s) with the value of aa and LL that satisfy the following equation is(are) ∫04πet(sin⁡6at+cos⁡4at)dt∫0π(sin⁡6at+cos⁡4at)dt=L?\dfrac{\mathop{∫}_{0}^{4\pi}{e}^{t}\left(\sin^{6} at + \cos^{4} at\right)dt}{\mathop{∫}_{0}^{\pi}\left(\sin^{6} at + \cos^{4} at\right)dt} = L ?

Options

  1. Aa=2,L=e4π−1eπ−1a = 2, L = \dfrac{{e}^{4\pi} - 1}{{e}^{\pi} - 1}
  2. Ba=2,L=e4π+1eπ+1a = 2, L = \dfrac{{e}^{4\pi} + 1}{{e}^{\pi} + 1}
  3. Ca=4,L=e4π−1eπ−1a = 4, L = \dfrac{{e}^{4\pi} - 1}{{e}^{\pi} - 1}
  4. Da=4,L=e4π+1eπ+1a = 4, L = \dfrac{{e}^{4\pi} + 1}{{e}^{\pi} + 1}

Correct answer

A. a=2,L=e4π−1eπ−1a = 2, L = \dfrac{{e}^{4\pi} - 1}{{e}^{\pi} - 1}

Step-by-step solution

l=∫04πet(sin62t+cos42t)dt=l1+l2+l3+l4l = {∫}_{0}^{4\pi}{e}^{t}\left(si{n}^{6}2t + co{s}^{4}2t\right)dt={l}_{1} + {l}_{2}{+ l}_{3}{+ l}_{4}
Where{l}_{1} = {∫}_{0}^{\pi}{e}^{t}\left(\sin^{6} 2t + \cos^{4} 2t\right)dtand{l}_{2} = {∫}_{\pi}^{2\pi}{e}^{t}(sin⁡62t+cos⁡42t)dtandl3=∫2π3πet(sin⁡62t+cos⁡42t)dtandl4=∫3π4πet(sin⁡62t+cos⁡42t)dt\left(\sin^{6} 2t + \cos^{4}2t\right)dtand{l}_{3} = {∫}_{2\pi}^{3\pi}{e}^{t}\left(\sin^{6} 2t + \cos^{4} 2t\right)dtand{l}_{4} = {∫}_{3\pi}^{4\pi}{e}^{t}\left(\sin^{6}2t + \cos^{4}2t\right)dt
t=x+πt = x + \pi
dt=dxdt = dx
l2=∫0πex+π(sin⁡62x+cos⁡42x)dx=eπl1{l}_{2} = {∫}_{0}^{\pi}{e}^{x + \pi}\left(\sin^{6}2x + \cos^{4}2x\right)dx = {e}^{\pi}{l}_{1}
t=2π+xt = 2\pi + x
l3=e2πl1{l}_{3} = {e}^{2\pi}{l}_{1}
l4=e3πl1{l}_{4} = {e}^{3\pi}{l}_{1}
l=l1(1+eπ+e2π+e3π)l = {l}_{1}\left(1 + {e}^{\pi} + {e}^{2\pi} + {e}^{3\pi}\right)
ll1=e4π−1eπ−1\dfrac{l}{{l}_{1}} = \dfrac{{e}^{4\pi} - 1}{{e}^{\pi} - 1}
The same for a = 4.
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