Let f′(x)=ax2+bx+c⇒f(x)=3ax3+2bx2+cx+d ⇒f(x)=61(2ax3+3bx2+6cx+6d), Now f(1)=f(0)=d, then according to Rolle's theorem ⇒f′(x)=ax2+bx+c=0 has at least one root in (0,1)
JEE Main 2004 — Mathematics Calculus
If 2a+3b+6c=0, then at least one root of the equation ax2+bx+c=0 lies in the interval
Held on 30 Apr 2004 · Verified 6 Jul 2026.
(0,1)
(1,2)
(2,3)
(1,3)
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