Note that it satisfies the following recursive formula: $$a_{n+2}=3a_{n+1}+2a_n\tag{$\star$}$$ where $a_n=\left(\frac{3+\sqrt{17}}2\right)^n+\left(\frac{3-\sqrt{17}}2\right)^n$. Thus, if $a_{n+1}$ is odd, then $a_{n+2}$ is odd. $(\star)$ comes from noting that $$a^2=3a+2\implies a=\frac{3\...

Note that the derivative of $$\frac{x+k}{x^2+k}$$ is $$\frac{-2kx+k-x^2}{(x^2+k)^2}$$ By the quotient rule. Since $(x^2+k)^2 \ge 0$, if the function is decreasing we have $$-x^2-2kx+k \le 0 \iff x^2+2kx-k \ge 0$$ For all $x$. Since this polynomial is always positive, the discriminant must not be ...

Note that it satisfies the following recursive formula: $$a_{n+2}=3a_{n+1}+2a_n\tag{$\star$}$$ where $a_n=\left(\frac{3+\sqrt{17}}2\right)^n+\left(\frac{3-\sqrt{17}}2\right)^n$. Thus, if $a_{n+1}$ is odd, then $a_{n+2}$ is odd. $(\star)$ comes from noting that $$a^2=3a+2\implies a=\frac{3\...

We are allowed to use a calculator in our linear algebra exam. Luckily, my calculator can also do matrix calculations. Let's say there is a task like this: Calculate the rank of this matrix: $$M =\begin{pmatrix} 5 & 6 & 7\\ 12 &4 &9 \\ 1 & 7 & 4 \end{pmatrix}$$ The problem with thi...