Daily Math · 2026-07-23

Question 1

A particle's position is $x(t) = t^3 - 6t^2 + 9t$ (in metres, $t$ in seconds). The velocity at $t = 2$ is:

Accepted Answer

$v(t) = 3t^2 - 12t + 9$. At $t = 2$: $v = 12 - 24 + 9 = -3$ m/s (the particle is moving backward).

Answer

$-3$ m/s

Answer

$0$ m/s

Answer

$3$ m/s

Answer

$9$ m/s

Question 2

For $\vec{u} = 3\vec{i} - 2\vec{j}$, the vector $2\vec{u}$ equals:

Accepted Answer

Multiply each component by $2$: $2\vec{u} = 6\vec{i} - 4\vec{j}$.

Answer

$5\vec{i} - 2\vec{j}$

Answer

$3\vec{i} - 4\vec{j}$

Answer

$6\vec{i} - 4\vec{j}$

Answer

$6\vec{i} + 4\vec{j}$

Question 3

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

Accepted Answer

$\int_0^1 e^x \, dx = e^1 - e^0 = e - 1$.

Answer

$1$

Answer

$e$

Answer

$e - 1$

Answer

$e + 1$

Question 4

The numbers $a, b, c$ are in GP and $a + b + c = 14$, $abc = 64$. The middle term $b$ equals:

Accepted Answer

Since $b^2 = ac$, the product $abc = b \cdot ac = b \cdot b^2 = b^3 = 64$, so $b = 4$.

Answer

$2$

Answer

$3$

Answer

$4$

Answer

$6$

Question 5

By Rolle, the equation $x^4 - 6x^2 + 8 = 0$ has how many distinct real solutions?

Accepted Answer

$y^2 - 6y + 8 = 0 \Rightarrow y \in \{2, 4\}$. Then $x^2 = 2 \Rightarrow x = \pm\sqrt{2}$ and $x^2 = 4 \Rightarrow x = \pm 2$. Four distinct real solutions.

Answer

$0$

Answer

$1$

Answer

$2$

Answer

$4$

Question 6

The value of $\sin\dfrac{\pi}{3} + \cos\dfrac{\pi}{6}$ is:

Accepted Answer

$\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$ and $\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$. Sum: $\sqrt{3}$.

Answer

$1$

Answer

$\dfrac{\sqrt{3}}{2}$

Answer

$\sqrt{3}$

Answer

$\dfrac{\sqrt{3} + 1}{2}$

Question 7

A line is tangent to the circle $x^2 + y^2 = 25$ at $(3, 4)$. Its equation is:

Accepted Answer

$3x + 4y = 25$.

Answer

$3x + 4y = 5$

Answer

$3x - 4y = 25$

Answer

$3x + 4y = 25$

Answer

$4x + 3y = 25$

Question 8

$(\mathbb{Z}, \cdot)$ — integers under multiplication — is:

Accepted Answer

Associative ✓, identity $1$ ✓. But $2$ has no integer multiplicative inverse. Hence a monoid but not a group.

Answer

a group

Answer

a monoid (but not a group)

Answer

a semigroup but not a monoid

Answer

not associative

Question 9

In $	riangle ABC$, $a = 7$, $b = 5$, $c = 3$. The measure of $\angle A$ is:

Accepted Answer

$\cos A = \dfrac{25 + 9 - 49}{2 \cdot 5 \cdot 3} = \dfrac{-15}{30} = -\dfrac{1}{2}$. Since $A \in (0, \pi)$, $A = \dfrac{2\pi}{3}$.

Answer

$\dfrac{\pi}{6}$

Answer

$\dfrac{\pi}{3}$

Answer

$\dfrac{2\pi}{3}$

Answer

$\dfrac{5\pi}{6}$

Question 10

The solution of $\log_2(2x^2 + x + 1) - \log_2(x^2 - x + 2) = 1$ is:

Accepted Answer

$2x^2 + x + 1 = 2(x^2 - x + 2) \Rightarrow 2x^2 + x + 1 = 2x^2 - 2x + 4 \Rightarrow 3x = 3 \Rightarrow x = 1$.

Verify the domain: at $x = 1$, $2x^2 + x + 1 = 4 > 0$ and $x^2 - x + 2 = 2 > 0$, so both logs are defined. The solution set is $\{1\}$.

Answer

$\{1\}$

Answer

$\{-1\}$

Answer

$\{1, -3\}$

Answer

No real solution

Problems & worked solutions

Problem 1 — Applications of Derivatives

Problem 2 — Vectors in the Plane

Problem 3 — Definite Integrals

Problem 4 — Geometric Sequences

Problem 5 — Rolle's Sign Method

Problem 6 — The Unit Circle

Problem 7 — Conic Sections

Problem 8 — Semigroups & Monoids

Problem 9 — Solving Triangles

Problem 10 — Powers, Radicals, Logarithms