Daily math problems for August 20, 2026 — Derivatives, Calculus, Systems of Linear Equations & more

$f'(x) = \dfrac{(x^2+1)'}{x^2+1} = \dfrac{2x}{x^2+1}$.

Polynomial functions are continuous everywhere on $\mathbb{R}$.

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

By Vieta's relations: $r_1+r_2+r_3 = 6$ and $r_1 r_2 + r_1 r_3 + r_2 r_3 = 11$. So $\sum r_i^2 = 6^2 - 2\cdot 11 = 36 - 22 = 14$.

$b_2 = 2$, $b_4 = 8$, $b_6 = 32$. The sum is $2 + 8 + 32 = 42$. (These three form a GP with first term $2$ and ratio $4$.)

$\sqrt{n+1} - \sqrt{n} = \dfrac{1}{\sqrt{n+1} + \sqrt{n}} \to 0$.

$\dfrac{1+i}{1-i} = \dfrac{(1+i)^2}{(1-i)(1+i)} = \dfrac{2i}{2} = i$, whose argument is $\dfrac{\pi}{2}$.

$(x - 3)(x + 3) \le 0 \Leftrightarrow x \in [-3, 3]$. The integers in this interval are $\{-3, -2, -1, 0, 1, 2, 3\}$ — seven values.

With $a = 2$: $\!\left(1 + \dfrac{2}{n}\right)^n \to e^2$.

$\bullet$ Evaluate $A(3)$ and its determinant:

$$A(3)=\begin{pmatrix}4&6\\3&-4\end{pmatrix}$$

$$\det A(3)=4\cdot(-4)-6\cdot 3=-34$$

$\bullet$ Evaluate $A(2)$ and its determinant:

$$A(2)=\begin{pmatrix}3&4\\3&-5\end{pmatrix}$$

$$\det A(2)=3\cdot(-5)-4\cdot 3=-27$$

$\bullet$ $A(3)$ has trace $0$, so by Cayley–Hamilton $A(3)^2=34I$. Inverting:

$$A^{-2}(3)=\tfrac{1}{34}I$$

$\bullet$ Invert $A(2)$:

$$A^{-1}(2)=\tfrac{1}{27}\begin{pmatrix}5&4\\3&-3\end{pmatrix}$$

$\bullet$ Square it:

$$A^{-2}(2)=\tfrac{1}{729}\begin{pmatrix}37&8\\6&21\end{pmatrix}$$

$\bullet$ One more product:

$$A^{-3}(2)=\tfrac{1}{19683}\begin{pmatrix}209&124\\93&-39\end{pmatrix}$$

$\bullet$ Add the matrices and apply $\det(sI+M)=s^2+s\,\operatorname{tr}M+\det M$ with $s=\tfrac{1}{34}$, $\operatorname{tr}M=\tfrac{170}{19683}$, $\det M=-\tfrac{1}{19683}$:

$$\det=\tfrac{1}{1156}+\tfrac{170}{34\cdot 19683}-\tfrac{1}{19683}$$

$$=\tfrac{1}{1156}+\tfrac{4}{19683}$$

$$=\tfrac{24307}{1156\cdot 19683}$$