Core 3 functions question

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    Hello, having some trouble with part iii on this question. The mark scheme says "Attempt arrangement of f(x)=x or f^-1(x)=x" and "Apply discriminant to resulting quadratic equation", but I'm not sure how to get the quadratic equation in the first place.

    Thanks in advance.
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    (Original post by NA LUL)


    Hello, having some trouble with part iii on this question. The mark scheme says "Attempt arrangement of f(x)=x or f^-1(x)=x" and "Apply discriminant to resulting quadratic equation", but I'm not sure how to get the quadratic equation in the first place.

    Thanks in advance.
    Can't see the image.
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    (Original post by NA LUL)


    Hello, having some trouble with part iii on this question. The mark scheme says "Attempt arrangement of f(x)=x or f^-1(x)=x" and "Apply discriminant to resulting quadratic equation", but I'm not sure how to get the quadratic equation in the first place.

    Thanks in advance.

    Can you post all your thoughts and working for this question? It will be useful to see what you've tried and how far you got.


    You should use the direct link to the image when using Imgur images on TSR otherwise it won't be displayed. So you should have used http://i.imgur.com/jpubfj1.png instead of http://imgur.com/a/fadLB.
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    (Original post by notnek)

    Can you post all your thoughts and working for this question? It will be useful to see what you've tried and how far you got.


    You should use the direct link to the image when using Imgur images on TSR otherwise it won't be displayed. So you should have used http://i.imgur.com/jpubfj1.png instead of http://imgur.com/a/fadLB.
    I'm not too sure where to being really, I understand that the curves are reflections of each other through y=x meaning they wont meet. I'm unsure as to why you would use f(x)=x or f^-1(x)=x and also to get a quadratic equation by doing this. Sorry if this isn't making much sense, I'm not too sure about it myself.

    Also apologies about the image situation, I'm new and still finding my way around here
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    (Original post by NA LUL)
    I'm not too sure where to being really, I understand that the curves are reflections of each other through y=x meaning they wont meet. I'm unsure as to why you would use f(x)=x or f^-1(x)=x and also to get a quadratic equation by doing this. Sorry if this isn't making much sense, I'm not too sure about it myself.

    Also apologies about the image situation, I'm new and still finding my way around here
    Either would work but since you are given the function and not its inverse, you should be considering the equation f(x) = x.

    Do you understand that if f(x) and f^{-1}(x) don't meet then f(x) = x has no solutions?

    f(x) = x \Rightarrow \sqrt{mx+7} - 4 = x

    Move the 4 to the other side and then square both sides. You should end up with a quadratic after simplifying. Post your working if you get stuck.
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    (Original post by notnek)
    Either would work but since you are given the function and not its inverse, you should be considering the equation f(x) = x.

    Do you understand that if f(x) and f^{-1}(x) don't meet then f(x) = x has no solutions?

    f(x) = x \Rightarrow \sqrt{mx+7} - 4 = x

    Move the 4 to the other side and then square both sides. You should end up with a quadratic after simplifying. Post your working if you get stuck.
    f(x) = x \Rightarrow \sqrt{mx+7} - 4 = x
    f(x) = x \Rightarrow \sqrt{mx+7}  = x + 4
    f(x) = x \Rightarrow mx+7  = x^2 + 8x + 16
    f(x) = x \Rightarrow mx = x^2 + 8x + 9

    Not sure how to simplify from here, if I divide by x then I won't have a quadratic anymore? The mark scheme says I need to get (m – 2)(m – 14) < 0.
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    (Original post by NA LUL)
    f(x) = x \Rightarrow \sqrt{mx+7} - 4 = x
    f(x) = x \Rightarrow \sqrt{mx+7}  = x + 4
    f(x) = x \Rightarrow mx+7  = x^2 + 8x + 16
    f(x) = x \Rightarrow mx = x^2 + 8x + 9

    Not sure how to simplify from here, if I divide by x then I won't have a quadratic anymore? The mark scheme says I need to get (m – 2)(m – 14) < 0.
    mx = x^2+8x+9

    x^2+8x-mx+9=0

    x^2+(8-m)x+9=0

    This is a quadratic with a = 1, b  = 8-m and c = 9. You need to find values of m where this quadratic has no solutions. You probably would have done questions like this at C1.
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    (Original post by notnek)
    mx = x^2+8x+9

    x^2+8x-mx+9=0

    x^2+(8-m)x+9=0

    This is a quadratic with a = 1, b  = 8-m and c = 9. You need to find values of m where this quadratic has no solutions. You probably would have done questions like this at C1.
    Thanks so much! Finally get it. Huge mental block but it's all coming back to me now
 
 
 
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