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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 xR is

A
17123
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B
3281
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C
82125
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D
45143
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Solution

The correct option is B 3281
Here, x2+bx+c>0 xR
D<0
b2<4c

Value of bPossible values of c11<4cc>14{1,2,3,4,5,6,7,8,9}24<4cc>1{2,3,4,5,6,7,8,9}39<4cc>94{3,4,5,6,7,8,9}416<4cc>4{5,6,7,8,9}525<4cc>6.25{5,6,7,8,9}636<4cc>9Impossible7Impossible8Impossible9Impossible
Number of favourable cases = 9 + 8 + 7 + 5 + 3
=32
Total ways =9×9=81
Required probability = 3281

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