HSC 2015 MX1 Marathon (archive) (5 Viewers)

davidgoes4wce · Jul 10, 2015

Re: HSC 2015 3U Marathon

I should get to know GeoGebra better.

VBN2470 · Jul 10, 2015

Re: HSC 2015 3U Marathon

Construct a circle with diameter AB and let C be a point lying on the circle. Let P and Q lie on the minor arcs AC and BC respectively. Construct the chords AP, PC, CQ and BQ. Construct another chord PB so that $\angle{APQ} = 90^{\circ}$ . Now, since BPCQ is a cyclic quadrilateral, it follows that $\angle{BPC}$ and $\angle{CQB}$ are supplementary angles, hence $\angle{BPC}+\angle{CQB} = 180^{\circ}$ . But, $\angle{APB}+\angle{BPC}=\angle{APC}$ (by construction) and hence $\angle{APC} + \angle{CQB} = \angle{APB}+\angle{BPC}+\angle{CQB} = 90^{\circ} + 180^{\circ} = 270^{\circ}$

Never mind, just realised I got beaten above

davidgoes4wce · Jul 10, 2015

Re: HSC 2015 3U Marathon

I'd actually be curious do they teach GeoGebra in the NSW high schools? I could understand why it would be beneficial to use it. Fitzpatrick, Aus , Curran textbook do make heavy emphasis on using GeoGebra.....

rand_althor · Jul 10, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I'd actually be curious do they teach GeoGebra in the NSW high schools? I could understand why it would be beneficial to use it. Fitzpatrick, Aus , Curran textbook do make heavy emphasis on using GeoGebra.....

It isn't really taught in my school, but two of the maths teachers I've had have used it in class and do recommend us to use it when doing questions.

davidgoes4wce · Jul 11, 2015

Re: HSC 2015 3U Marathon

Anyone can get me started on this one?

VBN2470 · Jul 11, 2015

Re: HSC 2015 3U Marathon

Consider $\frac{\text{d}}{\text{d}x}(1+x)^{50}$ by expanding out $(1+x)^{50}$ and differentiating it w.r.t. x term by term. Substitute x = 1 and the result should appear.

The correct answer is D $50 \times 2^{49} = 2^{50} \times 5^2$

InteGrand · Jul 11, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Anyone can get me started on this one?

That differentiation trick posted by VBN2470 is a good one to know. But even if you don't know it, you can deduce the answer from the given options. I am assuming you know that the sum of binomial coefficients $\binom{n}{k}$ from $k=1$ to $n$ equals $2^n$ (this you should be aware of).

Clearly the given sum is greater than the sum of all $\binom{50}{k}$ 's. The only option greater than $2^{50}$ is option (D), which must therefore be the answer.

davidgoes4wce · Jul 11, 2015

Re: HSC 2015 3U Marathon

Spent half an hour thinking about this question and still couldn't get it.

integral95 · Jul 11, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Spent half an hour thinking about this question and still couldn't get it.

Since BAC = BDC it's a cyclic quadrilateral since the angles subtended from chord BD are equal

DPC = APD = 48 (alternate angles of the parallel lines)

APD = ACB = 48 (angles subtended from the same arc)

therefore ACB is 48

VBN2470 · Jul 11, 2015

Re: HSC 2015 3U Marathon

(A) $48^{\circ}$ ?

Never mind, got beaten above.

davidgoes4wce · Jul 11, 2015

Re: HSC 2015 3U Marathon

This might be another stupid question but ABCD part of a cyclic quadrilateral?
I have drawn a circle around all the 4 points : A,B,C,D and find AP=PC

InteGrand · Jul 11, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
This might be another stupid question but ABCD part of a cyclic quadrilateral?
I have drawn a circle around all the 4 points : A,B,C,D and find AP=PC

$Yes, $ABCD$ is cyclic because $\angle CAB = \angle CBD$ (converse of ``angles in the same segment'').$

davidgoes4wce · Jul 11, 2015

Re: HSC 2015 3U Marathon

Similar to the one before

davidgoes4wce · Jul 11, 2015

Re: HSC 2015 3U Marathon

This is my honest attempt

rand_althor · Jul 11, 2015

Re: HSC 2015 3U Marathon

I haven't really done Binomials yet, but I think you'd expand each term, simplify, then sub in x=1.

InteGrand · Jul 11, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Similar to the one before

$We don't need to use calculus for this one, just the binomial theorem. By the binomial theorem,$

$$(1+x)^{100} - (1-x)^{100}=\left(\binom{100}{0}+\binom{100}{1}x + \binom{100}{2}x^2 + \binom{100}{3}x^3 + \cdots + \binom{100}{99}x^{99} + \binom{100}{100}x^{100} \right)-\left(\binom{100}{0}-\binom{100}{1}x + \binom{100}{2}x^2 - \binom{100}{3}x^3 + \cdots - \binom{100}{99}x^{99} + \binom{100}{100}x^{100} \right)$ (all odd powers of $x$ have a negative coefficient for the expansion of $(1-x)^{100}$, because $(1-x)^{100}=\sum _{k=0} ^{100} \binom{100}{k} (-x)^{k}=\sum _{k=0} ^{100} \binom{100}{k} (-1)^kx^{k}$, and $(-1)^k=-1$ when $k$ is odd, and +1 when $k$ is even)$

$$=2\binom{100}{1}x+2\binom{100}{3}x^3 + 2\binom{100}{5}x^5 + \cdots + 2\binom{100}{97}x^{97} + \binom{100}{99}x^{99}$ (since all terms with even powers of $x$ are cancelled out, whilst those with negative powers of $x$ are not, since a double negative is a positive)$

$In summary, $(1+x)^{100} - (1-x)^{100}=2\binom{100}{1}x+2\binom{100}{3}x^3 + 2\binom{100}{5}x^5 + \cdots + 2\binom{100}{97}x^{97} + 2\binom{100}{99}x^{99}$.$

$Let $x=1$:$

$$2^{100}=2\binom{100}{1}+2\binom{100}{3} + 2\binom{100}{5} + \cdots + 2\binom{100}{97} + 2\binom{100}{99}$ (since $1^k=1$ for any integer $k$).$

$Now subtract $2\binom{100}{1}=200$ from both sides then divide by $2$, noting that $2^{100}\div 2 = 2^{99}$, and we get that the answer is (D).$

leehuan · Jul 11, 2015

Re: HSC 2015 3U Marathon

I wonder if any 3u students can get this out...

$$It is known that $\int _{ a }^{ b }{ f\left( x \right) dx } =\int _{ a }^{ b }{ f\left( u \right) du } \\ $Find $ \int _{ 0 }^{ 1 }{ \frac { \ln { \left( 1+x \right) } }{ 1+{ x }^{ 2 } } dx } $ using the substitution $x=\frac { 1-u }{ 1+u }$

Source: James Ruse 4U trial, 2012 Q16 highlight to read

rand_althor · Jul 12, 2015

Re: HSC 2015 3U Marathon

leehuan said:
I wonder if any 3u students can get this out...

$$It is known that $\int _{ a }^{ b }{ f\left( x \right) dx } =\int _{ a }^{ b }{ f\left( u \right) du } \\ $Find $ \int _{ 0 }^{ 1 }{ \frac { \ln { \left( 1+x \right) } }{ 1+{ x }^{ 2 } } dx } $ using the substitution $x=\frac { 1-u }{ 1+u }$

$\begin{align*}x&= \frac{1-u}{1+u} \\\frac{\mathrm{d}x}{\mathrm{d}u}&=\frac{\frac{d}{dx}(1-u)\times(1+u)-\frac{d}{dx}(1+u)\times(1-u)}{(1+u)^2} \\&= \frac{-(1+u)-(1-u)}{(1+u)^2} \\&= \frac{-1-u-1+u}{(1+u)^2} \\&= \frac{-2}{(1+u)^2} \\\mathrm{d}x&= \frac{-2}{(1+u)^2}\times\mathrm{d}u \\ \int^1_0\frac{\ln{(1+x)}}{1+x^2}\,\mathrm{d}x&=\int^1_0\frac{\ln{(1+\frac{1-u}{1+u}})}{1+(\frac{1-u}{1+u})^2}\times\frac{-2}{(1+u)^2}\,\mathrm{d}u \\ &= -2\int^1_0 \frac{\ln({\frac{1+u+1-u}{1+u}})}{\frac{(1+u)^2+(1-u)^2}{(1+u)^2}}\times\frac{1}{(1+u)^2}\,\mathrm{d}u \\&= -2\int^1_0 \frac{\ln\frac{2}{(1+u)}}{1+2u+u^2+1-2u+u^2} \,\mathrm{d}u \\ &= -2\int^1_0 \frac{\ln2-\ln(1+u)}{2+2u^2} \,\mathrm{d}u \\&= \int^1_0\frac{\ln(1+u)}{1+u^2} - \frac{\ln2}{1+u^2}\,\mathrm{d}u \\&= \int^1_0\frac{\ln(1+x)}{1+x^2}\,\mathrm{d}x - \ln2\int^1_0\frac{1}{1+u^2}\,\mathrm{d}u \\ &\textup{Not sure what to do with the first integral.} \\ &\textup{Have I made a mistake somewhere?} \end{align*}$

RealiseNothing · Jul 12, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$\begin{align*}x&= \frac{1-u}{1+u} \\\frac{\mathrm{d}x}{\mathrm{d}u}&=\frac{\frac{d}{dx}(1-u)\times(1+u)-\frac{d}{dx}(1+u)\times(1-u)}{(1+u)^2} \\&= \frac{-(1+u)-(1-u)}{(1+u)^2} \\&= \frac{-1-u-1+u}{(1+u)^2} \\&= \frac{-2}{(1+u)^2} \\\mathrm{d}x&= \frac{-2}{(1+u)^2}\times\mathrm{d}u \\ \int^1_0\frac{\ln{(1+x)}}{1+x^2}\,\mathrm{d}x&=\int^1_0\frac{\ln{(1+\frac{1-u}{1+u}})}{1+(\frac{1-u}{1+u})^2}\times\frac{-2}{(1+u)^2}\,\mathrm{d}u \\ &= -2\int^1_0 \frac{\ln({\frac{1+u+1-u}{1+u}})}{\frac{(1+u)^2+(1-u)^2}{(1+u)^2}}\times\frac{1}{(1+u)^2}\,\mathrm{d}u \\&= -2\int^1_0 \frac{\ln\frac{2}{(1+u)}}{1+2u+u^2+1-2u+u^2} \,\mathrm{d}u \\ &= -2\int^1_0 \frac{\ln2-\ln(1+u)}{2+2u^2} \,\mathrm{d}u \\&= \int^1_0\frac{\ln(1+u)}{1+u^2} - \frac{\ln2}{1+u^2}\,\mathrm{d}u \\&= \int^1_0\frac{\ln(1+x)}{1+x^2}\,\mathrm{d}x - \ln2\int^1_0\frac{1}{1+u^2}\,\mathrm{d}u \\ &\textup{Not sure what to do with the first integral.} \\ &\textup{Have I made a mistake somewhere?} \end{align*}$

If I said solve for x:

$x=x-(\ln(2))A$

What would you do?

InteGrand · Jul 12, 2015

Re: HSC 2015 3U Marathon

RealiseNothing said:
If I said solve for x:

$x=x-(\ln(2))A$

What would you do?

That doesn't help him much since the desired integral cancels and it would follow that the second integral must be 0 (which it isn't). I think there's a mistake somewhere in the working, but I don't know where since I haven't looked at it carefully.

Edit: found the error — the limits of integration weren't changed.

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what is that?It is Cowpea

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