If \(x+\displaystyle\frac{1}{x}=2\), then the principal value of \(\sin^{-1}x\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{\pi}{2}\)
Multiplying the given equation by x:
\( x^2+1=2x \)
\( x^2-2x+1=0 \)
\( (x-1)^2=0,\quad x=1 \)
The principal value branch of \(\sin^{-1}\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), and the angle in it whose sine is 1 is \(\frac{\pi}{2}\):
\( \displaystyle \sin^{-1}(1)=\frac{\pi}{2} \)