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C2 24th May 2012 REVISION THREAD

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    All of these I consider to be relatively difficult questions and some I got stuck on, please answer these questions and show working.

    Thank you everyone!
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    pleaseeeee
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    okay, q.6:

    a) complete the square:
    (x-3)^2 -9 + (y+2)^2 -4 =12
    (x-3)^2 + (y+2)^2 =25
    centre at (3,-2)
    radius is root25 = 5 (+ve value because it's a length)

    b) P at (-1,1) and Q at (7,-5)
    midpoint PQ should be (3,-2) if a diameter
    midpoint= (7-1/2, -5-1/2) = (3,-2)

    c) I'm not really sure how you'd do this. There's a right-angled triangle between RPQ and you can draw the whole thing onto a set of axes but although R is (0,y), I'm not sure how you'd find the y coordinate
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    (Original post by jameshewitt)
    All of these I consider to be relatively difficult questions and some I got stuck on, please answer these questions and show working.

    Thank you everyone!
    For the second question, the (ii):

    4sinx=3tanx

    For this question you have to know the trigonometric identity that tanx is equal to sinx/cosx. So we substitute tanx for sinx/cosx, which transforms the equation into:

    4sinx = 3(sinx/cosx)

    We then multiply both sides by cosx:

    4sinx(cosx) = 3 sinx

    There is sinx on both sides so we can cancel them out so the equation becomes:

    4cosx = 3

    Then we use our calculator to find the value of x so this becomes:

    x = 41.4 (1 decimal place)

    Then I think there are various ways to do this next step, but I use a CAST graph. I could draw it out if you want me to, but I'm assuming you know what this is/know what to do as this stage.

    So the answer is 41.4 and 318.6 degrees.

    If you have any queries, quote me.
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    (Original post by jameshewitt)
    All of these I consider to be relatively difficult questions and some I got stuck on, please answer these questions and show working.

    Thank you everyone!
    For the third question, question 8 about logs:

    Part a:

    You have to know the log rule:

    logan = x is the same as ax = n

    So if we apply this rule to log2y = -3 then this becomes:

    2-3=y

    Work out 2-3 and we get the value for y and hence the answer to this question.

    The value of y = 0.125


    Part b:

    Firstly, I would multiply both sides by log2x to get rid of that fraction so the equation becomes:

    log232 + log216 = (log2x)2

    For this question you will have to know the log multiplication rule: logax + logay = logaxy

    So the equation becomes:

    log2(32*16) = (log2x)2

    Simplifying it:

    log2(512) = (log2x)2

    We then square root both sides. You can find the square root of log2(512) on your calculator, which equals 3 and -3 so the equation becomes:

    3 = log2x and -3 = log2x

    We can then apply the rule from part a: logan = x is the same as ax = n

    So the equation then becomes:

    23 = x and 2-3 = x

    And the final answer:

    x = 8 and x = 1/8


    Note that you can check if your answers are correct by substituting your answer back into the equation.
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    (Original post by MrJames16)
    For the third question, question 8 about logs:

    Part a:

    You have to know the log rule:

    logan = x is the same as ax = n

    So if we apply this rule to log2y = -3 then this becomes:

    2-3=y

    Work out 2-3 and we get the value for y and hence the answer to this question.

    The value of y = 0.125


    Part b:

    Firstly, I would multiply both sides by log2x to get rid of that fraction so the equation becomes:

    log232 + log216 = (log2x)2

    For this question you will have to know the log multiplication rule: logax + logay = logaxy

    So the equation becomes:

    log2(32*16) = (log2x)2

    Simplifying it:

    log2(512) = (log2x)2

    We then square root both sides. You can find the square root of log2(512) on your calculator, which equals 3 so the equation becomes:

    3 = log2x

    We can then apply the rule from part a: logan = x is the same as ax = n

    So the equation then becomes:

    23 = x

    And the final answer:

    x = 8


    Note that you can check if your answers are correct by substituting your answer back into the equation.
    There are two answers to the 2nd part,  x =\dfrac18 \ and \ x = 8

    I will do it in this way,
     \displaystyle \frac{log_2 32 + log_2 16}{log_2 x} = log_2 x

     log_2 32 + log_2 16 = log_2 2^5 + log_2 2^4 = 5log_2 2 +4 log_22 = 5+4=9

     \displaystyle \frac{log_2 32 + log_2 16}{log_2 x} = log_2 x \implies \frac{9}{log_2 x} = log_2 x \implies 9 = (log_2 x)^2 \\ \implies log_2 x = \pm \sqrt9 \implies log_2 x = \pm 3

     log_2 x =3 gives  x = 2^3 = 8
     log_2 x = -3 gives  x = 2^{-3} = \dfrac18

    So  \displaystyle \boxed{ x = 8 \ and \ \frac18 }
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    (Original post by raheem94)
    There are two answers to the 2nd part,  x =\dfrac18 \ and \ x = 8

    I will do it in this way,
     \displaystyle \frac{log_2 32 + log_2 16}{log_2 x} = log_2 x

     log_2 32 + log_2 16 = log_2 2^5 + log_2 2^4 = 5log_2 2 +4 log_22 = 5+4=9

     \displaystyle \frac{log_2 32 + log_2 16}{log_2 x} = log_2 x \implies \frac{9}{log_2 x} = log_2 x \implies 9 = (log_2 x)^2 \\ \implies log_2 x = \pm \sqrt9 \implies log_2 x = \pm 3

     log_2 x =3 gives  x = 2^3 = 8
     log_2 x = -3 gives  x = 2^{-3} = \dfrac18

    So  \displaystyle \boxed{ x = 8 \ and \ \frac18 }

    Damn, I always forget about the negative solution when square rooting lol. Thanks for that
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    (Original post by jameshewitt)
    All of these I consider to be relatively difficult questions and some I got stuck on, please answer these questions and show working.

    Thank you everyone!
    (Original post by ohtasha)
    okay, q.6:

    a) complete the square:
    (x-3)^2 -9 + (y+2)^2 -4 =12
    (x-3)^2 + (y+2)^2 =25
    centre at (3,-2)
    radius is root25 = 5 (+ve value because it's a length)

    b) P at (-1,1) and Q at (7,-5)
    midpoint PQ should be (3,-2) if a diameter
    midpoint= (7-1/2, -5-1/2) = (3,-2)

    c) I'm not really sure how you'd do this. There's a right-angled triangle between RPQ and you can draw the whole thing onto a set of axes but although R is (0,y), I'm not sure how you'd find the y coordinate
    For the last part, draw a diagram.

    Here is the diagram:




     |PR|^2 = (0+1)^2 + (y-1)^2 = y^2 -2y+2 \\ |RQ|^2 = (7-0)^2 + (-5-y)^2 = y^2 +10y +74

    As it is a right angled triangle, so applying Pythagoras theorem gives,  |PQ|^2 = |PR|^2 + |RQ|^2 \\ 10^2 = y^2-2y+2 + y^2 +10y+74 \implies y^2+4y-12=0 \\ \implies (y-2)(y+6)=0 \implies y= 2 \ or  -6 \implies \boxed{y=2}
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    (Original post by MrJames16)
    For the second question, the (ii):

    4sinx=3tanx

    For this question you have to know the trigonometric identity that tanx is equal to sinx/cosx. So we substitute tanx for sinx/cosx, which transforms the equation into:

    4sinx = 3(sinx/cosx)

    We then multiply both sides by cosx:

    4sinx(cosx) = 3 sinx

    There is sinx on both sides so we can cancel them out so the equation becomes:

    4cosx = 3

    Then we use our calculator to find the value of x so this becomes:

    x = 41.4 (1 decimal place)

    Then I think there are various ways to do this next step, but I use a CAST graph. I could draw it out if you want me to, but I'm assuming you know what this is/know what to do as this stage.

    So the answer is 41.4 and 318.6 degrees.

    If you have any queries, quote me.
    I didn't think to cancel sinx each side, thank you for your help!
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    (Original post by MrJames16)
    For the third question, question 8 about logs:

    Part a:

    You have to know the log rule:

    logan = x is the same as ax = n

    So if we apply this rule to log2y = -3 then this becomes:

    2-3=y

    Work out 2-3 and we get the value for y and hence the answer to this question.

    The value of y = 0.125


    Part b:

    Firstly, I would multiply both sides by log2x to get rid of that fraction so the equation becomes:

    log232 + log216 = (log2x)2

    For this question you will have to know the log multiplication rule: logax + logay = logaxy

    So the equation becomes:

    log2(32*16) = (log2x)2

    Simplifying it:

    log2(512) = (log2x)2

    We then square root both sides. You can find the square root of log2(512) on your calculator, which equals 3 and -3 so the equation becomes:

    3 = log2x and -3 = log2x

    We can then apply the rule from part a: logan = x is the same as ax = n

    So the equation then becomes:

    23 = x and 2-3 = x

    And the final answer:

    x = 8 and x = 1/8


    Note that you can check if your answers are correct by substituting your answer back into the equation.
    didn't think +/-3 just assumed +3 hence i got 1 value of x. Thank you again sir!
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    (Original post by raheem94)
    For the last part, draw a diagram.

    Here is the diagram:




     |PR|^2 = (0+1)^2 + (y-1)^2 = y^2 -2y+2 \\ |RQ|^2 = (7-0)^2 + (-5-y)^2 = y^2 +10y +74

    As it is a right angled triangle, so applying Pythagoras theorem gives,  |PQ|^2 = |PR|^2 + |RQ|^2 \\ 10^2 = y^2-2y+2 + y^2 +10y+74 \implies y^2+4y-12=0 \\ \implies (y-2)(y+6)=0 \implies y= 2 \ or  -6 \implies \boxed{y=2}
    part (c) was the bit i couldn't do, "R is on the positive y axis" i thought this meant positive quadrant rather than literally on the axis but i understand now, thank you!
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    (Original post by jameshewitt)
    Also Click image for larger version. 

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ID:	149696 please help
     y =sin(ax-b)

    We know the coordinates of P, Q and R.

    Sub in P,
     \displaystyle 0 = sin \left( \frac{\pi a }{10} - b \right) \implies \frac{\pi a }{10} - b = 0 \implies \frac{\pi a }{10} = b \longrightarrow (1)

    Sub in Q,
    \displaystyle 0 = sin \left( \frac{3 \pi a }{5} - b \right) \implies \frac{3 \pi a }{5} - b = \pi \longrightarrow (2)

    Sub in R,
    \displaystyle 0 = sin \left( \frac{11 \pi a }{10} - b \right) \implies \frac{11 \pi a }{10} - b = 2 \pi \longrightarrow (3)

    (3) - (2) gives,
     a=2
    Sub 'a=2' in (1), you will get,  b = \dfrac{\pi}5
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    (Original post by raheem94)
     y =sin(ax-b)

    We know the coordinates of P, Q and R.

    Sub in P,
     \displaystyle 0 = sin \left( \frac{\pi a }{10} - b \right) \implies \frac{\pi a }{10} - b = 0 \implies \frac{\pi a }{10} = b \longrightarrow (1)

    Sub in Q,
    \displaystyle 0 = sin \left( \frac{3 \pi a }{5} - b \right) \implies \frac{3 \pi a }{5} - b = \pi \longrightarrow (2)

    Sub in R,
    \displaystyle 0 = sin \left( \frac{11 \pi a }{10} - b \right) \implies \frac{11 \pi a }{10} - b = 2 \pi \longrightarrow (3)

    (3) - (2) gives,
     a=2
    Sub 'a=2' in (1), you will get,  b = \dfrac{\pi}5
    All makes sense now! Thank you, hopefully i'll remember to apply this on Thursday
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    (Original post by jameshewitt)
    All makes sense now! Thank you, hopefully i'll remember to apply this on Thursday
    Good luck for the exam
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    (Original post by MrJames16)
    For the second question, the (ii):

    4sinx=3tanx

    For this question you have to know the trigonometric identity that tanx is equal to sinx/cosx. So we substitute tanx for sinx/cosx, which transforms the equation into:

    4sinx = 3(sinx/cosx)

    We then multiply both sides by cosx:

    4sinx(cosx) = 3 sinx

    There is sinx on both sides so we can cancel them out so the equation becomes:

    4cosx = 3

    Then we use our calculator to find the value of x so this becomes:

    x = 41.4 (1 decimal place)

    Then I think there are various ways to do this next step, but I use a CAST graph. I could draw it out if you want me to, but I'm assuming you know what this is/know what to do as this stage.

    So the answer is 41.4 and 318.6 degrees.

    If you have any queries, quote me.
    Hey, this is how I've done it and now seeing your method I'm not sure if mine is right. Can you check mine?

    4sin(x) = 3tan(x) which is the same as:

    4sin(x) = 3sin(x) / cos(x)
    Multiply by cos(x) which gives: 4sin(x)cos(x) = 3sin(x)
    Subtract the 3sin(x), this gives: 4sin(x)cos(x) - 3sin(x) = 0

    Simplify by collecting the sin's together: sin(x)(4cos(x)-3) = 0

    Now use CAST or the graph method to work it out so..

    Sin(x) = 0 is 0 so the answers for this are 0 degrees and 180 degrees (its also 360, -360 and -180 but these are outside the range they ask for).

    Then solve this part of the equation: 4cos(x)-3 = 0

    Cos(x) = 3/4 is 41.4 degrees and the answers that fit the range are 41.4 and 318.6.

    So all my answers are: 0, 41.4, 180 and 318.6.

    I understand why you divided the sin(x) as that's what I used to do until I looked at my revision book. So does anyone know which is correct?
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    Guys you can't discuss the exam yet!!!!!!!


    This was posted from The Student Room's iPhone/iPad App
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    (Original post by Bantersaurus Rexx)
    Guys you can't discuss the exam yet!!!!!!!


    This was posted from The Student Room's iPhone/iPad App
    Huh? But they're discussing questions from past papers


    This was posted from The Student Room's iPhone/iPad App

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