Few Q from Trial (1 Viewer)

lyounamu · Aug 4, 2008

http://www.boredofstudies.org/courses/maths/3u/1139495242_2004_Mathematics_Extension_1_Trial_Paper.pdf

Can anyone help me with Q 3 c) ii)?

Also, can anyone help me with Q7 b)?

Thanks in advance!

Aerath · Aug 4, 2008

Only could do half of 7b.

midifile · Aug 4, 2008

question 3c ii)

In part A you should get A=2 and B=3
Therefore 3(2sinx + cosx) + 2(2cosx - sinx) = sinx + 8cosx

In the intergral sub 3(2sinx + cosx) + 2(2cosx - sinx) for the sinx + 8cos x on the top

Im gonna use I to represent the integral sign to make it clearer

I(sinx + 8cosx)/(2sinx + cosx) dx
=I(2(2sinx + cosx) + 3(2cosx - sinx))/(2sinx + cosx) dx
=I((2(2sinx + cosx)/(2sinx + cosx)) + (3(2cosx -sinx))/(2sinx + cosx)) dx
=I(2 + (3(2cosx -sinx))/(2sinx + cosx)) dx
= 2x + 3ln(2sinx + cosx) + C

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midifile · Aug 4, 2008

Question 7b.

For circle C2

You know that the circle touches the x axis.
So let y = 0 and you get
x^2 - 6x + 3k = 0
∆ = 36 - 12k
But because it only touches the x axis (does not pass through)
∆ = 0
therefor k = 3
Sub this into the equation and you get
x^2 + y^2 - 6x + 6y + 9 = 0

Aerath · Aug 5, 2008

midifile said:
Question 7b.

For circle C2

You know that the circle touches the x axis.
So let y = 0 and you get
x^2 - 6x + 3k = 0
∆ = 36 - 12k
But because it only touches the x axis (does not pass through)
∆ = 0
therefor k = 3
Sub this into the equation and you get
x^2 + y^2 - 6x + 6y + 9 = 0

Hmmm, nice work.

Is this sorta question 2 or 3U?

midifile · Aug 5, 2008

Aerath said:
Is this sorta question 2 or 3U?

No idea. I'm assuming 3 unit coz its in a 3u test, but its probably considered a harder 2u question.

lyounamu · Aug 5, 2008

Thanks.

Few Q from Trial (1 Viewer)

lyounamu

Reborn

Aerath

Retired

midifile

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midifile

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Aerath

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midifile

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lyounamu

Reborn

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