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Question

If a,b R, a> 0 and the quadratic equation ax2bx+1=0 has imaginary roots, then
a+b+1 is

A
positive
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B
negative
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C
zero
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D
depends on the sign of b.
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Solution

The correct option is A positive
f(x)=ax2bx+1
f(0)=1 (positive)
i.e. f(x)=ax2bx+1>0 (given ax2bx+1=0 has imaginary roots)
f(1)=a+b+1>0

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