Question

The probability of selecting integers $a \in[-5,30]$ such that $x^{2}+2(a+4) x-5 a+64 > 0$, for all $x \in R$, is:

• A. $\frac{7}{36}$
• B. $\frac{2}{9}$
• C. $\frac{1}{6}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$D < 0$
$\Rightarrow 4(a+4)^{2}-4(-5 a+64)<0$
$\Rightarrow a^{2}+16+8 a+5 a-64<0$
$\Rightarrow a^{2}+13 a-48<0$
$\Rightarrow(a+16)(a-3)<0$
$\Rightarrow a \in(-16,3)$
$\therefore $ Possible a : $\{-5,-4, \ldots \ldots, 3\}$
$\therefore $ Required probability $=\frac{8}{36}$
$=\frac{2}{9}$