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The domain of the function \(f(x) = \sin^{-1}(5 - 4x^{2}) + \cos^{-1}(5 - 4x^{2})\) is

  • A.\(\left[1, \sqrt{\tfrac{3}{2}}\right] \cup \left[-\sqrt{\tfrac{3}{2}}, -1\right]\)
  • B.\(\left[1, \sqrt{\tfrac{5}{2}}\right] \cup \left[-\sqrt{\tfrac{3}{2}}, -1\right]\)
  • C.\(\left[1, \sqrt{\tfrac{3}{2}}\right] \cup \left[-\sqrt{\tfrac{5}{2}}, -1\right]\)
  • D.\(\left[2, \sqrt{\tfrac{3}{2}}\right] \cup \left[-\sqrt{\tfrac{3}{2}}, -1\right]\)
  • E.\(\left[1, \sqrt{\tfrac{3}{2}}\right] \cup \left[-\sqrt{\tfrac{3}{2}}, -2\right]\)

Answer: A

The domain is \(\left[1, \sqrt{\tfrac{3}{2}}\right] \cup \left[-\sqrt{\tfrac{3}{2}}, -1\right]\).

Both \(\sin^{-1}\) and \(\cos^{-1}\) need their argument in \([-1, 1]\):

\[ \begin{aligned} -1 \leq 5 - 4x^{2} &\leq 1 \\ 1 \leq x^{2} &\leq \tfrac{3}{2} \end{aligned} \]

So \(1 \leq |x| \leq \sqrt{\frac{3}{2}}\), which gives the two intervals in the answer. Options with \(\sqrt{\frac{5}{2}}\), 2 or -2 as an endpoint come from solving one of the two inequalities wrongly.

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