Question

If $aϵ[-6,12]$, then the probability of that graph of $y=-x^2+2(a+4)-(3a+40)$ is strictly below x-axis is

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{2}$

The total length of the interval $=12-(-6)=18$. If graph of $y=-x^2+2(a+4)-(3a+40)$ is entirely below $x$-axis, the value of discriminant of the above quadratic expression must be negative.
$\therefore 4{(a+4)}^2-4\left(I\right)-(3a+40)<0$
$\Rightarrow a^2+5a-24<0$
$\Rightarrow (a+8)(a-3)<0$
$\Rightarrow -8<a<3$ but $aϵ[-6,12]$
$\therefore -6<a<3$ for event to happen
$\therefore$ length of interval $=3-(-6)=9$
Hence the required probability $=\frac{9}{18}=1/2$