Mathematics: Post your doubts here!

[anonymous123]

(x-3a)^2 > (x-a)^2

x^2 -6ax +9a^2 > x^2 -2ax +a^2
-4ax + 8a^2 > 0
ax - 2a^2 < 0
a(x-2a) < 0
a<0 , x-2a < 0
a<0 rejected...so x<2a

 

[Kumkum]

(x-3a)^2 > (x-a)^2

x^2 -6ax +9a^2 > x^2 -2ax +a^2
-4ax + 8a^2 > 0
ax - 2a^2 < 0
a(x-2a) < 0
a<0 , x-2a < 0
a<0 rejected...so x<2a

thnx....but just one question, the sign changed because you divided through out by -4, right?

 

[anonymous123]

thnx....but just one question, the sign changed because you divided through out by -4, right?

its a rule: whenever u divide or multiply by a negative number, the equality is inverted.

 

[Kumkum]

its a rule: whenever u divide or multiply by a negative number, the equality is inverted.

yep...thnx

 

[aksameerkhan27]

Can any one give me solution to complex number and vector questions from summer and winter 2011 and 2012 papers please. it is very urgent as i am preparing for the exam. also any tips for locus in argand diagram.

 

[anonymous123]

Can any one give me solution to complex number and vector questions from summer and winter 2011 and 2012 papers please. it is very urgent as i am preparing for the exam. also any tips for locus in argand diagram.

.its difficult to help like that. post the questions here

 

[salvatore]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_12.pdf
No. 2

Please help me solve the question above.. an explanation will be appreciated.
Thanks!

 

[anonymous123]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_12.pdf
No. 2

Please help me solve the question above.. an explanation will be appreciated.
Thanks!

v=pi integral(y^2) dx
v=pi integral(a^2/x^2) dx
v=pi (-a^2/x)

now put limits and set it equal to 24pi..u will get the ans

 

[salvatore]

v=pi integral(y^2) dx
v=pi integral(a^2/x^2) dx
v=pi (-a^2/x)

now put limits and set it equal to 24pi..u will get the ans

Thank you for your reply..
I don't understand the last part i.e v=pi(-a²/x). How did u get the integral as (-a²/x) ?
Sorry for bothering..

 

[anonymous123]

Thank you for your reply..
I don't understand the last part i.e v=pi(-a²/x). How did u get the integral as (-a²/x) ?
Sorry for bothering..

thats basic integration, what is the integral of 1/x^2?
1/x^2 = x^-2
so if we integrate this: x^(-2+1) /-1 => -1/x

 

[salvatore]

thats basic integration, what is the integral of 1/x^2?
1/x^2 = x^-2
so if we integrate this: x^(-2+1) /-1 => -1/x

Omg! How could I not know that? Thanks man..

 

[anonymous123]

Omg! How could I not know that? Thanks man..

no problem

 

[soumayya]

but its not in the form (ax+b)^2+c, isnt the ans (2x-3)^2-6?

Ooops..sorry..didn't notice zat..

4(x - 3/2)^2 - 6 = (2^2 *(x - 3/2)^2) - 6
= ( 2(x - 3/2) )^2 - 6
= (2x -3)^2 - 6

 

[soumayya]

can sum1 plz help me asap
-Express 4x2-12x+3 in the form (ax+b)2+c where a, b and c are constants and a is greater than 0
-The line y=5x-3 is a tangent to the curve y=kx2-3x+5 at the point A. Find the cordinates of A

Since u got k = 2 , the eq:
y = 2x^2 -3x +5
y = 5x-3

By equating (simultaneous eq):
2x^2 -3x +5 = 5x-3
2x^2 -3x -5x +5 +3 =0
2x^2 - 8x + 8 =0
2 (x^2 - 4x + 4) =0
2(x-2)(x-2) = 0
Hence, x= 2

When x=2 , y = 5(2) - 3 = 7
(2 , 7)

 

[Iffat]

Ooops..sorry..didn't notice zat..

4(x - 3/2)^2 - 6 = (2^2 *(x - 3/2)^2) - 6
= ( 2(x - 3/2) )^2 - 6
= (2x -3)^2 - 6

hehe thanx

 

[Iffat]

Since u got k = 2 , the eq:
y = 2x^2 -3x +5
y = 5x-3

By equating (simultaneous eq):
2x^2 -3x +5 = 5x-3
2x^2 -3x -5x +5 +3 =0
2x^2 - 8x + 8 =0
2 (x^2 - 4x + 4) =0
2(x-2)(x-2) = 0
Hence, x= 2

When x=2 , y = 5(2) - 3 = 7
(2 , 7)

thanx

 

[georges]

can anyone help with the friction in mechanics m1?
it has been a pain in the ass
so plz if there are any revision sheets provide me with them

 

[whitecorp]

can anyone help with the friction in mechanics m1?
it has been a pain in the ass
so plz if there are any revision sheets provide me with them

Just remember the basic idea of frictional force as one opposing the motion of an object.

However, if you are looking at a stationary object, make sure you consider the static coefficient of friction. Do also note
the direction in which the object has a tendency to move so as to determine how the frictional force would act.

I have solved some problems on my site related to mechanics, you may wish to take a look:

http://www.whitegroupmaths.com/2011/10/problem-2-mechanics.html

http://www.whitegroupmaths.com/2011/10/problem-6-double-incline-aka-wedge.html

Hope these help. Peace.

 

[rafayet]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_12.pdf


I need help with Q.9(i). i got coor of M, but how can i get coor of D? can you please help me?

 

[anonymous123]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf


I need help with Q.9(i). i got coor of M, but how can i get coor of D? can you please help me?

(Coordinates of B + Coordinates of D)/2 = Coordinates of M