Daily Math · 2026-07-26

Question 1

The number of ways to arrange $4$ distinct books on a shelf is:

Accepted Answer

$4! = 24$.

Answer

$4$

Answer

$16$

Answer

$24$

Answer

$256$

Question 2

$\displaystyle\int_0^1 x \, dx$ equals:

Accepted Answer

$F(x) = \dfrac{x^2}{2}$. $\int_0^1 x \, dx = F(1) - F(0) = \dfrac{1}{2}$.

Answer

$0$

Answer

$\dfrac{1}{2}$

Answer

$1$

Answer

$2$

Question 3

Calculați $\log_6 4 + \log_6 9$.

Accepted Answer

Aplicând regula produsului: $\log_6 4 + \log_6 9 = \log_6 36 = \log_6 6^2 = 2$.

Answer

$1$

Answer

$2$

Answer

$\log_6 13$

Answer

$3$

Question 4

If $r, s, t$ are the roots of $x^3 - 6x^2 + 11x - 6 = 0$, then $r^2 + s^2 + t^2$ equals:

Accepted Answer

By Viète's formulas, $r+s+t = 6$ and $rs+rt+st = 11$. Hence $r^2+s^2+t^2 = 36 - 22 = 14$.

Answer

$14$

Answer

$36$

Answer

$22$

Answer

$50$

Question 5

If the data set $\{x_1, \ldots, x_n\}$ has standard deviation $\sigma$, the standard deviation of $\{2x_1, \ldots, 2x_n\}$ is:

Accepted Answer

$\sigma(c \cdot x) = |c| \sigma(x)$. With $c = 2$: $\sigma(2x) = 2\sigma$.

Answer

$\sigma$

Answer

$\sigma + 2$

Answer

$2\sigma$

Answer

$4\sigma$

Question 6

The solution set of the inequality $\log_{1/3}(x^2 - 4) > 0$ is:

Accepted Answer

**Domain constraint:** $x^2 - 4 > 0 \Rightarrow x \in (-\infty, -2) \cup (2, +\infty)$. **Inequality:** $\log_{1/3}(x^2 - 4) > 0 = \log_{1/3} 1$. Since the base $\dfrac{1}{3} < 1$, $\log_{1/3}$ is decreasing, so the inequality flips: $x^2 - 4 < 1$, i.e. $x^2 < 5$, so $x \in (-\sqrt{5}, \sqrt{5})$. **Intersection with the domain:** $\left[(-\infty, -2) \cup (2, +\infty)\right] \cap (-\sqrt{5}, \sqrt{5}) = (-\sqrt{5}, -2) \cup (2, \sqrt{5})$.

Answer

$x \in (-\infty, 0)$

Answer

$x \in (0, 2)$

Answer

$x \in (-\sqrt{5}, -2) \cup (2, \sqrt{5})$

Answer

$x \in (\sqrt{5}, \infty)$

Answer

$x \in (2, \infty)$

Answer

$x \in (0, \sqrt{5})$

Question 7

The curve $x^2+xy+y^2=3$ passes through $(1,1)$. Find $\dfrac{dy}{dx}$ at that point.

Accepted Answer

$\bullet$ Differentiate both sides implicitly (product rule on $xy$):

$$2x + y + xy' + 2yy' = 0$$

$\bullet$ Substitute $(x,y)=(1,1)$:

$$2 + 1 + y' + 2y' = 0$$

$\bullet$ Solve for $y'$:

$$3y' = -3 \Rightarrow y' = -1$$

Answer

$-1$

Answer

$1$

Answer

$-2$

Answer

$0$

Answer

$-\dfrac{1}{2}$

Answer

$-3$

Question 8

The set of all complex solutions of $z^2 = -4$ is:

Accepted Answer

$z^2 = -4 \Rightarrow z^2 + 4 = 0 \Rightarrow (z - 2i)(z + 2i) = 0$. Solutions: $z = 2i$ or $z = -2i$.

Answer

$\{2, -2\}$

Answer

$\{2i\}$

Answer

$\{2i, -2i\}$

Answer

$\{1 + i, -1 - i\}$

Question 9

Two boxes — Box A has $3$ red and $1$ blue; Box B has $1$ red and $3$ blue. A box is chosen at random and a ball is drawn. Given the ball is red, the probability it came from Box A is:

Accepted Answer

$P(A) = P(B) = \dfrac{1}{2}$, $P(R|A) = \dfrac{3}{4}$, $P(R|B) = \dfrac{1}{4}$.

$P(A|R) = \dfrac{\dfrac{3}{4} \cdot \dfrac{1}{2}}{\dfrac{3}{4} \cdot \dfrac{1}{2} + \dfrac{1}{4} \cdot \dfrac{1}{2}} = \dfrac{3/8}{4/8} = \dfrac{3}{4}$.

Answer

$\dfrac{1}{4}$

Answer

$\dfrac{1}{2}$

Answer

$\dfrac{3}{4}$

Answer

$\dfrac{3}{8}$

Question 10

The determinant $\det \begin{pmatrix} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{pmatrix}$ equals (Vandermonde):

Accepted Answer

$\det = (b - a)(c - a)(c - b)$. (It's zero iff at least two of $a, b, c$ are equal.)

Answer

$abc$

Answer

$a + b + c$

Answer

$(b - a)(c - a)(c - b)$

Answer

$(a - b)(b - c)(c - a)$

Problems & worked solutions

Problem 1 — Permutations & Combinations

Problem 2 — Definite Integrals

Problem 3 — Logarithms

Problem 4 — Algebra

Problem 5 — Descriptive Statistics & Sampling

Problem 6 — Logs

Problem 7 — Derivatives

Problem 8 — Complex Numbers

Problem 9 — Probability

Problem 10 — Determinants