Need Help with Maths [Functions & Trigonometry]

[HybridStasis]

Hello,
I need help for this question in Nov 2001 P1 Q3, I'm trying really hard to think of the solution but to no avail. Can someone please thoroughly explain this to me:

Given that f(x)=cos x, for the domain 0=<x=<k, find the largest value of k for which f has an inverse.

 

[CaptainDanger]

For it to have an inverse, it has to be 1:1 function. 1:1 function occurs for the part

0 ≤ x ≤
π
So k = π

 

[HybridStasis]

thank you so much

 

[iKhaled]

i need help with this question in mechanics, please help meeee

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s06_qp_4.pdf

question 5, the first part!

 

[HybridStasis]

Let square root = |

4|2 is the resultant of the two tensions (tension made by Q and made by P), note that the two tensions are equal in magnitude.

so
(4|2)^2= T^2 +T^2
32=2T^2
T=4N

Assuming Q is in equillibrium,

mg - 4 = 0 (where g=10)
m = 0.4 kg.

 

[iKhaled]

Let square root = |

4|2 is the resultant of the two tensions (tension made by Q and made by P), note that the two tensions are equal in magnitude.

so
(4|2)^2= T^2 +T^2
32=2T^2
T=4N

Assuming Q is in equillibrium,

mg - 4 = 0 (where g=10)
m = 0.4 kg.

Mechanics is a piece of cake for me Core maths and Statistics is my weakness......

in the question it says " the force exerted on the string by the pulley is 4√2 " can u show me the direction of this force in a diagram and y is this force the resultant of both tensions ?

thanks.

 

[HybridStasis]

Assuming that the tension made by object P goes to the east direction and tension made by object Q goes to the south direction. Logic dictates that the resultant of both tensions will go somewhere to the south-east direction (assuming the only forces that applies onto the pulley is the tensions). Since both tensions are from the same string, we can say that the tensions are equal. Now, the question says " the force exerted on the string by the pulley is4√2". The only forces that are involved with the pulley as the focus of the object is the tensions it self. So, we can safely assume that the force is in fact the resultant which is 4√2.

Hope this explanation helps. Sorry for the long time to reply, I have been busy for the past few days.

 

[iKhaled]

Assuming that the tension made by object P goes to the east direction and tension made by object Q goes to the south direction. Logic dictates that the resultant of both tensions will go somewhere to the south-east direction (assuming the only forces that applies onto the pulley is the tensions). Since both tensions are from the same string, we can say that the tensions are equal. Now, the question says " the force exerted on the string by the pulley is4√2". The only forces that are involved with the pulley as the focus of the object is the tensions it self. So, we can safely assume that the force is in fact the resultant which is 4√2.

Hope this explanation helps. Sorry for the long time to reply, I have been busy for the past few days.

since mechanics is a piece of cake for u pls explain question 6 oct/nov 2003 when there is vertical wire and there is a ring on it arghhh hate these questions
pls draw diagrams showing forces if u don't mind

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w03_qp_4.pdf

 

[HybridStasis]

6(i)

First, we list out what forces are involved in this question. We got tension and applied force (5N). Since tension came from a same string, the tension is equal. Remember, the system is in equilibrium so the resultant force is equal to 0.

So, in the x-direction:
5cos(0) - T(cos 60) - T(cos 60) = 0
5 - T = 0
T= 5N

6(ii)

The question says that the ring is ON THE POINT OF SLIDING UP THE WIRE. This doesn't mean that the ring is moving. Thus, the resultant force is still equal to 0. When ring and friction is involved, we must consider two things. The direction of Friction (Fr) and the direction of Normal Reaction (Nr). You were asked to find the coefficient so the formula Fr=uNr is expected to be used. Don't forget the 0.2g force from the weight!

So, in the x-direction:
5cos(60) + Fcos(90) - Ncos(0) +2cos(90) = 0
2.5 - N = 0
Nr = 2.5

In the y-direction:
5sin(60) - Fsin(90) + Nsin(0) - 2sin(90) = 0
2.33 - F - 2 = 0
Fr = 2.33

Fr=uNr
2.33 = u (2.5)
u = 0.932

6(iii)

(sorry, this one is a bit blurry)

First, list out the forces involved. We have the tension (5N), Friction, Normal Reaction, weight of ring (2N) and weight of the new particle (10m). The question says that the coefficient stays the same. This means that the Friction (2.33) and Normal Reaction (2.5) is equal to the previous question. It stills says that it is ON THE POINT OF SLIDING DOWN THE WIRE. So, It is still in equilibrium.

So, in the y-direction:
5sin(60) + 2.33sin(90) + 2.5sin(0) - 10msin(90) - 2sin(90) = 0
2.33 + 2.33 - 10m - 2 = 0
m = 0.466 kg.

Again, sorry for the late reply.