Mr M's OCR (not OCR MEI) Core 1 answers May 2012

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  1. Mr M's Avatar
    321 17-05-2012  07:10
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by AspiringDoctor)
    Have to say a massive thank you to Mr M for your time and effort I don't think I did very well but at least the answers can let my mind rest and not fret about what the answers could have been.
    You are welcome
  2. Mr M's Avatar
    322 17-05-2012  07:11
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by Kabelan1)
    Hi Mr M,

    Approximately how much from the other two AS units e.g C2 and D1,
    would be required if I got an average B in this June 2012 C1 paper.
    Required for what?
  3. Mr M's Avatar
    323 17-05-2012  07:12
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by BillyFord188)
    For some odd reason on 10iv) I formed the quadratic by subbing in y=2x; but instead of then finding the discriminant as normal I just factorised using the formula and got two horrible surd answers, would I still get the 3/5 marks for forming the correct quadratic?
    Probably, might be 2/5
  4. Mr M's Avatar
    324 17-05-2012  07:14
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by pzoDe)
    Ah Mr M, if you're looking at this, I see what he means. I have repeated 9(ii) twice. The first 9(ii) needs to become 9(i).
    Ok changed
  5. Millyshyn's Avatar
    325 17-05-2012  07:54
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by Mr M)
    100 UMS is calculated from a mathematical formula that depends on the position of the other grades. Just wait and see.
    Sorry for bothering you again but I just realised what I did wrong for 9(i) after trying to work through it again. Can't believe I didn't spot it while checking, I had half an hour :/
    So.. I did something along the lines of:
    4x(x+3) <112
    4x^2 + 12 < 112 Missed out the x by accident here ugh.
    X^2 < 25
    I drew a graph of y = x^2 - 25 and got the intercepts as 5, -5
    then stated that -5 < x < 5

    Any idea how many marks I would get? Just one for forming the inequality they asked for?
    Last edited by Millyshyn; 17-05-2012 at 07:57.
  6. PeaceLoveKindness's Avatar
    326 17-05-2012  09:28
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    I can't remember what I got for all my questions but a lot of those figures seem familiar, heres hoping I got an A this time around xD thanks Mr M!
  7. Holz888's Avatar
    327 17-05-2012  09:44
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by pzoDe)
    It depends on what you want to get.
    Your final OCR Maths grade is determined by your average percentages in AS and A2 (as far as I am aware; this may only be strictly for A*).
    A*: You need an average of 80% in your AS modules and an average of 90% in your A2 modules.
    A: You need an average of 80% in both AS and A2 modules
    B: You need an average of 70% in both AS and A2 modules
    Hi, I don't think this is quite right. I was told that to get the A*, you need an average of 80% over BOTH AS and A2 (which could work out as 70% and 90%, 60% and 100%), as long as you have 90% or above in your A2 modules. Obviously, it is quite unlikely that someone getting a low B will actually proceed to get 90% in their A2, unless there was some particular reason for their B, but it is possible.
  8. Mr M's Avatar
    328 17-05-2012  10:23
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by Millyshyn)
    Sorry for bothering you again but I just realised what I did wrong for 9(i) after trying to work through it again. Can't believe I didn't spot it while checking, I had half an hour :/
    So.. I did something along the lines of:
    4x(x+3) <112
    4x^2 + 12 < 112 Missed out the x by accident here ugh.
    X^2 < 25
    I drew a graph of y = x^2 - 25 and got the intercepts as 5, -5
    then stated that -5 < x < 5

    Any idea how many marks I would get? Just one for forming the inequality they asked for?
    You might get 3 marks if you are lucky.
  9. pzoDe's Avatar
    329 17-05-2012  10:38
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by Holz888)
    Hi, I don't think this is quite right. I was told that to get the A*, you need an average of 80% over BOTH AS and A2 (which could work out as 70% and 90%, 60% and 100%), as long as you have 90% or above in your A2 modules. Obviously, it is quite unlikely that someone getting a low B will actually proceed to get 90% in their A2, unless there was some particular reason for their B, but it is possible.
    You are probably correct, I wasn't 100% - I mean 90% - sure about what I said.

    Just to clarify the 90% average is required in C3 and C4 (for example 80% and 100%).
  10. mcquitmp's Avatar
    330 17-05-2012  10:52
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    Hi Mr M,

    Earlier you confirmed that I would get some marks for an answer of Y=2x - 5 - for one of the questions as I got the gradient right. The question was out of 4 marks - Are you able to say how many marks you think I might get ?

    Thanks...
  11. Mr M's Avatar
    331 17-05-2012  11:13
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by mcquitmp)
    Hi Mr M,

    Earlier you confirmed that I would get some marks for an answer of Y=2x - 5 - for one of the questions as I got the gradient right. The question was out of 4 marks - Are you able to say how many marks you think I might get ?

    Thanks...
    2 I expect
  12. raheem94's Avatar
    332 17-05-2012  12:36
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by Mr M)
    100 UMS is calculated from a mathematical formula that depends on the position of the other grades. Just wait and see.
    The grade boundaries are only decided for A and E, right?

    100UMS means A+2(A-B), can you please confirm it.
  13. Intriguing Alias's Avatar
    333 17-05-2012  12:42
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by raheem94)
    The grade boundaries are only decided for A and E, right?

    100UMS means A+2(A-B), can you please confirm it.
    I think so but it doesn't always look like that, there might be a difference of 1 raw mark between each boundary sometimes (due to the real boundaries being decimals and then rounded).

    e.g. 30 for an E, 37.4 for a D (rounded to 37), and then 44.8 (rounded to 45) . So the differences in grades look different.
  14. Mr M's Avatar
    334 17-05-2012  12:43
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by raheem94)
    The grade boundaries are only decided for A and E, right?

    100UMS means A+2(A-B), can you please confirm it.
    A, C and E I think.

    The other formula is right from memory but I don't have the time now to look it up.
  15. pzoDe's Avatar
    335 17-05-2012  12:57
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    Working with allocated marks:

    1.
    (x^3-5x^2+3x-15)-(x^2+3x-4) [1]
    x^3-5x^2+3x-15-x^2-3x+4 [1]
    \boxed{x^3-5x^2-11} [1]
    2.
    (Original post by (i))
    \boxed{7^\frac{1}{4}} [1]
    (Original post by (ii))
    \frac{1}{7^\frac{3}{2}} [1]
    \boxed{7^{-\frac{3}{2}}} [1]
    (Original post by (iii))
    7^4 x (7^2)^{10} [1]
    \boxed{7^{24}} [1]
    3.
    (Original post by (i))
    5y=3x-20
    y=\frac{3}{5}x-4
    ∴ Gradient=\boxed{\frac{3}{5}} [1]
    (Original post by (ii))
    P\Rightarrow5(0)=3x-20
    3x=20
    x=\frac{20}{3}
    P=(\frac{20}{3},0) [1]
    Q\Rightarrow5y=3(0)-20
    y=-4
    Q=(0,-4) [1]
    MidpointPQ=(\frac{x1+x2}{2},\frac{y1+y2}{2 }) [1]
    MidpointPQ=(\frac{\frac{20}{3}+0}{2},\frac {0+(-4)}{2})
    MidpointPQ=\boxed{(\frac{10}{3},-2)} [1]

    Not sure about the 3rd mark I allocated here
    4.
    (Original post by (i))
    2[x^2-10x]+49 [1]
    2[(x-5)^2-25]+49 [1]
    2(x-5)^2-50+49 [1]
    \boxed{2(x-5)^2-1} [1]
    (Original post by (ii))
    \boxed{(5,-1)} [2]

    or

    \frac{dy}{dx}=4x-20
    4x-20=0
    x=5
    y=2(5)^2-20(5)+49
    y=-1
    Vertex=\boxed{(5,-1)} [2]

    Note: If you got (i) wrong, then you will should get ecf for either [1] or [2] marks
    5.
    (Original post by (i))
    Click image for larger version.  Name: root.gif  Views: 4  Size: 1.0 KB  ID: 148671

    Marks for following:
    Curve has a descreasing gradient and only in quadrant positive for x and y [1]
    Curve goes through (0,0)[1]
    (Original post by (ii))
    \boxed{Curve is translated 4 units in the positive x-direction} (owtte)

    Marks for following:
    Translated 4 units [1]
    Positive x-direction [1]

    Note: you may get [1] mark for 'in the right direction'
    (Original post by (iii))
    \boxed{y=\sqrt{\frac{x}{5}}} or the same in a different format [2]

    or

    y=5\sqrt{x} [1]

    Not sure whether the incorrect equation gains a mark or not.
    6.
    y=6x^{-2}-5
    \frac{dy}{dx}=-12x^{-3} [1]
    Gradienttangent=-12(2)^{-3} [1]
    Gradienttangent=-\frac{3}{2}
    Gradientnormal=\frac{2}{3} [1]
    y=6(2)^{-2}-5 [1]
    y=-\frac{7}{2} [1]
    y+\frac{7}{2}=\frac{2}{3}(x-2) (or alternative method of finding the equation of the line) [1]
    6y+21=4x-8
    \boxed{4x-6y-29=0} [1]

    Not sure about some of the allocated marks. I gave method over answer where necessary.
    7.
    Stating x^\frac{1}{2}=y (or other variable instead of y, e.g a, b)

    y^2-6y+2=0

    Complete the square:

    (y-3)^2-9+2=0 [1]
    (y-3)^2=7
    y-3=\pm \sqrt{7}
    y=3\pm \sqrt{7} [1]

    Quadratic equation:

    y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
    y=\frac{-(-6)\pm \sqrt{(-36)^2-4(1)(2)}}{2(1)} [1]
    y=\frac{6\pm \sqrt{28}}{2}
    y=3\pm \frac{\sqrt28}{\sqrt{4}} [1]
    y=3\pm \sqrt{7} [1]

    x=y^2
    x=(3\pm \sqrt{7})^2 [1]
    \boxed{x=16\pm 6\sqrt{7}} [2] ([1] if only +6 or only -6 coefficient of \sqrt{7} / one of the values of 16 or 6 is incorrect)

    (can be done without substitution of y (or a different variable) for x^{\frac{1}{2}})

    Not completely sure about allocated marks
    Last edited by pzoDe; 17-05-2012 at 18:37.
  16. olivers16's Avatar
    336 17-05-2012  13:07
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by pzoDe)
    Working with allocated marks:

    1.


    2.




    3.



    4.




    5.




    6.


    7.


    Will finish tomorrow evening after C2.
    For 5ii) Would you get the two marks if you put "translated to units to the right" and also in 5iii) I put 5 to the power of -1 , is that right? Thanks.
  17. pzoDe's Avatar
    337 17-05-2012  13:45
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by olivers16)
    For 5ii) Would you get the two marks if you put "translated to units to the right" and also in 5iii) I put 5 to the power of -1 , is that right? Thanks.
    If you mean "translated two units to the right" you will only get one of the marks (if they allow 'to the right' instead of positive x-direction) as it was translated four units, not two.

    As for 5^{-1}, if you did any of these you will get the marks:

    y=\sqrt{5^{-1}x}
    y=\sqrt{x*5^{-1}}
    y=\sqrt{(\frac{5}{x})^{-1}}
    y=\sqrt{\frac{5^{-1}}{x^{-1}}}
    y=\sqrt{5^{-1}}\sqrt{x}
    Last edited by pzoDe; 17-05-2012 at 14:31.
  18. olivers16's Avatar
    338 17-05-2012  14:23
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by pzoDe)
    If you mean "translated two units to the right" you will only get one of the marks (if they allow 'to the right' instead of positive x-direction) as it was translated four units, not two.

    As for 5^{-1}, if you did any of these you will get the marks:

    y=\sqrt{5^{-1}x}
    y=\sqrt{x*5^{-1}}
    y=\sqrt{(\frac{5}{x})^{-1}}
    y=\sqrt{\frac{5^{-1}}{x^{-1}}}
    Thank you. How do you know this? Have u seen the mark scheme?
  19. raheem94's Avatar
    339 17-05-2012  14:28
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    Here are the solutions to the paper:

    Q1:

    (x-5)(x^2+3)-(x+4)(x-1) = x^3 + 3x -5x^2 -15 - (x^2 - x +4x -4) \\ = x^3 +3x-5x^2 -15 -x^2 -3x+4 = \boxed{x^3 - 6x^2 - 11}

    Q2:
    (i) \sqrt[4]7 = \boxed{7^{\frac14}}
    (ii) \displaystyle \frac1{7\sqrt7 } = \frac1{7^1 \times 7^{\frac12}} = \frac1{7^{1+ \frac12} } = \frac1{7^{\frac32}} = \boxed{7^{- \frac32}}

    Q3:
    (i)
    \displaystyle 3x-5y-20=0 \implies 3x-20=5y \implies y = \frac35 x - \frac{20}5 \\ \begin{array}{cccc} y & = & \frac35 x & - 4 \\ \uparrow & & \uparrow \uparrow & \uparrow \\ y & & mx & c \end{array}

    \boxed{m = \dfrac35}

    (ii)
    When l crosses the x axis, y=0.
    \displaystyle \frac35 x -4 =0 \implies 3x = 20 \implies x = \dfrac{20}3 \\ \boxed{ P \left( \frac{20}3, 0 \right) }

    When l crosses the y axis, x=0.
    \displaystyle y = \frac35 \times 0 -4 \implies y = -4 \\ \boxed{Q\left( 0, -4\right) }

    Midpoint of PQ is
    \displaystyle \left( \frac{ \frac{20}3 + 0}{2} , \frac{0-4}2 \right) = \left( \frac{20}6 , -2 \right) \\ \boxed{\left( \frac{10}3 , -2 \right)}


    Q4:
    (i)
    \displaystyle 2x^2-20x+49 = 2(x^2 -10x)+49 = 2[(x-5)^2 -25] +49 = 2(x-5)^2 -50 + 49 = \boxed{2(x-5)^2 -1 }
    (ii)
    The vertex of the parabola is its max/min point.
    The minimum of this curve is at x = 5
    Sub in x= 5 in y=2(x-5)^2 -1 = 2(5-5)^2 -1 = 0 -1 = -1

    Coordinates of the vertex \boxed{(5, -1)}


    Q5:

    (i)


    (ii)
    \text{Let} f(x) = \sqrt{x}
    f(x-4) = \sqrt{x-4}
    Horizontal shift of 4 units in the x direction.

    (iii)
    \displaystyle f \left( \frac15 x \right) = \sqrt{ \frac15 x } \\ \boxed{y =\sqrt{ \frac15 x }}

    Q6:

    \displaystyle y=\frac{6}{x^2}-5 = 6x^{-2} - 5
    Differentiate it.
    \displaystyle \frac{dy}{dx} = -2 \times 6x^{-2-1} -0 = -12x^{-3} = - \frac{12}{x^3}
    Sub in x = 2
    \displaystyle \frac{dy}{dx} = - \frac{12}{x^3} = - \frac{12}{2^3} = - \frac{12}8 = - \frac32

    Gradient of normal: \displaystyle m = \frac{-1}{- \frac32} = \frac1{ \frac32} = \frac23

    Find the y-coordinate at x = 2 , \displaystyle y=\frac{6}{x^2}-5 = = \frac64 - 5 = - \frac72

    Use y - y_1 = m(x-x_1)
    (x_1, y_1) = \left(2, - \dfrac72 \right)
    \displaystyle y - \left( - \dfrac72 \right) = \frac23 (x-2) \implies y + \frac72 = \frac23 (x-2) \implies 3y + \frac{21}2 = 2x -4 \implies \boxed{4x-6y-29=0}


    Q7:

    x-6x^\frac{1}{2}+2=0
    \text{Let } y = x^{\frac12}
    \displaystyle y^2 -6 y +2 = 0 \implies (y-3)^2 - 9 +2 = 0 \implies (y-3)^2 = 7 \\ \implies y-3 = \pm \sqrt7 \implies y = 3 \pm \sqrt7 \implies x^{\frac12} = 3 \pm \sqrt7
    \displaystyle (x^{\frac12})^2 = (3 \pm \sqrt7)^2 \implies x = 9 + 2 \times 3 \times (\pm \sqrt7) + 7 \implies \boxed{ x = 16 \pm 6\sqrt7}


    Q8:

    (i)
    y = x^4 +32x
    Differentiate with respect to x
    \displaystyle \frac{dy}{dx} = 4x^3 + 32
    At stationary point gradient equals zero.
    \displaystyle 4x^3 + 32 =0 \implies x^3 = - \frac{32}4 = -8 \implies x^3= -2^3 \implies x = -2
    Sub in x = -2 in y = x^4 +32x = (-2)^4 +32(-2) = 16 - 64 = -48
    Coordinates of stationary point are: \boxed{(-2, -48)}

    (ii)
    Find the second derivative.
    \displaystyle \frac{d^2y}{dx^2} = 12x^2 +0 = 12x^2
    Sub in x = -2 ,
    \displaystyle \frac{d^2y}{dx^2} = 12x^2 = 12(-2)^2 = 12 \times 4 = 48&gt;0 \boxed{ \therefore \text{minimum} }

    (iii)
    Find \displaystyle \frac{dy}{dx} &gt; 0
    4x^3 + 32 &gt; 0 \implies x^3 &gt; -8 \implies \boxed{x&gt;-2}


    Q9:

    Area = length x width = 4x(x+3)
    \displaystyle 4x(x+3) &lt; 112 \implies x(x+3) &lt; 28 \implies x^2 +3x -28 &lt; 0

    Solve
    x^2 +3x -28 = 0 \implies x^2 ++7x-4x-28 = 0 \implies (x-4)(x+7) = 0 \implies x = 4, -7
    Draw the graph of y = x^2 +3x -28



    The region of x^2 +3x-28 &lt; 0 is -7 &lt; x &lt; 4
    Length is always positive, hence, the range is \boxed{ 0 &lt; x &lt; 4 }

    (ii)
    \text{Perimeter} = 4y+y+3+2y+3+y+2y = 10y+6
    \displaystyle 20 \le 10y+6 \le 54 \implies 14 \le 10y \le 48 \implies \frac{14}{10} \le y \le \frac{48}{10} \implies \boxed{ 1.4 \le y \le 4.8 }


    Q10:

    (i)
    (x-5)^2+(y+2)^2=25 \implies (x-5)^2 + (y-(-2))^2 = 5^2
    \text{Diameter} = 2 \times \text{radius} = 2 \times 5 = \boxed{10}
    \boxed{C(5, -2)}

    (ii)

    Find the gradient of the line
    \displaystyle m = \frac{2+2}{7-5} = \frac42 = 2
    Use y - y_1 = m(x-x_1)
    y - 2 = 2(x-7) \implies y-2 = 2x -14 \implies \boxed{y =2x -12}

    (iii)
    Use d^2 = (y_2-y_1)^2 + (x_2-x_1)^2
    |CP| = \sqrt{(5-7)^2 + (-2-2)^2} = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4 +16} = \sqrt{20} = 2\sqrt5 \implies \boxed{|CP| = 2 \sqrt5}
    2\sqrt5 &lt; 5 hence P lies inside circle.

    (iv)

    Sub in y = 2x in (x-5)^2+(y+2)^2=25
    (x-5)^2 + (2x+2)^2 = 25 \implies x^2 -10x + 25 +4x^2 +8x +4 = 25 \implies 5x^2 -2x+4=0
    \text{Discriminant} = b^2 - 4ac = (-2)^2 - 4 \times 5 \times 4 = 4 - 80 = -76 &lt; 0
    The equation has no roots, hence it doesn't meets the circle.

    I haven't revised my post, if there is any mistake please tell me.
    Post rating:
    1
  20. raheem94's Avatar
    340 17-05-2012  14:31
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    Re: Mr M's OCR (not OCR MEI) Core 1 answers May 2012
    (Original post by pzoDe)
    ...
    Oh i didn't saw that you are also doing the solutions!
    I was in the process of typing my post that time, hence didn't noticed it.
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