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\textrm{for part b} \textrm{from part a you should get }\frac{d}{dx} xtan^{-1}x = tan^{-1}x + \frac{x}{1+x^{2}} texture{That means }\int tan^{-1}x + \frac{x}{1+x^{2}} dx = \int tan^{-1}x dx + \int \frac{x}{1+x^{2}} dx = x tan^{-1} \textrm{The second integral is easy to find...

\textrm{Base case} , n=1, \textrm{then} \ \varnothing \ \textrm{and the whole set are the only subsets and the statement follows.} \textrm{Suppose it is true for} \ n=k: \ \textrm{the set A with k elements has } 2^{k} \ \textrm{subsets} \textrm{where } A = \{a_{1},a_{2},\dots, a_{k}...

\textrm{Q22. If you draw what's happening this is what you get:} \textrm{Then observe triangles BCQ , ABC and APB are isosceles,} AP = AB =1, BC = CQ \ \textrm{and} \ BC= AB =1. \textrm{Also APQ, ABC and PBQ are right angle triangles. } \textrm{since the angle CAB is 45...

\textrm{I am going to do the first part in details then the second part is going to be exactly the same.} \textrm{ Write out the equations of the motion:} x = 50 \ cos\theta \ t , y = 50\ sin\theta \ t - \frac{1}{2} g \ t^{2} = 50\ sin\theta \ t - 4.9 \ t^{2} \ \textrm{where g =9.8}...

\textrm{To find the lower bound you need to use the following inequality:} \sum_{k=1}^{2016} k! = 2016! (1 + \frac{1}{2016} + \frac{1}{2016 \times 2015} + \cdots + \frac{1}{2016 \times 2015 \times 2014 \times \cdots \times 1}) < 2016!(1 + \frac{1}{2016}+ \frac{1}{2016 \times 2015}+...

\textrm{Note in part d} \ x^{2} = sin^{2}\theta + 2sin\theta \ cos\theta + cos^{2}\theta \textrm{and} \ y^{2} = sin^{2}\theta -2sin\theta \ cos\theta + cos^{2}\theta \textrm{Hence}\ x^{2} + y^{2} = 2(sin^{2}\theta + cos^{2}\theta) = 2 \textrm{Second method is to observe} \ x+y = 2...

\textrm{In order to understand the bisection method you need to understand the following idea:} \textrm{Let's use an easy example} \ y=x^{2}-9 \ \textrm{on the interval [1,7]} \textrm{at} \ x=1, y=-8< 0 \ \textrm{but at} \ x=7, y=40 > 0 \ \textrm{so we have two points, A=(1,-8) and...

\textrm{This is another way of looking at it} \textrm{Well, it's obvious that if } \ f(x) \textrm{or} \ g(x) \ \textrm{is a zero function then h(x) is also zero and the statement follows} h(x) = h(-x) = 0 \textrm{we can make the following observation that if f(x) is odd then } f(x) f(-x)...

\textrm{This is another way of solving the question.Let Albert =B, Brian =B and last boy = C} \textrm{Let the girl to be denoted by} \ G_{1}, G_{2} \ \textrm{and} \ G_{3} \textrm{i) Note if Albert is infant of Brian then we have something like} \ .. B ... A ... \textrm{but if we...

\textrm{The base case is obvious.} \textrm{Assume the statement is true for n=k.} 7 + 77 + 777+ \dots + \underset{k \ \textrm{times}}{ 777 \dots 7} = \frac{7}{81}(10^{k+1} - 9k -10 ) \textrm{Note} \ 10(7+ 77 +\dots + \underset{k \ \textrm{times}}{ 777 \dots 7} ) + 7k + 7 = 7 + 77 +...

Re: HSC 2017 Maths (Advanced) Marathon x=\frac{6}{5} \Rightarrow y = 1+\frac{1}{x} = 1 + \frac{1}{\frac{6}{5}} = 1 + \frac{5}{6} = \frac{11}{6}

\textrm{well, long division is the fastest way but this is a way that does not use long division.} Q(1) = 0 \Rightarrow p+q = 3 Q(x) = (x^{2} +2)T(x) +1-7x = 0 , \ \textrm{where} \ T(x) = ax+bx+c Q(0) = 1 = 2T(0) +1 \Rightarrow T(0) = 0 \Rightarrow c= 0 Q(1) = 3T(1)-6 = 0...

\textrm{Well, if Megan receives the bonus every year then each she gets :} 83000 + 83000*0.05 = 830000*(1.05) = 87150 \ \textrm{every year} \textrm{let's see what Paul's Salary is going to be after 1 and 2 years :} \textrm{After one year:} \ 83000+83000*0.03 = 83000*(1.03) = 85490...

\textrm{let} \ t= t_{1}, x =x_{1} \ \textrm{such that } \ v_{t_{1}} = v_{x_{1}} = 0 note a= \frac{dv}{dt} = v \frac{dv}{dx} to get the distance: a= v \frac{dv}{dx} = -(100 + v^{2}) \Rightarrow \frac{dv}{dx} = - \frac{100}{v}- v \Rightarrow \int_{0}^{x_{1}} \frac{v'}{\frac{100}{v}...

This is what I found, hopefully the link works. http://www.pasthsc.com.au//HSC_Mathematics_files/Maths2U04QuickAnswers.pdf

\textrm{a)} \ \frac{d}{dx}(x^{\frac{5}{3}}) = \lim_{h\to 0} \frac{(x+h)^{\frac{5}{3}} - x^{\frac{5}{3}} }{h} = \lim_{h\to 0} \frac{(x+h)(x+h)^{\frac{2}{3}} - x x^{\frac{2}{3}}}{h} = \lim_{h \to 0} x (\frac{(x+h) ^{\frac{2}{3}} - x^{\frac{2}{3}} }{h}) + \frac{h (x+h) ^{\frac{2}{3}} }{h} =...

Well T_{0} = 20, T_{f} = 120 , k =5 Using the Newton's law: T(t) = T_{f} + Ae^{-5t} \Rightarrow T(0) = 20 = 120 + Ae^{0} \Rightarrow A=-100 We want to find the time it takes for for the body temp reaches 100, so assume for some t, T(t) =100: T(t) = 120 - 100 e^{-5t} =100...

Your method is quite elegant and smart, but there is a slightly simpler method which uses this identity: (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac) =a^{3}+b^{3}+c^{3}-3abc \textrm{Let} a= 2^{x} -4, b =4^{x}-2, c= 6-4^{x}-2^{x} hence a+b+c= 0, note a^{3}+b^{3}+c^{3} = 0 \therefore...

which question are you referring to and what is confusing about the question?

If \frac{1}{x} + x = A then find the below expression in terms of A x^{7}+ \frac{1}{x^{7}}