A) \[f''(x)=2\forall x\in (1,3)\]
B) \[f''(x)=f'(x)=5\] for some \[x\in (2,3)\]
C) \[f''(x)=3\,\forall \,x\in (2,\,\,3)\]
D) \[f''(x)=2\] for some \[x\in (1,3)\]
Correct Answer: D
Solution :
[d] Let \[g(x)=f(x)-{{x}^{2}}.\] We have \[g(1)=0,\,\,g(2)=0,\,\,g(3)=0\] \[[\therefore f(1)=1,\,\,f(2)=4,\,f(3)=9]\]. From Rolle's theorem on \[g(x),g'(x)=0\] for at least \[x\in (1,2).\] Let \[g'({{c}_{1}})=0\] where \[{{c}_{1}}\in (1,2)\]. Similarly, g(x) =0 for at least one \[x\in (2,3).\] Let \[g'({{c}_{2}})=0\] Where \[{{c}_{2}}\in (2,3)\]. Therefore, \[g'({{c}_{1}})=g'({{c}_{2}})=0\] By Rolle's Theorem at least one \[x\in ({{c}_{1}},{{c}_{2}})\]such that \[g''(x)=0\] or \[f''(x)=2\] for some\[x\in (1,3)\]
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