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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

If α = ∫ _ 0 ^ 1 ( e ^ 9x + 3 ^ - 1 x )( 12 + 9 x ^ 2 1 + x ^ 2 )dx...

If α=∫01(e9x+3tan⁡−1x)(12+9x21+x2)dx\alpha = \mathop{∫}_{0}^{1}\left({e}^{9x + 3\tan^{- 1} x}\right)\left(\dfrac{12 + 9{x}^{2}}{1 + {x}^{2}}\right)dx where tan⁡−1x\tan^{- 1} x takes only principal values, then the value of \left(\log_{e} \left|1 + \alpha\right| - \dfrac{3\pi}{4}\right) is

Correct answer

0

Step-by-step solution

α=∫01(e9x+3tan⁡−1x)(12+9x21+x2)dx\alpha = \mathop{∫}_{0}^{1}\left({e}^{9x + 3\tan^{- 1} x}\right)\left(\dfrac{12 + 9{x}^{2}}{1 + {x}^{2}}\right)dx
=∫01(e9x+3tan⁡−1x)(9+31+x2)dx= \mathop{∫}_{0}^{1}\left({e}^{9x + 3\tan^{- 1} x}\right)\left(9 + \dfrac{3}{1 + {x}^{2}}\right)dx
=(e9x+3tan⁡−1x)∣01= \left({e}^{9x + 3\tan^{- 1} x}\right){|}_{0}^{1}
=e9+3π4−1= {e}^{9 + \dfrac{3\pi}{4}} - 1
⇒ln⁡∣(1+α)∣−3π4=ln⁡∣(1+e9+3π4−1)∣−3π4=ln⁡∣e9+3π4∣−3π4=9⇒ \ln \left|( 1 + \alpha )\right| - \dfrac{3\pi}{4} = \ln \left|\left(1 + {e}^{9 + \dfrac{3\pi}{4}} - 1\right)\right| - \dfrac{3\pi}{4} = \ln \left|{e}^{9 + \dfrac{3\pi}{4}}\right| - \dfrac{3\pi}{4} = 9 .

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