Competition - post a photo you've taken of nature >>

AQA MPC4 Unofficial Mark Scheme 2016

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/showpost.php?p=65880978&postcount=718

Scroll to see replies

Reply 1

Parhomus

Can you tell me how you did part C.

Reply 2

gcsekid

Original post by bencurnow

Some of these r rong mate

Reply 3

bencurnow

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.

Reply 4

Parhomus

How much do you think I would lose if my sin3y was negative instead of positive but got the rest of it correct?

Reply 5

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!

Posted from TSR Mobile

(edited 7 years ago)

1466196466641.jpg39.9KB

1466196479605.jpg45.4KB

Reply 6

bencurnow

Original post by gcsekid

Some of these r rong mate

Can you elaborate?

Reply 7

irMike

Idk how It's 13/35, I got 23/35 somehow

Reply 8

bencurnow

Original post by C0balt

My working I posted in the main C4 thread for the last question and parametric gradient one

Posted from TSR Mobile

It looks like you've mistakenly put 2+9x^2 rather than 4+9x^2 in your solution. Just a typo though it seems

Reply 9

C0balt

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

Reply 10

Applicant75235

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!

Posted from TSR Mobile

Did we have to simplify the gradient to that form

Reply 11

C0balt

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

Reply 12

Applicant75235

Original post by bencurnow

Why dd I get -sin 3y -3ycos3y - pi = (1/6)tan^(-1)(3x/2)
Did u do it integration by parts?

Reply 13

Yo12345

COS(THETA) = 23/35!!!!

Posted from TSR Mobile

Reply 14

Simsllama

For the first question, isn't the valid range supposed to be -1 << x << 1?

Reply 15

bencurnow

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

Reply 16

bencurnow

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

Reply 17

Samtheoneson

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!

Posted from TSR Mobile

I used quotient rule for dy/dt but got slightly diffrent answer