Question

The domain of the function
\(f(x) = sin^-1 [2x^2-3]+log2(log_{\frac{1}{2}}(x^2-5x+5))\)
where [t] is the greatest integer function, is

• A. \((-\sqrt\frac{5}{2},\frac{5-√5}{2})\)
• B. \((\frac{5-√5}{2},\frac{5+√5}{2})\)
• C. \((1,\frac{5-√5}{2})\)
• D. \((1,\frac{5+√5}{2})\)

Answer 1

The Correct Option is C

To solve the problem of finding the domain of the given function \(f(x) = \sin^{-1} [2x^2-3] + \log_2(\log_{\frac{1}{2}}(x^2-5x+5))\), we need to individually analyze the constraints on each term within the expression:

1. Analyzing \(\sin^{-1} [2x^2 - 3]\):

The range of the inverse sine function, \(\sin^{-1}(y)\), is defined only for \(y \in [-1, 1]\). Therefore, we need:

\(-1 \leq [2x^2 - 3] \leq 1\)

The greatest integer function, denoted as \([t]\), returns the greatest integer less than or equal to \(t\). Thus, for \([2x^2 - 3]\) to take values -1, 0, or 1, we deduce:

• \(-1 \leq 2x^2 - 3 < 0 \rightarrow 1 \leq x^2 < 1.5\)
• \(0 \leq 2x^2 - 3 < 1 \rightarrow 1.5 \leq x^2 < 2\)

1. Analyzing \(\log_2(\log_{\frac{1}{2}}(x^2 - 5x + 5))\):

The inner logarithmic function, \(\log_{\frac{1}{2}}(x^2 - 5x + 5)\), is defined and positive (since the base is \(\frac{1}{2}\)) for:

\(x^2 - 5x + 5 > 1\)

Let's solve the inequality \(x^2 - 5x + 5 > 1\):

\(x^2 - 5x + 4 > 0\)

Factoring gives: \((x - 4)(x - 1) > 0\)

By testing intervals determined by the roots, we find the solution is \((1, 4)\).

The function is well-defined when both conditions are met simultaneously. Therefore, combining the intervals from both steps, we find:

The domain of the function is \((1, \frac{5-\sqrt{5}}{2})\).

Hence, the correct answer is:

\((1,\frac{5-√5}{2})\)