Question

If \( f(x) = \max(\sin x, \cos x) \) and \( g(x) = \min(\sin x, \cos x) \), then \[ \int_0^{\pi/2} f(x) dx + \int_0^{\pi/2} g(x) dx = \]

• A. \( 2\sqrt{2} + 2 \)
• B. \( 2\sqrt{2} - 2 \)
• C. 2
• D. \( 2\sqrt{2} \)

Hint:

Sum of max and min of two values equals their total, which simplifies integrals.

Answer 1

The Correct Option is C

Since max and min functions cover both \( \sin x \) and \( \cos x \), their sum is \( \sin x + \cos x \), so integral becomes: \[ \int_0^{\pi/2} (\sin x + \cos x) dx = [-\cos x + \sin x]_0^{\pi/2} = (1 + 1) = 2 \]