Mathematics: Post your doubts here!

[Chooi1993]

Can anyone explain how to find the greatest and smallest of argument ( complex number) especially when the question is about circle ?
I a little bit confused

 

[smzimran]

Can anyone explain how to find the greatest and smallest of argument ( complex number) especially when the question is about circle ?
I a little bit confused

http://www.xtremepapers.com/community/threads/mathematics-post-your-doubts-here.9599/page-63
SCROLL DOWN to post no.1249

 

[Chooi1993]

Pls anyone know how to solve oct/nov 2008 p3 q7&10

 

[Scafalon40]

A particle P moves along the x axis. At time t seconds (where t>0) the velocity of P is 3t^2-12t+5 m/s in the direction of x increasing. When t=0, P is at the origin O Find
a)the velocity of P when the acceleration is zero (Ans: 7)
b)the values of t when P is again at O (Ans: 1,5)
c)the distance travelled by P in the interval 3<t<4 (Ans:48)
I've done the first two parts just can't do the last one
PLEASE HELP I'LL BE VERY GRATEFUL PLEASE!!!!!!!!!!!
P.S the < stands for 'is lesser than or equal to', same goes for >
HELP!

 

[AhmedNasser]

OCT/NOVEMBER 2008 (q7)

 

[smzimran]

This is the answer to a question asked by @"USMAN ALI (MANI)"
s11/31/q10

dN/dt = N(1800 - N)/3600
Separating variables,
(1/N(1800 - N)) dN = (1/3600) dt
Using partial fractions, we get the result that 1/N(1800 - N) = 1/1800 (1/N + 1/(1800 - N))
So
1/1800 (1/N + 1/(1800 - N)) dN = (1/3600) dt
(1/N + 1/(1800 - N)) dN = (1800/3600) dt
(1/N + 1/(1800 - N)) dN = (1/2) dt
Integrating
lnN - ln(1800 - N) = t/2 + c
ln (N/(1800 - N)) = t/2 + c
Substitute N = 300, t = 0 to get
c = ln(1/5) 0r c = - ln5
Back to equation substitute value of c
ln (N/(1800 - N)) = t/2 + - ln5
ln (N/(1800 - N)) + ln5 = t/2
ln (5N/(1800 - N)) = t/2
5N/(1800 - N) = e^(t/2)
5N + Ne^(t/2) = 1800e^(t/2)
N(5 + e^(t/2)) = 1800e^(t/2)
N = (1800e^(t/2)) / (5 + e^(t/2))
Solved!

 

[anonymous123]

A rectangle has sides of (2x+3)cm and (x+1)cm.State the domain of x if the area of the rectangle lies between 10cm^2 and 36cm^2 inclusive?
PLZ HELP

 

[bamteck]

URGENT HELP REQUIRED PLEASE !


1. (a) If A is the acute angle such that sin A = 3/5 and B is the obtuse angle such that sin B = 5/13, find without using a calculator the values of cos(A+B) and tan(A-B).

(b) Find the solutions of the equation tan x + 3 cot x = 5 sec x for which 0 < x < 2π.

(Ans: (a) -63/65, 56/33 (b) π/6, 5π/6)


2. (a) Solve, in radians, the equation sin( x + π/6) = 2cos x giving all solutions in the range 0 < x < 2π.
(b)(i) Use the formula tan(A-B) = tanA - tanB/1+tanAtanB to show that tan(x - π/4) = tanx - 1/tanx +1
(ii) Hence, or otherwise, solve, in radians to 3 s.f, the equation tan(x-π/4) = 6tanx, giving all solutions in the range -π < x < π.

(Ans: (a) π/3 , 4π/3 (b)(ii) -0.464, 2.68, -0.322, 2.82)

3. Find, as x varies in [0,π], the greatest and the least values of x and of y, where
x = 17 + 5sin2x + 12cos2x, y = 17 + 5sin^2 x + 12cos^2 x (ans: 30, 4 ; 29, 22)

4. Given that A and B are acute angles such that sin A = 24/25 and cos B = 12/13, find, without using tables or a calculator the exact values of sin(A + B) and cos(A+B). Deduce that π/2 < A +B < π/ (Ans: 323/325, 36/325)

5. Given that cos (A-B) = 3 cos(A+B), show that tan A = 1/2 cot B. Hence or otherwise find, correct to 0.1 of a degree the solution of the equation cos(A-60) = 3 cos(A +60), where 0 < A < 360. ( Ans : 16.1, 196.1)

 

[anonymous123]

A rectangle has sides of (2x+3)cm and (x+1)cm.State the domain of x if the area of the rectangle lies between 10cm^2 and 36cm^2 inclusive?
PLZ HELP

HELP PLZZZZZZZZZZZZ

 

[Zephyro]

Need help in P3 Winter 2008

http://www.xtremepapers.com/CIE/ind...el/9709 - Mathematics/&file=9709_w08_qp_3.pdf

Q10 - All parts except the first one.

Q6 - part 2

 

[Aqsa Mohd]

need help in doings maths p1.. i have forgoten doing evrything.. specially quadratics..

 

[Chooi1993]

Anyone know how to do oct/nov 2008 (P3) Q 10 ?

 

[smzimran]

Peoplee, whenever you ask a question of a past paper, please post the link also, it helps a lot.
And your questions will get answered more quickly

 

[s1294]

Hi i have another question on binomial expansion

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

 

[s1294]

Hi i need help with this question

Relative to the origin O the point A has a poistion vector 4i+7j-pk and the point B has a position vector 8i-j-pk, where p is constatnt
i) Find OA.OB
ii) hence show that there are no real values of p for which OA and OB are perpendicular to each other
iii) Find the values of p for which the angle AOB= 60 degrees
(p1 oct/nov 2011 9709)
please explain with steps
your help will be appreciated!!

 

[s1294]

HELP PLZZZZZZZZZZZZ

this is how you solve it

area of rectangle= length *width
so you have to multiply both side out

(2x+3)*(x+1)= 2x^2 +5x+3
from this equation you set the outcome to 10cm^2 and 36cm^2
so you have

2x^2 + 5x+3=10
2x^2 +5x-7=0

2x^2+5x+3= 36
2x^2+5x- 33= 0

now you solve both equation for x using the quadratic formula
= -b+/- (b^2 - 4ac)^0.5 / 2a

so you get
a=2
b=5
c=-7 or -33

you solve it so get:

-5+/- (5^2 - (4*2*-7))^0.5 / 2*2
so you get -5 +/- (81)^0.5/ 4= -5+/- 9 divide by 4
x= 1 or -3.5..... the length of sides can never be negative so you eliminate the negative answers

you do the same but this time change the c in the formula from -7 to -33

-5+/- (5^2- (4*2*-33))^0.5 / 4
-5+/- (289)^0.5 / 4= -5+/- 17 divide by 4
x= 3 or -5.5
you eliminate the negative x's and this leaves you with the answer= 1<= x=<3 - x is greater than or equal to 1 and less than or equal to 3

i hope this helps

 

[Scafalon40]

Peoplee, whenever you ask a question of a past paper, please post the link also, it helps a lot.
And your questions will get answered more quickly

But mine doesn't even require a LINK!!!!!
Its right there IN FRONT OF YOU!!!!!!!!!!!!!
Please solve the last part

 

[leadingguy]

Hi i need help with this question

Relative to the origin O the point A has a poistion vector 4i+7j-pk and the point B has a position vector 8i-j-pk, where p is constatnt
i) Find OA.OB
ii) hence show that there are no real values of p for which OA and OB are perpendicular to each other
iii) Find the values of p for which the angle AOB= 60 degrees
(p1 oct/nov 2011 9709)
please explain with steps
your help will be appreciated!!

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)

 

[bamteck]

URGENT HELP REQUIRED PLEASE !


1. (a) If A is the acute angle such that sin A = 3/5 and B is the obtuse angle such that sin B = 5/13, find without using a calculator the values of cos(A+B) and tan(A-B).

(b) Find the solutions of the equation tan x + 3 cot x = 5 sec x for which 0 < x < 2π.

(Ans: (a) -63/65, 56/33 (b) π/6, 5π/6)


2. (a) Solve, in radians, the equation sin( x + π/6) = 2cos x giving all solutions in the range 0 < x < 2π.
(b)(i) Use the formula tan(A-B) = tanA - tanB/1+tanAtanB to show that tan(x - π/4) = tanx - 1/tanx +1
(ii) Hence, or otherwise, solve, in radians to 3 s.f, the equation tan(x-π/4) = 6tanx, giving all solutions in the range -π < x < π.

(Ans: (a) π/3 , 4π/3 (b)(ii) -0.464, 2.68, -0.322, 2.82)

3. Find, as x varies in [0,π], the greatest and the least values of x and of y, where
x = 17 + 5sin2x + 12cos2x, y = 17 + 5sin^2 x + 12cos^2 x (ans: 30, 4 ; 29, 22)

4. Given that A and B are acute angles such that sin A = 24/25 and cos B = 12/13, find, without using tables or a calculator the exact values of sin(A + B) and cos(A+B). Deduce that π/2 < A +B < π/ (Ans: 323/325, 36/325)

5. Given that cos (A-B) = 3 cos(A+B), show that tan A = 1/2 cot B. Hence or otherwise find, correct to 0.1 of a degree the solution of the equation cos(A-60) = 3 cos(A +60), where 0 < A < 360. ( Ans : 16.1, 196.1)

 

[smzimran]

A particle P moves along the x axis. At time t seconds (where t>0) the velocity of P is 3t^2-12t+5 m/s in the direction of x increasing. When t=0, P is at the origin O Find
a)the velocity of P when the acceleration is zero (Ans: 7)
b)the values of t when P is again at O (Ans: 1,5)
c)the distance travelled by P in the interval 3<t<4 (Ans:48)
I've done the first two parts just can't do the last one
PLEASE HELP I'LL BE VERY GRATEFUL PLEASE!!!!!!!!!!!
P.S the < stands for 'is lesser than or equal to', same goes for >
HELP!

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