A Level Maths - C3 (Mei) OCR - June 19 2018

Isn't it out of 90??

It is, but it is also segmented.
There will be grade boundaries for the paper too.

You are right, though. In total, coursework + paper is out of 90.

Reply 41

DarthRoar

20

How did you get k = 0.048? I got k = 0.69 because

e^{0.69*1} = 0.5

and

e^{0.69*2} = 0.25

Nevermind it was because t was in years not half-lives.

Reply 42

emmabrookes

2

Original post by Zgogol18

1. -3y/4x as dy/dx
dy/dt=-6
2. 1. Odd 2.neither, periodic 3. Even
3. 25%, k=0.048
4.x=0.55, y=0.29
5. Modulus graph x=-1/3
6. 5pi/6, y=sin(2x-pi)/4
7. Equation of the line x-4y+3=0, gradient=1/4. Point Q was 3/4 for me, the area was 23/8-4ln2
8. When integrated lnx had to times my 1
9. For the last part of the last question I have found the area for lnx between 4 and 1 I think, than found the area between 3 and 4 for 2ln(x-2), though I did a really weird thing. I have integrated 2ln(x-2) by parts and within that I had to integrate udv/dx by substitution. Then subtracted the areas.
10. For point Q for question 9 (not the one with trapezium) I got something like (4,2ln2) can’t remeber and for P I got 3

For question 7 do you think you’d get the mark for -x+4y-3=0

Reply 43

Daribig98

4

Original post by Fatemeh2018

I got 2ln2-5? did anyone get the same?

I got 4ln4 - 3. It's meant to be 2ln but im not sure about the -3. Wasnt it -4 + 1 so you got -3?

Reply 44

Zgogol18

8

Original post by emmabrookes

For question 7 do you think you’d get the mark for -x+4y-3=0

Yeah

Reply 45

AV_Luke

5

Original post by Zeem

I think question 6i) was 5pi/6

You're right that was a typo lol.

Reply 46

JaredzzC

19

Original post by DarthRoar

How did you get k = 0.048? I got k = 0.69 because

e^{0.69*1} = 0.5

and

e^{0.69*2} = 0.25

Why did you use

0.69?

The trick was to use

t=28.8

years and using

\frac{1}{4} M =M_0

as you worked out in the previous question.

(edited 5 years ago)

Reply 47

Zgogol18

8

Original post by JaredzzC

Why did you use 0.69? The trick was to use t=28.8 years and using (1/4)M=Minitial as you worked out in the previous question.

It wasn’t a nice question
Got it only cause I do physics

Reply 48

Daribig98

4

Original post by H4mbo

think this was the graph

I solved x for -1/3 but while re reading my answers i felt the need to solve x for when the modulus would be -ve so i also put down x can be 1... stupidl doubting my first answer so i put both down. How many marks do u reckon that'll cost me? if it's outta 3 i shud get 2/3 right?

Reply 49

Daribig98

4

Original post by Zgogol18

It wasn’t a nice question
Got it only cause I do physics

you'll get marks just for ln'ing the e and whatever you had on the other side of the "=" sign. it shud only cost you 1-2 marks dw about it

Reply 50

benjamin6822

4

Original post by Rose24

I got 4ln2-3 for that?

Well that’s wrong because it gives a negative area

Reply 51

captainslow12

11

Original post by Zeem

I think question 6i) was 5pi/6

What was question 6?

Reply 52

JaredzzC

19

Original post by Zgogol18

It wasn’t a nice question
Got it only cause I do physics

Yea, it was definitely one designed to catch people out.
It almost worked with me because I had to think for a minute how to approach it.

I do Physics too and this kinda question did seem relatable, but I wouldn't be surprised if people somehow tried to use their physics to help them and went into the whole

(1-x_0)e

or thought the half-life was when

x= e^{-1}

or something along those lines.

Reply 53

Daribig98

4

Original post by captainslow12

What was question 6?

Q6 was the composite function question. So it was like 2arcsin(.5) + 1/2 pie

Reply 54

VivChav

6

Original post by Rose24

I got 4ln2-3 for that?

Hi, yeah I got 4ln2 - 2 for some reason. I read the question and thought it not asking for the area between the two curves with x limits of 3 and 4, but for the area beneath the lower curve. I.e. the wording for that last question was atrocious

Reply 55

captainslow12

11

Original post by Daribig98

Q6 was the composite function question. So it was like 2arcsin(.5) + 1/2 pie

Oh yeah. Yeah that was fine. Cheers

Reply 56

JaredzzC

19

Original post by captainslow12

What was question 6?

using

h(x) = fg(x)

and plugging in

\frac{1}{2}

to

2arcsinx + \frac{\pi}{2}

I think.
and also finding the inverse of h(x)

(edited 5 years ago)

Reply 57

ariesstyx

3

Original post by Daribig98

I solved x for -1/3 but while re reading my answers i felt the need to solve x for when the modulus would be -ve so i also put down x can be 1... stupidl doubting my first answer so i put both down. How many marks do u reckon that'll cost me? if it's outta 3 i shud get 2/3 right?