OCR FSMQ Additional Maths 6th June 2016 Official Thread

How would you go about solving part (ii) of this?

Hi, I am preparing for my exam next week. Unfortunately, I cannot get to the answers for question 11ii/iii/iv on the 2015 June paper. I have put the MS answers below: (ii) -2x^2+10x-8 (2 marks) (iii) 2.5 (4 marks) (iv) 9 (4 marks)
Here is the question:
11
Two curves, S1 and S2 have equations y = x^2 - 4x + 7 and y = 6x - x^2 -1 respectively. The curves meet at A and at B.
GRAPH
Points P and Q lie on S2 and S1 between A and B. P and Q have the same x coordinate so that PQ is parallel to the y-axis.
(ii) Find an expression, in its simplest form, for the length PQ as a function of x. [2]
(iii) Use calculus to find the greatest length of PQ. [4]
(iv) Find the area between the two curves. [4] Does anyone have any clues??? Thanks

All your answers are correct.
I checked the mark scheme earlier

oh those aren't my answers, they are from the mark scheme... I just don't know how to get to them...

Hi, I am preparing for my exam next week. Unfortunately, I cannot get to the answers for question 11ii/iii/iv on the 2015 June paper. I have put the MS answers below: (ii) -2x^2+10x-8 (2 marks) (iii) 2.5 (4 marks) (iv) 9 (4 marks)
Here is the question:
11
Two curves, S1 and S2 have equations y = x^2 - 4x + 7 and y = 6x - x^2 -1 respectively. The curves meet at A and at B.
GRAPH
Points P and Q lie on S2 and S1 between A and B. P and Q have the same x coordinate so that PQ is parallel to the y-axis.
(ii) Find an expression, in its simplest form, for the length PQ as a function of x. [2]
(iii) Use calculus to find the greatest length of PQ. [4]
(iv) Find the area between the two curves. [4] Does anyone have any clues??? Thanks

Sorry, I didn't read your thing properly.
For part II, you're looking for the difference between the two curves, to give the length of the line.
You subtract the positive curve away from the negative curve, so you do (6x - x^2 -1) - (x^2-4x+7) which gives you -2x^2+10x-8

For part III, you differentiate your answer from (11) which gives you dy/dx = -4x+10 and then you rearrange and then solve, so you have 4x=10 so x=2.5

For iv, you integrate the answer from II, so you have -2/3(x)^3+5x^2 -8x, and then sub in your two x co-ordinates (4 and 1). You subtract your result from x=4 (16/3) with x=1 (-11/3) and you have 27/3 which leaves you with 9 units^2


Sorry if this didn't make much sense...

H

Hi I did this paper last year and this was definitely the most problematic question on the paper. Hints
For part ii) do one curve minus the other.
For part iii) Take the derivative of the function of part ii to find maximum point.
for part iii) Area under one curve minus the area under the other curve.
The spoiler has my working. Try it first before you look.

Can someone explain how to do 13iv its probably easy but i got 1.75 so idk

can u post the question

https://drive.google.com/folderview?id=0B-r2fgcRzAs_YWRNaFFidGpKb2M

question 13 iv

toughie!
I think you need to use the property of similar shapes. You know that you are going 1m up from the ground. You need to turn this into a fraction of the total vertical height (found in part i) (ie 1/2.35 i think)
multiply this fraction by 0.5 and then subtract the answer from 0.5. (this fives you the width of one of the sides)
Finally multiply this answer by 6 (6 sides)
The logic in the above is you are using the proportional properties of similar shapes.
tell me if it helps

Yeh that works thank you, i just need to figure out your processing, if don't i'll ask. Taking this test on my gcse year

feel free to come back if you need further assistance

Yes please if possible can u draw it because i learn things easier that way i find it easier to understand

tried drawing it, but was a bit rubbish, so i will give you a simple example instead (using easier numbers than are in the question)
Imagine that V in the diagram was 5m high. If you had to fix a wire 1m high, you would be going 1/5th up the shape. That means you would be qoing 1/5th up the triangular slope. But since the bottom of the slope is 0.5m wide, going up 1/5th will mean the width now becomes 0.4m (you have lost 1/5 of the width).
This is the same idea as in the question but with easier numbers.
Does that help at all, please refer to the picture in the question when reading the explanation

okay thank you that helps a lot but why do you minus that number from 0.5