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Edexcel A2 C3 Mathematics 12th June 2015

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Original post by suyoof123
Thanks and you wouldnt put >/ (equals to ) or /< (equals to ) on this because if it were when equals or bigger/smaller than 0, then you would do that?.

Also originally I meant what happens if you have two mods on either side of the equation, would the squaring method still be viable.


no, you see if the roots of x fell below then it would have the more than or equal to sign as a proper inequality as:

a</=x</=b
is there a trick or a way of knowing how the assymptotes shift with an Ln or e^x graph?

i'm talking about the compex ones such as ln(5x+3)+4 or e^3X^2-3 +4
does anyone know how to do part a and bi? i dont understand the markscheme
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Original post by SeanFM
Think about cos and the values it takes between 0 and 2pi, and try to spot the values that only come up once, and then see how you can apply that to find the correct answer.



If you're asking about fg(1), then mod(-1) = 1.


Ok, so basically cos the values it can take are 0, 1 and -1 and for cos-1 0 there are two solutions in that range same for cos-1 1 but for cos-1 just 1 so that means k/root2 = 1? how come markscheme has it +/-?

and for 4b, but don't we get a solution of 1 and 5?
Original post by Ripper Phoenix
no, you see if the roots of x fell below then it would have the more than or equal to sign as a proper inequality as:

a</=x</=b


Got it , thanks.
Original post by adorablegirl1202
Ok, so basically cos the values it can take are 0, 1 and -1 and for cos-1 0 there are two solutions in that range same for cos-1 1 but for cos-1 just 1 so that means k/root2 = 1? how come markscheme has it +/-?

and for 4b, but don't we get a solution of 1 and 5?


Almost for q3 :smile: Take a careful look at the range they specify in the question.

for 4b, you're using that mod(x) = max(x,-x) instead of taking both cases (where one is positive and one is negative to give you 1 and 5 as you've done).

If you put in -1 into mod(x) then you get out 1. I think you only consider the positive and negative solutions when you're working backwards (so if f(x) = 9, then either x = 3 or -3). But as you're putting in -1 into 2mod(x) you only get one solution, which is 2 (then add 3).
Original post by Ripper Phoenix
is there a trick or a way of knowing how the assymptotes shift with an Ln or e^x graph?

i'm talking about the compex ones such as ln(5x+3)+4 or e^3X^2-3 +4


Well if lnx = f(x) then ln(5x+3) + 4 = f(5x+3) + 4 so x would be divided by 5, and shift 3 to the left, and y would move up by 4
Original post by Phenylethanone
does anyone know how to do part a and bi? i dont understand the markscheme


have u learnt triangles? like using the triangles to find how SOHCAHTOA are related so for this q
COS A= 3/4 hence you use the triangle to find what sinA is...

once you get that we know that SIN2A is 2SINACOSA

so you put in the sinA and cosA to get your answer :smile:

for part bi im sure you just open it up using the cos (A+ B) AND COS (A-B) identity for each and im sure it will equal to that..

try it and let me know otherwise i might just do it all for you on paper :smile:
Original post by Ripper Phoenix
have u learnt triangles? like using the triangles to find how SOHCAHTOA are related so for this q
COS A= 3/4 hence you use the triangle to find what sinA is...

once you get that we know that SIN2A is 2SINACOSA

so you put in the sinA and cosA to get your answer :smile:

for part bi im sure you just open it up using the cos (A+ B) AND COS (A-B) identity for each and im sure it will equal to that..

try it and let me know otherwise i might just do it all for you on paper :smile:


i did that, and i got sinA = 5/4 but it says sinA = -root7/4
What do I do from here??? Anyone?


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For the domain and range sometimes the mark scheme doesn't include x/f(x) is equal to all real numbers. I f you do include it do you loose marks?
Original post by Phenylethanone
i did that, and i got sinA = 5/4 but it says sinA = -root7/4


thats because look at the domain its from 270 to 360 so you cannot use 5/4, it will be neg root 7/4
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Original post by Ripper Phoenix
thats because look at the domain its from 270 to 360 so you cannot use 5/4, it will be neg root 7/4


how do you work that out from sinA = 5/4?
Original post by Phenylethanone
how do you work that out from sinA = 5/4?


sin A = root 7 / 4
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Original post by Ripper Phoenix
no its c3

also i have a strong feelin that we may be asked to draw modulus graphs like

y=IX+2I +2 or something like that lol

i'm paranoid



oh yh i just checked :P Got confused coz i was revising c3 and c4 together
Can some on help. Normally to find the range you have to draw the graph. However this question is 3 marks and they differentiate it and get x=2.

Is there a way that you can differentiate the function and find the range ?
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Original post by SeanFM
Almost for q3 :smile: Take a careful look at the range they specify in the question.

for 4b, you're using that mod(x) = max(x,-x) instead of taking both cases (where one is positive and one is negative to give you 1 and 5 as you've done).

If you put in -1 into mod(x) then you get out 1. I think you only consider the positive and negative solutions when you're working backwards (so if f(x) = 9, then either x = 3 or -3). But as you're putting in -1 into 2mod(x) you only get one solution, which is 2 (then add 3).


Thanks I got the first one in the end! Now for 4b if I understand correctly we sub it in and then you're saying we get -1 and since it's mod we make it 1. But if we were to work out f(x) then we find two values???
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Original post by Ripper Phoenix


Why did you square both sides? Does squaring get rid of the mod?

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