STEP support programme assignment 1

how do you type maths like that, sorry, new to the site and wan to know how i couldhave presented my response better, as they have the same content more or less, is thre a guide on the forum or somehting.

For part (ii):
This part of the question uses the knowledge and skill you developed in part (i). Remember: STEP is NOTORIOUS for this sort of structure with its questions. You need to pay particular attention to what you did in (i) and be able to deftly apply that to part (ii).

You can initially rewrite the expression as a quadratic. By multiplying by $(x-a)(x-b)$, given that $x\neq a,b$ (since that would make the denominator of the original expression equal to $0$, which is impossible), you find that you obtain the equation $2x^2 - (a+b)x = (c+1)(x^2-(a+b)x+ab)$. This can be rearranged to be a quadratic with coefficients in terms of a,b,c: $(c-1)x^2 - c(a+b)x + (c+1)ab=0$.

This is now in a form we can work with. We want to show that if there is only one solution, then c satisfies that condition. Let's run back to the logic of one solution:
If a quadratic has one solution, then its discriminant is equal to zero. Therefore, if $(c-1)x^2 - c(a+b)x + (c+1)ab=0$ has one solution, then its discriminant is zero. Thus, by taking the discriminant and setting it equal to zero, we find that

$c^2(a+b)^2 - 4 \times (c-1) \times (c+1) \times ab = 0$

And hence by rearranging the result we see that

$c^2\left [ (a+b)^2 - 4 \times ab\right ] = -4ab$
$\implies c^2= \dfrac{-4ab}{(a-b)^2}$.

Therefore, if $(c-1)x^2 - c(a+b)x + (c+1)ab=0$ has one solution, then $c^2= \dfrac{-4ab}{(a-b)^2}$. This is the required result.

The next bit takes a bit of spotting, and you could work 'backwards' to get the result and then rewrite your solution 'in reverse'---that is, working back to the result you obtained, and then writing out your solution the correct way round. Moving on:

Rewriting the conditions for c gives us $c^2= 1-\dfrac{a^2+2ab+b^2}{a^2-2ab+b^2}$. Since $a,b,c \in \mathbf{R}$, it is evident $c^2\geqslant 0$ and $\left(\dfrac{a+b}{a-b}\right)^2 \geqslant 0$. Therefore $0\leqslant c^2\leqslant 1$.

However we are not done. If $c^2 = 0$ then it follows $a=0,\quad b=0$, which is not allowed. Thus $0<c^2\leqslant 1$

We never provide full worked solutions in the study help forum. As @Doonesbury said, 'It's study help, not do this question for me.'

It's more helpful in STEP to guide through the question.