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Re: MX2 2016 Integration Marathon $\noindent Another way to do that integral would be to split it up between $-1$ to $0$ and $0$ to $1$, use a substitution for the one from $-1$ to $0$ of $u=-x$, and make use of the identity $\tan^{-1} \frac{1}{t}=\frac{\pi}{2}-\tan^{-1} t$ for $t>0$. This...

$\noindent For part (iii), we should do it in three parts, based on where $P$ is. Let $y$ be the angle $AXB$, where $X$ is a point in the major arc $AB$ in the diagram (note $y$ is a constant, due to ``angles in the same segment'' and as $AB$ is fixed). The geometry is slightly different based...

$\noindent A polynomial that is odd contains only odd-power terms. So $a=0=c$.$

$\noindent If $f$ is odd, then $a$ must be 0 and also, $c$ needs to be 0. So $b=5$, as $f(5)=25$.$

$\noindent That's not really necessary for solving the inequation. Are you just asking out of interest? The function is $g(x):= \frac{1}{x+3}-2$. Its domain is $\mathbb{R}\backslash \{-3\}$ (i.e. the set of all real numbers excluding $-3$; for HSC purposes, it is sufficient to write this as...

$\noindent If $x+3$ is \textit{not} greater than 0 (i.e. if it is less than 0, since it can't actually equal 0, as it is a denominator), there's no way $\frac{1}{x+3}$ can be greater than 2 (as it would be negative).$

Re: HSC 2016 3U Marathon $\noindent Newton's method is used to find roots to an equation of the form $f(x) = 0$ (i.e. need to make the R.H.S. 0). So for your equation, it's equivalent to $\sin x -\frac{x}{5}=0$. So just use Newton's method with $f(x)=\sin x -\frac{x}{5}$, and initial guess the...

$\noindent We have $\frac{1}{x+3}-2 >0$ (by the way, if writing this without \LaTeX, it would be less ambiguous if we put brackets around the denominator, i.e. write it as $1/(x+3)$). Note that $x\neq -3$.$ $\noindent Rearranging, $\frac{1}{x+3}>2$. So $0<x+3 < \frac{1}{2}$ $\Big{(}$to see...

Re: HSC 2016 3U Marathon Either that, or there's not too many 2016'ers who can be bothered to answer Q's on these marathons. :p

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Observe that $y^\prime =-\frac{1}{\left(x+a\right)^2}$. Since $y^\prime =-1$ when $x=6$, we have $-1 = -\frac{1}{\left(6+a\right)^2}$. This is the equation to solve.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent We know the value of the derivative at a given $x$-value, so sub. in this $x$-value to the derivative expression and equate it to what we're told its value is here. This gives us an equation to solve for $a$.$

Re: HSC 2016 3U Marathon $\noindent There are nine people: $*********$. Now, imagine the three children together as one unit: $[C_1 C_2 C_3] ******$, where each $*$ represents an adult. In that orientation, there are $3!\cdot 6!$ arrangements ($3!$ for the children and $6!$ for the adults)...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent There is a simpler way. We have $12(4x-7)^2= 108$. So $(4x-7)^2 = 9$ (dividing both sides by 12). So $4x-7 = \pm 3$. I think you can do it from here.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent We first find the expression for the derivative (this tells us the slope at any point $x$). We then equate this to the slope we want and solve for $x$. For (a), the slope we want is clearly 108. For (b), the given...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent To find the point, we need to sub. $x=1$ into the function (the thing given to us). This tells us the $y$-value for a given $x$-value. So for 5(a), the point in question has $y=(5\times 1-4)^4 = 1$. So the point is...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent This is because we don't want negative numbers under the square roots. In general, if $y=\sqrt{h(x)}$, this function is defined (assuming we want a real-valued function with domain a subset of $\mathbb{R}$) at real...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent For this one, $y=\sqrt{x^2 - 2x}$. This function is defined at points where $x^2-2x \geq 0$, and nowhere else. Since $x^2 - 2x = x(x-2)$, the function is defined where and only where $x\leq 0$ or $x\geq 2$. So $x=1$...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Are you sure you mean 4 (f)? That is a sextic polynomial that looks basically like a steep parabola. It has a stationary point (point with horizontal tangent) at $x=5$.$

Depends how much you've memorised and the nature of the subject. If the subject is rote-based like the HSC Sciences for instance, there's probably not much point doing past papers closed book if you haven't memorised the content yet, because your answers would be sub-standard or incomplete etc...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent There are no $x$-intercepts because $y$ can't equal 0, as it is equal to the reciprocal of something without singularities.$