Mathematics: Post your doubts here!

[iKhaled]

can someone help me with question 2, 5 and 6 of the paper october november 2011 pr 53 M2

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_53.pdf

 

[snowbrood]

f(x) = a(x - p)^2 + k
Now plug in values from the graph:
4 = a(0.1 - 0.3)^2 + 8
a = -100

f(x) = -100(x - 0.3)^2 + 8

i did tried to solve this way but it gives only vertex plug in 0 or 0.6

 

[Khurram Memon]

Assalamoalaikum Wr Wb!

Post your doubts here. Make sure you give the link to the question paper when posting your doubts.


May Allah give us all success in this world as well as the HereAfter...Aameen!!

And oh yeah, let me put up some links for A level Maths Notes here.

Check this out! - Nice website, with video tutorials for everything! MUST CHECK

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Maths Notes by destined007

Some Notes for P1 and P3 - shared by hamidali391

A LEVEL MATHS TOPIC WISE NOTES

MATHS A LEVEL LECTURES

compiled pastpapers p3 and p4 - by haseebriaz

Permutations and Combinations (my explanation)- P6

Permutations and Combinations - P6

Vectors - P3

Complex No. max/min IzI and arg(z) - P3

Sketcing Argand Diagrams - P3 (click to download..shared by ffaadyy)

Range of a function. - P1

Need pat papers from 1990- 2000 with marking scheme

 

[Yousif Mukkhtar]

Can some one help me in q11 ) iii) and iv)
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_12.pdf

 

[Jspake]

What's stated in the question is the length of the side of triangle ABC. Length AB is not equal to length AQ. You have to use Pythagoras Theorem to find r or AQ.

I'm aware that you are supposed to find the length AQ using pythagoras theorem.. but I don't understand why AQ is the radius of sector APQC. The radius is 2 cm right??

 

[Dug]

I'm aware that you are supposed to find the length AQ using pythagoras theorem.. but I don't understand why AQ is the radius of sector APQC. The radius is 2 cm right??

Radius is not 2cm.

h^2 = p^2 + b^2
AB^2 = AQ^2 + BQ^2
2^2 = AQ^2 + 1^2
AQ = √3

 

[Iffat]

Help plz!
The position vectors,relative to an origin O, of 3 points P, Q and R are i+3j, 5i+11j and 9i-9j respectively.
a) By finding the magnitude of the vectors PR, RQ and QP, show that the ange PQR is 90 degreees
b) Find the unit vector parallel to PR
c) Given that OQ=mOP+nOR, where m and n are constants, find the value of m and of n

 

[salvatore]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_11.pdf

Can someone please help me with no. 11 (ii & iv)?
I totally don't understand how the range becomes 2 ≤ f(x) ≤ 10
In addition, please explain how to sketch the graph..

Thanks a lot!

 

[unseen95]

Hello, I need some help with question number 4 (ii) got 53.13 and 126.87 but could not get 20.16. How do we get 20.16 please give me detailed explanation so that i can solve similar problems next time. The link to question paper is http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_3.pdf

 

[Iffat]

I think you made a mistake while copying the question because the first angle comes out to be 37.8 and that is not a mistake but the actual answer using your values. Would you mind checking or perhaps attach the paper?

the question isnt 4rm a paper, its a homework we were given n i rechecked it, itsexactly the way i wrote it

 

[unseen95]

the question isnt 4rm a paper, its a homework we were given n i rechecked it, itsexactly the way i wrote it

The question is in http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_3.pdf and it is 4(ii) detailed solution please.

 

[unseen95]

people help me with the above mentioned question please.

 

[TERMINATOR]

An S1 normal distribution question. I think it will be like this 8.<X <32=.94. But the ans dies not match. The correct answer is 6.38. Any help will be highly appreciated.

 

[~`Heba`~ :)]

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

 

[Minato112]

The question is in http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_3.pdf and it is 4(ii) detailed solution please.

I havent done the question but I'll try to be brief on it. You have expressed 7 cosθ+24 sinθ in the form Rcos(θ−α), right?. For the second part, instead of solving 7 cosθ+24 sinθ = 15, solve Rcos(θ−α) = 15.

Then it'll be as follows,

Rcos(θ−α) = 15​

cos(θ−α) = 15/R​

(θ−α) = *cos inverse* (15/R) = x (Let's call it to be x)​

or (θ−α) = 360 - *cos inverse* (15/R) = y (Let's call it to be y)​

​

Therefore (θ−α) = x and (θ−α) = y​

Thus θ = x + α and θ = y + α​

​

Hope It Helps!​

 

[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..