Mathematics: Post your doubts here!

[redrecession]

can u tell me how i can use the quadratic equation to find turning point of a curve? Thx

 

[iKhaled]

need help with mechanics question 06 oct nov 2003

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w03_qp_4.pdf

 

[bamteck]

whitecorp please help me for the no. 10 on funtions !

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_11.pdf

 

[Minato112]

whitecorp please help me for the no. 10 on funtions !

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_11.pdf

Which part?

 

[bamteck]

Which part?

(i) (ii) (iv)

 

[Minato112]

(i) (ii) (iv)

(i) : gf(x) --> U take f(x) and then U replace in g(x)
f(x) = 2x + 1
g(x) = (2x - 1)/ (x +3)

gf(x) = (2(2x+1) + 1) / ((2x + 1) +3)
= (4x + 3) / (2x + 4)

Now, gf(x) = x
that is, (4x +3) / (2x +4) = x
4x + 3 = 2x^2 + 4x
2x^2 - 3 = 0
x ^ 2 = (3/2)
x = *root* (3/2)

(ii) : To find f inverse, U let the function f(x) to be equal to y

y = f(x)
y = 2x + 1
Now make x the subject of formula

x = (y -1)/ 2
Therefore f inverse : (x - 1)/2 (U replace the y and make it become x)

For g(x)

y = g(x)
(2x - 1)/ (x +3) = y
2x -1 = xy + 3y
2x - xy = 3y + 1
(2 - y)x = 3y + 1
x = (3y +1)/(2 - y)

Therefore g inverse : (3x +1) / (2 - x)

(iii) To sketch the graphs, first U sketch the graph of f(x).. (I believe this one is quite easy)

Now for the inverse, all that we have to do is to reflect the graph in the line y = x.

The reflected part will show the graph of f inverse relative to f(x).

Note : Any calculations were done mentally so am not sure if the answers are correct but I've explained all my workings... Hope it helps.

 

[Soldier313]

Aoa wr wb, i need urgent help with this integration.
Integrate:


minato112 parthrocks and everybody else

 

[Minato112]

Aoa wr wb, i need urgent help with this integration.
Integrate:
View attachment 15510

minato112 parthrocks and everybody else

I got the ans : - 16 cos ((1/2)x) + (1/2) ln | cos ((1/2)x)| + c

Is that the ans.? If yes, I shall explain my workings.

 

[Soldier313]

I got the ans : - 16 cos ((1/2)x) + (1/2) ln | cos ((1/2)x)| + c

Is that the ans.? If yes, I shall explain my workings.

yea it's right!!! thanx!

 

[Minato112]

yea it's right!!! thanx!

I'll use "{" as the integration sign.. srry I couldnt find any better

= { 8 sin ((1/2)x) - tan ((1/2)x) dx

= { 8 sin ((1/2)x) dx - { tan ((1/2)x) dx

= 8 { sin ((1/2)x) dx - { tan ((1/2)x) dx

= 8 * 2 * - cos ((1/2)x) - { [sin ((1/2)x)/ (cos ((1/2)x)] dx

= -16 cos ((1/2)x) - (-1/2) [ ((-1/2) (sin ((1/2)x))/ (cos ((1/2)x)] dx

= -16 cos ((1/2)x) + (1/2) ln |cos ((1/2)x)| + c

Srry it appears bulky but hope U understand.

 

[Soldier313]

= -16 cos ((1/2)x) - (-1/2) [ ((-1/2) (sin ((1/2)x))/ (cos ((1/2)x)] dx

= -16 cos ((1/2)x) + (1/2) ln |cos ((1/2)x)| + c

Srry it appears bulky but hope U understand.

JazakAllah khair....thanx a ton! but this is where i didn't understand

 

[Minato112]

JazakAllah khair....thanx a ton! but this is where i didn't understand

Mention not

Well I corrected it... The dot meant multiply

 

[Soldier313]

Mention not

Well I corrected it... The dot meant multiply

um well i understood that actually, it's the second part after the minus sign that i didn't quite understand:/

= -16 cos ((1/2)x) - (-1/2) [ ((-1/2) (sin ((1/2)x))/ (cos ((1/2)x)] dx

= -16 cos ((1/2)x) + (1/2) ln |cos ((1/2)x)| + c

the parts in purple are the ones i didn't understand sorry and thanx....

 

[Minato112]

um well i understood that actually, it's the second part after the minus sign that i didn't quite understand:/

Oh k

the parts in purple are the ones i didn't understand sorry and thanx....

U don't know this rule :

1) { (f ' (x))/ (f (x)) dx = ln | f(x) | + c

2) { 1 / (ax + b) dx = (1/a) { a/ (ax + b) dx
= (1 /a) ln | ax +b | + c

 

[Soldier313]

U don't know this rule :

1) { (f ' (x))/ (f (x)) dx = ln | f(x) | + c

2) { 1 / (ax + b) dx = (1/a) { a/ (ax + b) dx
= (1 /a) ln | ax +b | + c

nop i didn't know those rules! i understood it all now!
JazakAllah khair sooooo much! May Allah bless you inshaAllah.

 

[Minato112]

nop i didn't know those rules! i understood it all now!
JazakAllah khair sooooo much! May Allah bless you inshaAllah.

Oh glad U understood it
Ameen and same to U also

 

[Soldier313]

Oh glad U understood it
Ameen and same to U also

ameen inshaAllah

 

[whitecorp]

can u tell me how i can use the quadratic equation to find turning point of a curve? Thx

You will have to complete the square such that it is in the form y= a(x-b)^2 +c.
Then if a>0, the minimum point is given by (b,c) ; if a<0, the maximum point is given by (b,c) .
Alternatively, you can use calculus, ie find dy/dx and set this derivative =0 to obtain the coordinates of the turning point.
Hope this helps. Peace.

 

[shanky631]

Minato
whitecorp
leadingguy



can u plz help me with this question.

 

[parthrocks]

Minato
whitecorp
leadingguy

View attachment 15513

can u plz help me with this question.

is the answer this if yes then i will post you a scan!!



N= (40 – 30e^-0.02t)^2
see ya!!