Daily math problems for September 4, 2026 — Calculus, Algebra, Vectors & more

Rolle: there exists $c\in (a,b)$ with $f^{\mathrm{\prime }}(c)=0$.

$\mathrm{\mid }x-2\mathrm{\mid }=4⟺x-2=4$ or $x-2=-4⟺x=6$ or $x=-2$. The solution set is $\{-2,6\}$.

$\vec{u}\cdot \vec{v}=1\cdot 3+2\cdot 4=3+8=11$.

$\vec{v}=k\vec{u}$ for some $k$ ⇒ $a=k$ and $-2=k$. So $a=-2$.

$d={\displaystyle \frac{\mathrm{\mid }3\cdot 2-4\cdot 3+5\mathrm{\mid }}{\sqrt{9+16}}}={\displaystyle \frac{\mathrm{\mid }-1\mathrm{\mid }}{5}}={\displaystyle \frac{1}{5}}$.

$f(2)=12-14+2=0$. So $f(2025)\cdot f(2)=f(2025)\cdot 0=0$, regardless of $f(2025)$.

$X^2-2$ has irrational roots $\pm \sqrt{2}$, which are not in $\mathbb{Q}$. So it's irreducible over $\mathbb{Q}$. (It splits over $\mathbb{R}$ as $(X-\sqrt{2})(X+\sqrt{2})$.)

Let $L$ be the trail length. Define

- $u(t)$: the monk's distance from the base on Monday at time $t\in [7,17]$ - $d(t)$: her distance from the base on Tuesday at the same clock time

Both functions are continuous on $[7,17]$ (a hiker doesn't teleport).

Now consider $f(t)=u(t)-d(t)$. By the problem statement:

- $u(7)=0$ and $d(7)=L$, so $f(7)=-L<0$. - $u(17)=L$ and $d(17)=0$, so $f(17)=+L>0$.

Since $f$ is continuous and changes sign on $[7,17]$, the Intermediate Value Theorem guarantees some $t^{\ast }\in (7,17)$ with $f(t^{\ast })=0$, i.e. $u(t^{\ast })=d(t^{\ast })$. At that clock time, the monk stands at the same point on the trail on both days. $\mathrm{■}$

The physical intuition ("imagine two monks: one ascending Monday, one descending Tuesday at the same time — they must meet") collapses into a one-line IVT argument once you let $f$ do the work.

By induction, $A^n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$, so $(A^n{)}_{12}=n$.

${\displaystyle \frac{n(n-1)}{2}}=45\Rightarrow n(n-1)=90\Rightarrow n=10$.