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C3 January 2013 25/01/2013

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Reply 1880
Original post by TPJY
what do you do if you get a double transforation affecting the same thing?

like this: 2f(x)+3

or f(3x+5)

how do you know which order to do them in??? in the exams so far they have only given one of each, for example:

2f(x-5)
D:





you two seem to be contradicting each other :s


You would start with what is in the brackets first, hence, what is affecting x first. They aren't contradicting. Just think BIDMAS/BODMAS and apply the effect to the function in that order and you'll be fine
Reply 1881
Original post by lanky
If the question does not state how many decimal places or significant figures to answer to, what is the best amount to go for? :confused:


I always do three decimal places (on Rsin(theta+alpha) questions for example it has these in the ms)

Just to be safe if I'm using the numbers again i'll use the STO=>A button on my calculator though :P

also - make sure when it says "as an exact answer" it wants it in terms of ln or e or a root (not a rounded number basically), so they would want ln3 instead of 1.099
Original post by Henry.Lister
Someone help please! Question C. I get the wrong answer :/



3cos2x-sin2x = 0

tan2x=3

I'm sure you can work it out from there :smile:
Original post by Henry.Lister
Someone help
please! Question C. I get the wrong answer :/



Seems like they want you to use f'(x) in the interval 0<x<1 to see if there is a change in sign.
Original post by TPJY
what do you do if you get a double transforation affecting the same thing?

like this: 2f(x)+3

or f(3x+5)

how do you know which order to do them in??? in the exams so far they have only given one of each, for example:

2f(x-5)
D:


with 2f(x-5) the 2 is affecting the vertical and the 5 affects horizontal so it doesn't matter which order. As for 2f(x) + 3, think of it like this: if you first move it up by 3, then multiply by 2 that would be 2(f(x) + 3) which would be 2f(x) + 6. Which is wrong of course. So the only other option is to multiply by 2 and then do add 3 :smile:
Original post by JJMills
sin x / cos x = tan x
sin^2 x + cos^2 x = 1
1 / cos x = sec x
1 / sin x = cosec x
1 / tan x = cot x
sin 2x = 2sinxcosx
cos 2x = cos^2 x - sin^2 x = 1 - 2 sin^2 x = 2 cos^2 x - 1
tan 2x = 2tan x / (1 - tan^2 x)



thank you soo much!!!!! this is great! :biggrin:
Reply 1886
Original post by This Excellency
Seems like they want you to use f'(x) in the interval 0<x<1 to see if there is a change in sign.


It says find, not show that.

The interval is just indicating that you'll need to be using radians and to take and answer between 0 and 1.

3cos2x-sin2x = 0

tan2x=3

I'm sure you can work it out from there


Did i do A incorrectly? I used e^3x(3cos2x-2sin2x)=0

I therefore got 3cos2x-2sin2x=0
Reply 1888
Original post by Henry.Lister
Did i do A incorrectly? I used e^3x(3cos2x-2sin2x)=0

I therefore got 3cos2x-2sin2x=0


Your answer is correct.
Your answer is correct.


I don't understand where I've gone wrong then :frown:
3cos2x-2sin2x=0
3cos2x=2sin2x
3=2tanx
3/2=tanx
2x= 56. I need it between 0 and 1?
(edited 11 years ago)
What are the formulae in the book section 'trigonometric identities' beneath the top 3? Never used those! :redface:
Original post by Henry.Lister
- How do i do part c?

I tried linking the first differentiation with 0 but i don't get the right answer!


6a) use product rule to differentiate
b) product rule again to differentiate the differential
c) at the stationary point, dy/dx = 0, then solve
d)sub your x into d^2y/dx^2 and if its > 0 then min and if its < 0 then max
image201301250001.jpg
Reply 1892
Original post by Corvus
I have a quick question, does y= ln(x)+3 has a asymptote at y=3? Thanks


The curve y = ln(x) has an asymptote at x = 0. When the graph is translated by 3 units in the positive direction, the asymptote is still at x = 0 and y = 0.

For y = ln(x), x is a member of the reals and greater than 0. The range of y is infinite. It has no limits. So translating the graph upwards by 3 units will not mean the graph has no solutions when y = 3.

3 = ln(x) + 3
0 = ln(x)
x = 1

However, if the equation was y = ln(x) / (x - 3), there would be an asymptote at x = 3 because:

y = ln(3) / 0 = no value as a constant divided by 0 has not been mathematically defined yet.

This graph would thus have an asymptote at x = 3 and y = 0 as ln(0) doesn't exist or any negative values of ln(x).

(I graphed this here: https://www.desmos.com/calculator to make sure I was right :wink: )
Reply 1893
Original post by Henry.Lister
I don't understand where I've gone wrong then :frown:
3cos2x-2sin2x=0
3cos2x=2sin2x
3=2tanx
3/2=tanx
x= 56. I need it between 0 and 1?


Well it firstly should be tan 2x = 1.5 but I can only assume the interval they gave you was indicating radians.

arctan(1.5) / 2 = 0.491

Is that the answer?
Original post by Henry.Lister
I don't understand where I've gone wrong then :frown:
3cos2x-2sin2x=0
3cos2x=2sin2x
3=2tanx
3/2=tanx
2x= 56. I need it between 0 and 1?


use your calc in radians, check out my reply, I worked it out for you
Ahhh it was as the answer to c is in radians. I see, cheers for doing the calculations!

Thanks everyone :smile:

Just over an hour till exam time ... :wink:
Original post by Henry.Lister
I don't understand where I've gone wrong then :frown:
3cos2x-2sin2x=0
3cos2x=2sin2x
3=2tanx
3/2=tanx
2x= 56. I need it between 0 and 1?


sorry, made a mistake in my working, what I got for x should say 2x = 0.983 and therefore x = 0.983/2
Good luck to everyone doing this paper :smile::biggrin:
GOOD LUCK EVERYONE!!! IM SURE YOU ALL WILL GET AN A AND ABOVE! :smile:
:thumbs up:
"You have reached the limit of how many posts you can rate today!"

Not happening I'm afraid :frown:

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