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Interpret this as |z-w| = |z-0|, which means the distance from w must be the same as the distance from the point 0. The midpoint of 0 and w (0.5 + 1.5i) is one such point, and it follows that the line perpendicular line passing through this point retains this property.

Make sure you know all three versions of cos(2x) cos(2x)= cos^2(x)-sin^2(x)=1-2sin^2(x)=2cos^2(x)-1

I always like to make eco notes by syllabus dot point so that I know what I'm doing, but the case study description in the syllabus is extremely brief. Can anyone suggest a criteria or subheadings under which I can perform case study research and point me in the right direction? So that I know...

i) The diagram shows that the 4 vectors enclose a quadrilateral, so the resultant is 0, therefore the sum of the vectors is 0. ii) MN = (a + b/2) - (a/2) MN = 1/2(a + b) QP = (a + b + c/2) - (a + b + c - d/2) = -1/2(c + d) Now c + d = - (a + b) (using part i) So MN = QP...

After proving true for initial value.... Assumption Prove true for n=k+1 Use calculus to show that this last line is >0, hence LHS>RHS and conclude your proof.

It is very beneficial to write notes by syllabus point. I also like to use the syllabus 'students learn to' as literal broad practice questions

Primary technique is foreshadowing. Also, it is a part of the dramatic irony that Anne believes Richard genuinely loves her, despite it being a facade.

Not quite... 'area under the curve' is used to mean area between the x-axis and the line itself. For example, 'area under the curve' between 0 and 2 seconds is the triangle bound by (0,0) , (1,0) and (1,2) = 4, giving a displacement of 4m in this interval.

The car stops when v=0 (i.e. the graph crosses the x-axis). The displacement at that point is the area under the curve between 0 and 8 seconds, which should give you 29m. One more thing, the displacement at 2 seconds is actually 4m, not 8. Always look for the area under the curve, because the...

Show that d/dx LHS > d/dx RHS for the domain given, and then prove the inequality true for the start of the domain by substitution (x=0).

Put more simply; Any lemonade at the start of the day has either been sold or remains in the jug. Now the question says 1/3 of a jug (no particular relation to the initial amount) was sold, and 1/2 of a jug remains. So the lemonade at the start is 1/3 + 1/2 = 5/6. Interpret the values as...

Think about the range of values for which someone can shake someone's hand. For any person, they may shake 1,2,3...23 hands. But , if one person shakes 0 hands, there cannot be a person that shakes 24 hands, and vice versa (because there is a contradiction between one person not shaking anyone's...

38 is the maximum amount of ways in which the condition can be avoided. By picking another ball (making a total of 39) you ensure that the condition must be satisfied.

The process + an example is explained in depth here

I do select questions from medium past papers now (basically all schools that are not the top schools) then near trials/hsc I will do harder/better papers in full (bos trials, past hscs, jrahs and bhhs trials etc.)

Show AB and AC are parallel, because then it follows that B lies along AC (hence, ABC are collinear)

Multiply the numerator and denominator by the denominator's conjugate (root 3). This gives you 4/3 x root(3) - 3/2 = a + b x root(3), so you equate the rational and irrational coefficients.

@5uckerberg can you expand on the link between these two steps? Is it adding/multiplying the inequalities or performing AM-GM on them?

@Octavius old syllabus had lots of 'proof topic' questions (despite the proof topic not existing at the time) that are basically the same for practice. For example, the 2019 4u hsc has induction for question 14c, and an integration logic question for 15a) part i. The 2018 hsc had inequalities...

Assuming you could do the 'prove true for n=1' part... Assume true for n=k; 2^r(r+1) + 2^(r+1)(r+2) + ... + 2^k(k+1) = k2^(k+1) Prove true for n=k+1; LHS = 2^r(r+1) + 2^(r+1)(r+2) + ... + 2^k(k+1) + 2^(k+1)(k+2) = k2^(k+1) +2^(k+1)(k+2) by induction hypothesis = (2k+2)2^(k+1) =(k+1)2^(k+2) =...