Re: HSC 2015 3U Marathon
could someone please explain this to me, last question from 2010 hsc
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i] if you choose one ball, there are two possibilities, a red ball or a blue ball
if you choose two balls, there are three possibilities, 2 reds, 2 blues, or 1 red 1 blue
if you choose three balls, there are four possibilities, 3 reds, 3 blues, 2 red 1 blue, or two blue 1 red
following the pattern, choosing r balls, there are r+1 possibilities
ii] choosing (n-r) balls from n balls is nC(n-r) because groups of balls doesn't work on arrangements
using the rule, nC(n-r) = nCr we get the desired result
iii] the question asks n red balls and n blue balls (identical) and the labelled white balls
a selection of 'r' identical red and blue balls can be chosen in (r+1) ways, form part i]
therefore there are n-r white balls left, chosen in nCr ways part ii]
therefore selecting both can be made in (r+1) * nCr ways
the number of red or blue balls can go from 0 to n, so the total is sum from 0 to n [ (r+1) nCr ]
split that up into two different sums and use the answer from part b
Would it be ok to say ' angles at the circumference are equal when subtended by the same chord AC' (insteading of saying the word arc'? (to express angle AEC and angle ABC)
Sometimes I see some solution say 'chord' or 'arc'.
Yes, arc or chord, same concept
^n )
=x^3 - 6x^2 + 11x - 6$ are $\alpha$, $\beta$ and $\gamma$. Without finding the values of these roots, find the polynomial with roots: \\ \indent (i) $\alpha + \beta$, $\alpha + \gamma$, $\beta + \gamma$ \\ \indent (ii) $\alpha \beta$, $\alpha \gamma$, $\beta \gamma$ \\ \indent (iii) $\frac{1}{\alpha + \beta}$, $\frac{1}{\alpha + \gamma}$, $\frac{1}{\beta + \gamma}$ \\ )
=1 + \frac{x}{1!} + \frac{x(x+1)}{2!} + ... + \frac{x(x+1)...(x+n-1)}{n!}$, for $n=1,2,3,...$ where for $n=1$: $f_1(x)=1 + \frac{x}{1!} = x+1$ which has a zero at $x=-1$. \\ \indent (i) Show that for $n=2$: $f_2(x)=\frac{1}{2!}(x+1)(x+2)$ and state the zeroes of $f_2(x)$. \\ \indent (ii) Hence complete the proof by mathematical induction that the zeroes of the polynomial function $f_n(x)$ are $-1,-2,-3,...,-n$ for $n=1,2,3,...$, that is prove that $f_n(x)=\frac{1}{n!}(x+1)(x+2)(x+3)...(x+n)$ for $n=1,2,3,...$.)
 = \frac{1}{k!} (x+1)(x+2)\ldots (x+k)$.$)
$ was. It wasn't an arbitrary polynomial, it was a polynomial whose leading coefficient was $\frac{1}{k!}$, since $f_k (x)$ is defined in the question as $1+ \frac{x}{1!} + \frac{x(x+1)}{2!} + \cdots + \frac{x(x+1)\ldots (x+k-1)}{k!}$. This clearly is of degree $k$ and has leading coefficient $\frac{1}{k!}$ (see the last term in that sum). So \emph{by definition} of $f_k (x)$, your $a$ (which is the leading coefficient of $f_k (x)$) is actually just $\frac{1}{k!}$.$)
