Practicing Success
Practicing Success
CUET Preparation Today
If $x+\frac{1}{x}=2,$ then principal value of $sin^{-1}x$ is : |
$\pi $ $\frac{\pi }{4}$ $\frac{\pi }{2}$ $\frac{3\pi }{2}$ |
$\frac{\pi }{2}$ |
The correct answer is Option (3) → $\frac{\pi }{2}$ $x+\frac{1}{x}=2$ so $\frac{x^2+1}{x}=2$ so $x^2+1=2x$ $x^2-2x+1=0⇒(x-1)^2=0⇒x=1$ $\sin^{-1}(1)=\frac{\pi }{2}$ |
