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If the maximum value of $2 x-7-a x^2$ cannot exceed 20 , then the minimum value of $a$ is
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1237 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{27}$
We have,
$$
\begin{aligned}
& f(x)=2 x-7-a x^2 \\
& f(x)=-7-a\left(x^2-\frac{2}{a} x\right) \\
& f(x)=\frac{1}{a}-7-a\left(x^2-\frac{2 x}{a}+\frac{1}{a^2}\right) \\
& f(x)=\frac{1}{a}-7-a\left(x-\frac{1}{a}\right)^2
\end{aligned}
$$
Maximum value of $f(x)$ is $\frac{1}{a}-7$
But given $f(x) \leq 20$
$$
\begin{gathered}
\therefore \quad \frac{1}{a}-7 \leq 20 \\
a \geq \frac{1}{27}
\end{gathered}
$$
$\therefore$ Minimum value of $a=\frac{1}{27}$
$$
\begin{aligned}
& f(x)=2 x-7-a x^2 \\
& f(x)=-7-a\left(x^2-\frac{2}{a} x\right) \\
& f(x)=\frac{1}{a}-7-a\left(x^2-\frac{2 x}{a}+\frac{1}{a^2}\right) \\
& f(x)=\frac{1}{a}-7-a\left(x-\frac{1}{a}\right)^2
\end{aligned}
$$
Maximum value of $f(x)$ is $\frac{1}{a}-7$
But given $f(x) \leq 20$
$$
\begin{gathered}
\therefore \quad \frac{1}{a}-7 \leq 20 \\
a \geq \frac{1}{27}
\end{gathered}
$$
$\therefore$ Minimum value of $a=\frac{1}{27}$
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