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# AQA MPC4 Unofficial Mark Scheme 2016

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Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016
1. An unofficial mark scheme for MPC4:

1) A = -5/2, B = 9/2
-1 + 7x - (22/3)x2
-1/2 < x < 1/2

2) x = 99.6, 206.4

3) p = 2/15
r = 1/3, s = 4/15

4) -3x + 2 + 9/(2x-3)
-69/2, 9/2

5) 330 days

6) cos = 13/35
(2, 0, -1)
C = (0, 10/3, -5/3) D = (4, -10/3, -1/3)

7) (1/2)e^4
y = e / (2sqrt(1-x))(1-ln(2sqrt(1-x)))

8) k = 6
sin3y - 3ycos3y - pi = (1/6)tan^(-1)(3x/2)

The questions can be be found here:

http://www.thestudentroom.co.uk/show...&postcount=718
2. Can you tell me how you did part C.
3. (Original post by bencurnow)
An unofficial mark scheme for MPC4:

1) A = -5/2, B = 9/2
-1 + 7x - (22/3)x2
-1/2 < x < 1/2

2) x = 99.6, 206.4

3) p = 2/15
r = 1/3, s = 4/15

4) -3x + 2 + 9/(2x-3)
-69/2, 9/2

5) 330 days

6) cos = 13/35
(2, 0, -1)
C = (0, 10/3, -5/3) D = (4, -10/3, -1/3)

7) (1/2)e^4
y = e / (2sqrt(1-x))(1-ln(2sqrt(1-x)))

8) k = 6
sin3y - 3ycos3y - pi = (1/6)tan^(-1)(3x/2)

The questions can be be found here:

http://www.thestudentroom.co.uk/show...&postcount=718
Some of these r rong mate
4. I assume you mean 6)c)

As ACBD was a parallelogram, we know the point of intersection of the two lines is the same distance from a as b, and the same distance from c as d. I called this point O.

Then:
|AO| = 3 |CO|

As we know A and O, I solved for the parameter used for the point C in L2. In fact, this gives two solutions, which correspond to C and D.
5. How much do you think I would lose if my sin3y was negative instead of positive but got the rest of it correct?
6. My working I posted in the main C4 thread for the last question and parametric gradient one

I accidentally wrote 2+9x^2 instead of 4+9x^2 sorry!

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7. (Original post by gcsekid)
Some of these r rong mate
Can you elaborate?
8. Idk how It's 13/35, I got 23/35 somehow
9. (Original post by C0balt)
My working I posted in the main C4 thread for the last question and parametric gradient one

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It looks like you've mistakenly put 2+9x^2 rather than 4+9x^2 in your solution. Just a typo though it seems
10. (Original post by bencurnow)
It looks like you've mistakenly put 2+9x^2 rather than 4+9x^2 in your solution. Just a typo though it seems
Oh crap well spotted lol
11. (Original post by C0balt)
My working I posted in the main C4 thread for the last question and parametric gradient one

I accidentally wrote 2+9x^2 instead of 4+9x^2 sorry!

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Did we have to simplify the gradient to that form
12. (Original post by koolgurl14)
Did we have to simplify the gradient to that form
No, you could just sub t=2/3 to both dy/dt and dx/dt expression and numerically divide, or keep it unsimplified. As long as you got the right answer at the end you have nothing to worry about
13. (Original post by bencurnow)
An unofficial mark scheme for MPC4:

1) A = -5/2, B = 9/2
-1 + 7x - (22/3)x2
-1/2 < x < 1/2

2) x = 99.6, 206.4

3) p = 2/15
r = 1/3, s = 4/15

4) -3x + 2 + 9/(2x-3)
-69/2, 9/2

5) 330 days

6) cos = 13/35
(2, 0, -1)
C = (0, 10/3, -5/3) D = (4, -10/3, -1/3)

7) (1/2)e^4
y = e / (2sqrt(1-x))(1-ln(2sqrt(1-x)))

8) k = 6
sin3y - 3ycos3y - pi = (1/6)tan^(-1)(3x/2)

The questions can be be found here:

http://www.thestudentroom.co.uk/show...&postcount=718
Why dd I get -sin 3y -3ycos3y - pi = (1/6)tan^(-1)(3x/2)
Did u do it integration by parts?
14. COS(THETA) = 23/35!!!!

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15. For the first question, isn't the valid range supposed to be -1 << x << 1?
16. (Original post by koolgurl14)
Why dd I get -sin 3y -3ycos3y - pi = (1/6)tan^(-1)(3x/2)
Did u do it integration by parts?
Yes I integrated by parts
17. (Original post by Simsllama)
For the first question, isn't the valid range supposed to be -1 << x << 1?
As it was (1+2x) we have -1 < 2x < 1 so -1/2 < x < 1/2
18. (Original post by C0balt)
My working I posted in the main C4 thread for the last question and parametric gradient one

I accidentally wrote 2+9x^2 instead of 4+9x^2 sorry!

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I used quotient rule for dy/dt but got slightly diffrent answer
19. (Original post by Yo12345)
COS(THETA) = 23/35!!!!

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No = 13/35
20. (Original post by gcsekid)
Some of these r rong mate
im not sure about the vector one. Shoulddn't the distance CD be 1/3 if Ab which was root560 ?

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