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A relation on A is reflexive only when every element a of A satisfies (a,a) ∈ R. Since A has three elements, the relation must contain (1,1), (2,2) and (3,3). The set in the correct option is the only one containing all three required pairs.
The principal value of sin^{-1} lies in [-π/2, π/2]. Since sin(π/6)=1/2 and π/6 belongs to this interval, the principal value is π/6. The other angles either give a different sine value or lie outside the principal range.
A relation from A to B is any subset of A × B. Since A × B has 2×3=6 elements, it has 2^6=64 subsets. Hence there are 64 relations from A to B.
If f(a)=f(b), then 3a+2=3b+2, so a=b; hence f is one-one. For any y in R, choosing x=(y−2)/3 gives f(x)=y, so f is onto. Therefore f is a bijection.
By definition, (f ∘ g)(1)=f(g(1)). Now g(1)=2(1)−3=−1, and f(−1)=(−1)^2+1=2. Hence the composite value is 2.
The argument of cos^{-1} must lie in [−1,1]. Thus −1 ≤ 2x−1 ≤ 1. Solving gives 0 ≤ 2x ≤ 2, hence 0 ≤ x ≤ 1, so the domain is [0,1].
Since sin(2π/3)=√3/2, we need the angle in [−π/2,π/2] whose sine is √3/2. That angle is π/3. The angle 2π/3 itself is outside the principal range.
Write y=(2x+3)/5 and solve for x. Multiplying by 5 gives 5y=2x+3, so 2x=5y−3 and x=(5y−3)/2. Replacing y by x gives f^{-1}(x)=(5x−3)/2.
Put x=tan θ with θ∈(−π/2,π/2). Then (√(1+x^2)−1)/x = (sec θ−1)/tan θ = (1−cos θ)/sin θ = tan(θ/2). Hence the expression is tan^{-1}(tan(θ/2)) = θ/2 = (1/2)tan^{-1}x.
Using tan^{-1}a+tan^{-1}b=tan^{-1}((a+b)/(1−ab)) when ab<1, we get (5x)/(1−6x^2)=1. Thus 6x^2+5x−1=0, giving x=1/6 or x=−1. Checking the original equation, x=−1 makes the left side negative, so only x=1/6 works.
A reflexive relation on A must contain all three diagonal pairs (a, a), one for each element. The remaining 9 − 3 = 6 ordered pairs of A × A may each be included or left out independently, so the number of such relations is 2^6 = 64.
If n + 1 = m + 1 then n = m, so f is one-one. But the value 1 is never attained, because n ≥ 1 gives n + 1 ≥ 2, so f is not onto N.
The principal value branch of cos⁻¹ has range [0, π], and cos(2π/3) = −1/2. The angle −2π/3 has the same cosine but lies outside the principal range, so it cannot be the principal value.
Count the ordered pairs a | b inside A: a = 1 gives 6 pairs, a = 2 gives (2,2), (2,4), (2,6) i.e. 3 pairs, a = 3 gives 2 pairs, and a = 4, 5, 6 give one pair each. The total is 6 + 3 + 2 + 1 + 1 + 1 = 14.
The equivalence classes are the residue classes modulo 5, namely [0], [1], [2], [3] and [4]. Two integers are related exactly when they leave the same remainder on division by 5, so exactly 5 classes arise.
Substituting gives f(f(x)) = [x/(x − 1)] ÷ [x/(x − 1) − 1] = [x/(x − 1)] ÷ [1/(x − 1)] = x. Hence f is its own inverse on its domain.
The first element of A can be mapped to any of 5 elements of B, the second to any of the remaining 4, and the third to any of the remaining 3. Distinct choices keep the function one-one, so the count is 5 × 4 × 3 = 60.
If f(x₁) = f(x₂), then g(f(x₁)) = g(f(x₂)), and injectivity of g∘f forces x₁ = x₂. Therefore f is injective, while g need only behave injectively on the image of f, so g itself may fail to be injective.
For x > 0 the angles tan⁻¹x and tan⁻¹(1/x) are complementary, so their sum is π/2; for x < 0 both are negative and the sum is −π/2. Since x ≠ 0, only these two values occur, so the range is the two-element set {−π/2, π/2}.
The sum tan⁻¹2 + tan⁻¹3 has tangent (2 + 3)/(1 − 6) = −1, and because that sum exceeds π/2 it equals 3π/4. Adding tan⁻¹1 = π/4 gives 3π/4 + π/4 = π.
A matrix of order m × n has m rows and n columns, so the total number of entries is m × n. Here the order is 3 × 4, giving 3 × 4 = 12 entries in all.
A skew-symmetric matrix satisfies Aᵗ = −A, so every entry obeys aᵢⱼ = −aⱼᵢ. Putting i = j gives aᵢᵢ = −aᵢᵢ, that is 2aᵢᵢ = 0, so each diagonal element must be 0.
For a 2 × 2 determinant the rule is ad − bc, using the diagonal products. Here ad = 2 × 4 = 8 and bc = 3 × 1 = 3, so the determinant equals 8 − 3 = 5.
For a 2 × 2 matrix, A⁻¹ = (1/|A|) adj A, where adj A is obtained by interchanging the diagonal entries and changing the signs of the off-diagonal entries. Here |A| = 3 × 4 − 1 × 2 = 10 and adj A = [[4, −1], [−2, 3]], so A⁻¹ = (1/10)[[4, −1], [−2, 3]].
For a square matrix of order n, multiplying by a scalar k multiplies the determinant by kⁿ, because each of the n rows contributes a factor k. With n = 3 and k = 2, |2A| = 2³ × |A| = 8 × 4 = 32.
A square matrix is singular exactly when its determinant is zero. Here the determinant is x·x − 2·8 = x² − 16, and x² − 16 = 0 gives x = 4 or x = −4, so both values make the matrix singular.
The second equation is exactly twice the first, so both equations describe the same straight line in the plane. Every point of that line satisfies both equations, so the system is consistent and has infinitely many solutions.
The area is ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. Substituting gives ½|1(4 − 6) + 3(6 − 2) + 5(2 − 4)| = ½|−2 + 12 − 10| = 0, so the three points are collinear and enclose no area.
For an n × n matrix, |adj A| = |A|ⁿ⁻¹, so with n = 3 we get |adj A| = |A|² = 4. Applying the same rule again, |adj(adj A)| = |adj A|² = 4² = 16. Equivalently the result is |A|⁴ = 2⁴ = 16.
Transposing gives (AB − BA)ᵗ = BᵗAᵗ − AᵗBᵗ. Using Aᵗ = A and Bᵗ = −B, this becomes (−B)A − A(−B) = −BA + AB = AB − BA, so the matrix equals its own transpose and is symmetric.
For the product AB to be defined, the number of columns of A must equal the number of rows of B; here 3 = 3. The product then has order (rows of A) × (columns of B), which is 2 × 4.
Double each entry of A to get 2A = [[2, 4], [6, 8]], then subtract the corresponding entries of B. This gives [[2 − 0, 4 − 1], [6 − 2, 8 − 3]] = [[2, 3], [4, 5]].
Taking the transpose gives (A − Aᵀ)ᵀ = Aᵀ − A = −(A − Aᵀ), which is exactly the condition for a skew-symmetric matrix. It need not be zero unless A itself is symmetric.
Expanding along the first row gives 1(1·0 − 4·6) − 2(0·0 − 4·5) + 3(0·6 − 1·5). This simplifies to −24 + 40 − 15 = 1.
Adding the two equations eliminates y: (x + y) + (2x − y) = 3 + 0, so 3x = 3 and x = 1. Substituting x = 1 into x + y = 3 gives y = 2, and this also satisfies 2x − y = 0.
For an n × n matrix, |adj A| = |A|^(n − 1), provided |A| is nonzero. Here n = 3 and |A| = 3, so |adj A| = 3^2 = 9.
Compute A² = [[7, 10], [15, 22]], while 5A = [[5, 10], [15, 20]] and 2I = [[2, 0], [0, 2]]. Subtracting gives the zero matrix, consistent with the Cayley–Hamilton theorem for this 2 × 2 matrix.
Taking determinants of A² = A gives |A|² = |A|, so |A|(|A| − 1) = 0. Since |A| ≠ 0, the only possible value is |A| = 1.
Interchanging two rows changes the sign of the determinant, while multiplying a single row by 3 multiplies the determinant by 3. The combined factor is −3, so |B| = −3 × 4 = −12.
Multiply A by itself using row-by-column multiplication: A² = [[0·0 + 1·(−1), 0·1 + 1·0], [−1·0 + 0·(−1), −1·1 + 0·0]]. This simplifies to [[−1, 0], [0, −1]], which is −I.
Continuity at x = 3 requires lim(x→3) f(x) = f(3) = k. Since x² − 9 = (x − 3)(x + 3), cancelling the factor (x − 3) for x ≠ 3 gives the limit 3 + 3 = 6. Hence k must equal 6, the value of the removable limit.
The chain rule gives d/dx sin(u) = cos(u) · du/dx with u = x². Since du/dx = 2x, the derivative is 2x cos(x²). The inner derivative must be multiplied in; omitting it or misplacing the factor x gives an incorrect result.
Differentiability at a point always implies continuity at that point, because the difference quotient having a finite limit forces f(x) → f(a) as x → a. The converse is false: f(x) = |x| is continuous at 0 but not differentiable there. So continuity is a necessary consequence of differentiability.
At x = 2 the left-hand limit and the value of the function are both 2k + 1, while the right-hand limit is 3(2) − 1 = 5. Continuity requires these to be equal, so 2k + 1 = 5, giving k = 2.
The modulus function is continuous everywhere, so f is continuous at x = 2. However the left-hand derivative there equals −1 and the right-hand derivative equals +1; since these one-sided derivatives differ, f′(2) does not exist and the graph has a corner at x = 2.
Taking logarithms gives ln y = x ln x. Differentiating both sides with respect to x yields (1/y)·dy/dx = ln x + 1, so dy/dx = x^x(1 + ln x). The power rule cannot be applied directly because the exponent itself depends on x.
Here f(1) = 0 and f(3) = 0, and f is a polynomial, so it is continuous on [1, 3] and differentiable on (1, 3). Solving f′(x) = 2x − 4 = 0 gives c = 2, which lies inside (1, 3) as the theorem requires.
Differentiating the equation implicitly with respect to x gives 2x + y + x·dy/dx + 2y·dy/dx = 0. Collecting the derivative terms yields dy/dx·(x + 2y) = −(2x + y), so dy/dx = −(2x + y)/(x + 2y). Each y term contributes an extra dy/dx factor through the product and chain rules.
Continuity at x = 1 requires 1² = a + b, so a + b = 1. Differentiability requires the left-hand derivative 2x at x = 1, namely 2, to equal the right-hand derivative a, so a = 2; then b = 1 − 2 = −1. Both conditions must hold, since differentiability is stricter than continuity.
At the origin the difference quotient equals x sin(1/x), which tends to 0 because sin(1/x) is bounded, so f′(0) = 0. For x ≠ 0 the derivative is 2x sin(1/x) − cos(1/x), and cos(1/x) has no limit as x → 0, so f′ is discontinuous at 0. Thus differentiability at a point does not force the derivative to be continuous there.