That polynomial does not vanish at 1 and so is not in the span of either of your bases (nor does it need to be).
Your basis is equivalent to his since (x-1)^2=x(x-1)-(x-1).
(Also you probably shouldn't use the double implication symbol as informally as in the above post, you should only use...
Re: HSC 2017 MX2 Marathon
$Two circles $\Gamma_1,\Gamma_2$ of differing radii share a common point of tangency $P$ and lie on the same side of the tangent line $\ell$. Let $M$ be the point on $\ell$ such that $\angle Q_1MQ_2$ is maximised, where $Q_j$ is the point other than $P$ on $\Gamma_j$...
This is not correct, it is a parabola.
Direct calculation:
(x-2)^2+(y-3)^2=(y+1)^2
=> 8y=x^2-4x+12.
You could also immediately conclude that it is a parabola from the geometric definition of a parabola, and read off the focal length a by looking at the distance between the given focus and...
There are other factors that such statistics will be impacted by though. Are the same proportion of these "less talented" students truly interested and driven in their studies? Note also that I am not claiming that it will be as easy for the student as it would be with some talent, but rather...
x^{-1/2}=\frac{1}{3}\left(\frac{x-9}{9}+1\right)^{-1/2}=\frac{1}{3}\sum_{k=0}^\infty 3^{-2k}\binom{-1/2}{k}(x-9)^k.
You can also express the binomial coefficient in terms of factorials and powers of 2 if you like, but I leave that to you.
Super old post, but now that the thread has been bumped I should mention that I disagree with this. You can talk all you want about natural ability in math and limitations of how much you can achieve, but I personally think that a very large proportion of people are capable of understanding most...
$Hint: \\ If $\triangle ABC$ is a triangle, and $X$ is a point on the line segment $BC$ with $BX:CX=\lambda$, then $|\triangle ABX|:|\triangle ACX|=\lambda.$ $
Try to figure this out yourself, it is just a matter of definitions and you will learn 100x more deciding between the two options yourself.
Reasoning for the others:
2. We want our plane to pass through the origin, because the system is homogeneous. The correct answer is the plane through the...
What you need to know:
The number of free parameters is the dimension of the solution space.
This is equal to the dimension of the kernel of your linear operator/matrix. (Or complementary to the rank of your linear operator/matrix.)
You can read these things off directly from a row echelon form.
Because working with power series is pretty low level analysis, you don't need much theory development before you can start saying meaningful things about functions defined via power series. (Eg it is smooth and invariant under differentiation, which almost uniquely characterises the...
How does that make it any less cheating?
You should use the definition of the exponential explicitly (presumably it is the power series centred at 0, so the argument is much the same). Don't assume anything about the differentiability of the exponential if the limit you are asked to evaluate...
They are both true, your statement is just weaker. IBP usually gains you stuff when you are analysing oscillatory expressions...if you just look at the size of things via absolute values, you ignore a lot of the "cancellation" that comes from the oscillation. IBP picks this up.
Edit: Note that...
Let f(n) be the quantity for a given n. It is pretty easy to show that f(mn)=f(m)f(n) for coprime m and n (a and b each uniquely factorise into the product of something that divides m and something that divides n), and also that f(p^n)=(n+1)(n+2)/2 by manual counting.
Hence...
The last two terms vary in meaning between different authors etc. One of them could very well have a nonempty requirement in your course ("region" sometimes has this in particular, "domains" are a little more standard and usually just refer to open connected subsets of C in this context.)
This kind of question makes me wish that geometric transformations were discussed more in HS. The reduction of a geometric fact about conics to a geometric fact about circles using parallel projection is kinda cute.