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Discussion in 'International A And AS Level' started by XPFMember, Sep 8, 2011.
Compiled P3 Attached .....
Compiled P4 (M1) Attached...
We know that f(x) = 2x + 1
let f-1(x) = y
then f = x
f = 2 + 1 = x
y = (x - 1)/2
So f-1(x) = (x - 1)/2
f-1(g(x)) = ((x^2 - 2)-1)/2
f-1(g(x)) = (x^2 - 3)/2
The answer in the mark scheme is incorrect. Mine is correct, check the examiner report, my answer matches with that correct one.
Assalamoalaikum wr wb...
I have a doubt in Q:7 (iii) and Q:10 (b) of Nov 2011 Paper 3 variant 2....
Could you please help me with that?
in 7 (iii) how do I do it...it's like just one mark...
and in 10 (b) I don't know how to find the greatest value... :S
Still not clear?
Well for 7 III, it is the ratio between AP and PB which is equal to the ratio between OA and OB, now you just need to put in the values of AP, PB, OA, and OB and get he value of lambda
nopes...sorry i wasnt here...
Just checked...It's perfectly clear now...JazakAllah Khairen....
Thx got this as well but i was kinda confused by the ms
1) Question no. 9(iv) of Oct/Nov 2006 Paper3/9709.
2) Question 09 (ii) of May/June 2007. P3.
3)Question 07(iii) of May/jUNE 2006.
If p= 2i-j+3k, q=5i+2j and r=4i+j+k, find the set of numbers f, g, and h such that fp+gq+hr=0. What does this tell you about the translations represented by p, q and r?
If any one of you would be so kind enough as to give me workings of this problem and solve this problem will get a BIG thanks from me...
November 2006, Q9(iv):
First of all, we'll sketch an argand diagram showing the point representing the complex number 'u' which's '1 + i'. Next, we'll draw a circle of radius '1'' with its centre at '1 + i'. Now to calculate the least value of |z|, you'll be needing to take a look at this diagram (its not drawn to scale).
To calculate the least/greatest value of |z|, we should first look for the point from which we have to find it. For example, if we have to calculate the least/greatest value of |z + 3 - 2i|, we will draw a line from (-3 , 2) to the centre of the circle and then do all the calculations to find out the least/greatest value of |z + 3 - 2i|. Coming back to this question, they've asked us to calculate the least value of |z|. First of all, we'll draw a line from '0 + 0i', which is the origin, to the centre of the circle and then draw a vertical line from the centre of the circle to the x-axis. This sort of creates a right angle triangle as depicted in the above diagram in red colour. If your diagram is perfect and accurate, then the adjacent and opposite sides will both be equal to 1 unit. Next, using the pythagora's theorem, we will find the length of the hypotenuse.
A^2 = B^2 + C^2
A^2 = (1)^2 + (1)^2
Once we have calculated the length of the hypotenuse, we can easily calculate the least value of |z|. As I've mentioned in the diagram, the least value of |z| can be found out by subtracting '1', which's is the radius, from the length of the hypotenuse.
Length of the Hypotenuse - Radius
√2 - 1
Therefore, the least value of |z| is '√2 - 1'.
June 2007, Q9(ii):
In this part, we need to find the acute angle between the plane ABC and the plane OAB. We'll be needing the common perpendiculars of both of these planes and then using the formula "(common perpendicular of ABC) x (common perpendicular of OAB) = (modulus of the common perpendicular of ABC) x (modulus of the common perpendicular of OAB) x cos Θ", we can easily find out the angle. We've already found out the common perpendicular of the plane ABC in the first part of this questions which comes out to be '- 4i - 2j - k'. Similarly, we'll find the common perpendicular of the plane OAB. First we'll find the direction vectors of AO and AB. Then using the common perpendicular formula, we'll find the common perpendicular of the plane OAB.
AO = ( -2, 0, 0 )
AB = ( -1, 2, 0 )
i j k
-2 0 0
-1 2 0
The common perpendicular of the plane OAB will come out to be '-4k'. Next, we'll just plug in these values in the formula and find out the angle.
(common perpendicular of ABC) x (common perpendicular of OAB) = (modulus of the common perpendicular of ABC) x (modulus of the common perpendicular of OAB) x cos Θ
(-4) x (0) =
(-2) x (0) = (√21) x (4) x (cos Θ)
(-1) x (-4) =
4 = (√21) x (4) x (cos Θ)
Θ = 77.4°
hi ! helpppp!
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s03_qp_4.pdf
Q.3 iii part. how to calculate the distance between P and Q when Q’s speed reaches 10 m s.
Area under the graph of P minus area under the graph of Q. I hope you get it.
nope. if it was dat simple i wudn't have posted it. marking scheme says area of parallelogram OPQT is the distance between p n q. but how?
Don't know how to explain, but at the the speed 10m/s the Cyclist both have constant velocity hence the distance between them is fixed and the the difference between the two graphs shows the distance apart between them. That is 4 X 10, area of parallelogram formula
hey itx a littlle simple bro.
as p has started his journey 4 minutes before Q. so P will reach the final velocity 4min. before Q.
now P will cover the distance till Q reaches 10ms-1.
Q having same acc. so will definately reach till 10ms-1 aftr 4min.
now in the span of 4 min. with the velocity of 10ms-1 p will cover the distance....... s = vt
S= 10 * 4 = 40m (P has covered 40m of distance till Q attains a speed of 10ms-1)
simply muliply the T found in part 2 to the velocity (10)
It's in seconds.
Thank you Mani and Megusta!
Q3 ii) I do get the answer by just squaring the equation (x-2)^4 and integrating it...
But what if I open the equation y= (x-2)^2, take its square (x^2-4x+4)^2 and now integrate it? I don't end up getting the same answer. WHY?
If I further open it (x^2-4x+4)^2 and then integrate it I do get the answer... But why not without opening the second equation?
This might be a senseless question but how did you take square of (x^2-4x+4)?
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