Mathematics: Post your doubts here!

[ALM&B]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_32.pdf Q 10 part iii how to do it !! i know its easy but i dont get it

first, when they told you the tangent at point P passes through the origin and if you draw a straight line ..to find the gradient you will do
(y-0)/(x-0) and so you will the gradient for the tangent at P = y/x

then, from the dy/dx equation you got above, make it equal to y/x
so ..
e^-x (-x^2+2x) = y/x
and multiply the x.. you will end up with y= e^-x (2x^2-x^3) ..this is your y!
substitute this y in the equation y=x^2e^-x to get
x^2e^-x = e^-x(2x^2 - x^3)
cancel the e^-x

you will get
x^2 = 2x^2 - x^3
-1x^2 = -1x
-1x=-1
so x=1

 

[angelicsuccubus]

Use the iterative formula

xn+1 = tan−1(xn + p) to determine x
correct to 2 decimal places. Give
the result of each iteration to 4 decimal places.
how would I know what x1 is?

 

[angelicsuccubus]

nope,380 is greater than 360 so is incorrect.

hmm okies

 

[ALM&B]

Use the iterative formula

xn+1 = tan−1(xn + p) to determine x
correct to 2 decimal places. Give
the result of each iteration to 4 decimal places.
how would I know what x1 is?

I actually use 1 whenever they dont tell me the initial value..UNLESS they tell us the values which will show a change of sign in the equation, I use the first value given (as the root is between these 2 values)

 

[Abdulrab]

thanku

 

[Abdulrab]

first, when they told you the tangent at point P passes through the origin and if you draw a straight line ..to find the gradient you will do
(y-0)/(x-0) and so you will the gradient for the tangent at P = y/x

then, from the dy/dx equation you got above, make it equal to y/x
so ..
e^-x (-x^2+2x) = y/x
and multiply the x.. you will end up with y= e^-x (2x^2-x^3) ..this is your y!
substitute this y in the equation y=x^2e^-x to get
x^2e^-x = e^-x(2x^2 - x^3)
cancel the e^-x

you will get
x^2 = 2x^2 - x^3
-1x^2 = -1x
-1x=-1
so x=1

thanku

 

[redapple20]

There would b another part b4 this asking u to verify that x or wateva it is lies between this and that?
Take any value btw them

 

[angelicsuccubus]

This is how you do it:

∫1/(1 + x²) dx

suppose x = tan Ө so Ө = tan^(-1) x

so 1 + x² = 1 + tan² Ө = sec² Ө

and dx/dӨ = sec² Ө

dx = sec² Ө . dӨ

So ∫1/(1 + x²) dx = ∫(1/sec² Ө) . sec² Ө dӨ

= ∫ 1 dӨ

= Ө + k

= tan^(-1) x + k

credit: http://malaysia.answers.yahoo.com/question/index?qid=20090903194558AAkgNsU

so basically what I had in my head was complete insanity? .. why was I thinkin of ln?

Also, why are we "supposing" that x=tan theta? (I get that this is some ancient dervied thing.. but could x have been supposed as anything?)

 

[angelicsuccubus]

I actually use 1 whenever they dont tell me the initial value..UNLESS they tell us the values which will show a change of sign in the equation, I use the first value given (as the root is between these 2 values)

ty .. and if there are two values, like in sign change.. you take the average (as far as I was taught.. )

 

[kaz619]

sqr(10).cos [2 θ - 71.6'] = 2 ; 0' < θ < 90'
>> 2 θ - 71.6' = cos–1 [2 /sqr(10)].....
......................= 50.8'
>> θ = (50.8' + 71.6'] / 2 = 61.2'
also,


2 θ - 71.6' ~ 360' - 50.8' [since, cosθ = cos(360-θ) = cos(-θ)]
>> 2θ = 380.8' ~ 20.8' [since, 380.8' = 360' (1 complete rotation) + 20.8']
>> θ = 10.4'

θ = 61.2', 10.4'

Thanks!!

 

[hm12]

first, when they told you the tangent at point P passes through the origin and if you draw a straight line ..to find the gradient you will do
(y-0)/(x-0) and so you will the gradient for the tangent at P = y/x

then, from the dy/dx equation you got above, make it equal to y/x
so ..
e^-x (-x^2+2x) = y/x
and multiply the x.. you will end up with y= e^-x (2x^2-x^3) ..this is your y!
substitute this y in the equation y=x^2e^-x to get
x^2e^-x = e^-x(2x^2 - x^3)
cancel the e^-x

you will get
x^2 = 2x^2 - x^3
-1x^2 = -1x
-1x=-1
so x=1

oh okay got it thanks a lot for your help

 

[ALM&B]

ty .. and if there are two values, like in sign change.. you take the average (as far as I was taught.. )

wlcm ya the average value also works, but I usually use the 1st value given..works for me! (A)

 

[user]

Assalamoalaikum wr wb!

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf​

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_33.pdf​

Q:9 ii ...i don't understand why do we only substitute the point only in place of x, y and z?

 

[Sajana Meddaduwage]

Please guys big help!!! Tell me a difficult p3 past paper please quickly.

 

[RGBM211]

Prove the identity cos4teta + 4cos2teta = 8 cos^4 teta -3
(^May/june 2011 31 question 9 i)

mmm stuck

 

[angelicsuccubus]

Prove the identity cos4teta + 4cos2teta = 8 cos^4 teta -3
(^May/june 2011 31 question 9 i)

mmm stuck

I feel like I just did that and now I can't find the paper, hang on..

 

[angelicsuccubus]

Prove the identity cos4teta + 4cos2teta = 8 cos^4 teta -3
(^May/june 2011 31 question 9 i)

mmm stuck

And here you go... apparently I saw it in june 2005 xD
https://www.dropbox.com/s/26xt3wmgc9jp3zj/IMAG0276.jpg

 

[hm12]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_31.pdf
question 7 part (iv) can someone please draw the diagram for me

 

[angelicsuccubus]

someone please teach me how to do 7b
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf

 

[hm12]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_31.pdf
question 5 part (i) please its the 3RD TIME I AM POSTING THIS SOMEONE PLEASE HELP