Will they allow you to take 1141 with just your MX1? If so, I can help you learn 1141 contents during your December/January break after your HSC, so that you won't be badly handicapped. But you must be pretty good in your MX1.
z^5 + a = 0
is a polynomial equation (of degree 5) whose roots are the points on the circle. Since these points are symmetric about the x-axis, it means every (of the 4) complex roots appear in complex-conjugate pairs. Therefore the coeffs of this polynomial eqn must be all reals; therefore 'a'...
i.e |z - 0| = |z -(1+i)|
It means the distances of (all) z from the points O(0,0) and from A(1,1) are equal. So the equation defines the locus of a point which is equidistant from these 2 given points O & A, viz. the perpendicular bisector of the line segment OA. Draw this perpendicular...
No solutions because:
The smallest each of the terms: e^{sin^2} $ and $ e^{cos^2} can be is 1. Therefore the sum of these 2 terms cannot be less than 2. But this sum equals 2 only when each term equals 1 simultaneously. Each term equals 1 only when sin^2 = 0 $ or $ cos^2 = 0; but this...
You make study notes for your own needs. Different students should have different study notes, tailored to the needs of each. I suggest you prepare your notes by topics. For each, write down materials/items that are likely to be important to the course; include solutions of special relevance...
Q12
Following same approach as for Q19:
\overrightarrow {OD} = \overrightarrow {OA} + \overrightarrow {OC} - \overrightarrow {OB}\\ \\ = [3,-8,-2] + [-2,-2,1] - [2,4,5] \\ \\ = [-1,-14,-6](the position vector of point D)
Q19
Don't know how to write a column matrix. So will use [a,b,c] instead.
For case A-B-C:
\overrightarrow {BD} = \overrightarrow {BA} + \overrightarrow {BC}\\ \\ \therefore \overrightarrow {OD} - \overrightarrow {OB} = (\overrightarrow{OA}-\overrightarrow{OB}) + (\overrightarrow {OC} -...
Q19
If A, B, C are supposed to be 3 consecutive vertices, in cyclic order, (adjacent sides BA and BC), then there is only one D. In cyclic order, B-A-C (adjacent sides AB & AC) will generate another D and finally A-C-B (adjacent sides CA & CB) a 3rd D. A-B-C and C-B-A are treated as the...
Assuming suggested answer is correct, here's my effort:
\int \frac{1} {\sqrt{u^2 - a^2}}du = ln (u + \sqrt {u^2 - a^2}) + C \\ \\ sin 2x = 2sinxcosx = (sin x + cos x)^2 - 1 \\ \\ -d(sin x + cos x) = (cos x - sin x)dx \\ \\ \therefore I = \int \frac {sin x - cos x}{\sqrt {sin 2x}} dx = -\int...
Another way to look at it:
\frac{49\pi}{30} = 2\pi - \frac{11\pi}{30} $ and $ \frac {41\pi}{30} = \pi + \frac {11\pi}{30}\\ \\ So: sin \frac {49\pi}{30} $ and $ sin \frac {41\pi}{30}
have the same related angles (11pi/30) and they are both negative, being in the 3rd & 4th quadrants resp...