Search results

$\noindent Why do you have $xy$ in the integral? One possible approach is as follows.$ $\noindent Note that our sample space $\Omega$ is the part of the unit square below the line $y = 1-x$ in the $x$-$y$ plane. That is, the interior and boundary of the triangle with vertices $(0,0)$, $(1,0)$...

$\noindent You could find the CDF of $Y$ using $F_{Y}(y) = \int_{\text{all }x} \mathbb{P}\left(Y\leq y\mid X = x \right)f_{X}(x)\, \mathrm{d}x$, where $f_{X}(x)$ is the density of $X$ and the $\mathbb{P}$ in the integral is the CDF of $Y\mid X=x$. Then differentiation would yield the PDF of $Y$...

Equate real parts: x + 2y = 10. (1) Equate imaginary parts: x – 3y = 0. (2) Solving simultaneously yields x = 6 and y = 2.

$\noindent Note that the given limit is just the limit of a Riemann sum for $\int_{0}^{1}x^{3}\,\mathrm{d}x$, so has value $\frac{1}{4}$.$ $\noindent Just in general we have the following results. Fix a positive integer $d$ and let $S_{d}(n) = 1^{d}+2^{d} + \cdots + n^{d}$, for positive...

Answer: 1/4.

Re: Very mediocre first internal HSC marks - is a 99+ ATAR already out of the questio Yes, it's still possible to get 99+ ATAR.

$\noindent The answer is $V = \pi\int_{0}^{1}x^{2}\left(x^{3} -3\right)^{4}\,\mathrm{d}x$. We can substitute $u=x^{3}-3$, $\frac{1}{3}\mathrm{d}u = x^{2}\,\mathrm{d}x$. When $x=0$, $u=-3$, and when $x = 1$, $u=-2$. So$ $$\begin{align*}V &= \pi \int_{-3}^{-2}\frac{1}{3}u^{4}\,\mathrm{d}u \\ &=...

$\noindent I just did that so that the RHS would be $24$ like it was in the equation $3x_{0}x + 4y_{0}y = 24$, so we could compare coefficients of $x$ and $y$.$

$\noindent Let the desired external point be $\left(x_{0}, y_{0}\right)$. The chord of contact is $\frac{x_{0}x}{8} + \frac{y_{0}y}{6} = 1 \iff 3x_{0}x + 4y_{0}y = 24$. The desired chord of contact is $x-y = -2\iff -12x + 12y = 24$. So $3x_{0} = -12$, i.e. $x_{0} = -4$, and $4y_{0} = 12$, i.e...

$\noindent Observe that $\cos \phi -\sin \phi \equiv \sqrt{2}\cos \left(\phi +\frac{\pi}{4}\right)$ and $\cos \phi + \sin \phi \equiv \sqrt{2}\cos\left(\phi -\frac{\pi}{4}\right) \equiv \sqrt{2}\sin\left(\phi + \frac{\pi}{4}\right)$. So $R$'s coordinates satisfy$ $$\begin{align*}x = 2\sqrt{2}...

$\noindent By the reverse chain rule, we have $\int \frac{f'(x)}{(f(x))^{2}}\,\mathrm{d}x = -\frac{1}{f(x)} + c$. So if $f(x) = x^2 -1$, we have $f'(x) = 2x \Rightarrow x = \frac{1}{2}f'(x)$. Hence$ $$\begin{align*}\int \frac{x}{\left(x^{2}-1\right)^{2}}\,\mathrm{d}x &= \int...

$\noindent The curve was $y = f(x) = -\frac{\frac{1}{2}}{x^2 -1} + C$ for some constant $C$. By definition, the point $(2,-1)$ lies on the graph of $y=f(x)$ if and only if $f(2) = -1$. So we must have $-1 = -\frac{\frac{1}{2}}{2^2 -1} +C$. Now just solve for $C$.$

It's because the curve has to pass through (2, -1). When you integrated you probably ended up with a y = something "+C". You should then sub. in x = 2 and y = -1 to find what the constant C must be in order for (2, -1) to lie on the curve.

Also to find asymptotes for hyperbolas "centred" at the origin: just set the RHS of the hyperbola equation to 0 and rearrange to get the asymptotes' equations. (This technique also works if the hyperbola is of the form y^2/(A^2) - x^2/(B^2) = 1 so you don't need to learn different formulas for...

You could possibly get one (would probably depend on the year and how 'high' a 96 you got).

Didn't say it couldn't.

Was the reason that the exam was relatively easy this year and made it hard to separate students at the top?

Re: HSC 2017 4U Marathon $\noindent Show that the roots of the quadratic equation $az^{2} + bz + c = 0$ ($a,b,c$ real, $a\neq 0$) both have negative real parts if and only if $a,b,c$ are either all positive or all negative.$

We need the augmented matrix to end up with a zero row in order for the system of equations to have a solution (since there's more rows than columns in the left-hand matrix). This means the un-augmented matrix needs to have determinant 0. So if we were given that determinant formula (which is...

Re: HSC 4U Integration Marathon 2017 $\noindent Let $t\in \mathbb{R}$. Find $\int_{0}^{t} e^{-2x}\sin^{2} x\, \mathrm{d}x$.$