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Question

The integer k for which the inequality x2−2(4k−1)x+15k2−2k−7>0 is valid for x, is

A
2
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B
3
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C
4
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D
None of these
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Solution

The correct option is B 3
The inequality will be greater than zero
if D is less than zero.
D<0

4(4k1)24.1.(15k22k7)<0
Simplifying this we get
K26k+8<0

(k4)(k2)<0

k lies in (2,4)

In this interval only one integer is present i.e 3

k=3

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