Mathematics: Post your doubts here!

[Dug]

done but no help why dont u try to solve it and see if u can get an equation?>

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

 

[salvatore]

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

Please help me solve question no. 3, I do not understand how to do it.
An explanation will be appreciated.
Thanks a lot!

 

[Dug]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_11.pdf

Please help me solve question no. 3, I do not understand how to do it.
An explanation will be appreciated.
Thanks a lot!

i) Fit 2 cosine cycles in the stated domain. y = 1/2 is a straight horizontal line passing through (0, 1/2).
ii) Number of roots = Number of Points of Intersection of the line and the cosine function.
iii) Simply use ratio and proportion. 2pi = 4 roots, 10pi = 20 roots

 

[Jspake]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_11.pdf

In no. 3, why is the value of AQ the radius of the sector APR? As stated in the question, isn't the radius 2 cm?
Please help!

 

[Dug]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf

In no. 3, why is the value of AQ the radius of the sector APR? As stated in the question, isn't the radius 2 cm?
Please help!

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.

 

[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

My P1 Notes! - Only few chapters available at the moment!

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!​