Daily Math · 2026-07-15

Question 1

Let $f(x) = 2\ln x + 3x^2$ and $A = \begin{pmatrix} f(1) & \log_2(f'(2)+3) \ \sin 7\pi & |3-4i| \end{pmatrix}$. Find $\det(A^2)$.

Accepted Answer

$\bullet$ Compute each entry:

$$f(1) = 2\ln 1 + 3 = 3$$

$$f'(x) = 	frac{2}{x} + 6x \;\Rightarrow\; f'(2) = 13$$

$$\log_2(13+3) = \log_2 16 = 4$$

$\bullet$ Also $\sin 7\pi = 0$ and $|3-4i| = 5$, giving:

$$A = \begin{pmatrix}3 & 4\ 0 & 5\end{pmatrix}$$

$\bullet$ Use $\det(A^2) = (\det A)^2$:

$$\det A = 15 \;\Rightarrow\; \det(A^2) = 225$$

Answer

$15$

Answer

$30$

Answer

$225$

Answer

$625$

Answer

$75$

Answer

$-225$

Question 2

$\displaystyle\int 2x \cdot \cos(x^2) \, dx$ equals:

Accepted Answer

$u = x^2 \Rightarrow du = 2x \, dx$. So $\int \cos u \, du = \sin u + C = \sin(x^2) + C$.

Answer

$\sin(2x) + C$

Answer

$2\sin(x^2) + C$

Answer

$\sin(x^2) + C$

Answer

$\cos(x^2) + C$

Question 3

What is the sum of the solutions to $x^2 - 5x + 6 = 0$?

Accepted Answer

Factoring gives $(x-2)(x-3)=0$, so the roots are $2$ and $3$. Their sum is $5$. (Equivalently, $-b/a = 5/1$.)

Answer

$-5$

Answer

$-1$

Answer

$1$

Answer

$5$

Question 4

Are the four points $(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1)$ coplanar?

Accepted Answer

The tetrahedron with these vertices has volume $1/6 
e 0$, so the four points are NOT coplanar — they form a non-degenerate tetrahedron.

Answer

Yes

Answer

No

Answer

Only three of them are

Answer

Cannot determine

Question 5

For $\begin{cases} x + y + z = 6 \ x + 2y + 3z = 14 \ x + 4y + 9z = 36 \end{cases}$, the solution is:

Accepted Answer

R2 − R1: $y + 2z = 8$. R3 − R1: $3y + 8z = 30$. From the first: $y = 8 - 2z$. Substituting: $3(8 - 2z) + 8z = 30 \Rightarrow 24 + 2z = 30 \Rightarrow z = 3$. Then $y = 2$, $x = 1$. Solution: $(1, 2, 3)$.

Answer

$(1, 2, 3)$

Answer

$(2, 1, 3)$

Answer

$(0, 1, 5)$

Answer

$(3, 2, 1)$

Question 6

For the $4$-cycle $\sigma = (1\;2\;3\;4)$, the permutation $\sigma^2$ equals:

Accepted Answer

$\sigma$: $1 	o 2 	o 3 	o 4 	o 1$. So $\sigma^2$: $1 	o 3$, $3 	o 1$, $2 	o 4$, $4 	o 2$. That's $(1\;3)(2\;4)$.

Answer

$(1\;2\;3\;4)$

Answer

identity

Answer

$(1\;3)(2\;4)$

Answer

$(1\;4)(2\;3)$

Question 7

Three numbers in arithmetic progression have sum $15$ and the sum of their squares is $83$. The largest of them is:

Accepted Answer

Set the terms as $5 - r,\, 5,\, 5 + r$. Then $(5-r)^2 + 25 + (5+r)^2 = 50 + 2r^2 + 25 = 83$, so $2r^2 = 8 \Rightarrow r = 2$. The largest term is $5 + 2 = 7$.

Answer

$5$

Answer

$6$

Answer

$7$

Answer

$8$

Question 8

The inverse of the rotation matrix $R(	heta) = \begin{pmatrix} \cos	heta & -\sin	heta \ \sin	heta & \cos	heta \end{pmatrix}$ is:

Accepted Answer

$R(	heta) R(-	heta) = I_2$, so $R(	heta)^{-1} = R(-	heta)$.

Answer

$R(	heta)$

Answer

$R(2	heta)$

Answer

$R(-	heta)$

Answer

$R(\pi - 	heta)$

Question 9

The number of real roots of $x^3 - 3x + 1 = 0$ is:

Accepted Answer

$f'(x) = 3x^2 - 3$ vanishes at $x = \pm 1$. $f(-1) = -1 + 3 + 1 = 3 > 0$ and $f(1) = 1 - 3 + 1 = -1 < 0$. The local max is positive and the local min is negative, so the cubic crosses the $x$-axis three times.

Answer

$3$

Answer

$1$

Answer

$2$

Answer

$0$

Question 10

The volume of the tetrahedron with vertices $(0,0,0), (3,0,0), (0,4,0), (0,0,5)$ is:

Accepted Answer

The three edges from origin are $(3,0,0), (0,4,0), (0,0,5)$. Determinant: $3 \cdot 4 \cdot 5 = 60$. Volume: $60/6 = 10$.

Answer

$5$

Answer

$10$

Answer

$30$

Answer

$60$

Problems & worked solutions

Problem 1 — Matrices

Problem 2 — Antiderivatives

Problem 3 — Quadratic Equations

Problem 4 — 3D Coordinate Geometry

Problem 5 — Systems of Linear Equations

Problem 6 — Permutations & Symmetric Groups

Problem 7 — Arithmetic Sequences

Problem 8 — Matrix Equations

Problem 9 — Algebra

Problem 10 — 3D Coordinate Geometry