Check sibling questions

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/2 

(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


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)

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo