Author Topic: MATH A2 P2  (Read 1784 times)

Offline thecandydoll

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MATH A2 P2
« on: August 09, 2010, 09:36:32 am »
Please help me with this!
 i have seen this question in all the papers!
please help me solve this!

2007 CIE PAPER 2 M/J

nid404

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Re: MATH A2 P2
« Reply #1 on: August 09, 2010, 10:22:59 am »
which question? :P

Offline Saladin

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Re: MATH A2 P2
« Reply #2 on: August 09, 2010, 11:15:20 am »
LOL! Yes, which question?

Offline astarmathsandphysics

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Re: MATH A2 P2
« Reply #3 on: August 09, 2010, 01:59:56 pm »
I may have answered this on astarmathsandphysics.com go see

Offline Ghost Of Highbury

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Re: MATH A2 P2
« Reply #4 on: August 09, 2010, 03:13:25 pm »
Anyone having CIE AS math P1 book.

Pg 48, Q 9e) and 9f)

Pg 50 Q 20h) and 20i)

Thank you.

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Offline Saladin

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Re: MATH A2 P2
« Reply #5 on: August 09, 2010, 04:17:28 pm »
Anyone having CIE AS math P1 book.

Pg 48, Q 9e) and 9f)

Pg 50 Q 20h) and 20i)

Thank you.



Sorry, do not have the book, can you please give me a scanned copy to work with?

Offline Ghost Of Highbury

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Re: MATH A2 P2
« Reply #6 on: August 09, 2010, 04:46:24 pm »
Sorry, do not have the book, can you please give me a scanned copy to work with?

Will try..
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Offline Ghost Of Highbury

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Re: MATH A2 P2
« Reply #7 on: August 09, 2010, 05:09:46 pm »
Question : Suggest a possible equation for each of the graphs.

here

http://img707.imageshack.us/img707/103/scan0001cn.jpg

http://img197.imageshack.us/img197/4797/byhbyh.jpg

plz explain the steps to solve them too.
thanks

divine intervention!

Offline Saladin

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Re: MATH A2 P2
« Reply #8 on: August 09, 2010, 06:56:54 pm »
Question : Suggest a possible equation for each of the graphs.

here

http://img707.imageshack.us/img707/103/scan0001cn.jpg

http://img197.imageshack.us/img197/4797/byhbyh.jpg

plz explain the steps to solve them too.
thanks



I love the names you give to your images man! Keep up the good work! :D

(e) (x+1)(x+2)(x)

The line passes through two negative points and the origin.

(f) -(x+1)(x-1)(x-2)

The left side of the curve is going downward. This means that it is a negative curve. There are two interceptions where the x-axis is positive. Therefore two (x-z)s.

(g) -(x+2)^2(x-3)

The left side is negative, so it the curve is negative. It has a standing point with 2 interceptions on the x-axis, so there is a (x\pm z) somewhere. And it cuts the x-axis in the positive area. Thus, it has an (x-z).

(h) x^2(x+2)

There is a stationary point on (0,0) meaning that there is x^2 and there is an intersection on the negative part of the x-axis. So, there is a (x+z).

(i) Although this may seem intimidating, it is actually based on the simple rules as previous. -(x+3)(x+2)(x+1)(x-3)(x-2)(x-1)

You can see that there are 6 straight line sections. So the total power of x with be x^6.

Now, there are 3 intersections in the positive and the negative, so you will have a repeated brackets of these: (x+c) and (x-v)

Now, determining the the slope is very important. Before you do such big ones, just plot them in your head. Now even powers e.g x^2 will always have their negative x values on the positive y-axis. So, x^6 is no exception. But the graph we see shows that the negative x values are on the negative y-axis.

So, we put in a negative sign, and problem solved! :)
« Last Edit: August 09, 2010, 07:03:38 pm by Engraved »

Offline Ghost Of Highbury

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Re: MATH A2 P2
« Reply #9 on: August 10, 2010, 04:33:26 am »
I love the names you give to your images man! Keep up the good work! :D

(e) (x+1)(x+2)(x)

The line passes through two negative points and the origin.

(f) -(x+1)(x-1)(x-2)

The left side of the curve is going downward. This means that it is a negative curve. There are two interceptions where the x-axis is positive. Therefore two (x-z)s.

(g) -(x+2)^2(x-3)

The left side is negative, so it the curve is negative. It has a standing point with 2 interceptions on the x-axis, so there is a (x\pm z) somewhere. And it cuts the x-axis in the positive area. Thus, it has an (x-z).

(h) x^2(x+2)

There is a stationary point on (0,0) meaning that there is x^2 and there is an intersection on the negative part of the x-axis. So, there is a (x+z).

(i) Although this may seem intimidating, it is actually based on the simple rules as previous. -(x+3)(x+2)(x+1)(x-3)(x-2)(x-1)

You can see that there are 6 straight line sections. So the total power of x with be x^6.

Now, there are 3 intersections in the positive and the negative, so you will have a repeated brackets of these: (x+c) and (x-v)

Now, determining the the slope is very important. Before you do such big ones, just plot them in your head. Now even powers e.g x^2 will always have their negative x values on the positive y-axis. So, x^6 is no exception. But the graph we see shows that the negative x values are on the negative y-axis.

So, we put in a negative sign, and problem solved! :)
Thanks a lot man. +rep
divine intervention!

Offline Saladin

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Re: MATH A2 P2
« Reply #10 on: August 10, 2010, 08:24:50 am »