Mathematics: Post your doubts here!

[externityxzx]

sin 3x + 2cos 3x = 0

divid the equation by cos x and u will get tan 3x +2 = 0
tan 3x = -2
3x = -64.3 and 116.5
x= 158.9 and 38.9 but i have no idea how the mark scheme got 98.9 as one of x values sorry :/

you can substitute 3x to something else like y (anything) so 3x = y
so you have tan y =-2, you have to change the range 0<x<180 to 0<y(which is 3x)<540
then you have y= -63.4 + 180 (keep adding 180)
y2= 116.6 and you put it back into y =3x which gives you 38.9
y3= 296.6 => 98.9
y4=476.6 => 158.9

 

[Rutzaba]

ok here goes to find the limits of this curve y= sin cube 2x cos cube 2x we will put y=0 as on the x axis y=o
0= cos cube2x sin cube 2x o=sin cube 2x
sin cube 2x = 0 0 = cos cube 2x
x=0 2x=pi/2

x= pi/4

so the limits be pi/4 and 0

by substituting u= sin2x
du/dx = 2cos2x
du= 2cos2x dx

now substituting

(cos cube 2x)( u^3) (1/ 2cos2x) <----------- where this is du
one cos would be cut by the cos in the denominator
u^3 (cos square 2x)/2
remember that cos square x = 1-sin square x
so cos square x = 1-sin square x
so
cos square 2x = 1-sin square 2x
and sin 2x =u
so
cos square 2x= 1- u^2
then the total thing wud be
1/2 integral of u^3(1-u^2)
1/2 integral of u^3 - u^5
1/2 (( u^4)/4 - (u^6)/6)
now open u as sin2x
1/2 (( sin 2x^4)/4 - (sin 2x^6)/6)
1/2 (1/4- 1/6)
1/2 (1/12)
1/24

No need to thank me
because i didn't solve it
Rutzaba solved it for me

Silent Hunter

 

[iKhaled]

you can substitute 3x to something else like y (anything) so 3x = y
so you have tan y =-2, you have to change the range 0<x<180 to 0<y(which is 3x)<540
then you have y= -63.4 + 180 (keep adding 180)
y2= 116.6 and you put it back into y =3x which gives you 38.9
y3= 296.6 => 98.9
y4=476.6 => 158.9

oh that shows how to get it, thank you

 

[abcde]

I'd really appreciate if someone can direct me to some vectors notes that cover all or most of the A'level p3 sy. I need them for someone else.

 

[Alice123]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_31.pdf
N0 10i Dug PhyZac or any1 having p3 please help

 

[ahmed abdulla]

@@@@@@
If f(x) > 2x+1
and g(x) > x^2-2
find f^-1 g(x) ??? ::::::

 

[Alice123]

@@@@@@
If f(x) > 2x+1
and g(x) > x^2-2
find f^-1 g(x) ??? ::::::

find f inverse(x) and put x^2-2 in place of x

 

[Alice123]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf
N0 10i Dug PhyZac or any1 having p3 please help

got it

 

[iKhaled]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf
N0 10i Dug PhyZac or any1 having p3 please help

da/dt = KV
da/dt = k(4/3πr^3)
da/dt = 4/3πkr^3

A=4πr^2 V = 4/3πr^3

da/dr = 8πr dv/dr = 4πr^3

dr/dt = dr/da x da/dt
dr/dt = (1/8πr) x 4/3πkr^3 substitute r and dr/dt to find K then u will end up with the given equation

did u get it ?

 

[iKhaled]

got it

oops sorry i guess i was too late to answer ur question..u already got it -_-

 

[Alice123]

oops sorry i guess i was too late to answer ur question..u already got it -_-

da/dt = KV
da/dt = k(4/3πr^3)
da/dt = 4/3πkr^3

A=4πr^2 V = 4/3πr^3

da/dr = 8πr dv/dr = 4πr^3

dr/dt = dr/da x da/dt
dr/dt = (1/8πr) x 4/3πkr^3 substitute r and dr/dt to find K then u will end up with the given equation

did u get it ?

u givin p3 this may?

 

[iKhaled]

u givin p3 this may?

inshAllah yes i will.

 

[Alice123]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf
N0 10i Dug PhyZac or any1 having p3 please help

cant do 10iii iKhaled

 

[ahmed abdulla]

find f inverse(x) and put x^2-2 in place of x

well .. i did that ... and mark scheme shows a different answerr!

 

[iKhaled]

cant do 10iii iKhaled

we have r = 5/ 1-0.4t

find t when the denomerator is equal to 0 so 1-0.4t = 0 t = 2.5 now any value higher than 2.5 will give u a negative value and we cant have radius in negative so the maximum value of t is just below 2.5 and the lowest one is t = 0 so 0 < t < 2.5​

 

[Shreeram]

hi can anyone give me a detailed explanation for q3 may june 2012 p13 thanks
Alice123 or abruzzi

 

[Alice123]

well .. i did that ... and mark scheme shows a different answerr!

is it (x^2-3)/2????

 

[abruzzi]

sin 3x + 2cos 3x = 0

divid the equation by cos x and u will get tan 3x +2 = 0
tan 3x = -2
3x = -64.3 and 116.5
x= 158.9 and 38.9 but i have no idea how the mark scheme got 98.9 as one of x values sorry :/

Thanks I tried using the same method in a different way. This is what I did.. hope it'll help you:

tan 3x = -2
Let 3x = x,
tan x = -2
(Ignoring negative); x = tan-1 (2)
x = 63.435

2nd quadrant:
180 - 63.435 = 116.565
3x = 116.565
x = 38.8

4th quadrant:
360 - 63.435 = 296.565
3x = 296.565
x = 98.8

2nd revolution:
360 + 116.565 = 476.565
3x = 476.565
x = 158.8
Angles => 38.8, 98.8, 158.8

 

[salvatore]

Then we add 3, since -2tan(1/2x) + 3 , so when y is 0 becomes y 3
View attachment 24379


And that is the final shape!

I was going through this again and failed to understand something.. does it matter where the curve cuts the x-axis in the final shape? Like in your case, it cuts slightly after π/2..

 

[PANDA-]

I was going through this again and failed to understand something.. does it matter where the curve cuts the x-axis in the final shape? Like in your case, it cuts slightly after π/2..

According to the ms, no. But it would make your graph more accurate I guess.... and it's very simple, just replace y=0 and solve the eqn.