Mathematics: Post your doubts here!

[abruzzi]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_12.pdf
Someone please help me with question no. 10(a) iii..

 

[syed1995]

MustafaMotani , A star , daredevil , PANDA-

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_11.pdf

Question No. 7.. whole question, I will try and solve it but you guys solve it as well.. so that I can tally my working.. since I don't exactly understand the question over here...

 

[falcon678]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf
Someone please help me with question no. 10(a) iii..

u just find till wich term are negative...take their sum n add it to 525..as their sum will b negative will be subtracted frm original in the start..
hope u got it?

 

[PANDA-]

MustafaMotani , A star , daredevil , PANDA-

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf

Question No. 7.. whole question, I will try and solve it but you guys solve it as well.. so that I can tally my working.. since I don't exactly understand the question over here...

a)
an = a1 + (n-1)d
cos^2x = 1 +d
d= cos^2x -1

Sn = 10/2[2(1) + (10-1)(cos^2x - 1)]
Sn = 10/2[2(1) + 9(-sin^2x)]
Sn = 5[ 2 - 9sin^2x]
Sn = 10 - 45sin^2x
a = 10 ... b = 45

 

[PANDA-]

syed1995

b) i) r = 1/3tan^2θ
convergent means r < 1

1/3tan^2θ < 1
tan^2θ < 3
tanθ < √(3)
θ < tan-1(√3)
θ < 60
0 < θ < 60

Extra answer I came up with and proved right, but not in ms

180 < θ < 240
This answer because tan is valid in the third quadrant as well

 

[aroojshahzad]

thanx

 

[syed1995]

a)
an = a1 + (n-1)d
cos^2x = 1 +d
d= cos^2x -1

Sn = 10/2[2(1) + (10-1)(cos^2x - 1)]
Sn = 10/2[2(1) + 9(-sin^2x)]
Sn = 5[ 2 - 9sin^2x]
Sn = 10 - 45sin^2x
a = 10 ... b = 45

Yea I solved it.. the exact same way.. what about b part i ? I solved that as well but unsure of the working.. since mark scheme shows literally no working.

MustafaMotani , A star , daredevil , PANDA-

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_11.pdf

8 iv ..

How does one know when taking an inverse whether a + sign or a - sign would come before the √ equation?

like (x-2)^2 = y+4-k
x-2 = ±√(y+4-k)
How'd you know whether we will take + or - here??

 

[Sarah22]

tan inverse of gradient .. is the angle the line makes with x-axis , so when ever u need to find the angle of two lines intersecting ! just find their gradients and do tan^-1(gradient)
the one with larger angle minus the smaller angle so u get the angle ... this angle can be acute or obtuse ..
some time cie makes it thougher for us .. wen we get acute angle ! they might ask for an obtuse angle, so at that time you need to 180-(what ever angle u got) to get the obtuse angle

Thank you very much..

 

[PANDA-]

Yea I solved it.. the exact same way.. what about b part i ? I solved that as well but unsure of the working.. since mark scheme shows literally no working.

MustafaMotani , A star , daredevil , PANDA-

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf

8 iv ..

How does one know when taking an inverse whether a + sign or a - sign would come before the √ equation?

like (x-2)^2 = y+4-k
x-2 = ±√(y+4-k)
How'd you know whether we will take + or - here??

Check above I already posted it... I'll try this one as well..

 

[PANDA-]

Yea I solved it.. the exact same way.. what about b part i ? I solved that as well but unsure of the working.. since mark scheme shows literally no working.

MustafaMotani , A star , daredevil , PANDA-

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf

8 iv ..

How does one know when taking an inverse whether a + sign or a - sign would come before the √ equation?

like (x-2)^2 = y+4-k
x-2 = ±√(y+4-k)
How'd you know whether we will take + or - here??

Yup, I had the same doubt, and got stuck there too..
But the ms had it pretty clearly explained, the range of f(x) is the domain of f^-1(x)
In f^-1(x) --- x > k-4
f(x) = (x-2)^2 -4 + k

Take it as if f^-1(x) is y and switch as always

x = (y-2)^2 - 4 + k
x+4-k = (y-2)^2
y-2 = ±√(x+4-k)
f^-1(x) = ±√(x+4-k) + 2

Replace x with anything less than k-4, like k-5

f^-1(x) = ±√(k-5+4-k) + 2
y - 2 = ±√(k-5+4-k)

The two k's cancel each other (FINALLY)

y-2 = ±√(5+4)
y-2 = ±√(9)
y - 2 = ±3
y = 5 or y = -1

The range of f^-1(x) is... f >= 2
So y = 5, so +.
f^-1(x) = +√(x+4-k) + 2

 

[~`Heba`~ :)]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_41.pdf #6 (ii) please help , how to get H :'(

 

[syed1995]

Yup, I had the same doubt, and got stuck there too..
But the ms had it pretty clearly explained, the range of f(x) is the domain of f^-1(x)
In f^-1(x) --- x > k-4
f(x) = (x-2)^2 -4 + k

Take it as if f^-1(x) is y and switch as always

x = (y-2)^2 - 4 + k
x+4-k = (y-2)^2
y-2 = ±√(x+4-k)
f^-1(x) = ±√(x+4-k) + 2

Replace x with anything less than k-4, like k-5

f^-1(x) = ±√(k-5+4-k) + 2
y - 2 = ±√(k-5+4-k)

The two k's cancel each other (FINALLY)

y-2 = ±√(5+4)
y-2 = ±√(9)
y - 2 = ±3
y = 5 or y = -1

The range of f^-1(x) is... f >= 2
So y = 5, so +.
f^-1(x) = +√(x+4-k) + 2

Why did ya take k-5 as x in f^-1 when x is supposed to be greater than k-4 in f^-1 ? That part I didn't understand.. everything else is clear cut.

 

[PANDA-]

Why did ya take k-5 as x in f^-1 when x is supposed to be greater than k-4 in f^-1 ? That part I didn't understand.. everything else is clear cut.

I did that one wrong
I was supposed to replace with k-3 AND I did -5+4 = 9 It's 1:40 here, half asleep
So the correct one would be to replace it with k-3 so...
f^-1(x) = ±√(k-3+4-k) + 2
y - 2 = ±√(k-3+4-k)

The two k's cancel each other

y-2 = ±√(1)
y - 2 = ±1
y = 3 or y = 1

The range of f^-1(x) is... f >= 2
So y = 3, so +.
f^-1(x) = +√(x+4-k) + 2

... Sorry about that

 

[A star]

um how did we know x was suppose to be greater then x

 

[MustafaMotani]

syed1995 just a way to remeber which sign to take...
If yu are taking right hand side of the curve or in the domain x >a yu will take positive sign and vice versa... reasonjng is quite simple btw

 

[mominzahid]

Hi... can someone please help me with this problem.. Would really appreciate it.
I only want the explanation of the 2nd and third part. i dont get how the volume is being calculated.. please explain what cylinder does the MS talk about?
Its from w08.
Thanks

 

[Silent Hunter]

Any mechanics people here? !

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_41.pdf

Q6 (i) (c) and (ii) ?

JazakAllah

 

[Beaconite007]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_12.pdf
Question 9 i)a)
I always get lost when I see questions like these :c
Help please?

 

[daredevil]

Hi... can someone please help me with this problem.. Would really appreciate it.
I only want the explanation of the 2nd and third part. i dont get how the volume is being calculated.. please explain what cylinder does the MS talk about?
Its from w08.
Thanks

volume of rotation of this graph will be (pie) x integration of the square of the equation in terms of y

the equation will be (y^2 - 1)/3 ........... check if its ryt.
take the square of this equation and integrate it.
use integration limits to calculate integrant and then multiply it by pie.
if u dont' get it then tell me and i'll c wat i can do to further explain

 

[A star]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_12.pdf
Question 9 i)a)
I always get lost when I see questions like these :c
Help please?

well as an aid roughly sketch the graph and you will find graph varies between -1 and 3 which is the range