Year 13 Maths Help Thread

it annoys me too

I think that should be on your exam board formula booklet?

Show that $\displaystyle \sum_{r=1}^n \sin r =\frac{1}{2} (\sin n +\cot \frac{1}{2}-\cot \frac{1}{2} \cos n)$.

Show also that $\displaystyle \sum_{r=1}^n \sin(r+1)\cos (r+3) =\frac{1}{4}( \cos (5) \csc (1) -\cos (2n+5)\csc (1) -2n\sin (2))$.

Show that $\displaystyle \sum_{r=1}^n \sin r =\frac{1}{2} (\sin n +\cot \frac{1}{2}-\cot \frac{1}{2} \cos n)$.

Show also that $\displaystyle \sum_{r=1}^n \sin(r+1)\cos (r+3) =\frac{1}{4}( \cos (5) \csc (1) -\cos (2n+5)\csc (1) -2n\sin (2))$.

Is there really any way of preparing for STEP other than doing it 'step by step': This is my strategy:

1. Do all the easier STEP I questions first.

2. Check the solutions to see if you could have done it more efficiently.

3. When you have done enough of the easier questions, you will have picked up enough tricks to take a second look at the questions you originally struggled with.

4. Climb your way up the difficulty level...

I'll be resitting Core 1 & Core 2 this year as I ****ed it up last year. Anyone who has experience of resitting maths modules, when is a good time to start practicing/revising for them?

Shouldn't be too much of a hassle, if any at all, after doing C3 and C4. Just keep fresh on topics outside these modules.

I need help on a C3 domain and range questions .... again.
Ive been over it twice and thought i understood but then got confused when i got this question wrong.

In simple terms is the domain the x value and the range the y value you get out of the function?

f(x) = (2x +5) / (x - 3)
x E R, x is not equal to 3

for the first part you are told to find the range. I got it as

0< f(x)<= 13 but it should be f(x) E R, f(x) is not equal to 2.

Where have i gone wrong?? (this isnt related to the question but how is R different to x E R)

I'll be resitting Core 1 & Core 2 this year as I ****ed it up last year. Anyone who has experience of resitting maths modules, when is a good time to start practicing/revising for them?

Well, how did you work out the range?

If you say $\frac{2x+5}{x-3}$ =
$\frac{2x - 6 + 11}{x-3}$...





C2 requires very little practice.

Just be aware of the topics that aren't in C3 (and maybe C4) and the usual rules of not cancelling solutions, checking range of solutions etc and then work through the textbook close to exam time, you should be fine.

im not sure why you put -6 +11

To find the range i subbed in x=3 to get 0 so f(x) must be bigger than 0 because x is not equal to 3. The i subbed in x=4 (that would give 1 as the denominator - largest value of f(x)) = 13

?

Sometimes that method works (for an increasing/decreasing function) but this isn't one of them. As you can see from sketching a graph on it.

It's a way to manipulate the fraction so that you end up with (A - B/(x+c)) which is much easier to find the range of.

Notice that if you add 11 then take away 6 next to the x, then you can divide the three terms by the denominator to give a constant (A) as well as B and C.

Another (different) question

If y = 1 + sinx is stretched in the x-direction, scale factor 1/2
why is it y = 1 + 2sinx
instead of
y = 1 + sin2x

??

Another (different) question

If y = 1 + sinx is stretched in the x-direction, scale factor 1/2
why is it y = 1 + 2sinx
instead of
y = 1 + sin2x

??

y = 1 + sin x isn't a stretch, it just shifts the graph of sin x up by one. So it makes it range from 0 <-> 2 rather than -1 <-> 1.
Sin 2x is a compression in the x direction, scale factor of 1/2, sin x/2 is a stretch of scale factor of 2 though. You can rationalise that by thinking of values for x and comparing to sin x. For example, sin(x) = 1/2 when x = 45, however sin (x/2) when x = 90 is the same as sin(x) when x = 45 and sin(x/2) therefore doesn't reach it's maxima at f(x) = 1 until x = 180. Not the easiest thing to get your head around at first and seems a bit counter intuitive, but if you can learn to understand it and explain it to yourself as it were then it becomes a lot more simple

Thanks but why is it 2sinx instead of sin2x - in my textbook it says that a stretch in the x-direction means the transformation is y=f(x) to y=f(x/c)

I understand where your coming from: sin 2x is a compression and sin(x/2) is a stretch <--- did you mean 2sinx or is that the same thing??