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Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: preity on May 28, 2009, 08:56:23 am

Title: Pure math(p3)
Post by: preity on May 28, 2009, 08:56:23 am: Can someone help me with this questions

M/J 02 Q1
M/J 03 Q1
0/N 04 Q4
Title: Re: Pure math(p3)
Post by: astarmathsandphysics on May 28, 2009, 09:36:22 am: Can someone try this cos i wont be home for 3 hours.
Title: Re: Pure math(p3)
Post by: xcon89 on May 28, 2009, 09:51:33 am: for j02

open the right hand side
and then

2/tan2@

then
2/(2tan@/1-tan^2@)

then > 2(1 - tan^@)/2Tan@

(1 - tan^2@)/tan@

1/tan@ - tan^2@/tan@

cot@ - tan@
Title: Re: Pure math(p3)
Post by: xcon89 on May 28, 2009, 09:58:33 am: others are easy aswell but since they involve alot of latex and i am unfamiliar with latex, I can't write it here
sorry
Title: Re: Pure math(p3)
Post by: twilight on May 28, 2009, 12:41:51 pm: hang on .. i'll solve them
Title: Re: Pure math(p3)
Post by: twilight on May 28, 2009, 12:44:09 pm: Quote from: preity on May 28, 2009, 08:56:23 am
Can someone help me with this questions

M/J 02 Q1
M/J 03 Q1
0/N 04 Q4
for m/j 2003 Q1(i)

sin(x-60) - cos(30-x) = 1
use the formula sheet here
sin x cos 60 - sin 60 cos x - ( cos 30 cos x + sin 30 sin x ) = 1
sin x cos 60 - sin 60 cos x - cos 30 cos x - sin 30 sin x = 1
(1/2)sin x - ( $\surd$ 3/2)cos x - ( $\surd$ 3/2)cos x - (1/2)sin x = 1

(- $\surd$ 3)cos x = 1

cos x = -1/ $\surd$ 3
Title: Re: Pure math(p3)
Post by: twilight on May 28, 2009, 12:49:05 pm: Quote from: preity on May 28, 2009, 08:56:23 am
Can someone help me with this questions

M/J 02 Q1
M/J 03 Q1
0/N 04 Q4
m/j 2003 Q1(ii)

cos x = -1/ $\surd$ 3

reference angle = cos^-1 (1/ $\surd$ 3) = 54.7^o

therefore x = 180 + 54.7 = 125.3^o
Title: Re: Pure math(p3)
Post by: candy on May 28, 2009, 01:02:36 pm: The polynomial x^4 - 2x^3 - 2x^2 +a is denoted by f(x). It is given that f(x) is divisible by x^2 - 4x +4.
(i) Find the value of a.
(ii) When a has this value, show that f(x) is never negative.

can someone please explain part (ii)
i don't understand the question :(
Title: Re: Pure math(p3)
Post by: astarmathsandphysics on May 28, 2009, 01:05:11 pm: Quote from: candy on May 28, 2009, 01:02:36 pm
The polynomial x4 ?2x3 ?2x2 +a is denoted by f(x). It is given that f(x) is divisible by x2 ?4x +4.
(i) Find the value of a.
(ii) When a has this value, show that f(x) is never negative.

can someone please explain part (ii)
i don't understand the question :(

can you modify this post? the question is not clear
Title: Re: Pure math(p3)
Post by: ITS MEE... on May 28, 2009, 01:11:15 pm: can someone solve the last part please
An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is
empty. At time t hours after filling begins, the volume of liquid is V m3 and the depth of liquid is h m.
It is given that V = 4
3h3.
The liquid is poured in at a rate of 20m3 per hour, but owing to leakage, liquid is lost at a rate
proportional to h2. When h = 1,
dh
dt = 4.95.
(i) Show that h satisfies the differential equation
dh
dt = 5
h2 ? 1
20
. [4]
(ii) Verify that
20h2
100 ?h2 ? ?20 + 2000
(10 ?h)(10 +h). [1]
(iii) Hence solve the differential equation in part (i), obtaining an expression for t in terms of h.
Title: Re: Pure math(p3)
Post by: ITS MEE... on May 28, 2009, 01:12:35 pm: oops it looks so awkward but its nove 2008 p3 CIE
Title: Re: Pure math(p3)
Post by: astarmathsandphysics on May 28, 2009, 01:15:41 pm: Quote from: preity on May 28, 2009, 08:56:23 am
Can someone help me with this questions

M/J 02 Q1
M/J 03 Q1
0/N 04 Q4

M/J 02 Q1

cotx-tanx=cosx/sinx-sinx/cosx
=((cosx)^2-(sinx)^2)/sinxcosx=cos2x/(1/2sin2x)=2cot2x

M/J 03 Q1

sin(x ? 60 ) ? cos(30 ? x) = 1
sinxcos60-cosxsin60-cos30cosx-sin30sinx=1
sinx*1/2-cosx $sqrt(3)/2$ - $sqrt(3)/2$ cosx-1/2sinx=1
$sqrt(3)$ cosx=1 so cosx=1/ $sqrt(3)$

0/N 04 Q4
tan(45 + x) = 2 tan(45 ? x)

(tan45+tanx)/(1-tan45tanx)=2(tan45-tanx)/(1tan45tanx)

(1+tanx)/(1-tanx)=2(1-tanx)/(1+tanx)
cross multiply
(1+tanx)^2=2(1-tanx)(1-tanx)
1+2tanx+tan^2x=2-4tanx+2tan^2x
tan^2x-6tanx+1=0
Title: Re: Pure math(p3)
Post by: ITS MEE... on May 28, 2009, 01:20:43 pm: OK can someone do
nov08 question 7)ii
10)iv
8)iii
thank u
Title: Re: Pure math(p3)
Post by: ITS MEE... on May 28, 2009, 01:23:39 pm: please someone answer ???
Title: Re: Pure math(p3)
Post by: twilight on May 28, 2009, 01:41:06 pm: Quote from: preity on May 28, 2009, 08:56:23 am
Can someone help me with this questions

M/J 02 Q1
M/J 03 Q1
0/N 04 Q4
for 0/N 04 Q4
(i) tan(45+x) = 2tan(45-x) use tan(a+b)= (tan a + tan b)/(1- tan a tan b ) in the formula sheet
(tan 45 + tan x)/(1 - tan 45 tan x) = 2(tan 45 - tan x)/(1 + tan 45 tan x )
(1+tan x )/(1-tan x) = (2-2tan x)/(1+tan x)
cross multiply to get tan²x - 6tanx + 1 = 1

(ii) substitute tanx by y
then solve by y= -b+ $\surd$ (b²-4ac)
________________
2a
then y = -b- $\surd$ (b²-4ac)
_________________
2a

then substitute back with tanx and do the usual method to get the values for x for + and -
Title: Re: Pure math(p3)
Post by: astarmathsandphysics on May 28, 2009, 02:12:40 pm: Quote from: ITS MEE... on May 28, 2009, 01:11:15 pm
can someone solve the last part please
An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is
empty. At time t hours after filling begins, the volume of liquid is V m3 and the depth of liquid is h m.
It is given that V = 4
3h3.
The liquid is poured in at a rate of 20m3 per hour, but owing to leakage, liquid is lost at a rate
proportional to h2. When h = 1,
dh
dt = 4.95.
(i) Show that h satisfies the differential equation
dh
dt = 5
h2 ? 1
20
. [4]
(ii) Verify that
20h2
100 ?h2 ? ?20 + 2000
(10 ?h)(10 +h). [1]
(iii) Hence solve the differential equation in part (i), obtaining an expression for t in terms of h.

in this thread

https://studentforums.biz/index.php/topic,740.0.html
Title: Re: Pure math(p3)
Post by: ITS MEE... on May 28, 2009, 02:25:20 pm: thank u
Title: Re: Pure math(p3)
Post by: astarmathsandphysics on May 28, 2009, 02:25:49 pm: Quote from: candy on May 28, 2009, 01:02:36 pm
The polynomial x^4 - 2x^3 - 2x^2 +a is denoted by f(x). It is given that f(x) is divisible by x^2 - 4x +4.
(i) Find the value of a.
(ii) When a has this value, show that f(x) is never negative.

can someone please explain part (ii)
i don't understand the question :(

x^2 - 4x +4.=(x-2)^2 so put x=2 and solve f(x)=0 for a

16-16-8+a=a=0 to geta=8 Now do long division by x^2 - 4x +4.to getx^2+2x+2=(x+1)^2+1>0 and since(x-2)^2 is never -ve neither is the product of it and (x+1)^2+1
Title: Re: Pure math(p3)
Post by: ITS MEE... on May 28, 2009, 02:43:25 pm: can u integrate
sin^3xcosx
please :'(
Title: Re: Pure math(p3)
Post by: ITS MEE... on May 28, 2009, 02:49:27 pm: oh i got it no need to solve it but can u integrate this
(1+tanx)^4 sec^2x
Title: Re: Pure math(p3)
Post by: twilight on May 28, 2009, 02:50:55 pm: Quote from: ITS MEE... on May 28, 2009, 02:43:25 pm
can u integrate
sin^3xcosx
please :'(
how didya solv it?
Title: Re: Pure math(p3)
Post by: ITS MEE... on May 28, 2009, 03:00:21 pm: well wnever u see an ODD one ie unfamiliar and kind of compound just try to differentiate one of them and if u get the other (regardless of the power or something like a number multiplied or divided by it then u intergrate the one u differentiated
:-\ hope its a bit clear
Astarmathsandphysics can u explain it a bit more or make it less vague? please
Title: Re: Pure math(p3)
Post by: louis on May 28, 2009, 03:23:44 pm: ii) Find a vector equation for the line of intersection between two planes :
2x -y - 3z= 7 and x + 2y + 2z=0
Solution : The normal vectors of 2 planes are n1= ( 2,-1,-3) and n2= (1,2,2)
These 2 vectors form a new plane.
The normal vector of this new plane is ( 2,-1,-3) X (1,2,2) = (4, -7, 5)
The normal vector is parallel to the line of intersection of the 2 planes.
So the direction vector for the line of intersection of 2 planes is a ( 4, -7, 5 )

Now we have to find the position vector of a point on the line of intersection
In the equations of 2x - y -3z= 7 and x + 2y + 2z = 0,there are 3 unknowns.
Reduce to 2 equations with 2 unknowns only by putting x=0 or y=0 or z=o
We choose y=0
Solve for x and z in the equations 2x -3Z = 7
x + 2Z = 0
we get Z = -1 x = 2
So the vector equation for the line of intersection of 2 planes is
r= 2i -k + a( 4,-7 , 5 )
Title: Re: Pure math(p3)
Post by: twilight on May 28, 2009, 03:27:04 pm: ooo i think i get wut iu mean ITS ME..
so is it (1/4)(sin⁴x) ?
Title: Re: Pure math(p3)
Post by: astarmathsandphysics on May 28, 2009, 03:31:06 pm: Quote from: ITS MEE... on May 28, 2009, 02:49:27 pm
oh i got it no need to solve it but can u integrate this
(1+tanx)^4 sec^2x

Substitute u=1+tanx to get u^4dv ans=1/5(1+tanx)^5
Title: Re: Pure math(p3)
Post by: ITS MEE... on May 28, 2009, 03:33:32 pm: THANK U SOOOOOOOOOO MUCH LOUIS
Title: Re: Pure math(p3)
Post by: astarmathsandphysics on May 28, 2009, 03:34:58 pm: Quote from: twilight on May 28, 2009, 02:50:55 pm
Quote from: ITS MEE... on May 28, 2009, 02:43:25 pm
can u integrate
sin^3xcosx
please :'(
how didya solv it?

Substitute u=1+tanx to get u^4du ans=1/5(1+tanx)^5
Title: Re: Pure math(p3)
Post by: twilight on May 28, 2009, 03:36:39 pm: Quote from: astarmathsandphysics on May 28, 2009, 03:34:58 pm
Quote from: twilight on May 28, 2009, 02:50:55 pm
Quote from: ITS MEE... on May 28, 2009, 02:43:25 pm
can u integrate
sin^3xcosx
please :'(
how didya solv it?

Substitute u=1+tanx to get u^4du ans=1/5(1+tanx)^5

ummmm .. wher is the tan???
Title: Re: Pure math(p3)
Post by: ITS MEE... on May 28, 2009, 03:43:56 pm: THANK U Astarmathsandphysics AND TWIGHLIGHT IT RIGHT
Title: Re: Pure math(p3)
Post by: twilight on May 28, 2009, 03:44:54 pm: thanks evry1
Title: Re: Pure math(p3)
Post by: candy on May 28, 2009, 03:55:07 pm: second part pleasee!

The value of 'a' is 8
Title: Re: Pure math(p3)
Post by: sahrhp on May 28, 2009, 04:39:09 pm: SEE EARLIER PAGE 2
Title: Re: Pure math(p3)
Post by: louis on May 28, 2009, 04:44:12 pm: One way to get the other factor of (x^4) -2x³-2x² +8 is doing the long division.
Divided by x² – 4x + 4 , you get x² + 2x + 2

Now, (X^4 )- 2x³-2x² +8 =( x ²– 4x + 4)( x² + 2 x + 2)
=( x – 2 ) ² { ( x + 1 ) ² + 1 }
( x – 2 ) ² is always positive
( x + 1 ) ² + 1 is always positive too.
So (X^4 ) -2x³-2x² +8 cannot be negative
Title: Re: Pure math(p3)
Post by: crucio on May 28, 2009, 05:25:33 pm: another rather tricky way of doing it is:

$y=x^4 + 2x^3 -2x^2 +8$
${\frac{\delta y}{\delta x}}=4x^3 -6x^2 -4x$
${\frac{\delta y}{\delta x}}=0 \Rightarrow 4x^3 -6x^2 -4x=0\Rightarrow x=-0.5,0,2$
second derivative= $12x^2 -12x -4$
putting values of x in second derivative:

$-0.5 \rightarrow 5$
$0 \rightarrow -4$
$2 \rightarrow 20$
hence there is a min point at -0.5 and 2 and max at 0, putting these values of x gives values of min points (-0.5, 7.8) and (2, 0), therefore as none of y values are below x axis, $f(x)$ is always positive! :D
Title: Re: Pure math(p3)
Post by: SGVaibhav on May 28, 2009, 05:41:39 pm: this is an A level thread, hmm, i was thinking that why dont i even know the head and tail of this.
Title: Re: Pure math(p3)
Post by: astarmathsandphysics on May 28, 2009, 05:43:36 pm: Quote from: ITS MEE... on May 28, 2009, 02:43:25 pm
can u integrate
sin^3xcosx
please :'(

orry i missed this earlier sub u=sinx then du=cosxdx so dx=1/cosxdu so you integrate

sin^3xcosxdx=sin^3xcosx(1/cosxdu)=u^3du and the ans =1/4u^4=1/4sin^4x
Title: Re: Pure math(p3)
Post by: louis on May 28, 2009, 05:52:53 pm: ITS MEE,

Attachment ( complex number ) for your attention.
Title: Re: Pure math(p3)
Post by: twilight on May 28, 2009, 05:59:46 pm: Quote from: astarmathsandphysics on May 28, 2009, 05:43:36 pm
Quote from: ITS MEE... on May 28, 2009, 02:43:25 pm
can u integrate
sin^3xcosx
please :'(

orry i missed this earlier sub u=sinx then du=cosxdx so dx=1/cosxdu so you integrate

sin^3xcosxdx=sin^3xcosx(1/cosxdu)=u^3du and the ans =1/4u^4=1/4sin^4x

thksssss astar .. +rep
Title: Re: Pure math(p3)
Post by: louis on May 28, 2009, 06:05:01 pm: ITS mee,

Another attachment ( vertices of triangle ).
Title: Re: Pure math(p3)
Post by: LupeFiasco on May 28, 2009, 11:33:30 pm: anybody here taking bio P4, please send me what's coming for the exam at genetick2004@hotmail.com . Thanks and god speed.
Title: Re: Pure math(p3)
Post by: louis on May 29, 2009, 07:29:53 am: crucio,
I like your method by calculus calculation.
I make the graph based on your calculation.
Please look at the attachment.