Quadratic EquationHard
Question
Let the inequalities $y( - 1) > - 4,y(1) < 0$ and $y(3) > 5$ are known to hold for $y = {ax}^{2} + bx + c$, then
Options
A.$b > 3$
B.b $< 2$
C.$a < 0$
D.$a > \frac{1}{8}$
Solution
$a - b + c > - 4$
$$\begin{matrix} a + b + c < 0 & \Rightarrow & - a - b - c > 0 & \\ & \Rightarrow & & - 2b > - 4 \\ & \Rightarrow & & 9a + 3b + c > 5 \\ & & & - a - b - c > 0 \\ & \Rightarrow & & 8a + 2b > 5 \\ & & & 8a + 4 > 8a + 2b > 5 \end{matrix} \Rightarrow \ b < 2$$
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