Mathematics: Post your doubts here!

[sj0007]

Help please?
View attachment 57095

(i) f(x) = 4, so: 6 - 4cos(0.5x) = 4
cos(0.5x) = 0.5
0.5x = π / 3 (since it says exact, no need for basic angles and stuff)
x = 2π / 3

(ii) Max = a + b = 6+4 = 10 and Min = a - b = 6 - 4 = 2
So, range: 2 <f(x)<10

(iii) The graph is going to be upside down since the cos is -ve.............. so it would be something like this: (sorry for the crappy drawing though)


(iv) The inverse is just like the normal one

Hope it helps...............

 

[Sarosh Jameel]

Either my mind isnt working or the marking scheme is wrong................. can anyone tell me how the acceleration of P in the question below is gsin30? (I came to the conclusion that it is g/sin30)
View attachment 57097

Resultant force down the slope is .... mgsina=ma
m and m cancel out ... remaining is ... a=gsina

 

[sj0007]

Resultant force down the slope is .... mgsina=ma
m and m cancel out ... remaining is ... a=gsina

Oooooooooh oki....... (Hadn't studied that part yet)..........thnx

 

[FrostBite92]

HELP
A committee of 5 people is to be chosen from 6 men and 4 women. In how many ways can this be done
(iii) if there must be 3 men and 2 women, and one particular woman refuses to be on the committee with one particular man?

 

[qwertypoiu]

6C3 x 4C2 - 5C2 x 3C1 = 90

 

[sj0007]

Ummmm, I am having a prob in understanding the part (ii) (b) of Q7 in:
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_42.pdf
Thnx...

 

[qwertypoiu]

Ummmm, I am having a prob in understanding the part (ii) (b) of Q7 in:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_42.pdf
Thnx...

v1 = 0.02t + 1

v1 - v = 0.02t+1 - (0.04t − 0.000 05t^2)
= 0.02t + 1 - 0.04t + 0.00005t^2
= 0.00005t^2 - 0.02t + 1
= 0.00005(t^2 - 400t + 20000)
= 0.00005[t^2 - 400t + 200^2 - 200^2 + 20000]
= 0.00005[(t - 200)^2 - 20000]
= 0.00005(t - 200)^2 - 1

 

[sj0007]

v1 = 0.02t + 1

v1 - v = 0.02t+1 - (0.04t − 0.000 05t^2)
= 0.02t + 1 - 0.04t + 0.00005t^2
= 0.00005t^2 - 0.02t + 1
= 0.00005(t^2 - 400t + 20000)
= 0.00005[t^2 - 400t + 200^2 - 200^2 + 20000]
= 0.00005[(t - 200)^2 - 20000]
= 0.00005(t - 200)^2 - 1

Thnx for replying...........but I dont get the v1, how did u find it?

 

[qwertypoiu]

Thnx for replying...........but I dont get the v1, how did u find it?

v1 is the first part of the graph. So we simply need it's equation.
gradient = (9-1)/(400-0) = 0.02
y - intercept = 1
So v1 = 0.02t + 1

Or alternatively,
v = u+at,
Where u=1, a=0.02
v1 = 1 + 0.02t

 

[sj0007]

v1 is the first part of the graph. So we simply need it's equation.
gradient = (9-1)/(400-0) = 0.02
y - intercept = 1
So v1 = 0.02t + 1

Or alternatively,
v = u+at,
Where u=1, a=0.02
v1 = 1 + 0.02t

Thanku sooo much!!
God, it was that easy.................. I feel dumb!

 

[musiclover gurl]

(i) f(x) = 4, so: 6 - 4cos(0.5x) = 4
cos(0.5x) = 0.5
0.5x = π / 3 (since it says exact, no need for basic angles and stuff)
x = 2π / 3

(ii) Max = a + b = 6+4 = 10 and Min = a - b = 6 - 4 = 2
So, range: 2 <f(x)<10

(iii) The graph is going to be upside down since the cos is -ve.............. so it would be something like this: (sorry for the crappy drawing though)
View attachment 57098

(iv) The inverse is just like the normal one

Hope it helps...............

Sorry for the late reply. Could you explain how did you get the exact value of the inverse of cos 0.5?

 

[qwertypoiu]

Sorry for the late reply. Could you explain how did you get the exact value of the inverse of cos 0.5?

These are the common angles and their trigonometric ratios you should try to learn:

As you can see, cos^-1(0.5) = 60°, or π/3 radians.

 

[musiclover gurl]

These are the common angles and their trigonometric ratios you should try to learn:

As you can see, cos^-1(0.5) = 60°, or π/3 radians.

But thats the problem. I dont memorize these values but instead I calculate them, therefore I am unable to get the values of their inverse.

 

[musiclover gurl]

These are the common angles and their trigonometric ratios you should try to learn:

As you can see, cos^-1(0.5) = 60°, or π/3 radians.

Yesssss! Finally got a way to calculate these exact values!! thanks though qwertypoiu!

 

[musiclover gurl]

(i) f(x) = 4, so: 6 - 4cos(0.5x) = 4
cos(0.5x) = 0.5
0.5x = π / 3 (since it says exact, no need for basic angles and stuff)
x = 2π / 3

(ii) Max = a + b = 6+4 = 10 and Min = a - b = 6 - 4 = 2
So, range: 2 <f(x)<10

(iii) The graph is going to be upside down since the cos is -ve.............. so it would be something like this: (sorry for the crappy drawing though)
View attachment 57098

(iv) The inverse is just like the normal one

Hope it helps...............

Yeah I understood it. Thanks a lot for your help!

 

[Rizwan Javed]

Help please?
(9709/32)
View attachment 57109

year??

 

[zahra azam]

Can anybody help me with this qs :"(

 

[psychiatrist]

A small doubt..
According to grade threshold of 2015
To get A we should get at least 62 in paper 1 and 42 in paper 6
So my doubt is In paper 1 I scored 68 and in paper 6 I scored 30.
What will be my grade?? C or B or A ?

 

[psychiatrist]

Can anybody help me with this qs :"(
View attachment 57115

f(1) = -1/2 ?
A = 7/3 ?

 

[sj0007]

Yeah I understood it. Thanks a lot for your help!

Welcome................sorry didnt reply to ur msgs before (I just logged in now)..............glad i helped........