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Could you elaborate on your solution to part (ii) of the polynomials? Btw here is my solution for the part (i): x^3 + px + q = 0 (\beta - \gamma)^2 = \beta^2 - 2\beta\gamma + \gamma^2 = \alpha^2 + \beta^2 + \gamma^2 - \alpha^2 - 2\dfrac{\alpha\beta\gamma}{\alpha} =...

\text{Let } y = \ln|x| e^y = |x| \dfrac{\text{d}}{\text{d}x} e^y = \dfrac{\text{d}}{\text{d}x} |x| \dfrac{\text{d}}{\text{d}y} e^y \dfrac{\text{d}y}{\text{d}x} = \dfrac{|x|}{x} \text{Note that the gradient of } |x| \text{ is 1 for } x>0 \text{ and -1 for } x<0 \text{ just like }...

There are several useful hints that you can look for: 1.) These part (iii) style of questions nearly always use the previous part, so it is useful to think about how you can incorporate part (ii) into part (iii) 2.) On part (iii), there are 3 terms on LHS, but 3^2 = 9 and there is a 9 on RHS...

\text{The trick for question 1 is to use trig identities to make it into a telescoping sum so that the terms cancel out:} \text{Note that: } \cos((k-1)x) - \cos((k+1)x) = \cos(kx-x) - \cos(kx+x) = 2\sin(kx)\sin(x) \text{Therefore, } \sin(kx) = \dfrac{\cos((k-1)x) - \cos((k+1)x)}{2\sin(x)}...

\text{(a)} OP = OT \text{ (equal radii)} \angle OTA = 90^{\circ} \text{ tangent perpendicular to radius} \cos(\alpha) = OT / (OP + PA) = 1/2 \rightarrow \alpha = 60^{\circ} \text{Then, } \gamma = 30^{\circ} \text{Now, } \angle OPT = \angle OTP = 90^{\circ} \text{ (isoceles...

Yeah I also think the question is not true, counter examples can be illustrated here: https://www.desmos.com/calculator/fhj9angyd2

So I was doing this question from the 1967 4U HSC which is as follows: \text{Express } \dfrac{1-abx^2}{(1-ax)(1-bx)} \text{ in the form: } p + \dfrac{q}{1-ax} + \dfrac{r}{1-bx} \text{Given that } R_n(x) \text{ is a polynomial, and that:} 1-abx^2 = (1-ax)(1-bx)(1+u_1x + u_2x^2 + ...

A tad late, but in case anyone was wondering here is my solution: https://i.imgur.com/QO94Yx1.jpg https://i.imgur.com/5RkHkzc.jpg https://i.imgur.com/60EaPkc.jpg https://i.imgur.com/UCvTKrA.jpg https://i.imgur.com/it3NUDJ.jpg

Not sure what you meant by sec(phi), but here's what I got anway: Part (a) https://i.imgur.com/O2Ae7sY.png Part (b) https://i.imgur.com/0tKKBM9.png