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6086555670238378989670371734243169622657830773351885970528324860512791691264 12 The two numbers with the same thing in common that I find really cool (I cant say favourite, Ive got too many)

Are you doing the 4U HSC 2012 or 2013?

OH, I see that ok thank you. I feel so noob compared to others lol

But isnt it true that we can: \frac{1}{T_n}=\frac{1}{T_{n-1}}+n Which is then S_n=S_(n-1)+n? I dont get what you mean by shifting the subscript down for the n-term

My solution The coefficient in front of T for the denominator turns out double Realise's ?

Hmm, if this is the case then why is my answer wrong? (taking into account Realise's correction of my solution)

Are there non-integer rational solutions?

Ah yes of course, oops :/ Then yeah it would work out to your answer

\frac{1}{T_{n+1}}=\frac{1}{T_n}+n \\ \\ $let$ \ \ S_n=\frac{1}{T_n} \\ \\ \therefore S_{n+1}=S_n+n \\ \\ S_{n+1}=S_{n-1}+n+n\\ \\ S_{n+1}=S_0 n(n+1) \\ S_0=\frac{1}{T} \\ \\ S_{2012}=\frac{1}{T}(2011)(2012) \\ \\ \therefore T_{2012}=\frac{T}{(2011)(2012)} ??

The answer should be: \sqrt{7}+\frac{1}{\sqrt{7}}=\frac{8}{\sqrt{7}} Either the answer is wrong or you misread the answer (or misread question). I differentiated it myself to see if you differentiated incorrectly but it is correct

We can derive that the range of a ball is: x=\frac{V^2 \sin 2\theta}{g} The maximum range occurs at 45 degrees (can be shown mathematically) x_{max}=\frac{V^2}{10} We are going to assume g=10. Now to calculate V, we know that he can throw vertically upwards a ball for 40m, lets find the...

Ok so we find the cartesian equation of motion of the projectile and sub in x=a y=b: b=\frac{-ga^2}{2V^2 \cos^2 \theta}+a\tan \theta Now we are asked to find the greatest height given the restriction above. The greatest height in general for ANY V and any theta will be (after finding...

Also I will add to Spiral's explanation. Notice x^2-x(\alpha+\beta)+\alpha \beta =0 The fact is, all this time when you were evaluating quadratic equations by using the rule: 'Find two numbers which when we add them you get the middle term and when we multiply them we get the constant'...

Oh yes, of couse lol. well ax^2+bx+c=0. Do something to the x^2 term (and hence the whole expression) to get it into a desirable form in order to then utilise the equivalence to (x-alpha)(x-beta)

Introduction to Realise's question: i) Given that (x-\alpha)(x-\beta)=0 and (x-\alpha)(x-\beta) \equiv ax^2+bx+c Prove that: \alpha \beta = \frac{c}{a} \alpha + \beta =\frac{-b}{a} $(Where alpha and beta are roots to the equation)$ \ \ ax^2+bx+c=0 Then continue with realise's...

Have y10 learnt \alpha + \beta = \frac{-b}{a} \\ \\ for \ \ ax^2+bx+c=0 Where alpha and beta are roots.? I had the impression that they were introduced in y11.

I wouldnt call it trial and error simply because it only takes 1 'trial' if you know what you are doing in terms of triads In the HSC this rarely happens I would think, every hard question they give they lead you towards the answer through various parts. Like for example in one hard end...

Notice that they are asking for an integral answer. So one in the form of integers. But if we split our isosceles triangle in half, we get two right angled triangles. Each of these sides have an integer side (we can assume height is an integer since the area 12 is an integer, so height must be...

Good job =)

I originally thought of differentiating equating the gradient to 10, then finding the point of intersection of of the tangent and the parabola but it leads you nowhere really. My way is 2U and probably want they wanted you to do