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Mathematics: Post your doubts here!

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hey its very simple.........>
Tr+1 = 7Cr (x^7-r) (2^r) (1/x^2r)
Tr+1 = 7Cr (x^7-r-2r) (2^r)
Now,as we need coefficient of x^1..........so
x^7-3r = x^1
7-3r = 1
r=2............
now,putting value of r in general tern formula...........i-e,
Tr+1 = nCr (a^n-r) (b^r)
T2+1 = 7C2 (x^7-3(2)) (2^2)
T3 = (210 (x) (4)
so coefficient of x=21X4
x=84.............................answer
Hope u will get it(y):cool:
 
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Hello, i need help in p11 mj 11
q8 and 3ii

q3ii-: This should be alright if you know the formula. So you want the volume between two points. If you sketched the curve properly, you should get your limits: 0 and 2.
Now use the formulae π∫(y^2)dx. You know y= (x-2)^2. So substitute that into your formula:
1. π∫((x-2)^2)^2 dx = π∫((x-2)^4)dx.
2. Integrate that, and you should get: π [((x-2)^5)/5] (<---- I assume you know how to do that integration (y))
3. Now substitute in your limits in the place of x (remember, the bigger one first): π {((2-2)^5)/5) - ((0-2)^5)/5)}
4. That comes to ---> π[ (0) - (-32/5) ] = 32π/5

That's 3ii solved! Hope that helped. Remember to look at the mark schemes, those are usually helpful. (That's all I have time for today. Maybe I'll get back to you later)
 
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Solve the equation 3^x+2 = 3^x+ 3^2 giving your answer correct to 3 significant figures.
Guys please help me with this question.Can't figure out how to do this.I know it's related to something with logarithm but can't seem to understand how to get the answer.
This question is from 9709 october 2009 p31 q2.
Thanks
 
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Hello, i need help in p11 mj 11
q8 and 3ii


thanks



q3.2) to find the volume of any shape rotated around x axis we use the formula....... π∫y^2.dx
now here y = (x-2)^2 and y^2= [ (x-2)^2]^2 = (x-2)^4
now substitute the value(highlighted) of y^2 in the formula and integrate.

π∫(x-2)^4)dx

π [(x-2)^5/5] now use limits o and 2 which can be obtained by seeing frst part where U have drawn the graph.


π.[{(x-2)^5/5} - {(x-2)^5/5}]

place "2" in the frst part and "zero" in the second u will obtain (6.4) π Ans.
 
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Oct/Nov 2011 P61
The upper quartile for history is given as 50 in the mark scheme. However i get it as 49 marks. ( 13 x 75% ) This gives 9.75 and so i took the 10th reading. and also 50 is not given in the data.
Screenshot_2012-04-23-09-28-56-1.png
 
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i mean to find range we inject the limits and then chk if a vertex lies in the domain....so in dis case it wud be f(0) and f(pi) for limits and for vertex f(-b/2a) where did f(90) com frm?


s10 qp11 Q5 ii
its cos^2


no!!!!!!!!!!! here we are asked to find the greatest a nd the smallest possible value of "y" so we put all possible values of x that are o,90,180 in the eq. and found the largest and smallest one for the examiner.
apart If we are asked to determine the value of "y" at the extreme boundries of "x" then we would USE the o and pi. as o and pi will give the values of pi as the start and end value of Y axis

hopE U gt mY point??
 
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no!!!!!!!!!!! here we are asked to find the greatest a nd the smallest possible value of "y" so we put all possible values of x that are o,90,180 in the eq. and found the largest and smallest one for the examiner.
apart If we are asked to determine the value of "y" at the extreme boundries of "x" then we would USE the o and pi. as o and pi will give the values of pi as the start and end value of Y axis

hopE U gt mY point??
but ur ans is wrong :(
 
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Hello every one!

I have a really hard question on Mechanics 1, which I am dying for some help. Here it is:

3 Particles P and Q are attached to opposite ends of a light inextensible string which passes over a fixed smooth pulley. The system is released from rest with the string taut, with its straight parts vertical, and with both particles at a height of 2 m above horizontal ground. P moves vertically downwards and does not rebound when it hits the ground. At the instant that P hits the ground, Q is at the point X, from where it continues to move vertically upwards without reaching the pulley. Given that P has mass 0.9 kg and that the tension in the string is 7.2 N while P is moving, find the total distance travelled by Q from the instant it first reaches X until it returns to X. [6]

I hope I could learn something from you all, because Mechanics 1 is killing me :) Thank you! Have a nice day, everyone. Wish you best health for the upcoming exam!
 
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Hello every one!


I have a really hard question on Mechanics 1, which I am dying for some help. Here it is:

3 Particles P and Q are attached to opposite ends of a light inextensible string which passes over a fixed smooth pulley. The system is released from rest with the string taut, with its straight parts vertical, and with both particles at a height of 2 m above horizontal ground. P moves vertically downwards and does not rebound when it hits the ground. At the instant that P hits the ground, Q is at the point X, from where it continues to move vertically upwards without reaching the pulley. Given that P has mass 0.9 kg and that the tension in the string is 7.2 N while P is moving, find the total distance travelled by Q from the instant it first reaches X until it returns to X. [6]

I hope I could learn something from you all, because Mechanics 1 is killing me :) Thank you! Have a nice day, everyone. Wish you best health for the upcoming exam!
AoA wr wb

make a sketch of the problem first and give P the left side so the solution is identical

Considering P: Take downwards +
Resolving : 9 - 7.2 = 0.9a
a=2

Now considering Q: Take upwards +
s 2 ( because it travels same distance as P)
u 0 (strts frm rest)
v v (this we hav to find atm)
a 2 (see top)
t no need

v^2 = u^2 +2as
put values
v=√8

now make a sketch in ur mind the question says when P hits the ground Q is at X and continues to move up means it doesnt stop..use common sense when p is at ground and q still moving it means the string has become slack so gravity will act on it
at X:
s (we hav to find)
u √8 (the v found earlier becomes u)
v 0 (at max height v=0)
a -10
t noneed

v^2=u^2 + 2as
u get s = 0.4
use logic.. 0.4 when Q flies up when string is slack so it will be s = 0.4 x 2 = 0.8m for it to become taut again

that took a lot of tym pray for me now :cool:
 
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AoA wr wb

make a sketch of the problem first and give P the left side so the solution is identical

Considering P: Take downwards +
Resolving : 9 - 7.2 = 0.9a
a=2

Now considering Q: Take upwards +
s 2 ( because it travels same distance as P)
u 0 (strts frm rest)
v v (this we hav to find atm)
a 2 (see top)
t no need

v^2 = u^2 +2as
put values
v=√8

now make a sketch in ur mind the question says when P hits the ground Q is at X and continues to move up means it doesnt stop..use common sense when p is at ground and q still moving it means the string has become slack so gravity will act on it
at X:
s (we hav to find)
u √8 (the v found earlier becomes u)
v 0 (at max height v=0)
a -10
t noneed

v^2=u^2 + 2as
u get s = 0.4
use logic.. 0.4 when Q flies up when string is slack so it will be s = 0.4 x 2 = 0.8m for it to become taut again

that took a lot of tym pray for me now :cool:
Thank you so much for your time. I am really appreciate it. You save me! :love:
 
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Oct/Nov 2011 P61
The upper quartile for history is given as 50 in the mark scheme. However i get it as 49 marks. ( 13 x 75% ) This gives 9.75 and so i took the 10th reading. and also 50 is not given in the data.
View attachment 7086
Whenever there are odd number of possibilities you always do n+1 hence it will be 13+1=14. Then take 75% of it which is equal to 10.5. As the answer is not a whole number, you have to take the 10th and 11th number and then divide it by two:
(49+51)/2 =50
Hope it helped :)
 
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q3ii-: This should be alright if you know the formula. So you want the volume between two points. If you sketched the curve properly, you should get your limits: 0 and 2.
Now use the formulae π∫(y^2)dx. You know y= (x-2)^2. So substitute that into your formula:
1. π∫((x-2)^2)^2 dx = π∫((x-2)^4)dx.
2. Integrate that, and you should get: π [((x-2)^5)/5] (<---- I assume you know how to do that integration (y))
3. Now substitute in your limits in the place of x (remember, the bigger one first): π {((2-2)^5)/5) - ((0-2)^5)/5)}
4. That comes to ---> π[ (0) - (-32/5) ] = 32π/5

That's 3ii solved! Hope that helped. Remember to look at the mark schemes, those are usually helpful. (That's all I have time for today. Maybe I'll get back to you later)
Thanks
 
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Please help me this problems ;

1. Find the maximum value of 1/√cosx + sinx + 5 , and the value of x where the maximum value occurs.

2. Solve the equation 10sin^2 (1/2x) - 5sinx = 4 , giving the values of x between 0 and 360 to the nearest 0.1

Thank You in anticipation
 
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