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


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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

	$Thank you very much Sir!$
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