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# Edexcel C3 (20/06) Revision Thread watch

1. You're allowed to multiply top and bottom by the same number, leaving the RHS alone. It still works. Take a simpler example:

1/2 = a

If we multiply top and bottom of the LHS by 2, we get:

2/4 = a

Obviously, that's the same thing.

If you only want to multiply the top by 2, let's say, then you have to do both sides:

1/2 = a

2/2 = 2a
1 = 2a

Once again, we'll arrive to the same equation.
2. hi

how do u differenciate : cot²x ???

HELP!!
3. Can anyone who's done the A1 paper help me please?

Q5: Why can't the answer be simplified more to (x-3)/x(x-2) ?

Q6a: I get 225/32 but the answer is 255/32... How?

Q6c: How do you show it is an increasing function?

4. smatz: I think you would do:

cot²x = (cotx)² and use chain rule

2cotx*-cosec²x
-2cotxcosec²x

Anyone tell me if this is right? :/
5. (Original post by smatz01)
hi

how do u differenciate : cot²x ???

HELP!!
We use the chain rule.
6. Just to let people know there are 2 mistakes in the projectmaths c2 markscheme.
8. (Original post by SunGod87)
yep sorry.
9. (Original post by violet)
Can anyone who's done the A1 paper help me please?

Q5: Why can't the answer be simplified more to (x-3)/x(x-2) ?

Q6a: I get 225/32 but the answer is 255/32... How?

Q6c: How do you show it is an increasing function?

5. It can.

6. (a) You have done it wrong.

f ' ' (x) = 2x - 2x-3

so: f ' ' (4) = 8 - 2/64

=> 255/32

6. (c) There are a few ways.

Take f ' (x) and set it to less than zero and show there are no solutions.

Sketch the graph of f ' (x) showing that it is a tangent to the x axis so no solutions less than zero ==> not a decreasing function ==> an increasing function.
10. (Original post by samdavyson)
5. It can.

6. (a) You have done it wrong.

f ' ' (x) = 2x - 2x-3

so: f ' ' (4) = 8 - 2/64

=> 255/32

6. (c) There are a few ways.

Take f ' (x) and set it to less than zero and show there are no solutions.

Sketch the graph of f ' (x) showing that it is a tangent to the x axis so no solutions less than zero ==> not a decreasing function ==> an increasing function.
Erm, this might be a stupid question, but what is an increasing function?
I don't understand why when f ' (x) is less than 0 then it can't have no solutions.
Thanks for the other 2 btw
11. (Original post by violet)
Erm, this might be a stupid question, but what is an increasing function?
I don't understand why when f ' (x) is less than 0 then it can't have no solutions.
Thanks for the other 2 btw
If f(x) is an increasing function, its gradient f'(x) will be greater than zero for all values of x.

f'(x)
= x² - 2 + 1/x²
= (x-1/x)²

Factorising it in this form shows us that f'(x) > 0 for all values of x, since any number squared is positive.
12. (Original post by violet)
Erm, this might be a stupid question, but what is an increasing function?
I don't understand why when f ' (x) is less than 0 then it can't have no solutions.
Thanks for the other 2 btw
An increasing function as a fuction for which dy/dx > 0. [ C2 chapter 9 for more info. ]

(e.g. a line with a positive gradient).

Another answer that you might prefer is:

f ' (x) = x2 - 2 + 1/x2 = (x - 1/x)2

Square of anything ==> Positive.

So f ' (x) > or = to 0.

==> Increasing Function.
13. Hi,

Im stuck like nothing else with a c3 trig question...
Find, in terms of pie the solutions of the equation
sin5x+sinx=0
0<X<Pie
Any help would be much appreciated!
14. (Original post by Goody1)
Hi,

Im stuck like nothing else with a c3 trig question...
Find, in terms of pie the solutions of the equation
sin5x+sinx=0
0<X<Pie
Any help would be much appreciated!
sin5x + sinx
= sin(3x + 2x) + sin(3x-2x)
= sin3xcos2x + sin2xcos3x + sin3xcos2x - sin2xcos3x
= 2sin3xcos2x = 0

2sin3x = 0 OR cos2x = 0
sin3x = 0
3x = arcsin0, 180°-arcsin0, 360° + arcsin0
3x = 0°, 180°, 360°
x = 0°, 60°, 120°

cos2x = 0
2x = arccos0, 360° - arccos0,
2x = 90°, 270°
x = 45°, 135°

x = 45°, 60°, 120°, 135°

Ignore the values 0° and 180° since they aren't in range.
Unless the range is 0</=x</=pi
15. f ' (x) = x2 - 2 + 1/x2 = (x - 1/x)2

Is that completing the square?
16. (Original post by violet)
f ' (x) = x2 - 2 + 1/x2 = (x - 1/x)2

Is that completing the square?
factorising.

(x-1/x)(x-1/x)
= x² - x/x - x/x + 1/x²
= x² - 1 - 1 + 1/x²
= x² - 2 + 1/x²
17. (Original post by violet)
f ' (x) = x2 - 2 + 1/x2 = (x - 1/x)2

Is that completing the square?
It would be if it wasn't a perfect square.

But as it is there is nothing to "complete" (now addition or subtraction on the end) as it is a square.
18. (Original post by Widowmaker)
factorising.

(x-1/x)(x-1/x)
= x² - x/x - x/x + 1/x²
= x² - 1 - 1 + 1/x²
= x² - 2 + 1/x²
Oh I see!!!!
Ok so increasing function is when dy/dx is greater than 0... And decreasing is when it's less than 0 right?
19. (Original post by violet)
Oh I see!!!!
Ok so increasing function is when dy/dx is greater than 0... And decreasing is when it's less than 0 right?
yep, or in the range x = a and x = b
20. (Original post by violet)
Oh I see!!!!
Ok so increasing function is when dy/dx is greater than 0... And decreasing is when it's less than 0 right?
Yeah.

When it is equal to zero I think it is still classed as increasing. Someone confirm this??

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