Question

The domain of the function $f(x)=\frac{1}{\sqrt{\sin x}}$ $+\sqrt[3]{\sin x}+\log _{10} \frac{x-5}{x^2-10 x+24}$ is

• A. $\{(2 k \pi,(2 k+1) \pi): k \in I \}$
• B. $\{(2 k \pi,(2 k+1) \pi): k \in N \}$
• C. $(6, \infty) \cup(4,5)$
• D. $\{(2 k \pi,(2 k+1) \pi): k \in I \} \cup(6, \infty)$

Answer 1

Answer: (B)

The domain of $\sqrt[3]{\sin x}$ is $R$.
The domain of $\frac{1}{\sqrt{\sin x}}=\{x: \sin x>0\}$
$=\{(2 k \pi,(2 k+1) \pi): k \in I \}$
The domain of $\log _{10} \frac{x-5}{x^2-10 x+24}=\log _{10} \frac{x-5}{(x-6)(x-4)}$
$=\{x: x \neq 5, x >5, x >6, x>4 \text { or } $
$=(4,5) \cap(6, \infty)$
Thus the domain of $f(x)$
$=\{(2 k \pi,(2 k+1) \pi): k \in I\} \cap\{(4,5) \cup(6, \infty)\} $
$=\{(2 k \pi,(2 k+1) \pi): k \in N \}$
(since $(4,5)$ does not intersect $\{(2 k \pi, 2 k+1) \pi: k \in I\}$ )