Differentiating both the sides w.r.t. x
f(x)=2x−x2f(x)
⇒f(x)=1+x22x
⇒f′(x)=(1+x2)2(1+x2)×2−2x×2x
=(1+x2)22−2x2
Hence, f′(21)=2524
JEE Main 2019 — Mathematics Calculus
If ∫0xf(t)dt=x2+∫x1t2f(t)dt, then f′(21) is
Held on 10 Jan 2019 · Verified 6 Jul 2026.
2518
2524
54
256
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