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~3 hours/day and my desired uni course is 93.00 so anything above that is fine.

At home: 0-2 hours per day but before exam periods typically 3-6 hours. Most of my "study" is memorising.

Notice that the required answer includes the constant term c . That should be a hint - if we were to differentiate the polynomial, we would lose c , so not only would we have to expand the cube of a binomial (which can be really messy), we would also have to rewrite the result to include c...

Would it be valid to say that since there were r heads in n tosses, each toss has on average an r/n chance to be heads?

I remember learning this in Year 10, though not in the HSC course. But if you start from the observation that the perpendicular heights of the two similar triangles are in the same ratio as the sides then you could finish the question from there by using the area formula.

Is it actually true that Q must always be inside at least one of the ellipses? Suppose P lies on either of the co-ordinate axes. Then the chords of contact from P to both ellipses would be parallel (since the ellipses are 90 degree rotations of each other) and there would be no...

~70% and yes

i) Hint: the expression you need to prove can be rearranged to \cot \theta = \frac{1 + \tan n\theta\, \tan (n+1)\theta}{\tan(n+1)\theta - \tan n\theta} ii) Starting from the inductive hypothesis: \sum_{n= 1}^k \tan \theta\tan (n+1)\theta = -(k+1) + \cot \theta \tan (k+1)\theta...

Year 11 is still pretty early on, so you have a lot of time (many years) to decide. People will change their interests during high school, the HSC and even while at uni. For example, it might seem unlikely now, but later on you might develop a passion for math, science or art. Don't worry...

The horizontal range of the ball is maximised when \alpha = \pi / 4. This can be easily shown by examining the equation in ii). Case 1: The ceiling allows for an angle of projection of \pi / 4. If the ceiling allows for at least this angle of projection, then \alpha = \pi / 4 is the optimal...

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Well since there are still solutions that exist then there should be at least one complex number that works.

Also, I noticed that when you find the third (real) root using sum of roots and then sub these three roots into the product of roots, you seem to get an equation with certain solutions for k that contradict the question.

\cos \theta =\frac{1}{2\sqrt 2} \cos \alpha + \frac{1}{\sqrt 2}\sin \alpha Applying the auxiliary angle transformation: \begin{aligned} \cos \theta &= \sqrt{\left( \frac{1}{\sqrt 2}\right)^2 + \left( \frac{1}{2\sqrt 2} \right)^2} \cos \left(\alpha - \tan ^{-1} \frac {\left( \frac{1}{\sqrt2}...

89 -> 93

Consider \triangle ATP . \sin \alpha = \frac{h}{TP} \sin \frac{\pi}{4} = \frac{h}{AT}= \frac{1}{\sqrt2} By the cosine rule, AP^2 = 2h^2 + h^2 {\rm cosec}^2\, \alpha - 2\sqrt2\, h^2\, {\rm cosec\,} \alpha \cos\theta Equating and dividing out by h^2 (the height cannot be zero)...

Are those not triple roots?

I'm having some trouble with part (h). If one of the non-real roots had real part k , then by the complex conjugate root theorem, the sum of the two non-real roots should be 2k . Then, using the sum of roots, the last root should be k . But (k,\,P(k)) is the point of inflection of...

I didn't address those because they seemed trivial to me - my bad. If k is positive, then k + \sqrt{k^2-1} must be positive since two positive numbers are being added. If k is negative, then k - \sqrt{k^2-1} must be negative too since a positive number is being subtracted from a...

a) Consider the graph of y = P(z) on the cartesian plane, for real z . P(0) = -k P'(z) = 3(z^2-2kz+1) P'(0) = 3 P'(z) = 0,\, z = k \pm \sqrt{k^2-1} (Note that we require |k| \geq 1 for a turning point to exist - otherwise P(z) has exactly one real root) We will prove that...