Daily Math Problems — June 13, 2026: Sequences, Algebra, Trigonometry

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Problems & worked solutions

RO M1

Problem 1 — Arithmetic Sequences

The numbers $5, 8, 11, 14, \dots$ form an arithmetic progression. The common difference $r$ equals:

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A. $2$
B. $3$ ✓ correct
C. $4$
D. $5$

$r = a_{2} - a_{1} = 8 - 5 = 3$ .

RO M1

Problem 2 — Homomorphisms

The map $φ : (R, +) \to ((0, \infty), \cdot)$ defined by $φ (x) = e^{x}$ is:

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A. $not a homomorphism$
B. $a non-injective homomorphism$
C. $injective but not surjective$
D. $an isomorphism$ ✓ correct

$e^{a + b} = e^{a} e^{b}$ ✓ (homomorphism). $e^{x}$ is strictly increasing (injective) and surjective onto $(0, \infty)$ . Hence an isomorphism. (Inverse: $\ln$ .)

RO M1

Problem 3 — Trigonometric Equations

On $[0, 2 π)$ , the equation $\cos 2 x + \cos x = 0$ has exactly:

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A. $1 solution$
B. $2 solutions$
C. $3 solutions$ ✓ correct
D. $4 solutions$

$2 \cos^{2} x + \cos x - 1 = 0 \Rightarrow (2 \cos x - 1) (\cos x + 1) = 0$ . $\cos x = 1 / 2$ : 2 solutions ( $π / 3, 5 π / 3$ ). $\cos x = - 1$ : 1 solution ( $π$ ). Total: $3$ .

Problems & worked solutions

RO M1

Problem 1 — Arithmetic Sequences

The numbers $5, 8, 11, 14, \dots$ form an arithmetic progression. The common difference $r$ equals:

Show answer & worked solution

A. $2$
B. $3$ ✓ correct
C. $4$
D. $5$

$r = a_{2} - a_{1} = 8 - 5 = 3$ .

RO M1

Problem 2 — Homomorphisms

The map $φ : (R, +) \to ((0, \infty), \cdot)$ defined by $φ (x) = e^{x}$ is:

Show answer & worked solution

A. $not a homomorphism$
B. $a non-injective homomorphism$
C. $injective but not surjective$
D. $an isomorphism$ ✓ correct

$e^{a + b} = e^{a} e^{b}$ ✓ (homomorphism). $e^{x}$ is strictly increasing (injective) and surjective onto $(0, \infty)$ . Hence an isomorphism. (Inverse: $\ln$ .)

RO M1

Problem 3 — Trigonometric Equations

On $[0, 2 π)$ , the equation $\cos 2 x + \cos x = 0$ has exactly:

Show answer & worked solution

A. $1 solution$
B. $2 solutions$
C. $3 solutions$ ✓ correct
D. $4 solutions$

$2 \cos^{2} x + \cos x - 1 = 0 \Rightarrow (2 \cos x - 1) (\cos x + 1) = 0$ . $\cos x = 1 / 2$ : 2 solutions ( $π / 3, 5 π / 3$ ). $\cos x = - 1$ : 1 solution ( $π$ ). Total: $3$ .