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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).

the both have a common factor of (n-r-2)! so u can just multiply it by (n-r-1) like how 1/3 + 1/6 = 2/6 + 1/6 instead of 6/18 + 3/18

im pretty sure u can use the heaviside cover up method for partial fractions without proof

lol these emojis are so silly like has anybody ever used :chainsaw2:

yup 👍

yup basically just adding that limit notation to show that ur actually taking the infinite sum

if u do take the limit u get that the result would equal 1 + ln(2) + ln(2) > 1+ln2 u can take the limit from the start just by writing \lim_{k \rightarrow \infty} \sum_{n=2}^{k} \ln \left( \frac{n^2}{n^2-1} \right ) and if u want to expand the sum out to see the telescoping series the result...

infinity isnt a number most of the time when u see infinities in math ur actually just taking a limit to infinity. the infinity commonly used in math is just another way of saying what happens as this number gets really big. u can still make a telescoping sum if the symbol used is infinity it...