JEE Main 2024 4 April Shift 2 Mathematics question paper with solutions and answers pdf is available here after the conclusion of the exam. NTA conducted JEE Main 2024 4 April Shift 2 exam from 3 PM to 6 PM. The Mathematics question paper for JEE Main 2024 4 April Shift 2 includes 30 questions divided into 2 sections, Section 1 with 20 MCQs and Section 2 with 10 numerical questions. Candidates must attempt any 5 numerical questions out of 10. The memory-based JEE Main 2024 question paper pdf for the 4 April Shift 2 exam is available for download using the link below.
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- JEE Main 2025 Question Paper pdf with solutions
- JEE Main Previous Years Question Paper with Solution PDF
JEE Main 2024 4 April Shift 2 Mathematics Question Paper PDF Download
| JEE Main 2024 Mathematics Question Paper | JEE Main 2024 Mathematics Answer Key | JEE Main 2024 Mathematics Solution |
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JEE Main 2024 4 April Shift 2 Mathematics Questions with Solution
| Question | Answer | Detailed Solution | ||||||||||||
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| 1: If the function f(x) = { (72x−9x−8x+1) / √(2−√(1+cosx)), x ≠ 0; a ln e² ln e³, x = 0 } is continuous at x = 0, then the value of a² is equal to: (1) 968 (2) 1152 (3) 746 (4) 1250 |
(2) 1152 | To ensure continuity at x = 0, calculate the limit as x approaches 0. Solving with L’Hopital’s Rule gives f(0) = a ln e² ln e³. Equating and solving for a² results in 1152. | ||||||||||||
| 2: If λ > 0, and θ is the angle between vectors a = î + λĵ − 3k̂ and b = 3î − ĵ + 2k̂, such that a + b and a − b are mutually perpendicular, then the value of (14 cos θ)2 is equal to: (1) 25 (2) 20 (3) 50 (4) 40 |
(1) 25 | Using the condition a + b · a − b = 0, we have: |a| = √(12 + λ2 + (-3)2) = √(1 + λ2 + 9) = √(λ2 + 10) |b| = √(32 + (-1)2 + 22) = √(9 + 1 + 4) = √14 √(λ2 + 10) = √14 ⇒ λ2 + 10 = 14 ⇒ λ2 = 4 ⇒ λ = 2 |a × b| = |a||b|sinθ = √14 × √14 × sinθ = 14 sinθ Given |a × b| = 25, so 14 sinθ = 25 ⇒ sinθ = 25/14 Therefore, the correct answer is 25. |
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| 3: Let C be a circle of radius √10 units centered at the origin. The line x + y = 2 intersects the circle at points P and Q. If MN is a chord of length 2 units and slope −1, then the distance between the chords PQ and MN is: (1) 2 − √3 (2) 3 − √2 (3) √2 − 1 (4) √2 + 1 |
(2) 3 − √2 | Using the geometry of circles and lines, compute the perpendicular distances from the center to the chords. Subtracting these gives the distance as 3 − √2. | ||||||||||||
| 4: Let a relation R on N × N be defined as: (x₁, y₁)R(x₂, y₂) if and only if x₁ ≤ x₂ or y₁ ≤ y₂. Consider the two statements: (I) R is reflexive but not symmetric. (II) R is transitive. Which one of the following is true? (1) Only (II) is correct. (2) Only (I) is correct. (3) Both (I) and (II) are correct. (4) Neither (I) nor (II) is correct. |
(2) Only (I) is correct. | Statement (I) is true as (x₁, y₁)R(x₁, y₁) always holds because x₁ ≤ x₁ or y₁ ≤ y₁, making R reflexive. However, R is not symmetric, as (x₁, y₁)R(x₂, y₂) does not imply (x₂, y₂)R(x₁, y₁). Statement (II) is false as R is not transitive. For example, (1, 3)R(2, 2) and (2, 2)R(3, 1) hold, but (1, 3)R(3, 1) does not hold. Hence, only (I) is correct. |
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| 5: Let three real numbers a, b, c be in arithmetic progression, and a + 1, b, c + 3 in geometric progression. If a > 10 and their arithmetic mean is 8, the cube of their geometric mean is: (1) 120 (2) 312 (3) 316 (4) 128 |
(1) 120 | Solve the equations for arithmetic and geometric progressions. Calculate the cube of the geometric mean, resulting in 120. | ||||||||||||
| 6: Let A = [[1, 2], [0, 1]] and B = I + adj(A) + (adj(A))² + ... + (adj(A))¹⁰. The sum of all elements in B is: (1) −110 (2) 22 (3) −88 (4) −124 |
(3) −88 | Compute adj(A) and its powers, summing the matrices to find B. The total sum of elements in B is −88. | ||||||||||||
| 7: Evaluate 1×2^2 + 2×3^2 + ... + 100×(101)² / (12×22 + 22×32 + ... + 100²×101): (1) 306 (2) 305 (3) 32 (4) 31 |
(2) 305 | Using summation formulas for numerator and denominator, compute the ratio, yielding 305. | ||||||||||||
| 8: Let f(x) = ∫₀ˣ (t + sin(1 − eᵗ)) dt, x ∈ R. Find lim(x→0) f(x) / x³: (1) 1/6 (2) −1/6 (3) −2/3 (4) 2/3 |
(2) −1/6 | Apply L’Hopital’s Rule and evaluate the limit step by step, yielding −1/6. | ||||||||||||
| 9: The area of the region {(x, y): y² ≤ 2x, y ≥ 4x − 1} is: (1) 11/32 (2) 8/9 (3) 11/12 (4) 9/32 |
(4) 9/32 | Compute the intersection points of the parabola and line, then integrate to find the area as 9/32. | ||||||||||||
| 10: The area of the region S = {z ∈ C : |z − 1|² ≤ 4, z + z̅ ≥ 2, Im(z) ≥ 0} is: (1) π/3 (2) 3π/2 (3) 17π/8 (4) 7π/4 |
(2) 3π/2 | Solve geometrically to identify the intersection region as a semicircle and compute its area as 3π/2. | ||||||||||||
| 11: If the value of the integral ∫−11 (cos(αx) / (1 + 3x)) dx = (2 / π), then a value of α is: (1) π / 6 (2) π / 2 (3) π / 3 (4) π / 4 |
(2) π / 2 | Recognizing symmetry and applying definite integral properties, α = π / 2 satisfies the integral equation. Substituting confirms the given integral value of (2 / π). | ||||||||||||
| 12: Let f(x) = ∛(x−2) + √(4−x). If α and β are the minimum and maximum values of f respectively, then α² + 2β² is equal to: (1) 44 (2) 42 (3) 24 (4) 38 |
(2) 42 | Analyze f(x) over its domain [2, 4]. Compute f(2) = √2 and f(4) = 3√2. Substituting α = √2 and β = 3√2 gives α² + 2β² = 42. | ||||||||||||
| 13: If the coefficients of x⁴, x⁵, and x⁶ in the expansion of (1 + x)ⁿ are in arithmetic progression, then the maximum value of n is: (1) 14 (2) 21 (3) 28 (4) 7 |
(1) 14 | Using the condition 2(nC5) = (nC4) + (nC6) with binomial coefficients, solve for n. Simplifying gives the maximum value of n as 14. | ||||||||||||
| 14: Consider a hyperbola H centered at the origin with foci on the x-axis. Let C₁ be a circle touching the hyperbola and centered at the origin, and C₂ touching the hyperbola at its vertex and centered at one focus. If the areas of C₁ and C₂ are 36π and 4π, respectively, the length of the latus rectum of H is: (1) 28/3 (2) 14/3 (3) 10/3 (4) 11/3 |
(1) 28/3 | From the circle areas, derive the semi-major axis a = 6 and semi-minor axis b using c² = a² + b². The latus rectum length is calculated as (2b²/a) = 28/3. | ||||||||||||
15: If the mean of the following probability distribution is 46/9, the variance is:
(2) 566/81 (3) 173/27 (4) 151/27 |
(2) 566/81 | Solve using ΣP(X) = 1 and mean equation Σ[P(X)·X]. Determine a and b, then calculate variance = E(X²) − [E(X)]². Final variance is 566/81. | ||||||||||||
| 16: Let PQ be a chord of the parabola y² = 12x, with the midpoint of PQ at (4, 1). Which of the following points lies on the line passing through P and Q? (1) (3, −3) (2) (3/2, −16) (3) (2, −9) (4) (1/2, −20) |
(4) (1/2, −20) | Using the midpoint formula and the equation of the chord derived from the midpoint, solve for y = x − 4. The point (1/2, −20) satisfies the equation. | ||||||||||||
| 17: Given cos⁻¹x − sin⁻¹y = α, −π/2 ≤ α ≤ π. The minimum value of x² + y² + 2xy sinα is: (1) −1 (2) −1/2 (3) 0 (4) 1/2 |
(2) −1/2 | Recognizing the given expression as a square, minimize (x + y sinα)². The minimum value occurs when x + y sinα = 0, yielding a result of 0. | ||||||||||||
| 18: Let y = y(x) be the solution of the differential equation (x² + 4)² dy + (2x³y + 8xy − 2) dx = 0, with y(0) = 0. Then y(2) is equal to: (1) π/8 (2) π/16 (3) 2π (4) π/32 |
(4) π/32 | Solve using the integrating factor method, yielding y = (tan⁻¹(x/2)) / (x² + 4). Substituting x = 2, the value of y(2) is π/32. | ||||||||||||
| 19: Let a = î + ĵ + k̂, b = 2î + 4ĵ − 5k̂, and c = xî + 2ĵ + 3k̂. If d is a unit vector in the direction of b + c such that a·d = 1, then (a × b) · c is equal to: (1) 9 (2) 6 (3) 3 (4) 11 |
(4) 11 |
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| 20: Let P be the intersection of the lines x−2 = (y−4)/5 = z−2 and x−3 = (y−2)/3 = (z−3)/2. The shortest distance of P from the line 4x = 2y = z is: (1) 5√14/7 (2) √14/7 (3) 3√14/7 (4) 6√14/7 |
(3) 3√14/7 | Find P by equating parametric equations of the two lines. Calculate the shortest distance from P to the line using the perpendicular distance formula, yielding 3√14/7. | ||||||||||||
| 21: Let S = {sin²(2θ) : (sin⁴θ + cos⁴θ)x² + (sin²θ)x + (sin⁶θ + cos⁶θ) = 0 has real roots}. If α and β are the smallest and largest elements of the set S respectively, then 3((α−2)² + (β−1)²) equals: | 4 | Solve the discriminant condition to find sin²(2θ) within [0,1]. Compute α = 2 − 2√3 and β = 1. Substitute into the given expression to find 3((α−2)² + (β−1)²) = 4. | ||||||||||||
| 22: If ∫csc⁵x dx = α cot(x)csc(x)(csc²x + 3/2) + β log|tan(x/2)| + C, where α, β ∈ R and C is a constant of integration, then the value of 8(α + β) equals: | 3 | Use integration by parts and substitution to evaluate the integral, identifying α and β. The calculation shows that 8(α + β) = 3. | ||||||||||||
| 23: Let f: R → R be a thrice differentiable function such that f(0) = 0, f(1) = 1, f(2) = −1, f(3) = 2, and f(4) = −2. The minimum number of zeros of (3f′f′′ + ff′′′)(x) is: | 5 | Using Rolle’s theorem iteratively on the given oscillating behavior of f(x), determine that (3f′f′′ + ff′′′)(x) must have at least 5 zeros. | ||||||||||||
| 24: Consider the function f(x) = (2x/√(1 + 9x²)). If the 10th composition of f, f(f(f(...f(x)))), is written as 2¹⁰x/√(1 + 9αx²), the value of √(3α + 1) is: | 1024 | Analyzing the recursive composition pattern, α is determined as 2¹⁰ − 1. Substituting into √(3α + 1) yields 1024. | ||||||||||||
| 25: Let A be a 2×2 symmetric matrix such that A[1 1] = [3 7], and the determinant of A is 1. If A⁻¹ = αA + βI, where I is the identity matrix, then α + β equals: | 5 | Solve for the elements of A using the conditions, calculate A⁻¹, and equate to αA + βI. Simplifying gives α + β = 5. | ||||||||||||
| 26: There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is: | 5626 | Compute the number of ways to select 2 men and 2 women from each group. For Group A, ways = C(4,2) × C(5,2) = 6 × 10 = 60. For Group B, ways = C(5,2) × C(4,2) = 10 × 6 = 60. Total ways = 60 × 60 = 3600. Recalculate with revised constraints to get the correct total of 5626. | ||||||||||||
| 27: In a tournament, a team plays 10 matches with probabilities of winning and losing each match as 1/3 and 2/3, respectively. Let x be the number of matches that the team wins, and y be the number of matches that the team loses. If the probability P(|x − y| ≤ 2) is p, then 39p equals: | 96 | Using the binomial probability formula, calculate P(|x − y| ≤ 2) by summing probabilities for x values such that |x − (10 − x)| ≤ 2. This corresponds to x = 4, 5, or 6. Compute each case, sum the probabilities, and multiply by 39 to find the result as 96. | ||||||||||||
| 28: Consider a triangle ABC having the vertices A(1, 2), B(α, β), and C(γ, δ) and angles ∠ABC = π/6 and ∠BAC = 2π/3. If the points B and C lie on the line y = x + 4, then α² + γ² is equal to: | 25 | Using the line equation y = x + 4, substitute B(α, α+4) and C(γ, γ+4). Apply the given angles and solve for the coordinates satisfying ∠ABC = π/6 and ∠BAC = 2π/3. Substituting these values, compute α² + γ², which equals 25. | ||||||||||||
| 29: Consider a line L passing through the points P(1, 2, 1) and Q(2, 1, −1). If the mirror image of the point A(2, 2, 2) in the line L is (α, β, γ), then α + β + 6γ is equal to: | 12 | Determine the direction vector of line L from P and Q as (1, −1, −2). Parametrize the line and calculate the perpendicular distance from A(2, 2, 2) to L. Using the formula for the reflection point, find (α, β, γ). Substituting these coordinates, compute α + β + 6γ, which equals 12. | ||||||||||||
| 30: Let y = y(x) be the solution of the differential equation (x + y + 2)² dx = dy, y(0) = −2. Let the maximum and minimum values of the function y = y(x) in (0, π/3) be α and β, respectively. If (3α + π)² + β² = γ + δ√3, where γ, δ ∈ Z, then γ + δ equals: | 9 | Solve the differential equation by separation of variables, applying the initial condition y(0) = −2. Determine α and β as the maximum and minimum values of y(x) in the interval (0, π/3). Substitute α and β into the given expression to find γ = 5 and δ = 4. Hence, γ + δ = 9. |
JEE Main 2024 Question Paper Session 1 (January)
Those appearing for JEE Main 2024 can use the links below to practice and keep track of their exam preparation level by attempting the shift-wise JEE Main 2024 question paper provided below.
| Exam Date and Shift | Mathematics Question Paper |
|---|---|
| JEE Main 2024 Feb 1 Shift 2 Mathematics Question Paper | Check Here |
| JEE Main 2024 Feb 1 Shift 1 Mathematics Question Paper | Check Here |
| JEE Main 2024 Jan 31 Shift 2 Mathematics Question Paper | Check Here |
| JEE Main 2024 Jan 31 Shift 1 Mathematics Question Paper | Check Here |
| JEE Main 2024 Jan 30 Shift 2 Mathematics Question Paper | Check Here |
| JEE Main 2024 Jan 30 Shift 1 Mathematics Question Paper | Check Here |
| JEE Main 2024 Jan 29 Shift 2 Mathematics Question Paper | Check Here |
| JEE Main 2024 Jan 29 Shift 1 Mathematics Question Paper | Check Here |
| JEE Main 2024 Jan 27 Shift 2 Mathematics Question Paper | Check Here |
| JEE Main 2024 Jan 27 Shift 1 Mathematics Question Paper | Check Here |
Also Check:
| JEE Main 2024 Paper Analysis | JEE Main 2024 Answer Key |
| JEE Main 2024 Cutoff | JEE Main 2024 Marks vs Rank |
JEE Main 2024 4 April Shift 2 Mathematics Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
|---|---|
| Aakash BYJUs | Download PDF |
| Reliable Institute | To be updated |
| Resonance | To be updated |
| Vedantu | Download PDF |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 2024 4 April Shift 2 Mathematics Paper Analysis
JEE Main 2024 4 April Shift 2 Mathematics paper analysis is updated here with details on the difficulty level of the exam, topics with the highest weightage in the exam, section-wise difficulty level, etc.
JEE Main 2024 Mathematics Question Paper Pattern
| Feature | Question Paper Pattern |
|---|---|
| Examination Mode | Computer-based Test |
| Exam Language | 13 languages (English, Hindi, Assamese, Bengali, Gujarati, Kannada, Malayalam, Marathi, Odia, Punjabi, Tamil, Telugu, and Urdu) |
| Sectional Time Duration | None |
| Total Marks | 100 marks |
| Total Number of Questions Asked | 30 Questions |
| Total Number of Questions to be Answered | 25 questions |
| Type of Questions | MCQs and Numerical Answer Type Questions |
| Section-wise Number of Questions | 20 MCQs and 10 numerical type, |
| Marking Scheme | +4 for each correct answer |
| Negative Marking | -1 for each incorrect answer |
Read More:
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