| Updated On - Nov 27, 2024
CUET Mathematics Question Paper 2024 (Set A) is available here for download. 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.
CUET Mathematics Question Paper 2024 (Set A) PDF Download
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CUET 2024 Mathematics Question Paper with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| Question 1: 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: (1) 8a + 4 = b (2) a = 2b (3) b = 2a (4) 8b + 4 = a |
(2) a = 2b | The line segment AB has the points A(0,8) and B(4,0). The objective function Z = ax + by will have a maximum value on AB if a / b = -change in y / change in x Slope of AB = (0 - 8) / (4 - 0) = -2 Thus, the ratio a / b = 2 implies a = 2b. |
| Question 2: If t = e^(2x) and y = ln(t²), then d²y/dx² is: (1) 0 (2) 4t (3) 2t (4) 2t e^(2t(4t - 1)) |
(1) 0 | First, simplify y = ln(t²) as follows: Since t = e^(2x), we have: y = 2 * ln(t) = 4x. First derivative: dy/dx = 4 Second derivative: d²y/dx² = 0 Thus, d²y/dx² is 0. |
| Question 3: 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: (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, Z2 = 4a + 6b. Set the two equal: 8a + 2b = 4a + 6b. Simplify: 4a = 4b, so a = b. Using ab = 25, a² = 25, so a = 5 and b = 5. Substitute into Z = ax + by. At point (8, 2): Z = 8(5) + 2(5) = 50. |
| Question 4: The area of the region bounded by the lines x + 2y = 12, x = 2, x = 6, and the x-axis is: (1) 34 sq units (2) 20 sq units (3) 24 sq units (4) 16 sq units |
(4) 16 sq units | To find the area of the region, express y in terms of x from the equation x + 2y = 12: y = (12 - x) / 2. The area between x = 2 and x = 6 under the line y = (12 - x) / 2 is given by the integral: Area = ∫ from 2 to 6 of (12 - x) / 2 dx. Solving the integral gives the area = 16 sq units. |
| Question 5: A die is rolled thrice. What is the probability of getting a number greater than 4 in the first and second throw of the dice, and a number less than 4 in the third throw? (1) 1/3 (2) 1/6 (3) 1/9 (4) 1/18 |
(4) 1/18 | Probability of rolling greater than 4: P(greater than 4) = 2/6 = 1/3. Probability of rolling less than 4: P(less than 4) = 3/6 = 1/2. The combined probability for the three throws is: P(required outcome) = (1/3) * (1/3) * (1/2) = 1/18. |
| Question 6: Evaluate the integral ∫(π * xn+1 - x) dx: (1) πn * loge(xn - 1) (2) loge(xn + 1) / xn - 1 + C (3) πn * loge(xn + 1) / xn + C (4) π * loge(xn) / xn - 1 + C |
(1) πn * loge(xn - 1) | Begin with the integral: ∫(π * xn+1 - x) dx = π * ∫(xn+1 - x) dx. Factor the denominator and use substitution: The result simplifies to πn * loge(xn - 1) + C. |
| Question 7: Evaluate the integral ∫ from 0 to 1 of (a - bx²) / (a + bx²)² dx: (1) (a - b) / (a + b) (2) 1 / (a - b) (3) (a + b) / 2 (4) 1 / (a + b) |
(4) 1 / (a + b) | Let u = a + bx², so du = 2bx dx. Change limits: when x = 0, u = a; when x = 1, u = a + b. The integral becomes: ∫(1 / (a + b)) du. The result of solving this integral is 1 / (a + b). |
| Question 8: The second-order derivative of which of the following functions is 5x? (1) 5x * ln(5) (2) 5x * (ln(5))² (3) 5x (4) 5x * (ln(5))² |
(4) 5x * (ln(5))² | We need to determine which function’s second derivative equals 5x. For (1), (2), and (3), the second derivatives do not yield 5x. For (4), d²/dx² of 5x * (ln(5))² = 5x. |
| Question 9: The degree of the differential equation 1 - (dy/dx)² = k(d²y/dx²) is: (1) 1 (2) 2 (3) 3 (4) 3/2 |
(2) 2 | The degree of a differential equation is the highest power of the highest-order derivative. To remove fractional powers, raise both sides to the power of 3 to eliminate the fractional exponent. After simplifying, the highest power of the highest-order derivative d²y/dx² is 2. |
| Question 10: If A and B are symmetric matrices of the same order, then AB - BA is: (1) Symmetric matrix (2) Zero matrix (3) Skew-symmetric matrix (4) Identity matrix |
(3) Skew-symmetric matrix | Since A and B are symmetric matrices, we know AT = A and BT = B. Consider (AB - BA)T = B*A - A*B = -(AB - BA). This implies that AB - BA is a skew-symmetric matrix. |
| Question 11: If A is a square matrix of order 4 and |A| = 4, then |2A| will be: (1) 8 (2) 64 (3) 16 (4) 4 |
(2) 64 | For an n × n matrix, |kA| = k^n * |A|. Here, |2A| = 2^4 * 4 = 64. |
| Question 12: If [A]₃×₂ [B]ₓ×ᵧ = [C]₃×₁, then x and y are: (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, and [C]₃×₁, for matrix multiplication to be defined, x must equal 2. The resulting product [A][B] will have dimensions 3 × y, which must match [C]₃×₁, so y = 1. |
| Question 13: If a function f(x) = x² + bx + 1 is increasing in the interval [1, 2], then the least value of b is: (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 in [1, 2]. The derivative f'(x) = 2x + b. For f'(x) ≥ 0, check 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 is -2. |
| Question 14: Two dice are thrown simultaneously. If X denotes the number of fours, then the expectation of X will be: (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). E(X1) = 1/6, E(X2) = 1/6. Thus, E(X) = 1/6 + 1/6 = 2/6 = 1/3. |
| Question 15: 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 |
(4) (A)- (IV), (B)- (III), (C)- (I), (D)- (II) | Differentiate f(x) = 2x³ − 9x² + 12x − 5 to find f'(x) = 6x² − 18x + 12. Solve f'(x) = 0 to find critical points within the interval [0, 3]. Evaluate f(x) at the endpoints x = 0 and x = 3, and at the critical points, to determine the absolute maximum and minimum values. |
| Question 16: The rate of change (in cm²/s) of the total surface area of a hemisphere with respect to radius r at r = √3 is: (1) 66π (2) 6.6π (3) 3.3π (4) 4.4π |
(2) 6.6π | The total surface area S of a hemisphere is given by S = 3πr². Differentiate S with respect to r: dS/dr = 6πr. At r = √3, substitute r = 1.1 (since √3 ≈ 1.1). The rate of change is: dS/dr = 6π(1.1) = 6.6π. |
| Question 17: The area of the region bounded by the lines x + 7√3a + y/b = 4, x = 0, and y = 0 is: (1) 56√3ab (2) 56a (3) ab² (4) 3ab |
(1) 56√3ab | The equation of the line is x + 7√3a + y/b = 4. The intercepts are: when x = 0, y = 4b, and when y = 0, x = 28√3a. The area of the triangle formed is given by: Area = 1/2 × base × height. The base is 28√3a and the height is 4b, so the area is: Area = 1/2 × (28√3a) × (4b) = 56√3ab. |
| Question 18: If A is a square matrix and I is an identity matrix such that A² = A, then A(I − 2A)³ + 2A³ is equal to: (1) I + A (2) I + 2A (3) I − A (4) A |
(4) A | We are given that A² = A. Using this, simplify the expression A(I − 2A)³ + 2A³. First, simplify (I − 2A)³ = I − 6A. Then, the expression becomes A(I − 6A) + 2A. Simplify to get A. |
| Question 19: The value of the integral ∫ (1 − cot(x)) csc(x) + cos(x) dx is: (1) 0 (2) π/4 (3) ∞ (4) π/12 |
(1) 0 | The integrand is symmetric about x = π/4, and since the function is odd with respect to π/4, the integral evaluates to zero. Thus, the integral evaluates to 0. |
| Question 20: If vectors a, b, c are such that a + b + c = 0, |a| = |b| = 1, |c| = 2, then the angle between b and c is: (1) 60° (2) 90° (3) 120° (4) 180° |
(4) 180° | Given that a + b + c = 0, we can solve for c = −(a + b). Since a and b are perpendicular (a · b = 0), the magnitude of a + b is √2. Since c = −(a + b), the angle between b and c is 180°. |
| Question 21: Let [x] denote the greatest integer function. Then 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 |
(3) (A)- (II), (B)- (I), (C)- (III), (D)- (IV) | (A) |x − 1| + |x − 2| is continuous everywhere. Match: (A) → (II). (B) x − |x| is differentiable at x = 1. Match: (B) → (I). (C) x − [x] is not differentiable at x = 1. Match: (C) → (III). (D) x|x| is differentiable at x = 1. Match: (D) → (IV). |
| Question 22: Match List-I with List-II: List-I (A) Integrating factor of xdy − (y + 2x²)dx = 0 (B) Integrating factor of (2x² − 3y)dx = xdy (C) Integrating factor of (2y + 3x²)dx + xdy = 0 (D) Integrating factor of 2xdy + (3x³ + 2y)dx = 0 List-II (I) 1/x (II) x (III) x² (IV) x³ |
(2) (A)- (I), (B)- (IV), (C)- (III), (D)- (II) | Analyze each differential equation and find the corresponding integrating factor: (A) 1/x, (B) x³, (C) x², (D) x. |
| Question 23: If the function f: N → N is defined as f(n) = (A) f is injective (B) f is into (C) f is surjective (D) f is invertible Choose the correct answer from the options given below: (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 | The function is injective, surjective, and invertible. Hence, the correct answer is (A), (C), and (D). |
| Question 24: Evaluate ∫(1 − cot(x)) csc(x) + cos(x) dx from 0 to π/2: (1) 0 (2) π/4 (3) ∞ (4) π/12 |
(1) 0 | The integrand is odd with respect to x = π/4, and since the integral is symmetric about π/4, the value of the integral is 0. |
| Question 25: If the random variable X has the following distribution: X: 0 1 2 P(X): k 2k 3k Choose the correct answer from the options given below: (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 value of k is 1/6. (A) k = 1/6 (Match: (A) → (IV)). (B) P(X < 2) = 1/2 (Match: (B) → (III)). (C) E(X) = 4/3 (Match: (C) → (II)). (D) P(1 ≤ X ≤ 2) = 5/6 (Match: (D) → (I)). |
| Question 26: For a square matrix Aₙₓₙ: (A) |adj A| = |A|ⁿ⁻¹ (B) |A| = |adj A|ⁿ⁻¹ (C) A(adj A) = |A| (D) A⁻¹ = 1 / |A| Choose the correct answer from the options given below: |
(2) (A) and (D) only | For a square matrix Aₙₓₙ, the determinant of the adjugate of A is given by: |adj A| = |A|ⁿ⁻¹. This property confirms that (A) is correct. For the inverse of a matrix: |A⁻¹| = 1 / |A|. This property confirms that (D) is correct. (C) is valid, but it is not relevant to the determinant properties. (B) is incorrect because |A| ≠ |adj A|ⁿ⁻¹. |
| Question 27: The matrix [1 0 0] [0 1 0] [0 0 1] is a: Choose the correct answer from the options given below: |
(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 A = Aᵀ. It is not skew-symmetric because all diagonal elements are non-zero. |
| Question 28: The feasible region represented by the constraints 4x + y ≥ 80, x + 5y ≥ 115, 3x + 2y ≤ 150, x, y ≥ 0 of an LPP is: Choose the correct answer from the options given below: |
(3) Region C | To determine the feasible region, we plot the constraints: 4x + y ≥ 80 x + 5y ≥ 115 3x + 2y ≤ 150 x, y ≥ 0 (first quadrant). The region that satisfies all the constraints is Region C, which is bounded by the intersection of the lines. |
| Question 29: The area of the region enclosed between the curves 4x² = y and y = 4 is: Choose the correct answer from the options given below: |
(4) 16/3 sq. units | The curves 4x² = y and y = 4 intersect at y = 4. We find the area between these curves from y = 0 to y = 4. Using integration: Area = 2 * ∫ from 0 to 4 of √y dy = 16/3 sq. units. |
| Question 30: Evaluate ∫ e^x (2x + 1) / (2√x) dx: Choose the correct answer from the options given below: |
(4) e^x√x + C | The integrand simplifies to: e^x√x + 1 / (2√x) dx. After substitution and simplification, the result is e^x√x + C. |
| Question 31: If f(x) is defined as: f(x) = {kx + 1 if x ≤ π, cos(x) if x > π}, is continuous at x = π, then the value of k is: Choose the correct answer from the options given below: |
(4) -2/π | For continuity at x = π, the left-hand limit, right-hand limit, and the value of the function at x = π must all be equal. Equating the left-hand limit and right-hand limit gives k = -2/π. |
| Question 32: If P = [−1, 2, 1]ᵀ and Q = [2, −4, 1], then (PQ)ᵀ will be: Choose the correct answer from the options given below: |
(2) [-2 4 2, 4 −8 −4, −1 2 1] | Compute the product PQ: the result is a 3x3 matrix. Take the transpose of PQ to get (PQ)ᵀ = [-2 4 2, 4 −8 −4, −1 2 1]. |
| Question 33: If Δ = [1 cos(x) 1] [-cos(x) 1 cos(x)] [-1 -cos(x) 1] Then: Choose the correct answer from the options given below: |
(4) (B), (C), and (D) only | Expanding the determinant, we get: Δ = 2(2 − sin²(x)). The minimum value of Δ is 2, and the maximum value is 4. Thus, (B), (C), and (D) are correct. |
| Question 34: If f(x) = sin(x) + 1/2 cos²(x) in [0, π/2], then: Choose the correct answer from the options given below: |
(1) (A), (B), and (D) only | The derivative of f(x) is f'(x) = cos(x) − sin(2x). The critical points are x = π/6 and x = π/2. The maximum value of f(x) is 3/4, and the minimum value is 2. |
| Question 35: The direction cosines of the line which is perpendicular to the lines with direction ratios 1, -2, -2 and 0, 2, 1 are: Choose the correct answer from the options given below: |
(1) 2/3, −1/3, 2/3 | To find the direction cosines of the perpendicular line, compute the cross product of the direction ratios of the two given lines. Normalize the resulting vector to get the direction cosines. |
| Question 36: Let X denote the number of hours you play during a randomly selected day. The probability that X can take values x has the following form, where c is some constant: (0.1, if x = 0, cx, if x = 1 or x = 2, c(5 − x), if x = 3 or x = 4, 0, otherwise) Choose the correct answer from the options given below: |
(2) (A)- (IV), (B)- (III), (C)- (II), (D)- (I) | The constant c = 0.15. P(X ≤ 2) = 0.55, P(X = 2) = 0.3, P(X ≥ 2) = 0.75. Thus, the correct answer is (A)- (IV), (B)- (III), (C)- (II), (D)- (I). |
| Question 37: If sin y = x sin(a + y), then dy/dx is: (1) sin(a + y) / sin a (2) sin(a + y) / sin² a (3) sin(a + y) / sin a (4) sin²(a + y) / sin a |
(4) sin²(a + y) / sin a | Differentiate both sides with respect to x: cos y * dy/dx = sin(a + y) + x cos(a + y) * dy/dx. Isolate dy/dx and simplify to get: dy/dx = sin²(a + y) / sin a. |
| Question 38: The unit vector perpendicular to each of the vectors ⃗a + ⃗b and ⃗a − ⃗b, where ⃗a = î + ĵ + k̂ and ⃗b = î + 2ĵ + 3k̂, is: (1) 1/√6 î + 2/√6 ĵ + 1/√6 k̂ (2) −1/√6 î + 1/√6 ĵ − 1/√6 k̂ (3) −1/√6 î + 2/√6 ĵ + 2/√6 k̂ (4) −1/√6 î + 2/√6 ĵ − 1/√6 k̂ |
(4) −1/√6 î + 2/√6 ĵ − 1/√6 k̂ | The cross product of ⃗a + ⃗b and ⃗a − ⃗b gives the perpendicular vector. After computing the cross product, normalize it to find the unit vector: −1/√6 î + 2/√6 ĵ − 1/√6 k̂. |
| Question 39: The distance between the lines ⃗r = î − 2ĵ + 3k̂ + λ(2î + 3ĵ + 6k̂) and ⃗r = 3î − 2ĵ + k̂ + µ(4î + 6ĵ + 12k̂) is: (1) √328/7 (2) 7 (3) √199/7 (4) √421/7 |
(3) √328/7 | Use the formula for the distance between two skew lines: d = |(⃗d₁ × ⃗d₂) · (⃗r₂ − ⃗r₁)| / |⃗d₁ × ⃗d₂|. After calculating the cross product and dot product, the distance is √328/7. |
| Question 40: If f(x) = 2 tan⁻¹(ex) − π/4, then f(x) is: (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 (−∞, ∞) | f(x) = 2 tan⁻¹(ex) − π/4 is an odd function and is strictly increasing over the entire real line. Thus, the correct answer is (3). |
| Question 41: For the differential equation (x loge x) dy = (logex − y) dx: (1) Degree of the given differential equation is 1. (2) It is a homogeneous differential equation. (3) Solution is 2y loge x + A = (loge x)², where A is an arbitrary constant. (4) Solution is 2y loge x + A = loge(loge x), where A is an arbitrary constant. |
(1) (A) and (C) only | Rearranging the given differential equation shows it is a first-order linear equation with degree 1. The solution is 2y loge x + A = (loge x)². |
| Question 42: 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 a ball is drawn from Bag-2. If the ball drawn is not black in color, the probability that it was not drawn from Bag-2 is: (1) 4/9 (2) 3/8 (3) 2/7 (4) 4/19 |
(3) 2/7 | Using Bayes' theorem, we calculate the probability that the ball was not drawn from Bag-2, given that it is not black. The final result is 2/7. |
| Question 43: Which of the following cannot be the direction ratios of the straight line x − 3 / 2 = 2 − y / 3 = z + 4 / −1? (1) 2, −3, −1 (2) −2, 3, 1 (3) 2, 3, −1 (4) 6, −9, −3 |
(3) 2, 3, −1 | The direction ratios of the line x − 3 / 2 = 2 − y / 3 = z + 4 / −1 are 2, −3, −1. Option (3) does not match the correct direction ratios, as the second direction ratio should be negative. |
| Question 44: 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. Choose the correct answer from the options given below: |
(3) Region D | To solve this, plot the constraints: x + y ≥ 10, 2x + 2y ≤ 25, x ≥ 0, and y ≥ 0. The feasible region is the intersection of these constraints, which corresponds to Region D. |
| Question 45: Let R be the relation over the set A of all straight lines in a plane such that l₁Rl₂ ⇐⇒ l₁ is parallel to l₂. Then R is: (1) Symmetric (2) Transitive (3) An equivalence relation (4) Reflexive |
(3) An equivalence relation | The relation R is reflexive (a line is parallel to itself), symmetric (if l₁ is parallel to l₂, then l₂ is parallel to l₁), and transitive (if l₁ is parallel to l₂, and l₂ is parallel to l₃, then l₁ is parallel to l₃). |
| Question 46: The probability of not getting 53 Tuesdays in a leap year is: (1) 2/7 (2) 1/7 (3) 0 (4) 5/7 |
(4) 5/7 | In a leap year, the two extra days can be any combination of days. To avoid getting 53 Tuesdays, the extra days must not include a Tuesday. The probability of not getting 53 Tuesdays is 5/7. |
| Question 47: The angle between two lines whose direction ratios are proportional to 1, 1, −2 and (√3 − 1), (−√3 − 1), −4 is: (1) π/3 (2) π (3) π/6 (4) π/2 |
(1) π/3 | The angle θ between two lines is given by the formula cosθ = (⃗d₁ · ⃗d₂) / (|⃗d₁| |⃗d₂|). After calculating the dot product of the direction ratios and the magnitudes of the direction vectors, we find that θ = π/3. |
| Question 48: If (⃗a − ⃗b) · (⃗a + ⃗b) = 27 and |⃗a| = 2|⃗b|, then |⃗b| is: (1) 3 (2) 2 (3) 5/6 (4) 6 |
(1) 3 | From the equation (⃗a − ⃗b) · (⃗a + ⃗b) = 27, we simplify to obtain |⃗b| = 3 using the given condition |⃗a| = 2|⃗b|. |
| Question 49: If tan⁻¹(2 / 3 − x + 1) = cot⁻¹(3 / 3x + 1), then which one of the following is true? (1) There is no real value of x satisfying the above equation. (2) There is one positive and one negative real value of x satisfying the above equation. (3) There are two real positive values of x satisfying the above equation. (4) There are two real negative values of x satisfying the above equation. |
(2) There is one positive and one negative real value of x satisfying the above equation. | By using the identity cot⁻¹(y) = π/2 − tan⁻¹(y), we can solve the equation and find that there are two solutions for x, one positive and one negative. |
| Question 50: If A, B, and C are three singular matrices given by A = [3b, 5; a, 2], B = [1, 4; 3, 2], and C = [a + b + c, c + 1; a + c, c], then the value of abc is: (1) 15 (2) 30 (3) 45 (4) 90 |
(3) 45 | The determinant of each matrix is zero because they are singular. Solving the equations formed by the determinants, we find that the value of abc is 45. |
CUET Questions
1. Match List-I with List-II.List-I List-II (A) Confidence level (I) Percentage of all possible samples that can be expected to include the true population parameter (B) Significance level (III) The probability of making a wrong decision when the null hypothesis is true (C) Confidence interval (II) Range that could be expected to contain the population parameter of interest (D) Standard error (IV) The standard deviation of the sampling distribution of a statistic
Choose the correct answer from the options given below:
Match List-I with List-II.
| List-I | List-II |
|---|---|
| (A) Confidence level | (I) Percentage of all possible samples that can be expected to include the true population parameter |
| (B) Significance level | (III) The probability of making a wrong decision when the null hypothesis is true |
| (C) Confidence interval | (II) Range that could be expected to contain the population parameter of interest |
| (D) Standard error | (IV) The standard deviation of the sampling distribution of a statistic |
Choose the correct answer from the options given below:
- (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
- (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
- (A)-(I), (B)-(II), (C)-(IV), (D)-(III)
- (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
2. For the curve \( y(1 + x^2) = 2 - x \), if \(\frac{dy}{dx} = \frac{1}{A}\) at the point where the curve crosses the x-axis, then the value of \( A \) is:
For the curve \( y(1 + x^2) = 2 - x \), if \(\frac{dy}{dx} = \frac{1}{A}\) at the point where the curve crosses the x-axis, then the value of \( A \) is:
-5
5
- -1
- 0
3. If \( A = \begin{bmatrix} K & 4 \\ 4 & K \end{bmatrix} \) and \( |A^3| = 729 \), then the value of \( K^8 \) is:
If \( A = \begin{bmatrix} K & 4 \\ 4 & K \end{bmatrix} \) and \( |A^3| = 729 \), then the value of \( K^8 \) is:
- \( 9^8 \)
\( 5^8 \)
\( 3^8 \)
- \( (-3)^8 \)
4. Sanjay takes a personal loan of ₹6,00,000 at the rate of 12% per annum for 'n' years. The EMI using the flat rate method is ₹16,000. The value of 'n' is:
Sanjay takes a personal loan of ₹6,00,000 at the rate of 12% per annum for 'n' years. The EMI using the flat rate method is ₹16,000. The value of 'n' is:
3
- 6
5
- 4
5. A molecule X associates in a given solvent as per the following equation:
X ⇌ (X)n
For a given concentration of X, the van’t Hoff factor was found to be 0.80 and the
fraction of associated molecules was 0.3. The correct value of ‘n’ is:
A molecule X associates in a given solvent as per the following equation:
X ⇌ (X)n
For a given concentration of X, the van’t Hoff factor was found to be 0.80 and the
fraction of associated molecules was 0.3. The correct value of ‘n’ is:
X ⇌ (X)n
For a given concentration of X, the van’t Hoff factor was found to be 0.80 and the
fraction of associated molecules was 0.3. The correct value of ‘n’ is:
- 2
- 3
- 1
- 5
6. The shortest distance (in units) between the lines \(\frac{1 - x}{1} = \frac{2y - 10}{2} = \frac{z + 1}{1}\) and \(\frac{x - 3}{-1} = \frac{y - 5}{1} = \frac{z - 0}{1}\) is:
The shortest distance (in units) between the lines \(\frac{1 - x}{1} = \frac{2y - 10}{2} = \frac{z + 1}{1}\) and \(\frac{x - 3}{-1} = \frac{y - 5}{1} = \frac{z - 0}{1}\) is:
\(\frac{3}{\sqrt{14}}\)
- \(\frac{11}{3}\)
- \(\frac{14}{3}\)
\(\frac{\sqrt{11}}{\sqrt{3}}\)
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