# A question that give wrong answer?

• vkash
In summary: I'm probably wrong...In summary, the conversation is about a math question posed by the person's teacher. The question involves finding the value of "a" for a function that is "onto." Many students initially attempted to solve the question, but their answers were incorrect. The person who posted the question does not want anyone to provide the answer, as they want others to try and solve it themselves. A few different methods of solving the question are suggested, including using the quadratic formula and calculus. The person who posed the question eventually reveals that there is a "trick" to solving it and asks others to continue trying to solve it.
vkash
Hello!
Today my maths teacher ask a question to all student in class. that was
Code:
f:R-->R
f(x) = (a*x[SUP]2[/SUP]+6x-8) / (a+6x-8*x[SUP]2[/SUP])
f is onto then find the value of [b]a[/b]
after 10 minutes some students answer. every body was wrong.
this question was also answered wrong in many Indian writer's books for many years. Now my teacher tell where we all students were wrong and know i know the correct answer. I just want to show this question to all those persons who love mathematical beauty.I am not intended to get answer for homework help.
So guys try it.

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I'm sure if the question was worded properly, then some people would have attempted a solution. Exactly what does "f is onto then find the value of a" mean? Are we looking for a value of "a" such that both the numerator and the denominator can be factored over the reals?

I agree it could be worded more clearly, but it made perfect sense to me (though I could be wrong). "f is onto" means (afaik) "for every real number y, there is some real number x such that f(x)=y".

chogg said:
I agree it could be worded more clearly, but it made perfect sense to me (though I could be wrong). "f is onto" means (afaik) "for every real number y, there is some real number x such that f(x)=y".
Onto means co domain of function is equal to range of function. If there comes a number that is in co domain but not in range then it will not remain onto. Finally i mean co domain=range.(it is just hint to for onto not full definition)

If you try this question then post your answer. Not with full method but just tell what your procedure of doing question and final answer.

Let y be any number. If
$$y= \frac{ax^2+6x-8}{a+6x-8x^2}$$
then
$$y(a+ 6x- 8x^2)= ax^2+ 6x- 8$$
$$ay+ 6yx- 8yx^2= ax^2+ 6x- 8$$
$$(a+ 8y)x^2+ (6-6y)x- (8+ ay)= 0$$

You can solve that using the quadratic formula. Can you find a specific value of "a" so that has a real solution for any "y"?

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HallsofIvy said:
Let y be any number. If
$$y= \frac{ax^2+6x-8}{a+6x-8x^2}$$
then
$$y(a+ 6x- 8x^2)= ax^2+ 6x- 8$$
$$ay+ 6yx- 8yx^2= ax^2+ 6x- 8$$
$$(a+ 8y)x^2+ (6-6y)- (8+ ay)= 0$$

You can solve that using the quadratic formula. Can you find a specific value of "a" so that has a real solution for any "y"?

I think you mean
$$(a+ 8y)x^2+ (6-6y)x - (8+ ay)= 0$$
on the last line.

Yes, thanks. I will correct it.

HallsofIvy said:
Let y be any number. If
$$y= \frac{ax^2+6x-8}{a+6x-8x^2}$$
then
$$y(a+ 6x- 8x^2)= ax^2+ 6x- 8$$
$$ay+ 6yx- 8yx^2= ax^2+ 6x- 8$$
$$(a+ 8y)x^2+ (6-6y)x- (8+ ay)= 0$$

You can solve that using the quadratic formula. Can you find a specific value of "a" so that has a real solution for any "y"?

dear this is the way how every body do it(including me) but this will give you wrong answer. I don't know what is wrong with this method but it will give you wrong answer.If you are not believing me then find value of a mentioned by HallsofIvy and draw it's curve using any http://www.padowan.dk/graph/Download.php" . You will be confirmed.

Dude this amazing behavior of this question is it's beauty. I am not fool that i ask such easy question. Due to this reason this question was answered wrong in many books for many years. Everybody(who do it wrong) do this as you do.

HINT: think in different way.

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vkash said:
dear this is the way how every body do it(including me) but this will give you wrong answer. I don't know what is wrong with this method but it will give you wrong answer.If you are not believing me then find value of a mentioned by HallsofIvy and draw it's curve using any http://www.padowan.dk/graph/Download.php" . You will be confirmed.

Dude this amazing behavior of this question is it's beauty. I am not fool that i ask such easy question. Due to this reason this question was answered wrong in many books for many years. Everybody(who do it wrong) do this as you do.

HINT: think in different way.

HallsofIvy's approach seems perfectly sound to me; unless you have another valid solution, then you are wrong. Please do show us anything you have though, I'm sure a lot of us would like to see it.

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vkash said:
dear this is the way how every body do it(including me) but this will give you wrong answer. I don't know what is wrong with this method but it will give you wrong answer.If you are not believing me then find value of a mentioned by HallsofIvy and draw it's curve using any http://www.padowan.dk/graph/Download.php" . You will be confirmed.

Dude this amazing behavior of this question is it's beauty. I am not fool that i ask such easy question. Due to this reason this question was answered wrong in many books for many years. Everybody(who do it wrong) do this as you do.

HINT: think in different way.

OK, I'll bite. What is the correct way of solving it according to you?? And what do you get as solutions that HOI (perfectly sound) approach doesn't give?

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Wait i will give it's answer.

Let more people to try this question.

Answer will given after 12 hours.
Till then keep thinking.

REQUEST+> Don't post answer if you know it's trick.

Okay, so I'm just going to treat the top equation as a quadratic, and the bottom equation as a quadratic, and then multiply them together.

Now I replace the first fraction with a,b, and c with a,6, and -8. Then I replace the second fraction with a,b, and c with -8, 6, and a.

Now you know x, when y is zero, so you just plug into the original equation, and you might get the right answer?

Note: I think we have to do something with that y. I don't think the quadratic equation works when a,b, or c is not constant. I bet we have to use calculus to solve this.

Nvm, the above method is totally bogus.

a+6x−8x^2 cannot equal zero, so we just have to find a value of a that makes it impossible for a+6x−8x^2 to not equal zero.

So, just take the derivative: 6−16x = 0 so x =3/8 the max.
which means that value of a has to be when a+6(3/8)−8(3/8)^2) is less than zero.
I think it's when a is less than -9/8?

We can also just switch the signs on that equation so that we have -(a+6x−8x^2) on the bottom and -(ax^2+6x-8) on top.

The derivative is -6+16x = 0 so the minimum is now, 3/8 = x, which means that value of a has to be when -[a+6(3/8)−8(3/8)^2)] is greater than zero. (+9/8)

the fact that you said "math teacher" made me think calculus

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Oh, I think I know what it is know. Isn't it weird how there is an -8 on top where there is an a in the bottom and vise versa? You simply put in a=-8 and the answer is always 1!

((-8)*x^2+6x-8) / ((-8)+6x-8*x^2)=1

-8 is smaller than -9/8 so the calculus was true.
I don't quite get what the question is though.

given equation
$$y= \frac{ax^2+6x-8}{a+6x-8x^2}$$ and f:R-->R
so denominator will never zero OK.
using shredharanachrya principle D<0 only in that case this possible.
so 36-32*a<0 this implies $$a> \frac{9}{8}$$--------------------point 1.
In denominator coefficient of x2 is negative so all the values of a+6x-8x2 will smaller than 0.--------------------point 2.
since denominator value is smaller than 0 for all x so numerator should have all -ve as well as +ve values so that function became onto.
come to numerator how is it's D's nature if it will remain quadratic equation it will either give all the +ve or all the -ve values so a=0 so that it can give all the values +ve as well as negative from here we come to know that a=0.--------------------point 3.
is there any value that satisfy all the condition.

do you got the answer there is no solution for this question.

this is it's beauty friends. this way of thinking is easier and takes less time but everybody didn't think so. due this amazing nature of this question i post it here.
Now real question starts from now why these two methods give different answer even nothing is seeming wrong with method written by HallsofIvy.
I don't know what tukka(random method) does RandomMystery applies but my solution is correct.

If you did not know why first(given by HallsofIvy) method is wrong then ask this to all the friends(who can answer). If you got the solution then please message answer to me or post that here. I am also searching for it's reason.
This question came in IIT JEE 1996 entrance exam.

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I think this is a trick question. If you set a = x you get

f(x) = (x^3+6x-8) / (7x-8x^2)

and this is onto the reals.

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SteveL27 said:
I think this is a trick question. If you set a = x you get f(x) = (x^3+6x-8) / (7x-8*x2) and this is onto the reals.
first of all tell me what will happen when x=0. Will it finite. if not then it will not became onto.

every question in mathematics has a trick some are common trick those are called easy questions and some are rare tricks where question became tougher.
it's answer can't be 'x'. why see real question.

A function f:R-->R, where R is set of real numbers, is defined by $$f(x)= \frac{ax^2+6x-8}{a+6x-8x^2}$$. Find the interval values of a for which f is onto. Is function one to one for a=3? justify your answer.

this is the photo copy of question that come in 1996 entrance exam. Here a interval of a is asked not variable value.
if you got the reason why HallsofIvy wrong then post that reason here.

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HoI's method is perfectly correct. His method will give that the function f is onto if $a\geq 2$. However, since the function must be entire, it must also hold that $a+6x-8x^2$ has no roots. But no $a\geq 2$ satisfies this. So there are no solutions.

vkash said:

given equation
$$y= \frac{ax^2+6x-8}{a+6x-8x^2}$$ and f:R-->R
so denominator will never zero OK.
using shredharanachrya principle D<0 only in that case this possible.
so 36-32*a<0 this implies $$a> \frac{9}{8}$$--------------------point 1.
In denominator coefficient of x2 is negative so all the values of a+6x-8x2 will smaller than 0.--------------------point 2.
since denominator value is smaller than 0 for all x so numerator should have all -ve as well as +ve values so that function became onto.
come to numerator how is it's D's nature if it will remain quadratic equation it will either give all the +ve or all the -ve values so a=0 so that it can give all the values +ve as well as negative from here we come to know that a=0.--------------------point 3.
is there any value that satisfy all the condition.

do you got the answer there is no solution for this question.

this is it's beauty friends. this way of thinking is easier and takes less time but everybody didn't think so. due this amazing nature of this question i post it here.
Now real question starts from now why these two methods give different answer even nothing is seeming wrong with method written by HallsofIvy.
I don't know what tukka(random method) does RandomMystery applies but my solution is correct.

If you did not know why first(given by HallsofIvy) method is wrong then ask this to all the friends(who can answer). If you got the solution then please message answer to me or post that here. I am also searching for it's reason.
This question came in IIT JEE 1996 entrance exam.
I thought this expression had to be negative for there not to be any zero roots in the denominator.
$$a> -\frac{9}{8}$$

Why isn't that ^ the answer? I don't understand what onto and f:R-->R mean.
If you plug in any number less than -9/8 on any graphing calculator, you will get a continuous function.

What do you mean by "-ve as well as +ve" and what does "the function became onto" mean?
Does D mean denominator?

"how is it's D's nature if it will remain quadratic equation it will either give all the +ve or all the -ve values so a=0 so that it can give all the values +ve as well as negative from here we come to know that a=0"

What does this mean? Where did he get a=0?

I think you can still use the quadratic equation:

a has to be a value that makes this expression always true (since you can't take the square root of negative numbers):

[URL]http://latex.codecogs.com/gif.latex?0\leq&space;(6-6y)^2-4(a&plus;8y)(8&plus;ay)[/URL]

I'm not sure what you do next, but maybe you have to sub in y=a*x^2+6x-8) / a+6x-8*x^2 at some point.

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RandomMystery said:
I thought this expression had to be negative for there not to be any zero roots in the denominator.
$$a> -\frac{9}{8}$$

Why isn't that ^ the answer? I don't understand what onto and f:R-->R mean.
If you plug in any number less than -9/8 on any graphing calculator, you will get a continuous function.

What do you mean by "-ve as well as +ve" and what does "the function became onto" mean?
Does D mean denominator?

"how is it's D's nature if it will remain quadratic equation it will either give all the +ve or all the -ve values so a=0 so that it can give all the values +ve as well as negative from here we come to know that a=0"

What does this mean? Where did he get a=0?
I think either conventions used in your country for function notation is different from that of in INDIA
D is b*b-4ac discriminent of quadratic equation.
f:R->R this mean domain of function is all real number and co domain is also R all real number.

if you have ever drawn curve of $$f(x)=\frac{ax^2+bx+c}{}$$ then you must have found that it makes a parabola. A parabola can't cover all the values i mean a quadratic equation can't give all -ve as well all +ve values. If a =0 it became straight line which give all the -ve as well as +ve values. that's why i say a=0 for this to give all -ve as well as +ve values.

this is what i was saying.

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It's hard or impossible to catch the error in work I can't see. VKash, why don't you post your attempt & I can catch the mistake.

In the original post, it says
"f is onto"
Now, I read f is NOT onto.
So your question is just plain false. The original post asked the impossible.

As asked for the entrance exam, the empty set is a valid interval, a legitimate answer to the entrance exam variation. Please post questions verbatim whenever possible.

To withhold critical information, necessary to solve the question is disrespectful at best. If you want to play games, go to

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nickalh said:
It's hard or impossible to catch the error in work I can't see. VKash, why don't you post your attempt & I can catch the mistake.
that is too lengthy to type just do this on your own this is lengthy but not tooooo... lengthy to do in paper.

Errhhh, have you tried WolframAlpha.com?
It will do this level of complexity for you. One can even take results and

The best reason I have so far?
Simply solving the quadratic formula doesn't guarantee solutions for every y. It only guarantees solutions
Or maybe the value one gets for "a", generates a removable discontinuity in the original rational function.
If the discriminant is >0, we know there are two places where the graph crosses the x-axis, which isn't enough to be onto.

The main issue I ran into?
The function has horizontal asymptotes, but never seems to get to or cross the asymptote.

I hope I wasn't too harsh on you.
Bye.

I got a partial answer below, but stopped working when I saw the original post asked an invalid question.

Here's my stab at the answer.
The denominator has zero, one or two roots. The number of roots relates to the number of vertical tangents.
A. One Root, One Vertical Tangent:
... {see other work in this thread}
a = -9/8
A quick graph will show us clearly it is not onto the reals.

B. No Roots, No Vertical Tangents
...
a < -9/8
The end behavior, lim x->+- oo is a horizontal asymptote. Same asymptote for both +oo and -oo.
lim f = (-9/8)/ -8 = -9/64
Rational functions of polynomials only go to oo at the x=-oo or x=oo or at vertical asymptotes.
Unfortunately, with no vertical tangent, this means the function is bounded.

C. Two roots, Two vertical tangents
a>-9/8
I don't have a conclusive proof.
Let x = c & x = d, where c < d be the roots of the denominator. Vertical tangents appear at x = c & x = d.
(Assume a>0, because if -9/8 < a < 0 we have a similar argument with a vertical flip of the graph.)
The behavior for x<c and x>d is similar. Again both approach horizontal asymptotes as x->+-oo from above. (The rational function is (negative number)/ (negative number), for |x| sufficiently large.)

So we must rely on the graph between the vertical asymptotes, for everything below the horizontal asymptotes. However, choosing a, so the hump in the middle comes high enough is difficult or impossible.

vkash said:
Hello!
Today my maths teacher ask a question to all student in class. that was
Code:
f:R-->R
f(x) = (a*x[SUP]2[/SUP]+6x-8) / (a+6x-8*x[SUP]2[/SUP])
f is onto then find the value of [b]a[/b]
after 10 minutes some students answer. every body was wrong.
this question was also answered wrong in many Indian writer's books for many years. Now my teacher tell where we all students were wrong and know i know the correct answer. I just want to show this question to all those persons who love mathematical beauty.I am not intended to get answer for homework help.
So guys try it.

The quadratic in the denominator must have complex roots, which occur when a<-9/8. Then f(x) will be analytic, and onto iff the domain of f is ℝ.

I don't think there is such an a. First of all we need a < -9/8. We also need surjectivity, but this will ultimately imply that (a+8)^2 < 0 (if you do the calculations per Halls' method), which is impossible.

Another method is to note that such a function will have horizontal asymptotes, which denies surjectivity.

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nickalh said:
Unfortunately, with no vertical tangent, this means the function is bounded.

This is not true in general, consider f(x) = xsin(x). It will require some work to show that it is true for rational polynomial functions.

nickalh said:
Errhhh, have you tried WolframAlpha.com?
It will do this level of complexity for you. One can even take results and

The best reason I have so far?
Simply solving the quadratic formula doesn't guarantee solutions for every y. It only guarantees solutions
Or maybe the value one gets for "a", generates a removable discontinuity in the original rational function.
If the discriminant is >0, we know there are two places where the graph crosses the x-axis, which isn't enough to be onto.

The main issue I ran into?
The function has horizontal asymptotes, but never seems to get to or cross the asymptote.

I hope I wasn't too harsh on you.
Bye.
OK! solving quadratic equation does not involve solution always but one should always think about reasons. All rules(as i know) of mathematics are human assumptions so there is answer of all such things that why of why not. You can't say so in physics because there are rules which are made by GOD and we are studying those rules.

matphysik said:
The quadratic in the denominator must have complex roots, which occur when a<-9/8. Then f(x) will be analytic, and onto iff the domain of f is ℝ.
why are going to complex maths even thou i have mentioned that f is from real number to real number.If you are intended to just help me by saying all this is possible in complex math then OK! else. see real question .I posted that later in this thread.
Oh why some people create this complex mathematics? Is it branch of mathematics? I say so because it violate some rules of mathematics(I have read a very little complex maths).

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vkash said:
OK! solving quadratic equation does not involve solution always but one should always think about reasons. All rules(as i know) of mathematics are human assumptions so there is answer of all such things that why of why not. You can't say so in physics because there are rules which are made by GOD and we are studying those rules.why are going to complex maths even thou i have mentioned that f is from real number to real number.If you are intended to just help me by saying all this is possible in complex math then OK! else. see real question .I posted that later in this thread.
Oh why some people create this complex mathematics? Is it branch of mathematics? I say so because it violate some rules of mathematics(I have read a very little complex maths).
Hello. Suppose that the denominator quadratic DOES HAVE real roots, say for x=x1 and x=x2. Then if the domain of f is all of ℝ: f(x1),f(x2) →∞. So that there arent y1,y2∈ℝ s.t., f⁻¹(y1)=x1, f⁻¹(y2)=x2. Hence, f will not be onto.
NOTE: If f(x)=y then f∘f⁻¹(y)=y.

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matphysik said:
Hello. Suppose that the denominator quadratic DOES HAVE real roots, say for x=x1 and x=x2. Then if the domain of f is all of ℝ: f(x1),f(x2) →∞. So that there arent y1,y2∈ℝ s.t., f⁻¹(y1)=x1, f⁻¹(y2)=x2. Hence, f will not be onto.
NOTE: If f(x)=y then f∘f⁻¹(y)=y.

I don't think this is true. Just because f fails to be defined at a couple of points, how does that show there exists y1 and y2 that don't get hit by some value of x?

In the joke solution I gave in post #15, I said to take a = x. Of course that's a "what color is the bear?" answer, not a math answer.

If you do that, you get f(x) = (x^3+6x-8) / (7x-8x^2) which blows up at two points. But if you graph it, you'll see that it's onto! The three connected components of the graph overlap to cover the y-axis.

In general, rational function can be onto even if the denominator has real zeros and therefore blows up at a couple of points. I didn't prove that, I just looked at a graph ... so it's possible I'm totally wrong here.

Here's the graph I got from the Grapher app in OS X.

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SteveL27 said:
I don't think this is true. Just because f fails to be defined at a couple of points, how does that show there exists y1 and y2 that don't get hit by some value of x?

In the joke solution I gave in post #15, I said to take a = x. Of course that's a "what color is the bear?" answer, not a math answer.

If you do that, you get f(x) = (x^3+6x-8) / (7x-8x^2) which blows up at two points. But if you graph it, you'll see that it's onto! The three connected components of the graph overlap to cover the y-axis.

In general, rational function can be onto even if the denominator has real zeros and therefore blows up at a couple of points. I didn't prove that, I just looked at a graph ... so it's possible I'm totally wrong here.

Here's the graph I got from OS X Grapher.

First, a (parameter) cannot by chosen to be the independent variable (x). Second, the graph of f(x) = (x^3+6x-8) / (7x-8x^2) has pole at x=0. So as you can see from the graph there is no y∈ℝ s.t. y=f(0).

When the denominator quadratic has no real roots then f (a quotient of two polynomials) is onto. Just the monomial term in the numerator quadratic guarantees that.

vkash said:
Hello!
Today my maths teacher ask a question to all student in class. that was
Code:
f:R-->R
f(x) = (a*x[SUP]2[/SUP]+6x-8) / (a+6x-8*x[SUP]2[/SUP])
f is onto then find the value of [b]a[/b]
after 10 minutes some students answer. every body was wrong.
this question was also answered wrong in many Indian writer's books for many years. Now my teacher tell where we all students were wrong and know i know the correct answer. I just want to show this question to all those persons who love mathematical beauty.I am not intended to get answer for homework help.
So guys try it.

Just look at the essential part, and ignore the commentary

matphysik said:
First, a (parameter) cannot by chosen to be the independent variable (x).

I said about three times that this is a joke solution ... but the resulting function is very interesting anyway.

matphysik said:
Second, the graph of f(x) = (x^3+6x-8) / (7x-8x^2) has pole at x=0. So as you can see from the graph there is no y∈ℝ s.t. y=f(0).

Just look at the graph. There's clearly a zero at around x = 1 or so, where the right hand component of the graph crosses the x-axis. What do you make of that?

You keep saying that there's a pole at 0, which of course is true. But it does not therefore follow that NO value of x gives f(x) = 0.

f(x) = (x^3+6x-8) / (7x-8x^2) has pole at x=0, and for x very close to zero f(x)~ -8/7x. Hence, as x->0- f-> +∞ and as x->0+ f-> -∞. If the resolution of your graph were better you could see this behaviour more clearly.

matphysik said:
f(x) = (x^3+6x-8) / (7x-8x^2) has pole at x=0, and for x very close to zero f(x)~ -8/7x. Hence, as x->0- f-> +∞ and as x->0+ f-> -∞. If the resolution of your graph were better you could see this behaviour more clearly.

It has nothing to do with the behavior of f at x = 0. Better study the definition of onto. The graph crosses the x-axis at around x = 1. That means f takes the value zero for at least one value of x. So f is onto.

To see that f is onto, look at the y axis. You can see that you can pick any point on the y-axis and a horizontal line through it will intersect the graph of f.

You are fixating on the pole at 0 but that has nothing to do with whether f is onto. You are confusing domain and range.

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