Let f be twice differentiable function satisfying f(1)=1,f(2)=4,f(3)=9, then
A
f′′(x)=2,∀xϵR
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B
f′(x)=5=f′′(x). For some xϵ(1,3)
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C
There exists atleast one xϵ(1,3) such that f′′(x)=2
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D
None of the above
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Solution
The correct option is C There exists atleast one xϵ(1,3) such that f′′(x)=2 Let g(x)=f(x)−x2 ⇒g(x) has atleast 3 real roots which are x=1,2,3 (by mean value theorem) ⇒g′(x) has atleast 2 real roots in xϵ(1,3) ⇒g′′(x) has atleast 1 real root in xϵ(1,3) ⇒f′′(x)=2 for atleast one xϵ(1,3).