Mathematics: Post your doubts here!

[Rutzaba]

I agree that this thread should be on topic as much as possible but sometimes I too get carried away sometimes.. haha

like right now

 

[gary221]

gary221 three intervals are 0, pi/8, pi/4

u have to ensure that d is constant

bt how did u gt tht???
i know m being a bother..bt i kinda suck at maths!!

 

[MustafaMotani]

gary221 lol no yu arent bother.. .. I just got it by looking at it... it said we have to do it in three intervals... we knw two values 0 and pi/4 ... so to get third just find the mid -value ..

 

[sma786]

This is CIE making life difficult for us...

To get the vectors we will find the unit vector and multiply by the magnitude..

OA = Unit Vector .. 3i − 4k/√25

15/5(3i-4k)
=3(3i-4k)
=9i-12k

Unit Vector of OB = 2i + 3j − 6k / √(2^2)+3^2+(-6^2)
Unit Vector of OB = 2i+3j-6k/7

OB = 14/7(2i+3j-6k)
OB = 2(2i+3j-6k)
OB = 4i+6j-12k

lol, thanks

 

[sma786]

shafayat look, i just took pi-2.4 that gave me 0.74 . but i have 2 angles that are similar, so answer/2 : 0.37
means each angle is 0.37 rads, then i just use the sin formula
sin0.37/8 = sin2.4/x
and i got the correct answer

 

[MustafaMotani]

falcon678 daredevil.. here is a snapshot of the graph in the question u gave .. i just pinned it for yur understanding.. there is nothing to do with the answer.. u can see as we go across x (after x>0) y is decreasing... tis we had to prove..

 

[syed1995]

falcon678 daredevil.. here is a snapshot of the graph in the question u gave .. i just pinned it for yur understanding.. there is nothing to do with the answer.. u can see as we go across x (after x>0) y is decreasing... tis we had to prove..

How did ya draw the graph?

 

[MustafaMotani]

syed1995
from graph plotter.. xD

 

[PANDA-]

How did ya draw the graph?

http://rechneronline.de/function-graphs/

 

[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 :'(