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Year 13 Maths Help Thread

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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 ≤\leq 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.
14751719512691925047313.jpg
Original post by notnek
That's correct now. There's no ≤\leq 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)
(edited 7 years ago)
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.
Thank you in advance :yes:

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.
Original post by kiiten
14751719512691925047313.jpg

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.
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)
1475255993111.jpg26.5KB
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?
Posted from TSR Mobile

Oh I get it now ! Thanks
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
Screen Shot 2016-10-01 at 10.46.29.png7.3KB
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.
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?
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.
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 x2−4x^2 - 4 and equate coefficients.
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).
img984.jpg
Original post by NotNotBatman
How did you use the remainder theorem?
You can either use algebraic long division or multiply both sides by x2−4x^2 - 4 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.
Original post by doglover123
img984.jpg


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
(edited 7 years ago)
Original post by doglover123
img984.jpg







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
Original post by tiny hobbit
The end bit should be dx + e

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

The 24 is e


Original post by NotNotBatman
on the first line you just have +d, you should have +dx+e


I should have noticed that earlier! I've got the answer now, thank you!
Have i done something wrong ....

e^2x - e^x = 0
e^2x = e^x
lne^x = lne^2x
x=2x

??

Posted from TSR Mobile
Original post by kiiten
Have i done something wrong ....

e^2x - e^x = 0
e^2x = e^x
lne^x = lne^2x
x=2x

??

Posted from TSR Mobile


looks fine;

now just find the value of x for which x=2x \displaystyle x = 2x
Original post by kiiten
Have i done something wrong ....

e^2x - e^x = 0
e^2x = e^x
lne^x = lne^2x
x=2x

??

Posted from TSR Mobile


I think you should probably form a quadratic equation in e^x and go from there.
Doing it this way you may struggle to solve e2x−ex−1=0 e^{2x}-e^x-1=0 .

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