IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: thecandydoll on August 09, 2010, 09:36:32 am

Title: MATH A2 P2
Post by: thecandydoll 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
Title: Re: MATH A2 P2
Post by: nid404 on August 09, 2010, 10:22:59 am: which question? :P
Title: Re: MATH A2 P2
Post by: Saladin on August 09, 2010, 11:15:20 am: LOL! Yes, which question?
Title: Re: MATH A2 P2
Post by: astarmathsandphysics on August 09, 2010, 01:59:56 pm: I may have answered this on astarmathsandphysics.com go see
Title: Re: MATH A2 P2
Post by: Ghost Of Highbury 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.
Title: Re: MATH A2 P2
Post by: Saladin on August 09, 2010, 04:17:28 pm: Quote from: Ghost Of Highbury~ 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.

Sorry, do not have the book, can you please give me a scanned copy to work with?
Title: Re: MATH A2 P2
Post by: Ghost Of Highbury on August 09, 2010, 04:46:24 pm: Quote from: Engraved on August 09, 2010, 04:17:28 pm
Sorry, do not have the book, can you please give me a scanned copy to work with?

Will try..
Title: Re: MATH A2 P2
Post by: Ghost Of Highbury 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
Title: Re: MATH A2 P2
Post by: Saladin on August 09, 2010, 06:56:54 pm: Quote from: Ghost Of Highbury~ 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

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! :)
Title: Re: MATH A2 P2
Post by: Ghost Of Highbury on August 10, 2010, 04:33:26 am: Quote from: Engraved on August 09, 2010, 06:56:54 pm
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
Title: Re: MATH A2 P2
Post by: Saladin on August 10, 2010, 08:24:50 am: Quote from: Ghost Of Highbury~ on August 10, 2010, 04:33:26 am
Thanks a lot man. +rep

Just doin ma job! ;)