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    (Original post by FatimaHere)
    Couldn't do this one 🙂🙂🙂🙂🙂🙂🙃🙃 🙃🙃🙃


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    😟😟
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    (Original post by BethFM)
    Can you carry an error forward on core 2?
    Fairly sure you can i made mistake on the integration one to find area used the wrong limits but should still get marks
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    (Original post by SGHD26716)
    I got both 12 and -3, but said -3 is not valid since 2log base 3(-3) gives maths error. You could square the -3 but that's after manipulation and not the original equation. I really don't know. I wrote down 12 and -3 and then underlined 12
    As long as you didn't explicitly say -3 wasn't a valid answer, you should get the mark, assuming -3 is a valid answer (which i think it is, the equation is equivalent regardless of whether it's written in exactly the same way as the original)
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    Predicted Grade boundaries??
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    (Original post by MrCoolVille)
    Predicted Grade boundaries??
    Fairly easy paper, apart from question 9ii for me. I reckon around 59-60 because people make stupid mistakes as well. The difficulty was around the same as the June 14 paper
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    How were you meant to do the integration to find the area , was it with limits of 3 to -1 or do you do -1 to 0 and 0 to 3?
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    explanation of 8iv and 8v?? please please please ⭐️⭐️⭐️
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    To everyone who says that 9(i) was π/3a and 4π/3a, that is wrong.

    The real answer is π/3a and 7π/3a.

    This can be proven by substituting 7π/3a for x in the equation given;

    sin(ax)=√3cos(ax)
    sin(a*7π/3a)=√3cos(a*7π/3a)
    sin(7π/3)=√3cos(7π/3) [=√3/2]
    So 7π/3a is the correct answer.

    The mistake many people made is that the rearranged the equation, and ended up with tan(ax)=√3, and we know that tan is positive in the 1st and 3rd quadrant, so they put
    ax = tan-¹(√3) and π+tan-¹(√3)
    a = π/3a and 4π/3a.

    The first solution is right, as it's the principle value, but the second one is not, since the question wanted the first 2 positive answers for SIN(...)=...COS(...), not tan(...); I.E. the answers where both sin and cos are positive (the 1st and FIFTH quadrants)

    Therefore the solutions were:
    ax = tan-¹(√3) = π/3
    Principle value = π/3a
    x = π/3a and
    x=[period]+π/3a = (2π/a)+π/3a = 7π/3a

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    on 5bi) i misread the question and went on to work out two possible values for a, but i wrote down the correct answer before doing so, would i lose many marks?
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    what was the answer to the constant term question?
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    (Original post by anon326589)
    what was the answer to the constant term question?
    it was +/- sqrt(3)
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    (Original post by h3rmit)
    As long as you didn't explicitly say -3 wasn't a valid answer, you should get the mark, assuming -3 is a valid answer (which i think it is, the equation is equivalent regardless of whether it's written in exactly the same way as the original)
    However, when you use the equation as was given in the question (separated) then I think 12 would be the only valid answer by looking at the graph.
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    (Original post by CLGC98)
    I retook and would say that last years was similar in difficulty
    Okay fair enough
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    (Original post by abrahammurciano)
    To everyone who says that 9(i) was π/3a and 4π/3a, that is wrong.

    The real answer is π/3a and 7π/3a.

    This can be proven by substituting 7π/3a for x in the equation given;

    sin(ax)=√3cos(ax)
    sin(a*7π/3a)=√3cos(a*7π/3a)
    sin(7π/3)=√3cos(7π/3) [=√3/2]
    So 7π/3a is the correct answer.

    The mistake many people made is that the rearranged the equation, and ended up with tan(ax)=√3, and we know that tan is positive in the 1st and 3rd quadrant, so they put
    ax = tan-¹(√3) and π+tan-¹(√3)
    a = π/3a and 4π/3a.

    The first solution is right, as it's the principle value, but the second one is not, since the question wanted the first 2 positive answers for SIN(...)=...COS(...), not tan(...); I.E. the answers where both sin and cos are positive (the 1st and FIFTH quadrants)

    Therefore the solutions were:
    ax = tan-¹(√3) = π/3
    Principle value = π/3a
    x = π/3a and
    x=[period]+π/3a = (2π/a)+π/3a = 7π/3a

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    I guarantee the mark scheme will go with the tan ax method
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    please can someone explain 8iv???? also what's the answer to the last part of 9??
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    I think the first positive solution for the last question is 1/5 pi A and not 1/3 pi A.


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    (Original post by abrahammurciano)
    To everyone who says that 9(i) was π/3a and 4π/3a, that is wrong.

    The real answer is π/3a and 7π/3a.

    This can be proven by substituting 7π/3a for x in the equation given;

    sin(ax)=√3cos(ax)
    sin(a*7π/3a)=√3cos(a*7π/3a)
    sin(7π/3)=√3cos(7π/3) [=√3/2]
    So 7π/3a is the correct answer.

    The mistake many people made is that the rearranged the equation, and ended up with tan(ax)=√3, and we know that tan is positive in the 1st and 3rd quadrant, so they put
    ax = tan-¹(√3) and π+tan-¹(√3)
    a = π/3a and 4π/3a.

    The first solution is right, as it's the principle value, but the second one is not, since the question wanted the first 2 positive answers for SIN(...)=...COS(...), not tan(...); I.E. the answers where both sin and cos are positive (the 1st and FIFTH quadrants)

    Therefore the solutions were:
    ax = tan-¹(√3) = π/3
    Principle value = π/3a
    x = π/3a and
    x=[period]+π/3a = (2π/a)+π/3a = 7π/3a

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    I was thinking you were right but the period isnt 2pi over a. 2pi over a is the period of a sin ax graph not a tan ax graph
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    (Original post by iamDev)
    I was thinking you were right but the period isnt 2pi over a. 2pi over a is the period of a sin ax graph not a tan ax graph
    But the function was in terms of sine.
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    Can anyone remember what the sequences were?
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    (Original post by MrCoolVille)
    Can anyone remember what the sequences were?
    120*0.9^(n-1)

    and 3 + 1.5n (A = 5)
 
 
 
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