Question

Let f(x) be a quadratic polynomial such that f(–1) + f(2) = 0. If one of the roots of f(x) = 0 is 3, then its other root lies in :

(1) (–3, –1)

(2) (1, 3) 

(3) (–1, 0) 

(4) (0, 1)

Answer 1

Correct option is (c) (-1, 0)

Let \(f = a(x - \alpha) (x - \beta)\)

\(f(-1) + f(2) = 0 \)

\(\alpha |(-1-\alpha) (-1 - \beta) + (2 - \alpha)(2 - \beta)| = 0\)

⇒ \(\alpha = |5\alpha + 2| = 0\)

⇒ \(\alpha = \frac{-2}5 \in (-1, 0)\)

Answer 2

Answer is (3) (–1, 0)

f(x) = a(x – 3) (x – α)

f(2) = a(α – 2)

f(–1) = 4a(1 + α)

f(–1) + f(2) = 0 ⇒ a(α – 2 + 4 + 4α) = 0

a ≠ 0 ⇒ 5α = – 2

α = \(\frac{2}{5}\) = - 0.4

α  ∈ (–1, 0)