Question

F'(x)>0 f(x)=sin2x -

Answer 1

$f(x)=sin2x- \sqrt{3x}$
$f'(x)=(sin2x)'- (\sqrt{3x} )'= (2x)'*cos2x- \frac{(3x)'}{2 \sqrt{3x} } =2cos2x- \frac{3}{2 \sqrt{3x} }$
2cos2x-3/(2√3x)>0
2cos2x*x>√3/2
2x>π/6+2πk
x>π/12+πk   ⇒ x∈(π/12+πk;  +∞)  k∈Z