If a,b∈R, a≠0 and the quadratic equation ax2−bx+1=0has imaginary roots, then (a + b + 1) is

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positive

negative

zero

dependent on the sign of b

answer is A.

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Detailed Solution

D=b2−4a<0⇒a>0Therefore, the graph is concave upwards.f(x)>0,∀x∈R ⇒ f(−1)>0⇒a+b+1>0

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