A) \[[0,\pi ]\]
B) \[(2n+1)\pi /2,n\in Z\]
C) \[(0,\pi )\]
D) none of these
Correct Answer: D
Solution :
[d] \[f(x)=\frac{1}{\sqrt{\{sinx\}+\{sin(\pi +x)\}}}\] \[=\frac{1}{\sqrt{\{sinx\}+\{-sinx\}}}\] Now, \[\{\sin \,x\}+\{-\sin x\}=\left\{ \begin{matrix} 0,\,\,\,\,\,\,\,\sin \,x\,\,\text{is}\,\text{an}\,\text{integer} \\ 1,\,\,\sin x\,\text{is}\,\text{not}\,\text{an}\,\text{integer} \\ \end{matrix} \right.\]For \[f(x)\] to get defined, \[\{sinx\}+\{-sinx\}\ne 0\] Or \[\sin x\ne \operatorname{int}eger\] Or \[\sin x\ne \pm 1,0\] Or \[x\ne \frac{n\pi }{2},n\in I\] Hence, the domain is\[R-\left\{ \frac{n\pi }{2}/n\in I \right\}\].
You need to login to perform this action.
You will be redirected in
3 sec
