Number of solutions of the equation: $\sin^{-1}x+\cos^{-1}x^2=\frac{\pi}{2}$

Question

What is the number of solutions of the equation: $\sin^{-1}x+\cos^{-1}x^2=\frac{\pi}{2}$?

1. None
2. Greater than or equal to 1
3. Less than or equal to 1
4. Equal to 2

It is a single-option-correct MCQ. I tried using substitution, but that leads to a very lengthy expression that I cannot solve. Can anyone show me the way?

Answer 1

$$\sin^{-1}x+\cos^{-1}x^2=\frac{\pi}{2}$$ $$\cos^{-1}x^2=\frac{\pi}{2}-\sin^{-1}x=\cos^{-1}x$$ $$\cos^{-1}x^2=\cos^{-1}x$$

This implies, that for principal value of the argument, $$x^2=x$$ $$x(x-1)=0$$

The equation hence has $2$ roots.

The correct answer to the MCQ should be $\color{red}{\text{OPTION (D)}}$.

Answer 2

Forgive me... This was an easy question.

We know, $$ \sin^{-1}x+\cos^{-1}y=\frac{\pi}{2} $$ iff $x=y$

Therefore, in my question, $$x = x^2$$ $$or,x^2-x=0$$ $$or,x(x-1)=0$$ i.e. $x=1$ or $0$ The correct option is D. Thank you Paul and SchrodingersCat