Daily math problems for September 1, 2026 — Algebra, Trigonometry, Analytic Geometry & more

$(\mathbb{N},+)$ has no inverses (e.g., $1+?=0$ has no solution in $\mathbb{N}$). It's a monoid, not a group.

Reference angle $\frac{\pi }{4}$. In $[0,2\pi )$ cosine is $\frac{\sqrt{2}}{2}$ at $x=\frac{\pi }{4}$ (Q1) and $x=\frac{7\pi }{4}$ (Q4).

Reading off: $h=2$, $k=-3$. Center: $(2,-3)$.

$\det (2A)=2^3\det (A)=8\cdot 5=40$.

$a_5={\displaystyle \frac{a_3+a_7}{2}}={\displaystyle \frac{11+27}{2}}=19$.

$\tau (1)=3$, then $\sigma (3)=1$. So $(\sigma \circ \tau )(1)=1$.

Sorted: $\{1,3,4,5,6,8\}$. The two middle values are $4$ and $5$. Median $={\displaystyle \frac{4+5}{2}}=4.5$.

$\bullet$ Differentiate both sides implicitly (product rule on $xy$):

$$2x+y+x{y}^{\mathrm{\prime }}+2y{y}^{\mathrm{\prime }}=0$$

$\bullet$ Substitute $(x,y)=(1,1)$:

$$2+1+y^{\mathrm{\prime }}+2y^{\mathrm{\prime }}=0$$

$\bullet$ Solve for $y^{\mathrm{\prime }}$:

$$3y^{\mathrm{\prime }}=-3\Rightarrow y^{\mathrm{\prime }}=-1$$

$\cos u=-\frac{1}{2}$ on $[0,4\pi )$ at $u=\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3}$. Dividing by $2$: $x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}$. Four solutions.

Left branch ($x<0$): $f(x)=x+1\in (-\mathrm{\infty },1)$, strictly increasing. Right branch ($x\ge 0$): $f(x)=2x+1\in [1,+\mathrm{\infty })$, strictly increasing.

The two images $(-\mathrm{\infty },1)$ and $[1,+\mathrm{\infty })$ are disjoint and together cover all of $\mathbb{R}$, so $f$ is surjective. Each branch is strictly increasing, and the ranges don't overlap, so no two distinct $x$-values give the same $f(x)$ — $f$ is injective. Hence $f$ is bijective.

(Note: $f(-1)=0$ and $f(0)=1$, so distractor A is factually false.)