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# Year 13 Maths Help Thread Watch

1. (Original post by kiiten)
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??
Again, draw a number line and mark the values 1, 2 and 3. You will see you have created 4 regions.

What are they?

If the question was to draw y = I(x-1)(x-2)(x-3)I, then a better appoach here would be to draw y = (x-1)(x-2)(x-3) and then consider the transformation f(x) -> |f(x)|.
2. Posted from TSR Mobile

Can questions like this (questions 1&4) come up in the c3 edexcel exam? I kinda sorta understand q1 but haven't got a clue how to do q4.
Attached Images

3. (Original post by SeanFM)
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 <2
2< x< 3
x >3

Posted from TSR Mobile
4. (Original post by kiiten)
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 ??

Posted from TSR Mobile

Well, you tell me. You have to know why your answer is that. What does the graph show you? Refer back to the original question to answer this.
Hint: the shape of the graph is important, you need to figure out why that is and how to draw the full graph as a rough sketch.
5. (Original post by kiiten)
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 <2
2< x< 3
x >3

Posted from TSR Mobile
That's correct now. There's no symbols but they're not too important - just think about the four regions as you have done.

But this particular question should be done by considering transformations. Refer back to the (edited) end of my last post.
6. (Original post by notnek)
That's correct now. There's no symbols but they're not too important - just think about the four regions as you have done.

But this particular question should be done by considering transformations. Refer back to the (edited) end of my last post.
I know you just reflect the part that hits the x axis but i dont know how big to make the bump between the bounds (e.g. 1 and 2) - do you just draw it roughly - ill attach an image (its supposed to cross at 6 but its a little bit out)
7. (Original post by KappaRoss)
Posted from TSR Mobile

Can questions like this (questions 1&4) come up in the c3 edexcel exam? I kinda sorta understand q1 but haven't got a clue how to do q4.
Q1 could come up. Q4 is less likely and would be phrased in a different way.

For e.g. 4a, you have cot and cosec. This should ring a bell in your mind - there's an identity that involves both of these.

But in this identity the trig ratios are squared.

So the first step is to square both sides of each equation. Post your working if you get stuck.
8. (Original post by kiiten)

I know you just reflect the part that hits the x axis but i dont know how big to make the bump between the bounds (e.g. 1 and 2) - do you just draw it roughly - ill attach an image (its supposed to cross at 6 but its a little bit out)
That would be fine in a C3 exam unless the question asked you to mark the turning points (unlikely for a question like this but you never know).

To find the turning points you can find the turning points of the cubic function y = (x-1)(x-2)(x-3) and then think about how they change.
9. Posted from TSR Mobile
I have another problem. (See image attached) I understand part a) and I understand what f^-1(x) is, but I don't understand the the relationship between fg(x) and its importance in solving g(a)=f^-1(a), I think it might just be a common sense thing were the answers to a) should be used in b).

I just need the explanation. Thanks in advance as always.
a) x=-5, x=6
b) a=6 (still not sure but think its right)
Attached Images

10. (Original post by KappaRoss)
Posted from TSR Mobile
I have another problem. (See image attached) I understand part a) and I understand what f^-1(x) is, but I don't understand the the relationship between fg(x) and its importance in solving g(a)=f^-1(a), I think it might just be a common sense thing were the answers to a) should be used in b).

I just need the explanation. Thanks in advance as always.
a) x=-5, x=6
b) a=6 (still not sure but think its right)
as I'm sure you know, the way inverse functions work is that if f(x)=y, then f-1(y)=x

Now, if fg(x)=x, noting that fg(x) is the same as f(g(x)), what would f-1(x) be?
11. Posted from TSR Mobile

Oh I get it now ! Thanks
12. Hi, with questions like the one attached, how would I know whether to use long division or the remainder theorem? I started off using the remainder theorem but realised it didnt work. Is it just a case of trial and error? Thanks in advance
Attached Images

13. (Original post by doglover123)
Hi, with questions like the one attached, how would I know whether to use long division or the remainder theorem? I started off using the remainder theorem but realised it didnt work. Is it just a case of trial and error? Thanks in advance
Use long division.
14. (Original post by B_9710)
Use long division.
Thanks, I understand that, but how would I know to use long division instead of the remainder theorem?
15. (Original post by doglover123)
Thanks, I understand that, but how would I know to use long division instead of the remainder theorem?
How did you attempt to use the remainder theorem at first r when you tried it?
If you use long division here you should see that it's just straight forward - that's why you should use long division.
16. (Original post by doglover123)
Hi, with questions like the one attached, how would I know whether to use long division or the remainder theorem? I started off using the remainder theorem but realised it didnt work. Is it just a case of trial and error? Thanks in advance
How did you use the remainder theorem?
You can either use algebraic long division or multiply both sides by and equate coefficients.
17. (Original post by doglover123)
Thanks, I understand that, but how would I know to use long division instead of the remainder theorem?
Two reasons why not to use the remainder theorem:

It will only give you the value of the remainder (!) not the values of a, b and c

The bottom of the fraction isn't linear (you could put two values of x in and then combine the values you get, but it would be messy and more complicated than necessary).
18. (Original post by NotNotBatman)
How did you use the remainder theorem?
You can either use algebraic long division or multiply both sides by and equate coefficients.
(Original post by tiny hobbit)
Two reasons why not to use the remainder theorem:

It will only give you the value of the remainder (!) not the values of a, b and c

The bottom of the fraction isn't linear (you could put two values of x in and then combine the values you get, but it would be messy and more complicated than necessary).
(Original post by tiny hobbit)
Two reasons why not to use the remainder theorem:

It will only give you the value of the remainder (!) not the values of a, b and c

The bottom of the fraction isn't linear (you could put two values of x in and then combine the values you get, but it would be messy and more complicated than necessary).
(Original post by B_9710)
How did you attempt to use the remainder theorem at first r when you tried it?
If you use long division here you should see that it's just straight forward - that's why you should use long division.
That's what I did. I could find a,b, and c but not D or E. I worked it out correctly using long division so i'll just stick with that.
19. (Original post by doglover123)

That's what I did. I could find a,b, and c but not D or E. I worked it out correctly using long division so i'll just stick with that.
The end bit should be dx + e

You can then equate coefficients of the x terms to find d.

The 24 is e
20. (Original post by doglover123)

That's what I did. I could find a,b, and c but not D or E. I worked it out correctly using long division so i'll just stick with that.
on the first line you just have +d, you should have +dx+e

Updated: June 26, 2017
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