Daily Math · 2026-07-17

Question 1

Rotating $f(x) = \sqrt{R^2 - x^2}$ on $[-R, R]$ about the $x$-axis generates a:

Accepted Answer

The graph is a semicircle; rotating about the $x$-axis sweeps out a sphere of radius $R$. (Volume $= \dfrac{4}{3}\pi R^3$.)

Answer

cone

Answer

cylinder

Answer

sphere of radius $R$

Answer

ellipsoid

Question 2

Find $\displaystyle\lim_{n	o\infty}\sum_{k=1}^{n}\dfrac{1}{k(k+1)}+\log_{\sqrt{5}}\!\left(\dfrac{1}{25}ight)\cdot\cos(2024\pi)$.

Accepted Answer

$\bullet$ Telescope the sum:

$$\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$$

$\bullet$ The partial sum collapses:

$$\sum_{k=1}^{n}\!\left(\dfrac{1}{k}-\dfrac{1}{k+1}ight)=1-\dfrac{1}{n+1}	o 1$$

$\bullet$ Decode the second term:

$$\log_{\sqrt 5}\!\left(\dfrac{1}{25}ight)=\dfrac{-2}{1/2}=-4$$

$$\cos(2024\pi)=1$$

$\bullet$ Combine:

$$1+(-4)(1)=-3$$

Answer

$1$

Answer

$5$

Answer

$-3$

Answer

$-1$

Answer

$3$

Answer

$-5$

Question 3

How many solutions does the equation $\sin x = \dfrac{1}{2}$ have on $[0, 2\pi)$?

Accepted Answer

$\sin x = 	frac{1}{2}$ at $x = 	frac{\pi}{6}$ (Q1) and $x = 	frac{5\pi}{6}$ (Q2). Two solutions on $[0, 2\pi)$.

Answer

$1$

Answer

$2$

Answer

$3$

Answer

$4$

Question 4

The number of solutions of $\sin 2x = \sqrt{3}\,\cos x$ on $[0, 2\pi)$ is:

Accepted Answer

$2\sin x \cos x - \sqrt{3}\cos x = 0 \Rightarrow \cos x\,(2\sin x - \sqrt{3}) = 0$.

$\bullet$ $\cos x = 0 \Rightarrow x \in \!\left\{\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right\}$

$\bullet$ $\sin x = \dfrac{\sqrt{3}}{2} \Rightarrow x \in \!\left\{\dfrac{\pi}{3}, \dfrac{2\pi}{3}\right\}$

Total: $4$ solutions on $[0, 2\pi)$.

Answer

$2$

Answer

$3$

Answer

$4$

Answer

$5$

Question 5

$(1+i)^{10}$ equals:

Accepted Answer

$(1+i)^2 = 2i$, so $(1+i)^{10} = (2i)^5 = 32 \cdot i^5 = 32i$.

Answer

$32i$

Answer

$-32i$

Answer

$32$

Answer

$-32$

Question 6

The perpendicular bisector of the segment from $A(0, 0)$ to $B(4, 0)$ has equation:

Accepted Answer

Midpoint: $(2, 0)$. Segment $AB$ is horizontal, so the bisector is vertical through $(2, 0)$: $x = 2$.

Answer

$y = 2$

Answer

$y = 0$

Answer

$x = 2$

Answer

$x = 0$

Question 7

The area of the triangle with vertices $(0,0), (3,0), (1,4)$ is:

Accepted Answer

Base along the $x$-axis has length $3$; the third vertex sits at height $4$. Area $= 	frac{1}{2} \cdot 3 \cdot 4 = 6$.

Answer

$6$

Answer

$12$

Answer

$5$

Answer

$7$

Question 8

$(M_2(\mathbb{R}), +, \cdot)$ is:

Accepted Answer

Matrices form a ring (with $0$ matrix and identity matrix), but multiplication is non-commutative for $n \ge 2$. It's a non-commutative ring with unity.

Answer

a field

Answer

a commutative ring

Answer

a non-commutative ring with unity

Answer

not a ring

Question 9

Let $(x, y)$ be the solution of $\begin{cases} 3x + 4y = 15 \ 3x - y = 0 \end{cases}$. The value of $A = x^2 + y^2$ is:

Accepted Answer

Subtract: $5y = 15 \Rightarrow y = 3$. Then $3x = y = 3 \Rightarrow x = 1$. So $A = 1 + 9 = 10$.

Answer

$4$

Answer

$9$

Answer

$10$

Answer

$25$

Question 10

Find $\left|\left(\begin{pmatrix}3&2\0&i^2\end{pmatrix}^2+\begin{pmatrix}\lim_{x	o\infty}\frac{x^2}{x!}&\log_3 81\i^{2022}&	an225°\end{pmatrix}ight)ight|$.

Accepted Answer

$\bullet$ Square the first matrix using $i^2=-1$:

$$\begin{pmatrix}3&2\0&-1\end{pmatrix}^2=\begin{pmatrix}9&4\0&1\end{pmatrix}$$

$\bullet$ Evaluate the entries of the second matrix: $\lim x^2/x!=0$, $\log_3 81=4$, $i^{2022}=-1$, $	an 225°=1$:

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

$\bullet$ Add and take the determinant:

$$\begin{pmatrix}9&8\-1&2\end{pmatrix},\ \det=18-(-8)=26$$

Answer

$18$

Answer

$10$

Answer

$-26$

Answer

$26$

Answer

$20$

Answer

$36$

Problems & worked solutions

Problem 1 — Volumes of Revolution

Problem 2 — Limits

Problem 3 — Trigonometric Equations

Problem 4 — Trigonometric Identities

Problem 5 — Complex numbers

Problem 6 — Lines in the Plane

Problem 7 — Geometry

Problem 8 — Rings & Fields

Problem 9 — Linear Function

Problem 10 — Matrices