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Re: HSC 2016 4U Marathon You could obtain the answer to this new one from the answer to the old one given by Paradoxica. Just take conjugates, since the new equation is satisfied precisely by the conjugates of the old one.

Yeah the max. value is the y-value of the highest point on the graph.

Differentiate and find where the derivative is 0 for stationary points. Determine their nature using for instance the second derivative test. Sub. in the x-values of the stationary points to the f(x) expression to get the coordinates. Find where f"(x) = 0 for the inflection point. Last part...

Write b = 2a, c = 3a, then substitute into the given equation and solve for a.

Pretty much

They are clearly parallel and don't coincide or meet anywhere.

For the first two, show that the discriminant is a perfect square. For the last one, if we want distinct roots, we should set the discriminant to be positive and solve for m.

$\noindent Yeah you can do that (show the discriminant is negative), after noting the parabola is upwards-facing.$

$\noindent The roots are $p$ and $q$, and $q < p$, so when we sketch the parabola, because it's upwards-facing, it'll be above the $x$-axis to the left of the lower root (when $x < q$) and above the $x$-axis when we're to the right of the higher root ($x > p$). (Just sketch the parabola to see...

$\noindent a) Discriminant is $0$, i.e. $a^2 -4b =0 \Longleftrightarrow a = \pm 2b$.$ $\noindent b) Discriminant is positive, i.e. $a^2 > 4b$.$

$\noindent The roots are $p$ and $q$. So (noting the parabola is upwards-facing), the answer is $x \leq q$ or $x \geq p$.$

Yeah.

a) The sum of roots is 5, so k+2 = 5, whence k = 3. b) The product of roots is 12, so 4k = 12, which implies k = 3. c) Let the roots be a and a+2, then use sum and product of roots to obtain two equations in two unknowns.

$\noindent Since $2$ is a root, we have $4+2+m = 0 \Rightarrow m = -6$. Also, using sum of roots, the other root $a$ satisfies $a+2 = -1$, so the other root is $a = -3$. $

$\noindent This is a standard sort of question. The quadratic factors as $(3x-1)(x+5)$, so the roots are $-5$ and $\frac{1}{3}$. Therefore (sketch the graph, noting the parabola is upwards-facing), the quadratic is negative precisely when $-5< x < \frac{1}{3}$.$

The highest ATAR you can theoretically get is still 99.95.

$\noindent Let the roots be $a$ and $3a$. By sum of roots, $4a = -q$. By product of roots, $3a^2 = r$. So from these equations, $q^2 = 16a^2 = \frac{16}{3}r\Rightarrow 3q^2 = 16 r$.$

$\noindent $\Big{(}$The reason that is the coefficient of $x^2$ in the L.H.S. is that the coefficient of $x^2$ in a sum is the sum of the $x^2$ coefficients of each term, and each term has $x^2$ coefficient $\binom{r-1}{2}$, by the binomial theorem.$\Big{)}$$

$\noindent Assuming you got part i), we have $\sum _{r=1}^{n}\left(1+x\right)^{r-1} = \sum _{r=1}^{n} \binom{n}{r}x^{r-1}$. The fact we are asked to prove follows immediately by equating coefficients of $x^{2}$ of both sides of the part i) identity. Note that coefficient of $x^{2}$ on the L.H.S...

Which Maths is it? 2U? General?