Question

The domain of $\frac{1}{\sqrt{\left\{\sin x\right\}+\left\{\sin \left(\pi +x\right)\right\}}}$ (where {.} denotes the fractional part of $x$)

• A. $R-\left\{\frac{n\pi }{2}\right\}$
• B. $R$
• C. $R^{\prime }$
• D. $R-\left\{0\right\}$

Answer 1

Answer: (A)

$R-\left\{\frac{n\pi }{2}\right\}$

If fractional part of some number $z$ is $\left\{z\right\}=k$, where $0\le k<1$
$-\left\{z\right\}=1-k$
$\begin{array}{c}\therefore f\left(x\right)=\frac{1}{\sqrt{\left\{\sin x\right\}+\left\{\sin \left(\pi +x\right)\right\}}}\\ =\frac{1}{\sqrt{k+1-k}}=\frac{1}{1}\end{array}$
But at $k=\frac{\pi }{2}$
$\sin x=1$
$\sin (n+x)=-1$
$\therefore R-\left\{\frac{n\pi }{2}\right\}$