Mathematics: Post your doubts here!

[smzimran]

Q. For the GP 4,6,9.... find the value of n if the nth term is the first term greater than 100.
Help me plzz

a = 4
r = 1.5
nth term = 100
a [1.5^(n - 1)] = 100
4[1.5^(n - 1)] = 100
1.5^(n - 1) = 100/4
1.5^(n - 1) = 25
TAking ln of both sides
(n - 1) ln1.5 = ln25
n - 1 = ln25 / ln1.5
n - 1 = 7.94
n = 8.94
Round it off
n = 9

 

[HubbaBubba]

Oct/Nov 2011 - Question Paper 11:
Q3. (i) Sketch, on a single diagram, the graphs of y = cos 2q and y = 1
2 for 0 ≤ q ≤ 2p. [3]
(ii) Write down the number of roots of the equation 2 cos 2q − 1 = 0 in the interval 0 ≤ q ≤ 2p. [1]
(iii) Deduce the number of roots of the equation 2 cos 2q − 1 = 0 in the interval 10p ≤ q ≤ 20p. [1]

Can somebody please solve? Thanks

 

[ffaadyy]

Please remove the old link and put this one in place of Sketching Diagrams in the first post.

 

[ShudyShab]

The parametric equations of a curve are
x = a cos3 t, y = a sin3 t,
where a is a positive constant and 0 < t < 1
2
p.
(i) Express
dy
dx
in terms of t. [3]
(ii) Show that the equation of the tangent to the curve at the point with parameter t is
x sin t + y cos t = a sin t cos t.

 

[ffaadyy]

The parametric equations of a curve are
x = a cos3 t, y = a sin3 t,
where a is a positive constant and 0 < t < 1
2
p.
(i) Express
dy
dx
in terms of t. [3]
(ii) Show that the equation of the tangent to the curve at the point with parameter t is
x sin t + y cos t = a sin t cos t.

(i):

x = a cos^3 t
dx = -3a cos^2 t sin t

y = a sin^3 t
dy = 3a sin^2 t cos t

dy/dx = (3a sin^2 t cos t)/(-3a cos^2 t sin t)
dy/dx = - sin t / cos t

(ii):

x = a cos^3 t
y = a sin^3 t
m = - sin t / cos t

y - y1 = m ( x - x1)
y - a sin^3 t = (- sin t / cos t) ( x - a cos^3 t)
y cos t - a sin^3 t cos t = - x sin t + a cos^3 t sin t
x sin t + y cos t = a cos^3 t sin t + a sin^3 t cos t
x sin t + y cos t = a cos t sin t ( cos^2 t + sin^2 t)

Recall the identity 'cos^2 t + sin^2 t = 1'.

x sin t + y cos t = a cos t sin t ( cos^2 t + sin^2 t)
x sin t + y cos t = a cos t sin t ( 1 )
x sin t + y cos t = a cos t sin t

Therefore, the equation of the tangent to the curve at the point with parameter t is 'x sin t + y cos t = a cos t sin t '.

 

[salman haider]

I don't think that they've asked us to explain why is the angle twice the angle of arg z; In Q7(iii) N03 P3, they've asked us to find the greatest value of arg s for points on this locus. Are you referring to the same question?

yes the question is same......and i know they havent asked us to xplain this........but then for us to calculate the angle the diagram should be very accurate......or is there any other method.. using calculation

 

[salman haider]

thanks... this was very helpful

 

[ffaadyy]

yes the question is same......and i know they havent asked us to xplain this........but then for us to calculate the angle the diagram should be very accurate......or is there any other method.. using calculation

This is how you'll do q7.

 

[Ahsen420]

Excuse me... I need help in MJ 2011 P61 question 5.. Asap bcoz tmrw stats mock

 

[nonamehapazard]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

Assalamoalaikum!!

UPDATE: Link to Sequences Help by destined007 added!

please can u give me some complex no note which is so handy,i am totally worried about it

 

[smzimran]

Excuse me... I need help in MJ 2011 P61 question 5.. Asap bcoz tmrw stats mock

(a)
Let mean = m
standard deviation = s

3m = 7s^2
m = (7/3) s^2 ...... eq(1)

P(X > 2m) = 0.1016
P(X < 2m) = 1 - 0.1016
P(X < 2m) = 0.8984
(2m - m) / s = phi inverse (0.8984)
m / s = phi inverse (0.8984)
Read the table in reverse to get phi inverse
m /s = 1.272
m = 1.272s ..... eq(2)

Solving eq(1) and eq(2) simultaneously
(7/3)s^2 = 1.272s
(7/3)s = 1.272
s = 1.272 * (3/7)
s = 0.545

Substitute value of s in (2)
m = 1.272(0.545)
m = 0.693

So, mean = 0.693 and standard deviation = 0.545

(b)
P(X < a+33) = 0.75
P(Z < (a+33-33)/root21) = 0.75
a/root21 = phi inverse (0.75)
Read the table in reverse to get phi inverse
a/root21 = 0.674
a = 3.09

 

[XPFMember]

please can u give me some complex no note which is so handy,i am totally worried about it

check here: http://www.schoolworkout.co.uk/a_level.htm
there are some good notes here, for p3...i hope its for all chapters...but i used vectors and complex numbers, and they were good...

 

[anonymous123]

If the function " y=x^3+ax^2+3x -1" is always increasing,find the range of possible values of a.?
Plz help ..

 

[anonymous123]

i know the discriminant method but i want to do it using differentiation coz its frm that topic........

 

[Chaitanya Agrawal]

Compiled P3 Attached .....

Compiled P4 (M1) Attached...

can you give compiled for p1 and s1 also i mean pure mathematics 1 and statistics 1??please please

 

[anonymous123]

A circular cylinder has a diameter of 40 cm and is being filled with water at the rate of 1.5 litres per second. At what rate is the water level rising?

 

[unique840]

A circular cylinder has a diameter of 40 cm and is being filled with water at the rate of 1.5 litres per second. At what rate is the water level rising?

dV/dt = dV/dr * dr/dt (dV/dt will be taken in cm^3 because the radius is given in cm^3)
vol of cylinder = pie * r^2 * h
1500 = pie * 20^2 * h
h = 15/(4pie)
V = pie * r^2 * 15/(4pie)
pie cancels out
dV/dr = 15r/2
1500 = 15*20/2 * dr/dt
dr/dt = 10 cm/s

 

[anonymous123]

dV/dt = dV/dr * dr/dt (dV/dt will be taken in cm^3 because the radius is given in cm^3)
vol of cylinder = pie * r^2 * h
1500 = pie * 20^2 * h
h = 15/(4pie)
V = pie * r^2 * 15/(4pie)
pie cancels out
dV/dr = 15r/2
1500 = 15*20/2 * dr/dt
dr/dt = 10 cm/s

question asks for rate of change of water level means height..my ans is 1.19 but the book says 4.77

 

[JF thunder]

please help

june 2006 p3 q 7 (iii)

 

[Esme]

can anybody give me the link to download 'list of formulae(mf9) ?? please !