Search results

lloll yeh...but if my students ever have that mentality i whip them hard ;p

well from experience in tutoring it most importantly depends on the student.... The strength of the student will determine how they handle 3u or 4u.... However, all in all 4u is definitely not overated....The concepts you are faced with far more complex(get the pun) than in 3u.....the reason...

R has coordinates (x,0) as it lies on the real axis, and since we want PRQ to be a right angle triangle that means angle PRQ is 90 degrees which means we want the gradient of QR x gradient of PR = -1, and this will gives us a quadratic in x which when solved gives us x to be -5 and 3, try this...

Note the x/4 and 3x/4 should both be pie, this is from patel so i know which question it is The first thing you do here is find all the 4 roots of -1 and notice that they occur in conjugate pairs, im sure you can get that far so ill go straight to the rootss...

Another method if you are interested is to first convert into half angles: LHS= (1-(1-2sin^{2}\frac{x}{2})+4isin\frac{x}{2}cos\frac{x}{2})^{-1} this simplifies to: LHS= (2sin\frac{x}{2}(sin\frac{x}{2}+2icos\frac{x}{2}))^{-1} using conjugate properties this becomes: LHS=...

or by de moivre's theorem: let z = cosx + isinx taking both sides to the power of n: z^{n} = cos(nx) + isin(nx) now arg(z^{n}) = nx arg(z)= x hence we get: n[arg(z)]= nx Therefore: n[arg(z)]= arg(z^{n})

the problem with gradient multiplication here is that you want to prove angle OSP is 90 but if you use m1 x m2 = -1 it wont work here cause OS is a vertical line with infinite gradient, so you have to prove that PS has zero gradient. If you did everything right you should end up with p^2 = 1 so...

we want to make z the subject as follows; \frac{2z}{2+i}+(3-2i)= (1-i)z so first we dont want to have any fractions so we times everything by (2+i) 2z+(2+i)(3-2i)= (2+i)(1-i)z now we take the 2z to the other side so we can factor it out: (6+3i-4i-2i^{2})= (2+i-2i-i^{2})z -2z...

The answers to parts (i) and (ii) are just silly. An angle will not be the inverse sine of another angle. Part (ii) is not invariant under scaling. Only part (iii) deserves consideration. To see if the quadrilateral ABDE is always cyclic, one needs to do a calculation. Fix the position and...

lol obvious troll is obvious...this is a really good photoshopped question, it should be quite clear that this is fake, especially cause in i) it is two marks to explain why x is this, and besides an angle cant be the inverse of another angle lol

article made me lol hard...i looked at the paper and did it in less than an hour...i believe the paper was of moderate difficulty i reckon 2006 was still harder, but yeh major lol

yes...you would be looking at about 93-95 depending on how everyone else went

lol love it...you just made 4u seem like it was nothing

yeh even though the paper was pretty easy the problem was it was lengthy, which i think will be a recurring theme

i though it was very easy lol....only one mark in q8 was bad for me last yr when i did it, but i didn't get to finish the last few parts cause i wasted time in other sections

Think of Q and R as points and M is the midpoint so you average the coordinates

a better way...well for my preference..would be to change f(a-x) to f(-(x-a)) then just reflect in a after shifting

he does, but you would probs have to order it in

also if you wanna really excel...you should try and teach yourself most of the whole course by yr 12 and then go onto do like 100s of trial papers...well that's advice if you wanna get a rank lol

yeh that should be right...anyways if you just differentiate your answer..check it is the original eqn just to make sure it is right if you aren't sure