# If a is an integer lying in [-5,30], then the probability that the graph of y=x^2+2(a+4)x-5a+64 is strictly above the x-axis is

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

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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