Q52 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics
$\alpha$ is the maximum value of $1-2 x-5 x^2$ and $\beta$ is the minimum value of $x^2-2 x+r$. If $5 \alpha x^2+\beta x+6>0$ for all real values $x$, then the interval in which $r$ lies is
- A$(0,5)$
- B$(-5, \infty)$
- C$(-\infty, 7)$
- Correct answerD$(-11,13)$
Solution
We have, $f(x)=1-2 x-5 x^2$
$\begin{aligned}
& =-5\left[x^2+\frac{2}{5} x-\frac{1}{5}\right]=-5\left[\left(x+\frac{1}{5}\right)^2-\frac{1}{5}-\frac{1}{25}\right] \\
& =-5\left(x+\frac{1}{5}\right)^2+1+\frac{1}{5}=-5\left(x+\frac{1}{5}\right)^2+\frac{6}{5} \\
\therefore \quad f(x) & \leq \frac{6}{5} \quad \therefore \alpha=\frac{6}{5}
\end{aligned}$
Again, $g(x)=x^2-2 x+r=(x-1)^2+r-1$
$\begin{array}{rlrlrl}
\therefore g(x) \geq(r-1) \\
\therefore \beta =r-1
\end{array}$
So, $5 \alpha x^2+\beta x+6=6 x^2+(r-1) x+6$
Since, $5 \alpha x^2+\beta x+6>0$ and coefficient of $x^2>0$
$\begin{array}
\therefore D < 0 \\
\Rightarrow (r-1)^2-4 \times 6 \times 6 < 0 \\
\Rightarrow (r+11)(r-13) < 0 \\
\Rightarrow r \in(-11,13)
\end{array}$
Hence, the correct answer is option: 'D'.