Question

If a is an integer lying in $(-5,30]$, then probability that the graph of $y=x^2+2(a+4)x-5a+64$ is strictly above the x-axis is

• A. $\frac{7}{36}$
• B. $\frac{4}{5}$
• C. $\frac{2}{9}$
• D. $\frac{1}{5}$

Answer 1

The correct option is D $\frac{1}{5}$

For, $y=x^2+2(a+4)x-5a+64$ to be above X-axis discriminant has to be negative
$4(a+4{)}^2-4(-5a+64)<0$
$\Rightarrow a^2+13a-48<0$
$\Rightarrow (a+16)(a-3)<0$
$\Rightarrow -16<a<3($Since$-5<a\le 30)$
$\Rightarrow -4\le a\le 2$
Hence, favorable cases $n\left(E\right)=2-(-4)+1=7$
Hence, total cases $n\left(S\right)=30-(-5)=35$
Required probability $=\frac{7}{35}=\frac{1}{5}$