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AS Maths...


MulderMan

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In all fairness the whole course. I got an E for C1 in January, and I just can't seem to be able to do a lot of any of C2 or S1.

Guess its just a case of practice, practice, practice?

 

Pretty much yes, bytesize was ok I seem to recall, although I never really used it.

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Hasn't C2 already happened?

 

Past papers are always useful (if you haven't got any, go into school and beg for some).

 

Do exercises and read explanations on mathsnet. Or Bytesize, if you must.

 

Don't Panic.

 

Make notes of things that you have to remember (pattern for the binomial expansion IIRC) and if it comes to it, learn to recite them.

 

Go and talk to your teachers about anything you really don't understand, don't be shy about it.

 

EDIT: oh look, I found a PDF file from what I guess must be a year ago. I seem to remember there being a typo on it somewhere, if you find it, then give yourself a cookie. (I hereby release the attached document into the public domain)

useful-c2.pdf

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(I hereby release the attached document into the public domain)

 

I've got a few suggestions for that document.

 

Finding a distance

 

[math]\sqrt{(x_2-x_1)^2+(y_2+y_1)^2}[/math]

 

You've got a typo with the y-coordinates there.

 

Sum of a geometric series

 

[math]\sum_{n=1}^n ar^{n-1}=\frac{a(r^n-1)}{r-1}[/math]

 

You shouldn't use the symbol [math]n[/math] in both the index and the upper limit of the index. Let the index be something else, such as [math]j[/math]. Then we have the following.

 

[math]\sum_{j=1}^nar^{j-1}=\frac{a(r^n-1)}{r-1}[/math]

 

Quadrants and trig values

.

.

.

4th Quad. 270o-369o Only cosine is positive

 

I think you meant 359 degrees. But even so, why stop at 359 degrees? 359.5 degrees is also in Quad. 4, no? I think it would be better to express the angular range as either an inequality or an interval.

 

1st Quad: (0,90)

2nd Quad: (90,180)

3rd Quad: (180,270)

4th Quad: (270,360)

 

Identities with cosine

 

[math]\cos(360-\Theta)^o \equiv -cos(\Theta)^o[/math]

.

.

etc.

 

That one's wrong. You got it right two lines down from there though.

 

Stationary points

.

.

.

maximum point, minimum point, or point of inflexion

 

[math]\frac{dy}{dx}=0[/math] and

[math]\frac{d^2y}{dx^2}=0[/math]

 

Sure, but then what? I think it would be useful here to say: Second derivative test fails. Resort to first derivative test.

 

Definite integration

 

Area under a curve:

[math]Area=\int_a^bf(x)dx[/math]

 

Only if [math]f(x) \geq 0[/math] for all x in the interval of integration. Otherwise the integral doesn't give the area between f and the x-axis.

 

Continuing...

 

[math]Area=\left[\int f(x)dx\right]_a^b[/math]

 

Never seen that notation before. Did you perhaps mean this?

 

[math]Area=[ F(x) ]_{a}^{b}[/math],

 

where [math]F^{\prime}(x)=f(x)[/math].

 

Continuing...

 

[math]Area=\left(\int f(b)\right)-\left(\int f(b)\right)[/math]

 

You're subtracting something from itself. That would be zero, no matter what f is.

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O.k. lots of typos. And rather shoddy notation. And some notation that I just made up on the spot.

 

Really they were just revision notes from last year when I took the same exam as the OP, I made them in a few spare minutes and it was only really intended for my benefit and at a stretch the other people in my class.

 

I don't think I have the original, or if I do it is well hidden, so I apologies for the errors and hope that anyone who is unfortunate enough to use it will not take any incorrect information into account.

 

Actually, it's not all that useful to be honest, even if it were a shining beacon of accuracy. The notes at the end of each chapter probably have at least as much useful information, a revision guide brought from the high street can do a world of good and making your own posters is a worthy activity.

 

edit oh and Tom, you owe yourself a lot of cookies.

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