IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: T.Q on May 05, 2010, 10:28:37 am

Title: C3 and C4 DOUBTS HERE!!!
Post by: T.Q on May 05, 2010, 10:28:37 am: POST YOUR C3 DOUBTS HERE
Title: Re: C3 and C4 Doubts Here!!!
Post by: T.Q on May 05, 2010, 10:29:31 am: POST YOUR C4 DOUBTS HERE
Title: Re: C3 and C4 Doubts Here!!!
Post by: Saladin on May 05, 2010, 10:32:32 am: Good job man, this way, we can understand what they want!
Title: Re: C3 and C4 Doubts Here!!!
Post by: pjb13 on May 28, 2010, 02:11:38 pm: 1) Given h(x) = (x+1)(x-3). How do I know that h is differentiable on (-1,3) ?

2)How close to 2 must x be for the values of g(x) = x^2 to be within 0.2 from 4?

I know how to solve most of this...I reach an answer of -0.051<x-2<0.049
Then it says to insure both ends of the last compound inequality are satisfied take -0.049<x-2<0.049
My question is why cant we take -0.051<x-2<0.051....how do I know which one to take?

Thanx
Title: Re: C3 and C4 Doubts Here!!!
Post by: Saladin on May 28, 2010, 03:32:58 pm: Quote from: pjb13 on May 28, 2010, 02:11:38 pm
1) Given h(x) = (x+1)(x-3). How do I know that h is differentiable on (-1,3) ?

2)How close to 2 must x be for the values of g(x) = x^2 to be within 0.2 from 4?

I know how to solve most of this...I reach an answer of -0.051<x-2<0.049
Then it says to insure both ends of the last compound inequality are satisfied take -0.049<x-2<0.049
My question is why cant we take -0.051<x-2<0.051....how do I know which one to take?

Thanx

Where did you get this question from?
Title: Re: C3 and C4 Doubts Here!!!
Post by: pjb13 on May 29, 2010, 09:01:17 am: Ah just realized tht its not required...extra stuff given by my school >:(
Title: Re: C3 and C4 Doubts Here!!!
Post by: Saladin on June 10, 2010, 06:38:17 am: This exam is next!
Title: Re: C3 and C4 Doubts Here!!!
Post by: T.Q on June 10, 2010, 11:46:15 am: Quote from: Engraved on June 10, 2010, 06:38:17 am
This exam is next!

gd luck :)

i have done c3 in jan
Title: Re: C3 and C4 Doubts Here!!!
Post by: sweetest angel on June 11, 2010, 06:25:20 am: can anyone xplain to me Q 4 in June 2005 pleeaassee?? i got the ms but i dnt get the second step.
Thanks :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: vanibharutham on June 11, 2010, 06:52:48 am: For Q4 June 2005,

Integration by substitution...

x = sin A
x² = sin²A

1 - x² = 1 - sin²A

And since sin²A + cos²A = 1,

1 - sin²A = 1 - x² = cos²A
Title: Re: C3 and C4 Doubts Here!!!
Post by: sweetest angel on June 11, 2010, 09:39:42 am: umm ya i got that part its the part where the 1/(cos^2A)^3/2 X cosA is made into 1/cos^2A

isn't it supposed to be 1/cos^-1/2A?
Title: Re: C3 and C4 Doubts Here!!!
Post by: sweetest angel on June 11, 2010, 09:43:08 am: Jan '06 4-b :show did they do the last step??

i got till 1/8cos(arcsin(x/4)) but the next step is magic to me :/
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on June 11, 2010, 09:43:49 am: 1/(cos^2A)^3/2 X cosA =1/cos^3 A cosA= 1/cos^2 A
Title: Re: C3 and C4 Doubts Here!!!
Post by: sweetest angel on June 11, 2010, 10:02:38 am: Quote from: astarmathsandphysics on June 11, 2010, 09:43:49 am
1/(cos^2A)^3/2 X cosA =1/cos^3 A cosA= 1/cos^2 A

thank u thank u thank u!!! :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on June 11, 2010, 10:11:52 am: $x=4 sin(2y+6)$
differentiate wrt y
$\frac {dx}{dy} =8 cos(2y+6)$
$\frac {dy}{dx}=\frac {1}{8 cos(2y+6)} =\frac {1}{8 sqrt(1-sin^2 (2y+6))} =\frac {1}{8(1-(x/4)^2))$
Title: Re: C3 and C4 Doubts Here!!!
Post by: sweetest angel on June 11, 2010, 10:30:03 am: Quote from: astarmathsandphysics on June 11, 2010, 10:11:52 am
$x=4 sin(2y+6)$
differentiate wrt y
$\frac {dx}{dy} =8 cos(2y+6)$
$\frac {dy}{dx}=\frac {1}{8 cos(2y+6)} =\frac {1}{8 sqrt(1-sin^2 (2y+6))} =\frac {1}{8(1-(x/4)^2))$

from where did u get the square root at the end?
Title: Re: C3 and C4 Doubts Here!!!
Post by: cooldude on June 11, 2010, 10:38:44 am: Quote from: sweetest angel on June 11, 2010, 10:30:03 am
from where did u get the square root at the end?

(sinA)^2+(cosA)^2=1
therefore cos A=(1-(sinA)^2)^0.5
Title: Re: C3 and C4 Doubts Here!!!
Post by: sweetest angel on June 11, 2010, 10:41:09 am: oooh Thanks alot i got it :D
Title: Re: C3 and C4 Doubts Here!!!
Post by: sweetest angel on June 12, 2010, 06:43:39 am: hey can anyone xplain to me the domain and range of chapter 2??
i don't know how to get them in questions of state the domain and range of the function or inverse of function.
thanks :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: Saladin on June 12, 2010, 08:06:53 am: Quote from: sweetest angel on June 12, 2010, 06:43:39 am
hey can anyone xplain to me the domain and range of chapter 2??
i don't know how to get them in questions of state the domain and range of the function or inverse of function.
thanks :)

Please give a question.
Title: Re: C3 and C4 Doubts Here!!!
Post by: bilal920 on June 13, 2010, 06:21:17 am: i have a Question in the mok paper

How the hell did they do Q7b
Title: Re: C3 and C4 Doubts Here!!!
Post by: Hwa1 on June 14, 2010, 01:38:20 pm: Does anybody have a link to the Jan 09 mark scheme (edexcel)? The file on freeexampapers is damaged and the edexcel section on xtremepapers is under maintenance :(
Title: Re: C3 and C4 Doubts Here!!!
Post by: 7ooD on June 14, 2010, 09:16:02 pm: here is the markscheme u need http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202009/6665_01_rms_20090312.pdf
Title: Re: C3 and C4 Doubts Here!!!
Post by: T.Q on June 15, 2010, 02:14:19 pm: i need help in this

Use the substitution $x=sec\theta$ and integration to find the value of

$\int_{2}^{3}\frac{1}{(x^2-1)^ (\frac{3}{2} }$ $dx$

giving your answer to 3 decimal places

the answer in MS is $0.094$
Title: Re: C3 and C4 Doubts Here!!!
Post by: Meticulous on June 15, 2010, 03:42:00 pm: Is it (x² -1 )^3/2 ..?
Title: Re: C3 and C4 Doubts Here!!!
Post by: T.Q on June 15, 2010, 05:29:13 pm: Quote from: Meticulous on June 15, 2010, 03:42:00 pm
Is it (x² -1 )^3/2 ..?

yes it is
Title: Re: C3 and C4 Doubts Here!!!
Post by: S.M.A.T on June 15, 2010, 07:39:46 pm: Quote from: T.Q on June 15, 2010, 02:14:19 pm
i need help in this

Use the substitution $x=sec\theta$ and integration to find the value of

$\int_{2}^{3}\frac{1}{(x^2-1)^ (\frac{3}{2} }$ $dx$

giving your answer to 3 decimal places

the answer in MS is $0.094$

THE SOLUTION IS IN THE LINK BELOW
Title: Re: C3 and C4 Doubts Here!!!
Post by: T.Q on June 15, 2010, 11:11:13 pm: thanx asiftasfiq93

+rep
Title: Re: C3 and C4 Doubts Here!!!
Post by: M-H on September 10, 2010, 04:00:52 pm: i want to know how to solve these kind of Qs:
Eg.
Q: The functions l(x), m(x), n(x) and p(x) are defined by l(x)=2x+1, m(x)=x^2 -1, n(x)=1/(x+5) and p)x)=x^3.
Find in terms of l,m,n and p the functions:
a) 4x(^2) +4x.

I need a logical way, or atleast a way which can be used for solving all Qs of this type!!! Please help!

Thanks a million in advance!!
Title: Re: C3 and C4 Doubts Here!!!
Post by: S.M.A.T on September 10, 2010, 07:20:36 pm: 4x(^2) +4x
=4((x^2)+x)
=4(((x+0.5)^2)-(0.5)^2)
=4((x+0.5)^2)-1
=4(((2x+1)^2)/4)-1
=((2x+1)^2)-1
=l^2(x)-1
=ml(x)

When the equation is quadratic, always convert it to complete square form ,then simply the equation and write in terms of l(x),m(x),....etc.

If the quadratic equation is already written in complete square form,then all you have to do is write in terms of l(x),m(x),....etc.
Title: Re: C3 and C4 Doubts Here!!!
Post by: M-H on September 10, 2010, 09:29:57 pm: what for the cubic equations..?

And, i didn't want the solution. thanks tho. but i have it. what i really wanted to know how i'd solve for any kind of function i'm given ::)
Title: Re: C3 and C4 Doubts Here!!!
Post by: S.M.A.T on September 10, 2010, 10:10:37 pm: Quote from: M-H on September 10, 2010, 09:29:57 pm
what for the cubic equations..?

And, i didn't want the solution. thanks tho. but i have it. what i really wanted to know how i'd solve for any kind of function i'm given ::)

Give me a question i will show you how to do it with explanation
For different type of function there are different methods.

By the way don't worry with these type(write in terms of l(x),m(x),....etc.) of questions, :) at least i can assure you that :D
you see like you i use to worry with these type of question
ay type r question C3 exam a asa na, faw faw book a dia raksa, do the mixed exercise that is important.
ay type r question question paper a paba na not even in Solomon :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on September 10, 2010, 10:51:41 pm: Work down from the highest power of x.
Because you have 4x^2 in the question, part of the answer will be 4m(x)
Now do 4x^2 +4x -4m =4x^2+4x-4(x^2-1)=4x+4
Because you have x ihere part of the answer will be l
4x+4 =2(2x+1)+2=2l+2
4x^2+4=4m+2l+2
Title: Re: C3 and C4 Doubts Here!!!
Post by: M-H on September 11, 2010, 09:43:27 am: Sir, the ans is, as asif's mentioned, ml(x). They want the answers expressed only as m(x), l(x) etc.

Asif-- Doubt ase jokon, its gotta be solved. Irrespective of whether it comes in the exam or not ;).Anyway thanks for that info, i haven't seen the Q papers as yet. So that's removes part of my worries :D
Title: Re: C3 and C4 Doubts Here!!!
Post by: M-H on September 13, 2010, 10:18:15 am: Quote from: asiftasfiq93 on September 10, 2010, 07:20:36 pm
4x(^2) +4x
=4((x^2)+x)
=4(((x+0.5)^2)-(0.5)^2)
=4((x+0.5)^2)-1
=4(((2x+1)^2)/4)-1
=((2x+1)^2)-1
=l^2(x)-1
=ml(x)

Asif, how did reach the 4th step ???--as in, after x+.5 it becomes 2x+1..kibhabe??? :(
Title: Re: C3 and C4 Doubts Here!!!
Post by: S.M.A.T on September 13, 2010, 09:21:18 pm: Quote from: M-H on September 13, 2010, 10:18:15 am
Asif, how did reach the 4th step ???--as in, after x+.5 it becomes 2x+1..kibhabe??? :(

4((x+0.5)^2)
=4((x+(1/2))^2)
=4(((2x+1)/2)^2)
=4(((2x+1)^2)/(2^2))
=4(((2x+1)^2)/4)
=(2x+1)

If you don't understand then tell me :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on September 18, 2010, 11:29:31 am: I will look at that.
Title: Re: C3 and C4 Doubts Here!!!
Post by: M-H on September 24, 2010, 03:17:12 pm: Find dy/dx for-
x=1/(2t-1), y=(t^2)/(2t-1)

I don't know how to differentiate functions with reciprocals...

Thanks a bunch in advance.
Title: Re: C3 and C4 Doubts Here!!!
Post by: iluvme on September 24, 2010, 04:31:39 pm: Quote from: M-H on September 24, 2010, 03:17:12 pm
Find dy/dx for-
x=1/(2t-1), y=(t^2)/(2t-1)

I don't know how to differentiate functions with reciprocals...

Thanks a bunch in advance.

i) x=1/(2t-1),
dy/dx= 1(2t-1)^-1
=-1.(2t-1)^-1-1.(2)
=-1.(2t-1)^-2.(2)
=-2/(2t-1)²

ii) y=t²/(2t-1)
I'm sure you know the UV method, therefore,

dy/dx=2t(2t-1)-t².(2)/(2t-1)²
Title: Re: C3 and C4 Doubts Here!!!
Post by: S.M.A.T on September 24, 2010, 05:16:20 pm: Quote from: M-H on September 24, 2010, 03:17:12 pm
Find dy/dx for-
x=1/(2t-1), y=(t^2)/(2t-1)

I don't know how to differentiate functions with reciprocals...

Thanks a bunch in advance.

Here :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: S.M.A.T on September 24, 2010, 05:19:13 pm: Quote from: iluvme on September 24, 2010, 04:31:39 pm
i) x=1/(2t-1),
dy/dx= 1(2t-1)^-1
=-1.(2t-1)^-1-1.(2)
=-1.(2t-1)^-2.(2)
=-2/(2t-1)²

ii) y=t²/(2t-1)
I'm sure you know the UV method, therefore,

dy/dx=2t(2t-1)-t².(2)/(2t-1)²
I think that would be dx/dt :)

your answer is correct :D +rep
Title: Re: C3 and C4 Doubts Here!!!
Post by: M-H on September 24, 2010, 06:58:49 pm: Asif, I was expecting your photos :P :D
Thanks, +rep!!

Iluvme- ILY ;)
+rep :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: S.M.A.T on September 25, 2010, 05:04:20 am: You welcome :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: iluvme on September 25, 2010, 04:12:26 pm: Quote from: asiftasfiq93 on September 24, 2010, 05:19:13 pm
I think that would be dx/dt :)

your answer is correct :D +rep

Yeah, habit I guess.

Thank you.
Title: Re: C3 and C4 Doubts Here!!!
Post by: iluvme on September 25, 2010, 04:14:14 pm: Quote from: M-H on September 24, 2010, 06:58:49 pm
Iluvme- ILY ;)
+rep :)

<3 ya too. :-*
Glad to be of help. ;D
Title: Re: C3 and C4 Doubts Here!!!
Post by: break freak on November 21, 2010, 03:51:40 pm: I need answers for these question s pleaaase ( x2 means x square )?
1) The value of cos(sin-1x) is: A. 1 B. x C. 1/x D. 1/?1-x2 E. ?1-x2

2) The domain of f(x)= (3?x)/(1-x) is: A. (1,?) B. (0,1) U (1,?) C. (0,?) D. none of these
3)C is the circle (x-1)2 + (y-2)2 =144 and the point P(10,10), then: A. P is the center of C B. P is inside C but not the center C. P is outside C D. P is on C E. the location of P relative to C cannot be determined
Title: Re: C3 and C4 Doubts Here!!!
Post by: break freak on November 21, 2010, 03:53:19 pm: I need answers for these questions pleaaaaase today ( x2 means x square )?
1) The value of cos(sin-1x) is: A. 1 B. x C. 1/x D. 1/?1-x2 E. ?1-x2

2) The domain of f(x)= (3?x)/(1-x) is: A. (1,?) B. (0,1) U (1,?) C. (0,?) D. none of these
3)C is the circle (x-1)2 + (y-2)2 =144 and the point P(10,10), then: A. P is the center of C B. P is inside C but not the center C. P is outside C D. P is on C E. the location of P relative to C cannot be determined
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on November 24, 2010, 12:41:41 pm: 1)Put y=sin^-1 x then we need cos y
y=sin^-1 x so siny =x so cosy =sqrt(1-sin^2y) =sqrt(1-x^2)2) x=1 is not in domain f(x) since f(1) not defined so domain could be B
3)centre is (1,2)
distance from (1,2) to P os sqrt((1-10)^2 +(2-10)^2)=sqrt(81+64)=sqrt(145)>sqrt(144) so P is outside C,
Answer is C.
Title: Re: C3 and C4 Doubts Here!!!
Post by: qooqo111 on January 09, 2011, 08:43:58 am: Hey
I got this formula for integration by parts
i used it in loads of questions and they all seem to come true(even the past paper questions) but am still curious if it is correct if i used it in my C4 exam the formula goes like this :
u?vdx-?[du/dx.?vdx]dx
P.S

for some reason i cant put the integral mark it just comes as a question mark ..... so yeah the question mark is integral
the second integral is for all the bracket.
thank you
Title: Re: C3 and C4 Doubts Here!!!
Post by: mr.incredible on January 09, 2011, 10:43:58 pm: Quote from: sweetest angel on June 11, 2010, 09:43:08 am
Jan '06 4-b :show did they do the last step??

i got till 1/8cos(arcsin(x/4)) but the next step is magic to me :/

Since that the domain of sin(2y+6) is not stated, therefore you will have to take +ve/-ve values of cos(2y+6). You can obtain this value by either drawing a right-angled triangle and using the pythagoras theorem to get the unknown side and then cos(2y+6)=adj./hyp. OR by using the identity sin^2(2y+6) + cos^2(2y+6) = 1 and given that sin(2y+6)= x/4 you will be able to get the value of cos(2y+6).

The final answer is 1/(±2(16-x^2 )^2 )

:)
Title: Re: C3 and C4 Doubts Here!!!
Post by: Pias on January 18, 2011, 06:26:46 am: sin A= 3/5; find cos A;
after finding the value of cos A by formula, why is it that the negative value is taken of the root of the square of cos A.
Please could someone help me with that.
Title: Re: C3 and C4 Doubts Here!!!
Post by: Darkstar3000 on January 18, 2011, 10:24:56 am: How do they get 45 and 135 in question 8? I managed to get 22.5 and 112.5 but I have no idea how they got the other two values

Here is the question paper: http://www.scribd.com/doc/26846381/Edexcel-GCE-January-2010-Core-Mathematics-C3-QP (Jan 2010)
Here is the mark scheme : http://www.yourschoolmaster.com/mathematics/ks5/answers/2010web/C3_jan_2010ms.pdf

And also in question 3, why did they take their "alpha" to 4 decimal places, I took mine to 2 decimal places and my final answer has a 0.1 difference from theirs. Does this matter since they didn't state to what number of decimal places we should write alpha in ?
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on January 19, 2011, 09:43:21 pm: cosec^2 2x - cot 2x =1
1+cot^2 2x -cot 2x =1
cot^2 2x -cot 2x =0
cot 2x(cot 2x -1)=0
cot 2x =0 so tan 2x =infinity so 2x =90, 90+180=270 so x=22.5,135
or cot 2x -1=0 so cot 2x =1 so tan 2x =1 so 2x=45, 45+180=225 so x=22.5,112.5
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on January 19, 2011, 09:44:57 pm: 3 dp because in part b they want the answer to 2 dp
Title: Re: C3 and C4 Doubts Here!!!
Post by: Darkstar3000 on January 20, 2011, 04:13:51 am: Quote from: astarmathsandphysics on January 19, 2011, 09:43:21 pm
cosec^2 2x - cot 2x =1
1+cot^2 2x -cot 2x =1
cot^2 2x -cot 2x =0
cot 2x(cot 2x -1)=0
cot 2x =0 so tan 2x =infinity so 2x =90, 90+180=270 so x=22.5,135
or cot 2x -1=0 so cot 2x =1 so tan 2x =1 so 2x=45, 45+180=225 so x=22.5,112.5

how is 2x=90? That's the part I don't understand
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on January 20, 2011, 09:13:58 am: cos tan^{-1} infinity =90
Title: Re: C3 and C4 Doubts Here!!!
Post by: hmh on January 20, 2011, 03:16:56 pm: Q: Express as a single fraction in its simplest form

f(x)= ( 2x + 2 / x^2 - 2x - 3 ) - ( x + 1 / x - 3 )
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on January 20, 2011, 03:32:06 pm: f(x)= ( 2x + 2 / x^2 - 2x - 3 ) - ( x + 1 / x - 3 )
$f(x)=\frac {2(x+1)}{(x-3)(x+1)} -\frac {x+1}{x-3} =\frac{2}{x-3} -\frac{x+1}{x-3} =\frac{1-x}{x-3}$
Title: Re: C3 and C4 Doubts Here!!!
Post by: hmh on January 20, 2011, 03:34:49 pm: Thanks Astar :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on January 20, 2011, 03:55:00 pm: No prob
Title: Re: C3 and C4 Doubts Here!!!
Post by: Darkstar3000 on January 20, 2011, 06:22:12 pm: Quote from: astarmathsandphysics on January 20, 2011, 09:13:58 am
cos tan^{-1} infinity =90

Thank you my good sir :D
Title: Re: C3 and C4 Doubts Here!!!
Post by: qooqo111 on January 23, 2011, 02:06:39 pm: In chp 3 (binomial expansion)
what do they mean by state/find the range of values for which the expansion is valid ....??
Thank You :D
Title: Re: C3 and C4 Doubts Here!!!
Post by: elemis on January 23, 2011, 04:24:44 pm: Dude ? Seriously ?

There's a thread already in blue on the message board that says ALL C4 DOUBTS HERE !!!!

Why couldnt you post it there ?

Topic merged.
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on January 23, 2011, 08:01:15 pm: They mean the series convergences
Title: Re: C3 and C4 Doubts Here!!!
Post by: **RoRo** on February 12, 2011, 02:03:46 pm: I have a question, which formulas do we need to know for Trigonometry and which ones will be there in the Formula book?

Thanks in advance!
Title: Re: C3 and C4 Doubts Here!!!
Post by: elemis on February 12, 2011, 04:01:03 pm: Quote from: The Secret on February 12, 2011, 02:03:46 pm
I have a question, which formulas do we need to know for Trigonometry and which ones will be there in the Formula book?

Thanks in advance!

Attached is the formula booklet page for C3.
Title: Re: C3 and C4 Doubts Here!!!
Post by: **RoRo** on February 12, 2011, 04:04:14 pm: Quote from: Ari Ben Canaan on February 12, 2011, 04:01:03 pm
Attached is the formula booklet page for C3.

Thank you so much!! :D

Sorry to trouble you, but have you got that for C1, C2, C4, M1 and S1? That would be of great help!
Title: Re: C3 and C4 Doubts Here!!!
Post by: elemis on February 12, 2011, 04:19:59 pm: Quote from: The Secret on February 12, 2011, 04:04:14 pm
Thank you so much!! :D

Sorry to trouble you, but have you got that for C1, C2, C4, M1 and S1? That would be of great help!

http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/N38210A-GCE-Mathematical-Formulae-Statistical-Tables.pdf (http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/N38210A-GCE-Mathematical-Formulae-Statistical-Tables.pdf)
Title: Re: C3 and C4 Doubts Here!!!
Post by: **RoRo** on February 16, 2011, 08:41:21 am: Quote from: Ari Ben Canaan on February 12, 2011, 04:19:59 pm
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/N38210A-GCE-Mathematical-Formulae-Statistical-Tables.pdf (http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/N38210A-GCE-Mathematical-Formulae-Statistical-Tables.pdf)

Thanks a lot! :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: thenewkid™ on March 06, 2011, 02:07:32 pm: jan 2009 c3 edexcel q6

HELP
thnQ
Title: Re: C3 and C4 Doubts Here!!!
Post by: **RoRo** on April 03, 2011, 11:56:29 am: How do you integrate the following function sec²xtan²x by substitution method? I need the working please, thank you! :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: elemis on April 03, 2011, 02:10:36 pm: Quote from: The Secret on April 03, 2011, 11:56:29 am
How do you integrate the following function sec²xtan²x by substitution method? I need the working please, thank you! :)

10 minutes.
Title: Re: C3 and C4 Doubts Here!!!
Post by: elemis on April 03, 2011, 02:23:09 pm: $\int sec^2xtan^2x$

Let u=tanx

$\frac{du}{dx}=sec^2x$

Since du=sec²x dx :

$\int tan^2x~du$

Additionally u=tanx hence :

$\int u^2~du$

Integrating gives :

$\frac{u^3}{3}$

Substituting for u gives :

$\frac{tan^3x}{3}$
Title: Re: C3 and C4 Doubts Here!!!
Post by: **RoRo** on April 03, 2011, 02:58:28 pm: Quote from: Ari Ben Canaan on April 03, 2011, 02:23:09 pm
$\int sec^2xtan^2x$

Let u=tanx

$\frac{du}{dx}=sec^2x$

Since du=sec²x dx :

$\int tan^2x~du$

Additionally u=tanx hence :

$\int u^2~du$

Integrating gives :

$\frac{u^3}{3}$

Substituting for u gives :

$\frac{tan^3x}{3}$

THANK YOUUU!! ;D
Title: Re: C3 and C4 Doubts Here!!!
Post by: mdwael on April 05, 2011, 12:10:03 pm: Do we have to memorize the expansions of Sin(A+B), Cos(A+B), CosA CosB- SinA Sin B or they'll be given in the formula booklet?
Also does anyone know a helpful website to help me in chapters 6 and 7? (tirgnometry)
Title: Re: C3 and C4 Doubts Here!!!
Post by: **RoRo** on April 05, 2011, 02:18:36 pm: Quote from: mdwael on April 05, 2011, 12:10:03 pm
Do we have to memorize the expansions of Sin(A+B), Cos(A+B), CosA CosB- SinA Sin B or they'll be given in the formula booklet?
Also does anyone know a helpful website to help me in chapters 6 and 7? (tirgnometry)

They'll be given to you in the formula booklet.
Check this site: http://www.trigonometry-help.net/
Title: Re: C3 and C4 Doubts Here!!!
Post by: Arissa_04 on April 05, 2011, 06:10:08 pm: Hey guys, need help in a trig question.not sure which year it is is from.Here's how the question goes:

Using the expansion of cos(3x-x) and cos(3x+x) prove that:

1/2(cos2x - cos4x) = sin3x sinx

Its a 3 mark question so I'm pretty sure its not that complicated but I cant seem to get the answers.
Thanks in advance Smiley
Title: Re: C3 and C4 Doubts Here!!!
Post by: mdwael on April 06, 2011, 10:50:51 am: Anyone got a link or could upload the formula booklet which is provided when you do the Maths Edexcel exam?
Title: Re: C3 and C4 Doubts Here!!!
Post by: **RoRo** on April 06, 2011, 05:12:12 pm: Quote from: mdwael on April 06, 2011, 10:50:51 am
Anyone got a link or could upload the formula booklet which is provided when you do the Maths Edexcel exam?

I had asked the same question on Page 3, there you go with the link:
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/N38210A-GCE-Mathematical-Formulae-Statistical-Tables.pdf
Title: Re: C3 and C4 Doubts Here!!!
Post by: **RoRo** on April 30, 2011, 09:33:50 am: Can someone please explain to me how do you part b and c?
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on April 30, 2011, 10:00:55 am: Complete the square for the numerator to get (x+1/2)^2+3/4>=3/4
Title: Re: C3 and C4 Doubts Here!!!
Post by: Aadeez || Zafar on April 30, 2011, 10:08:06 am: b)

let x²+x+1 be equal to y
y = x²+x+1
= x²+x+.25+.75
= (x+.5)²+.75

there fore (x+.5)² is a positive number and so .75 added to positive number gives positive number

and so

y is positive
therefore y>0
=x²+x+1>0

C)²+x+1 is positive and (x+2)² is positive so positve number divided by positive is positive and so f(x) is positive

so therefor f(x)>0
Title: Re: C3 and C4 Doubts Here!!!
Post by: **RoRo** on April 30, 2011, 10:21:48 am: Quote from: Mr. MUST on April 30, 2011, 10:08:06 am
b)

let x²+x+1 be equal to y
y = x²+x+1
= x²+x+.25+.75
= (x+.5)²+.75

there fore (x+.5)² is a positive number and so .75 added to positive number gives positive number

and so

y is positive
therefore y>0
=x²+x+1>0

C)²+x+1 is positive and (x+2)² is positive so positve number divided by positive is positive and so f(x) is positive

so therefor f(x)>0

Okay, thank you very much! :D
Title: Re: C3 and C4 Doubts Here!!!
Post by: Blizz_rb93 on May 31, 2011, 12:01:44 pm: f(x) = x + 3/x-1 - 12/x^2 + 2x -3 [x is an element of all eral number, x>1]

Show that f(x) = x^2 + 3x + 3 / x+3

in the book page 11 mixed exercise 1e q4

Thank you
Title: Re: C3 and C4 Doubts Here!!!
Post by: thenewkid™ on June 08, 2011, 11:15:14 pm: anybody have any revision notes or pointers?
Title: Re: C3 and C4 Doubts Here!!!
Post by: thenewkid™ on June 08, 2011, 11:15:45 pm: anybody have any revision notes or pointers?
Title: Re: C3 and C4 Doubts Here!!!
Post by: deadlyelder on June 10, 2011, 06:43:49 am: Have attached the C3 Revision Sheet from http://www.r9paul.org/resources/a-level/ check the site for more notes on M1, M2, Physics, Computing and Religion Studies. They are quite handy notes.

Cheers
Title: Re: C3 and C4 Doubts Here!!!
Post by: thenewkid™ on June 11, 2011, 02:07:06 pm: THANKS :D
Title: Re: C3 and C4 Doubts Here!!!
Post by: studz on June 12, 2011, 06:31:12 pm: guys how are the last three chapters of c3 coming up for u all?
i am having alot of difficulty studying those last 3 chapters! :'(
any tips on how to get those chapters in my head? :/
i did those 3 chapters from the book twice but i still cant get them in my head!! :-\ :-[ :'(
Title: Re: C3 and C4 Doubts Here!!!
Post by: thenewkid™ on June 13, 2011, 06:53:50 pm: those chaps are plain nightmares, the only way out is practice :-\
Title: Re: C3 and C4 Doubts Here!!!
Post by: deadlyelder on June 14, 2011, 02:08:03 pm: The Trinometry and Further Trinometry are truly nightmare. Please share tips for those. Do not have any problem with last chapter. Am solving all the practice papers that are available. How are you guys remembering the trinometry formulas? ???
Title: Re: C3 and C4 Doubts Here!!!
Post by: mdwael on June 14, 2011, 05:24:39 pm: january 2007 question 4 ii

it says find the value of dy/dx:

In the end you get this.. 9 x 4^0.5

this should give +18 and -18 but why in the markscheme only +18 is given? They didnt mention anything that the value is greater than 0..
Title: Re: C3 and C4 Doubts Here!!!
Post by: elemis on June 15, 2011, 10:50:14 am: Think about the e^x graph. The question contains a graph that is very similar to the e^x graph.

Imagine it in your head... does the gradient ever have a negative value ? Nope, its an increasing function.
Title: Re: C3 and C4 Doubts Here!!!
Post by: deadlyelder on June 15, 2011, 08:38:19 pm: Good Luck to everyone for this paper. :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: hmh on June 19, 2011, 05:16:13 pm: Quote from: thenewkid™ on June 08, 2011, 11:15:45 pm
anybody have any revision notes or pointers?

Here are some..
Title: Re: C3 and C4 Doubts Here!!!
Post by: hmh on June 19, 2011, 05:22:36 pm: Can anyone please do C4 January 2011 Question 7 C.
Title: Re: C3 and C4 Doubts Here!!!
Post by: abuelzouz on June 19, 2011, 07:07:58 pm: Quote from: hmh on June 19, 2011, 05:22:36 pm
Can anyone please do C4 January 2011 Question 7 C.

i will solve this paper now and post solution
Title: Re: C3 and C4 Doubts Here!!!
Post by: abuelzouz on June 19, 2011, 08:04:45 pm: Quote from: hmh on June 19, 2011, 05:22:36 pm
Can anyone please do C4 January 2011 Question 7 C.

k so here is it
dx/du = 2(u-4)
dx= 2(u-4) du
x-1 = (u-4)^2
so root x-1 = u-4
and u= root x-1 +4

going back to equation
substitute
integral = integral 1/u & 2(u-4) du
= integral (2-8/u) du
= 2u- 8ln u
x=2 so u=5
x=5 so u=6
substitute these values to get area
thats it hope i helped
Title: Re: C3 and C4 Doubts Here!!!
Post by: hmh on June 19, 2011, 08:26:36 pm: Thanks, I just couldn't get the limits correct but now I know how to do it.
Title: Re: C3 and C4 Doubts Here!!!
Post by: WARRIOR on September 19, 2011, 01:58:45 pm: Hi guys , can anyone get me the answers of these questions ? all of them :P

thanks
Title: Re: C3 and C4 Doubts Here!!!
Post by: iluvme on September 19, 2011, 02:50:39 pm: Quote from: Kimo Jesus on September 19, 2011, 01:58:45 pm
Hi guys , can anyone get me the answers of these questions ? all of them :P

thanks

Looks like its your homework :P

You've got any doubts or are all these questions your doubts?
Title: Re: C3 and C4 Doubts Here!!!
Post by: WARRIOR on September 19, 2011, 05:05:24 pm: Quote from: iluvme on September 19, 2011, 02:50:39 pm
Looks like its your homework :P

You've got any doubts or are all these questions your doubts?

well ill be specific in a moment of which questions i need anyway. this is important how do u solve this

find the inverse function of f (x) = 2x + 3/x-1
Title: Re: C3 and C4 Doubts Here!!!
Post by: Arthur Bon Zavi on September 19, 2011, 05:14:24 pm: Quote from: Kimo Jesus on September 19, 2011, 05:05:24 pm
well ill be specific in a moment of which questions i need anyway. this is important how do u solve this

find the inverse function of f (x) = 2x + 3/x-1

Let f^-1(x) be y and y, x.

y = (2x + 3) / (x - 1)
y (x - 1) = (2x + 3)
yx - y = 2x + 3
yx - 2x = 3 + y
x (y - 2) = 3 + y
x = (3 + y) / (y - 2)

Now replace the values again to get :

f^-1(x) = (3 + x) / (x - 2)
Title: Re: C3 and C4 Doubts Here!!!
Post by: WARRIOR on September 19, 2011, 05:50:24 pm: Quote from: Arthur Bon Zavi on September 19, 2011, 05:14:24 pm
Let f^-1(x) be y and y, x.

y = (2x + 3) / (x - 1)
y (x - 1) = (2x + 3)
yx - y = 2x + 3
yx - 2x = 3 + y
x (y - 2) = 3 + y
x = (3 + y) / (y - 2)

Now replace the values again to get :

f^-1(x) = (3 + x) / (x - 2)

thanks it turned out to be really easy !
Title: Re: C3 and C4 Doubts Here!!!
Post by: WARRIOR on September 19, 2011, 06:02:24 pm: Quote from: Arthur Bon Zavi on September 19, 2011, 05:14:24 pm
Let f^-1(x) be y and y, x.

y = (2x + 3) / (x - 1)
y (x - 1) = (2x + 3)
yx - y = 2x + 3
yx - 2x = 3 + y
x (y - 2) = 3 + y
x = (3 + y) / (y - 2)

Now replace the values again to get :

f^-1(x) = (3 + x) / (x - 2)

anothr question

what is the range and domain of the inverse that you just got out?

for f(x) the domain is x>1 , x E R
range y >=7
right?

so for f^-1 = should be the opposite?
domain y>=7
range x>1?
Title: Re: C3 and C4 Doubts Here!!!
Post by: Arthur Bon Zavi on September 19, 2011, 06:30:00 pm: Quote from: Kimo Jesus on September 19, 2011, 06:02:24 pm
anothr question

what is the range and domain of the inverse that you just got out?

for f(x) the domain is x>1 , x E R
   range y >=7
right?

so for f^-1 = should be the opposite?
   domain y>=7
   range x>1?

Range : $0<$ f^-1 $(x)\le3$
Domain : $x\ge2$
Title: Re: C3 and C4 Doubts Here!!!
Post by: The Golden Girl =D on September 24, 2011, 04:47:42 pm: I've got THREE doubts that are Attached. I need Details/Explanations ONLY because I already know the Answers.

the first page attachment Q.7]d) and Q.8]c)

The THIRD Questions is :

Solve the following equations for values of theta in the interval 0 less than or equal theta less than or equal 2 pie

Equations is; cot theta = - square root 3.
{Got a Quiz tomorrow so please Answer ASAP}

Thanks in Advance =]
Title: Re: C3 and C4 Doubts Here!!!
Post by: The Golden Girl =D on September 24, 2011, 04:56:23 pm: I couldn't Attach it as usual T_T

Q.7]d) can be typed tbh so it's as follows :

write down the periods of the following functions. give your answers in terms of pie.

d) sec (- theta)
Title: Re: C3 and C4 Doubts Here!!!
Post by: Arthur Bon Zavi on September 24, 2011, 05:09:56 pm: Quote from: ~The Golden Girl~ on September 24, 2011, 04:56:23 pm
I couldn't Attach it as usual T_T

Q.7]d) can be typed tbh so it's as follows :

write down the periods of the following functions. give your answers in terms of pie.

d) sec (- theta)

What's the answer ? $2pie$ ?
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on September 24, 2011, 07:14:37 pm: ansering trig question above now
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on September 24, 2011, 07:19:05 pm: both answers in the attachment are solutions
Title: Re: C3 and C4 Doubts Here!!!
Post by: The Golden Girl =D on September 26, 2011, 07:06:16 pm: ^ Thank you Sir. + rep =]
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on September 26, 2011, 09:37:36 pm: No probs
Title: Re: C3 and C4 Doubts Here!!!
Post by: 7d on September 30, 2011, 09:22:38 am: Quote from: T.Q on May 05, 2010, 10:28:37 am
POST YOUR C3 DOUBTS HERE

hey! i really realllllyyyyyyyyyyyy need help in C3 chapter 3 - the exponential and log function chapter

im not able to get it any advice?
websites?
videos?
revision guides??

any help will be appreciated

thank you so much :)
Title: Re: C3 and C4 Doubts Here!!!
Post by: astarmathsandphysics on September 30, 2011, 09:27:08 am: Try this - all the notes are based on exam questions.
http://astarmathsandphysics.com/a_level_maths_notes/a_level_maths_notes_c3_menu.html
Title: Re: C3 and C4 Doubts Here!!!
Post by: Malak on November 11, 2011, 01:30:12 pm: Can anyone help me with this Q (attached)
Title: Re: C3 and C4 Doubts Here!!!
Post by: Arthur Bon Zavi on November 11, 2011, 01:42:21 pm: Quote from: Crooky :P on November 11, 2011, 01:30:12 pm
Can anyone help me with this Q (attached)

Is that $sec^2\theta$ ? Check.
Title: Re: C3 and C4 Doubts Here!!!
Post by: Malak on November 11, 2011, 01:45:38 pm: Quote from: Arthur Bon Zavi on November 11, 2011, 01:42:21 pm
Is that $sec^2\theta$ ? Check.
Yeah, it is :-\
Title: Re: C3 and C4 Doubts Here!!!
Post by: Arthur Bon Zavi on November 11, 2011, 01:47:49 pm: Quote from: Crooky :P on November 11, 2011, 01:45:38 pm
Yeah, it is :-\

I mean sec squared theta ?
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 11, 2011, 01:52:11 pm: Oops sorry, Its not squared just 2 theta ::)
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Arthur Bon Zavi on November 11, 2011, 02:04:00 pm: Quote from: Crooky :P on November 11, 2011, 01:52:11 pm
Oops sorry, Its not squared just 2 theta ::)

You are having trouble in sketching the graph then ?
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 11, 2011, 02:17:54 pm: Quote from: Arthur Bon Zavi on November 11, 2011, 02:04:00 pm
You are having trouble in sketching the graph then ?
No, the second part i.e. finding the points
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Arthur Bon Zavi on November 11, 2011, 02:25:25 pm: Quote from: Crooky :P on November 11, 2011, 02:17:54 pm
No, the second part i.e. finding the points

Intersection method : $sin \theta = sec 2\theta$
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 11, 2011, 02:30:32 pm: Quote from: Arthur Bon Zavi on November 11, 2011, 02:25:25 pm
Intersection method : $sin \theta = sec 2\theta$
Lol, I know that but I need a full solution cause I got stuck in between O_O
Thanks
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: astarmathsandphysics on November 11, 2011, 02:38:28 pm: theta =21.47 or 158.53
see attachment. Only sin theta =(-1+sqrt(3))/2 works cos(-1-sqrt(3))/2 is less than -1
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 11, 2011, 03:00:27 pm: Quote from: astarmathsandphysics on November 11, 2011, 02:38:28 pm
theta =21.47 or 158.53
see attachment. Only sin theta =(-1+sqrt(3))/2 works cos(-1-sqrt(3))/2 is less than -1
Oh god, I just missed one bit O_O

Thankyou
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 11, 2011, 03:09:29 pm: Another one:

Cotx = -1.5, find the value of cosecx
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Arthur Bon Zavi on November 11, 2011, 03:16:25 pm: Quote from: Crooky :P on November 11, 2011, 03:09:29 pm
Another one:

Cotx = -1.5, find the value of cosecx

Sorry. I was late for that.

1 + cot²x = cosec²x
1 + (-1.5)² = cosec²x
cosec x = 1.802775638 = 1.80
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: astarmathsandphysics on November 11, 2011, 03:39:47 pm: I forgot theta=0
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 19, 2011, 01:27:12 pm: For some reason, i cant work this out or more like simplify it O_O

Differentiate (product rule): x(1 + 3x^5)
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Arthur Bon Zavi on November 19, 2011, 01:35:34 pm: Quote from: Ang3l on November 19, 2011, 01:27:12 pm
For some reason, i cant work this out or more like simplify it O_O

Differentiate (product rule): x(1 + 3x^5)

15x⁵ + 3x⁵
=> 18x⁵
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Alpha on November 19, 2011, 01:36:15 pm: Angel, you must take care to differentiate what is in the bracket.

(1 + 3x^5)

Diff. but keep it in the bracket itself. Then, expand. It'll make your work easier.
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 19, 2011, 01:39:35 pm: Quote from: Arthur Bon Zavi on November 19, 2011, 01:35:34 pm
15x⁵ + 3x⁵
=> 18x⁵
Lol, I said product rule ::)

The answer is --> (1 + 3x)^4 (1 + 18x)

@Alpha: Yeah but I cant seem to understand how do we end up with (1 + 18x) :-\
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Alpha on November 19, 2011, 01:45:43 pm: Quote from: Ang3l on November 19, 2011, 01:39:35 pm
Lol, I said product rule ::)

The answer is --> (1 + 3x)^4 (1 + 18x)

@Alpha: Yeah but I cant seem to understand how do we end up with (1 + 18x) :-\

x(1 + 3x^5)

x(15x^4) + ((1 + 3x^5)(1) = 18x^5 + 1.

Did you copy the question correctly? :-\
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 19, 2011, 01:47:29 pm: ^ Yeah, the question is correct.

I got the answer now ;D

Thanks for the help anyway
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Arthur Bon Zavi on November 19, 2011, 02:00:45 pm: Quote from: Ang3l on November 19, 2011, 01:39:35 pm
Lol, I said product rule ::)

The answer is --> (1 + 3x)^4 (1 + 18x)

@Alpha: Yeah but I cant seem to understand how do we end up with (1 + 18x) :-\

Yes, I did with the product rule. I took 3x⁵ instead of 3x⁵ + 1 (which was a mistake I made). Isn't the answer 18x⁵ + 1 correct ?
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 19, 2011, 02:10:01 pm: ^ Okay, people don't KILL me ::)

I posted the question in a wrong way.
The question was: x(1 + 3x)^5 and not x(1 + 3x^5) :-[

Sorry O_O, I got the answer By the way.

Thanks
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Alpha on November 19, 2011, 02:24:31 pm: LOL, Angel. You cannot be killed. You are an Angel. :P
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 19, 2011, 02:42:19 pm: Quote from: ~Alpha on November 19, 2011, 02:24:31 pm
LOL, Angel. You cannot be killed. You are an Angel. :P
Aww, you are just so sweet that's why :D. I am not sure about anse :P*He will surely kill me now* ::)

Okay, I have another question

If we have something like (attached) and then we are supposed to take 3(2x + 1) ^(-0.5) as common then shouldnt the remaining be [x + (2x + 1)^-1]
:-\
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Arthur Bon Zavi on November 19, 2011, 02:45:05 pm: Quote from: Ang3l on November 19, 2011, 02:42:19 pm
Aww, you are just so sweet that's why :D. I am not sure about anse :P*He will surely kill me now* ::)

Okay, I have another question

If we have something like (attached) and then we are supposed to take 3(2x + 1) ^(-0.5) as common then shouldnt the remaining be [x + (2x + 1)^-1]
:-\

Was just about to but then I realized what will I get by killing an angel. ::)

Where is the attached part ?
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 19, 2011, 02:50:56 pm: ^ My wings ::)

Something is wrong with me today :o

Attached it now
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Arthur Bon Zavi on November 19, 2011, 03:44:03 pm: Quote from: Ang3l on November 19, 2011, 02:50:56 pm
^ My wings ::)

Something is wrong with me today :o

Attached it now

How does it feel to be a fallen angel ? Your wings are lying on the ground. How does it feel to be a fallen angel ? Forever is a long way down. :P

No, what you said is incorrect.
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 19, 2011, 04:34:20 pm: Quote from: Arthur Bon Zavi on November 19, 2011, 03:44:03 pm
How does it feel to be a fallen angel ? Your wings are lying on the ground. How does it feel to be a fallen angel ? Forever is a long way down. :P

No, what you said is incorrect.

Lol.

What is the correct answer then ::)
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: astarmathsandphysics on November 19, 2011, 09:12:17 pm: There is nothing in common there because the powers are not the same.
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Arthur Bon Zavi on November 20, 2011, 05:10:00 am: Quote from: astarmathsandphysics on November 19, 2011, 09:12:17 pm
There is nothing in common there because the powers are not the same.

All in my head. ::)
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on November 20, 2011, 03:03:04 pm: Quote from: astarmathsandphysics on November 19, 2011, 09:12:17 pm
There is nothing in common there because the powers are not the same.
What about the (2x + 1)^(0.5), is that not common? :-\
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: astarmathsandphysics on November 20, 2011, 10:56:56 pm: Because the power of one of them is negative, so is the same as 1/(2x+1)^(0.5)
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: WARRIOR on December 14, 2011, 07:47:51 am: COS ( 2x + PIE/3) + COS (2x - PIE/3 ) = COS2X

how is the above proven :O?
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: astarmathsandphysics on December 14, 2011, 03:17:44 pm: cos2xcos pi/3 -sin2x sin pi/3 +cos 2x cos pi/3 +sin 2x sin pi/3 = 2cos 2x cos pi/3 =cos 2x because cos pi/3 =1/2
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on January 14, 2012, 04:16:08 pm: Anyone?

I get the first step but then i am confused as to how they simplfy the numerator to 8 O_O
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: The Golden Girl =D on January 14, 2012, 04:19:52 pm: Give me a second ang3l :)

EDIT :

Using Quotient Rule , We differentiate : { => Which is : dy/dx = [ u'V - V'u ] / V² }

u = 5x² - 10x + 9 => u' = 10 x - 10

v = (x-1)² => v' = 2 (x-1)

dy/dx => [ { (10x-10) [ (x-1)² } - { 2 (x-1) [ 5x² - 10 x + 9] } ] / [( x-1)⁴ ]

I took (x-1) as a Common factor so I got ;

dy/dx => (x-1) { [ (10x-10)(x-1) ] - [2(5x² - 10x + 9)] } / [ (x-1)⁴ ]

=> I cancelled out the (x-1) from the numerator and then expanded what is inside the brackets (numerator) and got the following :

dy/dx => [ 10x² - 20x +10 - 18 - 10x² + 20x] / [ ( x-1 )³ ]

since the 10x and -10x , the 20x and -20x cancel out we have 10-18 = -8 ;)

So we get ;

dy/dx => -8 / [ (x-1 )³ ]

I hope I helped (:

Don't forget to include me in your prayers =D
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on January 14, 2012, 06:34:09 pm: Thanks GG :D

I have another doubt :(

Same paper c3 2005 June Q 5 part d (attached)

I get everything, except that in part d i am getting a negative answer for theta

for theta + 0.588 i get --> 0.429 and 0.5709..

and then for theta i get a negative answer :S
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: The Golden Girl =D on January 14, 2012, 06:44:29 pm: Since you knew how to solve the first part then I'll just go into the theta thing (:

the Equation you'll use is => sin theta [ squareroot52 sin (theta + 0.588)] =0 from part (c)

so sin theta = 0

we know that sin theta is 0 , at pie and 2pie .... only 0 is accepted => check the range ;)

squareroot52 sin (theta + 0.588) = 0

theta + 0.588 = 0 , pie , 2 pie (cuz I told you ,we know that sin theta is ZERO at these three angles )

so theta = pie - 0.588 = 2.553

theta =2pie -0.588 = 5.70 { rejected TOO big not within the range }

therefore Theta = 0 rad , 2.553 rad

This is one of the Tricks my teacher uses , I hope you got it (:

[I might have to go any minute cuz my bro's got a project -.- ]
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on January 14, 2012, 07:08:10 pm: but isnt the range greater han 0 and less than pie then why do we consider 2 pie :-\
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: The Golden Girl =D on January 14, 2012, 07:13:02 pm: I considered it but I didn't put it in the final answer you see :)
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on January 15, 2012, 11:04:29 am: Quote from: Miss Invisible on January 14, 2012, 07:13:02 pm
I considered it but I didn't put it in the final answer you see :)
you missed out 3 :S

Won't the equation be(x = theta) --> Sin x (4cosx + 6sinx -3) = 0

and later it will be Sinx [square root 52 sin(x + 0.588)] - 3 = 0

then, sin x = 0 and sqrt 52 sin(x + 0.588) - 3 = 0
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: The Golden Girl =D on January 15, 2012, 04:23:12 pm: Mr.Paul ,Can you please clear our doubt on this point =/

Thanks (:
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: astarmathsandphysics on January 16, 2012, 09:15:43 am: happpy too. Entire question answered in attachment
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: The Golden Girl =D on January 16, 2012, 01:09:58 pm: May God bless you Sir ..Thanks (:
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on January 16, 2012, 04:18:18 pm: Umm >_< can you explain the last 2 lines, I don't get where did the 3.142 come from :S

And thankyou sir.
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: astarmathsandphysics on January 17, 2012, 08:06:27 pm: 3.142 is pi!!!!!!!!!!!!
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on January 18, 2012, 12:53:49 pm: Quote from: astarmathsandphysics on January 17, 2012, 08:06:27 pm
3.142 is pi!!!!!!!!!!!!
OMG :o :o *epic fail...*

Thanks again..
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: -Cornea on January 20, 2012, 10:41:06 am: The chain rule used in differentiation is it a must to have two functions divided by each other so that i can use it what i mean is i can deffrentiate (x+3)/(x+5) this using the rule can i use the same way in deffrentiating (1)/(x+4)
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on January 20, 2012, 11:21:17 am: Quote from: -Cornea on January 20, 2012, 10:41:06 am
The chain rule used in differentiation is it a must to have two functions divided by each other so that i can use it what i mean is i can deffrentiate (x+3)/(x+5) this using the rule can i use the same way in deffrentiating (1)/(x+4)
I guess you meant the quotient rule.
Yes, i guess you can use it for the 2nd equation as well..but that isnt the best way. You can simply bring the (x + 4) up as (x+4)^-1 and solve it using the chain rule
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: -Cornea on January 21, 2012, 09:58:28 am: yeah i don't their names :p anyway thanks :)
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: -Cornea on January 21, 2012, 11:35:45 am: do i need to know how to prove that sin(A+B)=sinA cosB + cosA sinB =D because i can't get it from the book =(
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: The Golden Girl =D on January 21, 2012, 11:38:38 am: Give me a couple of minutes to type it down ..Okies ;)

[ Will be editing so check after some time ]

My teacher told me that there is no need for me to know How to prove this but if it's the Opposite of what my teacher said then please do tell me how we rpove it (:

I hope I helped :D
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: -Cornea on January 21, 2012, 01:06:48 pm: and i need to ask if u know what marks should i get in order to score an A* in my AL's if u have an idea about this ,i'll be grateful =}
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on January 21, 2012, 01:36:13 pm: ^It all depends on the grade boundary for that year.
But generally you need to get an A in AS in order to get an A* overall and ,I guess, you need like 90% or above in A2 to get an A*
summary: A in AS and 90% (or above) in A2 = A* overall ::)
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: -Cornea on January 22, 2012, 12:15:02 pm: y = 2 – e?^-x iam really confused in these stuff should i multiply it -1 and then add 2 or should i multiply it by 1 only and add 2 then :s
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on January 22, 2012, 12:54:04 pm: Quote from: -Cornea on January 22, 2012, 12:15:02 pm
y = 2 – e?^-x iam really confused in these stuff should i multiply it -1 and then add 2 or should i multiply it by 1 only and add 2 then :s
I dont understand the question :S
Can you be more clear? What does the question say?
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: The Golden Girl =D on January 22, 2012, 12:58:34 pm: Cornea ,if you're talking about how we Calculate the y = 2 - e^-x USING the calculator then Here is the Method :

You press shift on the ln button then Multiply it by -X , and THEN add 2 to it :)

I hope I helped (:
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: -Cornea on January 22, 2012, 02:22:04 pm: no i mean they asked me to sketch it :)
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: The Golden Girl =D on January 22, 2012, 04:33:39 pm: Cornea can you send me your e-mail ID via PMs cuz It seems my graph can't attach for some weird reason =/
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: -Cornea on January 22, 2012, 04:48:28 pm: done i sent you my email :D
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Malak on January 22, 2012, 07:58:53 pm: Good luck everyone for C3 2morrow :D
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Adzel on February 23, 2012, 03:28:58 pm: Solve for x :

4e^2x + 3e^x = 1

Anyone? :-\
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: astarmathsandphysics on February 24, 2012, 09:34:15 am: Sub y=e^x then y^2=e^2x
4y^2+3y-1=0
(4y-1)(y+1)=0
so 4y-1=0 so y=1/4 so e^x =1/4 so x=ln(1/4)
or y+1=0 so y=-1 so e^x=-1 so x=ln(-1) no solution
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: Adzel on February 24, 2012, 07:03:11 pm: I did the same thing, but in the answers; it has a solution and it's a negative number ???
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: astarmathsandphysics on February 25, 2012, 08:54:47 am: yes ln(1/4) is negative
-1.386
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: WARRIOR on March 11, 2012, 04:59:07 pm: The angle between the vectors i + 3j and j + xk is 60
Show that x = square root of (13/5)

i get squarte root of (26/10 )

can someone show me the right working?
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: astarmathsandphysics on March 11, 2012, 08:14:13 pm: you got the right answer.
Cancel it down
LAUGH AT YOURSELF
Title: Re: C3 and C4 DOUBTS HERE!!!
Post by: ~ Miss Relina ~ on October 12, 2012, 12:14:49 pm: how to differentiate this one :-[