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Mathematics: Post your doubts here!

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smzimran Assalamoalaikum wr wb!

I need help with the last part of the last question of Nov:2011 # 3

mark scheme

I do know how to do it...but i cant seem to get the answer...! :unsure:

it's like we take the root of − 2 ± 2 root 3i

how i do is...equate that to (a + bi)^2 but..if we simplify...i get a^2 + b^2 = -2.....and this cant be solved...ms uses a^2 - b^2.....can u tell me when exactly do we use '-' b^2 ?
 
smzimran

smzimran

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reina81 said:
oh so sorry. here it is
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_31.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_ms_31.pdf
Click to expand...
e^x = 1 + 6e^−x
Using u = e^x
The equation becomes
u = 1 + 6/u
u^2 = u + 6
u^2 - u - 6 = 0
(u+2)(u-3) = 0
u = -2 or u = 3
e^x = -2
This cannot be solved as ln or lg of a negative number is undefined!

The only solution is
e^x = 3
x = ln3 / lne
x = ln3
x = 1.10

We take lne^x
but according to power rule, the power becomes coefficient
so lne^x = xlne
and lne = 1
so
xlne = x

See the notes i posted on logs above !
 
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miss irfan

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smzimran said:
Wait i'll tell you in afternoon after i solve this paper myself!
Click to expand...

i have almost done it, but i just don't know how to remover "x" from the differential and put u into it. the differential is sec^2x and we have to use substitution u=tanx. du/dx is sec^2x but how can i just remove 'X" and replace it with U"?
 
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2pac

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smzimran said:
e^x = 1 + 6e^−x
Using u = e^x
The equation becomes
u = 1 + 6/u
u^2 = u + 6
u^2 - u - 6 = 0
(u+2)(u-3) = 0
u = -2 or u = 3
e^x = -2
This cannot be solved as ln or lg of a negative number is undefined!

The only solution is
e^x = 3
x = ln3 / lne
x = ln3
x = 1.10

We take lne^x
but according to power rule, the power becomes coefficient
so lne^x = xlne
and lne = 1
so
xlne = x

See the notes i posted on logs above !
Click to expand...
hey could you please help me with my doubt.I've posted it twice here but haven't gotten any reply.Since you've been helping a lot of people,I am asking you.
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_ms_32.pdf
On a sketch of an Argand diagram, shade the region whose points represent the complex numbers z which satisfy the inequality |z − 3i | ≤ 2. Find the greatest value of arg z for points in this region.
The question asks about greatest value of argument,sadly I don't even know the least argument value so could you just explain it to me or better if you are of any links,will u let me know?
Please do reply soon.thanks
 
R

reina81

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smzimran said:
e^x = 1 + 6e^−x
Using u = e^x
The equation becomes
u = 1 + 6/u
u^2 = u + 6
u^2 - u - 6 = 0
(u+2)(u-3) = 0
u = -2 or u = 3
e^x = -2
This cannot be solved as ln or lg of a negative number is undefined!

The only solution is
e^x = 3
x = ln3 / lne
x = ln3
x = 1.10

We take lne^x
but according to power rule, the power becomes coefficient
so lne^x = xlne
and lne = 1
so
xlne = x

See the notes i posted on logs above !
Click to expand...

But why didnt we take lne^3 as well?
 
smzimran

smzimran

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user said:
smzimran Assalamoalaikum wr wb!

I need help with the last part of the last question of Nov:2011 # 3

mark scheme

I do know how to do it...but i cant seem to get the answer...! :unsure:

it's like we take the root of − 2 ± 2 root 3i

how i do is...equate that to (a + bi)^2 but..if we simplify...i get a^2 + b^2 = -2.....and this cant be solved...ms uses a^2 - b^2.....can u tell me when exactly do we use '-' b^2 ?
Click to expand...

lknkin.png
 
smzimran

smzimran

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2pac said:
hey could you please help me with my doubt.I've posted it twice here but haven't gotten any reply.Since you've been helping a lot of people,I am asking you.
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_ms_32.pdf
On a sketch of an Argand diagram, shade the region whose points represent the complex numbers z which satisfy the inequality |z − 3i | ≤ 2. Find the greatest value of arg z for points in this region.
The question asks about greatest value of argument,sadly I don't even know the least argument value so could you just explain it to me or better if you are of any links,will u let me know?
Please do reply soon.thanks
Click to expand...
p.m me your email adress i will 4rward you the notes on least and greatest values of arguments and all
 
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reina81

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smzimran said:
Why would we ?
Click to expand...
ok so does this mean that any time we have a question like this, you know the last part, and i have to take ln of both sides, i simply add it to whatever constant they've given? what if its an equation like 2x or something?
 
smzimran

smzimran

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reina81 said:
ok so does this mean that any time we have a question like this, you know the last part, and i have to take ln of both sides, i simply add it to whatever constant they've given? what if its an equation like 2x or something?
Click to expand...
Plz read my notes on logs above
 
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2pac

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user said:
2pac
check the first post of this thread there are some links to notes for that..
Click to expand...
Hey first of all thanks for replying and those notes have really helped me but sadly there is no mention of greatest value of argument there.
 
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