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

(A) a/2ย 

(B) a/2 โˆซ 0^a f(x)ย  dx

(C) โˆซ 0^a f(x) dxย ย 

(D) aโˆซ 0^a f(x) dx

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

[Integrals] If ๐‘“ & ๐‘” are continuous functions [0, 1] satisfying f(x) - NCERT Exemplar MCQ

part 2 - Question 5 - NCERT Exemplar MCQ - Serial order wise - Chapter 7 Class 12 Integrals
part 3 - Question 5 - NCERT Exemplar MCQ - Serial order wise - Chapter 7 Class 12 Integrals

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Transcript

Question 5 If ๐‘“ "and" ๐‘” are continuous functions in [0, 1] satisfying ๐‘“(๐‘ฅ)=๐‘“(๐‘Žโˆ’๐‘ฅ) and ๐‘”(๐‘ฅ)+๐‘” (๐‘Žโˆ’๐‘ฅ)=๐‘Ž, then โˆซ1_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ). ๐‘”(๐‘ฅ)ใ€— ๐‘‘๐‘ฅ is equal to ๐‘Ž/2 (B) ๐‘Ž/2 โˆซ1_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ) ๐‘‘๐‘ฅใ€— (C) โˆซ1_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ) ๐‘‘๐‘ฅใ€— (D) ๐‘Žโˆซ1_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ) ๐‘‘๐‘ฅใ€— Let ๐‘ฐ =โˆซ_๐ŸŽ^๐’‚โ–’๐’‡(๐’™) ๐’ˆ(๐’™) ๐’…๐’™ Using g(๐‘ฅ)+๐‘”(๐‘Žโˆ’๐‘ฅ)=๐‘Ž I =โˆซ_0^๐‘Žโ–’๐‘“(๐‘ฅ) [๐‘Žโˆ’๐‘”(๐‘Žโˆ’๐‘ฅ)] ๐‘‘๐‘ฅ I = โˆซ_0^๐‘Žโ–’[๐‘Ž.๐‘“(๐‘ฅ)โˆ’๐‘“(๐‘ฅ)๐‘”(๐‘Žโˆ’๐‘ฅ)] ๐‘‘๐‘ฅ ๐‘ฐ =๐’‚โˆซ_๐ŸŽ^๐’‚โ–’ใ€–๐’‡(๐’™)๐’…๐’™โˆ’โˆซ_๐ŸŽ^๐’‚โ–’ใ€–๐’‡(๐’™) ๐’ˆ(๐’‚โˆ’๐’™) ใ€—ใ€— ๐’…๐’™ I =๐‘Žโˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ)๐‘‘๐‘ฅโˆ’โˆซ_๐ŸŽ^๐’‚โ–’ใ€–๐’‡(๐’‚โˆ’๐’™) ๐’ˆ(๐’‚โˆ’(๐’‚โˆ’๐’™)) ใ€—ใ€— ๐‘‘๐‘ฅ I =๐‘Žโˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ)๐‘‘๐‘ฅโˆ’โˆซ_0^๐‘Žโ–’ใ€–๐’‡(๐’‚โˆ’๐’™) ๐‘”(๐‘ฅ) ใ€—ใ€— ๐‘‘๐‘ฅ Using ๐‘“(๐‘ฅ)=๐‘“(๐‘Žโˆ’๐‘ฅ) I =๐‘Žโˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ)๐‘‘๐‘ฅโˆ’โˆซ_0^๐‘Žโ–’ใ€–๐’‡(๐’™) ๐‘”(๐‘ฅ) ใ€—ใ€— ๐‘‘๐‘ฅ I =๐‘Žโˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ)๐‘‘๐‘ฅโˆ’๐ˆใ€— I +I=aโˆซ_0^๐‘Žโ–’๐‘“(๐‘ฅ)๐‘‘๐‘ฅ 2I=aโˆซ_0^๐‘Žโ–’๐‘“(๐‘ฅ)๐‘‘๐‘ฅ ๐ˆ=๐š/๐Ÿ โˆซ_๐ŸŽ^๐’‚โ–’๐’‡(๐’™)๐’…๐’™ โˆด โˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ) ๐‘”(๐‘ฅ) ใ€— ๐‘‘๐‘ฅ=2โˆซ_0^๐‘Žโ–’๐‘“(๐‘ฅ)๐‘‘๐‘ฅ Hence Proved So, the correct answer is (b)

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