Year 13 Maths Help Thread

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    (Original post by Ano123)
    How did you solve and find the value of sin18 ?
    Second image, pi/10 radians = 18 degrees.
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    (Original post by RDKGames)
    Second image, pi/10 radians = 18 degrees.
    Yeah, but what I'm saying is how did you show that sin18 was the solution to the equation?
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    Please could someone check if im doing this right (c3 ques) - i dont have the answers :/

    f(x) = 2x - 3

    ff(2) = (2x-3)x-3

    then simplify
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    (Original post by kiiten)
    Please could someone check if im doing this right (c3 ques) - i dont have the answers :/

    f(x) = 2x - 3

    ff(2) = (2x-3)x-3

    then simplify
    Should be ff(x)=2(2x-3)-3
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    (Original post by kiiten)
    Please could someone check if im doing this right (c3 ques) - i dont have the answers :/

    f(x) = 2x - 3

    ff(2) = (2x-3)x-3

    then simplify
    No. The 2 within the brackets is what you plug into the equation, just like f(2)=1. ff(x) would require you to replace x with (2x-3), then plug in 2.
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    (Original post by Ano123)
    Yeah, but what I'm saying is how did you show that sin18 was the solution to the equation?
    Now that you mention it, I'm not sure. In the process I just found value of sin(x) and just arcsine'd it using the calc, which gave me pi/10, but that kinda contradicts the later part. How does one work out arcsin(x) by hand?
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    (Original post by RDKGames)
    No. The 2 within the brackets is what you plug into the equation, just like f(2)=1. ff(x) would require you to replace x with (2x-3), then plug in 2.
    Ohh so 4x-9 then plug in 2 which is -1 ?
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    (Original post by RDKGames)
    Now that you mention it, I'm not sure. In the process I just found value of sin(x) and just arcsine'd it using the calc, which gave me pi/10, but that kinda contradicts the later part. How does one work out arcsin(x) by hand?
    I should have said that you can't use a calculator.
    It's more the case of solving  \cos 3x=\sin 2x, \ \ 0<x<90^{\circ} , showing that it is a solution, and then forcing an exact expression for sin18 as you did correctly, but you just overlooked a step.
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    (Original post by kiiten)
    Ohh so 4x-9 then plug in 2 which is -1 ?
    That's correct.
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    (Original post by Ano123)
    That's correct.
    Thanks
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    (Original post by kiiten)
    Thanks
    Alternatively, since you are given a value to plug in, rather than doing 2(2x-3)-3, you can just find f(2), then plug that answer through the function again. It's faster this way with more compound functions involved.

    f(2)=1, then f(1)= -1

    Expanding brackets and whatnot is only really needed when you are required to find a funcion of a funcion of a function... and so on with a general variable rather than a value.
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    (Original post by Zacken)
    Here we go again.
    Since you were the top poster in the Year 12 Help Thread, am I allowed to add your name to the Year 13 Maths Help Thread helpers list? I know that you have university maths to do next year so if you are too busy, do let me know.
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    (Original post by Palette)
    Since you were the top poster in the Year 12 Help Thread, am I allowed to add your name to the Year 13 Maths Help Thread helpers list? I know that you have university maths to do next year so if you are too busy, do let me know.
    Yes, my post was a unconventional way of saying "I'll be here as a helper.".
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    (Original post by Zacken)
    Yes, my post was a unconventional way of saying "I'll be here as a helper.".
    Added

    I don't know why but I am so far finding the questions in the C3 textbook much easier than the ones in the C2 textbook when I did it in Year 12.
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    Planning to teach myself M2 over this summer and possibly FP1
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    This is an SMC problem which I solved, but I want to know if there is a way of solving it non-manually:

    Let  f(x)= \frac{x-1}{x+1}. Calculate  f^6(x).
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    (Original post by Palette)
    This is an SMC problem which I solved, but I want to know if there is a way of solving it non-manually:

    Let  f(x)= \frac{x-1}{x+1}. Calculate  f^6(x).
    Split function to give  f(x)=\frac{x+1}{x+1} -\frac{2}{x+1} = 1 - 2(x+1)^{-1} . From there it's just differentiation. Just manual - I hope you did not use quotient rule.

    You could consider that if  f(x)=(x+1)^{-1}, \ \ n>0 then  f^n (x) = (-1)^n \ n! (x+1)^{-n-1} .
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    (Original post by B_9710)
    Split function to give f(x)=\frac{x+1}{x+1}-\frac{2}{x+1} =1-2(x+1)^{-1} . From there it's just differentiation. Just manual - I hope you did not use quotient rule.
    Using the chain rule, f'(x)=2(x+1)^{-2} but I am confused as to why an SMC question expects you to use calculus. But again, this is from a 1997 paper so the syllabus may have changed.
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    (Original post by Palette)
    Using the chain rule, f'(x)=2(x+1)^{-2} but I am confused as to why an SMC question expects you to use calculus. But again, this is from a 1997 paper so the syllabus may have changed.
    What is SMC?
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    (Original post by B_9710)
    What is SMC?
    UK Senior Mathematics Challenge held annually to Year 12 and Year 13 students.
 
 
 
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