2008 HSC Quick Question (1 Viewer)

[imoO]

Hey guys,

Can I get solutions to these two questions from the 2008 Maths Extension 1 Paper (located here http://www.boardofstudies.nsw.edu.a...ms/pdf_doc/2008HSC-mathematics-extension1.pdf)

I'm dead for 3-u

Question 1 (e)

Question 3 (a) (ii)

Cheers

 

[lanesy101]

for 1e) use the substitution u = sin x (where x actually equals theta)

|2x-1|<|x-3| ........ i just squared and solved

(2x-1)^2 < (x-3)^2

4x^2 - 4x + 1 < x^2 - 6x + 9
3x^2 + 2x - 8 < 0
(x+2)(3x-4) < 0 ... and so on

 

[imoO]

for 1e) use the substitution u = sin x (where x actually equals theta)

|2x-1|<|x-3| ........ i just squared and solved

(2x-1)^2 < (x-3)^2

4x^2 - 4x + 1 < x^2 - 6x + 9
3x^2 + 2x - 8 < 0
(x+2)(3x-4) < 0 ... and so on

are you sure you can do that?

I thought we need to consider the +ve and -ve aspects of both (2x-1) and (x-3)

 

[Aplus]

For Question 3. (a) (ii) Sketch LHS and RHS on the same axis and then solve

 

[addikaye03]

1e)

int(pi/4-->0) cos@sin^2@ d@

Let u=sin@

du=cos@ d@

Change limits: when @=pi/4, u=1/rt2 when @=0, u=0

int(1/rt2-->0) u^2 du

|(1/3)u^3| (1/rt2-->0)

|(1/3)(1/2rt2)|-0

(1/6)rt2

Alternatively: This can just be done via inspection,

|(1/3)sin^3@| (pi/4-->0)

get the same answer

Q3aii)

|2x-1|<=|x-3|

-(x-3)<=2x-1<=x-3

-x+3<=2x-1<=x-3

-x+4<=2x<=x-2

-1/2(x+4)<=x<=1/2(x-2) #

 

[lanesy101]

are you sure you can do that?

I thought we need to consider the +ve and -ve aspects of both (2x-1) and (x-3)

Yeah you can, the squaring takes away the need to look at the positive and negative aspects