The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

AQA Core 3 - Tuesday 10th June 2014 (AM) - Official Thread

Scroll to see replies

Original post by SimiM
are u referring to the ones with the modulus function involved?


Yh
Original post by Math12345
ln[sin(5x+2)] but it's possible to do it with one application of the chain rule.

u=5x+2
v=sin(u)
y=ln[v]

How would you do that I've never seen that type of question before
Original post by The Monk
How would you do that I've never seen that type of question before


But would never pop up so don't worry
Reply 63
Avatar for mA*ths
mA*ths
Original post by Mahdifaily
But would never pop up so don't worry


Are you sure? The use of Integration by parts twice came up in Jan 2012. In the Cambridge University Press' pink AQA C3 book there is an example of the use of the 'extended chain rule' (using the chain rule twice) on page 101, meaning that it's definitely within the scope of the specification and therefore it would be unwise to completely rule out it coming up on Tuesday,
Original post by mA*ths
Are you sure? The use of Integration by parts twice came up in Jan 2012. In the Cambridge University Press' pink AQA C3 book there is an example of the use of the 'extended chain rule' (using the chain rule twice) on page 101, meaning that it's definitely within the scope of the specification and therefore it would be unwise to completely rule out it coming up on Tuesday,


Thing is you can differentiate the inner term in your head - at least if its a trigonometrical term. You can then just use that as u because you know du/dx. Worst case scenario is you do the chain rule individually twice. Its probably easier than using three letters.

Posted from TSR Mobile
Original post by ma*ths
are you sure? The use of integration by parts twice came up in jan 2012. In the cambridge university press' pink aqa c3 book there is an example of the use of the 'extended chain rule' (using the chain rule twice) on page 101, meaning that it's definitely within the scope of the specification and therefore it would be unwise to completely rule out it coming up on tuesday,

i am saying that a question like in(sin(2x+3)) would never pop up.
Original post by firsedge
If you want some hard questions - I got told to look at Edexcel and OCR past papers! As they are meant to be much harder
Just be selective with the questions you chose


ah really? i might have a look, although i'll probably stress myself out trying to do questions that aren't on our syllabus haha! thanks though :smile:
Reply 67
Avatar for Zozzy
Zozzy
2
Right silly question but for the graphs questions how do you work out the ranges this always stumbles me when it should be an easy 1/2 marks.
If it doesn't say to use radians or degrees which one should you use?


Posted from TSR Mobile
Original post by Scarlett44
If it doesn't say to use radians or degrees which one should you use?


Posted from TSR Mobile


Always use radians unless it asks for an answer to be in degrees
Reply 70
Avatar for Zozzy
Zozzy
2
Can somebody please explain to me why the modulus graph looks like it does for this question because I thought the modulus graphs were just making all the values positive.model of confusion.jpg
Thanks :smile:
Original post by Zozzy
Can somebody please explain to me why the modulus graph looks like it does for this question because I thought the modulus graphs were just making all the values positive.model of confusion.jpg
Thanks :smile:


Yes but the question says the modulus of x not the whole graph.

So therefore the modulus of -6 is 6 which gives a y value of 0(on the graph). Similarly the modulus of -5 is 5 which on the graph gives a y value of more than 0. Therefore you realise the shape of the graph.

Sorry if I've explained it badly as I am also struggling trying to explain lol
Original post by Zozzy
Can somebody please explain to me why the modulus graph looks like it does for this question because I thought the modulus graphs were just making all the values positive.model of confusion.jpg
Thanks :smile:


I think you just have to make sure that there's no negative values for the y axis, but you can have negative x values... If that makes sense :P


Posted from TSR Mobile
Original post by lukejohnston13
Always use radians unless it asks for an answer to be in degrees


Thanks :smile:


Posted from TSR Mobile
Original post by Scarlett44


The only exception to this rule is Vector angles, I think. But that's for Thursday,:wink:
What's the hardest type of questions in core 3?
Original post by mastermdc
What's the hardest type of questions in core 3?


Yes please post some

Posted from TSR Mobile
Original post by chloetheuincorn
The only exception to this rule is Vector angles, I think. But that's for Thursday,:wink:


Okay thanks :smile:


Posted from TSR Mobile
Reply 78
Avatar for Zozzy
Zozzy
2
Original post by the_googly
Yes but the question says the modulus of x not the whole graph.

So therefore the modulus of -6 is 6 which gives a y value of 0(on the graph). Similarly the modulus of -5 is 5 which on the graph gives a y value of more than 0. Therefore you realise the shape of the graph.

Sorry if I've explained it badly as I am also struggling trying to explain lol


No actually that makes a lot of sense so basically I just change all the negative x values back to positive and THEN stick them back into the equation. Thanks :biggrin:
I do this fun thing sometimes when I integrate and differentiate at the same time. It's so great. :s-smilie:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you