A) 1/6
B) 7/36
C) 2/9
D) 3/5
Correct Answer: C
Solution :
[c] \[{{x}^{2}}+2(a+4)x-5a+64\ge 0\] If\[D\le 0\], then \[{{(a+4)}^{2}}-(-5a+64)<0\] Or \[{{a}^{2}}+13a-48<0\] Or \[(a+16)(a-3)<0\] \[\Rightarrow -16<a<3\Leftrightarrow -5\le a\le 2\] Then, the favorable 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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