**Question **: If a is an integer lying in [-5,30], then the probability that the graph of y=x^{2}+2\left( a+4\right) x-5a+64 is strictly above the x-axis is

(A) 1/6

(B) 7/36

(C) 2/9

(D) 3/5

**Answer **: option (C)

**Solution **:

y=x^{2}+2\left( a+4\right) x-5a+64\geq 0, If D≤0 then,

\left( a+4\right) ^{2}-\left( -5a+64\right) <0
\begin{aligned}\Rightarrow a^{2}+13a-48 <0\ \Rightarrow \left( a+16\right) \left( a-3\right) <0\ \Rightarrow -16 <a <3\Rightarrow -5\leq a\leq 2\end{aligned}
Then, the favourable cases is equal to the number of integers in the interval [-5,2], i.e, 8

Total number of cases is equal to the number of integers in the interval [-5,30] i.e., 36

Hence, the required probability is 8/36=2/9.

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