Mathematics: Post your doubts here!

[Alice123]

does ne1 have permutations n combinations notes

 

[Iffat]

- 4[ x^2 - 3x + 3/4]
4 [ (x)^2 - 2(x)(3/2) + (3/2)^2 - (3/2)^ + 3/4]
4 [ (x-3/2)^2 - 9/4 + 3/4]
4(x-3/2)^2 -9 + 3
4(x-3/2)^2 -6

thanx wat bout the other 1

 

[Sadiqa Ali]

can anyone help me out for these questions?
nov2011 paper 11 question 10 part ii) how do we get the gradient..
nov 2011 paper 12 question 8 part i)

 

[french410]

can anyone help me out for these questions?
nov2011 paper 11 question 10 part ii) how do we get the gradient..
nov 2011 paper 12 question 8 part i)

CAN U PLZZ POST THE LINKS OF THE PAPERS

 

[Iffat]

thanx wat bout the other 1

sry i luked at the ans closely n didnt understand wat u did cud u plz explain?

 

[iKhaled]

can someone please explain question 10 (ii) in may/june 2005 mathematics paper 3

how in the mark scheme he got the vector parallel to the place. how can i find a vector parallel to a plane?!!

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s05_qp_3.pdf
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s05_ms_3.pdf

please help!

 

[Kandinsky]

can someone please explain question 10 (ii) in may/june 2005 mathematics paper 3

how in the mark scheme he got the vector parallel to the place. how can i find a vector parallel to a plane?!!

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s05_qp_3.pdf
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s05_ms_3.pdf

please help!

To solve this problem, you should be familiar with the following fact: a plane can be completely determined by 2 intersecting lines.

In this case, imagine: A and B are points that do not belong to the line l (obvious from (a))
now suppose that you already have a plane P that contains A and the line l.
Since P contains l, then it contains the point K with the position vector OK = 4i - 2j + 2k.
Since the points A and K are in the plane P, then the line AK is also in P: AK = (4i-2j+2k) - (2i+2j+k) = 2i - 4j + k.

Now we have two intersecting(at point K) lines AK and l. This completely determines the plane P.
Note: the line l is parallel to the vector (i+2j+k) hence we will use, to
find the normal vector of the plane as (i+2j+k)x(2i-4j+k) = 6i+j-8k
Now we can say the equation of plane P is 6(x-x0) + (y-y0) + 8(z-z0) = 0
Substitute the coordinates of A, or of K to obtain a final answer.

 

[Zsiddiqui]

help please
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_23.pdf
question 5

 

[Kandinsky]

help please
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_23.pdf
question 5

(a) If M is a minimum point, then dy/dx = 0 at this point.

dy/dx = d(4*e^(x/2) - 6x + 3)/dx = 4* d(e^(x/2))/dx - 6 = 2* e^(x/2) - 6;
dy/dx = 0 when e^(x/2) = 6/2
or, x/2 = ln(3)
x = ln(3) = ln(3*3) = ln(9). hence a = 9;

(b) The area enclosed by the curve is the definite integral of y wrt x:

Area = integral( from 0 to 2) {y dx} = 8*e^(x/2) - 3*x^2 + 3x;
insert limits:

8*e^(2/2) - 3*4 + 3*2 - 8*e^0 = 8e - 12 + 6 - 8 = 8e - 14;

Hence, the answers are a = 9, area = 8e - 14;
You could use the marking scheme for the papers to find out solutions.

 

[Zsiddiqui]

(a) If M is a minimum point, then dy/dx = 0 at this point.

dy/dx = d(4*e^(x/2) - 6x + 3)/dx = 4* d(e^(x/2))/dx - 6 = 2* e^(x/2) - 6;
dy/dx = 0 when e^(x/2) = 6/2
or, x/2 = ln(3)
x = ln(3) = ln(3*3) = ln(9). hence a = 9;

(b) The area enclosed by the curve is the definite integral of y wrt x:

Area = integral( from 0 to 2) {y dx} = 8*e^(x/2) - 3*x^2 + 3x;
insert limits:

8*e^(2/2) - 3*4 + 3*2 - 8*e^0 = 8e - 12 + 6 - 8 = 8e - 14;

Hence, the answers are a = 9, area = 8e - 14;
You could use the marking scheme for the papers to find out solutions.

thank you so much

 

[RadzMau]

SOMEONE PLS HELP!!! [Arithmetic Series]
Question:
A 62 m length of rope is cut into pieces whose lengths are in arithmetic progression with a common difference of d m. Given that the lengths of the shortest and longest pieces are 0.5m and 3.5m respectively, find the number pieces and the value of d.

 

[Binyamine]

SOMEONE PLS HELP!!! [Arithmetic Series]
Question:
A 62 m length of rope is cut into pieces whose lengths are in arithmetic progression with a common difference of d m. Given that the lengths of the shortest and longest pieces are 0.5m and 3.5m respectively, find the number pieces and the value of d.

Beautiful Question.

So, Sum = 62
First Term is going to be the shortest piece, a = 0.5
Last Term is the longest piece, L =3.5

So Sum =62
n ( a + L)/2 = 62
n ( 0.5 + L ) = 124


n ( 0.5 + 3.5 ) = 124
4n =124
n = 31

And we use the information that last term is 3.5 to find the value of d

a + d ( n-1) = 3.5
0.5 + d ( 31-1)= 3.5
30d = 3
d = 3/30
d = 0.1

 

[RadzMau]

Beautiful Question.

So, Sum = 62
First Term is going to be the shortest piece, a = 0.5
Last Term is the longest piece, L =3.5

So Sum =62
n ( a + L)/2 = 62
n ( 0.5 + L ) = 124


n ( 0.5 + 3.5 ) = 124
4n =124
n = 31

And we use the information that last term is 3.5 to find the value of d

a + d ( n-1) = 3.5
0.5 + d ( 31-1)= 3.5
30d = 3
d = 3/30
d = 0.1

Thank you very muchh!

 

[messi10]

Question of Parametric Equations:

Please see the attachment for the question. please help me asap, thank you.

 

[RadzMau]

I would be very grateful if someone could help!
Another question on Arithmetic Series:
The sum of the first and the second terms of an arithmetic series is x and the sum of the (n-1)th term and nth term is y. Show that the sum of the first n terms is n/4(x+y). Find the common difference in terms of x, y and n.

 

[Rahma Abdelrahman]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_42.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_ms_42.pdf
Please help in question 7, part ii b and c

 

[french410]

I would be very grateful if someone could help!
Another question on Arithmetic Series:
The sum of the first and the second terms of an arithmetic series is x and the sum of the (n-1)th term and nth term is y. Show that the sum of the first n terms is n/4(x+y). Find the common difference in terms of x, y and n.

THE SUM OF THE 1ST 2 TERMS IS EQUAL TO x. ILL USE THE 1ST TERM AS VARIABLE z AND THE COMMON DIFFERENCE AS d
THEREFORE : z+(z+d)=x
AND :[z+(n-2)d]+[z+(n-1)d]=y

(n-2)d=(n-1)d-d
THEREFORE: z+(n-1)d-d+z+(n-1)d=y
2z+2(n-1)d=y+d
2[z+(n-1)d]=y+d
z+(n-1)d=(y+d)/2

THE SUM OF ARITHMETIC SERIES: S=n/2[2z+(n-1)d]
WE SUBSTITUTE THE VALUE OF z+(n-1)d

THUS WE GET: S=n/2[z+{(y+d)/2}
BY SIMPLIFYING WE GET : S=n/4[2z+y+d] y+d=x
THUS S=n/4(x+y)

AS FOR FINDING COMMON DIFFERENCE IN TERMS OF x, y, n
ITS EQUAL to : d=x-y


hope it helped
if u get confused ask ill explain

 

[french410]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_42.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_ms_42.pdf
Please help in question 7, part ii b and c

ok listen up for part b:
v=0.04t−0.00005t*2 given

we found out in the previous part(i) that the acceleration from (0 to 400 s )=0.02m/s*2
to find the velocity we find the integral of a=0.02
it will be equal to v1=0.02t+constant
we dont know what the constant is but we can find it out: when v=1, t=0 therefore 1=0.02(0)+constant ;1=constant

thus we got v1=0.02t+1
we then form equation : v1−v=0.02t+1−0.04t−0.00005t*2

simplifying the equation gives : v1−v=0.000 05(t − 200)*2−1


as for part c it is quite easy:
the lowest value you can get is when t=200: v1−v=0.00005(200−200)*2−1
v1−v=0−1=−1

the maximum value you can get is when t=0: v1−v=0.00005(0-200)*2−1
v1−v=0.00005(40000)−1
v1−v=2−1=1

thus −1 ≤ v1− v ≤ 1

hope it helped; if u have any doubts in the solving i will be happy to help

 

[RadzMau]

THE SUM OF THE 1ST 2 TERMS IS EQUAL TO x. ILL USE THE 1ST TERM AS VARIABLE z AND THE COMMON DIFFERENCE AS d
THEREFORE : z+(z+d)=x
AND :[z+(n-2)d]+[z+(n-1)d]=y

(n-2)d=(n-1)d-d
THEREFORE: z+(n-1)d-d+z+(n-1)d=y
2z+2(n-1)d=y+d
2[z+(n-1)d]=y+d
z+(n-1)d=(y+d)/2

THE SUM OF ARITHMETIC SERIES: S=n/2[2z+(n-1)d]
WE SUBSTITUTE THE VALUE OF z+(n-1)d

THUS WE GET: S=n/2[z+{(y+d)/2}
BY SIMPLIFYING WE GET : S=n/4[2z+y+d] y+d=x
THUS S=n/4(x+y)

AS FOR FINDING COMMON DIFFERENCE IN TERMS OF x, y, n
ITS EQUAL to : d=x-y


hope it helped
if u get confused ask ill explain

"(n-2)d=(n-1)d-d" - I didnt understand this part!
Pls explain it again...

 

[french410]

"(n-2)d=(n-1)d-d" - I didnt understand this part!
Pls explain it again...

ill explain u how (n-2)d=(n-1)d-d

we expand (n-2)d : and get nd-2d
we change equation to nd-d-d. they both are equal its just that i changed -2d into -d-d
then we factor d outside and get : (n-1)d-d

hope that helped
if still not ask again by stating what part u didnt get.