Question

The ratio of areas bounded by the curves \[ y=\cos x \] and \[ y=\cos 2x \] between \[ x=0,\qquad x=\frac{\pi}{3} \] and the \(x\)-axis is:

• A. \(2:1\)
• B. \(1:2\)
• C. \(1:1\)
• D. \(1:3\)

Hint:

For functions of the form \(\cos(nx)\), increasing the value of \(n\) compresses the graph horizontally. As a result, the area under one arch generally decreases proportionally.

Answer 1

The Correct Option is A

Concept: The area bounded by a curve \(y=f(x)\), the \(x\)-axis, and the vertical lines \(x=a\) and \(x=b\) is given by: \[ \text{Area}=\int_a^b |f(x)|\,dx \] In the interval \[ 0\le x\le \frac{\pi}{3}, \] both \(\cos x\) and \(\cos 2x\) remain non-negative over the required portions, so direct integration can be used.

Step 1: Finding the area under \(y=\cos x\).
Let this area be \(A_1\). \[ A_1=\int_0^{\pi/3}\cos x\,dx \] Integrating: \[ A_1=[\sin x]_0^{\pi/3} \] Substituting limits: \[ A_1=\sin\left(\frac{\pi}{3}\right)-\sin(0) \] Using standard trigonometric values: \[ \sin\left(\frac{\pi}{3}\right)=\frac{\sqrt3}{2} \] Hence, \[ A_1=\frac{\sqrt3}{2} \]

Step 2: Finding the area under \(y=\cos 2x\).
Let this area be \(A_2\). \[ A_2=\int_0^{\pi/3}\cos 2x\,dx \] Integrating: \[ A_2=\left[\frac{\sin 2x}{2}\right]_0^{\pi/3} \] Substituting limits: \[ A_2= \frac12 \left( \sin\frac{2\pi}{3}-\sin0 \right) \] Since, \[ \sin\frac{2\pi}{3}=\frac{\sqrt3}{2}, \] we get: \[ A_2= \frac12\cdot\frac{\sqrt3}{2} = \frac{\sqrt3}{4} \]

Step 3: Calculating the ratio of the areas.
\[ A_1:A_2 = \frac{\sqrt3}{2}:\frac{\sqrt3}{4} \] Cancelling \(\sqrt3\): \[ = \frac12:\frac14 \] Multiplying both terms by \(4\): \[ =2:1 \] Hence, \[ \boxed{2:1} \]