Question

The ratio of the areas between the curves y = cos x and y = cos 2x and x-axis from x = 0 to x = π/3 is ________ .

Options

• 1 : 2

• 2 : 1

• \[\sqrt{3}\]

• none of these

Answer 1

none of these

\[\text{ The line }x = \frac{\pi}{3}\text{ meets the curve }y = \cos x \text{ at } B\left( \frac{\pi}{3}, \frac{1}{2} \right)\]
Area between the curve y = cos x and x-axis from x =0 and x =\[\frac{\pi}{3}\] is,
\[A_1 = \int_0^\frac{\pi}{3} y_1 d x .............\left[\text{Where, }y_1 = \cos\left( x \right) \right]\]
\[ = \int_0^\frac{\pi}{3} \cos\left( x \right) d x\]
\[ = \left[ \sin\left( x \right) \right]_0^\frac{\pi}{3} \]
\[ = \sin\left( \frac{\pi}{3} \right) - \sin\left( 0 \right)\]
\[ = \frac{\sqrt{3}}{2}\]

\[\text{ The line } x = \frac{\pi}{3}\text{ meets the curve }y = \cos 2x\text{ at }B'\left( \frac{\pi}{3}, - \frac{1}{2} \right)\]

Area between the curve y = cos 2x and x-axis from x =0 and x = \[\frac{\pi}{3}\] is,
\[A_2 = \int_0^\frac{\pi}{4} y_2 d x - \int_\frac{\pi}{4}^\frac{\pi}{3} y_2 d x ............\left[\text{Where, }y_2 = \cos \left( 2x \right) \right]\]
\[ = \int_0^\frac{\pi}{4} \cos \left( 2x \right) d x - \int_\frac{\pi}{4}^\frac{\pi}{3} \cos \left( 2x \right) d x\]
\[ = \left[ \frac{1}{2}\sin \left( 2x \right) \right]_0^\frac{\pi}{4} - \left[ \frac{1}{2}\sin \left( 2x \right) \right]_\frac{\pi}{4}^\frac{\pi}{3} \]
\[ = \frac{1}{2}\left[ \sin \left( \frac{\pi}{2} \right) - \sin \left( 0 \right) \right] - \frac{1}{2}\left[ \sin \left( \frac{2\pi}{3} \right) - \sin \left( \frac{\pi}{2} \right) \right]\]
\[ = \frac{1}{2} - \frac{1}{2}\left[ \frac{\sqrt{3}}{2} - 1 \right]\]
\[ = \frac{1}{2} - \frac{\sqrt{3}}{4} + \frac{1}{2}\]
\[ = 1 - \frac{\sqrt{3}}{4}\]
\[ = \frac{4 - \sqrt{3}}{4}\]
Therefore the ratios will be

\[A_1 : A_2 = \frac{A_1}{A_2} = \frac{\frac{\sqrt{3}}{2}}{\frac{4 - \sqrt{3}}{4}} = \frac{2\sqrt{3}}{4 - \sqrt{3}}\]