Mathematics: Post your doubts here!

[imanmalik]

ok...if u don't want to do it backwards think about what you have to do in order to get the req inequality..we do know that its in a^2 - b^2 form so if we multiply (|4+z| + |4-z|) to both sides of (i) we should get some results..
=> |4+z|^2 - |4-z|^2 = 6 (|4+z| + |4-z|)
6 (|4+z| + |4-z|) >=48
|4+z| + |4-z| >= 8

now u cant complain even though its the same thing repeated..where did u get this question anyway...doesn't seem like its for A level

Thank you so much. Our maths book is damn hard. CIE not Edexcel! :/

 

[Yousif Mukkhtar]

Guys can you explain how to prove the identity from simplifying left side to right side? Q2
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_12.pdf

 

[Ahmedraza73]

Can Anyone start The differentation and integeration tution For AS Level?
I want to learn it from the Begining

 

[french410]

Guys can you explain how to prove the identity from simplifying left side to right side? Q2
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_12.pdf

ok listen up: sin^2 means sin squared (^ means raised to the power of)

then we use common denominator

to obtain:

then we factor

out to obtain:

therefore:

hope that helped!!!!!

 

[Kumkum]

Can someone please help me with this:
Given that 'a' is a positive constant, solve the inequality
|x - 3a| > |x - a|

 

[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