The minimum value of f(a) = (2a^2 - 3) + 3(3 | vedclass

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(D) Given function is $f(a) = (2a^2 - 3) + 3(3 - a) + 4$. Simplifying the expression: $f(a) = 2a^2 - 3 + 9 - 3a + 4$ $f(a) = 2a^2 - 3a + 10$. To find

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$\frac{71}{8}$

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(D) Given function is $f(a) = (2a^2 - 3) + 3(3 - a) + 4$. Simplifying the expression: $f(a) = 2a^2 - 3 + 9 - 3a + 4$ $f(a) = 2a^2 - 3a + 10$. To find the minimum value,we find the derivative $f'(a)$: $f'(a) = \frac{d}{da}(2a^2 - 3a + 10) = 4a - 3$. Setting $f'(a) = 0$ for critical points: $4a - 3 = 0 \Rightarrow a = \frac{3}{4}$. Since the second derivative $f''(a) = 4 > 0$,the function has a local minimum at $a = \frac{3}{4}$. The minimum value is: $f\left(\frac{3}{4}\right) = 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) + 10$ $f\left(\frac{3}{4}\right) = 2\left(\frac{9}{16}\right) - \frac{9}{4} + 10$ $f\left(\frac{3}{4}\right) = \frac{9}{8} - \frac{18}{8} + \frac{80}{8} = \frac{71}{8}$.

The minimum value of $f(a) = (2a^2 - 3) + 3(3 - a) + 4$ is

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