Let-f-D-R-be-defined-as-f-x-x-2-2x-a-x-2-4x-3a-where-D-and-R-denote-the-domain-of-f-and-the-set-of-all-real-numbers-respectively-If-f-is-surjective-mapping-then-the-range-of-a-is

Question Number 33815 by rahul 19 last updated on 25/Apr/18

L e t f : D \to R b e d e f i n e d a s

f (x) = \frac{x^{2} + 2 x + a}{x^{2} + 4 x + 3 a} w h e r e D a n d R

d e n o t e t h e d o m a i n o f f a n d t h e s e t

o f a l l r e a l n u m b e r s r e s p e c t i v e l y .

I f f i s^{″} s u r j e c t i v e^{″} m a p p i n g t h e n

t h e r a n g e o f a i s ?

a) 0 ⩽ a ⩽ 1

b) 0 < a ⩽ 1

c) 0 < a < 1

d) 0 ⩽ a < 1

Answered by MJS last updated on 26/Apr/18

a = 0 \Rightarrow f (x) = \frac{x (x + 2)}{x (x + 4)} = \frac{x + 2}{x + 4} \land x \neq 0 \Rightarrow

\Rightarrow f (x) \neq \frac{1}{2} \Rightarrow not surjective

a = 1 \Rightarrow f (x) = \frac{{(x + 1)}^{2}}{(x + 1) (x + 3)} = \frac{x + 1}{x + 3} \land x \neq - 1 \Rightarrow

\Rightarrow f (x) \neq 0 \Rightarrow not surjective

{let}^{'} s try something in between

a = \frac{1}{2} \Rightarrow f (x) = \frac{2 x^{2} + 4 x + 1}{2 x^{2} + 8 x + 3}

\lim_{x \to \pm \infty} f (x) = 1

the question is, \exists x \in R ∣ f (x) = 1 ?

\frac{2 x^{2} + 4 x + 1}{2 x^{2} + 8 x + 3} = 1 \Rightarrow x = - \frac{1}{2} = - a (?)

\frac{x^{2} + 2 x + a}{x^{2} + 4 x + 3 a} = 1 \Rightarrow x = - a

so answer c should be right

Commented by rahul 19 last updated on 28/Apr/18

s o y o u t o o k o n e v a l u e b e t w e e n 0 & 1 .

w h a t i f i t a k e \frac{1}{4}, \frac{1}{3}, e t c .

a n d a l s o t e l l h o w f o r a n y v a l u e o f

0 < a < 1 t h e r a n g e i s R ?

Commented by MJS last updated on 28/Apr/18

good point

\frac{x^{2} + 2 x + a}{x^{2} + 4 x + 3 a} = \frac{(x + 1 + \sqrt{1 - a}) (x + 1 - \sqrt{1 - a})}{(x + 2 + \sqrt{4 - 3 a}) (x + 2 - \sqrt{4 - 3 a})}

the same problem as with a = 0 \lor a = 1 would

occur only if we can shorten this fraction .

in this case :

x + 1 + \sqrt{1 - a} = x + 2 + \sqrt{4 - 3 a} \lor

x + 1 + \sqrt{1 - a} = x + 2 - \sqrt{4 - 3 a} \lor

x + 1 - \sqrt{1 - a} = x + 2 + \sqrt{4 - 3 a} \lor

x + 1 - \sqrt{1 - a} = x + 2 + \sqrt{4 - 3 a}

\Rightarrow a = 0 \lor a = 1

\Rightarrow \forall a \in] 0; 1 [: \frac{x^{2} + 2 x + a}{x^{2} + 4 x + 3 a} \in R

Commented by rahul 19 last updated on 29/Apr/18

T hank you sir .