TS EAMCET 10 May 2024 Shift 2 Solved Paper

To find the domain of the function $f(x)=\sin^{-1}\left(\log_2\left(\frac{x^2}{2}\right)\right)$, we need to determine where the expression $\log_2\left(\frac{x^2}{2}\right)$ falls within the domain of the inverse sine function, which is $[-1,1]$.First, ensure that:$-1 \leq \log_2\left(\frac{x^2}{2}\right) \leq 1$This inequality translates to:$2^{-1} \leq \frac{x^2}{2} \leq 2$Solving the inequality:$\frac{1}{2} \leq \frac{x^2}{2} \leq 2$Multiply through by 2 to clear the fraction:$1 \leq x^2 \leq 4$This implies that $x$ must satisfy:$x \in[-2,-1] \cup[1,2]$We also need to ensure that $\frac{x^2}{2} \geq 0$, which is always true since $x^2 \geq 0$.Therefore, considering the intersection of the feasible solutions:$x \in[-2,-1] \cup[1,2]$Thus, the domain of the function is:$x \in[-2,-1] \cup[1,2]$