HSC 2015 MX1 Marathon (archive) (1 Viewer)

davidgoes4wce · Jul 22, 2015

Re: HSC 2015 3U Marathon

Find 'n' if the coefficients of the second, third and fourth terms in the expansion of (1+x)^n are successive terms of an arithmetic sequence.

integral95 · Jul 22, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Find 'n' if the coefficients of the second, third and fourth terms in the expansion of (1+x)^n are successive terms of an arithmetic sequence.

The property of an arithmetic series is $T_n - T_{n-1} = T_{n-1} - T_{n-2}$

so you solve nC4 - nC3 = nC3 - nC2. The opposite equation nC2 - nC3 = nC3 - nC4 would yield the same answer as you multiplied -1 on both sides

P.S: Bit curious as why you're on so early haha

davidgoes4wce · Jul 22, 2015

Re: HSC 2015 3U Marathon

Now Im not sure if there is an error in the text solutions.

my text book says the solution to that is angle XBT=angle YXA. I think this is wrong. Could someone confirm this?

InteGrand · Jul 22, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Now Im not sure if there is an error in the text solutions.

my text book says the solution to that is angle XBT=angle YXA. I think this is wrong. Could someone confirm this?

Your answer is correct. The textbook solutions probably just had a typo (T and Y are next to each other on a QWERTY keyboard).

davidgoes4wce · Jul 22, 2015

Re: HSC 2015 3U Marathon

integral95 said:
P.S: Bit curious as why you're on so early haha

I was bored and wanted to do more 3 Unit maths questions because I have nothing to do at 6am in the morning. (Yes I know I am a bit weird!)

davidgoes4wce · Jul 22, 2015

Re: HSC 2015 3U Marathon

integral95 said:
The property of an arithmetic series is $T_n - T_{n-1} = T_{n-1} - T_{n-2}$

so you solve nC4 - nC3 = nC3 - nC2. The opposite equation nC2 - nC3 = nC3 - nC4 would yield the same answer as you multiplied -1 on both sides

P.S: Bit curious as why you're on so early haha

I have just tried your method, the answer they have in the book is n=7.

Using your method 7C4-7C3= 0
7C3-7C2=14

so something is not quite right with your working...or i am not understand it right

dan964 · Jul 22, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I have just tried your method, the answer they have in the book is n=7.

Using your method 7C4-7C3= 0
7C3-7C2=14

so something is not quite right with your working...or i am not understand it right

correction:

reason: the first coeffecient is 7C0 not 7C1 thus the fourth coefficient is 7C3 not 7C4
7C3=35
7C2=21
7C1=7
(an arithmetic sequence of a=7, d=14)

davidgoes4wce · Jul 22, 2015

Re: HSC 2015 3U Marathon

someone remind me how to isolate the subject 'x' by itself.

I am not 100% sure how to do it.

davidgoes4wce · Jul 22, 2015

Re: HSC 2015 3U Marathon

Im doing a volume of solids questions but not sure how to transpose the x terms.

kawaiipotato · Jul 22, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
someone remind me how to isolate the subject 'x' by itself.

I am not 100% sure how to do it.

$y = \frac{4}{\pi}sin^{-1}x$

$\pi y = 4sin^{-1}x$

$\frac{\pi y}{4} = sin^{-1}x$

$sin(\frac{\pi y}{4}) = x $ Taking the sine of both sides $$

davidgoes4wce · Jul 23, 2015

Re: HSC 2015 3U Marathon

Grace has ten paintings, six of them are oil paintings and the rest are water colour paintings.

She wants to select at random four paintings to give as a gift.

What is the probability that she will select no more than two oil paintings?

davidgoes4wce · Jul 23, 2015

Re: HSC 2015 3U Marathon

Here is my attempt

kawaiipotato · Jul 23, 2015

Re: HSC 2015 3U Marathon

Is the answer
[6C0 * 4C4 + 6C1 * 4C3 + 6C2 * 4C2] all over 10C4

davidgoes4wce · Jul 23, 2015

Re: HSC 2015 3U Marathon

you are right 23/42 is the answer

davidgoes4wce · Jul 23, 2015

Re: HSC 2015 3U Marathon

Pretty hard to distinguish from using a binomial and using combinations . I wouldn't have predicted using combinations first up

Sy123 · Jul 24, 2015

Re: HSC 2015 3U Marathon

$\\ $i) By writing$ \ (1+x)^2 = 1 + x(2+x) \ $show that$ \ \binom{4n}{2n} \ $is equal to the coefficient of$ \ x^{2n} \ $in the expression$ \ \sum_{k=0}^{2n} \binom{2n}{k}x^k(2+x)^k \\ \\ $ii) Hence show that$ \\\\ \binom{4n}{2n} = \binom{2n}{n}\binom{n}{n} + 2\binom{2n}{n+1} \binom{n}{n-1} + 2^2 \binom{2n}{n+2}\binom{n}{n-2} + \dots + 2^n \binom{2n}{2n} \binom{n}{0}$

davidgoes4wce · Jul 26, 2015

Re: HSC 2015 3U Marathon

Not sure if this has been used in answering geometry questions . I know we have the reason 'Sum of the angles of a triangle is 180 degrees'

Could we have one for a quadrilateral , saying the 'Sum of the quadrilateral adds up to 360 degrees'?

InteGrand · Jul 26, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Not sure if this has been used in answering geometry questions . I know we have the reason 'Sum of the angles of a triangle is 180 degrees'

Could we have one for a quadrilateral , saying the 'Sum of the quadrilateral adds up to 360 degrees'?

Yes, angle sum of a quadrilateral is 360º (and you can use this in proofs if you need to).

davidgoes4wce · Jul 27, 2015

Re: HSC 2015 3U Marathon

I made this question up because I am bored of studies.

Find the directrix and Focus.

I know the formula for a parabola is (y-k)^2=4a(x-h)^2

where a =1

Knowing that the Vertex is (8,3)

Directrix x=7
Focus (9,3)

rand_althor · Jul 27, 2015

Re: HSC 2015 3U Marathon

$$Use the substitution $u=\frac{1}{\sqrt{3}}\tan{x}$ to evaluate $\int^{\frac{\pi}{4}}_0 \frac{\mathrm{d}x}{3-2\sin^2{x}}$

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