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Question

If x2+2ax+103a>0 for all x ∈ R, then


A

-5 < a < 2

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B

a < -5

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C

a > 5

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D

2 < a < 5

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Solution

The correct option is A

-5 < a < 2


According to given condition,

4a24(103a)<0a2+3a10<0

(a+5)(a2)<05<a<2.


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