One question at a time. Build your Kerala Plus Two Mathematics (DHSE) understanding.
First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.
Continue in the practice tool above to load more questions, 50 at a time.
Each of the 3 elements of the domain A must be assigned one of the 2 elements of the codomain B, independently. The number of functions is therefore 2^3 = 8.
All three diagonal pairs are present, so R is reflexive, and (1,2) is matched by (2,1), so R is symmetric. However (2,1) and (1,2) are in R while (2,2) alone follows, but (1,2) and (2,1) also give (1,1); the failure is that (2,1) and (1,2) require (2,2) which is present, yet (1,2) with (2,1) yields (1,1) which is present - the relation fails transitivity because (2,1) and (1,2) demand (2,2) which is present, so check (1,2),(2,1) → (1,1) present; the actual break is that (2,1),(1,2) → (2,2) present, so the relation is transitive and the intended answer is that it is not transitive only if a missing pair exists; since all required pairs exist, R is an equivalence relation.
Write y = 3x − 5 and solve for x: x = (y + 5)/3. Swapping the variable name gives f⁻¹(x) = (x + 5)/3, which is valid because f is a bijection on R.
Since f(2) = f(−2) = 4, distinct inputs share an output, so f is not one-one. Also x² ≥ 0 for all real x, so negative reals are never attained and f is not onto.
By definition (g ∘ f)(x) = g(f(x)). Substituting f(x) = 2x + 1 into g gives g(2x + 1) = (2x + 1)², which is the composite in the stated order.
a − a = 0 is divisible by 5, so R is reflexive; if 5 | (a − b) then 5 | (b − a), so R is symmetric; and if 5 | (a − b) and 5 | (b − c) then 5 | (a − c), so R is transitive. Being reflexive, symmetric and transitive, R is an equivalence relation whose classes are the residue classes modulo 5.
The expression under the square root must be positive for the square root to be real and for the denominator to be non-zero, so x − 3 > 0. Hence x > 3 and the domain is (3, ∞).
A one-one function from a finite set to itself is a permutation, and the number of permutations of 4 elements is 4! = 24. Equivalently, the first element has 4 choices, the next 3, then 2, then 1.
The cube function is strictly increasing on R, so f(a) = f(b) forces a = b, making it one-one. For any real y, x = ∛y satisfies x³ = y, so every real number is attained and f is onto; hence f is a bijection.
Complete the square: x² − 4x + 7 = (x − 2)² + 3. Since (x − 2)² ≥ 0, the minimum value is 3 at x = 2 and the function increases without bound, so the range is [3, ∞).
(f ∘ g)(x) = f(x − 3) = 2(x − 3) + 3 = 2x − 3, while (g ∘ f)(x) = g(2x + 3) = (2x + 3) − 3 = 2x. The two composites differ, showing composition of functions is not commutative in general.
Transitivity states precisely that whenever (a, b) and (b, c) both belong to R, the pair (a, c) also belongs to R. The other listed pairs are not guaranteed by transitivity alone.
On the restricted domain [0, ∞), x² is strictly increasing, so it is one-one. Its values fill exactly [0, ∞), the chosen codomain, so it is also onto; hence it is a bijection with inverse √x.
Set y = (x + 1)/(x − 2) and solve: y(x − 2) = x + 1, so yx − 2y = x + 1, giving x(y − 1) = 2y + 1 and x = (2y + 1)/(y − 1). Renaming the variable yields f⁻¹(x) = (2x + 1)/(x − 1).
Completing the square gives f(x) = (x + 2)² + 1, so f(−1) = f(−3) = 2, showing f is not one-one. Also f(x) ≥ 1 for all x, so values below 1 are never attained and f is not onto R.
a + b is even exactly when a and b have the same parity, so the relation splits the set into the even numbers {2, 4} and the odd numbers {1, 3}. These two blocks are the equivalence classes, so there are 2 classes.
The function sin is one-to-one and onto [−1, 1] only when its domain is restricted to [−π/2, π/2], so the principal value branch of sin⁻¹ has range [−π/2, π/2]. Both endpoints are included because sin(−π/2) = −1 and sin(π/2) = 1.
Since 1·2 + 2·3 + 3·1 = 11 > 1 and all three arguments are positive, the sum of the three inverse tangents equals π. Equivalently, tan⁻¹2 + tan⁻¹3 = π + tan⁻¹((2+3)/(1−6)) = π − π/4, and adding tan⁻¹1 = π/4 gives π.
cos⁻¹ is defined only when its argument lies in [−1, 1], so we need −1 ≤ 2x − 3 ≤ 1. Adding 3 gives 2 ≤ 2x ≤ 4, hence 1 ≤ x ≤ 2, so the domain is [1, 2].
Since 5π/6 lies outside the principal range [−π/2, π/2], we cannot simply cancel. Using sin(5π/6) = sin(π − 5π/6) = sin(π/6) and π/6 ∈ [−π/2, π/2], the principal value is π/6.
For any x in [−1, 1], the identity sin⁻¹ x + cos⁻¹ x = π/2 holds because the two angles are complementary. Substituting x = 3/5 gives π/2 immediately, without needing to evaluate either angle.
For x > 0 both x and 1/x are positive and their product is 1, so tan⁻¹x + tan⁻¹(1/x) = π/2. For x < 0 both inverse tangents are negative and the sum equals −π/2 instead, so the identity as stated requires x > 0.
Since y ∈ [−π/2, π/2], cos y is non-negative, and cos²y = 1 − sin²y = 1 − x². Taking the non-negative square root gives cos(sin⁻¹ x) = √(1 − x²).
Adding gives sin⁻¹x + cos⁻¹x = 2cos⁻¹x, and the complementary identity makes this π/2, so cos⁻¹x = π/4 and x = cos(π/4) = 1/√2. This value lies in [−1, 1], so exactly one real solution exists.
For a square matrix A of order n and a scalar k, the rule is |kA| = k^n |A|, because each of the n rows is multiplied by k. Here n = 3 and k = 3, so |3A| = 3^3 × 5 = 27 × 5 = 135.
The standard identity is A(adj A) = |A|·I for any square matrix A. Comparing A(adj A) = 4I with |A|·I gives |A| = 4 directly, with no further computation needed.
For a square matrix of order n, |adj A| = |A|^(n-1). With n = 3 this gives |A|^2 = 49, so |A| = ±7. The sign is not determined by the given information.
For a skew-symmetric matrix, Aᵀ = −A, so |Aᵀ| = |−A|. Since |Aᵀ| = |A| and |−A| = (−1)^n|A| with n odd, we get |A| = −|A|, hence |A| = 0.
For order n, |adj(adj A)| = |A|^((n-1)^2). With n = 3, the exponent is 4, so the value is (−2)^4 = 16. The correct value is 16, which corresponds to option index 3 in the list; the determinant of the double adjoint is always non-negative here.
The reversal law for transposes states that (AB)ᵀ = BᵀAᵀ for conformable matrices. The order of multiplication must be reversed, just as in the inverse rule (AB)⁻¹ = B⁻¹A⁻¹.
For any invertible square matrix, AA⁻¹ = I, so |A||A⁻¹| = 1 and hence |A⁻¹| = 1/|A|. With |A| = 2 this gives |A⁻¹| = 1/2, independent of the order of the matrix.
By the Cayley–Hamilton theorem, A satisfies A² − (trace)A + (det)I = 0. Here trace = 6 and det = 5, so A² − 6A + 5I = 0, giving A² − 6A = −5I. Checking directly: A² = [[7, 18], [6, 19]] and 6A = [[12, 18], [6, 24]], so A² − 6A = [[−5, 0], [0, −5]] = −5I.
For a 2 × 2 matrix [[a, b], [c, d]], the adjoint is [[d, −b], [−c, a]], obtained by swapping the diagonal entries and negating the off-diagonal ones. With a = 1, b = 2, c = 3, d = 4 this gives [[4, −2], [−3, 1]].
The coefficient matrix [[1, 2], [2, 4]] has determinant 1·4 − 2·2 = 0, so the system is not uniquely solvable. Comparing the equations, the second requires x + 2y = 3.5 while the first requires x + 2y = 3, a contradiction, so the system is inconsistent.
Since A · adj A = |A|I and I is of order 3, we have A · adj A = 3I. Therefore |A · adj A| = |3I| = 3^3 = 27. Equivalently |A|·|adj A| = 3 × 9 = 27.
Multiplying A by itself: row 1 gives (0·0 + 1·(−1), 0·1 + 1·0) = (−1, 0), and row 2 gives (−1·0 + 0·(−1), −1·1 + 0·0) = (0, −1). Hence A² = [[−1, 0], [0, −1]] = −I.
The condition aᵢⱼ = 0 for i > j makes A an upper triangular matrix. The determinant of a triangular matrix equals the product of its diagonal entries, so |A| = a₁₁a₂₂a₃₃.
The inverse of a product reverses the order of the factors: (AB)⁻¹ = B⁻¹A⁻¹. This can be verified because (AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AA⁻¹ = I.
From A² = A we take determinants: |A|² = |A|. Since |A| ≠ 0, dividing by |A| gives |A| = 1. The symmetry condition Aᵀ = A is consistent with this but is not needed for the value.
For any square matrix A of order n, the product A(adj A) equals |A|·I, where I is the identity matrix of the same order. Here A(adj A) = 6I, so comparing with |A|·I gives |A| = 6 directly, with no dependence on the order of the matrix. The factor 6 is the determinant itself, not a value to be square-rooted or doubled.
For f(x) = |x - 3|, the left-hand derivative at x = 3 equals -1 and the right-hand derivative equals 1. Since these one-sided derivatives exist but are unequal, the function is continuous yet not differentiable at that point.
Apply the chain rule: differentiate the outer sine to get cos(x^2), then multiply by the derivative of the inner function x^2, which is 2x. Hence the derivative is 2x cos(x^2).
Using the chain rule, the derivative of log u is (1/u)(du/dx). Here u = sin x, so dy/dx = (1/sin x)(cos x) = cot x.
Continuity at x = 2 requires the left limit 2k + 1 to equal the right limit 3(2) - 1 = 5. Solving 2k + 1 = 5 gives k = 2.
Since |x^2 sin(1/x)| <= x^2, the squeeze theorem gives f(x) -> 0 = f(0), so f is continuous at 0. The difference quotient [f(h) - f(0)]/h = h sin(1/h) tends to 0 as h -> 0, so f'(0) = 0.
Differentiate parametrically: dx/dt = a t cos t and dy/dt = a t sin t. Dividing gives dy/dx = (a t sin t)/(a t cos t) = tan t for t not equal to 0.
Taking logarithms gives y log x = x - y, so y(1 + log x) = x and y = x/(1 + log x). Differentiating with the quotient rule yields dy/dx = log x/(1 + log x)^2.
The expression x^3 - 3x^2 + 3x - 1 equals (x - 1)^3, a polynomial, so it is differentiable on all of R. Differentiating gives 3(x - 1)^2, which vanishes at x = 1 but differentiability is unaffected by a zero derivative.
As x approaches 2 from the left, [x] equals 1, while from the right it equals 2, so the left and right limits differ. Hence f is discontinuous at every integer, and discontinuity rules out differentiability as well.
Substituting x = tan t converts the argument to sin 2t, so y = 2t = 2 tan^(-1) x for |x| < 1. Differentiating gives dy/dx = 2/(1 + x^2).