The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$ are:

Question

=> The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$

takes all real values for $x\in R$ are:

My question is why we need to validate end points i.e. $1,7$ (Refer the last part of my attempt)

My attempt is as follows:-

$$y=\dfrac{ax^2+3x-4}{a+3x-4x^2}$$

$$ya+3yx-4yx^2=ax^2+3x-4$$ $$x^2(-4y-a)+x(3y-3)+ya+4=0$$

As $x$ can be any real, so $D\ge0$

$$9y^2+9-18y-4(ya+4)(-4y-a)\ge0$$ $$9y^2+9-18y+4(4y^2a+ya^2+16y+4a)\ge0$$ $$y^2(9+16a)+y(4a^2+64-18)+9+16a\ge0$$ $$y^2(9+16a)+y(4a^2+46)+9+16a\ge0$$

As range is $R$, so discriminant of quadratic in $y$ should be less than equal to zero

$$4(2a^2+23)^2-4(9+16a)^2\le0$$ $$(2a^2+23-9-16a)(2a^2+23+9+16a)\le0$$ $$(2a^2-16a+14)(2a^2+16a+32)\le0$$ $$(a^2-8a+7)(a^2+8a+16)\le0$$ $$a\in[1,7]$$

But in such type of questions, we always check at endpoints like here we need to check at $a=1$ and $a=7$. But I don't understand what is so special about endpoints.

From the above calculation I can only say at $a=1,7$ discriminant of quadratic in $y$ is zero, but what is so special about this. Please help me in this.

Answer 1

why at only endpoints {1,7} numerator and denominator are having common roots

Let $x=X$ be a common root. Then, we have $a+3X-4X^2=0$ and $aX^2+3X-4=0$. Subtracting the latter from the former gives $(a+4)(1-X^2)=0$. If $a=-4$, then $\dfrac{ax^2+3x-4}{a+3x-4x^2}=1$, so we have $a\not=-4$ from which $X=\pm 1$ follows. For $X=1$, we get $a=1$. For $X=-1$, we get $a=7$.

what is so special about these endpoints.

We see that $(1)$ is equivalent to $(2)$ where

$(1)$ For every $y$, there exists at least one $x$ such that $y=\dfrac{ax^2+3x-4}{a+3x-4x^2}$.

$(2)$ For every $y$, there exists at least one $x$ such that $(-4y-a)x^2+(3y-3)x+ya+4=0$ $\color{red}{\text{and}\ a+3x-4x^2\not=0}$.

(The red part is important. It seems that you are thinking that $(1)$ is equivalent to $(3)$, but note that $(3)$ is not equivalent to $(1)$ where

$(3)$ For every $y$, there exists at least one $x$ such that $(-4y-a)x^2+(3y-3)x+ya+4=0$.)

Here, let us take $y=-\frac a4$ for which the coefficient of $x^2$ is zero. Then, $(-4y-a)x^2+(3y-3)x+ya+4=0$ becomes $(a+4)(3x+a-4)=0$. Since $a\not=-4$, it follows from $x=\frac{4-a}{3}$ and $a+3x-4x^2\not=0$ that $a\not=1$ and $a\not=7$.

So, "$a\not=1$ and $a\not=7$" is a necessary condition.

If $a\not=1$ and $a\not=7$, then we have not only that the coefficient of $x^2$ is not zero, but also that the numerator and the denominator don't have any common root. So, we see that $x$ satisfying $y(a+3x-4x^2)=ax^2+3x-4$ also satisfies $a+3x-4x^2\not=0$ for every $y$, and that $(2)$ is equivalent to $(3)$ under the condition that $a\not=1$ and $a\not=7$.

In conclusion, under the condition that $a\not=1$ and $a\not=7$, we can consider $D$ safely as you did.

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Added :

if you explain this solution to anybody, then how will you do it. First you will calculate the interval of a which will come out [1,7],

No, I will not.

Considering $D\ge 0$ first is wrong because it is possible that $(-4y-a)x^2+(3y-3)x+ya+4=0$ is not a quadratic equation.

You cannot consider $D$ when the above equation is not a quadratic equation.

You have to deal with the case where the equation is not a quadratic equation first and separately, and then consider $D$.

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In the following, I'm going to write my solution to the question with some comments which should be helpful to understand why I take the step.

My solution :

We want to find $a$ such that for every $y$, there exists at least one $x$ satisfying $y=\dfrac{ax^2+3x-4}{a+3x-4x^2}$.

So, we want to find $a$ such that for every $y$, there exists at least one $x$ satisfying $y(a+3x-4x^2)=ax^2+3x-4$ and $a+3x-4x^2\not=0$.

[comments : This is an important step. Note that there is an additional condition $a+3x-4x^2\not=0$. This moment is not the time to consider $D\ge 0$. Before considering $D$, we have two points to consider. The first point is to consider the case where $y(a+3x-4x^2)=ax^2+3x-4$, i.e. $(-4y-a)x^2+(3y-3)x+ya+4=0$ is not a quadratic equation because you cannot consider $D$ when the equation is not a quadratic equation.]

Let us take $y=-\frac a4$ for which the equation is not a quadratic equation. Then, $(-4y-a)x^2+(3y-3)x+ya+4=0$ becomes $(a+4)(3x+a-4)=0$. Since $a\not=-4$, we get $x=\frac{4-a}{3}$, and so setting it into $a+3x-4x^2\not=0$ gives $$a+3\cdot\frac{4-a}{3}-4\bigg(\frac{4-a}{3}\bigg)^2\not=0\iff a\not=1\quad\text{and}\quad a\not=7$$

So, "$a\not=1$ and $a\not=7$" is a necessary condition.

[comments : The second point is to consider the part "$y(a+3x-4x^2)=ax^2+3x-4$ and $a+3x-4x^2\not=0$". We want to remove the part "and $a+3x-4x^2\not=0$". To do this, we want to find all $a$ such that $a+3x-4x^2$ and $ax^2+3x-4$ have a common root.]

Let $x=X$ be a common root of $a+3x-4x^2$ and $ax^2+3x-4$. Then, we have $a+3X-4X^2=0$ and $aX^2+3X-4=0$. Subtracting the latter from the former gives $(a+4)(1-X^2)=0$. If $a=-4$, then $\dfrac{ax^2+3x-4}{a+3x-4x^2}=1$, so we have $a\not=-4$ from which $X=\pm 1$ follows. For $X=1$, we get $a=1$. For $X=-1$, we get $a=7$.

So, we see that if $a\not=1$ and $a\not=7$, then $a+3x-4x^2$ and $ax^2+3x-4$ do not have any common root. From this, we see that if $a\not=1$ and $a\not=7$, then $x$ satisfying $y(a+3x-4x^2)=ax^2+3x-4$ also satisfies $a+3x-4x^2\not=0$ for every $y$.

This means that under the condition that $a\not=1$ and $a\not=7$, all we need is to find $a$ such that for every $y$, there exists at least one $x$ satisfying $(-4y-a)x^2+(3y-3)x+ya+4=0$.

[comments : Now, it is time to consider $D\ge 0$ because under the condition that $a\not=1$ and $a\not=7$, the equation is a quadratic equation, and also the additional condition "and $a+3x-4x^2\not=0$" is already removed. Now, you can consider $D$ safely.]

Considering $D\ge 0$, i.e.$$y^2(9+16a)+y(4a^2+46)+9+16a\ge0$$

we want to find $a$ such that this is true for every $y$. So, we have $$9+16a\gt 0\qquad\text{and}\qquad (4a^2+46)^2-4(9+16a)^2\le 0,$$ i.e. $$1\le a\le 7$$

With the condition that $a\not=1$ and $a\not=7$, the answer is $$\color{red}{1\lt a\lt 7}$$

Answer 2

The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$

$\dfrac{ax^2+3x-4}{a+3x-4x^2}$ will take all the values for $x \in \mathbb R$ only if the denominator is not equal to zero!

So, For $a + 3x - 4x^2 \neq 0; \iff -3 \pm \sqrt{9 +16a} \neq 0$

We need to check all the points for which the denominator will become equal to zero, and we need to remove those values of $a$ which includes $1, 7$.

Answer 3

When $a=1$, $\displaystyle y=\frac{x^2+3x-4}{1+3x-4x^2}=\frac{-x-4}{1+4x}\ne-\dfrac14$ for all $x\in\mathbb{R}$.

When $a=7$, $\displaystyle y=\frac{7x^2+3x-4}{7+3x-4x^2}=\frac{7x-4}{7-4x}\ne-\dfrac74$ for all $x\in\mathbb{R}$.

Answer 4

Let's denote $f(x)=ax^2+3x−4$, $g(x)=-4x^2+3x+a$, and $h(x)=f(x)/g(x)$.

Since $a>0>-9/16>-4$, $f(x)$ and $g(x)$ cannot have the same roots, and each of them has two distinct roots. However, they may have a common root. If $a=1,7$ then there exists $y_0\in \mathbb{\mathbb{R}}$ for which only one $x$ may give $h(x)=y_0$. This $x$ may be the root of the denominator, so we must check. Else, when $1<a<7$, there exists two distinct candidates $x_1,x_2$ which may give $h(x)=y$. In this case at least one of $x_1,x_2$ is not a root for $g$, since otherwise they will be the roots of $f$.