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# AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme] watch

1. (Original post by 2014_GCSE)
Can someone explain this to me without the use of a CAST diagram?
(My college never taught them)
Hello again. You don't know CAST (I know it as SACT, because it seems more logical to remember it that way)? Well, you should probably learn it !.

QUICK LESSON ON CAST/SACT:
Draw a big +, label the quadrants clockwise from the top-left in the order S-A-C-T. Each quadrant represents 90 degrees starting from the right - and going in an anti-clockwise direction. The letters represent what graph is positive in this range of degrees (i.e. S = sin, A = all, C = cos and T = tan).

Therefore, the reason you know tan is negative is that when the angle is obtuse is that an obtuse angle is from 90-180 and only sin is positive in that quadrant.

WITHOUT CAST/SACT:
Consider the graph of tanx. If an angle is obtuse then it must be somewhere on the portion of the graph for 90 degrees <= x < 180 degrees. From a tan graph you know that this region is from 0 <= tanx <= -1 (where y = tanx) and therefore tanx must be negative for an obtuse angle .

NOTE: I am extremely tired by this point and becoming somewhat delirious.
2. (Original post by amyrah)
Quick question! Does 2p^t = q^t equal

2 ln p^t = ln q^t

I'm just unsure what to do with the 2!

Thanks
It would be ln2 + ln p^t = ln q^t
3. (Original post by amyrah)
Quick question! Does 2p^t = q^t equal

2 ln p^t = ln q^t

I'm just unsure what to do with the 2!

Thanks
You have to ln both sides and then separate the LHS into ln2 + tlnp using the log rule that lnab = lna + lnb.

Also, it should be 2p^(t-10) .
4. i dont understand.. in the jan 2011 paper, Q1, we had to find R, but then didnt use it in the next part?? S
5. Could someone explain how to do part c

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6. (Original post by tiffanyyyyyyy)
i dont understand.. in the jan 2011 paper, Q1, we had to find R, but then didnt use it in the next part?? S
You do use R, as the maximum value is equal to R (the minimum would be equal to -R)? !
7. (Original post by gloria0816)
Could someone explain how to do part c

Posted from TSR Mobile
It has a different statting time (by 10 years) so you can either Add 10 years onto q or take 10 years away from p to make the times equal.
We choose the latter because we have a value for P^10 so we can eliminate it
8. (Original post by datpr0)
Hello again. You don't know CAST (I know it as SACT, because it seems more logical to remember it that way)? Well, you should probably learn it !.

QUICK LESSON ON CAST/SACT:
Draw a big +, label the quadrants clockwise from the top-left in the order S-A-C-T. Each quadrant represents 90 degrees starting from the right - and going in an anti-clockwise direction. The letters represent what graph is positive in this range of degrees (i.e. S = sin, A = all, C = cos and T = tan).

Therefore, the reason you know tan is negative when the angle is obtuse is that an obtuse angle is from 90-180 and only sin is positive in that quadrant.

WITHOUT CAST/SACT:
Consider the graph of tanx. If an angle is obtuse then it must be somewhere on the portion of the graph for 90 degrees <= x < 180 degrees. From a tan graph you know that this region is from 0 <= tanx <= -1 (where y = tanx) and therefore tanx must be negative for an obtuse angle .

NOTE: I am extremely tired by this point and becoming somewhat delirious.
Excellent response! Thank you So if they said cos(x) is obtuse, find sin(x), sin(x) would be positive and that would be almost a "trick question" by mentioning obtuse, as that only matters when working out cos(x) and tan(x)?
9. (Original post by datpr0)
Hello again. You don't know CAST (I know it as SACT, because it seems more logical to remember it that way)? Well, you should probably learn it !.

QUICK LESSON ON CAST/SACT:
Draw a big +, label the quadrants clockwise from the top-left in the order S-A-C-T. Each quadrant represents 90 degrees starting from the right - and going in an anti-clockwise direction. The letters represent what graph is positive in this range of degrees (i.e. S = sin, A = all, C = cos and T = tan).

Therefore, the reason you know tan is negative when the angle is obtuse is that an obtuse angle is from 90-180 and only sin is positive in that quadrant.

WITHOUT CAST/SACT:
Consider the graph of tanx. If an angle is obtuse then it must be somewhere on the portion of the graph for 90 degrees <= x < 180 degrees. From a tan graph you know that this region is from 0 <= tanx <= -1 (where y = tanx) and therefore tanx must be negative for an obtuse angle .

NOTE: I am extremely tired by this point and becoming somewhat delirious.
I have no idea how to cast
I just remember an algorithm
10. (Original post by yasmin221b)
A diagram would be very useful, thank you!

Here BC and BE have the same length, but you want the shorter length (5/7) because the original shape is a trapezium, not a parallelogram.
Attached Images

11. (Original post by ForgottenApple)
I have no idea how to cast
I just remember an algorithm
What's the "algorithm"? Always nice to have a back-up strategy.
12. (Original post by 2014_GCSE)
Excellent response! Thank you So if they said cos(x) is obtuse, find sin(x), sin(x) would be positive and that would be almost a "trick question" by mentioning obtuse, as that only matters when working out cos(x) and tan(x)?
That not quite correct, no; cosx wouldn't be obtuse, the angle would. If it were a "trick question" in that form they would tell you that for angle x, cosx is negative which would indicate the angle is obtuse .
13. (Original post by gloria0816)
Could someone explain how to do part c

Posted from TSR Mobile
Replace p^t with p^T-10 since the starting years have a difference of 10 years,
5000/2500p^T-10 = q^T
Ln2 - lnP^T-10 = lnQ^T
Then you can just put the power T-10 to the front of lnP and simplfy the rest, you will also get a value of 10lnP which is where the 5 comes from since you showed that p^10 = 5 earlier on
14. (Original post by datpr0)
You have to ln both sides and then separate the LHS into ln2 + tlnp using the log rule that lnab = lna + lnb.

Also, it should be 2p^(t-10) .
Lmao fk :| you have good memory! I didn't mention t-10 just for simplicity but thanks!!
15. (Original post by gloria0816)
How to you find the Cartesian equation in this form?
I got to the point where y=square root 36(1-cos^2thetacos^2theta)
Attachment 422403

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What exam papers it that question from , anyone knows? Which year?
16. (Original post by amyrah)
Lmao fk :| you have good memory! I didn't mention t-10 just for simplicity but thanks!!
Haha, no problem .

Just make sure to remember the three log rules, you'll probably need at least the third tomorrow; those being:

(1) lnab = lna + lnb
(2) lna/b = lna - lnb
(3) lna^b = blna
17. (Original post by datpr0)
That not quite correct, no; cosx wouldn't be obtuse, the angle would. If it were a "trick question" in that form they would tell you that for angle x, cosx is negative which would indicate the angle is obtuse .
You're brilliant, thank you!
18. i forgot to revise

φφδδιάέρπτββψλχλ σλέσπρξχ
19. This thread seems quite quiet compared to the Core 3 one the night before...
I wonder if people have just given up
20. Alright, guys, I am truly exhausted and need to sleep else I'll be too tired to add 1+1 :P.

So, good luck to everyone! Thanks for all your help!

(and prepare for this thread to be spammed once everyone gets home tomorrow)

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