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Question

Two numbers b and c are chosen at random with replacement from the numbers 1,2,3,4,5,6,7,8 and 9. The probability that x2+bx+c>0 for all xϵR is:

A
2381
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B
79
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C
3281
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D
65729
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Solution

The correct option is C 3281
Since b and c each can assume 9 values from 1 to 9.
So, total number of ways of choosing b and c is 9 × 9 = 81
Now, x2+bx+c>0 for all xϵR
D<0
b24ac<0b24c<0b2<4c
Alternatively: x2bx+c>0
(x+b2)2+4cb24>0
4cb2>0
b2<4c
Now, the following table shows the possible values of b and c for which b2<4c
cbTotal11,121,2231,2,3341,2,3351,2,3,4461,2,3,4471,2,3,4,5581,2,3,4,5591,2,3,4,5532
So, favourable number of cases = 32
Hence required probability = 3281

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