Addmaths paper 1

[bilawalss]

I think there will be no permutations in paper 2 and it will be full of vectors and circular measure.

What was the answer in p1 q 2 it was a permutation question

 

[Saad Mughal]

I think there will be no permutations in paper 2 and it will be full of vectors and circular measure.

It will be of Relative Velocity, Vectors, Circular Measure (Q.12), Applications of Integration, Modulus Graphs, Linear Law, Trigonometry.

 

[Ishrar Afrida]

It will be of Relative Velocity, Vectors, Circular Measure (Q.12), Applications of Integration, Modulus Graphs, Linear Law, Trigonometry.

Yep and hence it will be hard

 

[Saad Mughal]

Yep and hence it will be hard

Not really. Trigonometry, Modulus Graphs and Linear Law are all just graph sketching and plotting. They're kinda like the 'free marks' of the paper.
If vectors for the unknowns 'lambda' and something come, it'll be easy peasy.
Circular Measure and Applications of Integration are easy too.
SOME Relative Velocity questions are hard, I hope the hard ones don't come.

 

[Shadow]

I think there will be no permutations in paper 2 and it will be full of vectors and circular measure.

Dont forget RV

 

[Shadow]

wat was the length of the line?

 

[Saifyyy]

wat was the length of the line?

5/4 untis

 

[Ammarfasih]

4<k<12

yr i wrote k=4 N k=12 ... so how mch marks pssible out of 4 ..?

 

[Saad Mughal]

yr i wrote k=4 N k=12 ... so how mch marks pssible out of 4 ..?

Did you do b^2 - 4ac = 0?

 

[redd]

Can anyone please help me with these two relative velocity questions, i brought about the answer but can get no logic. please help me with these.


http://papers.xtremepapers.com/CIE/...ematics - Additional (4037)/4037_s06_qp_1.pdf (question 3)

http://papers.xtremepapers.com/CIE/...matics - Additional (4037)/4037_s11_qp_21.pdf (question 9)

 

[redd]

25/4 = nC2 (1/4) - nC1(1/2)
25/4 = n(n-1)(n-2)!/2!(n-2)! (1/4) - n(n-1)!/1!(n-1)! (1/2)
25/4 = n^2 - n/8 - n/2
50 = n^2 - n - 4n
n^2 - 5n - 50 = 0
Simplifying gives n=10 and n=-5 (rej.).

Please give the question for the trigonometry one, I'll solve.

well, its so surprising, u remember whole questions i find it so difficult to even remember my answers, but thank God, till now all answers are going perfect

 

[Muaztsu]

Did you do b^2 - 4ac = 0?

i also did the same thing

 

[Saad Mughal]

i also did the same thing

That's wrong.

 

[Muaztsu]

mar

That's wrong.

marks that i`ll get minimum????

 

[Saad Mughal]

mar
marks that i`ll get minimum????

I don't remember the number of marks for the question, but you would get 2 marks deducted.

 

[*Anonymous*]

25/4 = nC2 (1/4) - nC1(1/2)
25/4 = n(n-1)(n-2)!/2!(n-2)! (1/4) - n(n-1)!/1!(n-1)! (1/2)
25/4 = n^2 - n/8 - n/2
50 = n^2 - n - 4n
n^2 - 5n - 50 = 0
Simplifying gives n=10 and n=-5 (rej.).

Please give the question for the trigonometry one, I'll solve.

Hey...
The equation was (1-sinQ-cosQ)^2 - 2(1-cosQ)(1-sinQ)
And mante, i wrote the range of k wrong.. i wrote less than 12 and 4...
And in the question we had to find P.. I found the gradiend wrong... like i wrote it 2... it was 1/2 and so got it all wrong...
How many marks will I lose...?
And oh! the binomial one... I put in the values but was confused so did not solve it furthur.. LOL

 

[Saad Mughal]

Hey...
The equation was (1-sinQ-cosQ)^2 - 2(1-cosQ)(1-sinQ)
And mante, i wrote the range of k wrong.. i wrote less than 12 and 4...
And in the question we had to find P.. I found the gradiend wrong... like i wrote it 2... it was 1/2 and so got it all wrong...
How many marks will I lose...?
And oh! the binomial one... I put in the values but was confused so did not solve it furthur.. LOL

For the k, gradient and binomial, you'll only get partial credit for k and binomial. 1-2 marks for the gradient one.
Ok so, I let x = -sin Q - cos Q.
Then,
=(1+x)^2 - 2(1-cos Q)(1-sin Q)
=1 + 2x + x^2 - 2(1-sin Q - cos Q + sin Q cos Q).
=1 + 2(-sin Q - cos Q) + (-sin Q - cos Q)^2 - 2 + 2 sin Q + 2 cos Q - 2 sin Q cos Q
=1 - 2 sin Q -cos Q + sin^2 Q + 2 sin Q cos Q + cos^2 Q - 2 + 2 sin Q + 2 cos Q - 2 sin Q cos Q
Rearranging all the terms correctly,
=1-2 -2 sin Q + 2 sin Q - 2 cos Q + 2 cos Q + 2 sin Q cos Q - 2 sin Q cos Q + (sin^2 Q + cos^2 Q)
=-1 + (1)
=0

 

[Shadow]

For the k, gradient and binomial, you'll only get partial credit for k and binomial. 1-2 marks for the gradient one.
Ok so, I let x = -sin Q - cos Q.
Then,
=(1+x)^2 - 2(1-cos Q)(1-sin Q)
=1 + 2x + x^2 - 2(1-sin Q - cos Q + sin Q cos Q).
=1 + 2(-sin Q - cos Q) + (-sin Q - cos Q)^2 - 2 + 2 sin Q + 2 cos Q - 2 sin Q cos Q
=1 - 2 sin Q -cos Q + sin^2 Q + 2 sin Q cos Q + cos^2 Q - 2 + 2 sin Q + 2 cos Q - 2 sin Q cos Q
Rearranging all the terms correctly,
=1-2 -2 sin Q + 2 sin Q - 2 cos Q + 2 cos Q + 2 sin Q cos Q - 2 sin Q cos Q + (sin^2 Q + cos^2 Q)
=-1 + (1)
=0

Substitution ! Damn it ! Cudn't nail that one

 

[Saad Mughal]

Substitution ! Damn it ! Cudn't nail that one

It could be done without substitution, you just had to deal with (-sin Q - cos Q) as one whole. I did not find it appropriate to follow working in this way so I substituted x.

 

[*Anonymous*]

For the k, gradient and binomial, you'll only get partial credit for k and binomial. 1-2 marks for the gradient one.
Ok so, I let x = -sin Q - cos Q.
Then,
=(1+x)^2 - 2(1-cos Q)(1-sin Q)
=1 + 2x + x^2 - 2(1-sin Q - cos Q + sin Q cos Q).
=1 + 2(-sin Q - cos Q) + (-sin Q - cos Q)^2 - 2 + 2 sin Q + 2 cos Q - 2 sin Q cos Q
=1 - 2 sin Q -cos Q + sin^2 Q + 2 sin Q cos Q + cos^2 Q - 2 + 2 sin Q + 2 cos Q - 2 sin Q cos Q
Rearranging all the terms correctly,
=1-2 -2 sin Q + 2 sin Q - 2 cos Q + 2 cos Q + 2 sin Q cos Q - 2 sin Q cos Q + (sin^2 Q + cos^2 Q)
=-1 + (1)
=0

Would it get two or two would be cut...?
I wrote the gradient as 2 instead of 1/2 and then multiplying by 6 coz i was in a hurry lol...
And how much in The Binomial ques?