Let f be a real valued continuous function on 0 ,   1 and f x = x + ∫ 0 1 x - t f t d t . Then…

Let f be a real valued continuous function on 0, 1 and fx=x+01x-tftdt. Then which of the following points x, y lies on the curve y=fx?
  1. 2, 4
  2. 1, 2
  3. 4, 17
  4. 6, 8

Solution

Given,

fx=x1+01ftdt-01tftdt

fx=Ax-B ...i

A=1+01ftdt=1+01At-Bdt

A=21-B ...ii

Also B=01tftdt=01At2-Btdt

A=92B ...iii

From ii, iii

A=1813, B=413

So, fx=1813x-413

 f6=18×6-413=8

Asked in: JEE Main 2022 (29 Jun Shift 2)

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