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

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 ? a) 0≤a≤1 b) 0<a≤1 c) 0<a<1 d) 0≤a<1

$$\boldsymbol{{L}}{et}\:{f}:{D}\:\rightarrow\:\boldsymbol{{R}}\:{be}\:{defined}\:{as}\: \\ $$$${f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+{a}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}{a}}\:{where}\:{D}\:{and}\:{R} \\ $$$${denote}\:{the}\:{domain}\:{of}\:\boldsymbol{{f}}\:{and}\:{the}\:{set} \\ $$$${of}\:{all}\:{real}\:{numbers}\:{respectively}. \\ $$$${If}\:{f}\:{is}\:''\:{surjective}\:''\:\:{mapping}\:{then} \\ $$$${the}\:{range}\:{of}\:\boldsymbol{{a}}\:{is}\:? \\ $$$$\left.{a}\right)\:\mathrm{0}\leqslant{a}\leqslant\mathrm{1} \\ $$$$\left.{b}\right)\:\mathrm{0}<{a}\leqslant\mathrm{1} \\ $$$$\left.{c}\right)\:\mathrm{0}<{a}<\mathrm{1} \\ $$$$\left.{d}\right)\:\mathrm{0}\leqslant{a}<\mathrm{1}\: \\ $$

Answered by MJS last updated on 26/Apr/18

a=0 ⇒ f(x)=((x(x+2))/(x(x+4)))=((x+2)/(x+4))∧x≠0 ⇒ ⇒ f(x)≠(1/2) ⇒ not surjective a=1 ⇒ f(x)=(((x+1)^2 )/((x+1)(x+3)))=((x+1)/(x+3))∧x≠−1 ⇒ ⇒ f(x)≠0 ⇒ not surjective let′s try something in between a=(1/2) ⇒ f(x)=((2x^2 +4x+1)/(2x^2 +8x+3)) lim_(x→±∞) f(x)=1 the question is, ∃x∈R∣f(x)=1? ((2x^2 +4x+1)/(2x^2 +8x+3))=1 ⇒ x=−(1/2)=−a (?) ((x^2 +2x+a)/(x^2 +4x+3a))=1 ⇒ x=−a so answer c should be right

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

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

so you took one value between 0&1. what if i take (1/4) , (1/3),etc. and also tell how for any value of 0<a<1 the range is R ?

$${so}\:{you}\:{took}\:{one}\:{value}\:{between}\:\mathrm{0\&1}. \\ $$$${what}\:{if}\:{i}\:{take}\:\frac{\mathrm{1}}{\mathrm{4}}\:,\:\frac{\mathrm{1}}{\mathrm{3}},{etc}.\: \\ $$$${and}\:{also}\:{tell}\:{how}\:{for}\:{any}\:{value}\:{of} \\ $$$$\mathrm{0}<{a}<\mathrm{1}\:{the}\:{range}\:{is}\:\boldsymbol{\mathrm{R}}\:? \\ $$

Commented by MJS last updated on 28/Apr/18

good point ((x^2 +2x+a)/(x^2 +4x+3a))=(((x+1+(√(1−a)))(x+1−(√(1−a))))/((x+2+(√(4−3a)))(x+2−(√(4−3a))))) the same problem as with a=0 ∨ a=1 would occur only if we can shorten this fraction. in this case: x+1+(√(1−a))=x+2+(√(4−3a)) ∨ x+1+(√(1−a))=x+2−(√(4−3a)) ∨ x+1−(√(1−a))=x+2+(√(4−3a)) ∨ x+1−(√(1−a))=x+2+(√(4−3a)) ⇒ a=0 ∨ a=1 ⇒ ∀a∈]0; 1[: ((x^2 +2x+a)/(x^2 +4x+3a))∈R

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

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

Thank you sir.

$$\mathscr{T}\mathrm{hank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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