Question 5
If π "and" π are continuous functions in [0, 1] satisfying
π(π₯)=π(πβπ₯) and π(π₯)+π (πβπ₯)=π, then β«1_0^πβγπ(π₯). π(π₯)γ ππ₯ is equal to
π/2 (B) π/2 β«1_0^πβγπ(π₯) ππ₯γ
(C) β«1_0^πβγπ(π₯) ππ₯γ (D) πβ«1_0^πβγπ(π₯) ππ₯γ
Let
π° =β«_π^πβπ(π) π(π) π π
Using g(π₯)+π(πβπ₯)=π
I =β«_0^πβπ(π₯) [πβπ(πβπ₯)] ππ₯
I = β«_0^πβ[π.π(π₯)βπ(π₯)π(πβπ₯)] ππ₯
π° =πβ«_π^πβγπ(π)π πββ«_π^πβγπ(π) π(πβπ) γγ π π
I =πβ«_0^πβγπ(π₯)ππ₯ββ«_π^πβγπ(πβπ) π(πβ(πβπ)) γγ ππ₯
I =πβ«_0^πβγπ(π₯)ππ₯ββ«_0^πβγπ(πβπ) π(π₯) γγ ππ₯
Using π(π₯)=π(πβπ₯)
I =πβ«_0^πβγπ(π₯)ππ₯ββ«_0^πβγπ(π) π(π₯) γγ ππ₯
I =πβ«_0^πβγπ(π₯)ππ₯βπγ
I +I=aβ«_0^πβπ(π₯)ππ₯
2I=aβ«_0^πβπ(π₯)ππ₯
π=π/π β«_π^πβπ(π)π π
β΄ β«_0^πβγπ(π₯) π(π₯) γ ππ₯=2β«_0^πβπ(π₯)ππ₯
Hence Proved
So, the correct answer is (b)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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