Mathematics: Post your doubts here!

[ffaadyy]

Can anyone help me?
Question 3, May/June 2011 Paper 41.

Refer to the above diagram and solve the horizontal and the vertical components of forces.

Horizontal Component:

T cos θ + T sin θ = 15.5
T cos θ = 15.5 - T sin θ

Vertical Component:

T cos θ + 8 = T sin θ
15.5 - T sin θ + 8 = T sin θ
23,5 = 2 T sin θ
11.75 = T sin θ

Therefore, T sin θ = 11.75~12

Put back this value of 'T sin θ' in the equation 'T cos θ = 15.5 - T sin θ'.

T cos θ = 15.5 - T sin θ
T cos θ = 15.5 - 12
T cos θ = 3.5.

Therefore, T cos θ = 3.5,

To find the value of θ, use the formula 'tan θ = opposite/adjacent'.

tan θ = opposite/adjacent
tan θ = (T sin θ) / (T cos θ)
tan θ = 12/3,5
θ = 73.73o

 

[nonamehapazard]

check here: http://www.schoolworkout.co.uk/a_level.htm
there are some good notes here, for p3...i hope its for all chapters...but i used vectors and complex numbers, and they were good...

thank u so much

 

[anonymous123]

AoA wr wb
9709_w10_qp_11
q7 part i
how to find the range?? in ms its f(x) =< 3

9709_w10_qp_41
Q3
i need a detailed solution to this plz..

 

[hassam]

This question is about scaling of complex numbers and I don't think that its included in our syllabus. Still, this's how you'll do it:

arg (1/3 - z) = pi/6

Assume that z' = 1/3 - z.

z' = 1/3 - z
3z' = 1 - 3z
3z - 1 = -3z'

Scaling a complex number (by a positive real number) doesn't change the argument, just the modulus. Scaling by a negative real number just rotates the vector by 'pi' so you just need to account for this. In the case of '3z - 1 = -3z'', we can either add 'pi' to 'pi/6' or subtract 'pi' from 'pi/6'; the answer will be correct either way.

'pi/6' - 'pi'
-5pi/6

'pi/6' + 'pi''
7pi/6

Both the answers, '-5pi/6' and '7pi/6', are correct.

Check this diagram for a better understanding of this question.

View attachment 6763

thanks ...was a good question

 

[hassam]

@XPFmember

 

[anonymous123]

AoA wr wb
9709_w10_qp_11
q7 part i
how to find the range?? in ms its f(x) =< 3

9709_w10_qp_41
Q3
i need a detailed solution to this plz..

 

[confused123]

little help required:-
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w03_qp_1.pdf

Q.6 part iii) ans is 3.96 after applying cosine rule, why cudn't we use simply r sin theta in order to calculate the length of the chord PQ. thanks

 

[hassam]

last 2 parts of this questions please

 

[gadagada]

Salam, I am giving P1, this session inshallah. But I have found difficulty solving MJ 2010(11) Q1, (i).
Help would be really appreciated. Thank you in advance.

 

[2pac]

Hi guys,this going to be my first post in this forum and would really like if you all could help me clear my math doubts.
Can anyone help me with
Q6 and Q9 of CIE A2 Pure math May 2010.
Thanks

 

[leadingguy]

Hi guys,this going to be my first post in this forum and would really like if you all could help me clear my math doubts.
Can anyone help me with
Q6 and Q9 of CIE A2 Pure math May 2010.
Thanks

which varient??

 

[leadingguy]

little help required:-
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w03_qp_1.pdf

Q.6 part iii) ans is 3.96 after applying cosine rule, why cudn't we use simply r sin theta in order to calculate the length of the chord PQ. thanks

the formula U are talking abt is used for right angled triangles only. I guess so.

 

[confused123]

the formula U are talking abt is used for right angled triangles only. I guess so.

Oh all right. Thankyou for correcting

 

[leadingguy]

Salam, I am giving P1, this session inshallah. But I have found difficulty solving MJ 2010(11) Q1, (i).
Help would be really appreciated. Thank you in advance.

tan(A-B) =(tanA - tanB)/(1 + tanA.tanB)
use this identity to substitute values of A and b

it will be as follows

tan(pi - x) = [tan(pi) - tan(x)]/ [1+tan(pi).tan(x)]

= (0 - k )/(1 + 0)........................ remember { tanx = k }
= -k Ans. hope U gt it

 

[Jaf]

This is a statistics (S1) question.
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_61.pdf - Question 7 (iii).

Obviously we need to find the P(X and Y), P(X) and P(Y) to find the answer.
The marking scheme states P(X and Y) is 1/28. I believe this is wrong and P(X and Y) should be 13/140.
This is because A:2 B:N2 C:2 gives us 1/28. But there's another way of getting 2 from A and also exactly two same balls. This is when A:2 B:8 C:8. The probability for this is 1/4 * 2/5 * 4/7 = 2/35. So the total is 13/140.

Even though the final answer remains unchanged (X and Y are still not independent), the probability should be the 13/140 according to me. Thoughts?

 

[gadagada]

tan(A-B) =(tanA - tanB)/(1 + tanA.tanB)
use this identity to substitute values of A and b

it will be as follows

tan(pi - x) = [tan(pi) - tan(x)]/ [1+tan(pi).tan(x)]

= (0 - k )/(1 + 0)........................ remember { tanx = k }
= -k Ans. hope U gt it

THANK YOU, SO MUCH!!

 

[smzimran]

tan(A-B) =(tanA - tanB)/(1 + tanA.tanB)
use this identity to substitute values of A and b

it will be as follows

tan(pi - x) = [tan(pi) - tan(x)]/ [1+tan(pi).tan(x)]

= (0 - k )/(1 + 0)........................ remember { tanx = k }

THANK YOU, SO MUCH!!

= -k Ans. hope U gt it

This method is wrong for this question, check the examiner report!

 

[alphabravocharlie]

last 2 parts of this questions please

how'd you solve the first one? 14C12/2?

 

[2pac]

which varient??

Variant 1.Apologies for not mentioning it before.

 

[OrlandoBloom<3]

I need help with paper 1 may June 2005 question 7 . Also paper 1 October November 2005 question 5 thanks !