Question

The line \(x=\frac{\pi}{4}\) divides the area of the region bounded by \(y=\sin x\), \(y=\cos x\) and x-axis \((0\leq x\leq \frac{\pi}{2})\) into two regions of areas \(A_1\) and \(A_2\). Then \(A_1:A_2\) equals

• A. \(4:1\)
• B. \(3:1\)
• C. \(2:1\)
• D. \(1:1\)

Hint:

For \(\sin x\) and \(\cos x\), symmetry around \(x=\pi/4\) often makes areas equal in \([0,\pi/2]\).

Answer 1

The Correct Option is D

Step 1: Identify region.
For \(0\leq x\leq \frac{\pi}{2}\):
- On \([0,\frac{\pi}{4}]\), \(\cos x \ge \sin x\).
- On \([\frac{\pi}{4},\frac{\pi}{2}]\), \(\sin x \ge \cos x\).
So bounded region is between \(\sin x\), \(\cos x\) and x-axis.
Step 2: Compute area from 0 to \(\pi/4\).
Region above x-axis and below \(\sin x\) and \(\cos x\).
Minimum of \(\sin x,\cos x\) is \(\sin x\) on \([0,\pi/4]\).
\[ A_1=\int_0^{\pi/4}\sin x\,dx =[-\cos x]_0^{\pi/4}=1-\frac{1}{\sqrt2} \]
Step 3: Compute area from \(\pi/4\) to \(\pi/2\).
Minimum is \(\cos x\) on \([\pi/4,\pi/2]\).
\[ A_2=\int_{\pi/4}^{\pi/2}\cos x\,dx =[\sin x]_{\pi/4}^{\pi/2}=1-\frac{1}{\sqrt2} \]
Step 4: Compare.
\[ A_1=A_2 \Rightarrow A_1:A_2=1:1 \]
Final Answer:
\[ \boxed{1:1} \]