If-x-R-then-x-2-2x-a-x-2-4x-3a-can-take-all-real-values-if-1-a-0-2-2-a-0-1-3-a-1-1-4-None-of-these-

Question Number 20051 by Tinkutara last updated on 22/Aug/17

If x \in R then \frac{x^{2} + 2 x + a}{x^{2} + 4 x + 3 a} can take all

real values if

(1) a \in (0, 2)

(2) a \in [0, 1]

(3) a \in [- 1, 1]

(4) None of these

Answered by ajfour last updated on 22/Aug/17

l e t y = \frac{x^{2} + 2 x + a}{x^{2} + 4 x + 3 a}

y (x^{2} + 4 x + 3 a) = x^{2} + 2 x + a

(y - 1) x^{2} + 2 (2 y - 1) x + a (3 y - 1) = 0

y = e v e r y r e a l m u s t y i e l d a t l e a s t

o n e v a l u e o f r e a l x . S o D ⩾ 0

\Rightarrow 4 {(2 y - 1)}^{2} ⩾ 4 a (y - 1) (3 y - 1)

{(2 y - 1)}^{2} ⩾ a (y - 1) (3 y - 1)

4 y^{2} - 4 y + 1 ⩾ a (3 y^{2} - 4 y + 1)

(3 a - 4) y^{2} - 4 (a - 1) y + a - 1 ⩽ 0

\Rightarrow 3 a - 4 < 0 o r a < \frac{4}{3}

a n d d i s c r i m i n a n t ⩽ 0

\Rightarrow 16 {(a - 1)}^{2} ⩽ 4 (3 a - 4) (a - 1)

4 {(a - 1)}^{2} - (3 a - 4) (a - 1) ⩽ 0

4 a^{2} - 8 a + 4 - 3 a^{2} + 7 a - 4 ⩽ 0

a^{2} - a ⩽ 0

a (a - 1) ⩽ 0

\Rightarrow a \in [0, 1] .

Commented by Tinkutara last updated on 23/Aug/17

Thank you very much Sir!