Mathematics: Post your doubts here!

[Manoj Panigrahi]

examsolutions - This very good site, and it helped me even to finish my math P1 in 1 month and P4 in 1 month. But it need practice, if you won't, then Math is useless. All the best.
Here is your answer in my way :¬

So, as on your request here you go :¬
NOTE : I made two separate graphs so that to make sure how their graphs actually look like, now try them in one graph, do label it correctly, look at the points on cos 3x and cos x. You can also label it as 60 90 120 180 240 270 300 360 and also in radians, its our own wish, but I do use radians, as its related to trigonometry

View attachment 46123

If you still are not cleared. Do ask.

Please help me to solve Oct/Nov2011 Paper 4(Extended) 0580/42 Question No.4 (a,b,c)

 

[Thought blocker]

Awesome12

Please help me to solve Oct/Nov2011 Paper 4(Extended) 0580/42 Question No.4 (a,b,c)

 

[Thought blocker]

Rockstar RK
Like this :¬
int csc(4x)*cot(4x) dx

(1/sin(4x) )* cos(4x)/sin(4x) = cos(4x)/sin^2 (4x)

int cos(4x) dx/sin^2(4x)

u = sin(4x)

du = 4*cos(4x) dx

1/4 du = cos(4x) dx

then

int csc(4x)*cot(4x) dx = int 1/4u^2 du

-1/(4u) + C

-1/(4sin(4x)) + C

-csc(4x)/4 + C


-1/(4u)

-1/(4*sin(4x))

-csc(4x)/4

 

[Awesome12]

Awesome12

Yes?

 

[Thought blocker]

Yes?

Solve that IGCSE one.

 

[Tamara Tango]

hi! i just did may june 2010 paper 31 number 5 and saw the marking scheme,but still i did not get idea how to answer the question.
Given that y = 0 when x = 1, solve the differential equation
xy (dy/dx) = y2 + 4,
obtaining an expression for y2 in terms of x.

Can anyone help me? thank you anyway

 

[Thought blocker]

hi! i just did may june 2010 paper 31 number 5 and saw the marking scheme,but still i did not get idea how to answer the question.
Given that y = 0 when x = 1, solve the differential equation
xy (dy/dx) = y2 + 4,
obtaining an expression for y2 in terms of x.

Can anyone help me? thank you anyway

5)
xy dy/dy = y^2 + 4
⌡(y/ y^2 +4) dy = ⌡1/x dx
Diffrentiating y^2+ 4 = 2y
so 1/2⌡2y/y^2 +4 dy = ln x + c
1/2 ln |y^2+4| = lnx +c
when y=0, x= 1
1/2 ln4 = ln1 +c
2*1/2 ln 2 = c
c= ln 2

1/2ln |y^2+4| = lnx +ln2
ln (y^2+4)^1/2 = ln (2x)
cancel ln from both sides
(y^2+4)^1/2 =2x
y^2 = 4x^2 - 4

 

[Tamara Tango]

5)
xy dy/dy = y^2 + 4
⌡(y/ y^2 +4) dy = ⌡1/x dx
Diffrentiating y^2+ 4 = 2y
so 1/2⌡2y/y^2 +4 dy = ln x + c
1/2 ln |y^2+4| = lnx +c
when y=0, x= 1
1/2 ln4 = ln1 +c
2*1/2 ln 2 = c
c= ln 2

1/2ln |y^2+4| = lnx +ln2
ln (y^2+4)^1/2 = ln (2x)
cancel ln from both sides
(y^2+4)^1/2 =2x
y^2 = 4x^2 - 4

thank youuuuuuuuu! XDXDXD

 

[Thought blocker]

thank youuuuuuuuu! XDXDXD

I have a doubt. Will you solve it?

 

[Tamara Tango]

I have a doubt. Will you solve it?

if i understand about your doubtness, why not? ahaha

 

[Thought blocker]

if i understand about your doubtness, why not? ahaha

My mind is busted now. So getting weird answers, but I hope you can solve it.
Owen made 100 sandwiches which she sold for exactly $100. She sold caviar sandwiches for $5.00 each, the bologna sandwiches for $2.00, and the liverwurst sandwiches for 10 cents. How many of each type of sandwich did she make?

 

[Tamara Tango]

My mind is busted now. So getting weird answers, but I hope you can solve it.
Owen made 100 sandwiches which she sold for exactly $100. She sold caviar sandwiches for $5.00 each, the bologna sandwiches for $2.00, and the liverwurst sandwiches for 10 cents. How many of each type of sandwich did she make?

caviar 11
bologna 19
liverwurst 70
11+19+70=100
500x11+200x19+10x70=5500+3800+700=10000cents=$100
very interesting questions!

 

[Thought blocker]

caviar 11
bologna 19
liverwurst 70
11+19+70=100
500x11+200x19+10x70=5500+3800+700=10000cents=$100
very interesting questions!

^_^

 

[Jie Xi]

Help May /June 2010/31 question 4.and m/j/2010/32 question 10..thx!!!

 

[macfan]

Urgent Help Please
9709 June 2013 Paper 43 Question 5i and ii
9709 June 2012 Paper 42 Question 5 ii

Thanks in advance

 

[Thought blocker]

Help May /June 2010/31 question 4.and m/j/2010/32 question 10..thx!!!

What was hard in it? Go to examsolutions the person had explained all thing in deep. I prepared myself from this site. He is best. Hope you get benefit like me.

So here is your answer :
4)


10)
i)
Multiply by (2x-1) and then put x = 1/2 to get
D = {(1/4)-1}(1/4) = -3
[(2x^3-1)/x^2] = A(2x-1) +{B(2x-1)/x} +{C(2x-1)/x^2} -3 ------ (1) multiply by x^2 and put x = 0, to get
-1 = 0 + 0 -C or C = 1
Multiplying (1) by x^2 using C = 1 we have
2x^3 -1 = 2Ax^3 -Ax^2 +2Bx^2 -Bx +2x-1 -3x^2
Comparing coefficient of equal powers of x on both eisdes we get
A = 1, -A+2B -3 = 0 or B = 2

ii)
So given Integrand = 1 + (2/x) +(1/x^2) -3/(2x-1)
Which after integration, will give
x + 2lnx -1/x -(3/2)ln 2x-1)' which after putting limits 1 and 2 gives
(2-1) + 2ln2 +(1/2) -(3/2)ln3 = (3/2) + ln[{16/27}^(1/2)] = (3/2) + (1/2)ln(16/27)
Hence proved.

 

[Thought blocker]

Urgent Help Please
9709 June 2013 Paper 43 Question 5i and ii
9709 June 2012 Paper 42 Question 5 ii

Thanks in advance

IDK Mechanics much, but still I will try my best to clear your doubt.
So Here is your paper 42
5ii)
We will get two equations of the particles (you can get these equations from the basic knowledge of mechanics on Free Body Diagrams (FBD), still if you don't know, ask).
1) 3g - T - 1.6 = 3a
2) 2g + T - 4 = 2a
Now we have to find acceleration and the tension. We can equate and get both of them.
So, 3g - 3a - 1.6 = -2g + 2a + 4
-> 5g -5a = 5.6 (g = 10)
-> 5a = 44.4
-> a = 8.88 m/s^(2)

Now substitute the acceleration value in any two of the equations of particle, let say in 1 :
30 - T - 1.6 = 26.64
T = 1.76N

Paper 43
5
i)
let time taken for particle P to reach max height = t s
let time taken for particle Q to reach max height = t' s

=> 2t - 2t' = T or t - t' = T / 2 s --------------------> (i)

v = u + at => t = (v - u) / a

=> (0 - 17) / -9.8 = 1.73469 s

similarly t' = 7 / 9.8 = 0.71429 s

from (i) => 1.73469 - 0.71429 = T/2

=> (1.02040)2 = T

or T = 2.0408 s

Alternative :
For P, the time to reach highest point is give by
v = u + at
0 = 17 - 9.81 x t
9.81t = 17
t = 1.7329 s
So , total time of trajectory is 2 x t = 3.466 s.

Similarly, for Q, time to reach highest point is give by
0 = 7 - 9.81 x t
9.81t = 7
t = 0.7136 s
So , total time of trajectory is 2 x t = 1.427 s.
Since they reach ground at same instant, T = t(P) - t(Q) = 3.466 - 1.427 = 2.039 s
T = 2.039s

ii)
let s and s' represent the displacements of P and Q from the ground after time t s after Q is projected => s - s' = 5

s = ut + ½at² = 17 (t + 2.0408) - 4.9(t + 2.0408)²

s' = 7t - 4.9t²

s - s' = 17t - 7t + 34.6936 + 4.9t² - 4.9t² - 9.8t(2.0408) - (4.9)(4.16486464)

=> 5 = 10t - 19.99984t - 20.407836736 + 34.6936

=> 9.99984t = 9.28576

=> t = 0.929 s

=> P's velocity = v = u + at = 17 - (9.8)(0.929 + 2.0408)

= 17 - 9.8(2.9694) = -12.10012 m/s (downwards)

and Q's velocity = 7 - (9.8)(0.929) = 7 - 9.1042 = - 2.1042 m/s (downwards)

So here is the summary of ms :¬
i)
T = 2.04 s
ii)
the magnitude and direction of the velocities of P and Q at the instant P is 5m higher than Q are - 12.100 m/s and - 2.1042 m/s ( - indicating downwards) resp.

This is my best, I cannot help more then this. Sorry if I could not help you.

 

[Jie Xi]

Thx for ur help!! Appreciate it very much^^
Hmmm...on/9709/2008/03 question 8 and mj/9709/2008/03 quest8 too..(

 

[Thought blocker]

Thx for ur help!! Appreciate it very much^^
Hmmm...on/9709/2008/03 question 8 and mj/9709/2008/03 quest8 too..(

s08_3
8)
i)
Area of triangle PTN = tanx
1/2 x TN x PN = tanx
1/2 x TN x y = tanx

dy/dx = PN/TN
TN = PN x dx/dy
TN = y x dx/dy

Now substitute TN in area of triangle
1/2 ( y x dx/dy) y = tanx
1/2 y^2 (dx/dy) =tanx
dy/dx = (1/2 y^2)/tanx
dy/dx = 1/2 y^2 cotx

Alternative :
Gradient = dy/dx = PN/TN= tanx = 1/2(PN)(TN)

Rearrange PN/TN = tanx to give TN= PN/tanx
Substitute this into Area of the triangle (1/2 PN*TN) which gives 1/2* (PN)*(PN/tanx)
Note that PN= y.
So it goes down to 1/2. y^2. cotx

ii)
Separate the variables and integrate.
int(2/y^2) dy = int(cotx)dx
= -2/y = int(cosx/sinx) dx
= -2/y = ln(sinx) + c

Plug in (Pi/6, 2) to evaluate c. c= -0.31 Idk why they have it as 0.3 in the ms.
then rearrange to express y interms of x.

Will solve w08_3 later.

 

[jungly krinsky]

please help with question 7 (ii)
I can put it into the form -In(100-x) = 0.02t-In(k)
but I can not get K to equal 95 like they appear to do in the mark scheme.
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s03_qp_3.pdf