Mathematics: Post your doubts here!

[DaniyalK]

can somebody help me with this question

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_32.pdf
question 8b ???

the distance we need to find has 2 components, the purple and red distances.
If the purple distance is x then in triangle PQR
(1 + x)^2 = 1^ + 1^
x = sqrt2 - 1
In the brown triangle, the hypotenuse is sqrt2
the red distance is half of this
so the distance is (sqrt2 - 1) + sqrt2/2 = 1.12

 

[papajohn]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_32.pdf
Question 5 and question 9

Q9 a anyone??

 

[dubi2pro]

It's above B for 2.4 seconds. It is going up for half of that time to reach the max height. For the other half, it's falling back down to B. I just considered the first half of the motion, when it starts from t = 0 and reaches the maximum height at 3 seconds. Since I considered only half of the total time in air (6 seconds), I also used only half of the total time above B (2.4)

Excellent (Y)

Thanks

 

[DaniyalK]

Q9 a anyone??

the blue angle is 135. both the blue lines have length 2. Use cosine rule to find the length of the green line.
Edit:
for a part let w = x + iy
(x + iy) + 3(x-iy) = i(x+iy)^2
4x - 2iy = ix^2 - iy^2 - 2xy
equate real and imaginary parts
4x = -2xy
y = - 2
-2y = x^2 - y^2
put y = -2
x = sqrt 8

 

[crazytaylorfanXD]

the distance we need to find has 2 components, the purple and red distances.
If the purple distance is x then in triangle PQR
(1 + x)^2 = 1^ + 1^
x = sqrt2 - 1
In the brown triangle, the hypotenuse is sqrt2
the red distance is half of this
so the distance is (sqrt2 - 1) + sqrt2/2 = 1.12

Thank you so much

 

[crazytaylorfanXD]

Thank you so much

hey do you mind helping me with another question ?
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_32.pdf
question 10

 

[DaniyalK]

hey do you mind helping me with another question ?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_32.pdf
question 10

in the first part, use the triangle shown in the diagram.
tan60 = r/h
r = tan60 h
put this in V = pi*r^2*h
V = pih^3
dV/dh = 3pih^2
dV/dt = -k sqrt (h)
dV/dt = dV/dh * dh/dt
dh/dt = dV/dt / dV/dh
dh/dt = - (k/3pi) * h^(2/5)

in the second part, take h to the other side, integrate both sides and then use t = 0, h = H to get the constant of integration in terms of H. Then use t = 60, h = 0 to get k in terms of H. Rearrange it to make t the subject.
in the third part, in your equation with t as the subject, put h = 1/2H. H will cancel out and you should get the answer.

 

[dubi2pro]

It's above B for 2.4 seconds. It is going up for half of that time to reach the max height. For the other half, it's falling back down to B. I just considered the first half of the motion, when it starts from t = 0 and reaches the maximum height at 3 seconds. Since I considered only half of the total time in air (6 seconds), I also used only half of the total time above B (2.4)

Last question, i hope so but please help if u can..

i have drawn the graph for u, jus tell me how to do the a. part.

and is the area under for this specific graph above it (shaded region) or below it?

this is the question.

The brakes of a train which is travelling at 108 KM/H Are applied as the train passes a point A. The brakes produce a retardation of magnitude 3f ms/square until the speed of the train is reduced to 36km/h
The train travels at this speed for a distance and is then uniformly accelerated at f ms/square until it again reached the speed 108KM/H as it passes point B. The time taken by the train in travelling from A to B, a distance of 4 km, is 4 Minutes.

 

[crazytaylorfanXD]

in the first part, use the triangle shown in the diagram.
tan60 = r/h
r = tan60 h
put this in V = pi*r^2*h
V = pih^3
dV/dh = 3pih^2
dV/dt = -k sqrt (h)
dV/dt = dV/dh * dh/dt
dh/dt = dV/dt / dV/dh
dh/dt = - (k/3pi) * h^(2/5)

in the second part, take h to the other side, integrate both sides and then use t = 0, h = H to get the constant of integration in terms of H. Then use t = 60, h = 0 to get k in terms of H. Rearrange it to make t the subject.
in the third part, in your equation with t as the subject, put h = 1/2H. H will cancel out and you should get the answer.

thanks once again you sir are awesome

Do you mind helping me in a couple more??
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_32.pdf

question 7 part ii
question 8
question 10 part ii

 

[crazytaylorfanXD]

can some one help me with these questions?
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_32.pdf

question 7 part ii
question 8
question 10 part ii

i need to see the working too
i tried all of these like 3 to 4 times and still couldnt get it :/

 

[dubi2pro]

q 30! anyone?

 

[Dream.Eater]

Can someone please help me out with this question?

"When (1 + ax)^n is expanded as a series in ascending powers of x, the coefficients of x and x^2 are -6 & 9 respectively. Find a & n."

 

[DaniyalK]

Last question, i hope so but please help if u can..

i have drawn the graph for u, jus tell me how to do the a. part.

and is the area under for this specific graph above it (shaded region) or below it?

this is the question.

The brakes of a train which is travelling at 108 KM/H Are applied as the train passes a point A. The brakes produce a retardation of magnitude 3f ms/square until the speed of the train is reduced to 36km/h
The train travels at this speed for a distance and is then uniformly accelerated at f ms/square until it again reached the speed 108KM/H as it passes point B. The time taken by the train in travelling from A to B, a distance of 4 km, is 4 Minutes.

Let the three times for the three stages be t1,t2 and t3
Area under the graph = 4000
(0.5)(20)(t1) + (240)(10) + (0.5)(20)(t3) = 4000
t1 + t3 = 160
since t1 + t2 + t3 = 240
t2 = 80
Now, the remaining time is 160s. This is divided between t1 and t3 in the ratio 1:3 (we know this from the accelerations)
t1 = 1/4 * 160 = 40
t2 = 3/4 * 160 = 120
in the first part of the graph
u = 30, v = 10, a = -3f, t = 40
v = u + at
f = 1/6

can some one help me with these questions?
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_32.pdf

question 7 part ii
question 8
question 10 part ii

i need to see the working too
i tried all of these like 3 to 4 times and still couldnt get it :/

7) gradient is infinity so put the denominator of the derivative to 0
x(3y^3-1) = 0
3y^3 - 1 = 0
y = 0.693
use this in the equation ln(xy) - y^3 = 1 to get x

10) find the equation of line AB
AB = (3,-2,4) + s(-1,1,3)
express in parametric form
AB = (3 - s, -2 + s, 4 + 3s)
Subtract C
NC = (2-s, 3 + s, 7 + 3s) where N is the foot of the perpendicular
NC. AB = 0
(2-s, 3+s, 7 + 3s) . (-1,1,3) = 0
s = -2
NC = (4,1,1)
magnitude = sqrt(4^2 + 1^2 + 1^2)
= 3 sqrt2

8)i) dy/dx = e^(-.5x^2)* 4x(0.5)/(1+2x^2) -xe^(0.5x^2) * (1 + 2x^2)^0.5
take e^(0.5x^2) common and put dy/dx = 0
0 = 2x/sqrt(1+2x^2) - x* sqrt (1+ 2x^2)
2x = x(1 + 2x^2)
2x^3 - x = 0
x = sqrt2/2

8)ii) not sure about this
edit: I figured out how to get to the expression by working backwards
a ^2 = ln(4+8a^2)
e^(a^2) = 4 + 8a^2
e^(a^2) = (1+2a^2)/0.25
e^(-a^2) = 0.25/(1 + 2a^2) (taking reciprocal)
e^(-.5a^2) = 0.5/(1 + 2a^2)^0.5
0.5 = e^(-.5a^2) * sqrt(1+2a^2)


8)iii) use xn = 2 in the given equation and keep putting the previous answer in place of xn until the second decimal place stops changing.

Can someone please help me out with this question?

"When (1 + ax)^n is expanded as a series in ascending powers of x, the coefficients of x and x^2 are -6 & 9 respectively. Find a & n."

expand it till x^2
1 + anx + n(n-1)(ax)^2 /2
an = - 6
n(n-1)(a^2)/2 = 9
solve simultaneously.
a = -3, n =2

 

[Zain Salman Dar]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_32.pdf
Question 5 and question 9

For questions 5:

First differentiate the equation in terms of x
replace dy/dx as 0
You will get the equation: 3x^2+y^2-6ax=0
From original equation make y the subject
Replace in the equation above.
You will get the value of x= (3a)^1/2

 

[Zain Salman Dar]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_32.pdf
Question 5 and question 9

For question 9:

part 1)
w=x+iy
w*=x-iy
Substitue this in the equation
equate real and imaginary parts
get value of x= 2(2)^.5 and y=-2

part 2)
all drawing of circle and lines and distance :S. If you cannot calculate it,then reply below

 

[iYuuki]

Can someone please help me out with question 7(b)?? Much appreciated!

 

[DaniyalK]

Can someone please help me out with question 7(b)?? Much appreciated!

w = cos2X + isin2X

cos2X + isin2X - 1/cos2X + isin2X + 1
= cos^2(X) - sin^2(X) + i(2sinXcosX) - 1/ cos^2(X) - sin^2(X) + i(2sinXcosX) + 1
= -2sin^2(X) + i2sinXcosX/ 2cos^2(X) + i2sinXcosX
= 2sinX(icosX - sinX)/2cosX( cosX + isinX)
= tanX(icosX - sinX)/ (cosX + isinX)
= tanX.i(cosX + isinX)/(cosX + isinX)
= itanX

 

[papajohn]

can someone please help with question 9b) http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_33.pdf

Q9 B anyone? Please explain me how to calculate area of a triangle?

 

[papajohn]

Q9 B anyone? Please explain me how to calculate area of a triangle?

The main problem I do not understand is that how do we calculate the area of the triangle ABC while it is not making any triangle??
Please help me out of this confusion

 

[DaniyalK]

View attachment 51292

The main problem I do not understand is that how do we calculate the area of the triangle ABC while it is not making any triangle??
Please help me out of this confusion

use 1/2 * BC * OA