Rational functions inequality question

Hi all,

I am attempting to solve Q9 here.

https://ibb.co/MP3LhnP

Here is my thought process so far:

So I will first attempt to solve algebraically, as the unknown, p, makes the sketch difficult. So I am going to multiply both sides by (px+1)^2, to avoid multiplying by a negative.

This leads to the following inequality.

https://ibb.co/Hdr1n1r

So I can see that we can rearrange the function into a cubic which is less than 0. In other words we are interested in the parts of a cubic graph which are below the x axis.

We are told that the solutions are x<q, r<x<3.

So in a similar way to some of the questions I answered earlier in the week we can sketch roughly what the cubic may look like (note that r and q may be in different positions ie their signs might not be correct in the sketch, but this doesn't matter too much for now).

https://ibb.co/DKYztb3

So from my sketch I can get another expression for the function. (Note the constant p^2)

https://ibb.co/svShVkS

So like we did earlier in the week we can set these two expressions equal to each other and equate the coefficients.

https://ibb.co/JxhczYF


And this is as far as I am up to! I am struggling to solve the equations and get p, q and r.

I know that the equations I have are correct because they work when we substitute the answers in! Perhaps one of you can help here?

Good progress. No need to jump right into comparing coefficients however.

You have recognised that $p^2x^3 + (4p^2 -14p)x^2 + (7p-15)x+3$ can be written in the form $p^2(x-q)(x-r)(x-3)$.

Since $x=3$ is a root of $p^2(x-q)(x-r)(x-3)$, it automatically means it must be a root of $p^2x^3 + (4p^2 -14p)x^2 + (7p-15)x+3$. Imposing this allows you to deduce a quadratic in $p$ which you can solve.

Hopefully this gives you the nudge to finish it off by finding q,r.

Brilliant, thanks, 1 more question, how do I know to choose p=2 as opposed to p=-1/3

I think I've answered this one myself actually, using p=-1/3 leads to q=3, r=-3, which contradicts the inequality r<x<3

And how did you arrive at this conclusion?

The solutions stated at the start if the question are that x<q or r<x<3

q=-3 and r=3 do not work for this, how can r<x<3 if r =3
nor do
q=3 and r=-3.

These come from p=-1/3.


So therefore we must use p=2 which leads to q=-0.5, p=0.5.
Which do work for the inequalities stated at the start.

I was more interested in how you deduced the values q,r using p=-1/3

Substituted p=-1/3 into equations I had achieved earlier by equating the coefficients.

Can you post your working out?

Must have made a mistake.... I'm now getting 43.2 and-0.208 ???

While I have your attention can I have some help with Q10 aswell.

(check the original image)

I have solved part a:

https://ibb.co/wSKJ1n0

but at a loss where to go with part b

From part (a) it should be clear that for any value of $m>0$, there are two possible $c$ values. Hence, this is saying that there are two parallel tangents to the curve. One of them touches the curve at $P(p,q)$ and the other at $R(r,s)$.

So, the distance $d^2$ can be written in terms of $p,q,r,s$.

The point $P(p,q)$ lies on one of the tangents, hence you are able to express $q$ in terms of $p$ and $m$ via the equation $y=mx+c$.
Likewise, the point $R(r,s)$ lies on the other tangent, hence you are able to express $s$ in terms of $r$ and $m$.

Substituting these into $d^2$ allows you to express it in terms of $p,r,m$. Now you just need to somehow rewrite $p,r$ in terms of $m$.

Have a think... note that these points P,R lie on the curve itself as well.

I have the equation of the two tangents, and an expression for d^2,

Just struggling to put it all together.

https://ibb.co/n6Mz0Mh

As I said, you can sub in $q=pm + 2 - m + \sqrt{12m}$ into $d^2 = (r-p)^2 + (s-q)^2$. Likewise with the other one.

Hopefully it's then clear that you need to now focus on expressing p,r in terms of m. I gave you a hint about this in the last post, very last sentence.

Sorry it's just not clear, I have the following unknowns, d, r, p and m so I don't see how I can just express r,p in terms of m

https://ibb.co/CPRVG9p

The points lie on $y=mx+c$, and they also lie on $y=\dfrac{2x-5}{x-1}$.

Hence, the x coordinates $p,r$ satisfy the equation $mx+c = \dfrac{2x-5}{x-1}$, which you had rearranged into the more convenient form;

$mx^2 + (c-m-2)x + (5-c) = 0$

The points lie on $y=mx+c$, and they also lie on $y=\dfrac{2x-5}{x-1}$.

Hence, the x coordinates $p,r$ satisfy the equation $mx+c = \dfrac{2x-5}{x-1}$, which you had rearranged into the more convenient form;

$mx^2 + (c-m-2)x + (5-c) = 0$

Hold on no I don't understand how this helps at all...