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Question

The integer
k, for which the inequality x22(3k1)x+8k27>0 is valid for every x in R is :

A
3
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B
2
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C
4
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D
0
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Solution

The correct option is A 3
For x22(3k1)x+8k27>0
D<0
(2(3k1))24(8k27)<0
9k2+16k(8k27)<0
k26k+8<0
(k4)(k2)<0
2<k<4
then k=3

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