Search results

Why don’t you like doing it? Is it the way youre being taught or is it something inherent to maths?

Re: HSC 2018 MX2 Marathon i) \begin{aligned} z\alpha - \bar{z}\bar{\alpha} &= 0 \\ z\alpha &= \overline{z\alpha} \\ \mathrm{Im}(z\alpha) &= 0\end{aligned} i.e. z\alpha \in \mathbb{R} . But, \mathrm{Im}(z\alpha) = \mathrm{Im}((x+iy)(a+ib)) = bx + ay Therefore the locus of z\alpha -...

\int_1^2 \frac{2x^3-x^2+5x+3}{x}\, dx = \int_1^2 \frac{2x^3}{x} - \frac{x^2}{x} + ... \,dx

My area was a fraction because we still need to add the other area, the one under the curve from 3 to 2. Technically you only need to take the absolute value of \int_1^2 x^2-x-2 \, dx since the other part is known to be positive. \begin{aligned} A &= \left| \int_1^2 x^2-x-2\, dx\right|...

Let the dimensions be l,\,w,\,h. SA = 250 = 2lw + 2lh + 2 wh Since l = w = x , 250 = 2x^2 + 4xh \therefore h = \frac{250-2x^2}{4x}= \frac{125-x^2}{2x}

You can, but writing \int_1^2 x^2-x-2 \, dx = -\frac{7}{6} = \frac{7}{6} \mathrm{u^2} is a little questionable, and some markers may penalise you for this. Alternatively you could take the absolute value \left| \int_1^2 x^2-x-2 \, dx \right| = \frac{7}{6} or even swap the bounds...

b) i) use the arc length formula l = r \theta to find AC and DE . ii) use the area formula A = \frac{1}{2} r^2 \theta to find the areas of the large sector ABC and the small sector DBE . For d), be careful when notating angles using one letter ( \angle A) - it's technically...

Use the formula: A = \int^a_b f(x)-g(x)\, dx where f(x) is the curve that is on top (otherwise your integral will be negative). Since you drew the graph you should know that y = -x^2 + 2 will be on top. "First quadrant bounded by these two curves and the y-axis" means your...

https://www.wolframalpha.com/input/?i=x%5E2+%2B+pi((2(10-x))%2Fpi)%5E2,+x+%3D+40%2F(pi%2B4) Wolfram Alpha gives the answer \frac{400}{4+\pi} \approx 56.01...

The square has length x . Its perimeter, which is the amount of wire used to make the square, must therefore be 4x . The remaining amount of wire, 40 - 4 x cm (not 40 - x), is used to make the circumference of the circle ( C = 2 \pi r ). You now know the radius of the circle and the side...

Oh. Just expand the integrand then. You'll get x^2e^x + 2xe^x , and then the integration should be easy.

To be honest that doesn't look like an integral that could be given to a 2U student without first prompting them to differentiate x^2e^x. I might be missing something but I don't see how you could do that using only 2U methods.

Here’s another way to do it. \frac{^{10}P_5}{5} \times \frac{5!}{5} = 145152 For the first table, use ^{10}P_5 to obtain the number of ways you can arrange 5 people from a group of 10 (again, the concept of indistinguishable groups from before also applies here, but since the tables...

\binom{10}{5} \left(\frac{5!}{5}\right)^2 = 145152 There are \binom{10}{5} ways to pick 5 people from the group of 10. Then there are \frac{5!}{5} ways to arrange each group around the table. Normally we would have to consider the double counting case. Suppose the 10 people are...

(x+1)^4 = k(x^4+1) \frac{(x+1)^4}{x^4+1} = k (This equation is equivalent to the one before it, because (x^4+1) > 0 for all x \in \mathbb{R} . If this equation has two distinct real roots, what could be said about the graphs of y = \frac{(x+1)^4}{x^4+1}\, $and$\, \,y = k\, $?$

x^3 + y^3 = 1 When y = -x , x^3 - x^3 = 1 Which has no solution.

Yes, but it won't be too bad if you just do well.

i) z^7 - 1 = 0 Since z \neq 1 , \frac{z^7 - 1}{z-1} = 0 You can use long division and then divide by z^3 (since z \neq 0 ) to obtain the desired result. It's rather unintuitive, though...

\begin{aligned} \arg (z-1) &= 2 \arg z \\ \arg (z-1) &= \arg z+\arg z\end{aligned} Using \arg z + \arg w = \arg zw , \begin{aligned} \arg (z-1) &= \arg z^2 \\ \arg (z-1) - \arg z^2 &= 0\end{aligned} Using \arg z - \arg w= \arg \left(\frac{z}{w}\right) ...

Use the formula l = r \theta (where \theta is an angle, in radians) to calculate the length of the arc. Then you should be able to find the perimeter of the sector easily. I'm not sure what is meant by part b), as a ratio is comparing two quantities to each other and only one quantity was...