Mathematics: Post your doubts here!

[Tabi Sheikh]

This is from M/J 08 p4
View attachment 24331

For ii) can you explain why the speed at C is equal to the speed at A. Is it because both the points are at the same level? Even then, can someone provide the logic behind it? Maybe from a practical point of view?

at same point the potential energy gained or loss is equal for a body and here we assume that change in potenial energy is equal to change in kinetic energy

 

[PANDA-]

This is from M/J 08 p4
View attachment 24331

For ii) can you explain why the speed at C is equal to the speed at A. Is it because both the points are at the same level? Even then, can someone provide the logic behind it? Maybe from a practical point of view?

Since it is smooth, there is no loss of energy because there is no friction, so potential energy is directly converted to kinetic energy. Gain in potential energy = Loss in kinetic energy. So as it goes down from A to B, it loses P.E. and gains K.E., when it goes up to C, it gains the lost P.E., and loses the gained K.E., velocity returns to as it was in A.

 

[~`Heba`~ :)]

Lol.... i don't think it's possible to show a graph here
Unless we draw it scan etc...

oh okay

 

[ahmed abdulla]

ANY HELP >>> THIs question really conFUSES and u can call it hard question....
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_11.pdf q 7 (ii) .... ? with clear expaination plzz

 

[~`Heba`~ :)]

can anyone help me with #7 (ii)

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_11.pdf
how do i get the largest value?

 

[ahmed abdulla]

can anyone help me with #7 (ii)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf
how do i get the largest value?

same doubt

 

[PANDA-]

ANY HELP >>> THIs question really conFUSES and u can call it hard question....
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf q 7 (ii) .... ? with clear expaination plzz

can anyone help me with #7 (ii)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf
how do i get the largest value?

2cos^2(nθ) = 2-2sin^2(nθ)

2-2sin^2(nθ) = 3sin(nθ)

nθ = 30 or nθ = 150

Smallest value = 10
θ = 10

n(10) = 30
n=3

Now for the largest value, well I don't know, because logically, it would be 50 degrees... because

nθ = 150
3θ = 150
θ = 50 <--- But this isn't the correct answer according to the ms... PhyZac Could you help?

 

[PhyZac]

ANY HELP >>> THIs question really conFUSES and u can call it hard question....
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf q 7 (ii) .... ? with clear expaination plzz

can anyone help me with #7 (ii)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf
how do i get the largest value?

PANDA-
Okay, see.
From first part u get 150 as well as 30

Now they said smallest value now is 10 meaning
sin (n(10))= 1/2
10n = 30
n = 3

Okay now to find the largest value...first change the range
from 0 < q < 360
by multiplying with 3 (since now 3q)
to
0< q<1080

that is a three rounds in the quadrants [hope u get this point]


I drew a sketch, every round is different colour. So the largest value is on the third round, on the blue point, from the sketch u find, 2 rounds plus 150
(2 rounds, 360 x 2 = 720
3q = 720 + 150 = 870
q = 290

 

[PANDA-]

PANDA-
Okay, see.
From first part u get 150 as well as 30

Now they said smallest value now is 10 meaning
sin (n(10))= 1/2
10n = 30
n = 3

Okay now to find the largest value...first change the range
from 0 < q < 360
by multiplying with 3 (since now 3q)
to
0< q<1080

that is a three rounds in the quadrants [hope u get this point]


I drew a sketch, every round is different colour. So the largest value is on the third round, on the blue point, from the sketch u find, 2 rounds plus 150
(2 rounds, 360 x 2 = 720
3q = 720 + 150 = 870
q = 290

I see. Thanks for the explanation.

 

[iKhaled]

i need help with question 6(ii) may june 2009 paper 3

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_3.pdf

 

[ahmed abdulla]

PANDA-
Okay, see.
From first part u get 150 as well as 30

Now they said smallest value now is 10 meaning
sin (n(10))= 1/2
10n = 30
n = 3

Okay now to find the largest value...first change the range
from 0 < q < 360
by multiplying with 3 (since now 3q)
to
0< q<1080

that is a three rounds in the quadrants [hope u get this point]


I drew a sketch, every round is different colour. So the largest value is on the third round, on the blue point, from the sketch u find, 2 rounds plus 150
(2 rounds, 360 x 2 = 720
3q = 720 + 150 = 870
q = 290

i got the first part ... but is there any alternative way for the second part ?? i didnt get the quadrant way

 

[PANDA-]

i got the first part ... but is there any alternative way for the second part ?? i didnt get the quadrant way

Since we're having 3θ ... Then we have to multiply the range by 3, making it 0 < θ < 1080.
So we're going 3 cycles around the graph, now our minimum is within the first cycle, and our maximum is within the last cycle... So we start from the beginning of the second cycle, and add 150, our answer from the previous question... That's 720+150... Now you might ask why we don't start from the third cycle, as in 1080+150... Well that's because we'd be going into a fourth cycle, but we're limited to only the third cycle.... So 720+150 = 870. So maximum of 3θ = 870... So the maximum of θ is 870/3 = 290.

 

[salvatore]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_11.pdf
Could anyone please help me with question no. 7 iii? I don't understand how to draw the graph.
I'll appreciate if you could explain and draw it for me..

 

[PhyZac]

i got the first part ... but is there any alternative way for the second part ?? i didnt get the quadrant way

Read PANDA- reply, very nicely explain Ma Sha Allah!

Now I will just point something, always when you dont have x alone, change the range. For example.

sin(x)= 1/2 | 0< x< 360

sin(2x) = 1/2 | 0 < x < 720 (0x2 =0, and 360 x 2 = 720)

sin(x-30) = 1/2 | -30 < x < 330 ( 0 - 30 = -30 and 360 - 30 = 330)

So as you see, change the range according how the "x" is, and find the values, then change them. For example.

sin(2x) = 1/2 | 0< x< 360 will be 0 < x < 720

the values are

2x = 30, 150, 390, 510 and only! You see, all these values are below 720. Now we find "x"

x = 15, 75, 195, 255 , and you see, all values a below the actual range which is 0< x< 360

 

[PANDA-]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_11.pdf
Could anyone please help me with question no. 7 iii? I don't understand how to draw the graph.
I'll appreciate if you could explain and draw it for me..

Take a look at this... http://rechneronline.de/function-graphs/
Enter your formula and you will see the graph... I can't really explain it sorry

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_11.pdf
Could anyone please help me with question no. 7 iii? I don't understand how to draw the graph.
I'll appreciate if you could explain and draw it for me..

I solve such questions step wise.

My drawing isnt neat, but hope u get it.

First a normal tan from 0 to pi


Then, u see, 1/2x, since x is divided by 2 now multiply with 2, so what is in 1/2pi is now in pi




Then u see -2 tan (1/2x), the y is multiplied with -2 so it is now double and in opposite side,



Then we add 3, since -2tan(1/2x) + 3 , so when y is 0 becomes y 3



And that is the final shape!

 

[PANDA-]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_11.pdf
Could anyone please help me with question no. 7 iii? I don't understand how to draw the graph.
I'll appreciate if you could explain and draw it for me..

Well here's the explanation...
The equation is...
3-2tan(0.5x)
Refer it to
k+ntan(ax)

Now n is -2, in sin and cos curves, this would mean that the curve is switched, and the amplitude is 2, but since tan graphs have no amplitude, we'll just switch it, by switching I mean...
This is switched to this - Now remember, these are just examples, I'm merely referring to the shape of the graph.
Now k is the translation, you're gonna want to start your graph from 3, instead of where you usually start a graph at 0, when k is absent (0). Now you need to find the period, the period is 2π/a ... 2π/0.5 = 4π.
Now since our period is 4π, so the graph repeats every 4π units on the x axis, and they're only asking for 0 -> π, making it simpler... So you'll only draw this...

3 on the x axis here is the asymptote, which in our case, is π

 

[PANDA-]

Now PhyZac obviously explained it far better... So refer to his explanation

 

[syed1995]

Now where to start preparations for P1... I have no idea

 

[iKhaled]

Now where to start preparations for P1... I have no idea

maybe do past papers? ever heard of that?