KVPY 2019 SA Exam Previous Papers

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Question : 2 of 80

Marks: +1, -0

Let b be a non-zero real number. Suppose the quadratic equation

2x^2 + bx +1/b= 0

has two distinct real roots. Then

$b+1/b>5/2$
$b+1/b<5/2$
$b^2 - 3b > -2$
$b^2+1/b^2>4$

Solution:

Discriminant =

b^2-8/b>0

(for real roots)

={(b- 2)(b^3+ 2b+ 4)}/b>0

⇒

b ∊(-∞, 0) ∪ (2, ∞)

which is a subset of solution set of

b^2 - 3b + 2 > 0

Hence (C)

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