Mathematics: Post your doubts here!

[omaaaar]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_11.pdf
Q11 second and third part

 

[robinhoodmustafa]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_11.pdf

Someone help with question 1,2,3 PLEASEee. Need someone to show proper workingnot just explain how to do it

Edit in Q3.. Limits will be from 2 to 0 ,not 0 to 2



btw i see Thought blocker has already answered the query
I had this done but i was facing uploading error.

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_11.pdf
Q11 second and third part

ii)
from part i) f(g) = 2x^2- 3 and g(f) = 4x^2 + 4x - 1
so for this part put a instead of x, and solve the equation , you get a as -1
iii)
given g(b) = b ; b^2 - b - 2 = 0 ; you get b as 2 and -1 ; and b is not equal to a hence b is 2

 

[daredevil]

so may Allah help me i my studies and life n gimme successes bith here and the hereafter

Aameen!

 

[omaaaar]

ii)
from part i) f(g) = 2x^2- 3 and g(f) = 4x^2 + 4x - 1
so for this part put a instead of x, and solve the equation , you get a as -1
iii)
given g(b) = b ; b^2 - b - 2 = 0 ; you get b as 2 and -1 ; and b is not equal to a hence b is 2

U explain really well

 

[Thought blocker]

U explain really well

ty

 

[ahmed abdulla]

Can someone have a look at my doubt?
https://www.xtremepapers.com/commun...st-your-doubts-here.9599/page-597#post-760905

 

[Rutzaba]

Can someone have a look at my doubt?
https://www.xtremepapers.com/commun...st-your-doubts-here.9599/page-597#post-760905

Talha Khatri

 

[omaaaar]

Could anyone tell me which is the hardest P1 maths paper which I should practice for exam

 

[Thought blocker]

Could anyone tell me which is the hardest P1 maths paper which I should practice for exam

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_12.pdf

 

[omaaaar]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_12.pdf

Thanks for telling if I have any doubts then I will ask u

 

[Thought blocker]

Thanks for telling if I have any doubts then I will ask u

Sure

 

[Rutzaba]

i have to study my own maths and i daresay its quite pathetic

 

[Suchal Riaz]

hey this is a nice question from past papers.

finally a different type of question from binomial theorem.
I recommend you to try yourself. I will post the solutions tomorrow so everyone gets to do it himselves.

 

[istudent]

10i) u-w =4i ----1 and uw = 5--------2
u= 5/w
substitute this in eq 1
5/w -w =4i
5-w^2 = 4i
w^2 +4i -5 =0
w= -4i +/- √{(4i)^2 - 4*-5*1}
w= -41 +/- √(-16+ 20)
w = -4i +2/2 -4i-2/2
w= -2i+1 and -2i-1

u= 5/w
u= 5/1-2i
u= 5(1+2i) /4+1
u= 10i +5/5
u= 2i + 1
and it's conjugate u= 2i-1

Thank u

 

[omaaaar]

Sure

In Q3 part two it is saying to find the minimum area so why do we substitute the value of x at stationery point to find min value of area in the equation could you explain

 

[Thought blocker]

http://www.ittaleem.com/classes/326094-cie-november-2012-level-mathematics-paper-1-9709-12-a.html

if this works then regards to Binyamine

GCE As and a level
4i)
(x/k)+k = x^2/4
=>4((x/k)+k) = x^2
=>4x/k=x^2-4k
=>4x = kx^2 - 4k^2
=>kx^2-4x-4k^2 = 0
Use b^2 - 4ac = 0
16 +16k^3 = 0
k = -1

4ii)
kx^2-4x-4k^2 = 0 put k as -1 and get your x value as -2
now as we have y = x^(2)/4 put 2 as x, you get y as 1
so P is (-2,1)

6ii)
from part i) you got 2sin^(2)x + 3sinx -2 = 0
By solving this quadratic eqn, you get 2 values of sin theta i.e is 0.5 and -2 well sin inverse is not defined at -2 so consider sin theta to be 0.5 ; Now, sin inverse 0.5 is 30. Here we have 2y instead of x so for one value of y it will be 30/2 = 15 and for 2nd value of y will be (180-30)/ 2 = 75 .

7ii)
For AB to be unit vector it should equal to 1
hence,
(k-1)^2 + K^2 + (2k - 2)^2 = 1
6k^2 -10k + 4 = 0
now you got k as 1 or 0.67

8i)
Use ar^(n-1) formula here,
so for second term , ar^(2-1) = 24 and for fourth term, ar^(4-1) = 13.5
Now equate each other and get r as 3/4 and first term (a) as 32

8ii)
use a/(1-r) formula, you get answer as 128

8b)
ZaqZainab Suchal Riaz Idk this one ask 'em.


10ii)
When d2y/dx^2 is > 0 it has minimum value. Here it is > 0 i.e is minimum point


11)
Suchal Riaz ZaqZainab

Actually I am tired of typing. :/ Help him now, I have to do organic chemistry :/

 

[Thought blocker]

In Q3 part two it is saying to find the minimum area so why do we substitute the value of x at stationery point to find min value of area in the equation could you explain

Sorry I can't now I have to finish two chapters of chem, so sorry. @suchal help him too

 

[Thought blocker]

In Q3 part two it is saying to find the minimum area so why do we substitute the value of x at stationery point to find min value of area in the equation could you explain

come here : http://www.twiddla.com/1610154

 

[omaaaar]

Sorry I can't now I have to finish two chapters of chem, so sorry. @suchal help him too

No problem at all man clear my doubt whenever u are free
And yes could you tell me which is the hardest p2 paper of chemistry which I should practice