Mathematics: Post your doubts here!

[Oksey6]

Oh t

I pressumed you had problem with the (a) part so here's the solution.
First of all, convert dx in the du form;
u = sin2x
du/dx = 2cos2x (Differential of u)
therefore dx = du/2cos2x

y = sin^3 2x . cos^3 2x . du/2cos2x
Therefore after cancelling the cox2x, the remainder is

sin^3 2x . cos^2 x . du/2

Therefore this can also be written as:

sin^3x . (1-sin^2 2x) . du/2 (using the identity sin^2 2x + cos^2 2x = 1)

Now replace sin 2x with 'u'.

u^3 . (1-u^2) . du/2
u^3 -u^5 . du/2
Now integrate this and put the limits, you'll the the answer. But make sure to convert the limits into 'u' term.

Oh thankyou so much! but i am confused as to what limits are we supposed to put. One will '0' and the other will be?
Also can u please help me with part b).
Thanks!

 

[x-gamer-x]

anyone please help me
i am not getting this
thank u very much

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w12_qp_31.pdf

Q2

Ans: x = 1.14

 

[Oksey6]

anyone please help me
i am not getting this
thank u very much

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w12_qp_31.pdf

Q2

Ans: x = 1.14

It will be like this:
- taking 5^x = y

5^(x-1) can be written as 5^x * 5^-1

thus
y * 5^-1 = y -5
y/5=y-5
solving,
y=6.25

Now,
5^x=6.25
ln5^x=ln6.25
xln5=ln6.25
x=ln6.25/ln5
x=1.14

 

[x-gamer-x]

Oksey6
thank u sis

 

[TheZodiac]

Oh t


Oh thankyou so much! but i am confused as to what limits are we supposed to put. One will '0' and the other will be?
Also can u please help me with part b).
Thanks!

You can get limits by placing y = 0 and therefore sin^3 2x . cos^3 2x = 0
you'll get 'o' and '1/4pi'

Limits would be '0' and '1' in case of 'u'. You can get them by using the equation ' u = sin2x '
Place (0 and 1/4pi)
u = sin2(0)
u = sin2(1/4pi)

You'd get 0 and 1, now apply this in the integral in 'u' form.

Alternatively, you could also apply the inital limits obtained i.e: (0 and 1/4pi) But for that, you'd need to replace 'u' by sin2x which would be more complicated.

I'll solve part (b) as soon as I get time.

 

[Oksey6]

You can get limits by placing y = 0 and therefore sin^3 2x . cos^3 2x = 0
you'll get 'o' and '1/4pi'

Limits would be '0' and '1' in case of 'u'. You can get them by using the equation ' u = sin2x '
Place (0 and 1/4pi)
u = sin2(0)
u = sin2(1/4pi)

You'd get 0 and 1, now apply this in the integral in 'u' form.

Alternatively, you could also apply the inital limits obtained i.e: (0 and 1/4pi) But for that, you'd need to replace 'u' by sin2x which would be more complicated.

I'll solve part (b) as soon as I get time.

Thankyou so much!!

 

[ayaman Azmi]

Having trouble with May/June 2007 paper 01 question 1
PLS HELP!!!
P1 BTW

 

[queen of the legend]

for those who have done A level math ...is it possible to complete UNDERSTANDING and completing the course with a tutor of a level math in two months ?? just the completion of course. later i want to practice on my own
plss reply soon

 

[queen of the legend]

for those who have done A level math ...is it possible to complete UNDERSTANDING and completing the course with a tutor of a level math in two months ?? just the completion of course. later i want to practice on my own
plss reply soon

 

[SomeStudent]

Given that f : x  2x^2
+8x - 10 for the domain x greater than or equal to  k,
(iv) find the least value of k for which f is one-one

How will we do this part?

 

[Starlight97]

Could any one of you plz provide me the name of maths books and for M and S1

 

[salvatore]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_31.pdf
Could anyone please help me with question no. 5(ii)? I don't understand what to do..
Thanks!

I'm re-posting this.. anyone??

 

[x-gamer-x]

anyone plz help me
thank u

Q4 (i)
i have expanded both now how to prove that
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s10_qp_31.pdf

Thanks a lot

 

[x-gamer-x]

Plz help me with this also
thank u v much
Q6 (i) and Q7 (i)

 

[Relon]

Plz help me , how to solve this question >> May/June paper 41 2013 << ignore part (iv)

 

[TheZodiac]

Plz help me , how to solve this question >> May/June paper 41 2013 << ignore part (iv)

(i) Differentiate the displacement equation in order to get the velocity equation. Put the velocity '0' because it's at rest at point B.
You'd get time 0 and 100 seconds.
Put 100 in the original equation and you'll get distance of 1170 m

(ii) Differentiate the equation of velocity you found out in part (i) and then put acceleration = 0. This is because at max velocity, acceleration would be 'o'.
You'd get time 0 and 66.7 seconds. Place this in the velocity equation and you'll get speed of 20.8 m/s

(iii) (a) Here, the acceleration would be '0' as it's starting it's journey. (t=0)
(b) In this part, place t= 100 you obtained in the (i) part into acceleration equation and you'd get the acceleration of -1.404 m/s^2

Hope it helps.

 

[Mery95]

Assalamualikum,
Can you post notes or any links for probability and probability distribution (Alevels).
It will be really helpful.

 

[Relon]

(i) Differentiate the displacement equation in order to get the velocity equation. Put the velocity '0' because it's at rest at point B.
You'd get time 0 and 100 seconds.
Put 100 in the original equation and you'll get distance of 1170 m

(ii) Differentiate the equation of velocity you found out in part (i) and then put acceleration = 0. This is because at max velocity, acceleration would be 'o'.
You'd get time 0 and 66.7 seconds. Place this in the velocity equation and you'll get speed of 20.8 m/s

(iii) (a) Here, the acceleration would be '0' as it's starting it's journey. (t=0)
(b) In this part, place t= 100 you obtained in the (i) part into acceleration equation and you'd get the acceleration of -1.404 m/s^2

Hope it helps.

Thanks a lot , it is very helpful

 

[TheZodiac]

Plz help me with this also
thank u v much
Q6 (i) and Q7 (i)

Apply quotient rule to 1/cosx and you'd get secx as the answer. Now all you need to do is differentiate the given equation.
It can be done by the following way;

(1/(secx + tanx) ) . dy/dx (secx) + dy/dx(tanx)

secxtanx + sec^2x/secx + tanx

Solve it and you'd get the answer.

Q7(i)

 

[snowbrood]

guys please help me solve this