Question

If $\sin 2x.\cos 2x.\cos 4x=\gamma$ has a solution, then $\gamma$ lies in interval

• A. $[-\frac{1}{2},\frac{1}{2}]$
• B. $[-\frac{1}{4},\frac{1}{4}]$
• C. $[-\frac{1}{3},\frac{1}{3}]$
• D. $[-1,1]$

Answer 1

Answer: (B)

$[-\frac{1}{4},\frac{1}{4}]$

$\sin 2x\cos 2x\cos 4x=\gamma$
$\sin 4x\cos 4x=2\gamma$
$\sin 8x=4\gamma$

As $-1<\sin 8x<1$
$-1⩽4\gamma ⩽1$

$\frac{-1}{4}⩽\gamma ⩽\frac{1}{4}$