1. ## 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 $b<0$

2. ## Re: Range of b

Originally Posted by Punch
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.

3. ## Re: Range of b

Originally Posted by Prove It
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

4. ## 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.

5. ## Re: Range of b

Originally Posted by Prove It
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 $b>0$