JEE PYQ: Functions Question 31
Question 31 - 2019 (12 Apr Shift 1)
For $x \in \left(0, \frac{3}{2}\right)$, let $f(x) = \sqrt{x}$, $g(x) = \tan x$ and $h(x) = \frac{1-x^2}{1+x^2}$. If $\phi(x) = ((h \circ f) \circ g)(x)$, then $\phi\left(\frac{\pi}{3}\right)$ is equal to:
(1) $\tan \frac{\pi}{12}$
(2) $\tan \frac{11\pi}{12}$
(3) $\tan \frac{7\pi}{12}$
(4) $\tan \frac{5\pi}{12}$
Type: MCQ
Show Answer
Answer: (2) $\tan \frac{11\pi}{12}$
Solution
$\phi\left(\frac{\pi}{3}\right) = h(f(\sqrt{3})) = h(3^{1/4})$
$= \frac{1-\sqrt{3}}{1+\sqrt{3}} = \sqrt{3} - 2 = -\tan 15° = \tan \frac{11\pi}{12}$
BETA
