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Math Help - Range of b

  1. #1
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    Range of b

    Given the curve C has the equation y=-x+(a-2b)-\frac{2b(a-2b)}{x-2b} and  a=1+2b, find the range of values of b, such that the curve C has 2 stationary points.

    \frac{dy}{dx}=-1+\frac{(a-2b)(2b)}{(x-2b)^2}

    \frac{dy}{dx}=0, -1+\frac{(a-2b)(2b)}{(x-2b)^2}=0


    -(x-2b)^2+(a-2b)(2b)=0

    -(x^2+4bx+4b^2)+(2ab-4b^2)=0

    -x^2+4bx-4b^2+2ab-4b^2=0

    -x^2+4bx+(-8b^2+2ab)=0


    b^2-4ac>0, (4b)^2-4(-1)(-8b^2+2ab)>0

    16b^2+4(-8b^2+2(1+2b)(b))>0

    16b^2+4(-8b^2+2b+4b^2)>0

    16b^2+4(-4b^2+2b)>0

    8b>0

    b>0

    but ans is b<0
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  2. #2
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    Re: Range of b

    Quote Originally Posted by Punch View Post
    Given the curve C has the equation y=-x+(a-2b)-\frac{2b(a-2b)}{x-2b} and  a=1+2b, find the range of values of b, such that the curve C has 2 stationary points.

    \frac{dy}{dx}=-1+\frac{(a-2b)(2b)}{(x-2b)^2}

    \frac{dy}{dx}=0, -1+\frac{(a-2b)(2b)}{(x-2b)^2}=0


    -(x-2b)^2+(a-2b)(2b)=0

    -(x^2+4bx+4b^2)+(2ab-4b^2)=0

    -x^2+4bx-4b^2+2ab-4b^2=0

    -x^2+4bx+(-8b^2+2ab)=0
    I agree up to this point. For there to be two stationary points, the discriminant is positive, so

    \displaystyle \begin{align*} (4b)^2 - 4(-1)(-8b^2 + 2ab) &> 0 \\ 16b^2 + 4(-8b^2 + 2ab) &> 0 \\ 16b^2 - 32b^2 + 8ab &> 0 \\ -16b^2 + 8ab &> 0 \\ b^2 - \frac{1}{2}ab &< 0 \textrm{ since dividing by a negative changes the sign} \\ b^2 - \frac{1}{2}ab + \left(-\frac{1}{4}a\right)^2 - \left(-\frac{1}{4}a\right)^2 &< 0 \\ \left(b - \frac{1}{4}a\right)^2 - \frac{1}{16}a^2 &< 0 \\ \left(b - \frac{1}{4}a\right)^2 &< \frac{1}{16}a^2 \\ \left|b - \frac{1}{4}a\right| &< \frac{1}{4}|a| \\ -\frac{1}{4}|a| < b - \frac{1}{4}a &< \frac{1}{4}|a| \\ -\frac{1}{4}|a| - \frac{1}{4}a < b &< \frac{1}{4}|a| - \frac{1}{4}a  \end{align*}

    If you have been told that \displaystyle a > 0, then the answer \displaystyle b < 0 is (partially) correct.

    Edit: I realise now that you have written \displaystyle a = 1 + 2b. I'm sure you can finish.
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  3. #3
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    Re: Range of b

    Quote Originally Posted by Prove It View Post
    I agree up to this point. For there to be two stationary points, the discriminant is positive, so

    \displaystyle \begin{align*} (4b)^2 - 4(-1)(-8b^2 + 2ab) &> 0 \\ 16b^2 + 4(-8b^2 + 2ab) &> 0 \\ 16b^2 - 32b^2 + 8ab &> 0 \\ -16b^2 + 8ab &> 0 \\ b^2 - \frac{1}{2}ab &< 0 \textrm{ since dividing by a negative changes the sign} \end{align*}


    Edit: I realise now that you have written \displaystyle a = 1 + 2b. I'm sure you can finish.
    Combining them,

    b^2-\frac{1}{2}(1+2b)(b)<0

    2b^2-(b+2b^2)<0

    -b<0

    b>0

    ans agrees with my ans in post 1, but ans is b<0
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    Re: Range of b

    I have a feeling there must be a typo in the answer you have been given, because I get \displaystyle b > 0 after substituting \displaystyle a right at the beginning as well. I agree with \displaystyle b>0 as the answer.
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  5. #5
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    Re: Range of b

    Quote Originally Posted by Prove It View Post
    I agree up to this point. For there to be two stationary points, the discriminant is positive, so

    \displaystyle \begin{align*} (4b)^2 - 4(-1)(-8b^2 + 2ab) &> 0 \\ 16b^2 + 4(-8b^2 + 2ab) &> 0 \\ 16b^2 - 32b^2 + 8ab &> 0 \\ -16b^2 + 8ab &> 0 \\ b^2 - \frac{1}{2}ab &< 0 \textrm{ since dividing by a negative changes the sign} \\ b^2 - \frac{1}{2}ab + \left(-\frac{1}{4}a\right)^2 - \left(-\frac{1}{4}a\right)^2 &< 0 \\ \left(b - \frac{1}{4}a\right)^2 - \frac{1}{16}a^2 &< 0 \\ \left(b - \frac{1}{4}a\right)^2 &< \frac{1}{16}a^2 \\ \left|b - \frac{1}{4}a\right| &< \frac{1}{4}|a| \\ -\frac{1}{4}|a| < b - \frac{1}{4}a &< \frac{1}{4}|a| \\ -\frac{1}{4}|a| - \frac{1}{4}a < b &< \frac{1}{4}|a| - \frac{1}{4}a \end{align*}

    If you have been told that \displaystyle a > 0, then the answer \displaystyle b < 0 is (partially) correct.

    Edit: I realise now that you have written \displaystyle a = 1 + 2b. I'm sure you can finish.
    Why not get rid of a in the question.It is given that a-2b=1.So the equation of the curve can simply be written as

    y=-x-1-\frac{2b}{x-2b}

    and
    \frac{dy}{dx}=0
    yields
    -1+\frac{2b}{(x-2b)^2}=0
    and thus
    x=2b\pm\sqrt{2b}

    for which the necessary and sufficient condition is b>0
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