If the maximum value of 2x - 7 - ax^2 cannot | vedclass

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(D) Let $f(x) = -ax^2 + 2x - 7$. For the function to have a maximum value,we must have $a > 0$.Completing the square:$f(x) = -a(x^2 - \frac{2}{a}x) -

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$\frac{1}{27}$

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(D) Let $f(x) = -ax^2 + 2x - 7$. For the function to have a maximum value,we must have $a > 0$.Completing the square:$f(x) = -a(x^2 - \frac{2}{a}x) - 7$$f(x) = -a(x^2 - \frac{2}{a}x + \frac{1}{a^2} - \frac{1}{a^2}) - 7$$f(x) = -a(x - \frac{1}{a})^2 + \frac{1}{a} - 7$The maximum value of $f(x)$ is $\frac{1}{a} - 7$.Given that the maximum value cannot exceed $20$,we have:$\frac{1}{a} - 7 \leq 20$$\frac{1}{a} \leq 27$Since $a > 0$,we get $a \geq \frac{1}{27}$.Thus,the minimum value of $a$ is $\frac{1}{27}$.

If the maximum value of $2x - 7 - ax^2$ cannot exceed $20$,then the minimum value of $a$ is

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