KVPY2019MathematicsDefinite Integration
For x ∈ R , let f x = | sin x | and g x = ∫ 0 x f t d t . Let p x = g x - 2 π x . Then
Options
- Ap x + π = p x for all x
- Bp x + π ≠ p x for at least one but finitely many x
- Cp x + π ≠ p x for infinitely many x
- Dp is a one-one function
Correct answer
A. p x + π = p x for all x
Step-by-step solution
Given, for x ∈ R f x = | sin x | g x = ∫ 0 x f t d t , and p x = g x - 2 π x ∵   p x + π = g x + π - 2 π x + π = ∫ 0 x + π | sin t | d t - 2 π x - 2 = ∫ 0 π | sin t | d t + ∫ π π + x | sin t | d t - 2 π x - 2 = 2 ∫ 0 π 2 sin t d t + ∫ 0 x | sin t | d t - 2 π x - 2 [ ∵ | sin t | is periodic function having period π ] = 2 - cos t 0 2 + g x - 2 π x - 2 = 2 0 - - 1 + g x - 2 π x - 2