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Question

If xR, then 1x2+4x+9 lies in the interval

A
[5,)
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B
(0,15]
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C
(0,)
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D
(0,12]
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Solution

The correct option is B (0,15]
Here, xR
x2+4x+9=(x+2)2+5
=(x+2)2+5
Now
<x<<x+2<0(x+2)2<5(x+2)2+5<
0<1(x+2)2+515
​​​​​​​1x2+4x+9(0,15]

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