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Re: MX2 2016 Integration Marathon Not necessarily any mistake, try showing your answer differs from mine and Paradoxica's by a constant, which it looks like it does from a glance. Otherwise, you may have just made a silly mistake in arithmetic. Edit: yeah, your answer simplifies to ours...

Re: MX2 2016 Integration Marathon $\noindent \textbf{ALTERNATE METHOD}$ $\noindent The goal is to rewrite $a \cos x + b \sin x$ in the form $A \left( \cos x + \sin x\right)+B \left( \cos x + \sin x\right)^\prime$ so that we can easily compute the integral as it will be in the form $\int...

Re: MX2 2016 Integration Marathon $He used the rule $\int ^\alpha _0 f(x) \text{ d}x = \int _0 ^\alpha f\left( \alpha -x \right) \text{ d}x$.$

$The error in your working is that you double counted the terms (which is why you ended up with double the answer). The values of $a_2$ is going to be the sum of pairwise products of the roots, i.e. Take all the pairs of two distinct roots as products and add them up, where \textsl{the order of...

Re: HSC 2016 3U Marathon $This is not the case; rather, it is $f ^\prime (a) = \frac{1}{ \left( f^{-1} \right) ^\prime \left( b \right) }=\frac{1}{\left(f^{-1} \right) ^{\prime} \left(f(a)\right)}$, where $(a,b)$ is a point on the graph of $y=f(x)$, so $(b,a)$ lies on the graph of $y=f^{-1}...

Re: HSC 2016 3U Marathon $So to start that question about the arcsin derivative, we should let $y =\sin^{-1}x, x = \sin y$ \Big{(}so $x \in \left[-1,1\right]$ and $y \in \left[ -\frac{\pi}{2},\frac{\pi}{2}\right]$\Big{)}. Then we need to prove that $\frac{\mathrm{d}y}{\mathrm{d}x}...

Re: HSC 2016 3U Marathon $For example, $\mathrm{e}^x$ and $\ln x$ are inverse functions, but clearly their derivatives multiplied together don't give 1 in general. What \textsl{is} true though is that if $b =\mathrm{e}^a \Longleftrightarrow a = \ln b$, and $f(x) = \mathrm{e}^x$ and $g(y)=\ln y$...

Re: HSC 2016 3U Marathon $The first line of the working is incorrect. It is not true that if $f$ and $g$ are inverse functions, then $f^\prime (x)\cdot g^\prime (x) = 1$ (which is what you implied in your answer to (a)). What (a) states is that if $f$ is an invertible continuously...

$Since $\alpha$ is a root, it satisfies the polynomial equation given, so$ $$\begin{align*} \alpha^3 + 5\alpha -4 &=0 \\ \Rightarrow \alpha^3 &= -5 \alpha +4. \quad (1) \end{align*}$$ $Sinilarly for the other roots, we have$ $$\begin{align*} \beta^3 &= -5\beta + 4 \quad (2) \\ \gamma^3...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $(a) $y = -5x +2 \Rightarrow y^\prime = -5<0$, so the derivative is always negative, so the function is decreasing for all $x$.$ $(b) $y = x+7 \Rightarrow y^\prime = 1$, so the derivative is always positive (being equal to...

\text{Draw the complex number z on the unit circle represented by the point P (it is in the first quadrant). Let A be the point (1,0) in the complex plane, so A represents the number 1.} \text{Then} \ \overrightarrow{OQ} =\overrightarrow{OP} + \overrightarrow{OA}, \ \text{since z+1 = (z) + (1)...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Note that $z^2 - z -2=\left(z-2 \right)\left(z+1\right)$, so the given equation is $\left| \left(z-2 \right)\left(z+1\right) \right| = |z+1| \Longleftrightarrow |z-2||z+1| = |z+1|$. So if $z\neq -1$, we can divide through by...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $The locus is a ray starting at $i$ (excluding the point $i$) and making an angle of $\frac{\pi}{3}$ with the positive $x$-axis. The equation of the entire \textsl{line} can be written as $y=mx+b$. Clearly $b=1$, since the...

Yeah

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $As leehuan has said, don't bother doing it by letting $z=x+iy$, as this is an unnecessarily tedious method. Using leehuan's suggested method (which is the best way to go about this one), we have$ $$\begin{align*}...

$\textbf{ALTERNATE METHOD}$ $Since $\arg (z)$ is defined, we can assume $z\neq 0$. We will use the fact that a complex number $\omega \neq 0$ is real if and only if its reciprocal $\frac{1}{\omega}$ is real (this is easy to prove using polar form).$ $Denoting $\omega:= \frac{z}{z^2 + r^2}$...

Maybe you should post your working as it could just be a silly mistake in your algebra.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Here is a sketch of $y = -\frac{1}{2\left( x^2 + 1\right)}$, via WolframAlpha:$ http://wolframalpha.com/input/?i=plot+y+%3D+-1%2F%282%28x%5E2+%2B+1%29%29&x=0&y=0

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $I assume you mean $y = -\frac{1}{2\left( x^2 + 1\right)}$? This is an upside down bell-shaped sort of curve. If you meant $y = -\frac{1}{2} \left( x^2 + 1\right)$ (which is what what you wrote actually means with the way you...

Re: HSC 2016 4U Marathon $This is based on the arg rule $\arg \left(\frac{1}{\omega} \right) = -\arg \left( \omega \right) $ (can also be written as $\arg \left( \omega \right) = -\arg\left( \frac{1}{\omega}\right)$. So for example, $\arg \left( 1+i\right) =-\arg...