Daily math problems for August 31, 2026 — Trigonometry, Matrices, Calculus & more

$\mathrm{\mid }\sin x\mathrm{\mid }\le 1$ for any real $x$, so $\sin x=2$ has no solutions.

$\mid \begin{array}{cc}3& 1\\ 2& 2\end{array}\mid =3\cdot 2-1\cdot 2=6-2=4$.

$x-1\to 0^{+}$, so ${\displaystyle \frac{1}{x-1}}\to +\mathrm{\infty }$.

Base along the $x$-axis has length $3$; the third vertex sits at height $4$. Area $=\frac{1}{2}\cdot 3\cdot 4=6$.

$\bullet$ Find vertex $A=AB\cap AC$ by solving $x-2y=-3$ and $2x-y=3$:

$$A=(3,3)$$

$\bullet$ Find vertex $B=AB\cap BC$ by solving $x-2y=-3$ and $3x+2y=-1$:

$$B=(-1,1)$$

$\bullet$ Find vertex $C=AC\cap BC$ by solving $2x-y=3$ and $3x+2y=-1$:

$$C=(\frac{5}{7},-\frac{11}{7})$$

$\bullet$ Apply the shoelace-style area formula:

$$\mathcal{A}=\frac{1}{2}\mid x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\mid$$

$\bullet$ Plugging the coordinates gives $\frac{1}{2}\cdot \frac{96}{7}$:

$$\mathcal{A}=\frac{48}{7}$$

$(AB)(B^{-1}{A}^{-1})=A(B{B}^{-1})A^{-1}=AI{A}^{-1}=I$. So $(AB{)}^{-1}=B^{-1}{A}^{-1}$.

$f(\pm \sqrt{3})=0$, so Rolle applies. $f^{\mathrm{\prime }}(x)=3x^2-3=0\Rightarrow x=\pm 1$.

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }$. By even symmetry, ${\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx={\displaystyle \frac{\sqrt{\pi }}{2}}$.

$A^2=\left(\begin{array}{cc}0\cdot 0+1\cdot 0& 0\cdot 1+1\cdot 0\\ 0\cdot 0+0\cdot 0& 0\cdot 1+0\cdot 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$. ($A$ is nilpotent.)

$f^{\mathrm{\prime }}(x)={\displaystyle \frac{(1\mathrm{/}x)\cdot x-\ln x\cdot 1}{x^2}}={\displaystyle \frac{1-\ln x}{x^2}}$. At $x=1$: ${\displaystyle \frac{1-0}{1}}=1$.