Mathematics: Post your doubts here!

[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]

 

[syed1995]

PhyZac or anyone else..

Know of any place (online or via calculator) where one can get inverse of phi? without referring to the normal tables I mean...

 

[PANDA-]

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..

Honestly for P1 1hr 45 is plenty.

 

[PhyZac]

PhyZac or anyone else..

Know of any place (online or via calculator) where one can get inverse of phi? without referring to the normal tables I mean...

Wait, you mean phi of statistic.

Well, i dont think so there is a way, and if there is, in exam the only way is table. (what i think)

EDIT: check this

 

[A star]

PhyZac or anyone else..

Know of any place (online or via calculator) where one can get inverse of phi? without referring to the normal tables I mean...

there is a calculator function but i tried but didnt knoe its use properly

 

[iKhaled]

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]

thanks man that really helped!

 

[salvatore]

PhyZac and PANDA-
Thank you soooo much guys.. much appreciated

 

[minie23]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s11_qp_31.pdf
Please help for no. 7 (i)

 

[PhyZac]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_31.pdf
Please help for no. 7 (i)

OKaay..see..!

First we will do the limits.
in the first one it is 2 and 0

so, the sub is x = t^2 + 1
when t is 2, x is 5
when t is 0, x is 1

So now new limit are 5 and 1

Now to main thing!

First thing I do is find the dy/dx of substitution.

not of this >> x = t^2 + 1

make t the subject first.

t^2 = x - 1
t = (x - 1)^1/2 [power 1/2 is square root]

okay now, take dy/dx, but to be precise dt/dx

dt/dx = 1/2 ( x - 1)^(-1/2)

Now next thing!

We now should substitute the t's of formula.

since x=t^2 +1

then this
4t^3 ln(t^2 + 1)

is
4t^3 lnx

One more t is left! what is t alone?

x = t^2 +1
t = (x - 1)^1/2

so sub this and get

4((x-1)^1/2)^3 lnx

4(x-1)^3/2 ln x

Now add the differential to the top equation

4(x-1)^3/2 ln x 1/2 ( x - 1)^(-1/2)

4/2 (x-1) ^(3/2 -1/2) lnx

2 (x-1) ^1 ln x

2x - 2 ln x , with limits 5 to 1

 

[minie23]

OKaay..see..!

First we will do the limits.
in the first one it is 2 and 0

so, the sub is x = t^2 + 1
when t is 2, x is 5
when t is 0, x is 1

So now new limit are 5 and 1

Now to main thing!

First thing I do is find the dy/dx of substitution.

not of this >> x = t^2 + 1

make t the subject first.

t^2 = x - 1
t = (x - 1)^1/2 [power 1/2 is square root]

okay now, take dy/dx, but to be precise dt/dx

dt/dx = 1/2 ( x - 1)^(-1/2)

Now next thing!

We now should substitute the t's of formula.

since x=t^2 +1

then this
4t^3 ln(t^2 + 1)

is
4t^3 lnx

One more t is left! what is t alone?

x = t^2 +1
t = (x - 1)^1/2

so sub this and get

4((x-1)^1/2)^3 lnx

4(x-1)^3/2 ln x

Now add the differential to the top equation

4(x-1)^3/2 ln x 1/2 ( x - 1)^(-1/2)

4/2 (x-1) ^(3/2 -1/2) lnx

2 (x-1) ^1 ln x

2x - 2 ln x , with limits 5 to 1

Thank you

 

[PhyZac]

minie23
N/(1800-N) = e / 5

5N = e (1800- N)

5N = 1800 e - eN

5N + eN = 1800 e

N ( 5 + e) = 1800 e

N = 1800e/ 5 + e

[I made e^t/2 as e, to prevent errors]

 

[ahmed abdulla]

PhyZac .. do u think M2 is easier or S1 ?