Mathematics: Post your doubts here!

[littlecloud11]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_41.pdf i don't get #7 (iii) , please help. thanks

Ok, you have to get that A keeps moving up for a certain length of time even after B hits the ground owing to it's velocity at the time B comes to rest (Newton's first law). This is when the string gets slack. A then drops down from the max. height for the same length of time until the string gets taught again and stops it's movement.
So, first you have to find the velocity with which B hit the ground. Use the eq. v= u +at, initial velocity of B was 0 as it was at rest, the acceleration is the same as you calculated in part i and the time is 1.6 sec. The velocity of A is the same as B at the instant B hit the ground.
Now, you have to calculate the time it takes for A to come to rest. Use the same eq as before but take acceleration as -9.8 because gravity is the only force acting on it now. Final velocity is 0 and initial is the one you calculated. Then double this time to find your answer.

Hope this Helps!

 

[Minato112]

For this question there should be another value of θ when the base angle, x, is negative because as long as x is negative and less than 90 degree θ still lies in the 4th quadrant where Cos is positive. So another answer should be -x + α

The base angle x is negative?

 

[salvatore]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w06_qp_1.pdf

Please help me with question no. 3.
I understand how to find the area of the rectangle OCDB and sector AOB. But why are we supposed to find the area of the triangle? Isn't the area of the shaded part = (Area of rectangle - area of sector) ?

Thanks

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w06_qp_1.pdf

Please help me with question no. 3.
I understand how to find the area of the rectangle OCDB and sector AOB. But why are we supposed to find the area of the triangle? Isn't the area of the shaded part = (Area of rectangle - area of sector) ?

Thanks

No mate!

Area Of Trapezium - Area of Sector.. If you look at the diagram closely.

The answer would be Area Of Rectangle - ( Area Of Sector + Area Of Triangle)

I am going out for a bit.. so will solve it when I come back.

 

[salvatore]

No mate!

Area Of Trapezium - Area of Sector.. If you look at the diagram closely.

The answer would be Area Of Rectangle - ( Area Of Sector + Area Of Triangle)

I am going out for a bit.. so will solve it when I come back.

Thank you for your reply.. how come I don't see any trapezium there If possible, please help me identify it on the diagram.
Sorry for bothering..

 

[syed1995]

Forgive me for the vague drawing. Look at the 4 Arrows which I have placed. They show you the trapezium.. Honestly you can simply do it with the rectangle - ( Area of sector + triangle)... I just mentioned the Trapezium because you said Rectangle - Sector.

To do this we need to find the angle COA first. Which is marked as red in the next drawing




I will solve it with the Rectange method .. since it's both simle and easy to understand.

We need to find the Red Angle (COA)

COA = Pi/2 - Pi/3 = Pi/6

COA = Pi/6

Now We need to find the height and base of the triangle.

AC = 12 Sin (Pi/6) = 6 cm
OC = 12 Cos (Pi/6) = 6 √3


Area Of Rectangle = OB * OC = 12 * 6 √3
= 72 √3

Area Of Triangle = 1/2 * OC * AC = 1/2 * 6 * 6 √3
= 18 √3

Area Of Sector = 1/2 * r^2 * Angle AOB
= 1/2 * 12^2 * π/3
= 24 π

Shared Area= Area Of Rectangle - ( Area Of Unshaded Region) // Unshaded = Triangle + Sector

Therefore:
Rectangle - (Triangle + Sector)
=72 √3 - (18 √3 + 24 π)
=(72-18)√3 - 24 π
=54 (√3) - 24 π

a(√3) − bπ

a = 54 b = 24 Answer...

Also If we did the same with Trapezium .. then the height would be OC .. one side will be 12 .. other will be 12 - AC .. then find the area of trapezium with 1/2 * h * (Side 1 + Side 2)

= 1/2 * OC * (OB+AD)
= 1/2 * 6√3 * (12 + 6)
= 9 * 6√3
= 54√3

Area Of Trapezium - Area of Sector = Shaded

54√3 - 24π

same a = 54 b = 24

Check the answer and let me know if it's correct.

Hope this helps. Sorry again for the vague diagram and working .. I am not really good at explaining.

Cheers

 

[syed1995]

Thank you for your reply.. how come I don't see any trapezium there If possible, please help me identify it on the diagram.
Sorry for bothering..

The above was for you. Forgot to quote

 

[salvatore]

The above was for you. Forgot to quote

Thank you very very much for your help.. and you are gooooooood at explaining!

 

[sma786]

Salam,
I have my AS Level math mock paper on monday.
I'm very scared, please can anyone give me summarised notes for pure maths and statistics?
Thankyou!

 

[tanmaydube]

Summer 5 Q 7 ii) and Q8 ii)

Please help

 

[soumayya]

Summer 5 Q 7 ii) and Q8 ii)

Please help

Let f(x) = cosec x - 1/2 x - 1
f(0.5) = (1/sin 0.5 ) - 0.5(0.5) -1 =0.836
f(1) = (1/sin 1 ) - 0.5(1) -1 = -0.311

Since there is a sign change, a root lies between 0.5 and 1

 

[soumayya]

Summer 5 Q 7 ii) and Q8 ii)

Please help

 

[Iffat]

Ok, there definitely is a mistake. The position vector of R will be 9i+9j not minus. That's the only way you can solve this. If you doubt this just wait till your teacher confirms this.

a) PR = (9i+9j) - (i+3j) = 8i - 6j ; |PR|= 10
RQ = (9i+9j) -( 5i+11j) = 4i - 2j; |RQ| square root (20)
QP = (5i + 11j) - (i + 3j) = 4i + 8j; |QP| square root (80)
Use the Cos rule next.
Cos(theta) = (RQ^2+ QP^2- PR^2)/ 2*RQ*QP = (20+80 -10^2)/ 2* root 20* root 80 = 0
theta comes out as 90

b)PR/ |PR| = 8i-6j/ 10

c) (5i+ 11j) = m (i+3j) +n (9i+9j)
comparing two sides-
5= m +9n
11= 3m + 9n
solve and you get m=3 and n=2/9

Thanx u wer ryt it is plus ,
d photocopy i haf isnt clear

 

[sma786]

Can anyone give me a breif explanation of integration??

 

[tanmaydube]

Let f(x) = cosec x - 1/2 x - 1
f(0.5) = (1/sin 0.5 ) - 0.5(0.5) -1 =0.836
f(1) = (1/sin 1 ) - 0.5(1) -1 = -0.311

Since there is a sign change, a root lies between 0.5 and 1

Thanks a ton soumayya di!!!!!

 

[tanmaydube]

View attachment 21651

Cannot understand the 4 step clearly di,

it should be '4x+ln3' why is it '4x-ln3'

Thanks a lot again!

 

[Scafalon40]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w02_qp_3.pdf
Q5 part (iii)
It's only worth one mark, could you tell me how to solve it?
n.b I'd prefer a method which is worth only one mark, as the question indicates
thanks a million!

 

[agarveer]

paper 32

nov 04 q 7

 

[tanmaydube]

Winter 5
Q 5, Q 6, Q10
Please help!!

Thanks a lot!

 

[Minato112]

Winter 5
Q 5, Q 6, Q10
Please help!!

Thanks a lot!

5) 8 sin θ − 6 cos θ = R sin(θ−α)
= R sin θ cos α - R cos θ sin α
Therefore,
R cos α = 8 ----- (1)
and R sin α = 6 ------ (2)

(1)^2 + (2)^2 => (R cos α)^2 + (R sin α)^2 = (8)^2 + (6)^2
=> R^2 [ (cos α)^2 + (sin α)^2 ] = 100 [ (cos α)^2 + (sin α)^2 ] = 1 (Identity)
Thus R^2 = 100
R = 10

(2) / (1) => tan α = (3/4)
α = *tan inverse* (3/4)
= 36.9○

8 sin θ − 6 cos θ = 10 sin(θ−36.9)
.......................................................

8 sin θ − 6 cos θ = 7
i.e 10 sin(θ−36.9) = 7
sin(θ−36.9) = 0.7
(θ−36.9) = *sin inverse* (0.7) = 44.4○
or (θ−36.9) = 180○ - *sin inverse* (0.7) = 135.6○

Therefore θ = 44.4○ + 36.9○ = 81.3○
or θ = 135.6○ + 36.9○ = 172.5○

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For the next 2 questions, if noone answers it, i'll answer it. I have to go for now. Hope it helps