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Consider the following function $f(x)=\dfrac{x^2+2x+a}{x^2+4x+3a}$

Now the question states for us to find the constraint that limits $a$ so that $f(x)$ becomes surjective.

My Attempt

This can be further written as $f(x)=\dfrac{(x+1)^2+a-1}{(x+2)^2+3a-4}$
Now to make the function surjective, we can ensure two cases:-

CASE 1
The function on the top can have some $y$ $\in$ $\mathbb{Q}^{-}$ and for this $a-1<0$ and the function at bottom can have some $y$ $\in$ $\mathbb{Q}^{+}$ and for this $3a-4>0$ . But there is no solution.

CASE 2
The function on the top can have some $y$ $\in$ $\mathbb{Q}^{+}$ and for this $a-1>0$ and the function at bottom can have some $y$ $\in$ $\mathbb{Q}^{-}$ and for this $3a-4<0$. So $a\in(1,4/3)$

But this ans is wrong could someone point the mistake.

Angelo
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TheCuriousOne
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  • The answer is $;0<a<1,.$ – Angelo Jul 23 '23 at 08:41
  • TheCuriousOne, your method is completely incorrect. You should look for all the values of $a$ such that the equation $f(x)=y$ has real solutions $x$ for any $y\in\Bbb R,.$ Actually, what you wrote does not make sense. – Angelo Jul 23 '23 at 10:51
  • @Angelo, we want the function to be surjective, that is why I wrote the terms in that way. Now we want positive output as well as negative. Since a squared term would always mean more than or equal to zero. We can displace the graph so as to include negative terms so the whole real numbers are included. That was what I had tried. Would you mind explaining the flaw in my thinking. – TheCuriousOne Jul 23 '23 at 11:07
  • TheCuriousOne, $a-1<0$ is not equivalent to $(x+1)^2+a-1<0$. For example if $x=2$ and $a=0$, then $a-1=-1<0$ but $(x+1)^2+a-1=8>0$. Analogously, $3a-4>0$ is not equivalent to $(x+2)^2+3a-4>0$. Indeed if $x=2$ and $a=1$, then $(x+2)^2+3a-4=15>0$ but $3a-4=-1<0$. For that reason, I told you that your method is not correct. – Angelo Jul 23 '23 at 11:18
  • @Angelo Sir, but wont there be some x for which y is negative or positive? – TheCuriousOne Jul 23 '23 at 11:25
  • TheCuriousOne, if you want to prove that the function $f(x)$ is surjective, you have to prove that there exists some $x$ such that $f(x)=y$ for all $y\in\Bbb R$. It is not sufficient to prove that there exists some $x$ for which $f(x)=y$ is negative or positive. Indeed you have to prove that there exists some $x$ for which $f(x)$ is equal to every $y$ positive, zero or negative. It means the equation $f(x)=y$ has to have solutions for all $y$ positive, zero or negative. – Angelo Jul 23 '23 at 14:05

1 Answers1

2

The function

$\;f(x)\!=\!\dfrac{x^2+2x+a}{x^2+4x+3a}\!:\!\mathrm{dom}f\!\to\!\Bbb R\,$ is surjective if and only if

for any $\;y\in\Bbb R\;$ there exists $\;x\in\mathrm{dom}f\;$ such that $\;f(x)=y\,.$

So we have to find all the values of $\,a\,$ such that

$\begin{cases}\dfrac{x^2+2x+a}{x^2+4x+3a}=y\\[3pt] x^2+4x+3a\neq0\end{cases}$

has solutions $\,x\in\Bbb R\,$ for any $\,y\in\Bbb R\,.$

$\begin{cases}x^2+2x+a=yx^2+4yx+3ay\\[3pt] x^2+4x+3a\neq0\end{cases}$

$\begin{cases}(1-y)x^2+2(1-2y)x+a(1-3y)=0\\[3pt] x^2+4x+3a\neq0\end{cases}\qquad\color{blue}{(*)}$

If $\;y=1\;,\;$ then the system $\,(*)\,$ has a solution ($x=-a$) provided that $\color{brown}{\;a\neq0\;}$ and $\color{brown}{\;a\neq1\,}.\quad\color{blue}{(1)}$

If $\;y\neq1\;,\;$ then the system $\,(*)\,$ has solutions provided that

$\dfrac{\Delta}4=(1-2y)^2-a(1-3y)(1-y)\geqslant0\;\;,$

$1-4y+4y^2-a+ay+3ay-3ay^2\geqslant0\;\;,$

$(4-3a)y^2-4(1-a)y+1-a\geqslant0\;.$

If $\;\;\dfrac{\Delta’}4=4(1-a)^2-(4-3a)(1-a)>0\;\;,$

there would exist infinitely many $\,y\in\Bbb R\,$ such that

$\dfrac{\Delta}4=(4-3a)y^2-4(1-a)y+1-a<0$

and consequently there would be infinitely many $\,y\in\Bbb R\,$ for which the system $\,(*)\,$ would not have solutions.

Hence,

$\dfrac{\Delta’}4=4(1-a)^2-(4-3a)(1-a)\leqslant0\;\;,$

$4-8a+4a^2-4+4a+3a-3a^2\leqslant0\;\;,$

$a^2-a\leqslant0\;\;,$

$\color{brown}{0\leqslant a\leqslant1\,}.\quad\color{blue}{(2)}$

From $\,(1)\,$ and $\,(2)\,,\,$ it follows that

$0<a<1\,.$

Angelo
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