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OCR (not MEI) C1 - Wednesday 13th May 2015

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Hi, can anyone break down what they think the 4 marks would be for for question 7? Gradient of x^-1/3 when x=-8.
Reply 501
Somehow got 10root5 for last question but had the coordinates right... How many marks will I drop? :confused:
Original post by chloe-jessica
In the original post, click 'show spoiler'. The questions aren't in the right order as I can't remember the order (I have said people can correct me but nobody has) but the answers are there.

Sorry but I can't find it ???
Reply 503
Original post by Charlieukharris
Sorry but I can't find it ???


1. 1) 4 + 4root3 [3]
2) can't really describe the graph, if you know what y=1/x looks like then reflect in either axis. [2]
Translation of 2 in x was -1/(x-2) [1]
The translation they wanted you to state was a stretch of 1/3 in either axis. [2]
3) 5^8 [1]
5^(-1/4) [2]
5^(9/2) [2]
4) Disguised quadraticx = -8 and 27 [5?]
5) Line AB = ? [2]
Line was 6x-8y-29=0 [7]
5) differentiation? Can't remember the answer to this, was reasonably straight forward though if you expanded brackets.?
6) simultaneous equation - x=1, y=2 or x=14/3, y=-13/3 [5]
7) differentiate x^(-1/3)? Should get out -1/48 when you put x in. [4]
8) Draw the graph [4]
X < -1, X > 3/2 for y=0 [2]
k < - 25/8 [3]
9) a=13 [5]
Second derivative was positive therefore min point [2]?
X=1/3 [2]?
10) centre (5,-2) radius 5 [3]
Prove tangent [5]
Area of triangle = 40 [4]
Original post by Rr_rolo
I wrote 40 but no units squared would i lose marks


No, I highly doubt it
Original post by jamespaine
Hi, can anyone break down what they think the 4 marks would be for for question 7? Gradient of x^-1/3 when x=-8.


Yep.

y = x-1/3

dy/dx = -1/3 x-4/3 [2]
dy/dx|x=-8 = -1/3 (3√-8)4 [1]
= -1/3 times 1/16
= -1/48 [1]
Original post by TCO
625 root 5 I think

Posted from TSR Mobile


it was
(5xroot5) ^3

5 x root5 can be seen as 5 ^3/2

therefore using indicises laws, multiply the two and you get:

5^9/2 or 5^4.5
Original post by mountaindewwwww
That's wrong on so many levels, go back over your gcse work.


i think you need to go over gcses :tongue: it is definitely a third
Reply 508
probably 2 :smile: What did you guys get for the value of k?
Do you think I'd lose a mark for putting 4root3 + 4 rather than 4+4root3?

Nothing big but i'm aiming for full raw marks
Original post by ChoccyPhilly
Do you think I'd lose a mark for putting 4root3 + 4 rather than 4+4root3?

Nothing big but i'm aiming for full raw marks


Not at all haha
Original post by chloe-jessica
1. 1) 4 + 4root3 [3]
2) can't really describe the graph, if you know what y=1/x looks like then reflect in either axis. [2]
Translation of 2 in x was -1/(x-2) [1]
The translation they wanted you to state was a stretch of 1/3 in either axis. [2]
3) 5^8 [1]
5^(-1/4) [2]
5^(9/2) [2]
4) Disguised quadraticx = -8 and 27 [5?]
5) Line AB = ? [2]
Line was 6x-8y-29=0 [7]
5) differentiation? Can't remember the answer to this, was reasonably straight forward though if you expanded brackets.?
6) simultaneous equation - x=1, y=2 or x=14/3, y=-13/3 [5]
7) differentiate x^(-1/3)? Should get out -1/48 when you put x in. [4]
8) Draw the graph [4]
X < -1, X > 3/2 for y=0 [2]
k < - 25/8 [3]
9) a=13 [5]
Second derivative was positive therefore min point [2]?
X=1/3 [2]?
10) centre (5,-2) radius 5 [3]
Prove tangent [5]
Area of triangle = 40 [4]


I think the differentiation one was f'(x)=-3x^2+10x-3 and f(x)=-x^3+5x^2-3x+15? Which is f(x)=(5-x)(x^2+3)
Original post by Renzhi10122
I think the differentiation one was f'(x)=-3x^2+10x-3 and f(x)=-x^3+5x^2-3x+15? Which is f(x)=(5-x)(x^2+3)


Yeah that's right; how do you even remember that...
Reply 513
Original post by Renzhi10122
I think the differentiation one was f'(x)=-3x^2+10x-3 and f(x)=-x^3+5x^2-3x+15? Which is f(x)=(5-x)(x^2+3)


That sounds right to me :smile: thank you!
Original post by Madhutty
Yeah that's right; how do you even remember that...


Haha, I just worked backwards
Original post by Madhutty
Yep.

y = x-1/3

dy/dx = -1/3 x-4/3 [2]
dy/dx|x=-8 = -1/3 (3√-8)4 [1]
= -1/3 times 1/16
= -1/48 [1]


Would I get any marks if I misread it as y= x^1/3? so did

dy/dx = 1/3x^-2/3

1/3(-8)^-2/3 = 1/12
Original post by jamespaine
Would I get any marks if I misread it as y= x^1/3? so did

dy/dx = 1/3x^-2/3

1/3(-8)^-2/3 = 1/12


2/4 for correct differentiation, in theory 3/4 for correct substitution and arithmetic, but you'd definitely lose the final [A1] mark.
Original post by jamespaine
Would I get any marks if I misread it as y= x^1/3? so did

dy/dx = 1/3x^-2/3

1/3(-8)^-2/3 = 1/12


Possibly a mark for attempting differentiation but that's all i can think of
does anyone have an idea on how the 4 marks on the end question about area will be distributed?
Reply 519
Yeah I got that too but I didn't have time to simplify it so my final answer was root 8000 over 2 (I came out of the exam and simplified it and it would've been 20root5) ..though some people said that's wrong because I used the wrong height (for 1/2xbh area of triangle)...which is why I think some people are saying they got 40??

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