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OCR C1 Mark Scheme 16/05/12 (NOT MEI)

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Reply 20
Original post by As_Dust_Dances_
The total perimeter was 10y+3 and it had to be between 20 and 54 right?
so 20<10y+3<54
17<10y<51
1.7<y<5.1


no it was 10y +6 the perimeter!
Original post by As_Dust_Dances_
The total perimeter was 10y+3 and it had to be between 20 and 54 right?
so 20<10y+3<54
17<10y<51
1.7<y<5.1


total perimeter was 10Y+6
Reply 22
Original post by As_Dust_Dances_
The total perimeter was 10y+3 and it had to be between 20 and 54 right?
so 20<10y+3<54
17<10y<51
1.7<y<5.1


total perimeter was 10y+6
Reply 23
Original post by As_Dust_Dances_
The total perimeter was 10y+3 and it had to be between 20 and 54 right?
so 20<10y+3<54
17<10y<51
1.7<y<5.1


10y +6

check out http://www.thestudentroom.co.uk/showthread.php?t=2002389

unofficial mark scheme
Reply 24
i got -7<x<4 and 1.7<x<5.1 ...
Reply 25
Original post by I_am_god_123
total perimeter was 10Y+6


20< 10y + 6 >54!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Reply 26
Original post by max25
how did you find the last question?


it was ok, pretty sure i got it right

check out

http://www.thestudentroom.co.uk/showthread.php?t=2002389
Reply 27
Original post by As_Dust_Dances_
For the perimeter question I got 17/10<x<51/10


I thought it was in terms of y?

I think I got something like this...

34 <= 10y + 6 <= 54

When simplified gets:

28 <= 10y <= 48

then:

14/5 <= y <= 24/5
Reply 28
Original post by I_am_god_123
total perimeter was 10Y+6


eeek apologies, wrong question hehe!were both right:smile:
Original post by I_am_god_123
nah it was -7<x<4
but then x must be +ve


for part i) (x+3)(4x)<112 right?

so 4x^2 +12x-112<0

x^2+3x-28

(x+7)(x-4)

You're talking about part 1 right, I was on about part 2..

So yeah it is x<-7 and x>4
Reply 30
Original post by As_Dust_Dances_
for part i) (x+3)(4x)<112 right?

so 4x^2 +12x-112<0

x^2+3x-28

(x+7)(x-4)

You're talking about part 1 right, I was on about part 2..

So yeah it is x<-7 and x>4


But x cannot be negative - you cannot have a negative length as one of the sides was 4x negative dimensions are impossible so it had to be 0>x>4
100 ****ing percentaaah! :h:
Reply 32
Original post by max25
20< 10y + 6 >54!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


yaeh that's correct
Reply 34
Question 6

Equation of normal is

4x - 6y -29 = 0
Reply 35
would i lose a mark for having it as -4x + 6y + 29?
Reply 36
Original post by As_Dust_Dances_
for part i) (x+3)(4x)<112 right?

so 4x^2 +12x-112<0

x^2+3x-28

(x+7)(x-4)

You're talking about part 1 right, I was on about part 2..

So yeah it is x<-7 and x>4


this is right.
Original post by PhysicsMan
Question 6

Equation of normal is

4x - 6y -29 = 0


This rings a bell :smile: has the gradient was like 2/3 or something?
Reply 38
Original post by As_Dust_Dances_
This rings a bell :smile: has the gradient was like 2/3 or something?


Yep, the gradient of tangent was -3/2 therefore the gradient of normal is 2/3.
Reply 39
Original post by max25
20< 10y + 6 >54!!!!!!!!!!!!!!!!!!!!!!!!!!! !!


That can't be correct (I didnt sit the paper though).. You are saying 20 < value > 54... so 20 is bigger than 54? 20 >54: you needed two statements.

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