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Question

If the roots of the equation x22ax+a2+a3=0 are real and less than 3, then

A
a<2
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B
aa3
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C
3<a4
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D
a>2
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Solution

The correct option is A a<2


Solution : x22ax+a2+a3=0
D=4a24(a2+a3)0
a30
a3
f(3)=96a+a2+a3>0
a25a+6>0
(a2)(a3)>0
a<2 or a>3
2aa<3
a<3
(a<2)


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