Daily Math · 2026-07-30

Question 1

The degree of $(X + 1)(X^2 - 1)$ is:

Accepted Answer

$\deg = 1 + 2 = 3$.

Answer

$1$

Answer

$2$

Answer

$3$

Answer

$4$

Question 2

The function $f(x) = \dfrac{1}{x - 2}$ is discontinuous at:

Accepted Answer

$f$ is undefined at $x = 2$, where the denominator vanishes.

Answer

$0$

Answer

$2$

Answer

$-2$

Answer

nowhere

Question 3

In the expansion of $(x + 1)^4$, the coefficient of $x^2$ is:

Accepted Answer

The coefficient of $x^2$ corresponds to $k = 2$: $\binom{4}{2} = 6$.

Answer

$2$

Answer

$4$

Answer

$6$

Answer

$12$

Question 4

Find $\begin{vmatrix}\cos 2024\pi&	an\pi&i^{2024}\\sin 0&\log_2 16&0!\\ln 1&1&\cos 360°\end{vmatrix}$.

Accepted Answer

$\bullet$ Simplify each entry. Row 1: $\cos 2024\pi=1$, $	an\pi=0$, $i^{2024}=1$. Row 2: $\sin 0=0$, $\log_2 16=4$, $0!=1$. Row 3: $\ln 1=0$, $1$, $\cos 360°=1$.

$\bullet$ The matrix becomes:

$$\begin{pmatrix}1&0&1\0&4&1\0&1&1\end{pmatrix}$$

$\bullet$ Two zeros in column 1, so expand along it:

$$\det=1\cdot\begin{vmatrix}4&1\1&1\end{vmatrix}=4-1=3$$

Answer

$0$

Answer

$4$

Answer

$3$

Answer

$-3$

Answer

$12$

Answer

$1$

Question 5

Compute $f'(x)$ for $f(x) = x^2 \ln x$ (with $x > 0$).

Accepted Answer

$f'(x) = 2x \ln x + x^2 \cdot 	frac{1}{x} = 2x \ln x + x = x(2\ln x + 1)$.

Answer

$2x \ln x$

Answer

$x(2\ln x + 1)$

Answer

$x^2 \cdot 	frac{1}{x}$

Answer

$2x \ln x + x^2$

Question 6

A bag contains $a$ red, $b$ blue, $c$ green balls, with $a=\log_3 9$, $b=i^{100}+3$, $c=2!\cdot\cos 0$. Two balls are drawn without replacement. Find $P(	ext{both same color})$.

Accepted Answer

$\bullet$ Decode the counts:

$$\log_3 9=2,\ i^{100}+3=1+3=4,\ 2!\cos 0=2$$

$\bullet$ So $(a,b,c)=(2,4,2)$, total $8$ balls. Count favorable pairs:

$$\binom{2}{2}+\binom{4}{2}+\binom{2}{2}=1+6+1=8$$

$\bullet$ Total number of pairs:

$$\binom{8}{2}=28$$

$\bullet$ Probability:

$$P=\dfrac{8}{28}=\dfrac{2}{7}$$

Answer

$\dfrac{1}{7}$

Answer

$\dfrac{3}{14}$

Answer

$\dfrac{2}{7}$

Answer

$\dfrac{1}{4}$

Answer

$\dfrac{5}{14}$

Answer

$\dfrac{1}{2}$

Question 7

For $f(x) = \begin{cases} 2x + 1, & x \le 1 \ x^2 + a, & x > 1 \end{cases}$, find $a$ such that $f$ is continuous at $x = 1$:

Accepted Answer

LHS: $2 \cdot 1 + 1 = 3$. RHS: $1 + a$. Setting equal: $a = 2$.

Answer

$0$

Answer

$1$

Answer

$2$

Answer

$4$

Question 8

The number of permutations of $(1, 2, 3, 4)$ in which no element is in its original position is:

Accepted Answer

$!4 = 4!\left(1 - 1 + \dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{24}\right) = 24 \cdot \dfrac{9}{24} = 9$.

Answer

$9$

Answer

$6$

Answer

$12$

Answer

$24$

Question 9

The set of units (multiplicatively invertible elements) of $\mathbb{Z}_8$ is:

Accepted Answer

The units are the residues coprime to $8$: $\{1, 3, 5, 7\}$, four elements.

Answer

$\{\hat{0}\}$

Answer

$\{\hat{1}, \hat{2}, \hat{3}\}$

Answer

$\{\hat{1}, \hat{3}, \hat{5}, \hat{7}\}$

Answer

$\mathbb{Z}_8 \setminus \{\hat{0}\}$

Question 10

For the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$, the eccentricity equals:

Accepted Answer

$a = 5$, $b = 3$, $c = \sqrt{25 - 9} = 4$. So $e = \dfrac{4}{5}$.

Answer

$\dfrac{3}{5}$

Answer

$\dfrac{4}{5}$

Answer

$\dfrac{5}{4}$

Answer

$\dfrac{4}{3}$

Problems & worked solutions

Problem 1 — Polynomial Rings

Problem 2 — Continuity

Problem 3 — Binomial Theorem

Problem 4 — Matrices

Problem 5 — Differentiation

Problem 6 — Probability

Problem 7 — Continuity

Problem 8 — Combinatorics

Problem 9 — Rings & Fields

Problem 10 — Conic Sections