MCQ
The minimum value of $f(a) = (2{a^2} - 3) + 3(3 - a) + 4$ is
- A${{15} \over 2}$
- B${{11} \over 2}$
- C${{ - 13} \over 2}$
- ✓${{71} \over 8}$
$f'(a) = 4a - 3,f(a) = 4 > 0$
for exteremum, $f'(a) = 0 \Rightarrow a = \frac{3}{4}$
$\therefore$ $f(a)$ 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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