Submitted by jetay1 on Thu, 02/23/2017 - 17:27
I wonder if anyone could give me a hint on the best approach for Q5(ii) of 2012 Step 2. The question asks me to sketch \[g(x)=\frac{1}{((x-a)^2-1)((x-b)^2-1)}\] for the case of $b>a+2$. I've tried writing $g(x)$ as partial fractions in two different ways: \[\frac{Ax+B}{(x-a)^2-1}+\frac{Cx+D}{(x-b)^2-1}\] and \[\frac{A}{x-a+1}+\frac{B}{x-a-1}+\frac{C}{x-b+1}+\frac{D}{x-b-1}\] The second way leads to a sketch of a symmetrical curve, which is about right, but I am struggling to find the coordinates of the stationary points (which the question asks for).
I can tell from the symmetry that one of the stationary points is at $\frac{a+b}{2}$, but all attempts to find the other two have collapsed in really complicated algebra. I've tried differentiating the original function and the partial fractions, which all lead to a horrible cubic. Is that the right idea or am I missing something?
Translate?
I can't begin to post the algebra, but can you translate g(x) so it is symmetrical about x=0 and take it from there?
I used
\[h(x)=\dfrac{1}{\left(\left(x-3\right)^2-1\right)\left(\left(x+3\right)^2-1\right)}\]
to try to get the hang of what was going on. I'm still at it (I keep losing track of minus signs) but I think the idea is sound.
Is this a useful suggestion, or have I missed something blindingly obvious?
And thanks for yet another really interesting question.
I think you may be making a
I think you may be making a bigger deal of this than it is! The derivative is simply an involved application of the chain rule. You want to differentiate $[((x-a)^2 -1)((x-b)^2 - 1)]^{-1}$ so the chain rule gives $$g' = \frac{-2}{[((x-a)^2 -1)((x-b)^2 -1)]^{2}} \bigg[(x-b)((x-a)^2 - 1) + (x-a)((x-b)^2 - 1)\bigg]$$
Now the really handy fact here is that to solve $g' = 0$, we need only solve the numerator =0 since the denominator disappears. This gives us the relatively simple $$(x-b)[(x-a)^2 - 1] + (x-a)[(x-b)^2 - 1] = 0$$
Now factorising yields $$(x-a)(x-b) \big[x-a +x-b \big] + \big[x-a+x-b \big] = 0 \iff (2x-a-b)(x^2 - (a+b)x + (ab-1)) = 0$$ which has one linear factor and one quadratic factor that you can find the roots of straightforwardly by applying the quadratic formula. This will give you the stationary points and some analysis of the roots of the quadratic with the relationship between $a$ and $b$ will give you some insight into how the graph looks.
P.S: Can you see how I arrived at that factorisation? There are two terms with $(x-a)(x-b)$ as a common factor, pull that out and the rest follows straightforwardly.
Thanks
That's a very sneaky factorisation. It would have taken me a long time to spot it.
Thanks very much.
You're welcome
You were quite close! You had already discovered that $x = \frac{a+b}{2}$ was a root and so $2x - a -b$ must be a factor. You just had to rearrange things a little cleverly! But well done anyway. :-)
No worries!