The number of solutions of $\sin^{-1}\left(\frac{1+x^2}{2x}\right) = \frac{\pi}{2}\sec(x-1)$ is
Step-by-Step Solution
Key Concept: Domain restrictions of inverse trigonometric functions severely constrain the solution set; here the constraint $(|x|-1)^2 \leq 0$ forces a unique value.
For $\sin^{-1}\left(\frac{1+x^2}{2x}\right)$ to exist, the argument must satisfy $\left|\frac{1+x^2}{2x}\right| \leq 1$. This gives $1 + x^2 \leq 2|x|$, or $(|x|-1)^2 \leq 0$, which implies $|x| = 1$ (i.e., $x = \pm 1$). Checking: for $x = 1$ the original equation is satisfied, but for $x = -1$ it is not. Therefore, the only solution is $x = 1$.
Correct Answer: 1