A) 2
B) 3
C) 3/2
D) 1
Correct Answer: B
Solution :
\[f(x)=\left\{ \begin{align} & {{x}^{2}}+2,\,\,1\le x<2 \\ & \frac{{{x}^{2}}+2}{2},\,\,2\le x<3 \\ & \frac{{{x}^{2}}+2}{3},\,\,x=3 \\ \end{align} \right.\] \[f'(x)=\left\{ \begin{align} & 2x,\,1<x<2 \\ & x,\,2<x<3 \\ \end{align} \right.\] Thus, least value of\[f(x)\]in [1, 2] is 3 and least value of\[f(x)\]in [2, 3] is 3. \[f(3)=\frac{11}{3}\] Thus, least value of\[f(x)\]is 3.
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