Mathematics: Post your doubts here!

[rash1233]

combine the equations and then put discriminant equal to 0..its easy

I've done that, and gotten y^2+2(k-2y)-13=0 I don't know what to do next, because my 'b' contains k and y

 

[Dug]

I've done that, and gotten y^2+2(k-2y)-13=0 I don't know what to do next, because my 'b' contains k and y

i m used to making y the subject so lets go wid my method
y^2=13-2x-----------------1

y= (1/2)k -(1/2)x
y^2=(1/4)k^2 +(1/4)x^2 - (1/2)kx -------------2

set 1 and 2 equal u get
x^2 + (8-2k)x +k^2 - 52 = 0
now use b^2-4ac = 0
u get k=8.5

 

[rash1233]

i m used to making y the subject so lets go wid my method
y^2=13-2x-----------------1

y= (1/2)k -(1/2)x
y^2=(1/4)k^2 +(1/4)x^2 - (1/2)kx -------------2

set 1 and 2 equal u get
x^2 + (8-2k)x +k^2 - 52 = 0
now use b^2-4ac = 0
u get k=8.5

can you try the method using y? the x way is really confusing me..

 

[Dug]

can you try the method using y? the x way is really confusing me..

its even easier that way
ok lets see..
make x the subject in both eqs u get
x= (13-y^2)/2
x=k-2y

now set them equal
k-2y = (13-y^2)/2
cross multiply
2k-4y = 13 - y^2
y^2 -4y +2k -13=0
b^2-4ac=0

16 - 4(1)(2k-13) = 0
16 - 8k +52 = 0
k = 8.5

 

[hm12]

here range means the boundries in which the values of "f" lies.
the eq. is given
fx = 3-2tan(1/2x)
jxt put al the values of x (from 0 to π) and find values of "f" at each value of x

f(0) = 3-2tan(1/2(o)) = 3
f( π/2) = 3-2tan( 1/2* π/2) = 1
f( π) = 3-2tan(1/2* π) = not real

so the greatest value of f is 3 and least is 1 range of is 1<f<3.

fr drawing graph it is stated that fx = y
so what ever the vaues of fx are at x = 0 and π/2 use them to plot a graph
i.e
x 0 π/2
y 3 1 sketch I hope m not doing that wrong as nt seen ms yet

that is the range i calculated too but the mark scheme states the range to be f< and equal to 3??

 

[Dug]

if u still dont get it i can do it using calculus

 

[rash1233]

its even easier that way
ok lets see..
make x the subject in both eqs u get
x= (13-y^2)/2
x=k-2y

now set them equal
k-2y = (13-y^2)/2
cross multiply
2k-4y = 13 - y^2
y^2 -4y +2k -13=0
b^2-4ac=0

16 - 4(1)(2k-13) = 0
16 - 8k +52 = 0
k = 8.5

Thanks alot!! I got it now

 

[TaimoorK]

I have 2 questions! firstly, in J10 11 Q.5)(ii) how do u find the biggest and least value? ive found this question before and i dont know how to do it and secondly, Same paper Q.10)(ii) How do u find this angle?? Please reply

 

[hm12]

I have 2 questions! firstly, in J10 11 Q.5)(ii) how do u find the biggest and least value? ive found this question before and i dont know how to do it and secondly, Same paper Q.10)(ii) How do u find this angle?? Please reply

you put cosx=0 and get f(x)=2
cosx=1 f(x)=-3
cosx=-1 f(x)=-3 so the greatest value=2 and least value=-3

 

[rash1233]

Hi
Could you also helped me with
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_12.pdf
q 1 ii
Thanks

 

[leosco1995]

Hi
Could you also helped me with
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf
q 1 ii
Thanks

The expansion is 32 - 80y - 80y^2

According to the 2nd expression they provided, y = 2x - x^2

Replace y in the 1st expression with 2x - x^2 and you will get your answer.

 

[Dug]

w03 qp 4
I get 340m but the ms says 155
What am i doing wrong?

 

[sasukeuchiha]

yes it cocnsist of the essential frmulae needed fr calculating
mean, standard deviation, variance of groued and ungrouped data
some other formulae too lyk ... in binomial distribution we use meu = np variance = npq

th is all given in stats portion of mf9 and most importantly the table to calculate the probability is given in mf9

trig. formulae in p3 too? that'd be a great help

 

[leadingguy]

that is the range i calculated too but the mark scheme states the range to be f< and equal to 3??

ya so ??

 

[hm12]

ya so ??

what? shouldnt the range be 1<fx<3?

 

[leadingguy]

what? shouldnt the range be 1<fx<3?

taking x to be 0 gives 3
taking x to be pi/4 gives 1
taking x to be pi/2 gives 0
not surprisngly al values of "f" are getting smaller here as we are puting the values of x


so it is good to state that f< ,equal to 3

 

[PearlPeriwinklexo]

Ok i know this is weird... but i don't get why they divided by 2 and how in the world did they get c=1 | Q7...
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_11.pdf
Also, i don't understand why they didn't subtract 2x from the equation because its one whole plate; not two. | Q8...
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_11.pdf
I solved them many times and tried my best to understand the logic, but i don't get it
...
please and thank you!!! ^_^
P.S: I just need to understand the logic... you don't have to do the whole solution ^_^

 

[leadingguy]

Ok i know this is weird... but i don't get why they divided by 2 and how in the world did they get c=1 | Q7...
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_11.pdf
Also, i don't understand why they didn't subtract 2x from the equation because its one whole plate; not two. | Q8...
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_11.pdf
I solved them many times and tried my best to understand the logic, but i don't get it
...
please and thank you!!! ^_^
P.S: I just need to understand the logic... you don't have to do the whole solution ^_^

fr finding eq. we need to integrate dy/dx so we can get y in terms of x

now fr integrating a fraction of this type we use this formula ............ ( ax + b)^n


(ax + b)^n+1/ (n+1(a)) this is the formula we have to use in this qstn


3(1+2x)^-2+1/(-2+1)(2)

3(1+2x)^-1 /(-2) +c this c is the integration constant U always have to put it when ever U integrate any expresion
3/2(1+2x) +c = y now jxt substitute the value of x and y given in question and find c hope cleaR????

 

[leadingguy]

Ok i know this is weird... but i don't get why they divided by 2 and how in the world did they get c=1 | Q7...
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_11.pdf
Also, i don't understand why they didn't subtract 2x from the equation because its one whole plate; not two. | Q8...
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_11.pdf
I solved them many times and tried my best to understand the logic, but i don't get it
...
please and thank you!!! ^_^
P.S: I just need to understand the logic... you don't have to do the whole solution ^_^

wel i dont undrstnd wat u wanna ask here fr qstn 8 but it should be done this way have a look

the parameter of a rectangled plate is 2(x+y)
the parameter of circle is 2.pi.x

as it's one frth of the circle so (2.pi.x)/4 to get the parameter of the quadrant

now ad both of them and equate to 60

2(x+y) +(pi.x)/2 =60
2(x +y) + (pi.x)/2 = 60

( 4(x + y) + (pi.x))/2 = 60 send the 2 to R.H.s
4(x+y) + (pi.x) =120

4(x+y) = 120 - (pi.x) now send 4 at that side for divide

(x+y) = (120 - (pi.x)/4

x+y = 30 - pi.x/4

y = 30 -pi.x/4 -x Ans

 

[Aqsa Mohd]

oct.nov 02 p1 question 1 plzz any one ..
how to do it .. em blant at it .