Question

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

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

Answer 1

The correct option is D $\frac{2}{9}$

Since $x^2+2(a+4)x–5a+64>0,$ we have discriminant$\left(D\right)<0$
$\Rightarrow 4(a+4{)}^2–4(64–5a)<0$
$\Rightarrow a^2+8a+16–64+5a<0$
$\Rightarrow a^2+13a–48<0$
$\Rightarrow a^2+16a–3a–48<0$
$\Rightarrow (a+16)(a–3)<0$
$a\in (–16,3)$
in set $[–5,30]$ total integers $36$
favourable integers $8$
Required probability is $\frac{8}{36}=\frac{2}{9}$