Daily math problems for October 24, 2026 — Fundamental Algebra, Algebra, Calculus & more

$24=2^3\cdot 3$ and $36=2^2\cdot 3^2$.

Shared primes at their lowest powers: $2^2$ and $3$.

$\text{HCF}=4\cdot 3=12$.

$f(3)=2^3=8$.

$f^{\mathrm{\prime }}(x)=3x^2-3=3(x-1)(x+1)$.

$f^{\mathrm{\prime }}(x)<0$ when $(x-1)(x+1)<0$, i.e. on $(-1,1)$.

So $f$ is decreasing on $(-1,1)$.

$x^2-5x+6=(x-2)(x-3)$.

The product $(x-2)(x-3)<0$ when exactly one factor is negative — i.e. when $2<x<3$.

Solution: $(2,3)$.

$x=2\cos {\displaystyle \frac{\pi }{3}}=2\cdot {\displaystyle \frac{1}{2}}=1$.

$y=2\sin {\displaystyle \frac{\pi }{3}}=2\cdot {\displaystyle \frac{\sqrt{3}}{2}}=\sqrt{3}$.

So $(x,y)=(1,\sqrt{3})$.

$\bullet$ Simplify exponent and argument:

$${\log }_{2}16=4,i^4=1$$

$\bullet$ The function reduces to:

$$f(x)=x^4-\cos (\pi x)$$

$\bullet$ Differentiate:

$$f^{\mathrm{\prime }}(x)=4x^3+\pi \sin (\pi x)$$

$\bullet$ Evaluate at $x=1$, using $\sin \pi =0$:

$$f^{\mathrm{\prime }}(1)=4+\pi \cdot 0=4$$

$5$ is prime, so ${\mathbb{Z}}_5$ is a field. (Every non-zero element has a multiplicative inverse.)

$e^{a+b}=e^a{e}^b$ ✓ (homomorphism). $e^x$ is strictly increasing (injective) and surjective onto $(0,\mathrm{\infty })$. Hence an isomorphism. (Inverse: $\ln$.)

Vertical: $x=\pm 1$ (two). Horizontal: $\lim f(x)=1$ (one). Total: $3$. (No slant — the rational function has equal-degree numerator and denominator.)

$I_n+I_{n+1}={\int }_0^1{\displaystyle \frac{x^n+x^{n+1}}{1+x}}dx={\int }_0^1{x}^{n}dx={\displaystyle \frac{1}{n+1}}$.