Question:
The probability of selecting integers $a \in[-5,30]$ such that $x^{2}+2(a+4) x-5 a+64>0$, for all $x \in \mathbf{R}$, is:
Correct Option: , 2
Solution:
$D<0$
$\Rightarrow 4(a+4)^{2}-4(-5 a+64)<0$
$\Rightarrow a^{2}+16+8 a+5 a-64<0$
$\Rightarrow a^{2}+13 a-48<0$
$\Rightarrow(a+16)(a-3)<0$
$\Rightarrow a \in(-16,3)$
$\therefore$ Possible a : $\{-5,-4, \ldots \ldots . ., 3\}$
$\therefore$ Required probability $=\frac{8}{36}$
$=\frac{2}{9}$
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