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Question

Let aR. For possible real values of x, x2+ax+3x2+x+a takes all real values, then

A
4a3+39<0
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B
4a3+390
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C
a14
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D
a<14
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Solution

The correct option is D a<14
Let x2+ax+3x2+x+a=y
(1y)x2(ya)x+(3ay)=0
Since xR
(ya)24(1y)(3ay)0
(14a)y2+(2a+12)y+a2120 (1)

Eq.(1) is true for all yR
14a>0 and D0
a<14 and 4(a+6)24(a212)(14a)0
a<14 and 4a336a48
4a3<36(14)48 [a<14]
4a3+39<0

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