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Mathematics: Post your doubts here!

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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 .. :D

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

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

Now,
Using the identity
plot-formula.mpl
, if we subtract
plot-formula.mpl
from both sides, we obtain
plot-formula.mpl


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)
 
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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 :)
 
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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 .. :D
It is because haya ahmed the domain has restriction that x cannot exceed 0 thats it should be less than 0
 
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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
 
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then upload it. and about number 8
u still need it?
I dont know this method is correct or not :p
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 :)
 
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I dont know this method is correct or not :p
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?
 
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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 :)
 
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