Year 13 Maths Help Thread

part iii

Help - ques 6.b) iii) where did i go wrong?

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Your answer is correct, it just isn't in the 'simplest form' I guess. Multiply both the numerator and denominator by -1 and the rest should come quite easily.

$\dfrac{6-x}{-x+5} \Rightarrow \dfrac{x-6}{x-5} \Rightarrow \dfrac{x-5-1}{x-5} \Rightarrow \dfrac{x-5}{x-5} - \dfrac{1}{x-5} \Rightarrow 1 - \dfrac{1}{x-5}$

Ohh i see thanks

Would i still get the marks with my answer?

I'm not too sure about that I'm afraid, it's been a couple years since I did my A-levels, I don't quite remember how the markschemes go. The question doesn't say anything about what form you should leave your answer in, so my guess is you'd get the marks. At worst you'd be docked 1 mark.

Ohh i see thanks

Would i still get the marks with my answer?

Your answer is fine. There's no need to split up the fraction.

You could make it nicer if you multiply top and bottom by -1 but a mark scheme would allow either form.

Why did you think your answer was wrong?

First i thought my answer was different to the right answer. Then I confused myself by trying to simplify my answer to match the mark scheme but ended up with 1 + 6/5 (x cancel out) - I must have completely forgot that you cant do that :3 - just a silly mistake

Lack of understanding of cancelling fractions is very common at A Level and I've seen it cost many marks in exams.

If you're confident that you won't make this mistake again then ignore it. But if you're even slightly unsure then I recommend you rethink about why what you did was wrong. Feel free to ask if you want an explanation.

Some A Level students just don't like to admit to their teacher that they're unsure about cancelling

Im confused about drawing modulus graphs.

E.g. y = IxI + Ix-3I

I did -x + (-x+3)
Y = -2x +3

Then x + (-x+3) etc.

These are the lines you draw to make up the graph but how do you find the range/bounds for each of these. E.g. y=-2x+3 has the bound x <0 but how do you know that?

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You have to think about what values of x change the expressions inside the molulii from positive to negative.

For $|x|$, that point is $x = 0$.
For $|x-3|$, that points is $x =3$.

So you have two critical points, $x = 0$ and $x = 3$ which gives you three regions that you need to consider:

$x < 0$

$0 \leq x < 3$

$x \geq 3$.


If we consider the first region $x<0$:


Here $x<0$ and $x-3 < 0$ so the equation becomes

$y = -x -(x-3) = -2x + 3$

and so you draw the line $y=-2x + 3$ in the region $x<0$.

Ok i understand where you got the critical points from but how do you know x<0 and not x>0 for example?

The critical points x = 0 and x = 3 are where things change (either side of them). If you draw these two points on a number line then you'll see that you have created 3 distinct regions.

You have to consider these three regions to see what happens to the expression in each of the regions.

x>0 is considered but this is part of the middle region.

Oh I see. What if you had 3 critical values e.g. I(x-1)(x-2)(x-3)I - the values wold be 1, 2 and 3 but what about the bounds??

Oh I see. What if you had 3 critical values e.g. I(x-1)(x-2)(x-3)I - the values wold be 1, 2 and 3 but what about the bounds??

Have you tried sketching the graph, both in the 2 and 3 cv case?

Kinda - i put 1, 2, 3 on the x axis and -6 on the y axis for where the graph intersects. Im not sure but are the bounds x <1 1 <x <3 and x>3 ??

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