Let P(x)=x2-(2-p) x+p-2. If P(x) assumes both positive and negative values ∀ x ∈ R, then the range of p is

Question

Let P(x)=x2-(2-p) x+p-2. If P(x) assumes both positive and negative values ∀ x ∈ R, then the range of p is (A) (-∞, 2) ∪(6, ∞) (B) (2,6) (C) (-∞, 2) (D) (6, ∞)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/9395ca46c075e8ee69631772305fb33e-.png" /> Let f ( x )= x 2-(2- p ) x +( p -2) We must have D >0 ⇒ (p-2)(p-6)>0 ∴ p ∈(-∞, 2) ∪(6, ∞).

Q. Let $P (x) = x^{2} - (2 - p) x + p - 2$ . If $P (x)$ assumes both positive and negative values $\forall x \in R$ , then the range of $p$ is

$(- \infty, 2) \cup (6, \infty)$

$(2, 6)$

$(- \infty, 2)$

$(6, \infty)$

Let $f (x) = x^{2} - (2 - p) x + (p - 2)$
We must have $D > 0$
$\Rightarrow (p - 2) (p - 6) > 0$
$∴ p \in (- \infty, 2) \cup (6, \infty)$ .

Q. Let P(x)=x2−(2−p)x+p−2. If P(x) assumes both positive and negative values ∀x∈R, then the range of p is

(−∞,2)∪(6,∞)

(2,6)

(−∞,2)

(6,∞)