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# OCR (MEI) Mathematics C3 20/01/2010 watch

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1. (Original post by jakebv)
Ryt wat i did (not saying this is the best or only way) was to first let h(x)=gf(x), this just makes this neater.
Then it follows that h(-x)=gf(-x) but as f(x) is odd f(-x)=-f(x)
Therefore h(-x)=g(-f(x)) and as g(x) is even g(watevers in here)=g(-wateva that is) i.e g(-f(x))=gf(x)
Thus as required we have shown h(-x)=h(x) i.e. given that f(x) is odd and g(x) is even, the composite function gf(x) will be even.

If someone doesn't agree plz let me know but yeah out of all the q's i felt this was the one most people will drop marks.
I did it this way but without the function h(x).
2. . State conditions for f(x) to be an odd function and g(x) to be an even function. Hence, prove that gf(x) is always an even function
-----

f(-x) = -f(x)
g(-x)=g(x)
g(f(x)) = g(-(f-x))
erm, forgot what I did now... lol
Isn't it the idea that the point on the graph where the gradient is largest is at a point of inflexion, hence when f''(x)=0. I get what you mean though, I haven't seen it in previous papers or any of our worked examples.
I thought of it like f''(x) was the derivative of f'(x), so like you would use f'(x)=0 to find the turning points of f(x) you would use f''(x) to find the turning points of f'(x). That's another good way of thinking about it.
4. did anyone else get the same answers for P and Q as i did?? Look back up the thread, i posted them earlier xx
Isn't it the idea that the point on the graph where the gradient is largest is at a point of inflexion, hence when f''(x)=0. I get what you mean though, I haven't seen it previous papers or any of our worked examples.
Yeah, that's what I did. Personally, I liked that it was something a bit different.
6. (Original post by amulvanny)
did anyone else get the same answers for P and Q as i did?? Look back up the thread, i posted them earlier xx

pi/6 and pi/2
7. (Original post by Tallon)
. State conditions for f(x) to be an odd function and g(x) to be an even function. Hence, prove that gf(x) is always an even function
-----

f(-x) = -f(x)
g(-x)=g(x)
g(f(x)) = g(-(f-x))
erm, forgot what I did now... lol
f(-x) = -f(x) g(-x)=g(x)

Function gf(x) = g[f(x)]

g[f(-x)]=g[f(x)] as g(x) is an even function

g[-f(x)]=g[f(-x)]=g[f(x)] as g(x) is an even function

Therefore gf(x) is always an even function.

Meh something along those lines.
8. Solve e^(2x) - 5e^(x)

x = ln 5
9. 3i) 2x/cuberoot of (1+3x^2)^2 nd 3ii) you differentiate implicitly and you shoud get the same as you are told y = cube root of (1+3x^2)
10. (Original post by 2-Sweet)
3i) 2x/cuberoot of (1+3x^2)^2 nd 3ii) you differentiate implicitly and you shoud get the same as you are told y = cube root of (1+3x^2)

spend forever on this. Implicit differentiation was so much easier.
11. (Original post by Tallon)
pi/6 and pi/2
Hi Tallon, i got pi/2 for P and 3pi/2 for Q. im not sure if they are right tho - not feelin tooooo confident to be honest about the whole exam!!
i worked mine out by 0 = xcos3x
so x = 0 or cos3x = 0
from the cosine graph x = 0 at cos90 and cos270 which in radians = pi/2 and 3pi/2 xx
12. There's already a 3 there so it's pi/6 and pi/2 (I think)
13. Section B

8. y=xcos3x (graph, again cba)

P and Q have co-ordinates (x1,0) and (x2,0) respectively. Neither have x co-ordinates where x=0

i) Find co-ordinates of P and Q

ii) Find the exact gradient at P

iii) Show that at the turning point of the graph, xtan3x=1/3

iv) Integrate area between O (origin) and P

9. f(x)=y = (2x^2-1)/(x^2+1) domain 0< x <2 (should be greater/less than and equals to signs)

i) Show dy/dx=[6x/(x^2+1)^2] .Show that f'(x) is always increasing when x >0

ii) Find range of f(x)

iii) Given that d2y/dx2 = [6-18.(x^2)/[(x^2+1)^3]], find maximum value of f'(x)

iv) f(x) has an inverse g(x). Find domain and range of g(x). Hence sketch graph of g(x)

v) Show that g(x) = [(x+1)/(2-x)]^1/2

Again, please correct me if I'm wrong.
14. (Original post by Game_boy)
I thought of it like f''(x) was the derivative of f'(x), so like you would use f'(x)=0 to find the turning points of f(x) you would use f''(x) to find the turning points of f'(x). That's another good way of thinking about it.
That's what I did, I can't see why it is meant to have been hard... maybe I missed something that would have made it seem less obvious ahah
15. feeling so stupid atm, but just couldn't get the integration in Q4 part ii)! Could anyone please tell me how they did it?
16. (Original post by kennykwan)
There's already a 3 there so it's pi/6 and pi/2 (I think)
OMG ur right, can't believe i have done that
17. (Original post by FatZam)
feeling so stupid atm, but just couldn't get the integration in Q4 part ii)! Could anyone please tell me how they did it?
By substitution. u=x+1, therefore x=u-1

Integral was 2x/(x+1) with limits x=1 and x=0. Change limits appropriate to what you're integrating with respect to (i.e u). Differentiate u=x+1 and replace for dx (as dx=(du/dx).du)

You should get:

integral of [2(u-1)/u] du with limits x=2 and x=1

Get that far?
18. (Original post by FatZam)
feeling so stupid atm, but just couldn't get the integration in Q4 part ii)! Could anyone please tell me how they did it?

I vaguely remember.. was it the integrate x/(x+1) ?

This was a tricky one, since they misled you on the previous part in using a dx-->du to get rid of the numerator.

But anyways...

x/(x+1)

take u as x+1

= (u-1)/u

So you ll change limites etc.

integral of 1- 1/u

= [u - ln u]

sub in w.e values it was now.
19. (Original post by FatZam)
feeling so stupid atm, but just couldn't get the integration in Q4 part ii)! Could anyone please tell me how they did it?
I did:

Let u = x + 1 => x = u - 1

du/dx = 1 => du = dx

2(u - 1)du = 2xdx

= 2 - 2ln2
20. (Original post by MarkS2)
I did:

Let u = x + 1 => x = u - 1

du/dx = 1 => du = dx

2(u - 1)du = 2xdx

= 2 - 2ln2
I think I did exactly the same except for my answer was : 2 - ln4, which is I think the same as 2 - 2ln2 haha

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