If ∫ 0 x f t d t = x 2 + ∫ x 1 t 2 f t d t , then f ' 1 2 is

If ∫0xftdt=x2+∫x1t2ftdt, then f'12 is
  1. 1825
  2. 2425
  3. 45
  4. 625

Solution

Differentiating both the sides w.r.t. x

fx=2x-x2fx
⇒fx=2x1+x2
⇒f'x=1+x2×2-2x×2x1+x22
=2-2x21+x22

Hence, f'12=2425

Asked in: JEE Main 2019 (10 Jan Shift 2)

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