MCQ
$\int {\left( {\sin x\cos x\cos 2x\cos 4x\cos 8x} \right)dx}$ equal
- A$\frac{{ - 1}}{{128}}\cos 16x + C$
- B$\frac{{1}}{{256}}\cos 16x + C$
- C$\frac{{ - 1}}{{256}}\sin 16x + C$
- ✓$\frac{{ - 1}}{{256}}\cos 16x + C$
$ = \frac{1}{4}\int {\sin } \,4x\cos \,4x\cos \,8x\,dx$
$ = \frac{1}{8}\int {\sin } \,8x\cos \,8x\,dx = \frac{1}{{16}}\int {\sin } \,16x\,dx$
$=\frac{-1}{256} \cos 16 x+C$
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$R _{1}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \in Q \right\}$ and $R _{2}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \notin Q \right\}$
where $Q$ is the set of all rational numbers. Then