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Anyone need help with any *EDEXCEL* C3 and C4 past paper questions? watch

1. which formulaes do i need memorise for trigonometry (c3)?
ive done the double angle formula,
cot2 and cosec2,
tan2 and sec2,
sin2 and cos2...
anymore?
2. (Original post by obviouslystudying)
which formulaes do i need memorise for trigonometry (c3)?
ive done the double angle formula,
cot2 and cosec2,
tan2 and sec2,
sin2 and cos2...
anymore?
secx = 1/cosx

cosecx=1/sinx

cotx = 1/tanx
3. (Original post by gottastudy)
secx = 1/cosx

cosecx=1/sinx

cotx = 1/tanx
yeah, know them but sort of forgot since i think of them as a sort of definition for sec, cosec and cot.
thanks
4. Q5. part d)

Find the exact values of x for which |2/x-3| = 3

not sure how they obtained 3 2/3 and 2 1/3 fully.
i apprently lose marks if i multiply the 3 with x-3.
help?

thanks!
5. (Original post by Dyslex!c Duck)
Q5. part d)

Find the exact values of x for which |2/x-3| = 3

not sure how they obtained 3 2/3 and 2 1/3 fully.
i apprently lose marks if i multiply the 3 with x-3.
help?

thanks!
Getting rid of the modulus signs gives:

and

These can be solved individually to find both values of x.
6. sorry, my pc is freezing needs a format
7. AH ermm doesn't matter since i am doing AQA + its in june :P , the topic which i was iffy about was integration by substitution but i am alright now .
8. (Original post by sulexk)
Hello there,

need help with any edexcel c3/c4 past paper questions?

let me know

thank you so much!!!
Mate how do you solve the equation sin5x+sinx=0 for x in the interval between 0 and 180 degrees?? Thank you
9. (Original post by sulexk)
Hello there,

need help with any edexcel c3/c4 past paper questions?

let me know

thank you so much!!!

January 2007:

Given that: y = arccos x, -1 < x < 1 and 0 < y < pi,

(a) express arcsin x in terms of y. (2)

Edit: I know the answer to part (a), but I have no idea at all how to get the answer.

(b) Hence evaluate arccos x + arcsinx. Give your answer in terms of pi. (1)

________

10. (Original post by ViolinGirl)
You do-you start off by using the product rule.

y=uv where u= x and v = e^x^2

when you differentiate these you should get: du/dx = 1 and dv/dx= 2xe^x^2

Then dy/dx = vdu/dx + udv/dx

e^x^2 + 2x^2e^x^2

This can then be simplified.

As far as I can see, it only really involves the product rule. I may have omitted something. What is the answer?
yeah, thats how its shown, i just cant get dv/dx
well, how did you get that???
11. (Original post by ViolinGirl)
You do-you start off by using the product rule.

y=uv where u= x and v = e^x^2

when you differentiate these you should get: du/dx = 1 and dv/dx= 2xe^x^2

Then dy/dx = vdu/dx + udv/dx

e^x^2 + 2x^2e^x^2

This can then be simplified.

As far as I can see, it only really involves the product rule. I may have omitted something. What is the answer?

ok dv/dx

v = e^x^2

First you differentiate the power, with the chain rule, pretty much normal differentitation.

Power is x^2 ...this becomes 2x

But the rule with e^x is that dy/dx is e^x

so it becomes 2xe^x^2

Another example to make it clearer:

e^2x would become 2e^2x

Diferentiate power, and its that multiplies by e^ the orginial power.

Hope that helps.
12. (Original post by obviouslystudying)
well, how did you get that???
Above ^^
13. (Original post by obviouslystudying)
How do you differentiate when the base has got powers thats also got powers;
like: ?
This seems harder than C3 - but I think I figured it out.

Derivative of
therefore if we can find the derivative of , we can do the rest pretty easily.
So, is a bit tricky, but as far as I can see,
therefore we can differentiate using the rule above and the product rule:

Therefore;

I think

:P
14. (Original post by Mathematician!)
Getting rid of the modulus signs gives:

and

These can be solved individually to find both values of x.
silly me.
i solved it wrong before.
i basically moved the x-3 over to the 3
then divided the 2 and -2 with 3 whic is wrong.
haaa
thanks!
(:
15. (Original post by 4WhenImBanned)

January 2007:

Given that: y = arccos x, -1 < x < 1 and 0 < y < pi,

(a) express arcsin x in terms of y. (2)

Edit: I know the answer to part (a), but I have no idea at all how to get the answer.

(b) Hence evaluate arccos x + arcsinx. Give your answer in terms of pi. (1)

________

For this question you basically need to know cosx = sin (pi/2 -y) therefore arcsin = (pi/2 -y)

Then knowing this arc sinx + arccos x would be (pi/2 -y) + y = pi/2

----

Can anyone answer my previous question?

1. Using this as my dy/dx = -e^2 x^-2 +e^x (so in words: minus e to the power of two multiplied by x to the power of minus two plus e to the power of x)

How do I get the gradient of 3/4e^2 like it says in the MS if I insert in x=2?

2. For a cosec graph with a minimum at (pi/2, −1), a maximum at ( 3pi/2 , −5) and an asymptote with equation x = pi.

(a) Showing the coordinates of any stationary points, sketch the graph of y =
Given that
f : x → a + b cosec x, x ∈ , 0 < x < 2π, x ≠ π,
(b) find the values of the constants a and b

16. For 1) probably get a better response if you post the question. I don't have the paper.

2) Again think you need to type this out abit more clearly, y = ... ? is that (a+b) or a + bcosecx ? I'm guessing this is a case of a transformation of some kind.
17. Ahh I'm sorry I didn't realise it didn't paste the rest of the equation!

1) http://www.mathsandscience.org/resources/EC3sh_H.pdf

Question 8c.

2)This is my second question (Q4b) :

http://www.mathsandscience.org/resources/EC3sh_H.pdf

18. (Original post by Apple24)
For this question you basically need to know cosx = sin (pi/2 -y) therefore arcsin = (pi/2 -y)

Then knowing this arc sinx + arccos x would be (pi/2 -y) + y = pi/2
Hi, thanks but how do you know cosx = sin (pi/2 - y)?
19. (Original post by Toneh)
This seems harder than C3 - but I think I figured it out.

Derivative of
therefore if we can find the derivative of , we can do the rest pretty easily.
So, is a bit tricky, but as far as I can see,
therefore we can differentiate using the rule above and the product rule:

Therefore;

I think

:P
power of powers is in the c3 heinemann book

thats the rule i was searching for. thanks alot

the violin girl too, thanks
20. (Original post by obviouslystudying)
power of powers is in the c3 heinemann book

thats the rule i was searching for. thanks alot
o yer, powers of powers is alright, but x^y where y is not a constant is harder :P

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