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Ocr (mei) c2 exam paper 18/05

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Original post by Axion
y=100x^3


sounda fine :smile:
Reply 41
4. Point P=(6,3)

(i) y=3f(x)

Point P now= (6,9)

(ii) Y=f(4x)

Point P now= (3/2,3)
Reply 42
About question 6 (logs one) will I still get the marks (if yes how many ?) if my answer was
y = 10^(3log(x) + 2)

(which is working but there is simpler version of it which is y= 100x^3)
Reply 43
The sin equation - surely 7.02 isn't a solution as it's outside the range?

Thanks for the mark scheme though
Reply 44
Original post by mathsodoku
4. Point P=(6,3)

(i) y=3f(x)

Point P now= (6,9)

(ii) Y=f(4x)

Point P now= (3/2,3)


Isn't it

4. Point P=(6,3)

(i) y=3f(x)

Point P now= (6,9)

(ii) Y=f(4x)

Point P now= (1.5,3)

because it's one way strech parallel to the x-axis scale factor 1/4

correct me if im wrong
Original post by xayno
Isn't it

4. Point P=(6,3)

(i) y=3f(x)

Point P now= (6,9)

(ii) Y=f(4x)

Point P now= (1.5,3)

because it's one way strech parallel to the x-axis scale factor 1/4

correct me if im wrong


that's exactly the same but with a decimal :smile:
Reply 46
5. Area Of A Sector= 45cm^2

45cm^2= 1/2 *r^2 *1.6
(45)=0.8*r^2
(45)/0.8 = r^2
56.25=r^2
SQRT(56.25)= r
radius= 7.5

r*(1.6)= Arc length
7.5*(1.6)= 12
Arc length=12

12+(2*7.5)= perimeter

perimeter= 27
Anyone want me to write my solution for the last part or has someone else posted it?
Reply 48
Original post by AspiringGenius
that's exactly the same but with a decimal :smile:


yeah exactly the same.
Original post by Gilsenan
The sin equation - surely 7.02 isn't a solution as it's outside the range?

Thanks for the mark scheme though


I guess you are right, I didn't get it in the exam as well. I thought sin2theta should have 4 values. Lost about 17 marks.
I think its 3.52.
(edited 11 years ago)
Reply 50
Original post by mathsodoku
yeah exactly the same.


sry i missread it ;d


could anyone help me with that :


About question 6 (logs one) will I still get the marks (if yes how many ?) if my answer was
y = 10^(3log(x) + 2)

(which is working but there is simpler version of it which is y= 100x^3)
Original post by farhadarwin
I guess you are right, I didn't get it in the exam as well. I thought sin2theta should have 4 values. Lost about 17 marks.


It did... din;t it?

*tumbleweed*
Reply 52
Original post by AspiringGenius
It did... din;t it?

*tumbleweed*


It does - if you draw the graph of sin2x you'll see you get 4 points of intersection with the line y = 0.7 :smile:
Reply 53
q7. dy/dx = 6x^(1/2) -5
intergrate to.... y = 4x^(3/2) -5x + c passing thru (4,20)
20=4(4)^(3/2) -5(4) + c
c= -24

equation of line: y= 4x^(3/2) -5x -24
Original post by Gilsenan
It does - if you draw the graph of sin2x you'll see you get 4 points of intersection with the line y = 0.7 :smile:


phew.
Reply 55
Original post by mathsodoku
q7. dy/dx = 6x^(1/2) -5
intergrate to.... y = 4x^(3/2) -5x + c passing thru (4,20)
20=4(4)^(3/2) -5(4) + c
c= -24

equation of line: y= 4x^(3/2) -5x -24




That's how i got it

dy/dx = 6x^(1/2) - 5
integrated = 4x^(3/2) - 5x + c passing thru 4,20
20 = 4(4)^(3/2) -5(4) + c
20 = 32 - 20 + c
8 = c
Original post by xayno
sry i missread it ;d


could anyone help me with that :


About question 6 (logs one) will I still get the marks (if yes how many ?) if my answer was
y = 10^(3log(x) + 2)

(which is working but there is simpler version of it which is y= 100x^3)


In worst case you should lose one mark I think.
Reply 57
Original post by mathsodoku
q7. dy/dx = 6x^(1/2) -5
intergrate to.... y = 4x^(3/2) -5x + c passing thru (4,20)
20=4(4)^(3/2) -5(4) + c
c= -24

equation of line: y= 4x^(3/2) -5x -24


err no, its 8 not -24
did anyone get a=10 and r=0.6 and a=15 and r=0.4 for the first part of the last geometric qu? about 7marks i think, had no idea how to do part 2 about 3 marks????
Original post by enlighten25
did anyone get a=10 and r=0.6 and a=15 and r=0.4 for the first part of the last geometric qu? about 7marks i think, had no idea how to do part 2 about 3 marks????


I got the same and the last part was :s-smilie: Im hoping for low grade boundaries

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