Question

f : R → R, where f(x) = . Complete set of values of 'a' such that f(x) is onto is

• A. (-∞, ∞)
• B. (-∞, 0)
• C. (0, ∞)
• D. None

Answer 1

Answer: (D)

None

Answer 2

$y = \frac { x ^ { 2 } + a x + 1 } { x ^ { 2 } + x + 1 }$

⇒ x²(1 − y) + x(a − y) + (1 − y) = 0.

Since x is real

⇒ (a – y)² – 4(1 – y)² ≥ 0

⇒ −3y² + 2y(4 - a) + a² – 4 ≥ 0 ∀ y ∈ R

(for f(x) to be onto). But this is not possible as the leading coefficient of this quadratic is negative.