Daily math problems for October 2, 2026 — Determinants, Linear Function, Derivatives & more

$\det A=3\cdot 2-4\cdot 1=2$.

$2x>8\Rightarrow x>4$, so the set is $(4,+\mathrm{\infty })$.

$f^{\mathrm{\prime }}(x)=2e^{2x}$, so $f^{\mathrm{\prime }}(0)=2\cdot e^0=2$.

$\bullet$ Rewrite the logarithm using $\ln (1+u)\sim u$:

$$\ln \frac{n^2+an+1}{n^2+1}=\ln \text{\hspace{0.17em}⁣}(1+\frac{an}{n^2+1})\sim \frac{an}{n^2+1}$$

$\bullet$ Multiply by $n$ and take the limit:

$$\underset{n\to \mathrm{\infty }}{\lim }n\cdot \frac{an}{n^2+1}=a$$

$\bullet$ Require $a\ge 3$; the smallest integer satisfying this is:

$$a=3$$

$x+e-3=x\Rightarrow e=3$.

$A$ is $2\times 3$, so $A^T$ is $3\times 2$.

$\sin {\displaystyle \frac{7\pi }{6}}=-{\displaystyle \frac{1}{2}}$. Then $\arcsin \text{\hspace{0.17em}⁣}(-{\displaystyle \frac{1}{2}})=-{\displaystyle \frac{\pi }{6}}$ (in the principal range).

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.)

$\cos u=-\frac{1}{2}$ 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.

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

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