Results 1 to 5 of 5

Thread: Range of b

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Range of b

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

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

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


    $\displaystyle -(x-2b)^2+(a-2b)(2b)=0$

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

    $\displaystyle -x^2+4bx-4b^2+2ab-4b^2=0$

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


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

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

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

    $\displaystyle 16b^2+4(-4b^2+2b)>0$

    $\displaystyle 8b>0$

    $\displaystyle b>0$

    but ans is $\displaystyle b<0$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946

    Re: Range of b

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

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

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


    $\displaystyle -(x-2b)^2+(a-2b)(2b)=0$

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

    $\displaystyle -x^2+4bx-4b^2+2ab-4b^2=0$

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

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

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

    Edit: I realise now that you have written $\displaystyle \displaystyle a = 1 + 2b$. I'm sure you can finish.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    Posts
    755

    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 \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 $\displaystyle \displaystyle a = 1 + 2b$. I'm sure you can finish.
    Combining them,

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

    $\displaystyle 2b^2-(b+2b^2)<0$

    $\displaystyle -b<0$

    $\displaystyle b>0$

    ans agrees with my ans in post 1, but ans is b<0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946

    Re: Range of b

    I have a feeling there must be a typo in the answer you have been given, because I get $\displaystyle \displaystyle b > 0$ after substituting $\displaystyle \displaystyle a$ right at the beginning as well. I agree with $\displaystyle \displaystyle b>0$ as the answer.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318

    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

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

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

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

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

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

    for which the necessary and sufficient condition is $\displaystyle b>0$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. range
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 17th 2012, 02:30 AM
  2. Range
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Mar 7th 2011, 02:18 AM
  3. range for e^|f(x)|
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Mar 25th 2010, 03:35 PM
  4. Replies: 6
    Last Post: Sep 16th 2009, 06:25 AM
  5. range, inter-quartile range
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Jun 19th 2006, 05:06 AM

Search Tags


/mathhelpforum @mathhelpforum