Question

Let a ∈ (– ∞, 0).

Statement–1 : ax² – x + 4 < 0 for all x ∈ R

Statement–2 : If roots of ax² + bx + c = 0, b, c ∈ R are imaginary then signs of ax² + bx + c and a are same for all x ∈ R.

(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. 

(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.

(C) Statement – 1 is True, Statement – 2 is False. 

(D) Statement – 1 is False, Statement – 2 is True

Answer 1

Correct option (D) Statement – 1 is False, Statement – 2 is True

Explanation:

Statement – II is true as if ax² + bx + c = 0 has imaginary roots, then for no real x,

ax² + bx + c is zero, meaning thereby ax² + bx + c is always of one sign. Further  lim x→∞ (ax² + bx + c) = signum (a). ∞

statement – I is false, because roots of ax² – x + 4 = 0 are real for any a ∈(- ∞, 0) and hence ax² – x + 4 takes zero, positive and negative values.

Hence (d) is the correct answer.