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

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

$$\mathit{L}etf:D\to \mathit{R}bedefinedas$$$$f\left(x\right)=\frac{x^2+2x+a}{x^2+4x+3a}whereDandR$$$$denotethedomainof\mathit{f}andtheset$$$$ofallrealnumbersrespectively.$$$$Iffis{}^{″}surjective{}^{″}mappingthen$$$$therangeof\mathit{a}is?$$$$a)0⩽a⩽1$$$$b)0<a⩽1$$$$c)0<a<1$$$$d)0⩽a<1$$

Answered by MJS last updated on 26/Apr/18

$$a=0\Rightarrow f\left(x\right)=\frac{x(x+2)}{x(x+4)}=\frac{x+2}{x+4}\wedge x\ne 0\Rightarrow$$$$\Rightarrow f\left(x\right)\ne \frac{1}{2}\Rightarrow \mathrm{not}\mathrm{surjective}$$$$$$$$a=1\Rightarrow f\left(x\right)=\frac{{(x+1)}^2}{(x+1)(x+3)}=\frac{x+1}{x+3}\wedge x\ne -1\Rightarrow$$$$\Rightarrow f\left(x\right)\ne 0\Rightarrow \mathrm{not}\mathrm{surjective}$$$$$$$${\mathrm{let}}^{\prime }\mathrm{s}\mathrm{try}\mathrm{something}\mathrm{in}\mathrm{between}$$$$a=\frac{1}{2}\Rightarrow f\left(x\right)=\frac{2x^2+4x+1}{2x^2+8x+3}$$$$\underset{x\to \pm \mathrm{\infty }}{\lim }f\left(x\right)=1$$$$\mathrm{the}\mathrm{question}\mathrm{is},\mathrm{\exists }x\in \mathbb{R}\mid f\left(x\right)=1?$$$$\frac{2x^2+4x+1}{2x^2+8x+3}=1\Rightarrow x=-\frac{1}{2}=-a(?)$$$$\frac{x^2+2x+a}{x^2+4x+3a}=1\Rightarrow x=-a$$$$\mathrm{so}\mathrm{answer}\mathrm{c}\mathrm{should}\mathrm{be}\mathrm{right}$$

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

$$soyoutookonevaluebetween0\mathrm{\&}1.$$$$whatifitake\frac{1}{4},\frac{1}{3},etc.$$$$andalsotellhowforanyvalueof$$$$0<a<1therangeis\mathbf{R}?$$

Commented by MJS last updated on 28/Apr/18

$$\mathrm{good}\mathrm{point}$$$$$$$$\frac{x^2+2x+a}{x^2+4x+3a}=\frac{(x+1+\sqrt{1-a})(x+1-\sqrt{1-a})}{(x+2+\sqrt{4-3a})(x+2-\sqrt{4-3a})}$$$$\mathrm{the}\mathrm{same}\mathrm{problem}\mathrm{as}\mathrm{with}a=0\vee a=1\mathrm{would}$$$$\mathrm{occur}\mathrm{only}\mathrm{if}\mathrm{we}\mathrm{can}\mathrm{shorten}\mathrm{this}\mathrm{fraction}.$$$$\mathrm{in}\mathrm{this}\mathrm{case}:$$$$x+1+\sqrt{1-a}=x+2+\sqrt{4-3a}\vee$$$$x+1+\sqrt{1-a}=x+2-\sqrt{4-3a}\vee$$$$x+1-\sqrt{1-a}=x+2+\sqrt{4-3a}\vee$$$$x+1-\sqrt{1-a}=x+2+\sqrt{4-3a}$$$$\Rightarrow a=0\vee a=1$$$$\Rightarrow \mathrm{\forall }a\in ]0;1[:\frac{x^2+2x+a}{x^2+4x+3a}\in \mathbb{R}$$

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

$$\mathcal{T}\mathrm{hank}\mathrm{you}\mathrm{sir}.$$

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