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JEE Main 20154 April
Exam
JEE Main
Year
2015
Date
4 April
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2015 · 4 April · Mathematics · Inverse Trigonometric Functions

Let ^ - 1 y = ^ - 1 x + ^ - 1 ( 2x 1 - x ^ 2 ), where |x| < 1 3...

Let tan⁡−1y=tan⁡−1x+tan⁡−1(2x1−x2),where∣x∣<13\tan^{- 1}y = \tan^{- 1}x + \tan^{- 1}\left(\dfrac{2x}{1 - {x}^{2}}\right), where \left|x\right| < \dfrac{1}{\sqrt{3}},Then a value of yy is

Options

  1. A3x+x31+3x2\dfrac{3x + {x}^{3}}{1 + 3{x}^{2}}
  2. B3x−x31−3x2\dfrac{3x - {x}^{3}}{1 - 3{x}^{2}}
  3. C3x+x31−3x2\dfrac{3x + {x}^{3}}{1 - 3{x}^{2}}
  4. D3x−x31+3x2\dfrac{3x - {x}^{3}}{1 + 3{x}^{2}}

Correct answer

B. 3x−x31−3x2\dfrac{3x - {x}^{3}}{1 - 3{x}^{2}}

Step-by-step solution

Given tan⁡−1y=tan⁡−1x+tan⁡−1(2x1−x2),∣x∣<13\tan^{- 1}y = \tan^{- 1}x + \tan^{- 1}\left(\dfrac{2x}{1 - {x}^{2}}\right)\text{,}\left|x\right| < \dfrac{1}{\sqrt{3}}

=tan⁡−1x+2tan⁡−1x= \tan^{- 1}x + 2\tan^{- 1}x

=3tan⁡−1x= 3\tan^{- 1}x

⇒tan⁡−1y=tan⁡−1(3x−x31−3x2)⇒ \tan^{- 1}y = \tan^{- 1}\left(\dfrac{3x - {x}^{3}}{1 - 3{x}^{2}}\right)

∴y=3x−x31−3x2∴ y = \dfrac{3x - {x}^{3}}{1 - 3{x}^{2}}
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