Mathematics: Post your doubts here!

[Jelleh Belleh]

Ohh my god ! .. I'm confused more now !

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_13.pdf
Q9 (iii) part why is it (-5/2) <(or equal to) x < (or equal to) 0 ?

why it is not (-5/2) <(or equal to) x < (or equal to) (5/2)


Hadi Murtaza why are we talking about domain of f(x) and g(x) while we are dealing with gf(x) equation ? isn't it something else ? please explain what is related between all the 3 functions in the domain

Thought blocker If you can explain It will be helpful you know ..

f(x) and g(x) ARE related to each other by the function gf(x).

Solving (iii):
(2x+3)^2 - 6(2x+3)
= 4x^2 - 9 ≤ 16

= x^2 ≤ 25/4

x ≤ +/- 25/4

=> x ≤ +25/4 or x≤-25/4

The value of x must satisfy both the domains of g(x) and f(x)
Domain of g(x) is x must be less than or equal to 3, both +25/4 and -25/4 thus satisfy the domain of g(x)
However, the domain of f(x) is x must be less than or equal to 0. +25/4 obviously does not satisfy this domain and is therefore rejected.
=> -25/4 ≤ x ≤ 0

Hope that cleared things up.

 

[Hadi Murtaza]

f(x) and g(x) ARE related to each other by the function gf(x).

Solving (iii):
(2x+3)^2 - 6(2x+3)
= 4x^2 - 9 ≤ 16

= x^2 ≤ 25/4

x ≤ +/- 25/4

=> x ≤ +25/4 or x≤-25/4

The value of x must satisfy both the domains of g(x) and f(x)
Domain of g(x) is x must be less than or equal to 3, both +25/4 and -25/4 thus satisfy the domain of g(x)
However, the domain of f(x) is x must be less than or equal to 0. +25/4 obviously does not satisfy this domain and is therefore rejected.
=> -25/4 ≤ x ≤ 0

Hope that cleared things up.

dats exactly wat i wuz saying

 

[Faaiz Haque]

(1-(sin^2x/cos^2x))
--------------------------- =
(1+(sin^2x/cos^2x))

(cos^2x-sin^2x)
--------------------------
(cos^2x+sin^2x)

Now,
Using the identity

, if we subtract

from both sides, we obtain

so, 1-2sin^2x / 1 = 1-2sin^2x

I got everything except the beginning

How did ( (sin^2x) / (cos^2x) ) / ( (sin^2x) / (cos^2x) )

change to this :

(cos^2x-sin^2x)
--------------------------
(cos^2x+sin^2x)

 

[Haya Ahmed]

f(x) and g(x) ARE related to each other by the function gf(x).

Solving (iii):
(2x+3)^2 - 6(2x+3)
= 4x^2 - 9 ≤ 16

= x^2 ≤ 25/4

x ≤ +/- 25/4

=> x ≤ +25/4 or x≤-25/4

The value of x must satisfy both the domains of g(x) and f(x)
Domain of g(x) is x must be less than or equal to 3, both +25/4 and -25/4 thus satisfy the domain of g(x)
However, the domain of f(x) is x must be less than or equal to 0. +25/4 obviously does not satisfy this domain and is therefore rejected.
=> -25/4 ≤ x ≤ 0

Hope that cleared things up.

Ohh thanks .. of course cleared everything anyways you just forgot to square root the 25/4 ^_^ ... because It is +/- 5/2 ... but I got it thanks again

 

[hamza.k143]

Ohh my god ! .. I'm confused more now !

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_13.pdf
Q9 (iii) part why is it (-5/2) <(or equal to) x < (or equal to) 0 ?

why it is not (-5/2) <(or equal to) x < (or equal to) (5/2)


Hadi Murtaza why are we talking about domain of f(x) and g(x) while we are dealing with gf(x) equation ? isn't it something else ? please explain what is related between all the 3 functions in the domain

Thought blocker If you can explain It will be helpful you know ..

It is because haya ahmed the domain has restriction that x cannot exceed 0 thats it should be less than 0

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_13.pdf

Q6 (ii) can someone explain the midpoint of the two vectors thingy please and how to find it in the question & why is the mid point of AB is AC not OC directly .. .. Thanks ..

Jelleh Belleh

 

[Jelleh Belleh]

Ohh thanks .. of course cleared everything anyways you just forgot to square root the 25/4 ^_^ ... because It is +/- 5/2 ... but I got it thanks again

My bad, sorry about that!

 

[midha.ch]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_12.pdf
Question 2ii) , 4i),iii) , 5i), 6i), 7, 8i,ii) Please.
Some questions I am just stuck on formula like :¬
in question 4 i) dont we use : theta = cos inverse a^2 + b^2 - c^2 / 2(ab) ?
question 5i) how we got R.H.S as 10 ?
question 7) I got (3,9) what next ?

sorry was late but glad to see ur doubt was solved
about question number 7 i myself am confused. Idk how to do reflection sums

 

[Thought blocker]

sorry was late but glad to see ur doubt was solved
about question number 7 i myself am confused. Idk how to do reflection sums

I GOT IT HOW TO DAT TOO

 

[midha.ch]

I GOT IT HOW TO DAT TOO

then upload it. and about number 8
u still need it?

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_13.pdf

Q6 (ii) can someone explain the midpoint of the two vectors thingy please and how to find it in the question & why is the mid point of AB is AC not OC directly .. .. Thanks ..

Jelleh Belleh

this is why

 

[Thought blocker]

then upload it. and about number 8
u still need it?

I dont know this method is correct or not
We had to show a point of reflection, hence we have x1,y1 as (-1,3) and we found a midpoint (3,9) by sim. equation,
So, as to find midpoint we do (x1+x2)/2, and (y1+y1)/2,
similarly,
We know have to find x2 and y2, we have x1 y1 and midpoint
so (x1,y1, + x2,y2)/2 = midpoints
=(-1,3 + x2,y2) = 2(3,9)
=(-1,3 + x2,y2) = (6,18)
x2=(6-(-1))=7
y2=(18-3)=15

so reflection on (7,15)


T H I S R E A L L Y W O R K S


Yes I nees work solution on question 8

 

[abruzzi]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_3.pdf
Please help me understand how to solve question no. 9
I'm having a really tough time solving vector questions

 

[moonangel996]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_31.pdf
please help me Question 10
Mark scheme link
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_ms_31.pdf

 

[ZaqZainab]

I dont know this method is correct or not
We had to show a point of reflection, hence we have x1,y1 as (-1,3) and we found a midpoint (3,9) by sim. equation,
So, as to find midpoint we do (x1+x2)/2, and (y1+y1)/2,
similarly,
We know have to find x2 and y2, we have x1 y1 and midpoint
so (x1,y1, + x2,y2)/2 = midpoints
=(-1,3 + x2,y2) = 2(3,9)
=(-1,3 + x2,y2) = (6,18)
x2=(6-(-1))=7
y2=(18-3)=15

so reflection on (7,15)


T H I S R E A L L Y W O R K S


Yes I nees work solution on question 8

do you want me to do 8?

 

[Thought blocker]

do you want me to do 8?

Y are you guys asking ? Do it na, I want soln ASAP

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_11.pdf

Q3

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf

Q3

i'll try my best to explain

here we are asked to find 'rate at which area is increasing' so that means we have to find 'dA/dt'
we are given 'the rate at which the radius is changing' which is 'dr/dt = 3'
to find 'dA/dt' we also need 'dA/dr'

to find 'dA/dr' we need to differentiate an equation of A(area) in terms of r(radius)
the only equation we can use is the area of a circle, which is "A=(pi)r^2"
so differentiating this we get:
dA/dr = 2(pi)r

we can find the value of dA/dr when r=50
===> dA/dr = 2(pi) * 50
= 100(pi)

so now we can combine the rates i.e dA/dt , dA/dr, and dr/dt to find what we want(dA/dt)
so,
dA/dt = dA/dr * dr/dt
= 100(pi) * 3
=300(pi)

hope i was clear enough

 

[ZaqZainab]

Y are you guys asking ? Do it na, I want soln ASAP

 

[Thought blocker]

View attachment 40793

haww I suck at all of this type of questions. Help me na