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If f "and" g are continuous functions in [0, 1] satisfying f(x)=f(a-x) and g(x)+g (a-x)=a, then  ∫ 0 a f(x). g(x) dx is equal to

(A) a/2Β 

(B) a/2 ∫ 0^a f(x)  dx

(C) ∫ 0^a f(x) dx  

(D) a∫ 0^a f(x) dx

This question is similar to Ex 7.11, 19 - Chapter 7 Class 12 - Integrals



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Question 5 If 𝑓 "and" 𝑔 are continuous functions in [0, 1] satisfying 𝑓(π‘₯)=𝑓(π‘Žβˆ’π‘₯) and 𝑔(π‘₯)+𝑔 (π‘Žβˆ’π‘₯)=π‘Ž, then ∫1_0^π‘Žβ–’γ€–π‘“(π‘₯). 𝑔(π‘₯)γ€— 𝑑π‘₯ is equal to π‘Ž/2 (B) π‘Ž/2 ∫1_0^π‘Žβ–’γ€–π‘“(π‘₯) 𝑑π‘₯γ€— (C) ∫1_0^π‘Žβ–’γ€–π‘“(π‘₯) 𝑑π‘₯γ€— (D) π‘Žβˆ«1_0^π‘Žβ–’γ€–π‘“(π‘₯) 𝑑π‘₯γ€— Let 𝑰 =∫_𝟎^𝒂▒𝒇(𝒙) π’ˆ(𝒙) 𝒅𝒙 Using g(π‘₯)+𝑔(π‘Žβˆ’π‘₯)=π‘Ž I =∫_0^π‘Žβ–’π‘“(π‘₯) [π‘Žβˆ’π‘”(π‘Žβˆ’π‘₯)] 𝑑π‘₯ I = ∫_0^π‘Žβ–’[π‘Ž.𝑓(π‘₯)βˆ’π‘“(π‘₯)𝑔(π‘Žβˆ’π‘₯)] 𝑑π‘₯ 𝑰 =π’‚βˆ«_𝟎^𝒂▒〖𝒇(𝒙)π’…π’™βˆ’βˆ«_𝟎^𝒂▒〖𝒇(𝒙) π’ˆ(π’‚βˆ’π’™) γ€—γ€— 𝒅𝒙 I =π‘Žβˆ«_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯βˆ’βˆ«_𝟎^𝒂▒〖𝒇(π’‚βˆ’π’™) π’ˆ(π’‚βˆ’(π’‚βˆ’π’™)) γ€—γ€— 𝑑π‘₯ I =π‘Žβˆ«_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯βˆ’βˆ«_0^π‘Žβ–’γ€–π’‡(π’‚βˆ’π’™) 𝑔(π‘₯) γ€—γ€— 𝑑π‘₯ Using 𝑓(π‘₯)=𝑓(π‘Žβˆ’π‘₯) I =π‘Žβˆ«_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯βˆ’βˆ«_0^π‘Žβ–’γ€–π’‡(𝒙) 𝑔(π‘₯) γ€—γ€— 𝑑π‘₯ I =π‘Žβˆ«_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯βˆ’πˆγ€— I +I=a∫_0^π‘Žβ–’π‘“(π‘₯)𝑑π‘₯ 2I=a∫_0^π‘Žβ–’π‘“(π‘₯)𝑑π‘₯ 𝐈=𝐚/𝟐 ∫_𝟎^𝒂▒𝒇(𝒙)𝒅𝒙 ∴ ∫_0^π‘Žβ–’γ€–π‘“(π‘₯) 𝑔(π‘₯) γ€— 𝑑π‘₯=2∫_0^π‘Žβ–’π‘“(π‘₯)𝑑π‘₯ Hence Proved So, the correct answer is (b)

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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.