# Thread: HSC 2017 MX1 Marathon

1. ## Re: HSC 2017 MX1 Marathon

Originally Posted by Drsoccerball
What's f(x) ?
Any continuous, once differentiable function satisfying the given conditions.

2. ## Re: HSC 2017 MX1 Marathon

Originally Posted by InteGrand
That part was just a general statement.

3. ## Re: HSC 2017 MX1 Marathon

$\noindent a,b,c are in arithmetic progression \\\\ b,c,d are in geometric progression \\\\ \frac{1}{c},\frac{1}{d},\frac{1}{e} are in arithmetic progression \\\\ What progression is a,c,e ?$

4. ## Re: HSC 2017 MX1 Marathon

$\noindent It is given that \lim_{x \to 0} \frac{f(x)}{x} = k, where k is a real number. \\\\ Evaluate \lim_{x \to 1} \frac{f(x^n -1)}{x-1} where n is a positive integer.$

5. ## Re: HSC 2017 MX1 Marathon

Originally Posted by Paradoxica
$\noindent a,b,c are in arithmetic progression \\\\ b,c,d are in geometric progression \\\\ \frac{1}{c},\frac{1}{d},\frac{1}{e} are in arithmetic progression \\\\ What progression is a,c,e ?$
Alphabetic progression.

6. ## Re: HSC 2017 MX1 Marathon

Originally Posted by Paradoxica
$\noindent a,b,c are in arithmetic progression \\\\ b,c,d are in geometric progression \\\\ \frac{1}{c},\frac{1}{d},\frac{1}{e} are in arithmetic progression \\\\ What progression is a,c,e ?$
Geometric progression.

7. ## Re: HSC 2017 MX1 Marathon

Originally Posted by Paradoxica
$\noindent It is given that \lim_{x \to 0} \frac{f(x)}{x} = k, where k is a real number. \\\\ Evaluate \lim_{x \to 1} \frac{f(x^n -1)}{x-1} where n is a positive integer.$
The answer is nk.
$lim_{x\to \1 }\frac{f(x^{n}-1)(x^{n-1}+...+ 1)}{(x-1)(x^{n-1}+...+1)} = lim_{x\to \ 1}\frac{f(x^{n}-1)(x^{n-1}+..1)}{x^{n}-1}$

by letting $y = x^{n}-1$ and using assumption we get nk.

8. ## Re: HSC 2017 MX1 Marathon

Originally Posted by Mahan1
Geometric progression.
You're fun at parties...

9. ## Re: HSC 2017 MX1 Marathon

Originally Posted by mini8658
A, B are roots of x^2=5x-8.

Find an expression for a^1/3 + b^1/3
$\noindent Denote u = \sqrt[3]{\alpha} + \sqrt[3]{\beta} \\ Cube and replace to obtain: u^3 = \alpha + \beta + 3 u \sqrt[3]{\alpha \beta} = 5+6u \\ By inspection, u = -1 is a root of this cubic. Moving everything to one side and dividing gives: (u+1)(u^2 - u - 5) = 0 \\ This leads to: u = -1, u = \frac{1\pm \sqrt{21}}{2}$

I presently cannot think of a way to determine the value using only Extension 1 Methods (there is the subtle distinction of principal valued cube roots vs generalised cube roots) as there is a necessary step in deducing the magnitude of the complex cube roots from the arguments...

10. ## Re: HSC 2017 MX1 Marathon

$\\\text{1. Find }\lim_{a\to 0}\frac{\sin^{-1}a}{a}\\ \text{2. Evaluate }\lim_{h\to 0}\frac{(2x+h)h}{\left((x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}\right)h}\\ \text{3. Use the definition of the derivative to find }f^\prime(x)\text{, if }f(x)=\sin^{-1}x$

11. ## Re: HSC 2017 MX1 Marathon

\begin{align*}\text{1. }&\text{We have }\lim_{a \to 0}{\frac{\sin^{-1}{a}}{a}}\\ &\text{Let }a=\sin{u}\\ &\text{So }\sin^{-1}{a}=\sin^{-1}{(\sin{u})}=u\\ &\therefore\lim_{a \to 0}{\frac{\sin^{-1}{a}}{a}}=\lim_{a \to 0}{\frac{u}{\sin{u}}}\\ &\text{Now as }a \to 0\text{, }\sin^{-1}{a} \to 0\text{, and hence }u \to 0\\ &\therefore\lim_{a \to 0}{\frac{u}{\sin{u}}}\equiv\lim_{u \to 0}{\frac{u}{\sin{u}}}\\ &\text{We know that as }x \to 0\text{, }\sin{x} \to x\text{, and hence }\lim_{u \to 0}{\frac{u}{\sin{u}}}=\lim_{u \to 0}{\frac{u}{u}}=\lim_{u \to 0}{1}=1\\ &\therefore\lim_{a \to 0}{\frac{\sin^{-1}{a}}{a}}=1\end{align*}
\begin{align*}\text{2. }&\text{We have }\lim_{h \to 0}{\frac{(2x+h)h}{\left((x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}\right)h}}\\ &\text{Cancelling within a limit yields }\lim_{h \to 0}{\frac{2x+h}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}}\\ &\text{This allows us to simply substitute the limit as the function is now defined at }h=0\\ &=\frac{2x+0}{(x+0)\sqrt{1-x^2}+x\sqrt{1-(x+0)^2}}\\ &=\frac{2x}{x\sqrt{1-x^2}+x\sqrt{1-x^2}}\\ &=\frac{1}{\sqrt{1-x^2}}\end{align*}
\begin{align*}\text{3. }&\text{Let }y=f(x)=\sin^{-1}{x}\\ &\text{So }x=\sin{y}\\ &\frac{dx}{dy}=\lim_{h \to 0}{\frac{\sin{(y+h)}-\sin{y}}{h}}\\ &=\lim_{h \to 0}{\frac{\sin{y}\cos{h}+\sin{h}\cos{y}-\sin{y}}{h}}\\ &=\cos{y}\lim_{h \to 0}{\frac{\sin{h}}{h}}+\sin{y}\lim_{h \to 0}{\frac{cos{h}-1}{h}}\\ &=(\cos{y})(1)+(\sin{y})(0)\\ &\frac{dx}{dy}=\cos{y}\\ &\text{So, }\frac{dy}{dx}=\frac{1}{\cos{y}}\\ &\text{Now }\cos{y}=\cos{\sin^{-1}{x}}=\sqrt{1-x^2}\\ &\therefore f^{\prime}(x)=\frac{1}{\sqrt{1-x^2}}\end{align*}
Last one does feel like a cheat though.

12. ## Re: HSC 2017 MX1 Marathon

Originally Posted by Lugia101101
\begin{align*}\text{1. }&\text{We have }\lim_{a \to 0}{\frac{\sin^{-1}{a}}{a}}\\ &\text{Let }a=\sin{u}\\ &\text{So }\sin^{-1}{a}=\sin^{-1}{(\sin{u})}=u\\ &\therefore\lim_{a \to 0}{\frac{\sin^{-1}{a}}{a}}=\lim_{a \to 0}{\frac{u}{\sin{u}}}\\ &\text{Now as }a \to 0\text{, }\sin^{-1}{a} \to 0\text{, and hence }u \to 0\\ &\therefore\lim_{a \to 0}{\frac{u}{\sin{u}}}\equiv\lim_{u \to 0}{\frac{u}{\sin{u}}}\\ &\text{We know that as }x \to 0\text{, }\sin{x} \to x\text{, and hence }\lim_{u \to 0}{\frac{u}{\sin{u}}}=\lim_{u \to 0}{\frac{u}{u}}=\lim_{u \to 0}{1}=1\\ &\therefore\lim_{a \to 0}{\frac{\sin^{-1}{a}}{a}}=1\end{align*}
\begin{align*}\text{2. }&\text{We have }\lim_{h \to 0}{\frac{(2x+h)h}{\left((x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}\right)h}}\\ &\text{Cancelling within a limit yields }\lim_{h \to 0}{\frac{2x+h}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}}\\ &\text{This allows us to simply substitute the limit as the function is now defined at }h=0\\ &=\frac{2x+0}{(x+0)\sqrt{1-x^2}+x\sqrt{1-(x+0)^2}}\\ &=\frac{2x}{x\sqrt{1-x^2}+x\sqrt{1-x^2}}\\ &=\frac{1}{\sqrt{1-x^2}}\end{align*}
\begin{align*}\text{3. }&\text{Let }y=f(x)=\sin^{-1}{x}\\ &\text{So }x=\sin{y}\\ &\frac{dx}{dy}=\lim_{h \to 0}{\frac{\sin{(y+h)}-\sin{y}}{h}}\\ &=lim_{h \to 0}{\frac{\sin{y}\cos{h}+\sin{h}\cos{y}-\sin{y}}{h}}\\ &=\cos{y}\lim_{h \to 0}{\frac{\sin{h}}{h}}+\sin{y}\lim_{h \to 0}{\frac{cos{h}-1}{h}}\\ &=(\cos{y})(1)+(\sin{y})(0)\\ &\frac{dx}{dy}=\cos{y}\\ &\text{So, }\frac{dy}{dx}=\frac{1}{\cos{y}}\\ &\text{Now }\cos{y}=\cos{\sin^{-1}{x}}=\sqrt{1-x^2}\\ &\therefore f^{\prime}(x)=\frac{1}{\sqrt{1-x^2}}\end{align*}
Last one does feel like a cheat though.
Last one is meant to follow from the previous two. These aren't unconnected problems

13. ## Re: HSC 2017 MX1 Marathon

This is probably what the question wants, then:
\begin{align*}\text{3. }&\text{So by the definition of the derivative, }f^{\prime}(x)=\lim_{h \to 0}{\frac{\sin^{-1}{(x+h)}-\sin^{-1}{x}}{h}}\\ &\text{Let }y=\sin^{-1}{(x+h)}-\sin^{-1}{x}\\ &\sin{y}=\sin{(\sin^{-1}{(x+h)}-\sin^{-1}{x})}\\ &\sin{y}=\sin{\sin^{-1}{(x+h)}}\cos{\sin^{-1}{x}}-\sin{\sin^{-1}{x}}\cos{\sin^{-1}{(x+h)}}\\ &\sin{y}=(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}\\ &y=\sin^{-1}{((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})}\\ &\text{So }f^{\prime}(x)=\lim_{h \to 0}{\frac{\sin^{-1}{((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})}}{h}}\\ &=\lim_{h \to 0}{\frac{\sin^{-1}{((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})}}{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}}\cdot\lim_{h \to 0}{\frac{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}{h}}\\ &\text{Now as }h \to 0\text{, }(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2} \to 0\\ &\therefore\lim_{h \to 0}{\frac{\sin^{-1}{((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})}}{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}}=1\end{align*}
\begin{align*}.\,\quad\text{So }f^{\prime}(x)&=\lim_{h \to 0}{\frac{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}{h}}\\ &=\lim_{h \to 0}{\frac{\left((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}\right)\left((x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}\right)}{\left((x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}\right)h}}\\ &=\lim_{h \to 0}{\frac{(x+h)^2(1-x^2)-x^2(1-(x+h)^2)}{\left((x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}\right)h}}\\ &=\lim_{h \to 0}{\frac{(2x+h)h}{\left((x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}\right)h}}\\ &=\frac{1}{\sqrt{1-x^2}}\end{align*}

14. ## Re: HSC 2017 MX1 Marathon

$\text{Let }f(x)\text{ be a monotone function (increasing or decreasing).}$

$\text{Is }f(x)\text{ necessarily invertible? If yes, prove it. If no, provide further necessary and sufficient conditions.}$

15. ## Re: HSC 2017 MX1 Marathon

For a function to be invertable, it must be defined such that it is one-to-one. If a function is monotone, then every $x$ value will result in a single value of $f(x)$, however, the function must be strictly monotonic, else there could exist two values $x_1$ and $x_2$ such that $f(x_1)=f(x_2)$. So if the function is one-to-one, then there exists an inverse, and hence $f(x)$ will only have an inverse if it is strictly monotone.

16. ## Re: HSC 2017 MX1 Marathon

Originally Posted by Lugia101101
For a function to be invertable, it must be defined such that it is one-to-one. If a function is monotone, then every $x$ value will result in a single value of $f(x)$, however, the function must be strictly monotonic, else there could exist two values $x_1$ and $x_2$ such that $f(x_1)=f(x_2)$. So if the function is one-to-one, then there exists an inverse, and hence $f(x)$ will only have an inverse if it is strictly monotone.
A bit confused. What do you mean by 'strictly monotonic'? Or were you intending to say strictly increasing/decreasing (i.e. f'(x)≠0)

17. ## Re: HSC 2017 MX1 Marathon

A function with no two points $x_1$ and $x_2$ such that $f(x_1)=f(x_2)$. An example would be:
$f(x)=\begin{cases}(x+1)^3&x\leq -1\\ 0&-1\leq x\leq 1\\ (x-1)^3&1\leq x\end{cases}$
Which fits the definition for monotonic increasing, but isn't strictly monotonic increasing. (From Wikipedia)
A function can still have $f^{\prime}(x)=0$ at some point and have an inverse, such as $f(x)=x^3$, which is strictly monotonic increasing, has $f^{\prime}(x)=0$ at $x=0$, and has the inverse function $f^{-1}(x)=\sqrt[3]{x}$.

18. ## Re: HSC 2017 MX1 Marathon

$\noindent Yeah just monotonic isn't enough to imply invertibility, e.g. A constant function from \mathbb{R} to \mathbb{R} is monotonic but not invertible. `Strictly monotonic' would guarantee invertibility though (taking the codomain to be the range of the function).$

19. ## Re: HSC 2017 MX1 Marathon

Also, we don't need a function to be monotonic for it to be invertible, since we can easily form discontinuous functions that are not monotonic but pass the horizontal line test. However, you can prove as an exercise that a continuous one-to-one function must be monotonic.

20. ## Re: HSC 2017 MX1 Marathon

Such as a function defined as $f(x)=(-1)^{\lceil{x}\rceil}|x|$?

21. ## Re: HSC 2017 MX1 Marathon

Was basically the point of the question. In addition, the function has to be continuous.

$\text{Because it took me too long to realise }f(x)=\tan x, \, x \in \mathbb{R}-\left\{n\pi+\frac{\pi}{2}\right\}(n\in \mathbb{Z})\text{ caused troubles.}$

However, I haven't heard of the "strict monotonicity" grammar though

22. ## Re: HSC 2017 MX1 Marathon

The tan function defined above is continuous but not monotonic.

It's easy to come up with a discontinuous function that's not monotonic but is invertible. It suffices to take a graph that has two branches, one starting at the point (0, 2) and decreasing strictly, smoothly and asymptotically to y = 1, and the other branch a reflection of this about the y-axis and shifted down enough so that the overall graph passes the horizontal line test (and only include one of the points at the discontinuity at x = 0).

23. ## Re: HSC 2017 MX1 Marathon

$\\\text{1. By supplying a sufficient example, or otherwise, show that a tangent to a cubic polynomial}\\ \textit{can}\text{ intersect the curve again elsewhere.}\\ \text{2. Is this true for polynomials of degree 4 or higher?}\\ \text{3. Prove that the point of intersection between the tangent to a parabola and the parabola itself is unique}\\ \text{(i.e. there is only one point of intersection)}\\ \text{4. Where is this point of intersection?}$

Q4 is a trick question

24. ## Re: HSC 2017 MX1 Marathon

Originally Posted by leehuan
$\text{3. Prove that the point of intersection between the tangent to a parabola and the parabola itself is unique}\\ \text{(i.e. there is only one point of intersection)}\\ \text{4. Where is this point of intersection?}$

Q4 is a trick question
For question 3:
$\text{Let }f(x)=ax^2 + bx +c \\ f^{\prime}(x) = 2ax + b \\ \text{Sub in }x=k, f^{\prime}(k) = 2ak + b, f(k) = ak^2 + bk +c \\ \text{The equation of the line with that gradient and passing through that point it;} g(x)=x(2ak+b) - ak^2 +c. \\ \text{By solving simultaneously, to find the point of intersection we arrive at the following quadratic;} ax^2 -2akx + ak^2 = 0. \\ \text{But the discriminant for the equation of this line is = 0. Thus only one point of intersection}$

For question 4, as we showed there is only one point of intersection, this would occur when x=k.

25. ## Re: HSC 2017 MX1 Marathon

$\text{Prove that, for non-zero constants }a,b\\\ \ \sin \left(x+\tan^{-1}\frac{b}{a}\right)=\cos \left(x-\tan^{-1}\frac{a}{b}\right)$

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