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Q.
If $a \in[-20,0]$, then probability that the graph of the function $y=16 x^{2}+8(a+5) x-7 a-5$ is strictly above the $x$ - axis is
Probability - Part 2
Solution:
The graph of $y =16 x^{2}+8(a+5) x-7 a-5$ is strictly above the $x$ -axis
$\Rightarrow y >\, 0 \forall x \,\in R $
$\Rightarrow 16 x^{2}+8(a+5) x-7 a-5>\,0 \forall x \in R$
The above inequality holds if discriminant $<\,0$
$\left[\because\right.$ coefficient of $x^{2}>\,0$ ]
$\Rightarrow 64(a+5)^{2}-4.16(-7 a-5)<\,0$
$\Rightarrow a^{2}+17 a+30<\,0$
$ \Rightarrow (a+2)(a+15)<\,0$
$\Rightarrow -15<\,a<\,-2$
Given $-20 \leq a \leq 0$ and favourable cases $-15
$\therefore $ Required probability
$=\frac{\text { length of interval }(-15,-2)}{\text { length of interval }(-20,0)}=\frac{-2-(-15)}{0-(-20)}=\frac{13}{20}$
