Additional Math Help HERE ONLY...!

[ramezamgad]

lil^$tar

what is the relative velocity of two particles A and B moving

a) towards each other A with velocity 10 and B with velocity 15?
b) in the same direction A with velocity 10 and B with velocity 15?
c) in the same direction A with velocity 15 and B with velocity 15?

vanibharutham
what is the domain of the first eq.?

[vanibharutham]

x greater than or equal to 0...

sorry the symbol dint come up on there

[ramezamgad]

well
y=ln(4x-1)
u have to find what value of x makes y=0
ln1=0
so 4x-1=1
4x=2
x=1/2

u made 4x-1=0 which is a common mistake

[vanibharutham]

oh... silly me!

right, and then for the domain x > 1/2

you'll have to find the range, which is y > 0

THANKS SO MUCH...
i always make that mistake with domain and range when it comes to e^x and ln(x)

[ramezamgad]

an easy method to remember is that the domain of f(x) is the range of f-1
and the range of f(x) is the domain of f-1

[vanibharutham]

You are a genius!

thanks for that

remind me thank you over nd over agen

[lil^$tar]

lil^$tar

what is the relative velocity of two particles A and B moving

a) towards each other A with velocity 10 and B with velocity 15?
b) in the same direction A with velocity 10 and B with velocity 15?
c) in the same direction A with velocity 15 and B with velocity 15?

vanibharutham
what is the domain of the first eq.?

a)25 b)-5 c)0

[ramezamgad]

b is not -ve
u just need to know that B will move faster than A and the distance between them will decrease until they become side to side then the distance between them will increase
hope u find that easy

[lil^$tar]

thanks very much +rep 4 u

[Ghost Of Highbury]

Hi,

i was doing the following question:

The function f(x) = [ (e^x + 1) / 4 ] for the domain x ? 0

i) Obtain an expression for f^-1
ii) State the domain and range for f^-1

The first part was easy...
i got the answer, and when i checked it with the mark scheme i got the right answer, which was:
ln(4x-1)

The second part of my answer didn't match the mark scheme, and i dont seem to understand the why.

The domain, according to me should be:

x > 1/4

The range should be:

f^-1(x) > -3.2188

However, the mark scheme says that the answers are:

DOMAIN: x > 1/2
RANGE: f^-1(x) > 0

could some one please explain my mistake?

which paper did this question appear in..
year/month/paper..

[Ghost Of Highbury]

plzz answer,...which paper???

[vanibharutham]

Adi M...

May/ June 2003 Paper 2  Cambridge

[shan2391]

Guys i dont know how to find the amplitude and period of function plzz help.
Ref: Q.8 OCt/NOv08 Paper2

[vanibharutham]

Shan2391

The question states that

f(x) = 3 + 5 sin2x

Im sure you can easily rearrange that to make it simpler:

f(x) = 5sin2x + 3

The amplitude is always going to be the number infront the trignometric function...
In this case the amplitude will 5

simple,

if forexample f(x) was 25000sin2x, your amplitude would be 25000


Your amplitude involves the actual trignometric identity... in this case... 2x.

Always remember this rule:
To find the period of any trignometric graph, divide the coefficient of x by 360 (but in this case 2pi as we are dealing with radians)

Therefore our period is 180 degrees or simply pi.

Hope that helped

[shan2391]

Shan2391

The question states that

f(x) = 3 + 5 sin2x

Im sure you can easily rearrange that to make it simpler:

f(x) = 5sin2x + 3

The amplitude is always going to be the number infront the trignometric function...
In this case the amplitude will 5

simple,

if forexample f(x) was 25000sin2x, your amplitude would be 25000


Your amplitude involves the actual trignometric identity... in this case... 2x.

Always remember this rule:
To find the period of any trignometric graph, divide the coefficient of x by 360 (but in this case 2pi as we are dealing with radians)

Therefore our period is 180 degrees or simply pi.

Hope that helped

Thanks a lot man
but what does +3 show in a graph