A) \[f(6)<5\]
B) \[f(6)=5\]
C) \[f(6)\ge 8\]
D) \[f(6)<8\]
Correct Answer: C
Solution :
As \[f(1)=-2\] and \[f'(x)\ge 2\forall x\in [1,6]\] Applying lagrange?s mean value theorem, \[\frac{f(6)-f(1)}{5}\] \[=f'(c)\ge 2\] \[\Rightarrow f(6)\ge 10+f(1)\Rightarrow f(6)\ge 10-2\Rightarrow f(6)\ge 8\].
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