Daily math problems for July 25, 2026 — Polynomials, Combinatorics, Calculus & more

$x_1{x}_2{x}_3=-{\displaystyle \frac{-4}{2}}=2$.

$4\cdot 5\cdot 3=60$.

Cylinder of radius $2$ and height $5$: $V=\pi \cdot 4\cdot 5=20\pi$.

$\bullet$ Evaluate each of the nine limits to fill the matrix:

$$\left(\begin{array}{ccc}\sqrt{3}\mathrm{/}2& 2& 1\\ 3& 4& 4\\ 0& 3& 1\end{array}\right)$$

$\bullet$ Expand along row 1. The cofactor of the $(1,1)$ entry is $4\cdot 1-4\cdot 3=-8$:

$$\frac{\sqrt{3}}{2}\cdot (-8)=-4\sqrt{3}$$

$\bullet$ Cofactors of the $(1,2)$ and $(1,3)$ entries are $3\cdot 1-4\cdot 0=3$ and $3\cdot 3-4\cdot 0=9$:

$$-2\cdot 3+1\cdot 9=3$$

$\bullet$ Add the three contributions:

$$\det =3-4\sqrt{3}$$

By Pascal: $\left(\genfrac{}{}{0px}{}{7}{3}\right)+\left(\genfrac{}{}{0px}{}{7}{4}\right)=\left(\genfrac{}{}{0px}{}{8}{4}\right)={\displaystyle \frac{8!}{4!4!}}=70$.

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.)

$\sqrt{x+5}=2+\sqrt{x-3}$. Square: $x+5=4+4\sqrt{x-3}+x-3\Rightarrow 4=4\sqrt{x-3}\Rightarrow \sqrt{x-3}=1\Rightarrow x=4$. Check: $\sqrt{9}-\sqrt{1}=3-1=2$ ✓.

Invertibility ⇔ non-zero determinant.

Pair each subset $S$ with $S\mathrm{△}\{1\}$ (i.e., toggle the element $1$). This pairing has no fixed points and flips the parity of the sum, so even-sum and odd-sum subsets are equinumerous: $2^{10}\mathrm{/}2=512$.

$\bullet$ Eliminate the first column using $2R_2-R_1$ and $2R_3-R_1$:

$$2R_2-R_1=(0,3,2a-1,3)$$

$$2R_3-R_1=(0,-1,1,2b+1)$$

$\bullet$ For rank $2$, these two rows must be proportional. Matching the second components gives ratio $-3$:

$$2a-1=-3\cdot 1\Rightarrow a=-1$$

$$3=-3(2b+1)\Rightarrow b=-1$$

$\bullet$ So the answer is:

$$(a,b)=(-1,-1)$$