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$A line has$ \ x \ $and$ \ y \ $intercepts$ \ \ p \ $and$ \ q \ \ $respectively$ $What is the gradient of this line?$

Don't general maths students do algebra and making variables a subject sort of stuff

Re: HSC 2013 4U Marathon $Let$ \ x> 1 $Evaluate$ \frac{x}{x+1} + \frac{x^2}{(1+x)(1+x^2)} + \frac{x^4}{(1+x)(1+x^2)(1+x^4)} + \dots

$i) Show that if$ \ \ ax^2 + bx + c = 0 \ \Rightarrow \ \ \left(x+ \frac{b}{2a} \right)^2 + \frac{c}{a} = \frac{b^2}{4a^2} $ii) Hence prove by making$ \ \ x \ \ $the subject$ x= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Re: HSC 2013 4U Marathon First, we will prove this is true for distinct a,b,c $Without loss of generality, take$ \ \ a>b>c a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) = (a-b)(a-c)(b-c) \left(\frac{a}{b-c} - \frac{b}{a-c} + \frac{c}{a-b} \right) $The multiplier at the front is positive due...

Re: HSC 2013 4U Marathon $Let$ \ \ f(x) \ \ $be a differentiable function that satisfies$ f(x+y) = f(x) f(y) \ \ $for all$ \ \ x,y \in \mathbb{R} f'(0) = 1 $Find$ \ \ f(x)

Re: HSC 2013 3U Marathon Thread Yep

Re: HSC 2013 4U Marathon Why don't we just add 2014 posts on here I think its better to have a megathread rather than individual threads for each year Can someone post a question

Re: HSC 2013 3U Marathon Thread $Find$ \sum_{k=0}^m \binom{j}{k} \binom{i}{m-k}

Re: HSC 2013 4U Marathon $Let$ \ \ \lfloor z \rfloor \ \ $be the greatest integer less than or equal to$ \ z $That is$ \ \ \lfloor \pi \rfloor = 3 $Prove that$ \int_1^a \lfloor x \rfloor f'(x) \ dx = \lfloor a \rfloor f(a) - (f(1) + f(2) + \dots + f(\lfloor a \rfloor) )

Re: HSC 2013 2U Marathon $Consider the graph of$ \ \ y= \frac{1}{x} $i) Using the trapezium rule for$ \ \ n+1 \ \ $function values$ \ \ x=n \ \ $to$ \ \ x=2n $Show that$ \frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n} > \ln 2 + \frac{3}{2n} \ \ \ \fbox{3} $ii) State a way to...

Re: HSC 2013 2U Marathon Its like saying \frac{d}{dx} \sin x = \cos x THEREFORE \int \cos x = \sin x + c we are just reversing what we've done, in this case I re-arranged the integrals \frac{d}{dx} x\ln x = \ln x + 1 x\ln x = \int (\ln x + 1) \ dx x \ln x = \int \ln x \ dx + \int 1...

Re: HSC 2013 2U Marathon Do I really need to say, provide proof

Re: HSC 2013 2U Marathon $By making use of differentiation by first principles of$ \ \sin x $EVALUATE$ \lim_{h\to 0} \frac{\sin h}{h}

Re: HSC 2013 2U Marathon I know you've memorised the integral of ln x and you just want to take every opportunity to show off your supreme memorisation power

Re: HSC 2013 2U Marathon The proper answer is: \frac{d}{dx} x\ln x = \ln x + 1 Integrate both sides \therefore \ x\ln x = \int \ln x \ dx + \int 1 \ dx \therefore \ \ \int \ln x \ dx = x\ln x- x + c Then you just sub in values

Re: HSC 2013 2U Marathon Bumping for more exposure

Re: HSC 2013 2U Marathon nah mate just stick to your \int_1^2 2x \ dx

Differentiating does not necessarily hold the inequality Counter example: 2x + 100 > 5x - 100 (for certain values of x) but 2 > 5 is not true for any

Unfortunately there have been multiple occasions where I'd do the exam twice since I'd finish early And I'd make the same silly mistake twice =)