| Updated On - Nov 21, 2024
CUET Mathematics Question Paper 2024 (Set D) is available 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.
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CUET UG 2024 Mathematics Question Paper 319 E SET D with Solutions
| SL No. | Question | Answer | Solution |
|---|---|---|---|
| 1 | 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 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 and b = 5. Substitute a = 5 and b = 5 into Z = ax + by. At point (8, 2): Z = 8a + 2b = 8(5) + 2(5) = 40 + 10 = 50. Thus, the maximum value of Z is 50. |
| 2 | 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 bounded by x + 2y = 12, x = 2, x = 6, and the x-axis: Express y in terms of x: y = (12 − x) / 2. The area between x = 2 and x = 6 under the line y = (12 − x) / 2 is: Area = ∫(from 2 to 6) [(12 − x) / 2] dx. Simplify: Area = (1/2) ∫(from 2 to 6) (12 − x) dx. Evaluate the integral: (1/2) [12x − (x² / 2)] from 2 to 6. Substitute limits: (1/2) [(12 × 6 − 6² / 2) − (12 × 2 − 2² / 2)] = (1/2) [(72 − 18) − (24 − 2)]. Result: (1/2) [54 − 22] = (1/2) × 32 = 16. Therefore, the area of the region is 16 sq units. |
| 3 | Find the minimum value of Z = 3x + 4y, subject to x + 2y ≥ 8, 3x + 2y ≥ 12, and x, y ≥ 0: (1) 20 (2) 24 (3) 30 (4) 36 |
(2) 24 | Solve the inequalities to find the feasible region: 1. From x + 2y = 8: y = (8 − x) / 2. 2. From 3x + 2y = 12: y = (12 − 3x) / 2. Identify intersection points of the lines: - Solve x + 2y = 8 and 3x + 2y = 12 simultaneously: Subtract equations: (3x + 2y) − (x + 2y) = 12 − 8 ⟹ 2x = 4 ⟹ x = 2. Substitute x = 2 in x + 2y = 8 ⟹ 2 + 2y = 8 ⟹ 2y = 6 ⟹ y = 3. - Intersection point is (2, 3). Evaluate Z at vertices of the feasible region: - At (2, 3): Z = 3(2) + 4(3) = 6 + 12 = 24. - Other vertices yield higher Z values. Minimum value of Z = 24. |
| 4 | 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 points A(0, 8) and B(4, 0). Slope of AB = (0 - 8) / (4 - 0) = -2. Since Z = ax + by, for Z to have a maximum value on AB, the ratio a/b must equal 2. Therefore, a = 2b. |
| 5 | If t = e2x and y = ln(t2), then d²y/dx² is: (1) 0 (2) 4t (3) 2t (4) 2t/et(4t−1) |
(1) 0 | Simplify y = ln(t²): y = 2ln(t). Since t = e2x, y = 4x. First derivative: dy/dx = 4. Second derivative: d²y/dx² = 0. |
| 6 | 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 | For symmetric matrices A and B, AT = A and BT = B. (AB − BA)T = BA − AB = −(AB − BA). This confirms that AB − BA is skew-symmetric. |
| 7 | 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| = kn|A|. Here, |2A| = 24 × 4 = 64. |
| 8 | If [A]3×2[B]x×y = [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 | For matrix multiplication, the number of columns of A must equal the number of rows of B, so x = 2. The resulting matrix dimensions must match C (3 × 1), so y = 1. |
| 9 | 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 | For f(x) to be increasing, f′(x) ≥ 0 for x ∈ [1, 2]. f′(x) = 2x + b. At x = 1: 2(1) + b ≥ 0 ⇒ b ≥ −2. At x = 2: 2(2) + b ≥ 0 ⇒ b ≥ −4. The least b satisfying both conditions is b = −2. |
| 10 | 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 is the sum of the expectations for each die: E(X) = E(X1) + E(X2) = 1/6 + 1/6 = 2/6 = 1/3. |
| 11 | 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): f′(x) = 6x² − 18x + 12. Solve f′(x) = 0 to find critical points within [0, 3]. Evaluate f(x) at endpoints and critical points: Absolute maximum: 4 (at x = 3). Absolute minimum: -5 (at x = 0). Point of maxima: x = 3. Point of minima: x = 0. |
| 12 | The second-order derivative of which of the following functions is 5x? (1) 5xln(5) (2) 5x(ln(5))² (3) 5xln(5) (4) 5x(ln(5))² |
(4) 5x(ln(5))² | Verify derivatives for each option: For (4): f(x) = 5x(ln(5))²: f′(x) = 5x(ln(5))³. f′′(x) = 5x(ln(5))², which matches the given requirement. |
| 13 | 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 | Rewrite the equation to eliminate fractional powers: (1 − (dy/dx)²)3/2 = k(d²y/dx²). Raise both sides to the power of 2/3 to eliminate the fractional exponent: 1 − (dy/dx)² = (k(d²y/dx²))². The highest order derivative is d²y/dx², and its highest power is 2, so the degree is 2. |
| 14 | Evaluate the integral ∫π xn+1 − x dx: (1) π/n loge(xn−1/xn) + C (2) loge(xn+1/xn−1) + C (3) π/n loge(xn+1/xn) + C (4) π loge(xn/xn−1) + C |
(1) π/n loge(xn−1/xn) + C | Rewrite the denominator as x(xn − 1). Use substitution: u = xn − 1, du = nxn−1dx. Integral simplifies to (π/n) loge|xn − 1| + C. |
| 15 | Evaluate the integral ∫01(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) | Use substitution: u = a + bx², du = 2bx dx. Rewrite the integral in terms of u, change limits accordingly. Simplify to get 1/(a + b). |
| 16 | The unit vector perpendicular to each of the vectors ⃗a + ⃗b and ⃗a − ⃗b, where ⃗a = ˆi + ˆj + ˆk and ⃗b = ˆi + 2ˆj + 3ˆk, is: (1) √1/6 ˆi + √2/6 ˆj + √1/6 ˆk (2) −√1/6 ˆi + √1/6 ˆj − √1/6 ˆk (3) −√1/6 ˆi + √2/6 ˆj + √2/6 ˆk (4) −√1/6 ˆi + √2/6 ˆj − √1/6 ˆk |
(4) −√1/6 ˆi + √2/6 ˆj − √1/6 ˆk | Compute the vectors ⃗a + ⃗b = 2ˆi + 3ˆj + 4ˆk and ⃗a − ⃗b = 0ˆi − ˆj − 2ˆk. Calculate the cross product (⃗a + ⃗b) × (⃗a − ⃗b): Result: −2ˆi + 4ˆj − 2ˆk. Normalize the vector: Divide by its magnitude √24 = 2√6. Unit vector: −√1/6 ˆi + √2/6 ˆj − √1/6 ˆk. |
| 17 | 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: Given: P(X = 0) = 0.1 P(X = x) = cx for x = 1 or x = 2 P(X = x) = c(5 − x) for x = 3 or x = 4 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 |
(2) (A) - (IV), (B) - (III), (C) - (II), (D) - (I) | Sum of probabilities equals 1: 0.1 + c(1) + c(2) + c(2) + c(1) = 1. Solve: 6c = 0.9 ⇒ c = 0.15. P(X ≤ 2) = 0.1 + c(1) + c(2) = 0.55. P(X ≥ 2) = c(2) + c(2) + c(1) = 0.75. P(X = 2) = c(2) = 0.3. Matches: (A) → IV, (B) → III, (C) → II, (D) → I. |
| 18 | If sin y = x sin(a + y), then dy/dx is: (1) sin a / 2 sin(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 | Differentiate both sides of sin y = x sin(a + y): cos y (dy/dx) = sin(a + y) + x cos(a + y) (dy/dx). Rearrange and solve for dy/dx: dy/dx = sin(a + y) / (cos y − x cos(a + y)). Substitute x = sin y / sin(a + y): Simplify denominator to sin a / sin(a + y). Result: dy/dx = sin²(a + y) / sin a. |
| 19 | The distance between the lines ⃗r = ˆi − 2ˆj + 3ˆk + λ(2ˆi + 3ˆj + 6ˆk) and ⃗r = 3ˆi − 2ˆj + ˆk + μ(4ˆi + 6ˆj + 12ˆk) is: (1) √28 / 7 (2) √199 / 7 (3) √328 / 7 (4) √421 / 7 |
(3) √328 / 7 | Use the formula for the shortest distance between two skew lines: d = |(d⃗1 × d⃗2) · (⃗r2 − ⃗r1)| / |d⃗1 × d⃗2|. Compute cross product of direction vectors and the numerator dot product. Simplify to find d = √328 / 7. |
| 20 | If f(x) = 2tan⁻¹(ex) − π/4, then f(x) is: (1) even and is strictly increasing in (0,∞) (2) even and is strictly decreasing in (0,∞) (3) odd and is strictly increasing in (−∞, ∞) (4) odd and is strictly decreasing in (−∞, ∞) |
(3) odd and is strictly increasing in (−∞, ∞) | Analyze the function f(x): f(−x) = −f(x), proving it is odd. Differentiate f(x): f′(x) = 2ex / (1 + e2x). Since f′(x) > 0 for all x, the function is strictly increasing. |
| 21 | For the differential equation (x loge x)dy = (loge x − y)dx: (A) Degree of the given differential equation is 1. (B) It is a homogeneous differential equation. (C) Solution is 2y loge x + A = (loge x)2, where A is an arbitrary constant. (D) Solution is 2y loge x + A = loge(loge x), where A is an arbitrary constant. Choose the correct answer from the options below: 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 is a first-order linear differential equation, so (A) is correct. It is not homogeneous, so (B) is incorrect. The solution is 2y loge x + A = (loge x)2, confirming (C). Thus, (A) and (C) only are correct. |
| 22 | Two bags: Bag-1 contains 4 white and 6 black balls, Bag-2 contains 5 white and 5 black balls. A die is rolled. If it shows a number divisible by 3, draw from Bag-1; otherwise, draw from Bag-2. If the ball drawn is not black, what is the probability it was not drawn from Bag-2? Options: 1) 4/9 2) 3/8 3) 2/7 4) 4/19 |
(3) 2/7 | Using Bayes' theorem, calculate: P(¬B2 | A) = P(A | ¬B2)P(¬B2) / P(A). Total probability: P(A) = 7/15. P(A | ¬B2) = 2/5. Result: P(¬B2 | A) = 2/7. |
| 23 | Direction ratios of the line x-3/2 = (2-y)/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 | The correct direction ratios are 2, −3, −1. Check each option: (1), (2), and (4) are valid, but (3) is not scalar multiple of original ratios. |
| 24 | Feasible region determined by constraints 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 | Graph the constraints and identify intersection points. Region C satisfies all constraints and is the feasible region. |
| 25 | Relation R: l1 R l2 ⇐⇒ l1 is parallel to l2. R is: 1) Symmetric 2) An equivalence relation 3) Transitive 4) Reflexive |
(2) An equivalence relation | R satisfies reflexivity, symmetry, and transitivity. Hence, it is an equivalence relation. |
| 26 | 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 | In a leap year, the two extra days determine the possibility of having 53 Tuesdays. If these days do not include Tuesday, the probability is 5/7. |
| 27 | Angle between two lines with direction ratios 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 cos θ = (d1·d2) / (|d1||d2|). Calculate dot product and magnitudes, resulting in cos θ = 1/2. Hence, θ = π/3. |
| 28 | 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 | Using vector properties and given conditions, solve |b| = 3 through magnitude equations and dot product simplifications. |
| 29 | If tan⁻¹(2/(3−x+1)) = cot⁻¹(3/(3x+1)), then: Options: 1) No real value of x 2) One positive and one negative real value of x 3) Two real positive values of x 4) Two real negative values of x |
(2) One positive and one negative real value of x | Simplify using trigonometric identities. Solve for x to find one positive and one negative solution. |
| 30 | If A, B, C are singular matrices: A = [[1, 4], [3, 2]], B = [[3b, 5], [a, 2]], C = [[a+b+c, c+1], [a+c, c]]. Value of abc is: Options: 1) 15 2) 30 3) 45 4) 90 |
(3) 45 | Compute determinants for A, B, and C. Use singular matrix property (det = 0) to solve equations and find abc = 45. |
| 31 | Evaluate the integral ∫ (loge 3 to loge 2) [(e^(2x) - 1)/(e^(2x) + 1)] dx: Options: 1) loge 3 2) loge 4 − loge 3 3) loge 9 − loge 4 4) loge 3 − loge 2 |
(2) loge 4 − loge 3 | Simplify the integral using substitution u = e^(2x). Compute limits and solve, yielding loge 4 − loge 3. |
| 32 | If ⃗a, ⃗b, ⃗c are vectors such that ⃗a + ⃗b + ⃗c = 0, |⃗a| = |⃗b| = 1, |⃗c| = 2, find the angle between ⃗b and ⃗c: Options: 1) 60° 2) 90° 3) 120° 4) 180° |
(4) 180° | ⃗c = −(⃗a + ⃗b). Analyze resultant magnitude and direction to conclude that ⃗b and ⃗c are opposite, angle = 180°. |
| 33 | 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) Differentiable at x = 1 II) Continuous everywhere III) Not differentiable at x = 1 IV) Differentiable everywhere except x = 0 Options: 1) (A) - II, (B) - I, (C) - III, (D) - IV 2) (A) - I, (B) - III, (C) - II, (D) - IV 3) (A) - II, (B) - III, (C) - I, (D) - IV 4) (A) - III, (B) - II, (C) - IV, (D) - I |
(1) (A) - II, (B) - I, (C) - III, (D) - IV | Evaluate continuity and differentiability of each function using modulus, integer properties. Correct mappings verified. |
| 34 | Rate of change of surface area (in cm²/s) of a hemisphere with respect to radius r at r = √3:1.331 cm: Options: 1) 66π 2) 6.6π 3) 3.3π 4) 4.4π |
(2) 6.6π | Surface area formula S = 3πr². Differentiate to find dS/dr = 6πr. At r = 1.1, solve dS/dr = 6.6π. |
| 35 | Area of the region bounded by the lines x/(7√3a) + y/b = 4, x = 0, y = 0: Options: 1) 56√3ab 2) 56a 3) ab/2 4) 3ab |
(1) 56√3ab | Identify triangle vertices (0,0), (28√3a,0), (0,4b). Use area formula: 1/2 × base × height to find 56√3ab. |
| 37 | 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 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 | Analyze the differential equation and compute integrating factors to match each case. Correct matches verified. |
| 38 | If f: N → N is defined as f(n) = n−1 if n is even, n+1 if n is odd, then: Options: A) f is injective B) f is into C) f is surjective D) f is invertible 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 | Verify injectivity, surjectivity, and invertibility by analyzing mappings and inverses. Function satisfies A, C, and D. |
| 39 | Evaluate ∫ (0 to π/2) [(1−cot x)/(csc x+cos x)]dx: Options: 1) 0 2) π/4 3) ∞ 4) π/12 |
(1) 0 | Analyze symmetry of the integrand and verify that f(x) + f(π/2 − x) = 0. Integral over symmetric interval evaluates to 0. |
| 40 | For the random variable X with distribution X: 0, 1, 2, P(X): k, 2k, 3k, find the following: A) k B) P(X < 2) C) E(X) D) P(1 ≤ X ≤ 2) Match List-I with List-II: 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 | Normalize probabilities with ΣP(X) = 1, compute expected value E(X), and verify conditional probabilities to match. |
| 41 | For a square matrix An×n: (A) |adj A| = |A|n-1 (B) |A| = |adj A|n-1 (C) A(adj A) = |A|I (D) A-1 = 1/|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 | Property (A) holds for determinants of adjugates, and (D) for inverses. (B) is incorrect, as it misstates the determinant property. |
| 42 | 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 | Matrix is scalar (all diagonal elements equal), diagonal, and symmetric (A = AT). Not skew-symmetric (diagonal not zero). |
| 43 | The feasible region represented by constraints: 4x + y ≥ 80 x + 5y ≥ 115 3x + 2y ≤ 150 x, y ≥ 0 Options: 1) Region A 2) Region B 3) Region C 4) Region D |
(3) Region C | Plot all constraints. Feasible region is intersection of inequalities satisfying all constraints, verified as Region C. |
| 44 | The area of the region enclosed between 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 | Integrate √(y/4) over y = [0, 4] and multiply by 2 for symmetry. Compute definite integral to confirm result as 16/3 sq. units. |
| 45 | Evaluate ∫ (ex/(2x+1)²√x) dx: Options: 1) 1/2 √x ex + C 2) −ex√x + C 3) −1/2 ex + C 4) ex√x + C |
(4) ex√x + C | Simplify integrand and substitute √x = u. Solve step by step, confirming result as ex√x + C. |
| 46 | If f(x), defined by: f(x) = {kx + 1, if x ≤ π; cos x, if x > π} 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 continuity at x = π, LHL = RHL = f(π). Solve kπ + 1 = −1 to get k = −2/π. |
| 47 | If P =
[-1] [2] [1]and Q = [2 -4 1], then (PQ)′ is: 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] | Compute PQ as a 3x3 matrix. Transpose PQ to confirm result matches option (2). |
| 48 | If ∆ =
| 1 cos x 1 | | -cos x 1 cos x | | -1 -cos x 1 |then: (A) ∆ = 2(1 − cos²x) (B) ∆ = 2(2 − sin²x) (C) Minimum value of ∆ is 2 (D) Maximum value of ∆ is 4 Options: 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 determinant to show ∆ = 2(2 − sin²x). Minimum value is 2 (sin²x = 1), maximum value is 4 (sin²x = 0). |
| 49 | If f(x) = sin x + (1/2)cos 2x in [0, π/2], then: (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 Options: 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), solve for critical points. Verify that maximum value is 3/4. Minimum value of 2 is incorrect. |
| 50 | 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 | Compute cross product of direction ratios ⟨1, -2, -2⟩ and ⟨0, 2, 1⟩. Normalize to obtain direction cosines, confirming option (1). |
CUET Questions
2. 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:
- \( \frac{1}{4t\sqrt{t}} \)
- \({2t^2} \)
- \( \frac{\log_e 3}{4t\sqrt{t}} \)
- \( \frac{2t^2}{e^{\frac{1}{2}\log_e t}} \)
3. The cost of a machinery is ₹8,00,000. Its scrap value will be one-tenth of its original cost in 15 years. Using the linear method of depreciation, the book value of the machine at the end of the 10th year will be:
The cost of a machinery is ₹8,00,000. Its scrap value will be one-tenth of its original cost in 15 years. Using the linear method of depreciation, the book value of the machine at the end of the 10th year will be:
- ₹4,80,000
- ₹3,20,000
- ₹3,68,000
- ₹4,32,000
4. In a 600 m race, the ratio of the speeds of two participants A and B is 4:5. If A has a head start of 200 m, then the distance by which A wins is:
In a 600 m race, the ratio of the speeds of two participants A and B is 4:5. If A has a head start of 200 m, then the distance by which A wins is:
- 500 m
- 200 m
- 100 m
- 120 m
5. 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:
- \(\frac{609}{625}\)
- \(\frac{16}{625}\)
- \(\frac{513}{625}\)
- \(\frac{112}{625}\)
6. Vibhuti bought a car worth ₹10,25,000 and made a down payment of ₹4,00,000. The balance is to be paid in 3 years by equal monthly installments at an interest rate of 12% p.a. The EMI that Vibhuti has to pay for the car is:
(Use \( (1.01)^{-36} = 0.7 \))
Vibhuti bought a car worth ₹10,25,000 and made a down payment of ₹4,00,000. The balance is to be paid in 3 years by equal monthly installments at an interest rate of 12% p.a. The EMI that Vibhuti has to pay for the car is:
(Use \( (1.01)^{-36} = 0.7 \))
(Use \( (1.01)^{-36} = 0.7 \))
- ₹20,700.85
- ₹27,058.87
- ₹20,833.33
- ₹25,708.89
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