Mathematics: Post your doubts here!

[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?

 

[PANDA-]

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

I'm thinking of the same lol... P1 on 7th, and I barely started preparing. I do that by attempting to solve the questions posted here

 

[ahmed abdulla]

thanks PANDA- PhyZac !!

 

[~`Heba`~ :)]

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

Thanks!!

 

[syed1995]

maybe do past papers? ever heard of that?

Well I have heard of them.. but get bored while solving them... 1hr 45min on each paper isn't fair when doing at home .. It should be like chem and physics P2.. 1 hour and you're done

 

[syed1995]

I'm thinking of the same lol... P1 on 7th, and I barely started preparing. I do that by attempting to solve the questions posted here

haha that's all the practice that I have done.. Honestly I don't know functions one bit!

 

[PANDA-]

haha that's all the practice that I have done.. Honestly I don't know functions one bit!

Wow... same problem I'm having, I got a bit rusty in functions because I didn't practice for a long time.

 

[iKhaled]

Well I have heard of them.. but get bored while solving them... 1hr 45min on each paper isn't fair when doing at home .. It should be like chem and physics P2.. 1 hour and you're done

got bored of solving them and you finish before time by 45 mins means you r well prepared..am i wrong ?

 

[syed1995]

got bored of solving them and you finish before time by 45 mins means you r well prepared..am i wrong ?

naw.. I was talking about physics and chemistry.. they are 1 hour each for theory.. It takes 1 hour to solve them.. so one can solve like 3 or 4 papers in a day.. but with maths P1 I can't even solve 2 papers a day :\

And no one can solve the paper in 45 minutes! .. takes me about 1hr 30-40 minutes to solve a complete paper..

 

[PhyZac]

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

Okay, from first part you get that dy/dx = -sint/cost

Now they want the equation of tangent when t is t!!

Equation is in form

(y-y1) = m (x-x1)

M is the gradient, and when t is t, the gradient is -sint/cost ( the dy/dx)

the y1 is asin^3 (t)

and the x1 is acos^3(t)

Sub those value, and do some algebra and then you get the answer! [P.S, i am sure you know how to do this step, if you get any difficulty ask]