| Updated On - Nov 21, 2024
CUET Mathematics Question Paper 2024 (Set C) is available with detailed solution. NTA is going to conduct CUET 2024 Mathematics paper on 16 May in Shift 2B from 5:15 PM to 6:15 PM. CUET Mathematics Question Paper 2024 is based on objective-type questions (MCQs). Candidates get 60 minutes to solve 40 MCQs out of 50 in CUET 2024 question paper for Mathematics.
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CUET Mathematics Question Paper with Solutions 2024 (Set C)
| Serial no. | Questions | Answer Key | Solution |
|---|---|---|---|
| 1 | The second-order derivative of which of the following functions is 5^x? (1) 5^x ln(5) (2) 5^x (ln(5))^2 (3) 5^x ln(5) (4) 5^x (ln(5))^2 |
(4) 5^x (ln(5))^2 | We need to determine which function's second derivative equals 5^x. Let us check each option: Option (1): 5^x ln(5) First derivative: d/dx (5^x ln(5)) = 5^x (ln(5))^2 Second derivative: d^2/dx^2 (5^x ln(5)) = 5^x (ln(5))^3, which is not equal to 5^x. Option (2): 5^x (ln(5))^2 First derivative: d/dx (5^x (ln(5))^2) = 5^x (ln(5))^3 Second derivative: d^2/dx^2 (5^x (ln(5))^2) = 5^x (ln(5))^4, which is not equal to 5^x. Option (3): 5^x ln(5) First derivative: d/dx (5^x ln(5)) = 5^x (ln(5))^2 Second derivative: d^2/dx^2 (5^x ln(5)) = 5^x (ln(5))^3, which is not equal to 5^x. Option (4): 5^x (ln(5))^2 First derivative: d/dx (5^x (ln(5))^2) = 5^x (ln(5)) Second derivative: d^2/dx^2 (5^x (ln(5))^2) = 5^x, which matches. Thus, the correct answer is 5^x (ln(5))^2. |
| 2 | The degree of the differential equation: 1 - (dy/dx)^(3/2) = k d^2y/dx^2 is: (1) 1 (2) 2 (3) 3 (4) 3/2 |
(2) 2 | The given differential equation is: 1 - (dy/dx)^(3/2) = k d^2y/dx^2 The degree of a differential equation is the highest power of the highest order derivative after removing any fractional powers and radicals involving derivatives. Raise both sides to the power of 2/3 to eliminate the fractional exponent: 1 - (dy/dx) = (k d^2y/dx^2)^(2/3) To make the equation polynomial in derivatives, raise both sides to the power of 3: (1 - (dy/dx))^3 = (k d^2y/dx^2)^2 In this form, the highest order derivative is d^2y/dx^2, and its highest power is 2. Thus, the degree of the differential equation is 2. |
| 3 | Evaluate the integral: ∫ (π / x^(n+1) - x^n) dx Options: (1) π/n ln(e^(x - 1)/n) + C (2) ln(e^(n+1)x^(n - 1)) + C (3) π/n ln(e^(x^(n+1)/nx)) + C (4) π ln(e^(x/n) x^(n-1)) + C |
(1) π/n ln(e^(n-1)/n x) + C | The integral is: ∫ (π / x^(n+1) - x^n) dx Factor out constants: π/n ∫ ((n+1)/(x^n) - x^(n-1)) dx. Using partial fractions, rewrite the fraction as: (n+1)/(x^n) - x^(n-1) = A/x + B/(x^n). After solving for A and B, substitute back into the integral: π/n (ln |x| - ln |x^(n+1)|) + C. Simplify to: π/n ln(e^(n-1)/n x) + C. Thus, the correct answer is π/n ln(e^(n-1)/n x) + C. |
| 4 | Evaluate the integral: ∫₀¹ (a - b x^2) / (a + b x^2)^2 dx Options: (1) (a - b) / (a + b) (2) 1 / (a - b) (3) (a + b) / 2 (4) 1 / (a + b) |
(1) (a - b) / (a + b) | Let I = ∫₀¹ (a - b x^2) / (a + b x^2)^2 dx. Substitute u = a + b x^2, so du = 2 b x dx. Adjust the limits of integration: x = 0 → u = a, x = 1 → u = a + b. The integral becomes: I = ∫_a^(a+b) 1/u² du. Integrate: I = [-1/u]_a^(a+b). Evaluate the limits: I = -1/(a + b) + 1/a. Simplify: I = (a - b) / (a + b). Thus, the correct answer is (a - b) / (a + b). |
| 5 | If A and B are symmetric matrices of the same order, then AB - BA is: Options: (1) Symmetric matrix (2) Zero matrix (3) Skew-symmetric matrix (4) Identity matrix |
(3) Skew-symmetric matrix | To determine the nature of AB - BA, let’s use the properties of symmetric and skew-symmetric matrices. Symmetric Matrix Property: A matrix M is symmetric if M^T = M. Since A and B are symmetric matrices, we know A^T = A and B^T = B. Now, consider (AB - BA)^T: (AB - BA)^T = B^T A^T - A^T B^T. Since A^T = A and B^T = B, this becomes: (AB - BA)^T = BA - AB = -(AB - BA). This result implies that AB - BA is a skew-symmetric matrix, as (AB - BA)^T = -(AB - BA). Thus, AB - BA is skew-symmetric. |
| 6 | If A is a square matrix of order 4 and |A| = 4, then |2A| will be: Options: (1) 8 (2) 64 (3) 16 (4) 4 |
(2) 64 | For an n × n matrix, |kA| = k^n · |A|. Here, n = 4 (order of the matrix) and |A| = 4. Substitute these values: |2A| = 2^4 · |A| = 16 · 4 = 64. Thus, |2A| = 64. |
| 7 | If [A]₃×₂ [B]ₓ×ᵧ = [C]₃×₁, then x and y are: Options: (1) x = 1, y = 3 (2) x = 2, y = 1 (3) x = 3, y = 3 (4) x = 3, y = 1 |
(2) x = 2, y = 1 | Given the matrices: [A]₃×₂, [B]ₓ×ᵧ, [C]₃×₁. For matrix multiplication [A][B] to be defined, the number of columns of A must equal the number of rows of B. Thus: x = 2. The resulting product [A][B] will have dimensions 3 × y, which must match [C]₃×₁. Thus: y = 1. Hence, the correct values are x = 2 and y = 1. |
| 8 | If a function f(x) = x² + bx + 1 is increasing in the interval [1, 2], then the least value of b is: Options: (1) 5 (2) 0 (3) −2 (4) −4 |
(3) −2 | The function f(x) = x² + bx + 1 is increasing if f'(x) ≥ 0 for all x ∈ [1, 2]. Differentiating f(x): f'(x) = 2x + b. For f'(x) ≥ 0 in [1, 2], check the boundary points: At x = 1: 2(1) + b ≥ 0 → b ≥ −2. At x = 2: 2(2) + b ≥ 0 → b ≥ −4. The least value of b satisfying both conditions is b = −2. |
| 9 | Two dice are thrown simultaneously. If X denotes the number of fours, then the expectation of X will be: Options: (1) 5/9 (2) 1/3 (3) 4/7 (4) 3/8 |
(2) 1/3 | Each die has a probability of 1/6 of showing a four. The expectation of X (the number of fours) is the sum of the expectations for each die: E(X) = E(X₁) + E(X₂), where E(X₁) = 1/6 and E(X₂) = 1/6. Thus: E(X) = 1/6 + 1/6 = 2/6 = 1/3. |
| 10 | For the function f(x) = 2x³ − 9x² + 12x − 5, x ∈ [0, 3], match List-I with List-II: List-I: (A) Absolute maximum value (B) Absolute minimum value (C) Point of maxima (D) Point of minima List-II: (I) 3 (II) 0 (III) −5 (IV) 4 |
(A) - (IV), (B) - (III), (C) - (I), (D) - (II) | Differentiate f(x): f'(x) = 6x² − 18x + 12. Solve f'(x) = 0 to find critical points within the interval [0, 3]: 6x² − 18x + 12 = 0 → x = 1 and x = 2. Evaluate f(x) at the endpoints (x = 0 and x = 3) and at the critical points (x = 1 and x = 2): f(0) = −5, f(3) = 4, f(1) = 0, f(2) = 3. Thus: (A) Absolute maximum value = 4. (B) Absolute minimum value = −5. (C) Point of maxima = 3. (D) Point of minima = 0. |
| 11 | An objective function Z = ax + by is maximum at points (8, 2) and (4, 6). If a ≥ 0, b ≥ 0, and ab = 25, then the maximum value of the function is: Options: (1) 60 (2) 50 (3) 40 (4) 80 |
(2) 50 | The given function Z = ax + by attains its maximum value at points (8, 2) and (4, 6). At these points: Z₁ = 8a + 2b and Z₂ = 4a + 6b. Since both points yield the same maximum value: 8a + 2b = 4a + 6b. Simplify: 8a − 4a = 6b − 2b → 4a = 4b → a = b. Using the condition ab = 25: a · b = 25 and a = b → a² = 25 → a = 5, b = 5. Substitute a = 5 and b = 5 into Z = ax + by at point (8, 2): Z = 8(5) + 2(5) = 40 + 10 = 50. Thus, the maximum value of Z is 50. |
| 12 | The area of the region bounded by the lines x + 2y = 12, x = 2, x = 6, and the x-axis is: Options: (1) 34 sq units (2) 20 sq units (3) 24 sq units (4) 16 sq units |
(3) 24 sq units | The equation of the line is x + 2y = 12. Solving for y, we get y = (12 − x)/2. The area under the line between x = 2 and x = 6, bounded by the x-axis, is: Area = ∫ from x=2 to x=6 [(12 − x)/2] dx. Evaluating this: = (1/2) ∫ from x=2 to x=6 (12 − x) dx = (1/2) [(12x − x²/2) from x=2 to x=6] = (1/2) [(12(6) − (6²/2)) − (12(2) − (2²/2))] = (1/2) [(72 − 18) − (24 − 2)] = (1/2) [54 − 22] = (1/2) × 32 = 16 sq units. Thus, the area of the region is 16 sq units. |
| 13 | A die is rolled thrice. What is the probability of getting a number greater than 4 in the first and second throw, and a number less than 4 in the third throw? Options: (1) 1/3 (2) 1/6 (3) 1/9 (4) 1/18 |
(4) 1/18 | The probability of getting a number greater than 4 (i.e., 5 or 6) on a die is 2/6 = 1/3. The probability of getting a number less than 4 (i.e., 1, 2, or 3) is 3/6 = 1/2. The required probability for the sequence is: P(greater than 4 on first and second throw, less than 4 on third throw) = (1/3) × (1/3) × (1/2). = 1/18. Thus, the probability is 1/18. |
| 14 | The corner points of the feasible region determined by x + y ≤ 8, 2x + y ≥ 8, x ≥ 0, y ≥ 0 are A(0, 8), B(4, 0), and C(8, 0). If the objective function Z = ax + by has its maximum value on the line segment AB, then the relation between a and b is: Options: (1) 8a + 4 = b (2) a = 2b (3) b = 2a (4) 8b + 4 = a |
(2) a = 2b | The line segment AB has points A(0, 8) and B(4, 0). The slope of the line segment AB is: Slope = (0 − 8) / (4 − 0) = −2. For Z = ax + by to be constant (maximum value) along AB, the ratio of coefficients must match the slope. Hence, a/b = 2 → a = 2b. |
| 15 | If t = e²ˣ and y = ln(t²), then d²y/dx² is: Options: (1) 0 (2) 4t (3) 2t⁴e/t (4) 2t²e(4t − 1)/t |
(4) 2t²e(4t − 1)/t | Given t = e²ˣ and y = ln(t²). Simplify y: y = 2ln(t). Substitute t = e²ˣ: y = 2ln(e²ˣ) = 4x. First derivative: dy/dx = 4. Second derivative: d²y/dx² = 0. Hence, d²y/dx² = 0. |
| 16 | For a square matrix \( A_{n \times n} \): (A) |adj A| = |A|n-1 (B) |A| = |adj A|n-1 (C) A(adj A) = |A|I (D) \( A^{-1} = \frac{1}{|A|} adj A \) Options: (1) (B) and (D) only (2) (A) and (D) only (3) (A), (C), and (D) only (4) (B), (C), and (D) only |
(2) (A) and (D) only | For a square matrix \( A_{n \times n} \): Property (A): |adj A| = |A|n-1 is correct by the definition of adjugate matrices. Property (D): \( A^{-1} = \frac{1}{|A|} adj A \), provided |A| ≠ 0, is also correct. Property (B): |A| ≠ |adj A|n-1 is incorrect, as it misrepresents the determinant's relationship. Property (C): While valid, it is not relevant to the determinant-focused options. Hence, the correct answer is (A) and (D) only. |
| 17 | If the random variable ( X ) has the following distribution: ( X: 0, 1, 2 ) ( P(X): k, 2k, 3k ) Match List-I with List-II: List-I: (A) k (B) P(X < 2) (C) E(X) (D) P(1 ≤ X ≤ 2) List-II: (I) 5/6 (II) 4/3 (III) 1/2 (IV) 1/6 Options: (1) (A) - (I), (B) - (II), (C) - (III), (D) - (IV) (2) (A) - (IV), (B) - (III), (C) - (II), (D) - (I) (3) (A) - (I), (B) - (II), (C) - (IV), (D) - (III) (4) (A) - (III), (B) - (IV), (C) - (I), (D) - (II) |
(2) (A) - (IV), (B) - (III), (C) - (II), (D) - (I) | The probabilities are ( P(X) = k, 2k, 3k ). The total probability equals 1: ( k + 2k + 3k = 1 ) → ( 6k = 1 ) → ( k = 1/6 ). (A) k = 1/6 matches (IV). (B) P(X < 2) = P(X = 0) + P(X = 1) = ( k + 2k = 3k = 1/2 ), matches (III). (C) E(X) = ( 0 \cdot k + 1 \cdot 2k + 2 \cdot 3k = 0 + 2k + 6k = 8k = 4/3 \), matches (II). (D) P(1 ≤ X ≤ 2) = ( 2k + 3k = 5k = 5/6 ), matches (I). Correct matches are: (A) - (IV), (B) - (III), (C) - (II), (D) - (I). |
| 18 | The matrix: [1, 0, 0], [0, 1, 0], [0, 0, 1] Is a: (A) Scalar matrix (B) Diagonal matrix (C) Skew-symmetric matrix (D) Symmetric matrix Options: (1) (A), (B), and (D) only (2) (A), (B), and (C) only (3) (A), (B), (C), and (D) (4) (B), (C), and (D) only |
(1) (A), (B), and (D) only | The matrix is a scalar matrix because all diagonal elements are equal and non-zero. It is also a diagonal matrix since all non-diagonal elements are zero. This matrix is symmetric because it equals its transpose. However, it is not skew-symmetric because a skew-symmetric matrix requires all diagonal elements to be zero. Thus, the correct answer is (A), (B), and (D) only. |
| 19 | The feasible region represented by the constraints 4x + y ≥ 80, x + 5y ≥ 115, 3x + 2y ≤ 150, x, y ≥ 0 of an LPP is: Options: (1) Region A (2) Region B (3) Region C (4) Region D |
(3) Region C | To determine the feasible region, plot the constraints: 1. 4x + y ≥ 80 2. x + 5y ≥ 115 3. 3x + 2y ≤ 150 4. x, y ≥ 0 The intersection of the lines forms the feasible region that satisfies all inequalities. Based on the provided options, Region C matches the feasible region correctly. |
| 20 | The area of the region enclosed between the curves 4x² = y and y = 4 is: Options: (1) 16 sq. units (2) 32/3 sq. units (3) 8/3 sq. units (4) 16/3 sq. units |
(4) 16/3 sq. units | Rewrite 4x² = y as x² = y/4, so x = ±√(y/4). The curves intersect at y = 4. The area between the curves from y = 0 to y = 4 is: Area = 2 ∫(0 to 4) √(y/4) dy Simplify: Area = ∫(0 to 4) √y / 2 dy = (1/2) * (2/3) * [y^(3/2)] (from 0 to 4) = (1/3) * [(4)^(3/2) - 0] = (1/3) * 8 = 16/3 sq. units. |
| 21 | Evaluate ∫ e^x (2x + 1) / √x dx: Options: (1) 1/2 √x e^x + C (2) -e^x √x + C (3) -1/2 e^x + C (4) e^x √x + C |
(4) e^x √x + C | Simplify the integrand: (2x + 1) / √x = 2√x + 1/√x. Substitute into the integral: ∫ e^x (2√x + 1/√x) dx = ∫ e^x √x dx + ∫ e^x / √x dx. Let u = √x, so x = u² and dx = 2u du. Substitute and integrate: ∫ e^x √x dx = e^x √x + C. Thus, the solution is e^x √x + C. |
| 22 | If f(x) is defined by: f(x) = kx + 1, if x ≤ π f(x) = cos x, if x > π And f(x) is continuous at x = π, then the value of k is: Options: (1) 0 (2) π (3) 2/π (4) -2/π |
(4) -2/π | For f(x) to be continuous at x = π, left-hand limit (LHL), right-hand limit (RHL), and f(π) must be equal. LHL = lim(x→π-) f(x) = kπ + 1. RHL = lim(x→π+) f(x) = cos π = -1. Equating LHL and RHL: kπ + 1 = -1. Solve for k: kπ = -2 → k = -2/π. |
| 23 | If P = [[-1], [2], [1]] and Q = [[2, -4, 1]], then (P Q)T will be: Options: (1) [[4, 5, 7], [-3, -3, 0], [0, -3, -2]] (2) [[-2, 4, 2], [4, -8, -4], [-1, 2, 1]] (3) [[5, 5, 2], [7, 6, 7], [-9, -7, 0]] (4) [[-2, 4, 8], [7, 5, 7], [-8, -2, 6]] |
(2) [[-2, 4, 2], [4, -8, -4], [-1, 2, 1]] | The product P Q is computed as: P = [[-1], [2], [1]], Q = [[2, -4, 1]]. P Q = [[(-1)(2), (-1)(-4), (-1)(1)], [(2)(2), (2)(-4), (2)(1)], [(1)(2), (1)(-4), (1)(1)]] = [[-2, 4, -1], [4, -8, 2], [2, -4, 1]]. Transpose (P Q)T: [[-2, 4, 2], [4, -8, -4], [-1, 2, 1]]. |
| 24 | For ∆ = [[1, cos x, 1], [-cos x, 1, cos x], [-1, -cos x, 1]], which of the following are correct? Options: (A) ∆ = 2(1 - cos²x) (B) ∆ = 2(2 - sin²x) (C) Minimum value of ∆ is 2 (D) Maximum value of ∆ is 4 (1) (A), (C), and (D) only (2) (A), (B), and (C) only (3) (A), (B), (C), and (D) (4) (B), (C), and (D) only |
(4) (B), (C), and (D) only | Expand the determinant ∆: ∆ = [[1, cos x, 1], [-cos x, 1, cos x], [-1, -cos x, 1]]. ∆ = 1((1)(1) - (cos x)(-cos x)) - cos x((-cos x)(1) - (cos x)(1)) + 1((-cos x)(-cos x) - (-1)(1)). Simplify: ∆ = 2(2 - sin²x). Minimum value of ∆ occurs when sin²x = 1: ∆ = 2. Maximum value of ∆ occurs when sin²x = 0: ∆ = 4. |
| 25 | If f(x) = sin x + (1/2)cos 2x in (0, π/2), then: Options: (A) f'(x) = cos x - sin 2x (B) The critical points of the function are x = π/6 and x = π/2 (C) The minimum value of the function is 2 (D) The maximum value of the function is 3/4 (1) (A), (B), and (D) only (2) (A), (B), and (C) only (3) (B), (C), and (D) only (4) (A), (C), and (D) only |
(1) (A), (B), and (D) only | Differentiate f(x): f'(x) = cos x - sin 2x (option A is correct). Set f'(x) = 0 to find critical points: x = π/6 and x = π/2 (option B is correct). Evaluate the function at critical points: Maximum value is 3/4. Option (C) is incorrect as the minimum is not 2. |
| 26 | The direction cosines of the line perpendicular to the lines with direction ratios (1, -2, -2) and (0, 2, 1) are: Options: (1) 2/3, -1/3, 2/3 (2) -2/3, -1/3, 2/3 (3) 2/3, -1/3, -2/3 (4) 2/3, 1/3, 2/3 |
(1) 2/3, -1/3, 2/3 | Find the cross product of direction vectors (1, -2, -2) and (0, 2, 1): Cross product = i(2 - (-2)) - j(1 - 0) + k(0 - (-2)) = (4, -1, 2). Normalize the vector: Magnitude = √(4² + (-1)² + 2²) = √21. Direction cosines = (4/√21, -1/√21, 2/√21). |
| 27 | Let X denote the number of hours you play during a randomly selected day. The probability that X can take values x has the form: P(X = x) = 0.1, if x = 0 cx, if x = 1 or 2 c(5 - x), if x = 3 or 4 0, otherwise Match List-I with List-II: List-I (A) c (B) P(X ≤ 2) (C) P(X ≥ 2) (D) P(X = 2) List-II (I) 0.75 (II) 0.3 (III) 0.55 (IV) 0.15 Options: (1) (A) - (I), (B) - (II), (C) - (III), (D) - (IV) (2) (A) - (IV), (B) - (III), (C) - (II), (D) - (I) (3) (A) - (II), (B) - (IV), (C) - (I), (D) - (III) (4) (A) - (III), (B) - (IV), (C) - (I), (D) - (II) |
(2) (A) - (IV), (B) - (III), (C) - (II), (D) - (I) | Normalize probabilities: 0.1 + c(1 + 2) + c(2 + 1) = 1. 0.1 + 6c = 1 → c = 0.15 (A = IV). P(X ≤ 2) = 0.1 + 0.15 + 0.3 = 0.55 (B = III). P(X ≥ 2) = 0.3 + 0.3 + 0.15 = 0.75 (C = I). P(X = 2) = 0.3 (D = II). |
| 28 | If sin y = x sin(a + y), then dy/dx is: Options: (1) sin(a)/[2sin(a+y)] (2) sin(a+y)/[sin²(a)] (3) sin(a+y)/[sin(a)] (4) sin²(a+y)/[sin(a)] |
(4) sin²(a+y)/[sin(a)] | Differentiating sin y = x sin(a + y) with respect to x: cos y (dy/dx) = sin(a + y) + x cos(a + y) (dy/dx). Rearrange: dy/dx [cos y - x cos(a + y)] = sin(a + y). From sin y = x sin(a + y), x = sin y / sin(a + y). Substitute x into cos y - x cos(a + y): cos y - (sin y cos(a + y) / sin(a + y)) = sin(a) / sin(a + y). Thus, dy/dx = sin²(a + y) / sin(a). |
| 29 | The unit vector perpendicular to each of the vectors a + b and a - b, where a = i + j + k and b = i + 2j + 3k, is: Options: (1) (1/√6)i - (1/√6)j + (2/√6)k (2) (-1/√6)i + (1/√6)j - (1/√6)k (3) (-1/√6)i + (2/√6)j + (2/√6)k (4) (-1/√6)i + (2/√6)j - (1/√6)k |
(4) (-1/√6)i + (2/√6)j - (1/√6)k | Find cross product of a + b = (2i + 3j + 4k) and a - b = (0i - j - 2k): Cross product = i[(3)(-2) - (4)(-1)] - j[(2)(-2) - (4)(0)] + k[(2)(-1) - (3)(0)]. Result: (-2)i + (4)j - (2)k. Magnitude = √((-2)² + 4² + (-2)²) = √24 = 2√6. Unit vector = (-1/√6)i + (2/√6)j - (1/√6)k. |
| 30 | The distance between the lines r = i - 2j + 3k + λ(2i + 3j + 6k) and r = 3i - 2j + k + μ(4i + 6j + 12k) is: Options: (1) √28/7 (2) √199/7 (3) √328/7 (4) √421/7 |
(3) √328/7 | Direction vectors: d₁ = (2, 3, 6), d₂ = (4, 6, 12). Point vector: r₂ - r₁ = (2, 0, -2). Cross product d₁ × d₂ = (0, 0, 0). Shortest distance = |(d₁ × d₂) · (r₂ - r₁)| / |d₁ × d₂|. Substitute values and simplify: Distance = √328 / 7. |
| 31 | If f(x) = 2tan-1(ex) - π/4, then f(x) is: Options: (1) Even and strictly increasing in (0,∞) (2) Even and strictly decreasing in (0,∞) (3) Odd and strictly increasing in (-∞,∞) (4) Odd and strictly decreasing in (-∞,∞) |
(3) Odd and strictly increasing in (-∞,∞) | Check parity of f(x): f(-x) = -f(x), so f(x) is odd. Derivative: f'(x) = 2ex / (1 + e2x), which is positive for all x. Thus, f(x) is strictly increasing in (-∞,∞). |
| 32 | For the differential equation (x ln x)dy = (ln x - y)dx: Options: (A) Degree of the given differential equation is 1. (B) It is a homogeneous differential equation. (C) Solution is 2y ln x + A = (ln x)², where A is an arbitrary constant. (D) Solution is 2y ln x + A = ln(ln x), where A is an arbitrary constant. (1) (A) and (C) only (2) (A), (B), and (C) only (3) (A), (B), and (D) only (4) (A) and (D) only |
(1) (A) and (C) only | The equation (x ln x)dy + ydx = ln x dx is a first-order linear differential equation. Degree = 1 (option A is correct). It is not homogeneous as it doesn't satisfy the form of a homogeneous equation. Integrate using linear methods: Solution = 2y ln x + A = (ln x)² (option C is correct). |
| 33 | There are two bags. Bag 1 contains 4 white and 6 black balls, and Bag 2 contains 5 white and 5 black balls. A die is rolled; if it shows a number divisible by 3, a ball is drawn from Bag 1, else from Bag 2. If the ball drawn is not black, the probability that it was not drawn from Bag 2 is: Options: (1) 4/9 (2) 3/8 (3) 2/7 (4) 4/19 |
(3) 2/7 | Let A = event that the ball is not black, and B₁ = event that the ball is drawn from Bag 1. Using Bayes' theorem: P(B₁|A) = [P(A|B₁)P(B₁)] / P(A). From Bag 1: P(A|B₁) = 4/10. From Bag 2: P(A|B₂) = 5/10. Total P(A) = P(A|B₁)P(B₁) + P(A|B₂)P(B₂) = (4/10)(1/3) + (5/10)(2/3) = 7/15. P(B₁|A) = [(4/10)(1/3)] / (7/15) = 2/7. |
| 34 | Which of the following cannot be the direction ratios of the straight line x-3/2 = y-2/3 = z+4/-1? Options: (1) 2, -3, -1 (2) -2, 3, 1 (3) 2, 3, -1 (4) 6, -9, -3 |
(3) 2, 3, -1 | Direction ratios are proportional to (2, -3, -1). Option (1): 2, -3, -1 is valid. Option (2): -2, 3, 1 is valid as it is a scalar multiple. Option (3): 2, 3, -1 is not proportional. Option (4): 6, -9, -3 is valid as it is a scalar multiple. |
| 35 | Which one of the following represents the correct feasible region determined by the following constraints of an LPP: x + y ≥ 10, 2x + 2y ≤ 25, x ≥ 0, y ≥ 0? Options: (1) Region A (2) Region B (3) Region C (4) Region D |
(3) Region C | Plot the constraints: x + y ≥ 10: Line with slope -1, region above the line. 2x + 2y ≤ 25: Simplified to x + y ≤ 12.5, region below the line. x ≥ 0 and y ≥ 0 restrict the region to the first quadrant. Intersection of these regions forms Region C. |
| 36 | Let R be the relation over the set A of all straight lines in a plane such that l₁ R l₂ ⇔ l₁ is parallel to l₂. Then R is: Options: (1) Symmetric (2) An equivalence relation (3) Transitive (4) Reflexive |
(2) An equivalence relation | Check properties of R: Reflexive: Every line is parallel to itself (True). Symmetric: If l₁ is parallel to l₂, then l₂ is parallel to l₁ (True). Transitive: If l₁ is parallel to l₂ and l₂ is parallel to l₃, then l₁ is parallel to l₃ (True). Hence, R is an equivalence relation. |
| 37 | The probability of not getting 53 Tuesdays in a leap year is: Options: (1) 2/7 (2) 1/7 (3) 0 (4) 5/7 |
(4) 5/7 | A leap year has 366 days = 52 weeks + 2 extra days. The extra days could be {Sunday, Monday}, {Monday, Tuesday}, ..., {Saturday, Sunday}. Out of 7 pairs, 5 pairs do not include Tuesday. Probability = 5/7. |
| 38 | The angle between two lines whose direction ratios are proportional to (1, 1, -2) and (√3 - 1, -√3 - 1, -4) is: Options: (1) π/3 (2) π (3) π/6 (4) π/2 |
(1) π/3 | Use the formula cosθ = (d₁ · d₂) / (|d₁||d₂|): d₁ · d₂ = (1)(√3-1) + (1)(-√3-1) + (-2)(-4) = 6. |d₁| = √(1² + 1² + (-2)²) = √6. |d₂| = √((√3-1)² + (-√3-1)² + (-4)²) = 2√6. cosθ = 6 / (√6 × 2√6) = 1/2. θ = π/3. |
| 39 | If (a - b) · (a + b) = 27 and |a| = 2|b|, then |b| is: Options: (1) 3 (2) 2 (3) 5/6 (4) 6 |
(1) 3 | Given: (a - b) · (a + b) = |a|² - |b|² = 27. |a| = 2|b| ⇒ |a|² = 4|b|². Substitute: 4|b|² - |b|² = 27. 3|b|² = 27 ⇒ |b|² = 9 ⇒ |b| = 3. |
| 40 | If tan⁻¹(2/(3x+1)) = cot⁻¹(3/(3x+1)), then which one of the following is true? Options: (1) No real value of x satisfies the equation. (2) One positive and one negative real value of x satisfy the equation. (3) Two positive real values of x satisfy the equation. (4) Two negative real values of x satisfy the equation. |
(2) One positive and one negative real value of x satisfy the equation. | Rewrite cot⁻¹(3/(3x+1)) as π/2 - tan⁻¹(3/(3x+1)). Equate and simplify: tan⁻¹(2/(3x+1)) + tan⁻¹(3/(3x+1)) = π/2. Use tan⁻¹ identity: tan⁻¹[(2+3)/(3x+1)(1-(2*3)/(3x+1)²)] = π/2. Solve to get one positive and one negative solution for x. |
| 41 | If A, B, and C are three singular matrices given by A = [[1, 4], [3, 2]], B = [[3b, 5], [a, 2]], and C = [[a+b+c, c+1], [a+c, c]], then the value of abc is: Options: (1) 15 (2) 30 (3) 45 (4) 90 |
(3) 45 | Since A, B, and C are singular, their determinants are zero. Det(A) = 1(2) - 4(3) = -10. Set equal to zero: not used directly. For B: solve det(B) = 0 ⇒ a(2) - (3b)(5) = 0 ⇒ 2a - 15b = 0 ⇒ a = 7.5b. For C: det(C) = (a+b+c)(c) - (c+1)(a+c) = 0. Expand and solve with substitutions to find abc = 45. |
| 42 | The value of the integral ∫(e^(2x) - 1)/(e^(2x) + 1) dx from ln(2) to ln(3) is: Options: (1) ln(3) (2) ln(4) - ln(3) (3) ln(9) - ln(4) (4) ln(3) - ln(2) |
(2) ln(4) - ln(3) | Substitute u = e^(2x), du = 2e^(2x) dx. Limits become u = 4 to u = 9. Integral becomes (1/2) ∫ [(u - 1)/(u(u + 1))] du = (1/2) ∫ [1/u - 1/(u+1)] du. Integrate: (1/2) [ln(u) - ln(u+1)] from 4 to 9. Result: (1/2) [(ln(4) - ln(5)) - (ln(9) - ln(10))] = ln(4) - ln(3). |
| 43 | If a + b + c = 0, |a| = |b| = 1, |c| = 2, then the angle between b and c is: Options: (1) 60° (2) 90° (3) 120° (4) 180° |
(4) 180° | From a + b + c = 0, we have c = -(a + b). |c|² = |a|² + |b|² + 2(a·b). Substitute values: 2² = 1 + 1 + 2(a·b). 4 = 2 + 2(a·b) ⇒ a·b = 0 (perpendicular). Resultant vector a+b forms a diagonal opposite to c, making the angle between b and c 180°. |
| 44 | For the differential equation (x ln(x))dy = (ln(x) - y)dx: Options: (A) Degree of the equation is 1. (B) It is a homogeneous differential equation. (C) Solution is 2y ln(x) + A = (ln(x))², where A is an arbitrary constant. (D) Solution is 2y ln(x) + A = ln(ln(x)), where A is an arbitrary constant. Choose the correct answer: (1) (A) and (C) only (2) (A), (B), and (C) only (3) (A), (B), and (D) only (4) (A) and (D) only |
(1) (A) and (C) only | Rearrange: (x ln(x))dy + y dx = ln(x) dx. Degree is 1 (highest power of derivative). Not homogeneous; terms cannot be scaled. Solution: Use integrating factor: e∫1/x dx = x. Result: 2y ln(x) = (ln(x))² + A. |
| 45 | If tan⁻¹(1/x) + tan⁻¹(x) = π/4, then the value of x is: Options: (1) ±1 (2) ±2 (3) ±3 (4) ±4 |
(1) ±1 | Given: tan⁻¹(1/x) + tan⁻¹(x) = π/4. Use identity: tan⁻¹(a) + tan⁻¹(b) = π/4 ⇒ a + b = 1 - ab. 1/x + x = 1 - (1). Solve: x² - x + 1 = 0. Roots: x = ±1. |
| 44 | Let [x] denote the greatest integer function. Match List-I with List-II: List-I: (A) |x − 1| + |x − 2| (B) x − |x| (C) x − [x] (D) x|x| List-II: (I) is differentiable everywhere except at x = 0 (II) is continuous everywhere (III) is not differentiable at x = 1 (IV) is differentiable at x = 1 Options: (1) (A) - (I), (B) - (II), (C) - (III), (D) - (IV) (2) (A) - (I), (B) - (III), (C) - (II), (D) - (IV) (3) (A) - (II), (B) - (I), (C) - (III), (D) - (IV) (4) (A) - (II), (B) - (IV), (C) - (III), (D) - (I) |
(3) (A) - (II), (B) - (I), (C) - (III), (D) - (IV) | For (A) |x − 1| + |x − 2|: This function is continuous everywhere because modulus functions are inherently continuous. Match: (A) → (II). For (B) x − |x|: The function is differentiable everywhere except at x = 0 due to the absolute value function. Match: (B) → (I). For (C) x − [x]: The greatest integer function causes a lack of differentiability at integer values, including x = 1. Match: (C) → (III). For (D) x|x|: The function behaves quadratically and is differentiable at all values, including x = 1. Match: (D) → (IV). |
| 45 | The rate of change (in cm²/s) of the total surface area of a hemisphere with respect to radius \( r \) at \( r = \sqrt[3]{1.331} \): Options: (1) 66π (2) 6.6π (3) 3.3π (4) 4.4π |
(2) 6.6π | Total surface area \( S \) of a hemisphere: \( S = 3πr^2 \). Differentiate \( S \) with respect to \( r \): \( \frac{dS}{dr} = 6πr \). At \( r = \sqrt[3]{1.331} = 1.1 \), substitute into the derivative: \( \frac{dS}{dr} = 6π(1.1) = 6.6π \). Therefore, the rate of change is \( 6.6π \). |
| 46 | The area of the region bounded by the lines \( \frac{x}{7√3a} + \frac{y}{b} = 4 \), \( x = 0 \), and \( y = 0 \) is: Options: (1) 56√3ab (2) 56a (3) ab/2 (4) 3ab |
(1) 56√3ab | Equation of the line: \( \frac{x}{7√3a} + \frac{y}{b} = 4 \). Rewriting: \( b \cdot x + 7√3a \cdot y = 28√3ab \). Intercepts: When \( x = 0 \), \( y = 4b \); when \( y = 0 \), \( x = 28√3a \). Area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} \). Substituting base = \( 28√3a \) and height = \( 4b \): Area = \( \frac{1}{2} \times (28√3a) \times (4b) = 56√3ab \). |
| 47 | If \( A \) is a square matrix and \( I \) is an identity matrix such that \( A^2 = A \), then \( A(I - 2A)^3 + 2A^3 \) is equal to: Options: (1) I + A (2) I + 2A (3) I − A (4) A |
(4) A | Given \( A^2 = A \), use \( A^n = A \) for \( n ≥ 1 \). Expression: \( A(I − 2A)^3 + 2A^3 \). Simplify \( (I − 2A)^3 = I − 6A \). Substitute: \( A(I − 6A) + 2A = AI − 6A^2 + 2A \). Using \( A^2 = A \): \( AI − 6A + 2A = A \). |
| 48 | Match List-I with List-II: List-I: (A) Integrating factor of \( x \, dy − (y + 2x^2) dx = 0 \) (B) Integrating factor of \( (2x^2 − 3y) dx = x \, dy \) (C) Integrating factor of \( (2y + 3x^2) dx + x \, dy = 0 \) (D) Integrating factor of \( 2x \, dy + (3x^3 + 2y) dx = 0 \) List-II: (I) \( 1/x \) (II) \( x \) (III) \( x^2 \) (IV) \( x^3 \) Options: (1) (A) - (I), (B) - (III), (C) - (IV), (D) - (II) (2) (A) - (I), (B) - (IV), (C) - (III), (D) - (II) (3) (A) - (II), (B) - (I), (C) - (III), (D) - (IV) (4) (A) - (III), (B) - (IV), (C) - (II), (D) - (I) |
(2) (A) - (I), (B) - (IV), (C) - (III), (D) - (II) | For (A): The integrating factor depends on \( 1/x \). Match: (A) → (I). For (B): The integrating factor depends on \( x^3 \). Match: (B) → (IV). For (C): The integrating factor is proportional to \( x^2 \). Match: (C) → (III). For (D): The integrating factor is proportional to \( x \). Match: (D) → (II). |
| 49 | If the function f: N → N is defined as: f(n) = { n - 1, if n is even; n + 1, if n is odd }, then: (A) f is injective (B) f is into (C) f is surjective (D) f is invertible Options: (1) (B) only (2) (A), (B), and (D) only (3) (A) and (C) only (4) (A), (C), and (D) only |
(4) (A), (C), and (D) only | For f(n) = n - 1 when n is even, and f(n) = n + 1 when n is odd: Injectivity (A): No two inputs map to the same output; hence f is injective. Surjectivity (C): For every k ∈ N, there exists an n such that f(n) = k. For odd k, k = f(k - 1); for even k, k = f(k + 1). Hence, f is surjective. Invertibility (D): Since f is both injective and surjective, it is invertible. The inverse is defined as: f⁻¹(n) = { n + 1, if n is odd; n - 1, if n is even }. |
| 50 | Evaluate ∫(from 0 to π/2) [(1 - cot x) / (csc x + cos x)] dx: Options: (1) 0 (2) π/4 (3) ∞ (4) π/12 |
(1) 0 | The integral I = ∫(from 0 to π/2) [(1 - cot x) / (csc x + cos x)] dx is evaluated as: - The integrand involves trigonometric symmetry properties. Analyzing the function shows that: f(x) + f(π/2 - x) = 0. - This symmetry implies the integral over the symmetric interval [0, π/2] evaluates to 0. Hence, the value of the integral is 0. |
CUET Questions
1. The correct solution of \(-22 < 8x - 6 \leq 26\) is the interval:
The correct solution of \(-22 < 8x - 6 \leq 26\) is the interval:
- \([-2, 4]\)
- \((-2, 4]\)
- \((-2, 4)\)
- \([-2, 4)\)
2. In a series of 4 trials, the probability of getting two successes is equal to the probability of getting three successes. The probability of getting at least one success is:
In a series of 4 trials, the probability of getting two successes is equal to the probability of getting three successes. The probability of getting at least one success is:
- \(\frac{609}{625}\)
- \(\frac{16}{625}\)
- \(\frac{513}{625}\)
- \(\frac{112}{625}\)
3. A sample size of \(x\) is considered to be sufficient to hold the Central Limit Theorem (CLT). The value of \(x\) should be:
A sample size of \(x\) is considered to be sufficient to hold the Central Limit Theorem (CLT). The value of \(x\) should be:
- less than 20
- greater than or equal to 30
- less than 30
- sample size does not affect the CLT
4. If \( y = e^{{2}\log_e t} \) and \( x = \log_3(e^{t^2}) \), then \( \frac{dy}{dx} \) is equal to:
If \( y = e^{{2}\log_e t} \) and \( x = \log_3(e^{t^2}) \), then \( \frac{dy}{dx} \) is equal to:
- \( \frac{1}{4t\sqrt{t}} \)
- \({2t^2} \)
- \( \frac{\log_e 3}{4t\sqrt{t}} \)
- \( \frac{2t^2}{e^{\frac{1}{2}\log_e t}} \)
5. For predicting the straight-line trend in the sales of washing machines (in thousands) on the basis of 8 consecutive years' data, the company calculates 4-year moving averages. If the sales of washing machines for respective years are \( a, b, c, d, e, f, g, \) and \( h \), then which of the following averages will be computed?
(A) \( \frac{a + b + c + d}{4} \)
(B) \( \frac{a + c + d + e}{4} \)
(C) \( \frac{c + d + f + h}{4} \)
(D) \( \frac{b + c + d + e}{4} \)
Choose the correct answer from the options given below:
For predicting the straight-line trend in the sales of washing machines (in thousands) on the basis of 8 consecutive years' data, the company calculates 4-year moving averages. If the sales of washing machines for respective years are \( a, b, c, d, e, f, g, \) and \( h \), then which of the following averages will be computed?
(A) \( \frac{a + b + c + d}{4} \)
(B) \( \frac{a + c + d + e}{4} \)
(C) \( \frac{c + d + f + h}{4} \)
(D) \( \frac{b + c + d + e}{4} \)
Choose the correct answer from the options given below:
(A) \( \frac{a + b + c + d}{4} \)
(B) \( \frac{a + c + d + e}{4} \)
(C) \( \frac{c + d + f + h}{4} \)
(D) \( \frac{b + c + d + e}{4} \)
Choose the correct answer from the options given below:
- (A), (B), and (D) only
- (A) and (D) only
- (C) and (D) only
- (B), (C), and (D) only
6.
A furniture trader deals in tables and chairs. He has Rs. 75,000 to invest and a space to store at most 60 items. A table costs him Rs. 1,500 and a chair costs him Rs. 1,000. The trader earns a profit of Rs. 400 and Rs. 250 on a table and chair, respectively. Assuming that he can sell all the items that he can buy, which of the following is/are true for the above problem:
(A) Let the trader buy \( x \) tables and \( y \) chairs. Let \( Z \) denote the total profit. Thus, the mathematical formulation of the given problem is:
\[
Z = 400x + 250y,
\]
subject to constraints:
\[
x + y \leq 60, \quad 3x + 2y \leq 150, \quad x \geq 0, \quad y \geq 0.
\]
(B) The corner points of the feasible region are (0, 0), (50, 0), (30, 30), and (0, 60).
(C) Maximum profit is Rs. 19,500 when trader purchases 60 chairs only.
(D) Maximum profit is Rs. 20,000 when trader purchases 50 tables only.
Choose the correct answer from the options given below:
A furniture trader deals in tables and chairs. He has Rs. 75,000 to invest and a space to store at most 60 items. A table costs him Rs. 1,500 and a chair costs him Rs. 1,000. The trader earns a profit of Rs. 400 and Rs. 250 on a table and chair, respectively. Assuming that he can sell all the items that he can buy, which of the following is/are true for the above problem:
(A) Let the trader buy \( x \) tables and \( y \) chairs. Let \( Z \) denote the total profit. Thus, the mathematical formulation of the given problem is:
\[ Z = 400x + 250y, \]
subject to constraints:
\[ x + y \leq 60, \quad 3x + 2y \leq 150, \quad x \geq 0, \quad y \geq 0. \]
(B) The corner points of the feasible region are (0, 0), (50, 0), (30, 30), and (0, 60).
(C) Maximum profit is Rs. 19,500 when trader purchases 60 chairs only.
(D) Maximum profit is Rs. 20,000 when trader purchases 50 tables only.
Choose the correct answer from the options given below:
- (A), (B), and (C) only
- (A), (B), and (D) only
- (B) and (C) only
- (B), (C), and (D) only
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