If Rolle's theorem holds true for the function f(x)=2x3+bx2+cx,x∈[−1,1] at the point x=12, then (2b+c) is equal to
A
1
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B
-1
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C
2
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D
-3
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Solution
The correct option is B -1 Since, Rolle's Theorem holds true for f(x) in the interval [−1,1] ∴f(1)=f(−1) 2+b+c=−2+b−c⇒2c=−4or,c=−2 f′(12)=0 ∵f′(x)=6x2+2bx+c ∴f′(12)=64+b+c=0⇒b+c=−32 or,b=12 ∴2b+c=1−2=−1