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    (Original post by Tasha_140)
    Attachment 540689

    How would you go about solving part (ii) of this?
    Write it as 2x + 3y - x.
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    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
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    (Original post by EBworkin)
    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
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    (Original post by iwishicouldfly14)
    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...
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    (Original post by EBworkin)
    oh those aren't my answers, they are from the mark scheme... I just don't know how to get to them...
    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...
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    (Original post by EBworkin)
    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
    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.
    Spoiler:
    Show
    For part ii)
    \left[6x-x^{2}-1\right]-\left[x^{2}-4x+7\right]=-2x^{2}+10x-8
    For part iii)
    y=-2x^{2}+10x-8
    \frac{dy}{dx}=-4x+10
    For maximum \frac{dy}{dx}=0
    -4x+10=0
    So x=2.5
    For part iv)
    \int^{4}_{1}\left(-2x^{2}+10x-8\right)dx=\left[-\frac{2}{3}x^{3}+5x^{2}-8x\right]^{4}_{1}=9
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    (Original post by iwishicouldfly14)
    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...
    Thank you so much! It all makes sense now....
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    (Original post by Cryptokyo)
    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.
    Spoiler:
    Show
    For part ii)
    \left[6x-x^{2}-1\right]-\left[x^{2}-4x+7\right]=-2x^{2}+10x-8
    For part iii)
    y=-2x^{2}+10x-8
    \frac{dy}{dx}=-4x+10
    For maximum \frac{dy}{dx}=0
    -4x+10=0
    So x=2.5
    For part iv)
    \int^{4}_{1}\left(-2x^{2}+10x-8\right)dx=\left[-\frac{2}{3}x^{3}+5x^{2}-8x\right]^{4}_{1}=9
    Thanks!
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    Can someone explain how to do 13iv its probably easy but i got 1.75 so idk
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    (Original post by mlyke)
    Can someone explain how to do 13iv its probably easy but i got 1.75 so idk
    can u post the question
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    (Original post by candol)
    can u post the question
    https://drive.google.com/folderview?...WRNaFFidGpKb2M

    question 13 iv
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    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
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    (Original post by candol)
    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
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    (Original post by mlyke)
    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
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    That question was tricky, still don't understand part iv
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    (Original post by candol)
    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
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    (Original post by mlyke)
    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
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    Will I probably fail the exam if I haven't started revising yet?
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    (Original post by candol)
    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
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    (Original post by mlyke)
    okay thank you that helps a lot but why do you minus that number from 0.5
    Because in that example i still have 4/5 of the width left (i have gone 1/5th up from the bottom where the triangular plane is at its widest, leaving 4/5)
 
 
 
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