Daily math problems for July 16, 2026 — Functions, Sets, Systems of Linear Equations & more

$g(2)=5$, then $f(5)=2\cdot 5-1=9$. Hence $(f\circ g)(2)=9$.

$A\cap B=\{2,3\}$. The complement in $U=\{1,2,3,4,5\}$ is $\{1,4,5\}$.

Multiplying the first by $2$: $2x+2y=2\mathrm{\ne }5$. Hence the system is inconsistent — no solution.

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

$\mathrm{\mid }x-2\mathrm{\mid }<3\iff -3<x-2<3\iff -1<x<5$. The solution set is $(-1,5)$.

The integral of a non-negative continuous function over $[a,b]$ is non-negative (representing the area under the curve).

With $Y=X^2$: $Y^2-5Y+4=0\Rightarrow Y\in \{1,4\}$. Then $X^2=1\Rightarrow X=\pm 1$ and $X^2=4\Rightarrow X=\pm 2$. Four real solutions.

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

$R=\sqrt{1+3}=2$, so $f(x)=2\sin (x+\pi \mathrm{/}3)$. Max value: $2$.

$P(-1)=0$, so $(X+1)$ is a factor. Polynomial division: $P(X)=(X+1)(X^2-4)=(X+1)(X-2)(X+2)$.