Mathematics: Post your doubts here!

[s1294]

i. (4 7 -p).(8 -1 -p) = 32 -7 +p^2

25 + p^2 = O ANS

ii) for two vectors to be 90^o the eq. will be

OA.OB = (mag of OA).(mag of OB)cos(90

cos 90 is zero "O" though the remaining part will be OA.OB = O
means the eq. found for OA.OB is part (i) could be arranged as.... P^2 = -25 now check that is it possible to find value of "P" ????
I t is not as itx not a real. (itx complex) so proven.

iii) angle OAB is 60^o means that the eq. for angle will be as follows

OA.OB = (mag of OA).(mag of OB)cos(60

now put values
OA.OB = (mag of OA).(mag of OB)cos(60

25 + p^2 = [(4^2 + 7^2 + p^2)^1/2].[(8^2 + 1^2 +p^2)^1/2]cos(60

25 + p^2 = [(65 + p^2)^1/2].[(65 +p^2)^1/2]cos(60

25 + p^2 = [(65 + p^2)cos(60

25 + p^2 / 65 + p^2 = 1/2 now solve for P and U will obtain p= 3.87 (wid sign of plus and minus)

THANK YOU SOOO MUCH

 

[Scafalon40]

Integrate the equation of velocity to get equation of displacement in terms of t.
Use limits 3 and 4 while integrating..

Can't get the correct answer , you try it.

 

[smzimran]

Can't get the correct answer , you try it.

First, lets check if there are any values of t if v = 0 (which represents time at which particle changes direction)
3t^2 - 12t +5 = 0
Using quadratic formulae,
t = 3.53 and t = 0.47
At t = 3.53 direction changes and so we have to integrate and use limits of 3 to 3.53 separately and 3.53 to 4 separately
First lets integrate and obtain the result:
s = t^3 - 6t^2 +5t
Now lets use limits 3 and 3.53 to find s1
s1 = [ (3.53)^3 - 6(3.53)^2 + 5(3.53) ] - [ (3)^3 - 6(3)^2 + 5(3) ]
s1 = [-13.13] - [-12]
s1 = 1.13
Now lets use limits 3.53 and 4 to find s2
s2 = [ (4)^3 - 6(4)^2 + 5(4) ] - [ (3.53)^3 - 6(3.53)^2 + 5(3.53) ]
s2 = [-12] - [-13.13]
s2 = 1.13
So,
s = s1 + s2 = 1.13 + 1.13 = 2.26 m

 

[smzimran]

Need help in P3 Winter 2008

http://www.xtremepapers.com/CIE/index.php?dir=International A And AS Level/9709 - Mathematics/&file=9709_w08_qp_3.pdf

Q10 - All parts except the first one.

Q6 - part 2

Q6(ii)
From (i), we know that 5sin θ +12cos θ = 13sin( θ + 67.38)
So,
5sin 2θ +12cos 2θ = 11
means
13sin( 2θ + 67.38) = 11
sin( 2θ + 67.38) = 11/13
basic angle, x = sin`-1 (11/13)
x = 57.8
We know that 0◦< θ < 180◦
So for ( θ + 67.38)
67.38◦ < ( 2θ + 67.38) < 427.38◦
So,
( 2θ + 67.38) = 180 - 57.8 , 360 + 57.8
(sinx is positive means 1st and 2nd quadrants in two cycles but the 1st quadrant in 1st cycle and 2nd quadrant in 2nd cycle are out of range)
Solve to obtain
θ = 27.4 , 175.2

 

[smzimran]

Need help in P3 Winter 2008

http://www.xtremepapers.com/CIE/index.php?dir=International A And AS Level/9709 - Mathematics/&file=9709_w08_qp_3.pdf

Q10 - All parts except the first one.

Q6 - part 2

Q1o
(ii)
|wz| = |w||z| = (1)(R) = R
arg|wz| = arg|w| + arg|z| = 2π/3 + θ

|z/w| = |z|/|w| = R/1 = R
arg|z/w| = arg|z| - arg|w| = θ - 2π/3

(iii)
Modulus of all three are same meaning their lengths are equal ( = R)
All angles subtended are π/3

(iv)
z = 4 +2i
The other two vertices are zw and z/w

zw = (4 + 2i)(-0.5 +i√3/2)

zw = - (2 + √3) + (2√3 -1)i

z/w = (4 + 2i)/(-0.5 +i√3/2)
Rationalise the denominator to get
z/w = - (√3 + 2) + (2√3 + 1)i

 

[Scafalon40]

First, lets check if there are any values of t if v = 0 (which represents time at which particle changes direction)
3t^2 - 12t +5 = 0
Using quadratic formulae,
t = 3.53 and t = 0.47
At t = 3.53 direction changes and so we have to integrate and use limits of 3 to 3.53 separately and 3.53 to 4 separately
First lets integrate and obtain the result:
s = t^3 - 6t^2 +5t
Now lets use limits 3 and 3.53 to find s1
s1 = [ (3.53)^3 - 6(3.53)^2 + 5(3.53) ] - [ (3)^3 - 6(3)^2 + 5(3) ]
s1 = [-13.13] - [-12]
s1 = 1.13
Now lets use limits 3.53 and 4 to find s2
s2 = [ (4)^3 - 6(4)^2 + 5(4) ] - [ (3.53)^3 - 6(3.53)^2 + 5(3.53) ]
s2 = [-12] - [-13.13]
s2 = 1.13
So,
s = s1 + s2 = 1.13 + 1.13 = 2.26 m

i did the exact same thing!
and got the exact same answer!
even drew the graph to double check!
the 48 doesn't make sense to me, do you think i should just consider it to be a typing error on my worksheet?

 

[smzimran]

i did the exact same thing!
and got the exact same answer!
even drew the graph to double check!
the 48 doesn't make sense to me, do you think i should just consider it to be a typing error on my worksheet?

Yes, it must be, just think 48 seems to be illogical answer ; in 1 second how will it move 48 m !

 

[Scafalon40]

Yes, it must be, just think 48 seems to be illogical answer ; in 1 second how will it move 48 m !

If its travelling at 48 m/s

 

[smzimran]

If its travelling at 48 m/s

Logically, it wont

 

[Scafalon40]

Logically, it wont

Meant to be a joke, Einstein

 

[smzimran]

Meant to be a joke, Einstein

I know

 

[Scafalon40]

I know

 

[Zephyro]

Q1o
(ii)
|wz| = |w||z| = (1)(R) = R
arg|wz| = arg|w| + arg|z| = 2π/3 + θ

|z/w| = |z|/|w| = R/1 = R
arg|z/w| = arg|z| - arg|w| = θ - 2π/3

(iii)
Modulus of all three are same meaning their lengths are equal ( = R)
All angles subtended are π/3

(iv)
z = 4 +2i
The other two vertices are zw and z/w

zw = (4 + 2i)(-0.5 +i√3/2)

zw = - (2 + √3) + (2√3 -1)i

z/w = (4 + 2i)/(-0.5 +i√3/2)
Rationalise the denominator to get
z/w = - (√3 + 2) + (2√3 + 1)i

Thanks, understood everything. Really Helpfull.

 

[s1294]

Please help with this. urgent- please explain you steps

find the term independent of x in the expansion of (2x+ 1/x^2)^6
its 3 marks from paper 1 of oct nov 2011
unfortunately i dnt have a link cos i have on hard copy sorry

please help out

 

[leadingguy]

Please help with this. urgent- please explain you steps

find the term independent of x in the expansion of (2x+ 1/x^2)^6
its 3 marks from paper 1 of oct nov 2011
unfortunately i dnt have a link cos i have on hard copy sorry

please help out

it is [(2x) + (1/x^2)]^6 simply expand it using binomials

6C0. (2x)^6 . (1/x^2)^o + 6C1. (2x)^5 . (1/x^2)^1 + 6C2.(2x)^4.(1/x^2)^2 + 6C3.(2x)^3.(1/x^2)^3 +....

solve each term

64x^6 + 192x^3 + 240x^4/x^4 +.... now look this x^4 can be cancelled out leaving the only 240 without any power of x


so the term independant of x is 240 Ans

 

[s1294]

Need help with this as well (please draw and show as well)
i) sketch on a single diagram the graphs of y-cos2(theta) and y=1/2 for 0<= (theta)=<2(pi)

ii) write down the number of roots of the equation 2cos2(theta) - 1=0 in the interval 0<= (theta)=< 2(pi)

iii) deduce the number of roots of the equation 2cos2(theta) - 1 =0 in the interval 10(pi)<= (theta) =< 20(pi)

i really don't get the theta and pi stuff, pls explain

 

[s1294]

it is [(2x) + (1/x^2)]^6 simply expand it using binomials

6C0. (2x)^6 . (1/x^2)^o + 6C1. (2x)^5 . (1/x^2)^1 + 6C2.(2x)^4.(1/x^2)^2 + 6C3.(2x)^3.(1/x^2)^3 +....

solve each term

64x^6 + 192x^3 + 240x^4/x^4 +.... now look this x^4 can be cancelled out leaving the only 240 without any power of x


so the term independant of x is 240 Ans

isnt there a simpler way of doing it- writing out all the terms in my test will take me ages? but this is really helpful.... thank u

 

[leadingguy]

U jxt need to show a some imp. steps which are highlighted. but reaching there itx neccessary to go through other steps, whether in rough or fair. if U are doing these all then u surely come up with few or no errors as
no. of steps is inversly proportional to mistakes

 

[s1294]

U jxt need to show a some imp. steps which are highlighted. but reaching there itx neccessary to go through other steps, whether in rough or fair. if U are doing these all then u surely come up with few or no errors as
no. of steps is inversly proportional to mistakes

i know but it takes up my time in my practice exams that's why n sometimes it says ascending power of x or descending power of x- how do u figure out whether you start with 6C6 or 6C1?

 

[leadingguy]

jxt start with 6c1 always regardless of itx saying thats what i used to do and aftr getting some terms take ur required one whether ascending or decending