Question:
For $x \in\left(0, \frac{3}{2}\right)$, let $f(x)=\sqrt{x}, \mathrm{~g}(x)=\tan x$ and $h(x)=\frac{1-x^{2}}{1+x^{2}}$.
If $\phi(x)=((h o f) \log )(x)$, then $\phi\left(\frac{\pi}{3}\right)$ is equal to:
Correct Option: , 2
Solution:
$\because \quad \phi(x)=(($ hof $) \circ g)(x)$
$\because \phi\left(\frac{\pi}{3}\right)=h\left(f\left(g\left(\frac{\pi}{3}\right)\right)\right)=h(f(\sqrt{3}))=h\left(3^{1 / 4}\right)$
$=\frac{1-\sqrt{3}}{1+\sqrt{3}}=-\frac{1}{2}(1+3-2 \sqrt{3})=\sqrt{3}-2=-(-\sqrt{3}+2)$
$=-\tan 15^{\circ}=\tan \left(180^{\circ}-15^{\circ}\right)=\tan \left(\pi-\frac{\pi}{12}\right)=\tan \frac{11 \pi}{12}$
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