Q64 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics

The number of real roots of the equation $\sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0$ that are greater than or equal to one are

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

Solution

Given equation $\sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0$ $\Rightarrow 2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=n \pi, n \in \mathbf{Z}$ $\Rightarrow \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=\frac{n \pi}{2}$ $\because \quad \cos ^{-1} x \in[0, \pi]$ $\therefore \quad \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=0, \frac{\pi}{2}, \pi$ $\Rightarrow \cot \left(2 \tan ^{-1} x\right)=1,0,-1 \Rightarrow 2 \tan ^{-1} x=m \pi+\frac{\pi}{4}$ $m \pi+\frac{\pi}{2}$ or $m \pi-\frac{\pi}{4}, m \in \mathbf{Z}$ $\Rightarrow \tan ^{-1} x=\frac{m \pi}{2} \pm \frac{\pi}{8}$ or $\frac{m \pi}{2}+\frac{\pi}{4}, m \in \mathbf{Z}$ $\because \quad \tan ^{-1} x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ $\therefore \tan ^{-1} x= \pm \frac{\pi}{8}, \frac{\pi}{2}-\frac{\pi}{8},-\frac{\pi}{2}+\frac{\pi}{8}, \pm \frac{\pi}{4}$ $\therefore \quad x= \pm \tan \frac{\pi}{8}, \pm \cot \frac{\pi}{8}, \pm \tan \frac{\pi}{4}$ $\Rightarrow \quad x= \pm(\sqrt{2}-1), \pm \frac{1}{\sqrt{2}-1}, \pm 1$ $= \pm(\sqrt{2}-1), \pm(\sqrt{2}+1), \pm 1$ The roots which are greater than or equal to one are $\sqrt{2}+1,1$.

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

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