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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

Let f(x) = 7 ^ 8 x + 7 ^ 6 x - 3 ^ 4 x - 3 ^ 2 x for all x ∈ (-...

Let f(x)=7tan⁡8x+7tan⁡6x−3tan⁡4x−3tan⁡2xf\left(x\right) = 7\tan^{8} x + 7\tan^{6} x - 3\tan^{4} x - 3\tan^{2} x for all x∈(−π2,π2).x \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right) . Then the correct expression(S) is(are)

Options

  1. A∫0π4xf(x)dx=112\mathop{∫}_{0}^{\dfrac{\pi}{4}}xf\left(x\right)dx = \dfrac{1}{12}
  2. B∫0π4f(x)dx=0\mathop{∫}_{0}^{\dfrac{\pi}{4}}f\left(x\right)dx = 0
  3. C∫0π4xf(x)dx=16\mathop{∫}_{0}^{\dfrac{\pi}{4}}xf\left(x\right)dx = \dfrac{1}{6}
  4. D∫0π4xf(x)dx=1\mathop{∫}_{0}^{\dfrac{\pi}{4}}xf\left(x\right)dx = 1

Correct answer

A. ∫0π4xf(x)dx=112\mathop{∫}_{0}^{\dfrac{\pi}{4}}xf\left(x\right)dx = \dfrac{1}{12}

Step-by-step solution

f(x)=7tan⁡8x+7tan⁡6x−3tan⁡4x−3tan⁡2xf\left(x\right) = 7\tan^{8} x + 7\tan^{6} x - 3\tan^{4} x - 3\tan^{2} x
f(x)=sec⁡2x(7tan⁡6x−3tan⁡2x)f\left(x\right) = \sec^{2} x\left(7\tan^{6} x - 3\tan^{2} x\right)
Now,
∫0π4sec⁡2x(7tan⁡6x−3tan⁡2x)dx\mathop{∫}_{0}^{\dfrac{\pi}{4}}\sec^{2} x\left(7\tan^{6} x - 3\tan^{2} x\right)dx
=∫01(7t6−3t2)dt= {∫}_{0}^{1}\left(7{t}^{6} - 3{t}^{2}\right)dt
=7t77−3t33= \dfrac{7{t}^{7}}{7} - \dfrac{3{t}^{3}}{3}
=17−13=0= {1}^{7} - {1}^{3} = 0
Also
∫0π4xsec2x(7tan⁡6x−3tan⁡2x)dx{∫}_{0}^{\dfrac{\pi}{4}}x{sec}^{2} x\left(7\tan^{6} x - 3\tan^{2} x\right)dx
tanx=ttan x = t
∫01tan⁡−1t(7t6−3t2)dt{∫}_{0}^{1}\tan^{- 1} t\left(7{t}^{6} - 3{t}^{2}\right)dt
tan⁡−1t(t7−t3)∣01−∫0111+t2(t7−t3)dt\tan^{- 1} t\left({t}^{7} - {t}^{3}\right){|}_{0}^{1} - {∫}_{0}^{1}\dfrac{1}{1 + {t}^{2}}\left({t}^{7} - {t}^{3}\right)dt
=−∫0111+t2t3(t4−1)dt= - {∫}_{0}^{1}\dfrac{1}{1 + {t}^{2}}{t}^{3}\left({t}^{4} - 1\right)dt
=+∫01t3(1+t2)(1−t2)dt1+t2=∫01(t3−t5)dt= + \mathop{∫}_{0}^{1}\dfrac{{t}^{3}\left(1 + {t}^{2}\right)\left(1 - {t}^{2}\right)dt}{1 + {t}^{2}} = \mathop{∫}_{0}^{1}\left({t}^{3} - {t}^{5}\right)dt
=14−16=112= \dfrac{1}{4} - \dfrac{1}{6} = \dfrac{1}{12}
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