Daily Math · 2026-07-21

Question 1

Determine $n \in \mathbb{N}$, $n \ge 2$, such that $\binom{n}{2} = 15$.

Accepted Answer

$\dfrac{n(n-1)}{2} = 15 \Rightarrow n^2 - n - 30 = 0 \Rightarrow n \in \{6, -5\}$. Only $n = 6$ is admissible.

Answer

$4$

Answer

$5$

Answer

$6$

Answer

$15$

Question 2

The limit $\displaystyle\lim_{n 	o \infty} \dfrac{\sin n}{n}$ equals:

Accepted Answer

By the squeeze theorem, $-\dfrac{1}{n} \le \dfrac{\sin n}{n} \le \dfrac{1}{n}$ and both bounds tend to $0$, so the limit is $0$.

Answer

$0$

Answer

$1$

Answer

$\dfrac{1}{n}$

Answer

does not exist

Question 3

The solutions of $\cos x = 0$ on $[0, 2\pi)$ are:

Accepted Answer

$\cos x = 0 \iff x = \pi/2 + k\pi$. On $[0, 2\pi)$: $\pi/2$ and $3\pi/2$.

Answer

$x = 0$

Answer

$x = \pi$

Answer

$x = \pi/2$ and $x = 3\pi/2$

Answer

$x = \pi/4$ and $x = 5\pi/4$

Question 4

In a monoid $(M, *)$ with cancellation, if $a * b = a * c$ then:

Accepted Answer

The cancellation law states: $a * b = a * c \Rightarrow b = c$. (Even without inverses, this property may or may not hold; in a group it always does.)

Answer

$b = e$ (identity)

Answer

$a$ is invertible

Answer

$b = c$

Answer

$b * c = e$

Question 5

The image of $f: \mathbb{R} 	o \mathbb{R}$, $f(x) = -x^2 + 4x - 1$, is:

Accepted Answer

$\Delta = 16 - 4 = 12$, $a = -1$, so $f_{\max} = -\dfrac{\Delta}{4a} = -\dfrac{12}{-4} = 3$. The image is $(-\infty, 3]$.

Answer

$\mathbb{R}$

Answer

$[3, +\infty)$

Answer

$(-\infty, 3]$

Answer

$(-\infty, -1]$

Question 6

Compute $f'(x)$ for $f(x) = 	an(x^2)$.

Accepted Answer

Outer: $\sec^2(x^2)$. Inner derivative: $2x$. So $f'(x) = 2x\sec^2(x^2)$.

Answer

$\sec^2(x^2)$

Answer

$2x 	an(x^2)$

Answer

$2x \sec^2(x^2)$

Answer

$\sec^2(2x)$

Question 7

The set of complex numbers $z$ satisfying $|z - 1| = |z + 1|$ is:

Accepted Answer

$|z - 1| = |z - (-1)|$ describes the set of points equidistant from $1$ and $-1$. That's the perpendicular bisector of the segment between them, i.e. the imaginary axis $\operatorname{Re}(z) = 0$.

Answer

A circle centered at the origin

Answer

A circle centered at $1$

Answer

The imaginary axis

Answer

The real axis

Question 8

Find $\cos 20°\cdot\cos 40°\cdot\cos 80°$.

Accepted Answer

$\bullet$ Multiply numerator and denominator by $2\sin 20°$ and use $2\sin	heta\cos	heta=\sin 2	heta$:

$$\cos 20°\cos 40°\cos 80° = \dfrac{\sin 40°\cos 40°\cos 80°}{2\sin 20°}$$

$\bullet$ Apply the identity again:

$$= \dfrac{\sin 80°\cos 80°}{4\sin 20°}$$

$\bullet$ And once more:

$$= \dfrac{\sin 160°}{8\sin 20°}$$

$\bullet$ Use $\sin 160° = \sin(180°-20°) = \sin 20°$:

$$= \dfrac{\sin 20°}{8\sin 20°} = \dfrac{1}{8}$$

Answer

$\dfrac{1}{2}$

Answer

$\dfrac{1}{4}$

Answer

$\dfrac{1}{8}$

Answer

$\dfrac{\sqrt{3}}{8}$

Answer

$\dfrac{1}{16}$

Answer

$\dfrac{\sqrt{2}}{8}$

Question 9

For $I_n = \displaystyle\int_0^1 \dfrac{x^n}{1 + x} \, dx$, the inequality $0 \le I_n \le \dfrac{1}{n + 1}$ holds because:

Accepted Answer

On $[0, 1]$: $1 \le 1 + x \le 2$, hence $\dfrac{1}{2} \le \dfrac{1}{1 + x} \le 1$. So $\dfrac{x^n}{1 + x} \le x^n$, giving $I_n \le \int_0^1 x^n \, dx = \dfrac{1}{n + 1}$.

Answer

$\dfrac{1}{1 + x} \ge 0$ on $[0, 1]$

Answer

$\dfrac{x^n}{1 + x} \le 1$ on $[0, 1]$

Answer

$\dfrac{1}{1 + x} \le 1$ on $[0, 1]$

Answer

$\dfrac{x^n}{1 + x} \le x^n$

Question 10

The lines $y = 2x - 1$ and $y = -x + 5$ intersect at the point:

Accepted Answer

$2x - 1 = -x + 5 \Rightarrow 3x = 6 \Rightarrow x = 2$, $y = 3$. Intersection: $(2, 3)$.

Answer

$(0, -1)$

Answer

$(1, 4)$

Answer

$(2, 3)$

Answer

$(3, 2)$

Problems & worked solutions

Problem 1 — Permutations & Combinations

Problem 2 — Limits of Sequences

Problem 3 — Trigonometric Equations

Problem 4 — Semigroups & Monoids

Problem 5 — Quadratic Function

Problem 6 — Differentiation

Problem 7 — Complex Numbers

Problem 8 — Trigonometry

Problem 9 — Recursive Integrals

Problem 10 — Lines in the Plane