If α, β are the roots of the quadratic equation x2-2 p(x-4)-15=0, then the set of values of p for which one root is less than 1 the other root is greater than 2 is:

Question

If α, β are the roots of the quadratic equation x2-2 p(x-4)-15=0, then the set of values of p for which one root is less than 1 the other root is greater than 2 is: (A) ((7/3), ∞) (B) (-∞, (7/3)) (C) x ∈ R (D) (-∞, (11/4))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/d26cfb05f92acb762a650697eae25089-.png" /> x2-2 p x+(8 p-15)=0 f(1)<0 and f(2)<0 f(1)=1-2 p+8 p-15<0 p< 7 / 3 f(2)=4-4 p+8 p-15<0 4 p-11<0 ⇒ p<(11/4) p ∈(-∞, 7 / 3) Ans.

Q. If $α, β$ are the roots of the quadratic equation $x^{2} - 2 p (x - 4) - 15 = 0$ , then the set of values of $p$ for which one root is less than $1&$ the other root is greater than 2 is:

$(\frac{7}{3}, \infty)$

$(- \infty, \frac{7}{3})$

$x \in R$

$(- \infty, \frac{11}{4})$

$x^{2} - 2 p x + (8 p - 15) = 0$
$f (1) < 0$ and $f (2) < 0$
$f (1) = 1 - 2 p + 8 p - 15 < 0$
$p < 7/3$
$f (2) = 4 - 4 p + 8 p - 15 < 0$
$4 p - 11 < 0 \Rightarrow p < \frac{11}{4}$
$p \in (- \infty, 7/3)$ Ans.

Q. If α,β are the roots of the quadratic equation x2−2p(x−4)−15=0, then the set of values of p for which one root is less than 1& the other root is greater than 2 is:

(37​,∞)

(−∞,37​)

x∈R

(−∞,411​)