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Q.
Number of roots of the equation $\cos ^{-1}(\sqrt{3} x)+\cos ^{-1} x=\frac{\pi}{2}$ are
Inverse Trigonometric Functions
Solution:
$ \cos ^{-1}( x \sqrt{3})+\cos ^{-1}( x )=\frac{\pi}{2} \Rightarrow \cos ^{-1}( x \sqrt{3})=\frac{\pi}{2}-\cos ^{-1}( x )=\sin ^{-1}( x ) $
$\cos ^{-1}( x \sqrt{3})=\sin ^{-1}( x ) ; \text { Let } \sin ^{-1}( x )=\theta$
$\therefore x =\sin \theta$
$\text { and } \cos ^{-1}( x \sqrt{3})=\theta$
$ x \sqrt{3}=\cos \theta$
$\text { and } \sin ^2 \theta+\cos ^2 \theta=1 $
$\therefore x ^2+3 x ^2=1$
$ x =\frac{1}{2} \text { or }-\frac{1}{2}$
If $x=\frac{1}{2}$
L.H.S. of (1) $=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$
$=\frac{-1}{2}$
L.H.S. of (1) $=\frac{5 \pi}{6}+\frac{2 \pi}{3} \neq \frac{\pi}{2}$
Hence $x=\frac{1}{2}$ is the only solution. $\Rightarrow B$