VCE Specialist 3/4 Question Thread!

[Special At Specialist]

Can someone please explain what this question means?
"Prove that the median to the base of an isosceles triangle is perpendicular to the base".

What is the "median to the base"?

[dc302]

Can someone please explain what this question means?
"Prove that the median to the base of an isosceles triangle is perpendicular to the base".

What is the "median to the base"?

http://en.wikipedia.org/wiki/Median_(geometry) There's a nice picture too.

[abd123]

Hey, guys i really got stuck in this question :/, i tried yesterday night and this morning









this is where i got stuck in, please help

[dc302]

Question must be wrong. What does the answer say?

[dc302]

This question has been puzzling me for a while :/.

Part of the graph with equation   is shown below.

Find the area that is bounded by the curve and the x-axis. Give your answer in the form
, where a, b and c are the integers.

Do you know how to do linear substitution?







edit: hint: let u = x+1

[Planck's constant]

This question has been puzzling me for a while :/.

Have you tried the substitution u=x+1 ?


.

[dc302]

Yep, so you would sub in u=x+1 and expand it out. You should get some fractional powers.

The a,b,c refers to your answer. Your answer will be in this form. For example, if I found the area to be 2sqrt(5)/3, then a=2, b=5 and c=3. Of course, these combination of numbers should be in simplest form.

[brightsky]

- int^(1)_(-1) (x^2 - 1)sqrt(x+1) dx
let u = sqrt(x+1), du = 1/(2sqrt(x+1)) dx
so the integral becomes:
- int^(sqrt(2))_(0) (2u^2*u^2(u^2-2)) dx
= -2 int^(sqrt(2))_(0) (u^6 - 2u^4) dx
= -2 [u^7/7 - 2u^5/5]^(sqrt(2))_(0)
= -2 (8sqrt(2)/7 - 2*(4sqrt(2))/5)
= -16sqrt(2)/7 + 16sqrt(2)/5
= (-80sqrt(2) + 112sqrt(2))/35
= 32sqrt(2)/35

hopefully didn't make any careless mistakes..

[abd123]

Thanks Guys, I will put this up incase for someone who wants to tackle a question like this later on .

















=

[Planck's constant]

Perfect.


(PS. Just small thing about the limits of integration. Get into the early habit of not writing,

x=-1, u=0
x=1, u=2

but

x=1, u=2
x=-1, u=0

so that you dont have to do mental handstands to make sure that they are the right way up.
Yes, I know it sounds like weird advice ... but there you have it    )

[mr.politiks]

Hey, guys i really got stuck in this question :/, i tried yesterday night and this morning









this is where i got stuck in, please help

The integrand is not defined for x=-1 (vertical asympotote), hence definite integral with x=-1 is undefined

[Dominatorrr]

COMPLEX NUMBERS

If z=x+yi find the values of x and y such that (z-1)/(z+1)=z+2

Pretty simple question you would think, I know the process in how to do it but I'm just stuck in the algebra.

[brightsky]

z - 1 = (z+1)(z+2) = z^2 + 3z + 2
z^2 + 2z + 3 = 0
z^2 + 2z + 1 -1 + 3 =0
(z + 1)^2 = 2i^2
z + 1 = +- sqrt(2) i
z = -1 +- sqrt(2) i
hence x = -1 and y = +- sqrt(2)

[Dominatorrr]

Makes sense, thanks.

[Bhootnike]

http://imageshack.us/photo/my-images/528/spec2.jpg


Could someone please help me on this one?!