Mathematics: Post your doubts here!

[PhyZac]

Assalamu Alikum


Solve the equation

cos(q+60) = 2 sin q

giving all solutions in the interval 0 =< q >= 360 [5 marks]

 

[iKhaled]

ANY EASY WAY TO ATTEMPT THIS QUESTION!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_33.pdf
Q4
PLZ HELP!!!!!!!!!!!!!!

dx/dt = 1/x - x/4

make this into one fraction first:

dx/dt = (4-x^2)/4x then dt/dx = 4x/(4-x^2) and dt = 4x/(4-x^2) dx so t = ∫4x/(4-x^2) dx integrate this and u will get an answer of t = -2ln(4-x^2) + C ( if u dont know how to integrate it then tell me) anyway after u have integrated this u will sub the values of x = 1 when t = 0 and u will get C = 2ln 3 so..

t = -2ln(4-x^2) + 2ln3
t - 2ln3 = -2ln(4-x^2)
1/2t + ln3 = ln(4-x^2)
e^1/2t . e^ln3 = 4-x^2
3e^1/2t - 4 = x^2
x^2 = 4-3e^(1/2t)

i hope u got it..any questions about it pls ask me!

 

[iKhaled]

Assalamu Alikum


Solve the equation

cos(q+60) = 2 sin q

giving all solutions in the interval 0 =< q >= 360 [5 marks]

cos ( q + 60) = 2sinq
cosqcos60 - sinqsin60 = 2sinq
1/2cosq - √3/2 sinq = 2sinq
1/2cosq = 2sinq + √3/2 sinq
tan q = 0.1745
q = 9.9 and 189.9

check if its correct in the mark scheme or something and tell me !

 

[PhyZac]

cos ( q + 60) = 2sinq
cosqcos60 - sinqsin60 = 2sinq
1/2cosq - √3/2 sinq = 2sinq
1/2cosq = 2sinq + √3/2 sinq
tan q = 0.1745
q = 9.9 and 189.9

check if its correct in the mark scheme or something and tell me !

Oh thank you so much...!!!! Jazak Allah khairan!!!

Yes correct and i understood the method...May Allah reward you for helping me..

 

[Silver Wing]

Thank you for your help
We did it this way, and we THINK we got it correct (since we came second)
Let the 3 doors be A ( leading to safety in 2hrs)
B ( bringing man back in 3hrs) n
C ( " " " 5hrs)

possible pathway - probability - time taken
1. A - (1/3) - 2hrs
2. B,A - (1/3).(1/3)= 1/9 - 5hrs
3. C,A - 1/9 - 7hrs
4. B,C,A or C,B,A - 1/27 - 10hrs


so expected time= one with max probability= 2hrs

What do you think?

P(T=A) = P(T=B) = P(T=C) = 1/3
Options avalible are:
A, p = 1/3
BA, p = 1/6
CA, p= 1/6
BCA, p = 1/6
CBA, p = 1/6

Construct a prabability density function table:
x * 2hrs * 5hrs * 7hrs * 10hrs *
P(T=x)* 1/3 * 1/6 * 1/6 * 1/3 *

E(X) = sigma(xp)
= (2* 1/3) + (5* 1/6) + (7* 1/6) + (10* 1/3)
= 6 hrs

Hope this has helped

 

[iKhaled]

Oh thank you so much...!!!! Jazak Allah khairan!!!

Yes correct and i understood the method...May Allah reward you for helping me..

Ameen ya rab..

 

[Syed Mohammad Ali]

please can you solve me da followin quest ?
I will be grateful
Q:find the values of k for which equation (k2 X2) +2kX +1 = 0 have no roots.

 

[Syed Mohammad Ali]

can uh make meh understand da followin questions , i will be grateful to ya buddy...
Q : Find the values of k for which the equation k2X2 + 2kX +1 =0 have no roots.
Q 2 : Solve the following inequality :
X(X-2)<5

 

[Martee100]

Maths As Co Ordinate Geometry Questions

http://www.scribd.com/doc/121689460/AS-Mathematics-Co-ordinate-Geometry-Questions

Add this to the list

 

[Raiyan3]

can uh make meh understand da followin questions , i will be grateful to ya buddy...
Q : Find the values of k for which the equation k2X2 + 2kX +1 =0 have no roots.
Q 2 : Solve the following inequality :
X(X-2)<5

For question no. 2 you multiply inside the brackets first and then take the 5 from the other side, it becomes
X^2-2X-5<O
Now it becomes quadratic equation. Now you can solve that, using a calculator aswell.

 

[PhyZac]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.. iKhaled

5 The parametric equations of a curve are
x = ln(tan t), y = sin2t, (it is sin square t )
where 0 < t <1/2π.

(i) Express
dy/dx
in terms of t. [4]

(ii) Find the equation of the tangent to the curve at the point where x = 0.

 

[Syed Mohammad Ali]

For question no. 2 you multiply inside the brackets first and then take the 5 from the other side, it becomes
X^2-2X-5<O
Now it becomes quadratic equation. Now you can solve that, using a calculator aswell.

it give complex roots...

 

[Kandinsky]

For the question x(x-2) < 5:
x^2 - 2x - 5 < 0

x^2 - 2x + 1 - 6 < 0
(x-1)^2 - 6 < 0
(x-1)^2 < 6
|x-1| < sqrt(6)
hence,
1-sqrt(6) < x < sqrt(6)+1

 

[iKhaled]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.. iKhaled

5 The parametric equations of a curve are
x = ln(tan t), y = sin2t, (it is sin square t )
where 0 < t <1/2π.

(i) Express
dy/dx
in terms of t. [4]

(ii) Find the equation of the tangent to the curve at the point where x = 0.

w alikom al salam..

here is the solution to the question

(i) you have to know that whenever u have a parametric equation, dy/dx = dy/dt X dt/dx

dy/dt = 2sin t. cos t
dy/dt = 2sintcost

dx/dt = (1/tan t) x sec^2 t
dx/dt = (cos t / sin t ) x 1/cos^2 t
dx/dt = 1/sin t cos t

dy/dx = 2sintcost X sintcost
dy/dx = 2sin^2 t cos^2 t

(ii) here we need to find the equation of the tangent at the point where x is 0 so first lets find y so we have a coordinate and we know that dy/dx is out tangent

x = ln(tan t )
0 = ln (tan t)
e^0 = tan t
t = tan^-1(1)
t= 1/4π

dy/dx = 2sin^2(1/4π)cos^2(1/4π)
m(gradient) = 1/2

y = sin^2(1/4π)
y= 1

y-1 = 1/2(x-0)
2y-x = 2

is this how it is in the mark scheme? if its correct pls tell me and i hope u get it! any questions feel free to ask me

 

[PhyZac]

w alikom al salam..

here is the solution to the question

(i) you have to know that whenever u have a parametric equation, dy/dx = dy/dt X dt/dx

dy/dt = 2sin t. cos t
dy/dt = 2sintcost

dx/dt = (1/tan t) x sec^2 t
dx/dt = (cos t / sin t ) x 1/cos^2 t
dx/dt = 1/sin t cos t

dy/dx = 2sintcost X sintcost
dy/dx = 2sin^2 t cos^2 t

(ii) here we need to find the equation of the tangent at the point where x is 0 so first lets find y so we have a coordinate and we know that dy/dx is out tangent

x = ln(tan t )
0 = ln (tan t)
e^0 = tan t
t = tan^-1(1)
t= 1/4π

dy/dx = 2sin^2(1/4π)cos^2(1/4π)
m(gradient) = 1/2

y = sin^2(1/4π)
y= 1

y-1 = 1/2(x-0)
2y-x = 2

is this how it is in the mark scheme? if its correct pls tell me and i hope u get it! any questions feel free to ask me

Jazak Allah Khairan..thank you so much, May Allah reward you for the help ur providing other students with.

I fully understood the first part. Alhamdulilah. And your answer is exactly similar to markscheme

coming to the second part of the question..

t=1/4pi ..................i got it
m=1/2 ....................i got it

but in markscheme the equation seem different..... the "c"

here is a copy:

 

[iKhaled]

Jazak Allah Khairan..thank you so much, May Allah reward you for the help ur providing other students with.

I fully understood the first part. Alhamdulilah. And your answer is exactly similar to markscheme

coming to the second part of the question..

t=1/4pi ..................i got it
m=1/2 ....................i got it

but in markscheme the equation seem different..... the "c"

here is a copy:
View attachment 20543

oh sorry i made such a stupid error

i calculated y as in sin (2t) not sin^2 t so i got the answer y=1 instead of y=1/2

did u get me ?

 

[Syed Mohammad Ali]

For the question x(x-2) < 5:
x^2 - 2x - 5 < 0

x^2 - 2x + 1 - 6 < 0
(x-1)^2 - 6 < 0
(x-1)^2 < 6
|x-1| < sqrt(6)
hence,
1-sqrt(6) < x < sqrt(6)+1

wat bout the first 1. BETWEEN! Thanks . May Allah bless you.

 

[aleezay]

I still think the expected time owuld be the sum of all probabilities like it always is. Your solution would be completely correct only if the maximum probabilty was 1 i.e. A was certain to happen so the expected time would definitely be 2 hours.
P.S. Which contest was this?

LUMS PsiFi This was a question from Math Gauge.

 

[abcde]

LUMS PsiFi This was a question from Math Gauge.

I knew it!

 

[Kandinsky]

Q : Find the values of k for which the equation k2X2 + 2kX +1 =0 have no roots.

In fact, it will not have roots only if k = 0.
Here is the explanation:

k^2 * x^2 + 2k*x + 1 = 0
(kx)^2 + 2*(kx) + 1 = 0
it seems like it is full square
(kx+1)^2 = 0
so, kx+1 = 0
this equation does not have roots if and only if k = 0;