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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.