Let $S_1$ is the complete solution set of the inequality $\cos^{-1}(x) > \cos^{-1}\left(x^2\right)$ and $S_2$ is the complete solution set of the inequality $\left(\cos^{-1} x^2\right) > 0$, then $S_1 \cap S_2$ is
Step-by-Step Solution
Key Concept: Use the fact that $\cos^{-1}$ is a strictly decreasing function, so $f(a) > f(b)$ implies $a < b$
Since $\cos^{-1}x$ is a decreasing function, $\cos^{-1}x > \cos^{-1}x^2$ implies $x < x^2$. This gives $x^2 - x > 0$, or $x(x-1) > 0$. For $x \in [-1,1]$, this inequality holds when $x < 0$ or $x > 1$. However, we must also ensure both $x$ and $x^2$ are in the domain $[-1,1]$. Since $x^2 \leq x$ for $x \in [0,1]$ and $x^2 \geq x$ for $x \in [-1,0]$, the intersection with the domain constraint gives $x \in (0,1)$.
Correct Answer: 3