# Thread: HSC 2017 MX1 Marathon

1. ## Re: HSC 2017 MX1 Marathon

Originally Posted by leehuan
$\text{Prove that, for non-zero constants }a,b\\\ \ \sin \left(x+\tan^{-1}\frac{b}{a}\right)=\cos \left(x-\tan^{-1}\frac{a}{b}\right)$
$let \tan \theta=\frac{a}{b}, \tan \phi =\frac{b}{a} \\ \\\tan(\theta + \phi)=\frac{\tan \theta + \tan \phi}{1-\tan \theta \tan \phi} \\ = \frac{\frac{a}{b}+\frac{b}{a}}{1-1}= undefined/ \infty \\ \Rightarrow \theta + \phi = \frac{\pi}{2}\\ \therefore \tan^{-1}\frac{a}{b}+\tan^{-1}\frac{b}{a}=\frac{\pi}{2} (\ast) \\ \\ LHS = \sin\left (x+\tan^{-1}\frac{b}{a} \right ) \\ = \cos\left ( \frac{\pi}{2}-\tan^{-1}\frac{b}{a} -x\right ) \\ = \cos\left ( \tan^{-1}\frac{a}{b}-x \right ) ( from \ast ) \\ =\cos\left ( x-\tan^{-1}\frac{a}{b} \right )\\ =RHS$

2. ## Re: HSC 2017 MX1 Marathon

Originally Posted by jathu123
$let \tan \theta=\frac{a}{b}, \tan \phi =\frac{b}{a} \\ \\\tan(\theta + \phi)=\frac{\tan \theta + \tan \phi}{1-\tan \theta \tan \phi} \\ = \frac{\frac{a}{b}+\frac{b}{a}}{1-1}= undefined/ \infty \\ \Rightarrow \theta + \phi = \frac{\pi}{2}\\ \therefore \tan^{-1}\frac{a}{b}+\tan^{-1}\frac{b}{a}=\frac{\pi}{2} (\ast) \\ \\ LHS = \sin\left (x+\tan^{-1}\frac{b}{a} \right ) \\ = \cos\left ( \frac{\pi}{2}-\tan^{-1}\frac{b}{a} -x\right ) \\ = \cos\left ( \tan^{-1}\frac{a}{b}-x \right ) ( from \ast ) \\ =\cos\left ( x-\tan^{-1}\frac{a}{b} \right )\\ =RHS$
Note that arctan(t) + arctan(1/t) is only equal to pi/2 if t := a/b > 0. If t < 0, then it's -pi/2. The result in the question in the case t < 0 is similar, but with a minus sign in front of one of the trig. functions.

3. ## Re: HSC 2017 MX1 Marathon

The inspiration of that question was actually just the fact

a sin(x) + b cos(x) = b cos(x) + a sin(x)

4. ## Re: HSC 2017 MX1 Marathon

Originally Posted by leehuan
The inspiration of that question was actually just the fact

a sin(x) + b cos(x) = b cos(x) + a sin(x)
Auxiliary! wow thats clever

5. ## Re: HSC 2017 MX1 Marathon

This was the solution:

I had my own solution. With geometry proofs Im aware that there are multiple ways you can go about explaining step by step.

$\angle AED=180-a (supplementary angle)$

$\angle EAB=180-a (alternate angles, AB||DE)$

$\angle ABC= 180-(180-a+c) (Angle sum of a triangle)$

$=a-c$

Just want to check if my working out is right.

6. ## Re: HSC 2017 MX1 Marathon

- a quick inverse trig question
Q. let f(x) = sin^(-1) (x)
a) for what values of m does the line y = mx cut y = f(x) at three points
b) Investigate the concavity of the curve and find the coordinates of the point of intersection

7. ## Re: HSC 2017 MX1 Marathon

Originally Posted by JackPatel
- a quick inverse trig question
Q. let f(x) = sin^(-1) (x)
a) for what values of m does the line y = mx cut y = f(x) at three points
b) Investigate the concavity of the curve and find the coordinates of the point of intersection
a) $1 < m \leq \frac{\pi}{2}$
because $sin^{-1}x$ is an odd function. So to find all possible lines we have to make sure that it passes through the origin, which it always does, and one point on the first quadrant (x,y). and by oddness of $sin^{-1}x$, the line also passes through (-x,-y) as well. therefore
$m \leq \frac{\pi}{2}$
But at x=0 the tangent has a slope of 1 therefore
m > 1

b) Well, just take the second derivative and the intersection points are :
$(0,0),(x,sin^{-1}x), (-x,-sin^{-1}x)$

8. ## Re: HSC 2017 MX1 Marathon

Originally Posted by davidgoes4wce

This was the solution:

I had my own solution. With geometry proofs Im aware that there are multiple ways you can go about explaining step by step.

$\angle AED=180-a (supplementary angle)$

$\angle EAB=180-a (alternate angles, AB||DE)$

$\angle ABC= 180-(180-a+c) (Angle sum of a triangle)$

$=a-c$

Just want to check if my working out is right.
The solution is correct. But your solution is exactly the same as the other one:
In the textbook's solution, they subtracted two angles one at a time, meaning:
$\angle ABC = 180 - \angle ACB - \angle CBA$
But you subtracted the sum in one go, meaning:
$\angle ABC = 180 - (\angle ACB + \angle CBA)$

9. ## Re: HSC 2017 MX1 Marathon

$\textrm{1. Suppose}\ a,b,c \in \mathbb{R} \ \textrm{such that}$

$9a+11b+29c =0$

$\ \textrm{prove } \ ax^{3}+bx^{2}+c = 0, \textrm{has a root between } \ 0 \leq x \leq 2$

$\textrm{2.If polynomial} \ P(x) = ax^{3} + bx+ c \ \textrm{is divisible by} \ G(x) = x^{2} + tx+1$

$\textrm{Find a relation between the coefficients of P(x)}$

10. ## Re: HSC 2017 MX1 Marathon

If you have trouble with Mathematical Induction, can you always expand the equation? It works, but it's just really slow.

11. ## Re: HSC 2017 MX1 Marathon

Originally Posted by si2136
If you have trouble with Mathematical Induction, can you always expand the equation? It works, but it's just really slow.
Expand what equation? Do you have an example?

12. ## Re: HSC 2017 MX1 Marathon

Originally Posted by pikachu975
Expand what equation? Do you have an example?
Well just say you're solving an equation and you're at the RTP step.

You have to prove LHS = RHS, which is

k/6 (k+1)(2k+1) + (k+1)^2 = (k+1)/6 (k+2)(2k+3)

Can you simplify and expand left hand side, then simplify and expand the right hand side, so they're equal?

I'm not moving the equation to the other side, but just simplifying it. It works if you can't figure out how to manipulate the factors. Is it possible to do that though?

13. ## Re: HSC 2017 MX1 Marathon

Originally Posted by si2136
Well just say you're solving an equation and you're at the RTP step.

You have to prove LHS = RHS, which is

k/6 (k+1)(2k+1) + (k+1)^2 = (k+1)/6 (k+2)(2k+3)

Can you simplify and expand left hand side, then simplify and expand the right hand side, so they're equal?

I'm not moving the equation to the other side, but just simplifying it. It works if you can't figure out how to manipulate the factors. Is it possible to do that though?
Yeah, you can. But better to just factor k+1 from the LHS and go from there (saves you from having to expand so much).

14. ## Re: HSC 2017 MX1 Marathon

Originally Posted by Mahan1
$\textrm{1. Suppose}\ a,b,c \in \mathbb{R} \ \textrm{such that}$

$9a+11b+29c =0$

$\ \textrm{prove } \ ax^{3}+bx^{2}+c = 0, \textrm{has a root between } \ 0 \leq x \leq 2$

$\textrm{2.If polynomial} \ P(x) = ax^{3} + bx+ c \ \textrm{is divisible by} \ G(x) = x^{2} + tx+1$

$\textrm{Find a relation between the coefficients of P(x)}$
By relation, I suppose you mean a single equation...

a²=ab+c²

15. ## Re: HSC 2017 MX1 Marathon

Prove the following statement using mathematical induction
$7+77+777+...+77...77=\frac{7}{81}(10^{n+1}-9n-10)\quad\quad\text{where}\quad n\in\mathbb{Z},n\geq2$

The last term on the L.H.S has n-digits

16. ## Re: HSC 2017 MX1 Marathon

Originally Posted by wu345
Prove the following statement using mathematical induction
$7+77+777+...+77...77=\frac{7}{81}(10^{n+1}-9n-10)\quad\quad\text{where}\quad n\in\mathbb{Z},n\geq2$

The last term on the L.H.S has n-digits
$\textrm{The base case is obvious.}$

$\textrm{Assume the statement is true for n=k.}$

$7 + 77 + 777+ \dots + \underset{k \ \textrm{times}}{ 777 \dots 7} = \frac{7}{81}(10^{k+1} - 9k -10 )$

$\textrm{Note} \ 10(7+ 77 +\dots + \underset{k \ \textrm{times}}{ 777 \dots 7} ) + 7k + 7 = 7 + 77 + \dots + \underset{k+1 \ \textrm{times}}{ 777 \dots 7}$

$\textrm{By induction hypothesis }$

$7 + 77 + \dots + \underset{k+1 \ \textrm{times}}{ 777 \dots 7} = 10(7+ 77 +\dots + \underset{k \ \textrm{times}}{ 777 \dots 7} ) + 7k + 7 = 10 \frac{7}{81}(10^{k+1} - 9k -10 ) + 7k + 7$

$7 + 77 + \dots + \underset{k+1 \ \textrm{times}}{ 777 \dots 7} = \frac{7}{81}(10^{k+2} - 90k -100 ) + 7n + 7$

$= \frac{7}{81}(10^{k+2}) - \frac{70}{9}k -\frac{700}{81} + 7n + 7 = \frac{7}{81}(10^{k+2}) - \frac{7}{9}n - \frac{133}{81}$

$= \frac{7}{81}(10^{k+2} - 9n - 19) = \frac{7}{81}(10^{k+2} - 9(n+1) -10)$

$\textrm{That proves it is true for} \ n = k+1$

17. ## Re: HSC 2017 MX1 Marathon

Just need a quick 'yes' or 'no. Would it be correct to say

$\angle ABC=\angle DEC (alternate angles, AB \parallel ED)$ ?

$\angle BAC=\angle EDC (alternate angles, AB \parallel ED)$ ?

I want to get perfect with my reasoning in congruency of triangles.

18. ## Re: HSC 2017 MX1 Marathon

Originally Posted by davidgoes4wce

Just need a quick 'yes' or 'no. Would it be correct to say

$\angle ABC=\angle DEC (alternate angles, AB \parallel ED)$ ?

$\angle BAC=\angle EDC (alternate angles, AB \parallel ED)$ ?

I want to get perfect with my reasoning in congruency of triangles.
Yup, I don't see anything wrong in that and that's the exact way I would do it too.

19. ## Re: HSC 2017 MX1 Marathon

Originally Posted by davidgoes4wce

Just need a quick 'yes' or 'no. Would it be correct to say

$\angle ABC=\angle DEC (alternate angles, AB \parallel ED)$ ?

$\angle BAC=\angle EDC (alternate angles, AB \parallel ED)$ ?

I want to get perfect with my reasoning in congruency of triangles.
Yep it's perfect and it's also nice that you took into account the order in which you named the angles, as you named them correspondingly for each triangle while the book just did any order.

20. ## Re: HSC 2017 MX1 Marathon

Originally Posted by pikachu975
Yep it's perfect and it's also nice that you took into account the order in which you named the angles, as you named them correspondingly for each triangle while the book just did any order.
Different books do it slightly differently for some reason. I did all of the Year 10 5.3 Questions (Congruency triangles & Circle Geometry) from Cambridge text book and they did it in the a chronological order.

I also had a look at the Oxford textbook and their method of reasoning and the way they write it , I noticed it is slightly different to the Cambridge text.I have a personal preference towards the way they write in Cambridge.

21. ## Re: HSC 2017 MX1 Marathon

I also do notice a huge overlap in 5.3 Maths Year 10 and Year 11 Prelim Extension 1 maths. (Trig, Circle Geometry, Polynomials in particular)

22. ## Re: HSC 2017 MX1 Marathon

Consider the five functions:

$f_{1} (x)=x \ f_{2} (x)=x^2 \ f_{3} (x)=1 \ f_{4} (x)=x^3+x \ f_{5} (x) = cos (x) +x^4$

How many are even functions and how many are odd functions?

I was a bit unsure about $f_{3} (x)$ ?

23. ## Re: HSC 2017 MX1 Marathon

Originally Posted by davidgoes4wce
Consider the five functions:

$f_{1} (x)=x \ f_{2} (x)=x^2 \ f_{3} (x)=1 \ f_{4} (x)=x^3+x \ f_{5} (x) = cos (x) +x^4$

How many are even functions and how many are odd functions?

I was a bit unsure about $f_{3} (x)$ ?
$\noindent That one is even, but not odd.$

24. ## Re: HSC 2017 MX1 Marathon

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25. ## Re: HSC 2017 MX1 Marathon

$\noindent Hint: Expand using the Binomial Theorem to get (2x^3 - \frac{1}{x})^n \equiv \sum_{k=0}^n \binom{n}k 2^{k}(-1)^{n-k}x^{4k -n}. Consider x^j for j\in\mathbb{Z} where j is an expression of k,n. When j=0, what does this say about k and n?$

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