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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2018 · Mathematics · Inverse Trigonometric Functions

Consider the following values of x : 1. 8 2. -4 3. (1)/(6) 4....

Consider the following values of x : 1. 8 2. -4 3. 16\frac{1}{6} 4. −14-\frac{1}{4} Which of the above values of xx is/are the solutions of the equation tan⁡−1(2x)+tan⁡−1(3x)=π4\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}

Options

  1. A3 only
  2. B2 and 3 only
  3. C1 and 4 only
  4. D4 only

Correct answer

A. 3 only

Step-by-step solution

tan⁡−1(2x+3x1−2x.3x)=π4tan⁡−15x1−6x2=π45x1−6x2=11−6x2=5x6x2+5x−1=0(6x−1)(x+1)=0⇒x=16,−1⇒x=16\tan ^{-1}\left(\frac{2 x+3 x}{1-2 x.3 x}\right)=\frac{\pi}{4}\begin{gathered} \tan ^{-1} \frac{5 x}{1-6 x^2}=\frac{\pi}{4} \\ \frac{5 x}{1-6 x^2}=1 \\ 1-6 x^2=5 x \\ 6 x^2+5 x-1=0 \\ (6 x-1)(x+1)=0 \\ \Rightarrow x=\frac{1}{6},-1 \end{gathered}\Rightarrow \quad x=\frac{1}{6} Here x=−1x=-1 rejected.
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