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# Edexcel C3 (20/06) Revision Thread watch

1. (Original post by TheWolf)
Indeed. Btw, is it just me or does paper A3's first q not show up?
yeah, some of the questions are missing. I think that's the only one.
2. Im trying to figure out sin 3x = cos x now....
3. The normal to the curve at the point Q, with x coordinate q, passes through the origin. Show that x = q is a solution to the equation x² + ln3x = 0

can anybody help me with this please?!?!
4. i've found a mistake on markscheme for c3 paper 4.
It says 3² = 3, λ should be 5/3 not 5.
5. (Original post by luketennisboy)
The normal to the curve at the point Q, with x coordinate q, passes through the origin. Show that x = q is a solution to the equation x² + ln3x = 0

can anybody help me with this please?!?!
Not enough information. Can you give the full question?
6. It doesnt show up.
I need help with this.

f(x) = 1/(x+1)(x-1)

What is the range? The answers just say the range of f(x) is y where y>0
7. (Original post by sb1986)
It doesnt show up.
I need help with this.

f(x) = 1/(x+1)(x-1)

What is the range? The answers just say the range of f(x) is y where y>0
f(x) = 1/(x²-1)
f(x) = u/v
=> f'(x) = (v.du/dx - u.dv/dx)/v²
f'(x) = [(x²-1) x 0 - 1(2x)]/(x²-1)²
f'(x) = -2x/(x²-1)²

As x -> infinity, f'(x) --> zero
Therefore range is f(x) > 0

You can also see from the equation that as x becomes very large and positive, f(x) > 0 so doesn't cross the x-axis.
8. ok now can someone explain how to find the range, i didnt know there was a specific method i wasnt taught. Iv been guessing all this time man.

by the way thanks for sortin me out on that range question
9. (Original post by sb1986)
It doesnt show up.
I need help with this.

f(x) = 1/(x+1)(x-1)

What is the range? The answers just say the range of f(x) is y where y>0
That's right because for all values of x, y will always be positive, because denominator is always positive.
10. (Original post by Seth)
That's right because for all values of x, y will always be positive, because denominator is always positive.
what about f(1/2) = -3/4
11. (Original post by mik1w)
what about f(1/2) = -3/4
If I remember correctly, the domain of the function is x>1
12. (Original post by mik1w)
what about f(1/2) = -3/4
Sorry, my mistake still half asleep im afraid.. well spotted
13. I'v been tryin dat sin3x = cosx for 30 mins now lol, i ain gettin anywhere
14. Hi there,

I'm having a bit of trouble with Question 6a) Page 52.

f(x)≡e^0.8x - 1 / 3-2x

a) Show that the equation f(x)=0 can be written as x = 1.5 - 0.5e^-0.8x

Ive been trying it for a long time now but I can't do it. Maybe i'm missing something

Thanks In Advance
15. (Original post by Widowmaker)
If I remember correctly, the domain of the function is x>1
In that case, what I said above applies...

cheers
16. (Original post by PoseidonX)
Hi there,

I'm having a bit of trouble with Question 6a) Page 52.

f(x)≡e^0.8x - 1 / 3-2x

a) Show that the equation f(x)=0 can be written as x = 1.5 - 0.5e^-0.8x

Ive been trying it for a long time now but I can't do it. Maybe i'm missing something

Thanks In Advance
If f(x) = 0
e^0.8x - 1/3-2x = 0
e^0.8x = 1/3-2x
1/e^0.8x = 3 - 2x
e^-0.8x = 3 - 2x
3 - e^-0.8x = 2x
3/2 - 0.5e^-0.8x = x

so x = 1.5 - 0.5e^-0.8x
17. (Original post by PoseidonX)
Hi there,

I'm having a bit of trouble with Question 6a) Page 52.

f(x)≡e^0.8x - 1 / 3-2x

a) Show that the equation f(x)=0 can be written as x = 1.5 - 0.5e^-0.8x

Ive been trying it for a long time now but I can't do it. Maybe i'm missing something

Thanks In Advance
e^0.8x - 1 / 3-2x = 0
e^0.8x = 1 / 3-2x
e^0.8x = (3-2x)^-1
e^-0.8x = 3-2x (inverse of both sides)
-3+e^0.8x = -2x
x = 1.5 -0.5e^-0.8x

yep featherflare got to it first...
18. (Original post by PoseidonX)
Hi there,

I'm having a bit of trouble with Question 6a) Page 52.

f(x)≡e^0.8x - 1 / 3-2x

a) Show that the equation f(x)=0 can be written as x = 1.5 - 0.5e^-0.8x

Ive been trying it for a long time now but I can't do it. Maybe i'm missing something

Thanks In Advance
e^0.8x = 1 / 3-2x
e^0.8x(3-2x) = 1
3e^0.8x - 2xe^0.8x = 1
3 - 2x = 1 / e^0.8x
3 - e^-0.8x = 2x
1.5 - 0.5e^-0.8x = x

Beaten to the punch...
19. Can someone please help with this:
[1-cos2x]/{[1-cos2x]/sin2x}

Does this cancel into just sin2x?

I always get confused with this, is there some easy way of remembering how things cancel? thanks
20. Yup

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