Question

Value of the expression $2sinx-cos2x$ is always?

• A. greater than or equal to -3/2
• B. less than or equal to 3/2
• C. greater than or equal to -1/2
• D. none of these

Answer 1

The correct option is A greater than or equal to -3/2

Let $f\left(x\right)=2\sin x-\cos 2x\to 1$

For critical points,

$f\left(x\right)=0$

$2\cos x+2\sin x=0$

$2\cos x=-2\times 2\sin x\cos x$

$-2\sin x=1$

$\sin x=-\frac{1}{2}$

Putting value of $\sin x$ in equation $1$ for minimum value,

$f\left(x\right)=2\times \left(\frac{-1}{2}\right)-[1-2{\sin }^{2}x]$

$=-1-[1-2\times \frac{1}{4}]$

$=-1-\frac{1}{2}$

$=-\frac{3}{2}$

Thus $f\left(x\right)\ge -\frac{3}{2}$

Thus $2\sin x-\cos 2x$ always greater than or equal to $-\frac{3}{2}$