# Thread: Cambridge HSC MX1 Textbook Marathon/Q&A

1. ## 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

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

Originally Posted by Mongose528
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

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

How do I do this

Also

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

help.png

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

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

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

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

Originally Posted by infiniteeee
help.png

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... :P

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

Originally Posted by eyeseeyou
Also

Cosine Rule XOR Sine Rule

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

Originally Posted by eyeseeyou
How do I do this

...seriously just draw a diagram

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

Originally Posted by infiniteeee
help.png

anyone who can help me with question 12b and 13a of exercise 4F?
EX 4F Q13 a

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Originally Posted by 123ryoma12
$\noindent Recall that the binomial coefficients are symmetric about the central binomial coefficient \binom{2n}{n}. Hence if we take the sum of coefficients from r=0 to n-1, it's half what we'd get if we took the sum of \textsl{all} \binom{2n}{r} from r = 0 to 2n excluding the middle term r=n. In other words, \sum _{r=0}^{n-1} \binom{2n}{r} = \frac{1}{2}\left( \sum _{r=0}^{2n} \binom{2n}{r} - \binom{2n}{n}\right). Now, \sum _{r=0}^{2n}\binom{2n}{r} = 2^{2n} (by subbing x=1 into the binomial expansion), so we have \sum _{r=0}^{n-1}\binom{2n}{r} = \frac{1}{2}\left(2^{2n} - \binom{2n}{n}\right). Thus$

\begin{align*}\sum _{r=0}^{n}\binom{2n}{r} &= \frac{1}{2}\left(2^{2n} - \binom{2n}{n}\right) + \binom{2n}{n} \\ &= \frac{1}{2}\left(2^{2n} + \binom{2n}{n}\right)\\ &= 2^{2n-1} + \frac{(2n)!}{2(n!)^{2}}.\end{align*}

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

Hi, is anyone able to do question 17 d from exercise 10E ?
Thank you!
Capture.PNG

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

If N is to the left of U, the answers must all be in the form ____N____U____, where the underscores represent places the remaining four letters can go. So:
Consider _N_U_, where we are trying to insert M. There are 3 places for it to go.
Now consider what will be left, we have: _N_U_M_, or _N_M_U_, or _M_N_U_. Regradless of where we place the M, there are four places for the next letter.
Following this, wherever we place the next letter will take away one available space, but create two either side of it, netting one more. So for M there are 3, for B there are 4, for E there are 5, and for R there are 6.
Doing it this way gives an answer of 3*4*5*6=360, and with this method we cover the event of placing the letters down in different orders.

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

http://prnt.sc/dfvluh pls halp. what does this ques mean

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

It's asking what values of a, b, and c are required to change the degree of the polynomial. For example, if you had a polynomial $ax^3+bx^2+cx+d$, and you wanted a polynomial of degree 1, a and b would both have to be zero, so that the polynomial became $cx+d$, with the degree of a polynomial being the highest exponent.

17. ## Re: Cambridge HSC MX1 Textbook Marathon/Q&A

Can anyone help me question 23 of Exercise 1 E

18. ## Re: Cambridge HSC MX1 Textbook Marathon/Q&A

Originally Posted by sleepywood
Can anyone help me question 23 of Exercise 1 E
Yr 11 or Yr 12 text? Edition? Is it the one involving y = sin-1 x ??

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