Mathematics: Post your doubts here!

[Hazooba]

9709_s11_qp_31 guys can you help with q1 pleasex

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w06_qp_3.pdf

Q7 (ii)

 

[Zain Salman Dar]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_32.pdf

Can someone please solve question 2?

take 5^x=y
y/5=y-5
y=5y-25
y=25/4

5^x=25/4
xlog5=log25/4
x=(log25/4)/log5

 

[Zain Salman Dar]

9709_s11_qp_31 guys can you help with q1 pleasex

Post paper link here.

 

[Wolfgangs]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_4.pdf

Can someone please solve q6 and explain part iv as soon as possible

 

[3l_general]

Help me out please with the last question
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_41.pdf

 

[3l_general]

This is the question

 

[DaniyalK]

Help me out please with the last question
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_41.pdf

Find the 2 angles in the triangle.

There are 2 tension forces, one going towards A and the other towards C. By resolving forces at B, you can make 2 equations

8 = TaSin(53.13) + TcSin(36.87)
TaCos(53.13) = TcCos(35.87)
Solve simultaneously and you'll get Ta = 6.4 N

ii) Resolve the forces at point A.
Vertical: Friction + (0.2)(10) = 6.4Cos(53.13)
Friction = 1.84N
Use F = uR
1.84 = u*6.4Sin(53.13)
u = 0.359

 

[DaniyalK]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_4.pdf

Can someone please solve q6 and explain part iv as soon as possible

B will go up 0.36m because A has gone down 0.36m, but then it will move a small distance once A reaches the floor. So we need to find that distance. A's final velocity before reaching the ground is 1.2m/s, which is also the initial velocity of B when it starts to move the extra distance.
u = 1.2, v = 0, a = -10, s = ?
solve for s and you get 0.072m
Total distance is 0.36+0.36+0.072 = 0.792m

 

[DaniyalK]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w06_qp_3.pdf

Q7 (ii)

let the direction of the plane = n = (a,b,c)
since the plane is perp. to the given plane, their direction vectors are perpendicular, so the dot products of their direction vectors is 0.
(a,b,c) . (1,2,3) = 0
a + 2b + 3c = 0 Eq 1
since the plane is parallel to the line, n is perp. to the direction vector of the line
(a,b,c) . (1, -2, 1) = 0
a - 2b + c = 0 Eq 2

From eq 1
a = -2b -3c
put it in eq 2 and simplify, you get
c = -2b
put it in eq 1
a = 4b
This means that if b = 1 then c = -2 and a = 4
so n = (4,1,-2)
The equation of the plane is

(4,1,-2).(x,y,z) = (4,1,-2).(2,1,4)
4x + y -2z = 1

 

[subaz sarma]

Can you please help me find the pdf version for the book ?

Thank You

 

[papajohn]

Please anyone solve for me. I cant understand anything. Please anyone explain me briefly. Daniyal & Zain
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_32.pdf

Question number Q6(i)
MS --->http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_ms_32.pdf

 

[DaniyalK]

Please anyone solve for me. I cant understand anything. Please anyone explain me briefly. Daniyal & Zain
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_32.pdf

Question number Q6(i)
MS --->http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_ms_32.pdf

Triangle AOB is isosceles since OB = r.
Use law of cosine in AOB and you'll get r^2 = AB^2 + r^2 - 2ABrCosθ
Simplify and you'll get AB = 2rCosθ
Angle AOB is (pi - 2θ)
Shaded area = Area of sector BAC+2*Area of sector BOA−2*Area of triangle BOA
put this equal to half of the area of the circle (0.5 pi r^2)
while simplifying, you will notice that r^2 will be in every term on both sides, so you can cancel it out and will be left with θ only.
Use the double angle formula to change the Cosθ in AB to Cos2θ once you've simplified as much as you can and make it the subject.

 

[dubi2pro]

Some one Please help..

Mechanics doubt.

A Particle P is projected vertically upwards with a speed of 30 m/s from a point A.
The Point B is h metres above A. The particle moves freely under gravity and is above B for a time 2.4 seconds.

Calculate the Value of h

 

[crazytaylorfanXD]

Hey can someone help me out here??

/9709_s13_qp_32.pdf
q5)

 

[Zain Salman Dar]

Hey can someone help me out here??

/9709_s13_qp_32.pdf
q5)

Post the link to the paper please.

 

[crazytaylorfanXD]

Post the link to the paper please.

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

 

[DaniyalK]

Some one Please help..

Mechanics doubt.

A Particle P is projected vertically upwards with a speed of 30 m/s from a point A.
The Point B is h metres above A. The particle moves freely under gravity and is above B for a time 2.4 seconds.

Calculate the Value of h

Find the time to reach max height.
u = 30 v = 0 a = -10 t = ?
0 = 30 - 10t
t = 3 seconds
It reached B at 3 - 2.4/2 = 1.8 s
u = 30 a = -10 t = 1.8 s =?
s = 30(1.8) +(0.5)(-10)(1.8)^2
s = 37.8 m

 

[dubi2pro]

Find the time to reach max height.
u = 30 v = 0 a = -10 t = ?
0 = 30 - 10t
t = 3 seconds
It reached B at 3 - 2.4/2 = 1.8 s
u = 30 a = -10 t = 1.8 s =?
s = 30(1.8) +(0.5)(-10)(1.8)^2
s = 37.8 m

why did u divide 2.4 by 2?

 

[DaniyalK]

why did u divide 2.4 by 2?

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)