Q51 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics
If the maximum value of $2 x-7-a x^2$ cannot exceed 20 , then the minimum value of $a$ is
- A27
- B$\frac{1}{13}$
- C13
- Correct answerD$\frac{1}{27}$
Solution
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 $a=\frac{1}{27}$
Hence, the correct answer is option: 'D'.