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# Edexcel C3 (20/06) Revision Thread watch

1. (Original post by Featherflare)
Yea iv done the unthinkable.. given up. I dont think it can be done with c1- c4 knowledge which is all i have
Its like trying to solve 5x = 3x.

I have a question which should be easy but I don't understand it.

(Original post by C3)
The function f is defined for positive real values of x by

f(x) = 12lnx - x^3/2

Write down the set of values of x for which f(x) is an increasing function of x
Since its the differentiation chapter I guess you have to find f'(x) and then what?

2. (Original post by ntrik)
Its like trying to solve 5x = 3x.

I have a question which should be easy but I don't understand it.

Since its the differentiation chapter I guess you have to find f'(x) and then what?

Yes you need to find f'(x) and do f'(x) > 0 as it is an increasing function (ie gradient is more than zero)
3. f'(x) = 12/x - (3/2)x^0.5
f'(x) > 0
12/x - (3/2)x^0.5 > 0
(3/2)x^0.5 < 12/x
x^(3/2) < 8 (cross multiply)
x< 8^(2/3)
x<4

sorry i only just saw the x values should be positive so
0<x<4
4. Thanks.

Heres another question I'm having trouble on if anyone wants to have a go.

(Original post by C3)
The maximum point on the curve with equation x√sinx, 0<x<180, is the point A. Show that the x-cordinate of point A satisfies the equation 2tanx + x = 0.
5. Aight..in terms of formulas, we only need to know these:

cos²x + sin²x = 1
sec²x = 1+tan²x
cosec²x = 1 + cot²x

and

the R formula for e.g. AcosX + BsinX = c i.e. tanσ = b/a R = √a²+b²

the double angles can be worked out using the sin(A+B) = sinAcosB + cosAsinB formula, which is given in the formula booklet.
6. We also need to know the qoutient rule aswell i think. Is there anything else we need to know?
7. Hmm I never thought of the quotient rule as a formula =o, cus then you have chain rule..etc
8. (Original post by TheWolf)
Hmm I never thought of the quotient rule as a formula =o, cus then you have chain rule..etc
yep

chain rule: dy/dx = dy/du x du/dx
product rule: if y = uv => dy/dx = u.dv/dx + v.du/dx
quotient rule: if y = u/v => dy/dx = [v.du/dx - u.dv/dx]/v²

dy/dx = 1/[dx/dy]
I get that theta can be equal to 0 and tan theta = -1
I just don't understand the last step they did, I forgot this stuff
Thanks
10. (Original post by TheWolf)
Hmm I never thought of the quotient rule as a formula =o, cus then you have chain rule..etc
Quotient rule is given.

[ although it is a fairly ugly form: loads of g(x) and f(x) ]
11. (Original post by violet)
I get that theta can be equal to 0 and tan theta = -1
I just don't understand the last step they did, I forgot this stuff
Thanks
You get: 2sin(x/2)[cos(x/2) + sin(x/2)] = 0

For this to be true either 2sin(x/2) = 0

OR cos(x/2) + sin(x/2) = 0

==> Two solutions.

The second give tan(x/2) = -1

As: sin(x/2) = -cos(x/2)

=> tan(x/2) = -1
12. (Original post by Featherflare)
Yea iv done the unthinkable.. given up. I dont think it can be done with c1- c4 knowledge which is all i have
Ahhh noo u shouldn't have given up...I think I have the answer...

sin3x=cosx
sin(2x+x)=cosx
sin2xcosx+cos2xsinx=cosx
(2sinxcosx)cosx+(1-2sin²x)sinx=cosx
2sinxcos2x+sinx-2sin³x=cosx
2sinx(1-2sin²x)+sinx-2sin³x=cosx
2sinx-4sin³x+sinx-2sin³x=cosx
(3sinx-6sin³x)²=(cosx)² <----(square both sides)
9sin²x-36sin^4 x+36sin^6 x=1-sin²x
36sin^6 x-36sin^4 x+10sin²x-1=0

let y=sin²x:
36y³-36y²+10y-1=0
gives 3 solutions...2 of which are complex, 1 which is real.

The real one is y=0.6282

so...

sin²x=0.6282
sinx= √0.6282
x = 52.4°

You'd be extremely unlucky to get a question like this,especially because complex numbers aren't in C3(at least I don't think so), took me a long time to solve and got lots of help from a further mathmo..
13. (Original post by nas7232)
We also need to know the qoutient rule aswell i think. Is there anything else we need to know?
Only sin = (diff) => cos

and cos = (diff) => -sin

[ the only two not given in the booklet ]
14. (Original post by samdavyson)
You get: 2sin(x/2)[cos(x/2) + sin(x/2)] = 0

For this to be true either 2sin(x/2) = 0

OR cos(x/2) + sin(x/2) = 0

==> Two solutions.

The second give tan(x/2) = -1

As: sin(x/2) = -cos(x/2)

=> tan(x/2) = -1
Yep, I get that far, but I dunno how to get 3/2 pi

I get theta is 90 which is 1/2 pi
15. (Original post by violet)
Yep, I get that far, but I dunno how to get 3/2 pi

I get theta is 90 which is 1/2 pi
tan-1(-1) = -pi/4 = θ/2

But the interval is 0 < θ < 2pi ==> 0 < θ/2 < pi

So you do a CAST diagram or look at the graph (or whatever your method is) and as tan is negative it must be in the second or fourth quadrants.

But the limits restrict it to the second quadrant for θ/2.

=> Adding pi to the -pi/4 gives the value for θ/2

=> 3pi/4

So θ = 3pi/2

This isn't very clear but you should have some method for dealing with stuff like this from chapter 10 of C2.
16. Aside from

(Original post by C3)
The maximum point on the curve with equation x√sinx, 0<x<180, is the point A. Show that the x-cordinate of point A satisfies the equation 2tanx + x = 0.
Does anyone know how to solve:

cos2x + sin2x = 0

17. (Original post by ntrik)
Aside from

Does anyone know how to solve:

cos2x + sin2x = 0

Divide by cos2x to get tan2x = -1 and then solve
18. (Original post by samdavyson)
tan-1(-1) = -pi/4 = θ/2

But the interval is 0 < θ < 2pi ==> 0 < θ/2 < pi

So you do a CAST diagram or look at the graph (or whatever your method is) and as tan is negative it must be in the second or fourth quadrants.

But the limits restrict it to the second quadrant for θ/2.

=> Adding pi to the -pi/4 gives the value for θ/2

=> 3pi/4

So θ = 3pi/2

This isn't very clear but you should have some method for dealing with stuff like this from chapter 10 of C2.

Oh, yep, I see.. Whoops!
Thanks.
19. (Original post by RichE)
Divide by cos2x to get tan2x = -1 and then solve
tan2x = -1
2x = -45
x = -90

if x = -90

then cos(-180) + sin(-180) = 0
-1 + 0 = 0

The answers are 67.5 and 157.5 apparently
20. (Original post by ntrik)
tan2x = -1
2x = -45
x = -90

if x = -90

then cos(-180) + sin(-180) = 0
-1 + 0 = 0

The answers are 67.5 and 157.5 apparently
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