A) \[\frac{15}{2}\]
B) \[\frac{11}{2}\]
C) \[\frac{-13}{2}\]
D) \[\frac{71}{8}\]
Correct Answer: D
Solution :
\[f(a)=2{{a}^{2}}-3a+10\] \[f'(a)=4a-3,f''(a)=4>0\] for exteremum, \[f'(a)=0\Rightarrow a=\frac{3}{4}\] \ \[f(a)\] is minimum at \[a=\frac{3}{4}\] \[f{{(a)}_{\min }}=2\times {{\left( \frac{3}{4} \right)}^{2}}-3\times \left( \frac{3}{4} \right)+10=\frac{71}{8}\].
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