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Author Topic: [CIE] All Pure Mathematics (P1, P2 and P3) doubts here !  (Read 79987 times)

Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #135 on: March 20, 2011, 09:22:47 am »
Quote from: Vin on March 18, 2011, 03:46:44 pm
You've got to take the derivative of (x^2 + 3), therefore 2x.

Sorry, but I have to correct you :

9709/11/O/N/09

Q 7 (i)

dy/dx of 12/(x2 + 3) not only x2 + 3
= 12 (x2 + 3)-1
= -12 (x2 + 3)-2 (2x)
= -24x (x2 + 3) -2
= -24x/(x2 + 3) 2
That’s it for the first question.

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Offline Vin

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #136 on: March 20, 2011, 11:01:47 am »
Quote from: Fidato on March 20, 2011, 09:22:47 am
Sorry, but I have to correct you :

9709/11/O/N/09

Q 7 (i)

dy/dx of 12/(x2 + 3) not only x2 + 3
= 12 (x2 + 3)-1
= -12 (x2 + 3)-2 (2x)
= -24x (x2 + 3) -2
= -24x/(x2 + 3) 2
That’s it for the first question.



I know, but i remember him askin why "2x", therefore just a short explanation ;)
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Offline iluvme

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #137 on: March 23, 2011, 04:27:06 pm »
Please help

Question attached
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elemis

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #138 on: March 23, 2011, 05:03:36 pm »
Give me 10-15 minutes.
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elemis

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #139 on: March 23, 2011, 05:11:14 pm »
Vector OA is parallel to DE.

Vector AB is parallel to EF.

Hence, OA = DE and AB = EF.

OA = 6i

AB = 6j

DQ = 0.5*DF

Since DF = 6i + 6j

DQ = 3i + 3j

To find the position vector of Q we must go from the origin to D then to Q.

So its simply 6k + 3i + 3j
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Offline iluvme

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #140 on: March 23, 2011, 05:13:47 pm »
Oh right, got it, Thanks! :)

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Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #141 on: March 27, 2011, 02:02:48 pm »
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Offline iluvme

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #142 on: March 27, 2011, 07:34:16 pm »
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elemis

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #143 on: March 28, 2011, 04:47:08 am »
Quote from: i<3SL on March 27, 2011, 07:34:16 pm
Question 10

http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w07_qp_1.pdf

(ii)
<br />\frac{\begin{pmatrix}<br />2\\ <br />2\\ <br /><br />2\end{pmatrix}\cdot \begin{pmatrix}<br />-2\\ <br />2\\ <br />4<br />\end{pmatrix}}{\sqrt{2^2+2^2+2^2}\times\sqrt{-2^2+2^2+4^2} }

Solving the above gives : \frac{\sqrt{\3}}{3}

Doing the cos-1 of the above gives the angle : 61.9 degrees.


(iii)

If  PR = 2i + 2j + 2k
    PQ = -2i + 2j +4k

Then QR = PR - PQ

Find the modulus of PQ, PR and QR. Then add them all up; you should get the answer.
« Last Edit: March 28, 2011, 05:13:42 am by Ari Ben Canaan »
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Offline ashish

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #144 on: March 28, 2011, 06:19:15 am »
Quote from: Fidato on March 27, 2011, 02:02:48 pm
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s09_qp_1.pdf

(10) (ii) Please ...

when x=0, y=13
 the curve has turning point at (3,-5)
the line of symmetry is at x=3, since we have moved a length of 3 units from (0,0)(neglecting the y coordinate) till (3,0), to get the next part of the graph we ill have to move another 3 units, which will give (6,0)

or you could just solved for y=13 which worth more than one mark :D

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Offline iluvme

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #145 on: March 28, 2011, 06:57:58 pm »
Quote from: Ari Ben Canaan on March 28, 2011, 04:47:08 am
(ii)
<br />\frac{\begin{pmatrix}<br />2\\ <br />2\\ <br /><br />2\end{pmatrix}\cdot \begin{pmatrix}<br />-2\\ <br />2\\ <br />4<br />\end{pmatrix}}{\sqrt{2^2+2^2+2^2}\times\sqrt{-2^2+2^2+4^2} }

Solving the above gives : \frac{\sqrt{\3}}{3}

Doing the cos-1 of the above gives the angle : 61.9 degrees.


(iii)

If  PR = 2i + 2j + 2k
    PQ = -2i + 2j +4k

Then QR = PR - PQ

Find the modulus of PQ, PR and QR. Then add them all up; you should get the answer.


Thanks Ari.

How exactly do you find PR and PQ?
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Offline HUSH1994

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #146 on: March 29, 2011, 07:03:05 am »
O\N 2010 P.13 Q4 part 2 please?
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elemis

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #147 on: March 29, 2011, 07:09:09 am »
Quote from: i<3SL on March 28, 2011, 06:57:58 pm
Thanks Ari.

How exactly do you find PR and PQ?

PR = PA + 0.5 AE + 0.5 AB      Read the information given in the question CAREFULLY

PQ = PO + OD + 0.5 DG
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elemis

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #148 on: March 29, 2011, 07:16:36 am »
Quote from: ~~!$!HUSH!$!~~ on March 29, 2011, 07:03:05 am
O\N 2010 P.13 Q4 part 2 please?

<br />2\pi sinx - \pi = -x


\pi(2sinx-1)=-x

<br />2sinx -1 =\frac{-x}{\pi}


2sinx = 1-\frac{x}{\pi}


Hence, equation of the straight line is  : y = 1-\frac{x}{\pi}
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Offline HUSH1994

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #149 on: March 29, 2011, 07:22:13 am »
Thanks man +rep,and im really sorry for the trouble  :(
« Last Edit: March 29, 2011, 07:44:58 am by ~~!$!HUSH!$!~~ »
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