2011 hsc mx1 marathon (1 Viewer)

apollo1 · Oct 20, 2011

muzeikchun852 said:
got everything except the minus 1.

first you expand the binomial. integrate. let x = 7

the minus one is from the constant of integration.

barbernator · Oct 20, 2011

muzeikchun852 said:
got everything except the minus 1.

first you expand the binomial. integrate. let x = 7

u have to remember to evaluate C because u have integrated

barbernator · Oct 20, 2011

post a question muzeikchun

muzeikchun852 · Oct 20, 2011

barbernator said:
u have to remember to evaluate C because u have integrated

apollo1 said:
the minus one is from the constant of integration.

fail me. haha.

barbernator · Oct 20, 2011

heres a nice one from the 1998 HSC paper, integrate, using z=x-1/2

$\int \frac{1}{\sqrt{x(1-x))}}dx$

muzeikchun852 · Oct 20, 2011

barbernator said:
post a question muzeikchun

easy one: -
a stone is thrown horizontally with a velocity of 20ms-1 from the top of a tower 100m high. assuming no air resistance, and that the acceleration due to gravity, g = 10ms-2

find the velocity and direction of the stone on impact with the ground.

nightweaver066 · Oct 20, 2011

apollo1 said:
lol i couldnt get it either. but usually when they have squares of choose that means equating coefficients so i think nightweaver isnt giving us the expansion. the sly bastard.

Haha.. i'm not hiding anything.. that was the whole question.

I got as far as here and i'm unsure what to do or if i did anything wrong.

$(1+x)^n(nx(1+x)^{-1} + 1)(1+\frac{1}{x})^n = c_0^2 + 2c_1^2 + 3c_2^2 + 4c_3^2 + ... + (n+1)c_n^2\text{ .. expanded as far as constants}$

apollo1 · Oct 20, 2011

heres another question. the quadratic equation $ax^{2}+bx+c = 0$ has roots $x = \tan \alpha$ and $x = \tan \beta$ show that:

$\tan ^{2}(\alpha -\beta ) = \frac{b^{2}-4ac}{(a+c)^{2}}$

muzeikchun852 · Oct 20, 2011

barbernator said:
heres a nice one from the 1998 HSC paper, integrate, using z=x-1/2

$\int \frac{1}{\sqrt{x(1-x))}}dx$

$\int \frac{dz}{\sqrt{(z+0.5)(0.5-z)}}\\\\=\int \frac{dz}{\sqrt{(0.25-z^2)}}\\\\=sin^{-1}2z + c\\\\=sin^{-1}(2x-1) + c$

barbernator · Oct 20, 2011

apollo1 said:
heres another question. the quadratic equation $ax^{2}+bx+c = 0$ has roots $x = \tan \alpha$ and $x = \tan \beta$ show that:

$\tan ^{2}(\alpha -\beta ) = \frac{b^{2}-4ac}{(a+c)^{2}}$

I liked that question

barbernator · Oct 20, 2011

muzeikchun852 said:
$\int \frac{dz}{\sqrt{(z+0.5)(0.5-z)}}\\\\=\int \frac{dz}{\sqrt{(0.25-z^2)}}\\\\=sin^{-1}2z + c\\\\=sin^{-1}(2x-1) + c$

correct

barbernator · Oct 20, 2011

good luck all in 4u on monday, while i enjoy my 2u test

apollo1 · Oct 20, 2011

another interesting question:

Screen shot 2011-10-20 at 10.34.31 PM.png

muzeikchun852 · Oct 20, 2011

apollo1 said:
heres another question. the quadratic equation $ax^{2}+bx+c = 0$ has roots $x = \tan \alpha$ and $x = \tan \beta$ show that:

$\tan ^{2}(\alpha -\beta ) = \frac{b^{2}-4ac}{(a+c)^{2}}$

assume that we all know what is the quadratic formula.

$tan(\alpha -\beta )=\frac{ tan\alpha - tan\beta }{1+tan\alpha tan\beta } \\\\=\frac{\frac{\sqrt{b^2-4ac}}{a}}{1-\frac{b^2-4ac-b^2}{4a^2}}\\\\=\frac{\sqrt{b^2-4ac}}{a+c}$
then square it.

muzeikchun852 · Oct 20, 2011

apollo1 said:
another interesting question:
View attachment 23347

i (9,3)
e (14,7)

nightweaver066 · Oct 20, 2011

muzeikchun852 said:
i (9,3)
e (14,7)

i(6, 3)
e(14, 7)

apollo1 · Oct 20, 2011

the second part of the question is the more important one lol.

muzeikchun852 · Oct 20, 2011

apollo1 said:
the second part of the question is the more important one lol.

centre (10,5) ? radius root 20

therefore $(x-10)^2 + (y-5)=20$ ?

apollo1 · Oct 20, 2011

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