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Edexcel A2 C3 Mathematics 12th June 2015

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We know two points so we can form two equations:

1=a|6-b|-1This means a|6-b|=0 So a could equal zero and b could equal 6

We also have:

11=a|0-b|-1This means ab=1
This means that a cannot equal zero, so b equals 6 and a equals 1/6 (hope that is the right answer!)

A = 2 and i didn't get how you did it :/

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Reply 741

randlemcmurphy

Original post by lam12

A = 2 and i didn't get how you did it :/

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I realised I wrote 1 instead of 11, post edited.

Reply 742

aersh8

Original post by mismash

hello, i am really sorry to but in.. but im a year 12 student and i am really worried because i dont think i have done v well in my c1 and c2 maths exams.
would u guys recommend retaking those modules in y13 or simply is it going to waste time and increase work load too much...???

It really won't increase your workload too much since you need to know C1 and C2 to be able to do C3 and C4 anyway - 90% of my class are resitting them since they seem so much easier now so think they might as well improve their grade.

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Reply 743

cerlohee

Original post by randlemcmurphy

We know two points so we can form two equations:

1=a|6-b|-1This means a|6-b|=0 So a could equal zero and b could equal 6

We also have:

11=a|0-b|-1This means ab=12
This means that a cannot equal zero, so b equals 6 and a equals 1/6 (hope that is the right answer!)

If ab=12, that'd make b 2 which is the right answer! I didn't think of solving it this way (used that it had moved 6 units to the right) but might use in future! My problem was that I stupidly used (11,0) instead of (0,11)... :/ Thanks :biggrin:

Original post by lam12

The range changes because it's 3theta. You use 2pi... I don't exactly know what you're exactly asking tbh haha

For the first bit because it's a ^2 it can't be a negative and therefore the sin bit = 0

For the second part it would be 3 theta - 1.107 = 2pi as the range increases as its 3theta and now we can go up to 3pi but it's not = to 3pi and therefore we use just 2pi

I don't think I explained this well but I tried, I may be wrong but it's how it was explained to me :-)

Attachment not found

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Ahh thankyou! My problem was that I didn't see the 3theta, assuming it was theta so ended up witht eh wrong answer! Thanks for drawing it out and everything though :smile:

))

Original post by studentwiz

coordinates (6,-1)

a l6-bl -1 = -1
a l6-bl =0
divide by a
6-b=0
b=6

Hey, I had found b but couldn't find a, this is another way to approach it that I didn't think of though so thanks :biggrin:

Reply 744

TeeEm

useful material for the upcoming exams

http://www.thestudentroom.co.uk/showthread.php?t=3353445

http://www.thestudentroom.co.uk/showthread.php?t=3361867

http://www.thestudentroom.co.uk/showthread.php?t=3361905

(edited 8 years ago)

Reply 745

studentwiz

d and c thanks in advance nvm !

figured it out :/

(edited 8 years ago)

functions.PNG42.7KB

Reply 746

studentwiz

someone explain how you would go about getting the range?

range.PNG10.4KB

Reply 747

Gilo98

Original post by studentwiz

someone explain how you would go about getting the range?

consider range of y=e^x then consider what will happen to this range when the graph is shifted upward by k^2 units

Reply 748

Medicjug

Hey guys,

I've done all the real papers from 2005-2014 and also finished up the the Solomon papers. I'm hoping to finish all the Elmwood by tomorrow any suggestions for more papers that are more challenging but still relevant.

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Reply 749

studentwiz

Original post by Gilo98

consider range of y=e^x then consider what will happen to this range when the graph is shifted upward by k^2 units

1+k^2?

Reply 750

ridirahman

Can anyone help me with ii) ?? I'm confused..

Reply 751

math42

Original post by ridirahman

Can anyone help me with ii) ?? I'm confused..

Let the length of a side be denoted r. Express V in terms of r.
You want dr/dt; you can find this using dV/dt and dV/dr. For the specific value you'll need to find r from the given volume.

Reply 752

Krollo

Original post by ridirahman

Can anyone help me with ii) ?? I'm confused..

If the side length of the cube is x, then its volume will be x^3. We know dV/dt = 2, we can find dV/dx, and we can find dx/dt from there.

Reply 753

ridirahman

Original post by Krollo

If the side length of the cube is x, then its volume will be x^3. We know dV/dt = 2, we can find dV/dx, and we can find dx/dt from there.

Ohh okay thanks I got it now
and I have another question if you don't mind..

I've done part a) and got root over 109 cos(deta + 16.70)

How would you do part b) ?

Reply 754

Luciferr

Original post by ridirahman

Ohh okay thanks I got it now
and I have another question if you don't mind..

I've done part a) and got root over 109 cos(deta + 16.70)

How would you do part b) ?

if you sub root over 109 cos(30t + 16.7) in place of 10cos(30t) + 3sin(30t),
it becomes
H = 12 - root 109 cos (30t+16.7)
which is a transformation of the cos curve. Knowing the maximum in a normal cos curve is 1, you can work backwards to get the maximum in this one.

for part b, use your maxima and solve for t