Question

Number of roots of the equation ${\cos }^{-1}(\sqrt{3}x)+{\cos }^{-1}x=\frac{\pi }{2}$ are

• A. 0
• B. 1
• C. 2
• D. more than 2

Answer 1

Answer: (B)

${\cos }^{-1}(x\sqrt{3})+{\cos }^{-1}(x)=\frac{\pi }{2}\Rightarrow {\cos }^{-1}(x\sqrt{3})=\frac{\pi }{2}-{\cos }^{-1}(x)={\sin }^{-1}(x)$
${\cos }^{-1}(x\sqrt{3})={\sin }^{-1}(x);\text{ Let }{\sin }^{-1}(x)=\theta$
$\therefore x=\sin \theta$
$\text{ and }{\cos }^{-1}(x\sqrt{3})=\theta$
$x\sqrt{3}=\cos \theta$
$\text{ and }{\sin }^2\theta +{\cos }^2\theta =1$
$\therefore x^2+3x^2=1$
$x=\frac{1}{2}\text{ or }-\frac{1}{2}$
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) $=\frac{5\pi }{6}+\frac{2\pi }{3}\ne \frac{\pi }{2}$
Hence $x=\frac{1}{2}$ is the only solution. $\Rightarrow B$