HSC 2017 MX1 Marathon (1 Viewer)

JackPatel · Feb 8, 2017

- 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

Mahan1 · Feb 17, 2017

JackPatel said:
- 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 :
$(0,0),(x,sin^{-1}x), (-x,-sin^{-1}x)$

Mahan1 · Feb 17, 2017

davidgoes4wce said:
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:
$\angle ABC = 180 - (\angle ACB + \angle CBA)$

Mahan1 · Mar 5, 2017

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

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

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

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

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

si2136 · Mar 12, 2017

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

pikachu975 · Mar 12, 2017

si2136 said:
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?

si2136 · Mar 12, 2017

pikachu975 said:
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?

InteGrand · Mar 12, 2017

si2136 said:
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).

Paradoxica · Mar 12, 2017

Mahan1 said:
$\textrm{1. Suppose}\ a,b,c \in \mathbb{R} \ \textrm{such that}$

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

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

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

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

By relation, I suppose you mean a single equation...

a²=ab+c²

wu345 · Apr 6, 2017

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

Mahan1 · Apr 9, 2017

wu345 said:
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

$\textrm{The base case is obvious.}$

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

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

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

$\textrm{By induction hypothesis }$

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

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

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

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

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

davidgoes4wce · Jun 20, 2017

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.

Green Yoda · Jun 20, 2017

davidgoes4wce said:
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.

pikachu975 · Jun 20, 2017

davidgoes4wce said:
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.

davidgoes4wce · Jun 20, 2017

pikachu975 said:
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.

davidgoes4wce · Jun 20, 2017

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)

davidgoes4wce · Jul 23, 2017

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 $f_{3} (x)$ ?

InteGrand · Jul 23, 2017

davidgoes4wce said:
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 $f_{3} (x)$ ?

$\noindent That one is even, but not odd.$

pikachu975 · Oct 1, 2017

http://prntscr.com/grt308

Part ii please

kawaiipotato · Oct 1, 2017

HSC 2017 MX1 Marathon (1 Viewer)

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