Mathematics: Post your doubts here!

[The Sarcastic Retard]

Q2 again.....

 

[awesomaholic101]

There are so many silly mistakes done in P1 which can create a loss of about at least 6 marks of me in paper Plus I never worked smooth with S1.... Hope so I get a B in this exams... Pray for me. Please...

Lol same here ... I lost about 9 marks idiotically
S1 was good thou ... Got 47 ... I thought u were done with A level
Ur writing this o/n? I'm writing now ...
Yea all the best May God help us all

 

[The Sarcastic Retard]

Lol same here ... I lost about 9 marks idiotically
S1 was good thou ... Got 47 ... I thought u were done with A level
Ur writing this o/n? I'm writing now ...
Yea all the best May God help us all

Not over yet... :/
Good luck...

 

[abdxyz]

Just a simple stupid question:

 

[Serial_Ripper]

Paper 03 M/J 08 Q5 (ii) (sorry, dunno how to insert photos)
PLZ. HELP. ME!

 

[Marinaa1998]

Paper 3 June 2006 9 (i), can someone please show me the steps? Thank you

 

[Hala Ezzat]

Helllo Can someone Solve June 2012 Paper 13 Question 3 and October 2014 paper 13 Plzzzzz. Also who is sitting in variant maths 9709 p12?

 

[The Sarcastic Retard]

Helllo Can someone Solve June 2012 Paper 13 Question 3 and October 2014 paper 13 Plzzzzz. Also who is sitting in variant maths 9709 p12?

Mj12
3) First three terms of expansion of (1 + ax)^6 = 1 + 6ax + 15a^2x^2
Expanding (1 - 2x)^2 = 4x^2 - 4x + 1
Multiply both to get x^0 term, x^1 and x^2 term.
x^0 = 1
x^1 = -4x + 6ax
x^2 = -24ax^2 + 4x^2 + 15a^2x^2

We know coffiecients
x^0 = 1
x^1 = -1
x^2 = b

x^1 : -4 + 6a = -1 ; a = 1/2
x^2 : -12 + 4 + 15/4 = -17/4

so a = 1/2 and b = -17/4

 

[The Sarcastic Retard]

Helllo Can someone Solve June 2012 Paper 13 Question 3 and October 2014 paper 13 Plzzzzz. Also who is sitting in variant maths 9709 p12?

W14 also question 3??

 

[The Sarcastic Retard]

Helllo Can someone Solve June 2012 Paper 13 Question 3 and October 2014 paper 13 Plzzzzz. Also who is sitting in variant maths 9709 p12?

M sitting.

 

[Hala Ezzat]

W14 also question 3??

Yasss , sorry silly me I forgot to state which question. Its question three w14 p13 But omg thanks a buunch for Binomial expansion ! xx

 

[The Sarcastic Retard]

Yasss , sorry silly me I forgot to state which question. Its question three w14 p13 But omg thanks a buunch for Binomial expansion ! xx

(i)
9x^2 - 12x + 5
9(x^2 - 4/3x + 5/9)
9((x - 2/3)^2 + 1/9)
9(x - 2/3)^2 + 1
-> 1(3x - 2)^2 + 1
(ii)
d(3x^3 - 6x^2 + 5x - 12) / dx
9x^2 -12x + 5 = 9(x - 2/3)^2 +1
f'(x) always > O so increasing function.

 

[Hala Ezzat]

(i)
1) 9x^2 - 12x + 5
2) 9(x^2 - 4/3x + 5/9)
3) 9((x - 2/3)^2 + 1/9)
4) 9(x - 2/3)^2 + 1
-> 1(3x - 2)^2 + 1
(ii)
d(3x^3 - 6x^2 + 5x - 12) / dx
9x^2 -12x + 5 = 9(x - 2/3)^2 +1
f'(x) always > O so increasing function.

Hey how did u transform the thing from step 3 to step 4 (like where did the 1/9 go ? and then from step 4 to the answer (where did the 9 go and what happened to the values inside the brackets.)

 

[Hala Ezzat]

(i)
9x^2 - 12x + 5
9(x^2 - 4/3x + 5/9)
9((x - 2/3)^2 + 1/9)
9(x - 2/3)^2 + 1
-> 1(3x - 2)^2 + 1
(ii)
d(3x^3 - 6x^2 + 5x - 12) / dx
9x^2 -12x + 5 = 9(x - 2/3)^2 +1
f'(x) always > O so increasing function.

Hey how did u transform the thing from step 3 to step 4 (like where did the 1/9 go ? and then from step 4 to the answer (where did the 9 go and what happened to the values inside the brackets.)

btw i added nummbers to ur steps so you could know which step i am referring too. Also 9 (a) The first term of an arithmetic progression is −2222 and the common difference is 17. Find the value of the first positive term. [3] (b) The first term of a geometric progression is ï3 and the second term is 2 cos 1, where 0 < 1 < 0. Find the set of values of 1 for which the progression is convergent. Can you solve question 3b for mehh and explain )? This question is in http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s15_qp_13.pdf if you dont understand it here

 

[awesomaholic101]

http://justpastpapers.com/wp-content/uploads/2015/05/IMG_5219.jpg
12/m/j/15 - q11 ... Why should it be 1/2 <g(x)<13 and not 5 instead of 1/2 as calculated from domain?

 

[The Sarcastic Retard]

btw i added nummbers to ur steps so you could know which step i am referring too. Also 9 (a) The first term of an arithmetic progression is −2222 and the common difference is 17. Find the value of the first positive term. [3] (b) The first term of a geometric progression is ï3 and the second term is 2 cos 1, where 0 < 1 < 0. Find the set of values of 1 for which the progression is convergent. Can you solve question 3b for mehh and explain )? This question is in http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s15_qp_13.pdf if you dont understand it here

Sorry for the late reply I am busy with some work.
Here is the thing 9 * 1/9 = 1
and 9(x - 2/3)^2 + 1 = 3^2(x - 2/3)^2 + 1
-------------------------> (3(x - 2/3))^2 + 1
-------------------------> (3x - 2)^2 + 1

In 3(b) we know -1 < r < 1 use it, you will get the answer.

 

[The Sarcastic Retard]

http://justpastpapers.com/wp-content/uploads/2015/05/IMG_5219.jpg
12/m/j/15 - q11 ... Why should it be 1/2 <g(x)<13 and not 5 instead of 1/2 as calculated from domain?

f(-b/2a)
f(3/2) = 1/2

 

[Maayee]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w14_qp_31.pdf
how do i solve 8 iii)

 

[awesomaholic101]

f(-b/2a)
f(3/2) = 1/2

-b/2a??

 

[The Sarcastic Retard]

-b/2a??

Vertex y value, as it will be the lowest the range will be from there to the highest value of domain substituted in f(x).