Question

Question is in attachment

if possible someone can answer​

Answer 1

Answer:

y does not belong to (2,8)

Step-by-step explanation:

Question :

$y = \dfrac{2x^2 - 6x+5}{x^2-3x+2}$ ;  x is real

To find :

The range of y

Solution :

$y = \dfrac{2x^2 - 6x+5}{x^2-3x+2}$

$: \implies y \ (x^2 -3x + 2)= 2x^2 -6x + 5$

$: \implies yx^2 -3xy + 2y = 2x^2 -6x+5$

$: \implies x^2(y -2 ) -3x (y-2) + (2y+5) = 0$

It is of the form, ax²+by+c = 0

So, as x is real, Δ (discriminant) ≥ 0

$\Delta \equiv b^2 -4ac \geq 0$

Here, a = (y-2), b = -3(y-2); c = 2y + 5

$: \implies \Delta \equiv (-3(y-2))^2 - 4(y-2)(2y-5) \geq 0$

$: \implies 9(y-2)^2 -4(y-2)(2y-5) \geq 0$

$: \implies (y-2)(9y-18-8y + 10) \geq 0$

$: \implies (y-2)(y-8) \geq 0$

$: \implies y \in (-\infty , 2] \cup [8, \infty) \ (or) \ y \leq 2 \ or \ y\geq 8$

$\therefore y \notin (2,8)$

Hope it helps!!