Q185 / 270 · NDA 2020 · Mathematics

Paragraph: Read the following information and answer: Let $\frac{\tan 3 A}{\tan A}=K$, where $\tan A \neq 0$ and $K \neq \frac{1}{3}$.
Question: For real values of $\tan A, K$ cannot lie between

  • A
    $\frac{1}{6}$ and 5
  • Correct answer
    B
    $\frac{1}{2}$ and 2
  • C
    $\frac{1}{5}$ and 5
  • D
    $\frac{1}{7}$ and 7

Solution

$\tan A=\sqrt{\frac{(K-3)}{(3 K-1)}}$ For real value of $\tan A,\left(\frac{K-3}{3 K-1}\right) \gt 0$. $\therefore$ For, $\frac{1}{3} \lt K \lt 3$. $\tan \mathrm{A}$ is not real. Also, for $\frac{1}{2} \lt \mathrm{k} \lt 2$, $\tan \mathrm{A}$ is not real.

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

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