Year 13 Maths Help Thread

Original post by Ano123

How did you solve and find the value of sin18 ?

Second image, pi/10 radians = 18 degrees.

(edited 7 years ago)

Reply 21

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?

Reply 22

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

Reply 23

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

Reply 24

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.

Reply 25

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?

Reply 26

kiiten

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 ?

Reply 27

Original post by RDKGames

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.

Reply 28

Original post by kiiten

Ohh so 4x-9 then plug in 2 which is -1 ?

That's correct.

Reply 29

kiiten

Original post by Ano123

That's correct.

Thanks

Reply 30

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.

(edited 7 years ago)

Reply 31

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.

Reply 32

Zacken

Original post by Palette

Yes, my post was a unconventional way of saying "I'll be here as a helper.".

Reply 33

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.

Reply 34

metrize

Planning to teach myself M2 over this summer and possibly FP1

Reply 35

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)

Reply 36

B_9710

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}

(edited 7 years ago)

Reply 37

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, [tex\displaystyle]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.

Reply 38

B_9710

Original post by Palette

Using the chain rule, [tex\displaystyle]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?

Reply 39