Cambridge HSC MX1 Textbook Marathon/Q&A (2 Viewers)

davidgoes4wce · Jul 18, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

I know from Physics formulae with the equations of motions we have:

$v^2=u^2+2as$

$s=ut+\frac{1}{2} at^2$

I'm just struggling to find the link between

$180= V \times t$

InteGrand · Jul 18, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
I know from Physics formulae with the equations of motions we have:

$v^2=u^2+2as$

$s=ut+\frac{1}{2} at^2$

I'm just struggling to find the link between

$180= V \times t$

$\noindent Do you mean how did I arrive at $180 = Vt$ at the time of collision? This was done by equating the two equations of motion and cancelling common terms from each side.$

davidgoes4wce · Jul 18, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

OK I get it now.

I treated the origin at the base of the cliff.

$y_B=180-5t^2$

Where I went wrong was I assumed the stone had an initial velocity of 0, the question stated it was V.

$y_S=Vt-5t^2$

$By equating $ y_S = y_B $ I can then get the equation of 180=Vt$

davidgoes4wce · Jul 18, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

If I treated the top of the lookout as the mean line, my equations would obviously be different.

$y_B=5t^2$

$y_S=-Vt+5t^2+180$

(treating anything downwards as positive g=+10)

$$ By equating $ y_B=y_S , 5t^2=-Vt+5t^2+180$

$Vt=180$

AfroNerd · Jul 30, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

having trouble with this binomial question

HeroicPandas · Jul 31, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

AfroNerd said:
having trouble with this binomial question
View attachment 33393

$(1+x)^{n-1} = \binom{n-1}{0} + \binom{n-1}{1}x + \dots + \binom{n-1}{n-2} x^{n-2} + \binom{n-1}{n-1}x^{n-1} \\ $Put $x = 1, \\\\ 2^{n-1} = \binom{n-1}{0} + \underbrace{\binom{n-1}{1} + \dots + \binom{n-1}{n-2}}_{\text{isolate this}} + \binom{n-1}{n-1} \\\\ \begin{align*} \binom{n-1}{1} + \dots + \binom{n-1}{n-2} &= 2^{n-1} - \binom{n-1}{0} - \binom{n-1}{n-1} \\&= 2^{n-1} - 1 -1 \end{align*}$

Then multiply both sides by n

si2136 · Aug 2, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

I think there is a formula for this but I'm not sure what it's called.

Q: Four out of the seven letters of the word COCONUT are selected and arranged in a row. How many different arranges are possible?

InteGrand · Aug 2, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

si2136 said:
I think there is a formula for this but I'm not sure what it's called.

Q: Four out of the seven letters of the word COCONUT are selected and arranged in a row. How many different arranges are possible?

You may wish to consider cases based on things like how many ''double-letters'' are selected.

si2136 · Aug 2, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
You may wish to consider cases based on things like how many ''double-letters'' are selected.

I know how to do it if it was all of the letters, but not 4 out of the 7

InteGrand · Aug 2, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

si2136 said:
I know how to do it if it was all of the letters, but not 4 out of the 7

Yeah, that's why you may find it helpful to consider cases. E.g. The cases could be: no pairs of double-letters; exactly one pair of double-letters; exactly two pairs of double-letters.

Then count the no. of possibilities in each case and sum them up.

Mongoose528 · Aug 3, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Isn't the formula this?

For the number of ways to arrange n objects taken r at at a time

n!/(n-r)!

So it would be

7!/(7-4)! = 7!/3! = 7*6*5*4*3!/3! = 7*6*5*4=840

But since we have 2 c's and 2 o's, the answer would be 840/2!2! = 840/4 =210

si2136 · Aug 6, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Mongose528 said:
Isn't the formula this?

For the number of ways to arrange n objects taken r at at a time

n!/(n-r)!

So it would be

7!/(7-4)! = 7!/3! = 7*6*5*4*3!/3! = 7*6*5*4=840

But since we have 2 c's and 2 o's, the answer would be 840/2!2! = 840/4 =210

Wrong, you would need to take it by cases or Pork's Insertion Method

eyeseeyou · Aug 10, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

How do I do this

eyeseeyou · Aug 10, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Also

infiniteeee · Sep 27, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

anyone who can help me with question 12b and 13a of exercise 4F?

Drongoski · Sep 27, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

$\therefore x^3 - Ax + 0.x + 3A = 0\\ \\ \therefore $ sum of roots $ = \alpha + \beta + (\alpha + \beta) = -(-A) = A \implies 2(\alpha + \beta) = A \implies \alpha + \beta = \frac {1}{2} A \\ \\ $ sum of roots taken 2 at a time $ = \alpha \beta +\alpha (\alpha+\beta) + \beta (\alpha+\beta) = 0 \\ \\ \therefore $ $\alpha \beta = - (\alpha + \beta)^2 = - (\frac {1}{2} A)^2 = -\frac {1}{4} A^2 \\ \\ $product of roots $ = \alpha \beta (\alpha + \beta) -3A \implies \alpha \beta = \frac {-3A}{\frac {1}{2}A} = -6 \implies -\frac {1}{4}A^2 = -6 \\ \\ \implies A = 2\sqrt 6$

kashkow · Sep 27, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

infiniteeee said:
View attachment 33481

anyone who can help me with question 12b and 13a of exercise 4F?

The first question:

The "sum of the products of pairs of roots" refers to the x term which is absent, meaning the coefficient is 0. Therefore the "sum of the products of pairs of roots" = 0

Algebraically: ab + a(a+b) + b(a+b) = 0 [... a = alpha, b = beta]

Expanding: ab + a^2 + ab + ab + b^2 = 0

ab + a^2 + 2ab + b^2 = 0

Combining the perfect square:

ab + (a+b)^2 = 0

And you can use the results from the question before 12a what (a+b)^2 is.

That is substitute (a+b) = 1/2*A = A/2

ab + (A/2)^2 = 0

ab + (A^2)/4 = 0

Therefore: ab = -(A^2)/4 = -1/4 * A^2

Sorry if it's hard to read. Rats i was late again...

Paradoxica · Sep 27, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

eyeseeyou said:
Also

Cosine Rule XOR Sine Rule

Paradoxica · Sep 27, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

eyeseeyou said:
How do I do this

...seriously just draw a diagram

davidgoes4wce · Sep 30, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

infiniteeee said:
View attachment 33481

anyone who can help me with question 12b and 13a of exercise 4F?

EX 4F Q13 a

$Let the three roots be \alpha, \alpha \ $ and $ \beta$

$2 \alpha + \beta =\frac{8}{4}=2 \ \textcircled{1}$

$\alpha \beta +\alpha \beta +\alpha^2 =\frac{-3}{4} \implies 2 \alpha \beta +\alpha^2=\frac{-3}{4} \textcircled{2}$

$\alpha^2 \beta =\frac{-9}{4} \textcircled{3}$

$$ Rearranging \textcircled{1} \ \beta=2-2\alpha$

$$ Substituting $ \textcircled{1} into \textcircled{2}$

$2 \alpha (2-2 \alpha)+\alpha^2=\frac{-3}{4}$

$4 \alpha -4 \alpha^2+\alpha^2=\frac{-3}{4}$

$-3 \alpha^2+4\alpha=\frac{-3}{4}$

$-12 \alpha^2+16 \alpha =-3 \implies 0=12\alpha^2-16\alpha-3$

$$ By using the Quadratic Roots formula $ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} $ we look to solve $ \alpha$

$\alpha=\frac{16 \pm \sqrt{256-4(12)(-3)}}{2 \times 12}$

$\alpha= \frac{3}{2} \ and \ \alpha=\frac{-1}{6}$

$if \alpha =\frac{3}{2} \beta=-1$

$$ Substitute $ \alpha =\frac{3}{2} \ $ into equation $ \textcircled{1} \ \beta=-1$

$$ Substitute $ \alpha =\frac{3}{2} \ $ into equation $ \textcircled{3} \ \beta=-1$

$Therefore, $ \alpha =\frac{3}{2} $ and $ \beta=-1 $ are consistent solutions$

$$ if $ \alpha =\frac{-1}{6}$

$$ Substitute $ \alpha =\frac{-1}{6} \ $ into equation $ \textcircled{1} \ \beta=\frac{7}{3}$

$$ Substitute $ \alpha =\frac{-1}{6} \ $ into equation $ \textcircled{3} \ \beta=-81$

$Therefore, $ \alpha =\frac{-1}{6} $ is not consistent and is not a solution to the cubic polynomial.$

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