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you could also add that many questions assume the fundamental theorem of calculus, whilst the proof of the theorem is far beyond the syllabus

would euler's formula not be useful in solving equations such as \sin(\theta)=2 and \log(x)=-1? or are these not able to be assessed within the scope of the hsc

since u know that 0< sin(theta) < 1 since 0<theta<pi/2 , u can just put the expression for sin(theta) into the inequality. normally to solve the inequality u would multiply both sides by 14T but yk that T > 0 so u can just multiply everything by T without worrying about the sign switching.

real men go into the exam with no calculator and use taylor series approximations

are these really axioms though? from what i remember from ext 1 is that enumerative combinatorics its built upon the addition principle, multiplication principle and the inclusion-exclusion principle, which can be all be proven. https://math.stackexchange.com/questions/225265/prove-sum-product-rule

1) a rectangle is composed of 2 sides. in a 9x9 grid u can make a rectangle by drawing 2 lines vertically and 2 lines horizontally. theres 10 horizontal lines and 10 vertical lines and hence the number of rectangles is 10C2 x 10C2 2) https://boredofstudies.org/threads/probability-question.84117/...

personally for me i check my answers as soon as i finish the problem (checking computations and the logic behind each step). if i check every question for the first time after its abit overwhelming and sometimes i forget what i actually did in the problem.

surely if we use out of syllabus methods a competent examiner will give us full marks for it. like in this question for example if someone used tywebbs solution or even a contour integral we should be awarded full marks. iirc in last years ext 1 hsc people were still awarded marks for using the...

\text{Let points }A(x_1,y_1) \text{ and } B(x_2,y_2) \text{ be points on the curve } y=ae^{bx} \text{Substituting these in we have: } y_1=ae^{bx_1} \text{ and } y_2=ae^{bx_2} \text{Dividing y1 by y2 yields: }\frac{y_1}{y_2} = e^{b(x_1 - x_2)} b(x_1 - x_2) = \ln\left(\frac{y_1}{y_2}\right)...

i recently found another way to do this integral and thought it was pretty cool \int \frac{x^2 - 1}{x^2+1}\frac{dx}{\sqrt{1+x^4}} = \int \frac{1-\frac{1}{x^2}}{1+\frac{1}{x^2}}\frac{dx}{\sqrt{x^2\left(x^2+\frac{1}{x^2}\right)}} =\int \frac{ 1 - \frac{1}{x^2} }{\left( x + \frac{1}{x} \right)...

after using ibp try rationalising the numerator (multiplying by sqrt(1-x^2)/sqrt(1-x^2)). should be pretty obvious from there

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u guys are tweakin every sdehs task ive seen is harder than any putnam exam and the imo. not even terrence tao could score above a 50% these state rankers are a different breed.

shouldnt it be k^8 det(A) as when a row/column is multiplied by a scalar k then det(A) increases by a factor of k? so since its an 8x8 matrix it would be k^8?

u can write det(WM^-1) as \det (-4MM^{T}M^{-1}) = -4\det(MM^{T}M^{-1}) =-4\det(M)\det(M^{T}M^{-1}) =-4\det(M)\det(MM^{-1}) = -4\det(M) u could also split the product up at the start to get \det (-4MM^{T}M^{-1}) = -4\det(M)\det(M^{T})\det(M^{-1}) =-4\det(M)\det(M)\frac{1}{\det(M)} =-4\det(M)

u can consider cases as |x+1| = -(x+1) for x< -1 and |x-5| = -(x-5) for x < 5. then from there u just solve the inequality for each case (x<-1, -1<x<5 , x > 5) for example -1<x<5 we have: x+1-(x-5) > 7 => -4 > 7 which isnt true so theres no value of x between -1 and 5 that satisfies the...

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for c) the harmonic series diverges so it is bounded above and the lower bound will just be 1

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one way to approach it is using the sum to product identity (sin(A+B) - sin(A-B) = 2cosAsinB).