HSC 2015 MX1 Marathon (archive) (2 Viewers)

braintic · Jul 6, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
$$ Show that : $ \\\ \binom{2n}{0}^2 -\binom{2n}{1}^2+\binom{2n}{2}^2 -...+\binom{2n}{2n}^2 = (-1)^n \cdto \binom{2n}{n}$

Can anyone come up with a combinatoric proof for this one?

Speed6 · Jul 7, 2015

Re: HSC 2015 3U Marathon

Integrand is a computer himself

Drsoccerball · Jul 7, 2015

Re: HSC 2015 3U Marathon

braintic said:
Can anyone come up with a combinatoric proof for this one?

Can you do this ?

leehuan · Jul 7, 2015

Re: HSC 2015 3U Marathon

(i)
$\int _{ 0 }^{ x }{ { \left( 1+x \right) }^{ n }dx } =\int _{ 0 }^{ x }{ \sum _{ k=0 }^{ n }{ \left( \begin{matrix} n \\ k \end{matrix} \right) { x }^{ k } } dx } \\ { \left[ \frac { { \left( 1+x \right) }^{ n+1 } }{ n+1 } \right] }_{ 0 }^{ x }={ \left[ \sum _{ k=0 }^{ n }{ \left( \begin{matrix} n \\ k \end{matrix} \right) \frac { { x }^{ k+1 } }{ k+1 } } \right] }_{ 0 }^{ x }\\ \frac { { \left( 1+x \right) }^{ n+1 }-{ \left( 1+0 \right) }^{ n+1 } }{ n+1 } =\sum _{ k=0 }^{ n }{ \frac { 1 }{ k+1 } \left( \begin{matrix} n \\ k \end{matrix} \right) { x }^{ k+1 } } -\sum _{ k=0 }^{ n }{ \frac { 1 }{ k+1 } \left( \begin{matrix} n \\ k \end{matrix} \right) { 0 }^{ k+1 } } \\ \frac { { \left( 1+x \right) }^{ n+1 }-1 }{ n+1 } =\sum _{ k=0 }^{ n }{ \frac { 1 }{ k+1 } \left( \begin{matrix} n \\ k \end{matrix} \right) { x }^{ k+1 } } \\ $Q.E.D.$$

davidgoes4wce · Jul 7, 2015

Re: HSC 2015 3U Marathon

Sy123 said:
The first helpful thing to notice, is that the range of the cosine inverse function is between $0$ and $\pi$ . This means we can't do the old method of cancelling out the inverse cosine and the cosine because this would give us $\pi + \alpha$ which since $\alpha$ is acute would not be valid since it is outside the range.

However, we can still try to apply this by manipulating the expression. First, to give a proper definition of the trick, remember that:

$\cos^{-1}(\cos x) = x \ $for$ \ 0 \leq x \leq \pi \ (*)$

This is true precisely because of the definition of an inverse function, and precisely because that domain $0 \leq x \leq \pi$ is the domain of the original $y = \cos x$ function that we wish to 'invert'.

So proceeding from this, we want to manipulate the given expression into one in which we can use $(*)$ .

Remember that, $\cos (\pi + \alpha) = -\cos \alpha$ and $\cos (\pi - \alpha) = - \cos \alpha$ , which means $\cos(\pi - \alpha) = \cos(\pi + \alpha)$
(To see this fact more clearly, imagine drawing horizontal lines in a y = cos x graph below the x-axis, and see that when the line intersects one part of the cosine graph, it intersects the opposite side, symmetrical to $\pi$ )

So that, $\cos^{-1}(\cos (\pi + \alpha)) = \cos^{-1} (\cos (\pi - \alpha))$

$\\ $But since$ \ 0 \leq \pi - \alpha \leq \pi \ $since$ \ \alpha \ $is acute$$

$\\ \therefore \ \cos^{-1}(\cos (\pi - \alpha)) = \pi - \alpha \ $by using$ \ (*)$

$\\ \therefore \ \cos^{-1}(\cos (\pi + \alpha)) = \pi - \alpha$

After having a more deeper thought about this question last night. I decided to draw it out on the quadrant circle

Now I know a function like cos^(-1) (Cos x) exists for all real x. But in this question we should be looking at the domain as Sy123 said between 0 and Pi.

My thinking regards to this question is based on the trig identity we can derive '- cos a ' from the expansion (sorry I dont know how to write alpha on here just yet). From this, Pi subtract alpha can be derived from the unit circle graph at the solution must be between that range of 0 to Pi, which in this case is Pi-alpha.

Crisium · Jul 7, 2015

Re: HSC 2015 3U Marathon

Differentiate y = x^1/2 using first principles

I've done questions similar to these and I find that the only way to approach it is by rationalising the numerator (Once I've inserted everything into the first principles formula)

Can somebody confirm if there are any other ways to do these types of questions?

davidgoes4wce · Jul 7, 2015

Re: HSC 2015 3U Marathon

I am having trouble drawing this diagram for circle geometry.

"AB is a chord of a circle with centre O. The bisector of angle OAB meets the circle at D. Prove that OD||AB

I had a go at it.

rand_althor · Jul 7, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I am having trouble drawing this diagram for circle geometry.

"AB is a chord of a circle with centre O. The bisector of angle OAB meets the circle at D. Prove that OD||AB

Your diagram is correct, but why is angle DBA 90 degrees?

davidgoes4wce · Jul 7, 2015

Re: HSC 2015 3U Marathon

I think I should remove it.

I think I was getting my terms between perpendicular and bisector confused.

leehuan · Jul 7, 2015

Re: HSC 2015 3U Marathon

Triangle OAD is isosceles with OA = OD (equal radii) so OAD = ODA (base angles)
But OAD = BAD from our question
So BAD = ODA

Hence OD||AB (alternate angles equal)

Speed6 · Jul 7, 2015

Re: HSC 2015 3U Marathon

Crisium said:
Differentiate y = x^1/2 using first principles

I've done questions similar to these and I find that the only way to approach it is by rationalising the numerator (Once I've inserted everything into the first principles formula)

Can somebody confirm if there are any other ways to do these types of questions?

Bump

leehuan · Jul 7, 2015

Re: HSC 2015 3U Marathon

Not that I know of for d/dx sqrt(x) by first principles.

Crisium · Jul 7, 2015

Re: HSC 2015 3U Marathon

leehuan said:
Not that I know of for d/dx sqrt(x) by first principles.

Thanks for clarifying!

I was just curious as to whether there were quicker methods or not (It's still a relatively quick method though)

Chris_S · Jul 9, 2015

Re: HSC 2015 3U Marathon

Hey everyone I am just wandering how common are inverse functions in the paper? Like how many questions involving them will be in it?

enigma_1 · Jul 9, 2015

Re: HSC 2015 3U Marathon

Chris_S said:
Hey everyone I am just wandering how common are inverse functions in the paper? Like how many questions involving them will be in it?

Best way to find out is to do the papers

davidgoes4wce · Jul 10, 2015

Re: HSC 2015 3U Marathon

C is any point on a circle with a diameter AB. P and Q are points on the minor arcs AC and BC. Prove that angle APC + angle CQB= 3 right angles.

davidgoes4wce · Jul 10, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce · Jul 10, 2015

Re: HSC 2015 3U Marathon

Im not sure which side to draw the Point Q on.

davidgoes4wce · Jul 10, 2015

Re: HSC 2015 3U Marathon

Ill post the original question here.

rand_althor · Jul 10, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Ill post the original question here.

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