HSC 2017 MX1 Marathon (2 Viewers)

Lugia101101 · Nov 30, 2016

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

Paradoxica · Nov 30, 2016

Lugia101101 said:
$\begin{align*}\text{1. }&\text{We have }\lim_{a \to 0}{\frac{\sin^{-1}{a}}{a}}\\ &\text{Let }a=\sin{u}\\ &\text{So }\sin^{-1}{a}=\sin^{-1}{(\sin{u})}=u\\ &\therefore\lim_{a \to 0}{\frac{\sin^{-1}{a}}{a}}=\lim_{a \to 0}{\frac{u}{\sin{u}}}\\ &\text{Now as }a \to 0\text{, }\sin^{-1}{a} \to 0\text{, and hence }u \to 0\\ &\therefore\lim_{a \to 0}{\frac{u}{\sin{u}}}\equiv\lim_{u \to 0}{\frac{u}{\sin{u}}}\\ &\text{We know that as }x \to 0\text{, }\sin{x} \to x\text{, and hence }\lim_{u \to 0}{\frac{u}{\sin{u}}}=\lim_{u \to 0}{\frac{u}{u}}=\lim_{u \to 0}{1}=1\\ &\therefore\lim_{a \to 0}{\frac{\sin^{-1}{a}}{a}}=1\end{align*}$
$\begin{align*}\text{2. }&\text{We have }\lim_{h \to 0}{\frac{(2x+h)h}{\left((x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}\right)h}}\\ &\text{Cancelling within a limit yields }\lim_{h \to 0}{\frac{2x+h}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}}\\ &\text{This allows us to simply substitute the limit as the function is now defined at }h=0\\ &=\frac{2x+0}{(x+0)\sqrt{1-x^2}+x\sqrt{1-(x+0)^2}}\\ &=\frac{2x}{x\sqrt{1-x^2}+x\sqrt{1-x^2}}\\ &=\frac{1}{\sqrt{1-x^2}}\end{align*}$
$\begin{align*}\text{3. }&\text{Let }y=f(x)=\sin^{-1}{x}\\ &\text{So }x=\sin{y}\\ &\frac{dx}{dy}=\lim_{h \to 0}{\frac{\sin{(y+h)}-\sin{y}}{h}}\\ &=lim_{h \to 0}{\frac{\sin{y}\cos{h}+\sin{h}\cos{y}-\sin{y}}{h}}\\ &=\cos{y}\lim_{h \to 0}{\frac{\sin{h}}{h}}+\sin{y}\lim_{h \to 0}{\frac{cos{h}-1}{h}}\\ &=(\cos{y})(1)+(\sin{y})(0)\\ &\frac{dx}{dy}=\cos{y}\\ &\text{So, }\frac{dy}{dx}=\frac{1}{\cos{y}}\\ &\text{Now }\cos{y}=\cos{\sin^{-1}{x}}=\sqrt{1-x^2}\\ &\therefore f^{\prime}(x)=\frac{1}{\sqrt{1-x^2}}\end{align*}$
Last one does feel like a cheat though.

Last one is meant to follow from the previous two. These aren't unconnected problems

Lugia101101 · Dec 2, 2016

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

leehuan · Dec 7, 2016

Lugia101101 · Dec 7, 2016

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.

leehuan · Dec 7, 2016

Lugia101101 said:
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)

Lugia101101 · Dec 7, 2016

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 $f^{-1}(x)=\sqrt[3]{x}$ .

InteGrand · Dec 7, 2016

InteGrand · Dec 7, 2016

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.

Lugia101101 · Dec 8, 2016

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

leehuan · Dec 8, 2016

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

InteGrand · Dec 8, 2016

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).

leehuan · Dec 18, 2016

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

Q4 is a trick question

frog1944 · Dec 19, 2016

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

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.

leehuan · Jan 16, 2017

jathu123 · Jan 16, 2017

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

$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$

InteGrand · Jan 16, 2017

jathu123 said:
$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.

leehuan · Jan 17, 2017

The inspiration of that question was actually just the fact

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

jathu123 · Jan 17, 2017

leehuan said:
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

davidgoes4wce · Feb 6, 2017

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.

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