Question

The ratio of the areas bounded by y = cos x, y = cos 2x between x = 0 and x = $\frac{\pi }{3}$ and the x-axis is

• A. $2\sqrt{3}:4-\sqrt{3}$
• B. 1 : 2
• C. 1 1
• D. 3 : 1

Answer 1

The correct option is A $2\sqrt{3}:4-\sqrt{3}$


$A_1={\int }_0^{\frac{\pi }{3}}cosxdx=[sinx{]}_0^{\frac{\pi }{3}}=\frac{\sqrt{3}}{2}$
$y=cos2x\Rightarrow cos2x=0\Rightarrow 2x=-\frac{\pi }{2}.\frac{\pi }{2}$
$\Rightarrow x=-\frac{\pi }{4},\frac{\pi }{4}$
$A_2={\int }_0^{\frac{\pi }{4}}cos2xdx–{\int }_0^{\frac{\pi }{4}}cos2xdx$
$={\left[\frac{sin2x}{2}\right]}_0^{\frac{\pi }{4}}-{\left[\frac{sin2x}{2}\right]}_{\frac{\pi }{4}}^{\frac{\pi }{3}}=\frac{1}{2}[\frac{\sqrt{3}}{2}-1]=1-\frac{\sqrt{3}}{4}$
Ratio = $\frac{\sqrt{3}}{2}:-\frac{\sqrt{3}}{4}=2\sqrt{3}:4-\sqrt{3}$