Mathematics: Post your doubts here!

[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

 

[daredevil]

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

drawing the graph like we normally do for cos graphs?? taking 4 values on x-axis like (pie)/4 and stuff??

 

[iKhaled]

Any mechanics people here? !

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_41.pdf

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

JazakAllah

hey there!

for 6 c(i) you need to find the ratio of the speed at the ground to the speed at B. we have the kinetic energy at both points. At B the kinetic energy is 1.6j and at the ground it will be 6j so

1/2mv^2 = 1.6
v = 4 ( this is the speed of the object at B)

1/2mv^2 = 6
v = √60 ( this is the speed of the object on the ground)

ratio = v at G / v at B = √60/4 = 1.94

(c) initial energy at X is the potential energy = 2H all of this will be converted to kinetic energy when it reaches the ground so the final energy at the ground will be 2H which is all kinetic energy so 1/2m(VG)^2 = 2H
so VG^2 = 20H

now we will calculate the value of V at B (VB)

initial energy at x:

Ep = 0.2(10)(H) = 2H

final energy at B :
Ek = 1/2(0.2)(VB)^2
Ek = 0.1VB^2

EP = 2(H-2.2) = 2H - 4.4

2H = 0.1VB^2 + 2H - 4.4
VB^2 = 20H - 44

we got both VB^2 and VG^2 now we can use our ratio VG/VB= 2.55

VB^2/VG^2 = 6.5025
20H - 44/20H = 6.5025

solve for H and u will get it as 0.385 subtract it from 3 and u will get the answer 2.62

 

[qwerty123123]

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?

Just go to your calculator table function and type in:

f(x) = 3 − 4(cos(x))^2
type in your limits and step.
So start = 0
end = pi
step = pi/4
Now you can immediatly see the highest and lowest value's of f(x) i.e. your range so it is: -1=<f(x)<=3

OR

Since you know that the max value of cos^2(x) = 1 and the min is 0, you know that 3-4(0) = 3 and 3-4(1) = -1 Hence your range is -1=<f(x)<=3

 

[WayneRooney10]

Can someone please solve (Nov 2012, P13) Q2, and Q7 part (ii), with full working and explanations? Thank you. There seems to be some problem with the mark scheme.

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_13.pdf

 

[daredevil]

Can someone please solve (Nov 2012, P13) Q2, and Q7 part (ii), with full working and explanations? Thank you. There seems to be some problem with the mark scheme.

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_13.pdf

i don't know how to write so much maths here cz it gets all confusing but i'll try
look if f is a decreasing function then dy/dx should be <0
find dy/dx of this funciton which wen i calculated it was pretty wierd: -
dy/dx = -3(1-x^2)/x^4

is this wat u r getting too?

 

[Rutzaba]

I usually get confused when we are asked to find the range of trigonometric functions.. its just too complicated for me! What do I have to consider when solving such questions?

Look at these two functions:
f(x) = 3 - 2sinx for 0 ≤ x < pi
f(x) = 3 - 2sin(1/2x) for 0 ≤ x < pi.

How different would the range of the two functions be?
regards salvatore

 

[tom ed]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_63.pdf
q2

 

[mominzahid]

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

The answer I get if i do it this way is 4pi/9 or 1.89 whereas the marking scheme gives the answer 1.5Pi or 4.71. :/

 

[Rutzaba]

The answer I get if i do it this way is 4pi/9 or 1.89 whereas the marking scheme gives the answer 1.5Pi or 4.71. :/

mai solve karun? poora?

 

[Rutzaba]

I usually get confused when we are asked to find the range of trigonometric functions.. its just too complicated for me! What do I have to consider when solving such questions?

Look at these two functions:
f(x) = 3 - 2sinx for 0 ≤ x < pi
f(x) = 3 - 2sin(1/2x) for 0 ≤ x < pi.

How different would the range of the two functions be?
regards salvatore

HELLO SOLVE SOMEONE!
A star syed1995 iKhaled daredevil

 

[A star]

i don't know how to write so much maths here cz it gets all confusing but i'll try
look if f is a decreasing function then dy/dx should be <0
find dy/dx of this funciton which wen i calculated it was pretty wierd: -
dy/dx = -3(1-x^2)/x^4

is this wat u r getting too?

yup thats what i got and its not wierd is it .
i guess i solved wierder then this
i