Mathematics: Post your doubts here!

[abruzzi]

you dont have any intercept in this graph,look carefully the range is given from k<x<4 and where you found k=2
so,when x is 2,y is 8
when x is 3,y is 7 and when x is 4,y is 4.
plot them and join the curve

Oh yeah.. thanks

 

[syed1995]

How much do you get on average when solving P1? As in past papers.

around 60 or so .. but the 2012 ones were quite a disaster .. one of them went for 48

 

[sma786]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s04_qp_1.pdf
Question 4, (ii) , what is the basic rule for these types of questions?

 

[nafis shahRiyar]

Oh yeah.. thanks

anytime

 

[MustafaMotani]

j

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s04_qp_1.pdf
Question 4, (ii) , what is the basic rule for these types of questions?

just multiply those terms that give u your desired term (X^3) in this case ...
like in (1-3x)(1+2x)^6
multiply 1 by the term with x^3 in the expansion of (1+2x)^6 (prev part) and multiply -3x by the term with x^3 in the expansion of (1+2x)^6

 

[daredevil]

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

i dont understand question 11 :| can someone explain?

hey
I am solving ur question but hree's the deal. u need to have that paper and a page and pencil with u and u need to do it as I walk u through it otherwise u won't get it bcz making a diagram here won't be possible for mme and u hav to bear with the descriptions of the diagram nad make one for urself side by side.


see wat u need to do here is make a right angled triangle. consider OR to be the tangent and make a radius line from C to R. connect C to O also..... there u have ur right angle.

in part i) Sin(thita) = x/20
thita = 1.2/2 = 0.6 radians

solve it and u will get 7.218cm

in part ii) Area = (Area of sector) - (Area of circle)
= [(1/2)(1.2)(20)^2] - [(pie)(7.218)^2]
= 240 - 163.68
= 76.32 cm^2

got it??

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s04_qp_1.pdf
Question 4, (ii) , what is the basic rule for these types of questions?

adding to what Mustafa said.... actually just emphasising a point he already made.... use the expanded form instead of the secnd bracket.... meaning the answer of ur 1st part it makes the questions a lot less confusing and it becomes vivid as to wat u need to do there

 

[MustafaMotani]

adding to what Mustafa said.... actually just emphasising a point he already made.... use the expanded form instead of the secnd bracket.... meaning the answer of ur 1st part it makes the questions a lot less confusing and it becomes vivid as to wat u need to do there

she doesnt need to expand in first part.. -.-

 

[Rutzaba]

around 60 or so .. but the 2012 ones were quite a disaster .. one of them went for 48

out of 75? :O

 

[daredevil]

she doesnt need to expand in first part.. -.-

ohh oops *sleep deprivation error* better go now starting to be a liability now

 

[abcde]

I hope I'm not late.

Here: http://www.mediafire.com/?6hwfccjmuht9cah

Thank you. :]

 

[syed1995]

out of 75? :O

YEA! .. i am not that bad in maths to get 60% -,-

 

[MustafaMotani]

Rutzaba underestimated you syed1995

 

[syed1995]

Rutzaba underestimated you syed1995

I don't know why people do that .. Honestly getting underestimated is way better than having expectations and not being able to meet them So I am good.

 

[MustafaMotani]

I don't know why people do that .. Honestly getting underestimated is way better than having expectations and not being able to meet them So I am good.

agreed ..

 

[ZainH]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_12.pdf

Q4)ii) I'm clueless .___.

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf

Q4)ii) I'm clueless .___.

Equate both the equations.. you will have a equation in terms of k..

b^2 -4ac = 0 for the curve to be intersecting at only one point..

you will get the value of k.

 

[ahmed abdulla]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf

Q4)ii) I'm clueless .___.

make "y" as subject in BOTH cases ...and equate the two ... and arrane in form of quadratic eqn ... finally use b^2-4ac=0 since it intersect and find "K" value

 

[MustafaMotani]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf

Q4)ii) I'm clueless .___.

for line to be tangent it should just cut the curve at one point, in other words just touch the curve.
for such scenario b^2-4ac = 0

first make x the subject from the line equation and then substitute it to the curve equation. You will be left with quadratic equation of y .... now simply apply b^2-4ac=0

 

[ZainH]

Equate both the equations.. you will have a equation in terms of k..

b^2 -4ac = 0 for the curve to be intersecting at only one point..

you will get the value of k.

I tried that, I'm pretty sure I'm using b^2-4ac wrong.
This is the equation I'm getting when they're equated: y^2+2k-4y-13 = 0 ..

What's A , B and C... 4 terms. I forgot how to do these x_x