Find the set of values of $\alpha$ so that $f(x)=\dfrac{\alpha x^2+6x-8}{\alpha+6x-8x^2}$ is one one.

Question

Let $f$ be a function defined in its domain given by $f(x)=\dfrac{\alpha x^2+6x-8}{\alpha+6x-8x^2}$. Find the set of values of $\alpha$ so that $f(x)$ is one-one.

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My attempt

As $f(x)$ have to be one-one so any line $\parallel$ to $x$-axis must cut the graph only once. So, necessary condition is that $f(x)$ must be monotonic. $$f'(x)={(\alpha+6x-8x^2)(2\alpha x+6)-(\alpha x^2+6x-8)(6-16x)\over(\alpha+6x-8x^2)^2}$$ $$\implies f'(x)={6(\alpha+8)x^2+2(\alpha+8)(\alpha-8)x+6(\alpha+8)\over(\alpha+6x-8x^2)^2}=\frac{Q(x)}{(\alpha+6x-8x^2)^2}$$ In order to have $f'(x)\le0$ or $\ge0$ $\forall$ $x\in R$ , Discriminant of $Q(x)$ $\le0$. $$\implies (\alpha+8)^2(\alpha-14)(\alpha-2)\le 0 \\ {\implies \alpha \in [ 2 \; , \; 14] \tag{1}}$$

But, for this range $f(x) $ is not one one (I have projected it in desmos for verification) and I found it later that $(1)$ is also the condition for which $f(x)$ is onto i.e. Range of $f \in \mathbb R$.

$\boxed{y=\dfrac{\alpha x^2+6x-8}{\alpha+6x-8x^2} \\ \implies (\alpha+8y)x^2+(6-6y)x+(-8-\alpha y)=0 \\ \text{as}\; x\in \mathbb R \;; D\ge0 \\ \implies 6-6y)^2+4(\alpha+8y)(8+\alpha y)\ge 0 \\ \implies (9+8\alpha)y^2+(\alpha^2+46)y+(8\alpha+9)\ge0\; \; \text{now as} \; y\in R\\ \implies (\alpha^2+46)^2-4(8\alpha+9)^2\le0 \implies (\alpha+8)^2(\alpha-14)(\alpha-2)\le0 \; \text{....same as (1) above}}$

Desmos Graph shows that it is many one for this range as line line parallel to x axis cut it 2 times although function continuously increases for all $x\in \mathbb R$... Please tell what is happening here and how to solve this question...

Link for the diagram

Question Source: India's JEE math practice book.

Answer 1

Your strategy is good, and your computations are mostly correct, with a minor correction: $(1)$ implies that $$\color{red}{\alpha} \in [2, 14], \tag{1}$$ not $x$. And this is a correct conclusion, but monotonicity of $f$ does not guarantee that it is injective, because if there is a discontinuity of $f$, then $f$ can be monotone but not injective: for example, $$g(x) = x - \frac{1}{x}$$ implies $$g'(x) = 1 + \frac{1}{x^2} \ge 1 \quad \forall x \in \mathbb R \setminus \{0\}$$ but it is not injective since $g(1-\sqrt{2}) = g(1+\sqrt{2}) = 2$. This phenomenon is what is happening to your problem. Monotonicity is a necessary condition but not a sufficient one. So your reasoning has so far established that $(1)$ must be true, but we must go further. In particular, I recommend the following approach:

1. Assume there exist distinct $x, y$ such that $f(x) = f(y)$. Show that this implies $$6(xy+1) + (\alpha-8)(x+y) = 0. \tag{2}$$
2. Show that $(2)$ may also be written in the form $$(x+h(\alpha))(y+h(\alpha)) = k(\alpha), \tag{3}$$ for suitable functions $h(\alpha), k(\alpha)$ that are constants with respect to $x$ and $y$.
3. Conclude that for (almost) every $x$, there exists a $y \ne x$ that satisfies $(3)$, except when $k(\alpha) = 0$, hence we require $k(\alpha) = 0$ for $f$ to be injective.
4. Consider the cases satisfying $k(\alpha) = 0$ and simplify $f$ to show that these cases do actually result in $f$ being injective.

Answer 2

Hint Notice that if $\pm 1$ are both in the domain of $f$, then $$f(\pm 1) = \frac{\alpha (\pm 1)^2 + 6(\pm 1) - 8}{\alpha + 6(\pm 1) - 8(\pm 1)^2} = \frac{\alpha - 8 \pm 6}{\alpha - 8 \pm 6} = 1 .$$

In particular $f(1) = f(-1) = 1$, a contradiction. So, if $f$ is one-to-one, then at least one of $1$ and $-1$ is a root of the denominator, giving $$\alpha = 8 \mp 6 ,$$ and we can check these $2$ cases directly.

Remark More generally, suppose $f(x)$ is a ratio of a polynomial $p(x) = a x^2 + b x + c$ with $a \neq 0, c \neq 0$, and its reciprocal, $p^\ast(x) = a + b x + c x^2$. If $f$ is one-to-one, then $p^\ast(\pm 1) = 0$ and hence $p(\pm 1) = 0$, so $x \mp 1$ is a common factor of $p, p^\ast$. Thus, $f$ is actually a ratio of two linear functions, less a removable singularity at $\mp 1$: $$f(x) = \frac{a x \mp c}{c x \mp a}, \qquad x \neq \pm 1 .$$ This function being one-to-one requires further precisely that $c \neq \pm a$.

Answer 3

For $g(x)=-6x+8x^2$ and $h(x)=x^{-1}(\alpha -g(x))$ we have $f(x)={h(x^{-1})\over h(x)}.$ Therefore $f(x)$ should be undefined at $x=1$ or at $x=-1,$ as otherwise $f(1)=f(-1)=1.$ Hence $h(x)$ should vanish at $x=1$ or at $x=-1,$ i.e. $g(1)=\alpha$ or $g(-1)=\alpha.$

We can use the above for more complicated functions $g(x)$ defined on the whole real line. For example let $g(x)=\sin(\pi x)+xe^x.$ Then we must have $\alpha=e$ or $\alpha=e^{-1}.$