Mathematics: Post your doubts here!

[Soldier313]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_63.pdf
Q6 c)
plz explain how u get this answer..

okay look
let's assume these 8 lines below are the 8 tiles
__ __ __ __ __ __ __ __

the question states that each of the tiles has an equal probability so p(black) = p(white) = p(grey) = 1/3
now the restriction is that we can't have two tiles of the same colour placed next to each other.
so let's assume that the first tile is black, hence the next one can either be grey OR white ( 1/3 + 1/3) = 2/3
since there are 7 tiles we have to place after the first one we simply do (2/3)^7
(each of the 7 remaining tiles only have an option of 2 colours, the first could be any of the three)

Hope it helped !

 

[Alice123]

its 1 / 2*(x^2-3)

these r o level stuff

 

[iKhaled]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s09_qp_3.pdf q7i) can some1 give me detailed procedure of how we can obtain roots of 1-3^o.5i and -1-3^05i

ok we have z^2 + (2√3)i z -4 = 0 if u notice this is a quadratic equation so we use the formula -b ± √(b^2-4ac)/2a to solve it
a= 1 b = (2√3)i and c = -4

substitute this in the equation and dont forget that i^2 = -1

if u still didn't get it and want me to solve it pls let me know

 

[cool Asviva]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_11.pdf
q8. anyone please explain why A.P sum formula is not used?
a=1000 and d=50
thnx in advance

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s09_qp_3.pdf q7i) can some1 give me detailed procedure of how we can obtain roots of 1-3^o.5i and -1-3^05i

One very important note for these kind of equation where one or more of the coefficients (a, b and c in the quadratic formula) are imaginary, (in this case the coefficient of z = (2√3)i) is that the rule of conjugate pairs being roots to the equation doesn't follow.

To solve this equation, you have to just use the quadratic formula. For this solution, the values are:

a = 1
b = (2√3)i
c = -4
4ac = -16
b^2 = 2 * 2 * √3 * √3 * i * i = 4 * 3 * -1 = -12

Putting these in the quadratic equation gives us

z = (-(2√3)i +- √(-12 + 16))/2

z = (-(2√3)i +- 2)/2

z = -√3i + 1 OR -√3i - 1

Hope this helped!

Good Luck for your exams!

 

[Sanis]

One very important note for these kind of equation where one or more of the coefficients (a, b and c in the quadratic formula) are imaginary, (in this case the coefficient of z = (2√3)i) is that the rule of conjugate pairs being roots to the equation doesn't follow.

To solve this equation, you have to just use the quadratic formula. For this solution, the values are:

a = 1
b = (2√3)i
c = -4
4ac = -16
b^2 = 2 * 2 * √3 * √3 * i * i = 4 * 3 * -1 = -12

Putting these in the quadratic equation gives us

z = (-(2√3)i +- √(-12 + 16))/2

z = (-(2√3)i +- 2)/2

z = -√3i + 1 OR -√3i - 1

Hope this helped!

Good Luck for your exams!

thx alot .. it really really helped me

 

[HorsePower]

okay look
let's assume these 8 lines below are the 8 tiles
__ __ __ __ __ __ __ __

the question states that each of the tiles has an equal probability so p(black) = p(white) = p(grey) = 1/3
now the restriction is that we can't have two tiles of the same colour placed next to each other.
so let's assume that the first tile is black, hence the next one can either be grey OR white ( 1/3 + 1/3) = 2/3
since there are 7 tiles we have to place after the first one we simply do (2/3)^7
(each of the 7 remaining tiles only have an option of 2 colours, the first could be any of the three)

Hope it helped !

thanks alot

 

[PANDA-]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_11.pdf
q8. anyone please explain why A.P sum formula is not used?
a=1000 and d=50
thnx in advance

Well I have no idea what A.P sum formula is...
But model 1 here is an arithmetic sequence, and model 2 is a geometric sequence...

For model 1 you have to find the sum of the first 40 terms of the arithmetic sequence... That's the formula..
Sn = n/2[2a1+(n-1)d]
Sn = 40/2[2(1000)+(40-1)(1000)]
Sn = 20[2000+39000]
Sn = 20[41000]
Sn = 820000

Then multiply it by 5%
5/100(820000) = 41000

Now for the geometric sequence you use the sum of 40 terms with the respective formula... Which is..
Sn = a1(1-r^n)/1-r
5% is the increase, so we'll use 110% for ratio, which is 1.1 for short.
Sn = 1000(1-1.1^40)/1-1.1
Sn = 442592

Now do the same as you did above with the arithmetic, 5%...
5/100(442592)
= 22129
3 significant figures...
= 22100

 

[abruzzi]

They didn't ask for the unit vector, they just asked for the vector and its magnitude, I guess I forgot to mention the magnitude...
Well the magnitude is the square root of the sum of i^2, j ^2, k^2. Which is √(i^2+j^2+k^2).
Which is √(4^2 + 4^2 + 5^2) = √57.

Unit vector would be another story, where you divide the vector itself by the magnitude.

Oh alright
Thanks

 

[sagar65265]

Guys, could someone please help out with question No. 10 (b) for the following paper?
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_32.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_ms_32.pdf
Thanks in advance!

 

[sagar65265]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_1.pdf

can someone please solve Q 4 and 5


thanks in advance

i) Since
y = asin(bx) + c ====> we can set x = 0 to get
y = c (Because asin(b * 0) = asin(0) = 0)
Therefore, checking the graph when x = 0 gives us a value of 3, so c = 3.
The maximum value of the sine and cosine functions for all angles is equal to 1. So taking the maximum point on the graph and setting sin(bx) to 1, we get
y = a(1) + c = a + c = a + 3
Since the maximum value of f(x) on the graph = 9,
a + 3 = 9 so a = 6
Now we should be able to subsitute any other values of y, a, c and x into the equation to get b:
At x = pi/2, y = 3 so:
3 = 6sin(b * pi/2) + 3
6sin(b * pi/2) = 0
sin(b * pi/2) = 0
Since the sine function is zero only for the even multiples of pi/2, we can assume that b = 2.

ii) Since we know that in this condition y = 0, we can substitute that value straight into the equation:
0 = 6sin(2x) + 3
-3 = 6sin(2x)
sin(2x) = -0.5
Using "Sine Inverse", we get
2x = -30 degrees.
This translates to an angle of 330 degrees (360 + (-30) = 330).
However, we know that the sine function for 2x is negative, and the sine function is negative in the third and the fourth quadrants, so we can find another value (aside from 330 degrees) for 2x in the third quadrant. This value can be obtained by:
180 - (-30) = 210 degrees -- adding -30 degrees to 180 would give us 150; sin(150) is positive, so 2x cannot be 150 and we have to subtract fro 180 instead.
So out of 210 degrees and 330 degrees, 210 is the smaller one and this means that x = 210/2 = 105 degrees. This is equal to 1.83 radians.

5) If the perimeter of arc R1 is equal to the length of arc R2, the perimeter of arc R1 is:
r + r + r(theta) = 2r + r(theta) = r(2 + theta)
The length of arc R2 is
r(2pi - theta)
If these two expressions are equal, we have
r(2 + theta) = r(2pi - theta)
2 + theta = 2pi - theta
2 * theta = 2(pi - 1)
Cancelling out 2 from both sides,
theta = pi - 1

ii) If the area of R1 = 30,
0.5 * r^2 * theta = 30
r^2 * theta = 60
r^2 = 60/(pi -1) = 28.016
r = 5.29 cm

Area of R2 = 0.5 * r^2 * (2pi - (pi - 1))
0.5 * 28.016 * (pi + 1) = 58.01 cm^2

Hope this helped!

Good Luck for your exams!

 

[Tkp]

oct nov 11 p-11
6b

 

[PANDA-]

oct nov 11 p-11
6b

I did this one, but I can't find the paper in which I did it on... Anyway!

Sum to infinity is S∞..
S∞ = a1/(1-r)
Both S∞ of both sequences are equal.
a1/1-r = a1/1-r
1/1-r = 4/1-0.25r

P.S. I replaced 1/4(r) with 0.25r cuz it's easier to read and make out from the rest... anyway
Cross multiply

1-0.25r = 4-4r
1-4 = -4r +0.25r
-3 = -3.75r
r= 5/4

Now just replace r= 5/4 in any one of the above S∞

S∞= 1/1-5/4
S∞ = 5

 

[GorgeousEyes]

Please jun012 no.3 v.13 .. jun012 no.7 v.12 .. jun012 v.11 9)iii . Pure Math As . thanks

 

[GorgeousEyes]

Guys is it sufficient to do past papers and variants from year 2009 to 2012 to get an A ? I am talking about Math As .

 

[Soldier313]

Guys, could someone please help out with question No. 10 (b) for the following paper?
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_32.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_ms_32.pdf
Thanks in advance!

hence the final answer is 2.3 radians....
hope it helps!

 

[syed1995]

Guys is it sufficient to do past papers and variants from year 2009 to 2012 to get an A ? I am talking about Math As .

Nothing is ever enough. You can't guarantee that doing all the papers will get you an A. The main thing in maths is concepts.. If your concepts are clear, you'd get an A.. I'd say if you grasp all the concepts from the past papers and variants .. and you think you can score around 60 or so in P1 then sure..

And most importantly always remember Allah.. and offer prayers.. And surely he will reward you

 

[Dudu]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_13.pdf

Help please.

Q10)
iii) (did ii)
iv)
v)

 

[GorgeousEyes]

Nothing is ever enough. You can't guarantee that doing all the papers will get you an A. The main thing in maths is concepts.. If your concepts are clear, you'd get an A.. I'd say if you grasp all the concepts from the past papers and variants .. and you think you can score around 60 or so in P1 then sure..

And most importantly always remember Allah.. and offer prayers.. And surely he will reward you

You are right , thank youu i will incha'allah .

 

[Esme]

When an equation of a plane and equation of a line is given, how do we determine whether the line intersects the plane or not? I know how to find the point of intersection, but if we're asked to tell whether they intersect or not?