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A question that give wrong answer?

  1. Jul 13, 2011 #1
    Today my maths teacher ask a question to all student in class. that was
    Code (Text):

    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.
    Please don't answer this if you already know this questions truth.
    Last edited: Jul 13, 2011
  2. jcsd
  3. Jul 13, 2011 #2
    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?
  4. Jul 13, 2011 #3
    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".
  5. Jul 13, 2011 #4
    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.
  6. Jul 14, 2011 #5


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

    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"?
    Last edited: Jul 14, 2011
  7. Jul 14, 2011 #6
    I think you mean
    [tex](a+ 8y)x^2+ (6-6y)x - (8+ ay)= 0[/tex]
    on the last line. :smile:
  8. Jul 14, 2011 #7


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    Yes, thanks. I will correct it.
  9. Jul 14, 2011 #8
    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" [Broken]. 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.
    Last edited by a moderator: May 5, 2017
  10. Jul 14, 2011 #9
    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.
    Last edited by a moderator: May 5, 2017
  11. Jul 14, 2011 #10


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    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?
    Last edited by a moderator: May 5, 2017
  12. Jul 14, 2011 #11
    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.
  13. Jul 14, 2011 #12
    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 in to 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
    Last edited: Jul 14, 2011
  14. Jul 14, 2011 #13
    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.
  15. Jul 15, 2011 #14
    if this is your first visit to this thread then please try this question before seeing answer.(think different)

    see it's answer.

    given equation
    [tex]y= \frac{ax^2+6x-8}{a+6x-8x^2}[/tex] 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 [tex]a> \frac{9}{8}[/tex]--------------------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.
    Last edited: Jul 15, 2011
  16. Jul 15, 2011 #15
    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.
    Last edited: Jul 15, 2011
  17. Jul 15, 2011 #16
    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 [tex]f(x)= \frac{ax^2+6x-8}{a+6x-8x^2}[/tex]. 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.
    Last edited: Jul 15, 2011
  18. Jul 15, 2011 #17


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    HoI's method is perfectly correct. His method will give that the function f is onto if [itex]a\geq 2[/itex]. However, since the function must be entire, it must also hold that [itex]a+6x-8x^2[/itex] has no roots. But no [itex]a\geq 2[/itex] satisfies this. So there are no solutions.
  19. Jul 15, 2011 #18
    I thought this expression had to be negative for there not to be any zero roots in the denominator.
    [tex]a> -\frac{9}{8}[/tex]

    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?
  20. Jul 15, 2011 #19
    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):


    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.
    Last edited by a moderator: Apr 26, 2017
  21. Jul 15, 2011 #20
    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 [tex]f(x)=\frac{ax^2+bx+c}{}[/tex] 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.
    Last edited: Jul 15, 2011
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