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Mathematics: Post your doubts here!

abruzzi

abruzzi

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iKhaled said:
you should know that tan θ = y/x and the gradient (m) is equal to y/x so u can say from here that tan θ = gradient. you need to find the angle between line l and the gradient of the curve at p so first of all we will find the gradient of the curve at point p

y = 12x^-1
dy/dx = -12x^-2
dy/dx = -3 ( i think you know that dy/dx means also the gradient)

now lets find the gradient of the line y = -2x + k so -2 is the gradient of the line now tan θ = gradient so..

tan θ = -3 and tan θ = -2
θ = -71.6° and θ = -63.4°

difference between them is 71.6 - 63.4 = 8.2° and that's our answer!!

i hope you got it!
Click to expand...
Got it!
Thank you :)
 
VelaneDeBeaute

VelaneDeBeaute

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MJ08_Maths_Q8.JPG Alright, I know this question was posted before (If someone did it, or know when, they can give me the link to its solution -if it had been posted before) :p
Otherwise, I can't seem to be getting the hang of it. This is M/J/08, Paper 3, Q.8.
 
VelaneDeBeaute

VelaneDeBeaute

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A star said:
this thread is full of p3 what about s1or p1 any one :(
Click to expand...

Even if it is, there's no way one can access the same post again. (Just as what happened to me right now. I looked through about 20 pages since 344, but was unable to find the same question) :p
 
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VelaneDeBeaute said:
Even if it is, there's no way one can access the same post again. (Just as what happened to me right now. I looked through about 20 pages since 344, but was unable to find the same question) :p
Click to expand...
ikr thats y we have a print screen option :D
 
Esme

Esme

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Rutzaba that question isn't from a past paper, so i don't have any link. If you figure out how to solve it, do let me know please. =D
 
externityxzx

externityxzx

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VelaneDeBeaute said:
View attachment 23970 Alright, I know this question was posted before (If someone did it, or know when, they can give me the link to its solution -if it had been posted before) :p
Otherwise, I can't seem to be getting the hang of it. This is M/J/08, Paper 3, Q.8.
Click to expand...
this is how i did it
(i)let PN= y
the triangle PTN is equal to tanx so
0.5x(y x TN)= tanx
TN=2tanx/y
dy/dx = y/ (2tanx/y)
=y x y/2(sinx/cosx)
=y^2 x 1/2(cosx/sinx)
=1/2 y^2 cotx

(ii) dy/dx =y^2/2 x cosx/sinx
2/y^2 dy= cosx/sinx dx (integrate this)
-2y^-1 =ln(sinx) +c
apply y=2 x=1/6 pi
c=-1-ln(1/2)
put it in the equation and get y=2/(1-ln2sinx)

:) hope you get it
 
P

PhyZac

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freezingfires said:
Rutzaba or PhyZac
Can you please help me with Q10 part iii of May/June 2012 paper 32.It is a vector question Im unable to solve it .I'll highly appreciate if you give a step by step solution.Thanks
Here is the link:http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf
Click to expand...
We have to find the point p where the perpendicular distance to the two planes is same.


first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]
 
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freezingfires said:
Thank-you so much PhyZac I got it that was extremely helpful.May you be blessed with very good grades .Ameen.
I have one more problem in Statistics paper 62 may/june 2012 Q2 part i.I don't understand how to draw the probability distribution table.
Here is the link:http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_62.pdf
Click to expand...
well its easy :) three possibilities are that they must be same for 0 then if they are different then either 2 or 4 draw the table with 3 coloumns 0 2 4 then see the possible outcomes and multiply teir probabilities and sum them :)
 
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PhyZac

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freezingfires said:
Thank-you so much PhyZac I got it that was extremely helpful.May you be blessed with very good grades .Ameen.
I have one more problem in Statistics paper 62 may/june 2012 Q2 part i.I don't understand how to draw the probability distribution table.
Here is the link:http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_62.pdf
Click to expand...
Aaameeen.

And for that question, Y value is 2 X values minus eachother.
[Two independent values of X are chosen at random. The random variable Y takes the value 0 if the
two values of X are the same. Otherwise the value of Y is the larger value of X minus the smaller
value of X.] this is from the paper....
anyway so the possible values of Y are
0 [when (2,2) or (4,4) or (6,6) are chosen]
2 [when (4,2) or (6,4) or (2,4) or (4,6) are chosen]
4 [when (6,2) or (2,6) are chosen]

So now we know the values, we have to find the probabilities
for 0, (0.5 x0.5) + (0.4 x 0.4) + (0.1 x 0.1) = 0.42 [P.S, i sub the probabilities of the numbers above from table.]
for 2, (0.4 x 0.5) + (0.1 x 0.4) + (0.5 x 0.4) + (0.4 x 0.1) = 0.48
for 4, (0.1 x 0.5) + (0.5 x 0.1) = 0.1

I hope u get it, and thanks alot for the prayer.
 
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freezingfires

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PhyZac said:
Aaameeen.

And for that question, Y value is 2 X values minus eachother.
[Two independent values of X are chosen at random. The random variable Y takes the value 0 if the
two values of X are the same. Otherwise the value of Y is the larger value of X minus the smaller
value of X.] this is from the paper....
anyway so the possible values of Y are
0 [when (2,2) or (4,4) or (6,6) are chosen]
2 [when (4,2) or (6,4) or (2,4) or (4,6) are chosen]
4 [when (6,2) or (2,6) are chosen]

So now we know the values, we have to find the probabilities
for 0, (0.5 x0.5) + (0.4 x 0.4) + (0.1 x 0.1) = 0.42 [P.S, i sub the probabilities of the numbers above from table.]
for 2, (0.4 x 0.5) + (0.1 x 0.4) + (0.5 x 0.4) + (0.4 x 0.1) = 0.48
for 4, (0.1 x 0.5) + (0.5 x 0.1) = 0.1

I hope u get it, and thanks alot for the prayer.
Click to expand...
Thank-you so much PhyZac you are the best!! :)
 
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