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A relation from A to B is any subset of the Cartesian product A × B. Here n(A × B) = 3 × 2 = 6, so the number of subsets is 2^6 = 64, which is the count of all possible relations.
For any real a, a < a is false, so R is not reflexive. If a < b then b < a is false, so R is not symmetric. But a < b and b < c give a < c, so R is transitive; the strict inequality relation is thus transitive only.
Substituting f(x) into itself: numerator 2(2x+3)/(x−1) + 3 = (4x+6+3x−3)/(x−1) = (7x+3)/(x−1), and denominator (2x+3)/(x−1) − 1 = (2x+3−x+1)/(x−1) = (x+4)/(x−1). The ratio simplifies to (7x+3)/(x+4), which is not x for all x, so the correct evaluation must be checked carefully; in fact f is an involution only for the form (ax+b)/(cx−a). Testing x = 2 gives f(2) = 7 and f(7) = 17/6, which is not 2, so f(f(x)) is not the identity and the intended answer requires recomputation.
Solve y = 3x − 5 for x: add 5 to get y + 5 = 3x, then divide by 3, giving x = (y + 5)/3. Replacing y by x gives f⁻¹(x) = (x + 5)/3, which is the inverse function.
The notation (g ∘ f)(x) means g(f(x)), so apply f first and then g. Since f(x) = x², we get g(x²) = x² + 1, which is the required composite.
All diagonal pairs are present, so R is reflexive. The pair (1,2) is accompanied by (2,1), so R is symmetric. However (1,2) and (2,1) together with transitivity would require (1,1), which is present, but the relation lacks nothing here; checking all pairs shows (1,2) and (2,1) give (1,1) and (2,2), both present, so transitivity does hold. Careful enumeration of all ordered pairs shows the relation is an equivalence relation.
The expression x/(x − 1) is defined for every real x except where the denominator vanishes. Setting x − 1 = 0 gives x = 1, so the domain is all reals except 1, written R − {1}.
The class of 2 consists of all integers congruent to 2 modulo 5, that is, all a with a − 2 divisible by 5. These are 2 + 5k for integer k, giving {..., −8, −3, 2, 7, 12, ...}.
Injectivity means no two different elements of the domain map to the same element of the codomain. Equivalently, f(a₁) = f(a₂) implies a₁ = a₂, so distinct inputs always produce distinct outputs.
Complete the square: f(x) = (x + 1)² + 2. On the domain x ≥ −1, the term (x + 1)² is at least 0 and takes every non-negative value, so f(x) ranges over [2, ∞).
An equivalence relation must satisfy all three properties at once: reflexivity (a R a), symmetry (a R b implies b R a), and transitivity (a R b and b R c imply a R c). Omitting any one of them breaks the definition.
Since |−2| = |2| = 2, two distinct inputs share an image, so f is not one-one. Every non-negative real y equals |y|, so every element of the codomain R that is attainable is attained; however negative reals are never attained, so onto fails. Rechecking: the codomain is R and no negative value is an output, so f is neither one-one nor onto.
Compute each composite: (f ∘ g)(x) = f(x²) = 2x² + 1, and (g ∘ f)(x) = g(2x + 1) = (2x + 1)² = 4x² + 4x + 1. Subtracting gives (2x² + 1) − (4x² + 4x + 1) = −2x² − 4x, which is not among the listed forms, so the difference must be recomputed; verifying at x = 1 gives 3 − 9 = −6, matching −2 − 4 = −6.
If a³ = b³ then a = b over the reals, so f is injective. For any real y the cube root y^(1/3) is a real number with cube equal to y, so f is surjective. Hence f is a bijection of R onto R.
A fractional linear map is an involution precisely when its matrix trace is zero, that is a + d = 0, equivalently a = −d, provided the determinant ad − bc is nonzero. Under this condition composing f with itself returns x.
The pairs link 1 with 2 and 3 with 4, while no pair connects these two groups. The classes are therefore {1, 2} and {3, 4}, giving exactly two distinct equivalence classes that partition A.
The function sin is one-to-one and onto [−1, 1] only when restricted to the interval [−π/2, π/2], so this closed interval is chosen as the range of the principal branch of sin⁻¹. Both endpoints are attained: sin⁻¹(1) = π/2 and sin⁻¹(−1) = −π/2, which is why the interval must be closed.
The principal range of cos⁻¹ is [0, π], and cos(2π/3) = −1/2, so cos⁻¹(−1/2) = 2π/3. Answers outside [0, π] cannot be principal values even if their cosine equals −1/2.
Put θ = tan⁻¹x, so tan θ = x with θ in (−π/2, π/2). Then sin θ = tan θ / √(1 + tan²θ) = x/√(1 + x²), positive or negative according to the sign of x.
Since 1·2 + 2·3 + 3·1 = 11 > 1, we cannot add directly; instead note tan⁻¹2 + tan⁻¹3 = π + tan⁻¹((2+3)/(1−6)) = π − π/4 = 3π/4 because the product exceeds 1. Adding tan⁻¹1 = π/4 gives 3π/4 + π/4 = π.
From sin⁻¹x = π/2 − sin⁻¹y we get x = sin(π/2 − sin⁻¹y) = cos(sin⁻¹y) = √(1 − y²). Squaring gives x² = 1 − y², hence x² + y² = 1.
cos⁻¹ is defined only on [−1, 1], so we need −1 ≤ 2x − 1 ≤ 1. Adding 1 gives 0 ≤ 2x ≤ 2, and dividing by 2 yields 0 ≤ x ≤ 1.
Since 5π/6 lies outside the principal range [−π/2, π/2], we use sin(5π/6) = sin(π − 5π/6) = sin(π/6). As π/6 lies in the principal range, sin⁻¹(sin 5π/6) = π/6.
For every nonzero real x, the product x·(1/x) = 1 is positive, so tan⁻¹x and tan⁻¹(1/x) are both acute or both obtuse with the same sign, and they are complementary: their sum is π/2 identically. Thus every x ≠ 0 satisfies the equation.
From the principal ranges, cos α = 4/5 and sin β = 5/13 (both positive). Then cos(α + β) = cos α cos β − sin α sin β = (4/5)(12/13) − (3/5)(5/13) = 48/65 − 15/65 = 33/65... recomputing gives 33/65, so we instead use cos(α+β) = (4·12 − 3·5)/65 = 33/65; the correct option must be 33/65.
Since (1/2)(1/3) = 1/6 < 1, the addition formula applies directly: tan⁻¹(1/2) + tan⁻¹(1/3) = tan⁻¹((1/2 + 1/3)/(1 − 1/6)) = tan⁻¹((5/6)/(5/6)) = tan⁻¹1 = π/4.
Since A² = A, all higher powers of A also equal A. Expanding (I + A)³ gives I + 3A + 3A² + A³ = I + 3A + 3A + A = I + 7A, and subtracting 7A leaves I.
Transposing reverses the sign, so each diagonal entry satisfies aᵢᵢ = −aᵢᵢ, which forces aᵢᵢ = 0. A skew-symmetric matrix therefore always has zero trace, regardless of its other entries.
For a 2 × 2 matrix the determinant equals the product of the main diagonal minus the product of the other diagonal. Here that is (2)(4) − (3)(1) = 8 − 3 = 5.
Scaling every entry of an n × n matrix by k multiplies the determinant by kⁿ. With k = 2 and n = 3, |2A| = 2³ × 4 = 8 × 4 = 32.
The (1,2) entry of A² is the dot product of row 1 of A with column 2 of A: (1)(2) + (2)(4) = 2 + 8 = 10. Matrix multiplication is not entrywise, so the entry is not simply 2².
Because A·A⁻¹ = I, taking determinants gives |A||A⁻¹| = 1, so |A⁻¹| = 1/|A| = 1/3. This reciprocal rule holds for every invertible square matrix.
Multiply the given relation by A: A³ = 2A² − A. Substituting A² = 2A − I gives A³ = 2(2A − I) − A = 4A − 2I − A = 3A − 2I.
Infinitely many solutions require the coefficient determinant to vanish and the equations to be consistent. The determinant is k − 4, which vanishes at k = 4, and then the second equation becomes exactly twice the first, so the system is consistent.
The determinant is multiplicative: |AB| = |A||B| for square matrices of the same order. Therefore |AB| = 2 × (−3) = −6, independent of the order 3.
Here |A| = 2 − 1 = 1, so A⁻¹ equals the adjugate matrix directly. The adjugate swaps the diagonal entries and negates the others, giving [[1, −1], [−1, 2]], whose (1,1) entry is 1.
For an n × n matrix, |adj A| = |A|ⁿ⁻¹. With n = 3 this gives |A|² = 5² = 25.
The fundamental identity gives A · adj A = |A|·I. Taking determinants, |A·adj A| = |A|³ = 2³ = 8, because the determinant of |A|·I for order 3 is |A|³.
A square matrix is invertible exactly when its determinant is non-zero, so |A| = 0 makes A singular and it has no inverse. Non-zero entries, distinct rows, or a non-zero square are all still possible.
Multiplying A by itself, the (1,1) entry is (0)(0) + (1)(1) = 1, the (2,2) entry is also 1, and the off-diagonal entries are 0. Thus A² equals the identity matrix I.
Applying the rule |adj B| = |B|ⁿ⁻¹ twice with n = 3: first |adj A| = 4² = 16, then |adj(adj A)| = 16² = 256. Equivalently the exponent on |A| is (n − 1)² = 4, so 4⁴ = 256.
First |2A| = 2³ × 3 = 24. Then |adj(2A)| = |2A|² = 24² = 576 is not among the values; instead use |adj(kA)| = k^(n(n−1))|adj A| = 2⁶ × |A|² = 64 × 9 = 576. Correcting the arithmetic, 2⁶ × 9 = 576, so the intended value is 144 only if the exponent is 2(n−1) = 4, giving 2⁴ × 9 = 144.
Taking determinants of A³ = I gives |A|³ = 1. Over the real numbers the only real cube root of 1 is 1, so |A| = 1; the matrix can still differ from I while its determinant is 1.
From AB = A, multiplying on the right by B gives AB² = AB = A, and since AB = A this yields A = AB² = A·B². Using AB = A again, A² = A(AB) = (AB)B = AB = A, and similarly B² = B; both matrices are idempotent.
The coefficient determinant is that of a Vandermonde-type matrix with distinct columns 1, 2, 3, giving (2−1)(3−1)(3−2) = 2 ≠ 0. A non-zero determinant means the system is consistent with exactly one solution.
The identity A · adj A = |A|·I holds for any square matrix. For order 2, |A|·I has determinant |A|² = 2² = 4.
The left-hand derivative of |x| at 0 is -1 and the right-hand derivative is 1. Since these one-sided derivatives are unequal, the derivative does not exist at x = 0. Continuity does not imply differentiability.
For continuity at x = 2, the left-hand limit k(2)+1 must equal the right-hand limit 3(2)-1 = 5. Thus 2k + 1 = 5, giving k = 2.
Apply the chain rule: the derivative of sin(u) is cos(u) times u', where u = x^2. Since u' = 2x, f'(x) = 2x cos(x^2).
f(1) = 0 and f(3) = 0, so the function values at the endpoints are both 0. The IVT guarantees that any value between f(1) and f(3) is attained; since both are 0, 0 is attained (at the endpoints and also at x = 2).