If f:R→R is a twice differentiable function such that f′′(x)>0 for all x∈R, and f(12)=12,f(1)=1,then
A
f′(1)≤0
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B
0<f′(1)≤12
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C
12<f′(1)≤1
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D
f′(1)>1
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Solution
The correct option is Df′(1)>1 f(12)=12,f(1)=1Letg(x)=f(x)−x,x∈[12,1]g(12)=g(1)=0 Using Rolle's theorem, we get g′(c)=0,c∈(12,1)⇒f′(c)−1=0[∵g′(x)=f′(x)−1]⇒f′(c)=1,c∈(12,1)Asf′′(x)>0,⇒f′(x)is increasing onR∴f′(1)>f′(c)⇒f′(1)>1