Question

if the expression (x-p)÷x^-3x+2 takes all real values for x€R,then find the bounds for 'p'

Answer 1

it is given that expression, (x - p)/(x² - 3x + 2) takes real value for X belongs to R.

function , y = (x-p)/(x² - 3x + 2) will be undefined at x² - 3x + 2 ≠ 0

⇒x² - 2x - x + 2 ≠ 0

⇒(x - 2)(x - 1) ≠ 0

⇒ x ≠ 1 , 2

so, domain , x $\in$ R - {1, 2}

now, let's solve this equation,

y = (x - p)/(x² - 3x + 2)

⇒ y(x² - 3x + 2) = x - p

⇒ yx² - 3yx + 2y - x + p = 0

⇒yx² - (3y + 1)x + (2y + p) = 0

for real values of x ,

discriminant, D = (3y + 1)² - 4(2y + p)y ≥ 0

(3y + 1)² - (8y² + 4py) ≥ 0

or, 9y² + 6y + 1 - 8y² - 4py ≥ 0

or, y² + (6 - 4p)y + 1 ≥ 0 it is possible when D ≤ 0

or, (6 - 4p)² - 4 ≤ 0

or, (6 - 4p)² - 2² ≤ 0

or, (6 - 4p - 2)(6 - 4p + 2) ≤ 0

or, (4 - 4p)(8 - 4p) ≤ 0

or, (1 - p)(2 - p) ≤ 0

or, 1 ≤ p ≤ 2

hence, bounds for p : [1, 2]