Mathematics: Post your doubts here!

[asexamskillme111]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_31.pdf
Can someone please explain why the circle has to pass through the origin in 4 ii? I understand that there's a circle. What I did is I made the radius of the circle equal to the modulus of u since it made sense to me. Help please?

 

[WayneRooney10]

Why does (cosx)^4=5/8 have 4 answers? Can someone explain the logic behind it? Thanks!

 

[kiara15]

hey plz give me link for how to draw argand diagrams, i m totally messed up with this

 

[kiara15]

can anyone solve permut n comb question from 2011 p.paper, i had an ambiguity solving it.. paper 3

 

[kiara15]

c an u solve 2-3 quest of perm n comb from p.papr? i had ambiguity or anyone?

 

[Esme]

hey plz give me link for how to draw argand diagrams, i m totally messed up with this

These notes were uploaded by Phyzac

 

[kiara15]

These notes were uploaded by Phyzac

thank u

 

[Esme]

Why does (cosx)^4=5/8 have 4 answers? Can someone explain the logic behind it? Thanks!

That's because you'll write it as cos x = +- (5/8)^1/4
just like for square root, you write +-(square root)...
So you'll have x= cos^-1 ( positive (5/8)^1/4 ) and x=cos^-1 (negative (5/8)^1/4)
This should give you 4 different values of x.

 

[Esme]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_31.pdf
Can someone please explain why the circle has to pass through the origin in 4 ii? I understand that there's a circle. What I did is I made the radius of the circle equal to the modulus of u since it made sense to me. Help please?

Yeah you're right. The modulus of u is the radius of the circle. Automatically the circle will pass through the origin because the modulus is the distance from the origin to the vector i.e. your radius of cirlce

 

[Ashique]

c an u solve 2-3 quest of perm n comb from p.papr? i had ambiguity or anyone?

Which questions do you need help with?

 

[Ashique]

Also, can someone PLEASE help me out in question 10? http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf

Like, when we substitute we're going to get
(u^(n+2)+u^n)/sec^2 x du
right?

How would we integrate this? I urgently need help.

 

[kiara15]

Which questions do you need help with?

p63- q 3 and p62 - q5

 

[biba]

The complex number u is defined by u = (1 + 2i)^2 / (2+i)

(i) Without using a calculator and showing your working, express u in the form x + iy, where x and

y are real. [4]

(ii) Sketch an Argand diagram showing the locus of the complex number s such that |s − u| = |u|.
[3]


can sum1 help me with the argand diagram ?

anyone??? [^__^]

 

[SararaIH]

The answer in the marking scheme is INCORRECT for this question. The examiner report shows the correct answer which is {(x^2+4)^3}/2
I'm too lazy to type out the solution so here you go.
I used limits to for integration. You could also integrate and find a constant instead of using limits, which I find easier.

No wonder I couldn't understand the Mark Scheme -_-
Thanks so much

I'm confused with one thing
When can I use this formula---> (1/2a)*ln(a+b/a-b) ?

 

[Esme]

aah yu got confuse in functions ....
p(z) = z^3 + 6z^2 +24z +32
p(1) = 1^3 + 6(1)^2 +24(1) +32 =63
similarly
p(z^2) = (z^2)^3 + 6(z^2)^2 + 24z^2 + 32 (just replaced z by z^2)
now if yu make substitution x=z^2 to solve the equation ... u shall get the same equation as in part ii)a
x^3 + 6x^2 +24x + 32

now the solutions of this equation (values of x) will be same as that of values of z in previous parts

now z^2 = x
so z = x^(1/2)
ans so we square root all the values of x to get z ....
Hope yu get it ..

Yeah I got it thanks

 

[Ashique]

p63- q 3 and p62 - q5

For June 2012, p63, q3-
i) The question tells you that P and L and at the ends so
P *rest all the remaining 7 letters in between* L

So in how many way can yo arrange the letters in between?
You have 7 ! Since you have three E's, you should divide by 3!
So 7!/3!= 820 ways.
Now because you can switch the P and the L, you should multiply this by 2, so 820*2= 1680

ii) In this part they told you that there will be no E's. So you have 6 letters remains to choose from. Now because you'll have to choose 4 letters, your combination is going to be= 6C4= 15 ways.

iii)Here there told you that you already have one E in the combinations of four letter.
So
E *three other letters chosen from the remaining 6*

So you have to choose 3 from the 6, so 6C3= 20 ways

iv) In this question they told you that there will be no restrictions. You could go about three ways of doing this-
no E's= 15 ways (you found above)
1 E= 20 ways (again you found it above)
2 E = EE *two other letters from the remaing 6 letters*= 6C2=15 ways

3 E= EEE *one more letter from the remaining 6 letters*= 6C1= 6 ways




Now addiing them up= 15+20+15+6= 56 ways.


Hope you got it!


For p62 please see my post below

 

[talalz]

does any body have MATHS P3 VECTORS notes ?? please i need help urgent ??

 

[SararaIH]

does any body have MATHS P3 VECTORS notes ?? please i need help urgent ??

This might be really useful
Vectors Study Guide

 

[Ashique]

p63- q 3 and p62 - q5

For the p63 please see my post above

For June 2012, p62, q5-
i) Since there are no restrictions, the candidates have to choose 6 questions from a choice of total 11 questions, so= 11C6 =462

ii) Candidates are bound to choose atleast 4 questions from part A. So there are many ways, so go about this

*ALL questions chosen from part A*
OR
*5 questions chosen from part A* and *1 questions chosen from part B*
OR
*4 questions chosen from part A* and *2 questions chosen from part B*

So you can now put the numbers ins
8C6=28
OR
8C5*3C1=168
OR
8C4*3C2=210

Adding all of them up= 28+168+210= 406

iii) Now again a LOT of combinations:
If 1 and 2 from part A are chosen, there are 4 more questions to be selected. Here are the possible ways of choosing the questions:
all question from part A, 6c4 = 15 OR
3 questions from Part A and 1 question from part B 6c3*3c1= 60 OR
2 questions from Part A and 2 question from part B 6c2*3c2= 45 OR
1 questions from Part A and 3 question from part B 6c1*3c3 = 6

If 1 and 2 from part A are not selected, 6 more to be selected, here are the combinations =
all questions from from A= 6C6=1 OR
6 questions from Part A and 1 question from part B= 6C5*3C1 =18 OR
4 questions from Part A and 2 question from part B = 6C4*3C2= 45 OR
3 questions from Part A and 3 question from part B = 6C3*3C3 20

Adding all of them up= 15+60+45+6+1+18+45+20= 210 ways

Please do tell me if you didn't get it.

 

[Esme]

does any body have MATHS P3 VECTORS notes ?? please i need help urgent ??