For x ∈ R, let f(x) = |sinx| and g(x) =\int^{x}_{0}\f(x) dt. Let p(x) = g(x) -2/\pix. Then

Correct Answer is : p(x+π)=p(x) for all x

p(x + π) ≠ p(x) for at least one but finitely many x

p(x + π) ≠ p(x) for infinitely many x

KVPY 2019 SB SX Exam Previous Papers

Question : 88 of 120

Marks: +1, -0

For x ∈ R, let f(x) = |sinx| and g(x) =

∫↖{x}↙{0}\f(x) dt

. Let

p(x) = g(x) -2/πx

. Then

Solution:

p'(x) = g'(x) =2/π=\f(x)-2/π

={|\sinx|}-2/π

⇒ p(x) is not one-one

p(x + π) = g(x + π) -2/π(x+π)

∫↖{x+π} ↙{0}{|\sinx|}dt-{2x}/π-2

∫↖{x } ↙{0}{|\sinx|}dt-{2x}/π = g(x) -{2x}/π=p(x)

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