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AQA maths core 1, 13th May 2015 watch

1. I found it good thanks. The cylinder question involved you knowing the area of a circle equation and the circumference equation in terms of pi.
It was an optimisation question.

BTW the graph bounced off 0 under the x axis and then went down and then up to 3
2. (Original post by Moe21)
anyone know how we were supposed to sketch the graph?
I sketched a cubic with points at x-intercepts at -3, 0, 3
I think it was meant to be a double root at 0 and passes through x at 3... I think

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3. I think it will be around 58(+ or-)1
4. (Original post by Flangastic)
The formula for the area of a cylinder was in the formulae booklet...guessing no one thought of looking in there, during a core 1 paper 😂

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I could only see area of a cone not a cylinder. Luckily I remembered what it was anyway aha:')
5. (Original post by DexterMabel123)
i got 21.6 for the integration question anyone get this?
I got that!
6. (Original post by Lizzie33)
I think it was meant to be a double root at 0 and passes through x at 3... I think

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yeah, what i did was the graph starts at the bottom left hand side quadrant, touches the origin, goes back down, has a minimum at the bottom right quadrant, then goes back up crossing the x axis at +3
7. For the fourth question (with the equations of circles) what values did everyone get for k????
8. I found it ridiculously easy
9. In the circle eq it was =d so diameter
10. (Original post by Heffalump .)
There was a double '0' root. so it started under the x axis and to the left of the y axis, the went up to the origin, touched it, went back down, then went back up to '3' on the a axis and passed above to x axis to the right of the y axis.
would i lose all marks if i sketched 3 roots?
11. Pretty sure i got full marks. Wasn't that bad. If you didnt know the area of a cylinder equation, you could use the fact that the area of a circle is 2Pi * r and then multiply it by the height h.
12. (Original post by Anam)
i don't understand what you mean! i did it so that the curve touched the 3 on x axis and went up and down towards 0
As it's a positive cubic equation it starts bottom left and ends up top right, it has a single root at 3, so it goes through, and a double root at 0, so it touches
13. (Original post by Anam)
i don't understand what you mean! i did it so that the curve touched the 3 on x axis and went up and down towards 0
I think that's wrong I'm afraid, there was definitely a double root at 0, not 3
14. (Original post by tanyapotter)
yeah, what i did was the graph starts at the bottom left hand side quadrant, touches the origin, goes back down, has a minimum at the bottom right quadrant, then goes back up crossing the x axis at +3
yeah this is right
15. (Original post by Moe21)
would i lose all marks if i sketched 3 roots?
What do you mean by 3 roots? it depends on where they were, it was 3 marks, but if you plotted on of the roots correctly (the root 3) then you would have only lost two I think...
16. I thought it was very good actually. Did anyone get 7 for re shortest distance between C and the chord RQ or something??
As it's a positive cubic equation it starts bottom left and ends up top right, it has a single root at 3, so it goes through, and a double root at 0, so it touches
Aw man! i lost those 3 marks then but thanks for explaining
18. (Original post by RhysKirkwood)
I thought it was very good actually. Did anyone get 7 for re shortest distance between C and the chord RQ or something??
Yeah that's correct.

For anyone that wants to know how:

The radius of the circle was 5sqrt(2) and the chord length was 2.

You can turn this into a right angled triangle with the hypotenuse being the radius, and another side being length 1 (half chord). Then you can work out the distance between the chord and centre using Pythagoras.
19. Think this one was just 2 marks if I am not mistaken.

(Original post by Anam)
Aw man! i lost those 3 marks then but thanks for explaining
20. You put the co ordinates of x2 x1 and y2 and y1 into
Y2-y1 ÷ x2 - x1
Then multiply both top and bottom by x2 +x1
To get the denominator and numerator which cancel divides to get a simplified form of something like

7- square root of something
Forgot the digits sorry :/

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