For let =|\sin \boldsymbol{x}| \) and =\int_{0}^{x} f(t) d t . \) Let =g(x)-\frac{2}{\pi} x \) Then

Maths · Applications of derivatives: Rate of change of quantities, monotonic-Increasing and decreasing functions, Maxima and minima of functions of one variable

For let =|\sin \boldsymbol{x}| \) and =\int_{0}^{x} f(t) d t . \) Let =g(x)-\fra

For \( \boldsymbol{x} \in \boldsymbol{R} \) let \( \boldsymbol{f}(\boldsymbol{x})=|\sin \boldsymbol{x}| \) and \( g(x)=\int_{0}^{x} f(t) d t . \) Let \( p(x)=g(x)-\frac{2}{\pi} x \) Then

A. \( p(x+\pi)=p(x) \) for all \( x \)
B. \( p(x+\pi) \neq p(x) \) for all least one but finitely many \( x \)
C. \( p(x+\pi) \neq p(x) \) for infinitely many \( x \)
D. \( p \) is a one-one function

Step-by-step solution

Since f(x)=|sin x| has period π, ∫_0^{π} |sin t| dt = 2. Then g(x+π)=∫_0^{x+π} |sin t| dt = g(x)+2. Hence p(x+π)=g(x+π)- (2/π)(x+π) = g(x)+2 - (2/π)x -2 = g(x)-(2/π)x = p(x) for all x. Therefore p is periodic with period π, so option A is correct. Options B and C are false because equality holds for all x, and D is false because a periodic function cannot be one-one.

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Step-by-step solution

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