Question

If $a\in [-20,0]$, then probability that the graph of the function $y=16x^2+8(a+5)x-7a-5$ is strictly above the $x$ - axis is

• A. $\frac{7}{20}$
• B. $\frac{13}{20}$
• C. $\frac{17}{20}$
• D. $\frac{3}{20}$

Answer 1

Answer: (B)

The graph of $y=16x^2+8(a+5)x-7a-5$ is strictly above the $x$ -axis
$\Rightarrow y>0\forall x\in R$
$\Rightarrow 16x^2+8(a+5)x-7a-5>0\forall x\in R$
The above inequality holds if discriminant $<0$
$[\because$ coefficient of $x^2>0$ ]
$\Rightarrow 64(a+5{)}^2-4.16(-7a-5)<0$
$\Rightarrow a^2+17a+30<0$
$\Rightarrow (a+2)(a+15)<0$
$\Rightarrow -15<a<-2$

Given $-20\le a\le 0$ and favourable cases $-15<a<-2$
$\therefore$ Required probability
$=\frac{\text{ length of interval }(-15,-2)}{\text{ length of interval }(-20,0)}=\frac{-2-(-15)}{0-(-20)}=\frac{13}{20}$