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Question

The least integral value of k for which (k2)x2+8x+k+4>0 for all xϵR, is

A
5
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B
4
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C
3
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D
None of these
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Solution

The correct option is A 5
(k2).x2+8x+(k+4)>0

Here b24.a.c=0.

(8)24.(k2).(k+4)=0

or (k2)(k+4)=16.

or k2+2k816=0.

or k2+2k24=0.

or (k+6)(k4)=0.

k=6 or 4

K>4
Hence, the answer is 5.

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