Search results

Re: MX2 2016 Integration Marathon $\noindent Find $\int \frac{\sqrt{2+x^2}}{x}\, \mathrm{d}x$.$

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread Yeah, that's why you may find it helpful to consider cases. E.g. The cases could be: no pairs of double-letters; exactly one pair of double-letters; exactly two pairs of double-letters. Then count the no. of possibilities in...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread You may wish to consider cases based on things like how many ''double-letters'' are selected.

Probably. If it was 4U, we could differentiate P(x) and sub. in a root we've found already to see if it's also a root of the derivative, which would show us whether or not it's a double root (and if it was a double root, we could differentiate again and sub. in to the second derivative to check...

$\noindent You can also factorise by inspection. Let $Q(x) = x^3 -x^2 -5x-3=-P(x)$. Test factors of $-3$ for roots.$ $\noindent Note $Q(-1) = -1-1+5-3 = 0 \Rightarrow -1$ is a root. So $Q(x) = x^3 -x^2 -5x-3 = \left(x+1\right)\left(x^2 + ax -3\right)$, where (comparing $x^2$ terms), $a+1 =...

What was the question? There may be shortcuts available.

Re: Discrete Maths Sem 2 2016 Which thing was confusing you? Well the second function isn't well-defined because the given "codomain" isn't really a codomain at all (a codomain should contain all the function values, but it clearly doesn't here. I.e. if f: X -> Y is a well-defined...

$\noindent Did you get your last answer by doing $s = ut + \frac{1}{2}at^2$? Remember, this gives the change in \emph{displacement} only. For this time interval, the particle was moving backwards at the start, and then after a while, starts moving forward (ending up past where it started). To...

$\noindent Yeah, if we assume a convention of reporting the angle of inclination $\alpha$ to be in the range $0\leq \alpha < 180^\circ$. You can see why doing $180^\circ -34^\circ$ gives the desired answer by drawing a rough sketch. The line is sloping downwards and the angle under the $x$-axis...

For d), your answer is less than your answer for c) – how can this be possible? The total distance it travels would have to be at least whatever its change in displacement is over that time interval.

The significance of the negative sign is that the line makes an angle of 34 degrees below the horizontal (because it's sloping downwards).

For each of these, we just need to find the appropriate value of a. Here are some hints: a) What is the coefficient of x^2 of the given quadratic in terms of a? b) What is the constant term in terms of a? (Product of roots times a) c) The y-intercept is found by subbing x = 0 into the given...

For the first one, write 4^x as (2^2)^x (as 4 = 2^2) and use index laws. For the second one, write 2^(2x +1) as 2*(2^(2x)), then let u = 2^x and you'll get a quadratic equation in u, which you can solve via the quadratic formula (or whatever method you like). Once you find the values of u, the...

$\noindent You would need their numerical values. We can find the values of $\alpha +\beta$ and $\alpha \beta$ from the given quadratic equation using its sum and product of roots. Knowing these quantities, we can then find $3\alpha + 3\beta = 3\left(\alpha + \beta\right)$ and $3\alpha \cdot...

$\noindent Find the values of $3\alpha \cdot 3\beta$ (call this $P$ for product of the roots) and $3\alpha +3\beta$ (call this $S$ for the sum of the roots). Then a desired quadratic equation is $x^2 -Sx + P = 0$. (This is true for general quadratic equations when we want a sum of roots to be a...

$\noindent For simplicity, write $u\equiv|x+1|$. Then the given equation is$ $$\begin{align*}u^2 -4u-5 &= 0 \\ \iff \left(u+1\right)\left(u-5\right) &= 0 \\ \iff u = -1 &\text{ or } u=5.\end{align*}$$ $\noindent There are no solutions for $u=-1$ since $u\geq 0$, being an absolute value. So we...

$\noindent For simplicity, write $s\equiv \sin x$ and $c\equiv \cos x$. Note $s^{2}=1-c^2$ (trig. identity), so the equation becomes$ $$\begin{align*}2\left(1-c^2\right)&= 3\left(c+1\right) \\ \iff 2 - 2c^2 &= 3c +3 \\ \iff 2c^2 +3c +1 &= 0 \\ \iff...

You also don't need to do HSC Biology for Medicine. You should probably only do it if you think it will help your ATAR.

$\noindent And if you want to use a hyperbolic substitution, it is $x = 4\sinh t$. It may lead to a neater integral (hyperbolic substitutions often do), but you'll need to be familiar with hyperbolic identities and integrating hyperbolic analogs of the circular trig. functions.$

By the geometric definition of the parabola, it's a parabola with focus S(2,3) and directrix y = 5. So the vertex is halfway between these at the point (2,4) and the focal length is 1. The parabola is downward facing. This is now enough information to write down the equation of the parabola.