f(x)=2x3−9x2+12x+41
f′(x)=2−1(2x3−9x2+12x+4)23(6x2−18x+12)
f′(x)=−2(2x3−9x2+12x+4)236(x−1)(x−2)
f′(x)=0⇒x=1,2
f(1)=31 , f(2)=81
31<f(x)<81
⇒∫1231dx<∫12f(x)dx<∫1281dx
⇒31<I<81
⇒91<I2<81.
JEE Main 2020 — Mathematics Calculus
If I=∫122x3−9x2+12x+4dx, then
Held on 8 Jan 2020 · Verified 6 Jul 2026.
81<I2<41
91<I2<81
161<I2<91
61<I2<21
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