Question

If \(f(x)=\sin ^{-1}\left[\log _{1/2}\left(\frac{x^{2}}{2}\right)\right]\), then the domain of the function \(f(x)\) is:

• A. \([-2,-1]\cup [1,2]\)
• B. \([-\sqrt{2},-1]\cup [1,\sqrt{2}]\)
• C. \([-1,0)\cup (0,1]\)
• D. \([-2,2]-\{0\}\)

Hint:

Require the log to lie between -1 and 1, and remember that base one half reverses the inequality.

Answer 1

The Correct Option is A

Step 1: Test the End Points
At $x=1$, $\log_{1/2}\dfrac{1}{2}=1$. At $x=2$, $\log_{1/2}2=-1$. Both are allowed.

Step 2: Test an Inside Point
At $x=1.8$, $\dfrac{x^{2}}{2}=1.62$ and $\log_{1/2}1.62\approx -0.70$, which lies in $[-1,1]$. So $x=1.8$ is in the domain, which rules out (B) and (C).

Step 3: Conclusion
At $x=0.5$, $\log_{1/2}\dfrac{1}{8}=3>1$, so values near zero fail, which rules out (D). Option (A) remains.

Final Answer:
Testing points confirms the domain $[-2,-1]\cup[1,2]$. \[ \boxed{\text{(A) }[-2,-1]\cup[1,2]} \]