• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

A

A star

Messages
2,703
Reaction score
3,939
Points
273
abruzzi said:
Thank you for your reply..

Yeah, but that's the domain for the inverse.. not the function. I'm confused :(
Click to expand...
when we inverse a function its domain becomes the range and range the domain of the inverse BUT the defining range remains constant . take it like this that only and only for value of x greater then 0 is the curve following the formulae given so it will only have a valid inverse function after zero
 
Tkp

Tkp

Messages
1,660
Reaction score
11,026
Points
523
luniaz226 said:
need help! JazakAllah.

3. The first three terms in the expansion of (1 − 2x)^2 (1 + ax)^6, in ascending powers of x, are 1 − x + bx2. Find the values of the constants a and b. (9709/13, May/June 2012)
Click to expand...

(1-4x+4x2)(1+6c1*ax+6c2*(ax)2)
(1-4x+4 x2)(1+6ax+15a2x2)
For a u need the coefficient of x
6a-4=-1
A=.5
For b
4+15a2-24a=b(put a =.5)
B=-17/4
 
Y

Yousif Mukkhtar

Messages
373
Reaction score
125
Points
53
VelaneDeBeaute said:
When a function as such is written, the 'a' is the amplitude of the wave i.e. its maximum height/displacement from the mean position. Hence, a = 9 -3 =6.
For a normal sine curve, b = 1. However, if we compare a normal curve with the given, we find that this one completes two wave cycles in a time period of 2 pi. Therefore, b = 2.
'c' is the value of how high or low the mean position of the wave is when compared to the line y = 0. Here the graph is raised +3 respective of the line y = 0, hence c = 3.
Click to expand...
Thank you for your help as it really was useful.
 
parthrocks

parthrocks

Messages
821
Reaction score
231
Points
53
Hey can anyone help me with this
A(a,-3); B(-3,1);C(2,-9) are collinear,Find A....can someone plz show me how to solve this sum plz asap!!:)
 
abruzzi

abruzzi

Messages
128
Reaction score
112
Points
53
A star said:
when we inverse a function its domain becomes the range and range the domain of the inverse BUT the defining range remains constant . take it like this that only and only for value of x greater then 0 is the curve following the formulae given so it will only have a valid inverse function after zero
Click to expand...
Oh.. alright.. thanks :)
 
S

sagar65265

Messages
227
Reaction score
571
Points
103
tanmaydube said:
View attachment 23414




The answer to the first part is 1/x + 10/x^2 + 1/(10-x)


Cannot do part (ii) please help!
Click to expand...


Okay, the first thing to notice is that they've given you the rate of change of x with respect to t in terms of x; this is not a very easy form to work with,
it's like writing dy/dx in terms of y. What you can do now is take the reciprocal of each side; it is proved that

dx/dy = 1/(dy/dx)

so you can take the reciprocal on each side and write

dt/dx = 100/x^2(10-x)

so this is more useful, since this equation now tells us how t changes with respect to x in terms of x. Another thing about this new form is that it is now the same equation as the one they asked you to split up into partial fractions in part (i), so you can substitute the values you have obtained earlier:

dt/dx = 1/x + 10/x^2 + 1/(10-x)

now you have an integral that can be carried out more easily than the earlier one. Multiplying both sides by dx,

dt = 1/x + 10/x^2 + 1/(10-x) dx

This you can integrate to give

t = ln(x) - 10/x - ln(10-x) + k ------- (1)

This constant is still an issue, so substitute the given values :- t = 0 and x = 1 to obtain:

0 = 0 - 10 - ln(9) + k
10 + ln(9) = k

So you can plug this value of k back in your equation (1) to get

t = ln(x) - 10/x - ln(10-x) + 10 + ln(9)

And using the properties of logarithms ( ln(a) + ln(b) = ln(ab) and ln(a) - ln(b) = ln(a/b) )

t = ln((9 times x)/(10-x)) - 10/x + 10
t = ln(9x/(10-x)) - 10/x + 10

which is the answer in the marking scheme.
Hope this helped!

Note: Before multiplying both sides by dx, make sure that you have separated out the variables in the equation, i.e. all terms in t must be on the left side and all terms in x must be on the right side, because it may not be possible to integrate t in terms of x or vice versa; separate the terms to the correct side and integrate both sides to get a differential equation with the constant.

Good Luck!
 
iKhaled

iKhaled

Messages
870
Reaction score
374
Points
73
Beaconite007 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_42.pdf
Can anybody please help me with question 4 part ii)?
Click to expand...
in part i we got v= 40 + Tsin25 and x= Tcos25

we have been told that the system is in equilibrium so this means the frictional force (F) will be equal to horizontal component (x) of the tension and we know the equation F = μR
R is the contact force which is the force that the rod is acting on ring which is 40 + Tsin25

substitute these in our equation and we will get:

T cos25 = 0.4(40 + T sin25)
T cos 25 = 16 + 0.4T sin25
T cos25 - 0.4T sin25 = 16
T = 16/(cos 25-0.4sin25)
T= 21.7 N...thats it!

i hope u got it
 
A

A star

Messages
2,703
Reaction score
3,939
Points
273
Y

yubakkk

Messages
257
Reaction score
4
Points
28
Please answer this question
For a biased cubical dice the probability of any particular score between 1 and 6 (inclusive) being obtained is inversely proportional to that score. Find the probability of scoring a 1.​
 
You must log in or register to reply here.
Top