| Statement -1: The period of \[f(x)=\sin 2x\cos [2x]-\cos 2x\sin [2x]\]is \[\frac{1}{2}\] |
| Statement-2: The period of x-[x] is 1 |
A) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
B) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
C) Statement-1 is true, Statement-2 is false.
D) Statement-1 is false, Statement-2 is true.
Correct Answer: A
Solution :
\[f(x)=x-[x]\] \[f(x+1)=x+1-([x]+1)=x-[x]\] Period of x-[x] is 1 f(x)=sin(2x-[2x]) \[f\left( x+\frac{1}{2} \right)=\sin \left( 2\left( x+\frac{1}{2} \right)-\left[ 2\left( x+\frac{1}{2} \right) \right] \right)\] \[=\sin (2x+1-[2x]-1)=\sin (2x-[2x])\] Period is\[\frac{1}{2}.\]
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