Q195 / 270 · NDA 2018 (Phase 2) · Mathematics

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

  • Correct answer
    A
    3 only
  • B
    2 and 3 only
  • C
    1 and 4 only
  • D
    4 only

Solution

$\tan ^{-1}\left(\frac{2 x+3 x}{1-2 x .3 x}\right)=\frac{\pi}{4}$ $\begin{gathered} \tan ^{-1} \frac{5 x}{1-6 x^2}=\frac{\pi}{4} \\ \frac{5 x}{1-6 x^2}=1 \\ 1-6 x^2=5 x \\ 6 x^2+5 x-1=0 \\ (6 x-1)(x+1)=0 \\ \Rightarrow x=\frac{1}{6},-1 \end{gathered}$ $\Rightarrow \quad x=\frac{1}{6}$ Here $x=-1$ rejected.

Hence, the correct answer is option: 'A'.

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