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Re: International Baccalaureate Maths Marathon They mean the derivative is non-negative for all (real) x. We can thus use discriminants to obtain the desired inequality.

Re: International Baccalaureate Maths Marathon $\noindent It's a typo in the definition of the function, it should just be$ $$f(x) = x^{3} + px^{2} + qx,$$ $\noindent i.e. leading coefficient should be $1$.$

$\noindent Assume downwards is positive direction. We have $\ddot{x} = 2-2x$. Since $\ddot{x} = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{2}v^{2}\right)$, we get $ $$\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{2}v^{2}\right) &= 2 - 2x \\ \Rightarrow \frac{1}{2}v^{2} &= 2x...

Re: Several Variable Calculus Yeah it is a boundary point. Here's some hints of a possible method. Note that S is just the graph of y = f(x) := sin(1/x) (x =/= 0). Use (or show) the fact that f attains the value 1 for arbitrarily small values of x > 0 to deduce that (0, 1) is a boundary point...

Re: MATH2601 Linear Algebra/Group Theory Questions For a), we can do it like this: since T is from V -> V and < u, T(v) > = 0 for all u, v in V, for each v, just take u = T(v), and the result will follow.

Here is their solution with LaTeX:

Maybe the way LaTeX is used on this site has changed over the years.

Re: Several Variable Calculus 1) True 2) True 3) False 4) False 5) True 6) True

Re: Discrete Maths Sem 2 2016 $\noindent The definition of $\preceq$ being a reflexive relation on $S$ is $f\preceq f$ for all $f\in S$ (so \emph{same function both times}).$

Re: Discrete Maths Sem 2 2016 Let S be the set of all functions from ℝ -> ℝ. To show ⪯ is reflexive, we need to show that for every f in S, f ⪯ f. This is quite straightforward: Let f be any element of S. Then clearly for every x in ℝ, we have f(x) ≤ f(x) (this is implied by the trivial fact...

Re: Discrete Maths Sem 2 2016 If you do the Euclidean algorithm calculations correctly, you should end up with something like -31*79 + 25*98 = 1. Multiplying through by 5, we have the desired x as being (-31)*5 = -155 modulo 98, which is 41 modulo 98. So the solution to the congruence is x ≡...

Re: Discrete Maths Sem 2 2016 Note that solving 79x ≡ 5 (mod 98) is equivalent to finding integers x and y such that 79x = 5 + 98y <==> 79x - 98y = 5. Since -y is an integer (as y is an integer), this is equivalent to writing 79x + 98b = 5 for some integer b. So our goal is to express 5 in...

Maybe because it's very different and largely irrelevant in general.

$\noindent In two variables, the solution set can be written as $\left\{\left(a, \frac{4-a}{2}\right) : a \in \mathbb{R}\right\}$, or equivalently $\left\{\left(4-2a, a\right) : a \in \mathbb{R}\right\}$.$

The digital root of a non-negative integer is the single digit value we get if we find the sum of its digits and keep doing this on the result until we end up with a single digit answer. Examples: The digital root of 25 is 2+5 = 7. For 123, 1+2+3 = 6, so the digital root of 123 is 6. For...

$\noindent For example in the first one, the solution set is just $\left\{\frac{5}{2}\right\}$ if viewed as an equation in one variable, whilst it is $\left\{(x_{1}, x_{2})\in \mathbb{R}^{2}: x_{1} = \frac{5}{2}\right\}$ if viewed as an equation in two real variables (basically $x_{1} =...

It's asking us to write down the set of solutions to those equations (viewing them as an equation in the mentioned number of variables).

Re: International Baccalaureate Maths Marathon Yes, you can use (a, b). You can read about the two notations here: https://en.wikipedia.org/wiki/Interval_(mathematics)#Including_or_excluding_endpoints .

Re: International Baccalaureate Maths Marathon They are basically saying the limit is infinity (the numerator has higher degree than denominator).

Re: Statistics The reason for terms like "population" and "tagged" is that one place this distribution comes up is in ecology when we tag some members of an animal population (like a fish population) and then later draw (without replacement) a random sample from the animal population and count...