JEE Main 2024 Jan 29 Shift 2 Mathematics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Jan 29 Shift 2 exam from 3 PM to 6 PM. The Mathematics question paper for JEE Main 2024 Jan 29 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 Jan 29 Shift 2 exam is available for download using the link below.
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JEE Main 2024 Jan 29 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 29 Jan Shift 2 2024 Mathematics Questions with Solutions
| Question | Answer | Detailed Solution | ||||||||||||||||||
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1: Let A =
P =
The sum of the prime factors of |P−1A P − 2I| is equal to: (1) 26 (2) 27 (3) 66 (4) 23 |
(1) 26 | The determinant |P−1A P − 2I| is calculated using matrix algebra. Its prime factors are identified and summed to give the final result, which is 26. | ||||||||||||||||||
| 2: The number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to: (1) 18 (2) 16 (3) 12 (4) 15 |
(4) 15 | Distribute 8 books into 4 shelves: When no shelves are empty, (2, 2, 2, 2), (3, 3, 1, 1), (4, 2, 1, 1), (3, 2, 2, 1), (5, 1, 1, 1) – 5 ways. Thus, total number of ways = 15. | ||||||||||||||||||
| 3: Let P(3, 2, 3), Q(4, 6, 2), R(7, 3, 2) be the vertices of △PQR. Then, the angle △QPR is: (1) π/6 (2) cos⁻¹(7/18) (3) cos⁻¹(1/18) (4) π/3 |
(4) π/3 | The direction ratios of PR are (7 − 3, 3 − 2, 2 − 3) = (4, 1, −1). The direction ratios of PQ are (4 − 3, 6 − 2, 2 − 3) = (1, 4, −1). To find the cosine of θ, we use the dot product formula: cos θ = (4 · 1) + (1 · 4) + (−1 · −1) / (√(4² + 1² + (−1)²)) · (√(1² + 4² + (−1)²)) = 9 / 18 = 1/2. Therefore, θ = π/3. | ||||||||||||||||||
| 4: If the mean and variance of five observations are 24/5 and 194/25 respectively, and the mean of the first four observations is 7/2, then the variance of the first four observations is equal to: (1) 4/5 (2) 77/12 (3) 5/4 (4) 105/4 |
(3) 5/4 | Let the five observations be x₁, x₂, x₃, x₄, x₅. From the given mean of all observations (24/5) and the first four observations (7/2), we can calculate x₅ = 10. Using variance formula, we find the variance of the first four observations to be 5/4. | ||||||||||||||||||
| 5: The function f(x) = 2x + 3x^(2/3), x ∈ R, has: (1) Exactly one point of local minima and no point of local maxima. (2) Exactly one point of local maxima and no point of local minima. (3) Exactly one point of local maxima and exactly one point of local minima. (4) Exactly two points of local maxima and exactly one point of local minima. |
(3) Exactly one point of local maxima and exactly one point of local minima. | Given f(x) = 2x + 3x^(3/2), by finding the first and second derivatives, we determine that there is a local maximum at x = −1 and a local minimum at x = 0. | ||||||||||||||||||
| 6: Let r and θ respectively be the modulus and amplitude of the complex number z = 2 − i(2 tan(5π/8)). Then, (r, θ) is equal to: (1) (2 sec(3π/8), 3π/8) (2) (2 sec(3π/8), 5π/8) (3) (2 sec(5π/8), 3π/8) (4) (2 sec(11π/8), 11π/8) |
(1) (2 sec(3π/8), 3π/8) | Given z = 2 − i(2 tan(5π/8)), we calculate the modulus r and amplitude θ using the formula for complex numbers. We find r = 2 sec(3π/8) and θ = 3π/8. | ||||||||||||||||||
| 7: The sum of the solutions x ∈ R of the equation 3 cos(2x) + cos³(2x)/(cos⁶(x) − sin⁶(x)) = x³ − x² + 6 is: (1) 0 (2) 1 (3) −1 (4) 3 |
(3) −1 | The equation simplifies using trigonometric identities and Vieta's formulas. The sum of the roots of the cubic equation x³ − x² + 6 = 0 is −1. | ||||||||||||||||||
| 8: Let OA = a, OB = 12a + 4b, and OC = b, where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then the ratio of the area of the quadrilateral OABC to the area of S is equal to: (1) 6 (2) 10 (3) 7 (4) 8 |
(4) 8 | The area of parallelogram S is given by S = |a × b|. The area of quadrilateral OABC is found by splitting it into two triangles OAB and OBC. The total area of OABC is 8|a × b|, so the ratio is 8. | ||||||||||||||||||
| 9: If log a, log b, log c are in an A.P. and log a − log 2b, log 2b − log 3c, log 3c − log a are also in an A.P., then a : b : c is equal to: (1) 9 : 6 : 4 (2) 16 : 4 : 1 (3) 25 : 10 : 4 (4) 6 : 3 : 2 |
(1) 9 : 6 : 4 | By solving the equations derived from the arithmetic progression properties of log a, log b, log c, and the given A.P. differences, we find that a : b : c = 9 : 6 : 4. | ||||||||||||||||||
| 10: If ∫(sin^(3/2)x + cos^(3/2)x) / √(sin³x cos³x sin(x − θ)) dx = A cos θ sin x − B sin θ cos x + C, where C is the integration constant, then AB is equal to: (1) 4 csc(2θ) (2) 4 sec θ (3) 2 sec θ (4) 8 csc(2θ) |
(4) 8 csc(2θ) | After solving the integral by simplifying the trigonometric expressions and performing the required substitutions, we find that AB = 8 csc(2θ). | ||||||||||||||||||
| 11: The distance of the point (2, 3) from the line 2x − 3y + 28 = 0, measured parallel to the line √3x − y + 1 = 0, is equal to: (1) 4√2 (2) 6√3 (3) 3 + 4√2 (4) 4 + 6√3 |
(4) 4 + 6√3 | The distance of the point (2, 3) from the line 2x − 3y + 28 = 0, measured parallel to the line √3x − y + 1 = 0, is calculated by finding the perpendicular distance between the point and the line. The result is 4 + 6√3. | ||||||||||||||||||
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12: If sin(y/x) = ln|x| + α/2 is the solution of the differential equation x cos(y/x) dy/dx = y cos(y/x) + x, and y(1) = π/3, then α^2 is equal to: |
(1) 3 | Given the differential equation and using the method of separation of variables, solving for α yields α/2 = 3. | ||||||||||||||||||
| 13: If each term of a geometric progression a₁, a₂, a₃, ... with a₁ = 1/8 and a₂ ≠ a₁, is the arithmetic mean of the next two terms and Sₙ = a₁ + a₂ + ... + aₙ, then S₂₀ − S₁₈ is equal to: (1) 2^15 (2) −2^18 (3) 2^18 (4) −2^15 |
(4) −2^15 | Let r be the common ratio of the geometric progression. Using the condition 2a_r = a_{r+1} + a_{r+2}, we find r = -2. Calculating T₁₉ and T₂₀, the difference S₂₀ − S₁₈ = T₁₉ + T₂₀ = −2^15. | ||||||||||||||||||
| 14: Let A be the point of intersection of the lines 3x + 2y = 14 and 5x − y = 6, and B be the point of intersection of the lines 4x + 3y = 8 and 6x + y = 5. The distance of the point P(5, −2) from the line AB is: (1) 13/2 (2) 8 (3) 5/2 (4) 6 |
(4) 6 | Solving for the intersection points A(2, 4) and B(1/2, 2), we determine the equation of line AB as 4x − 3y + 4 = 0. Using the perpendicular distance formula, the distance from P(5, −2) to the line is 6. | ||||||||||||||||||
| 15: Let x = m/n (m, n are co-prime natural numbers) be a solution of the equation cos(2 sin⁻¹ x) = 1/9, and let α, β (α > β) be the roots of the equation mx² − nx − m + n = 0. Then the point (α, β) lies on the line: (1) 3x + 2y = 2 (2) 5x − 8y = −9 (3) 3x − 2y = −2 (4) 5x + 8y = 9 |
(4) 5x + 8y = 9 | Using cos(2 sin⁻¹ x) = 1/9, we calculate x = 2/3. Substituting into the given quadratic equation, we determine the roots α and β. These roots satisfy the equation 5x + 8y = 9. | ||||||||||||||||||
| 16: The function f(x) = x / (x² - 6x - 16), x ∈ R \ {−2, 8}, has: (1) decreases in (−2, 8) and increases in (−∞, −2) ∪ (8, ∞) (2) decreases in (−∞, −2) ∪ (−2, 8) ∪ (8, ∞) (3) decreases in (−∞, −2) and increases in (8, ∞) (4) increases in (−∞, −2) ∪ (−2, 8) ∪ (8, ∞) |
(2) decreases in (−∞, −2) ∪ (−2, 8) ∪ (8, ∞) | Using the quotient rule, we find that f'(x) < 0 for all x ∈ R \ {−2, 8}. Therefore, the function decreases in the given intervals. | ||||||||||||||||||
| 17: Let y = ln((1 − x²) / (1 + x²)), −1 < x < 1. Then at x = 1/2, the value of 225(y' − y'') is equal to: (1) 732 (2) 746 (3) 742 (4) 736 |
(4) 736 | Differentiating y, we find y' = −4x / ((1 − x²)(1 + x²)) and y'' = −4(1 + 3x²) / ((1 − x²)²(1 + x²)²). Substituting x = 1/2, 225(y' − y'') = 736. | ||||||||||||||||||
| 18: If R is the smallest equivalence relation on the set {1, 2, 3, 4} such that {(1, 2), (1, 3)} ⊆ R, then the number of elements in R is: (1) 10 (2) 12 (3) 8 (4) 15 |
(1) 10 | The minimal equivalence relation includes reflexivity, symmetry, and transitivity. Adding all such pairs results in the set containing 10 elements. | ||||||||||||||||||
| 19: An integer is chosen at random from the integers 1, 2, 3, ..., 50. The probability that the chosen integer is a multiple of at least one of 4, 6, 7 is: (1) 8/25 (2) 21/50 (3) 9/50 (4) 14/25 |
(2) 21/50 | Using inclusion-exclusion, the probability is calculated as P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C), resulting in 21/50. | ||||||||||||||||||
| 20: Let a unit vector u = xi + yj + zk make angles π/2, π/3, 2π/3 with the vectors p₁ = (1/√2)i + (1/√2)k, p₂ = (1/√2)j + (1/√2)k, and p₃ = (1/√2)i + (1/√2)j, respectively. If v = (1/√2)(i + j + k), then |u − v|² is equal to: (1) 11/2 (2) 5/2 (3) 9 (4) 7 |
(2) 5/2 | Using the dot product properties and vector subtraction, the unit vector u = (−1/√2)i + 0j + (1/√2)k is calculated. Subtracting v = (1/√2)(i + j + k), we find |u − v|² = 5/2. | ||||||||||||||||||
| 21: Let α, β be the roots of the equation x² − √6x + 3 = 0 such that Im(α) > Im(β). Let a, b be integers not divisible by 3 and n be a natural number such that αⁿ / β + α⁹⁹ + α⁹⁸ = 3ⁿ(a + ib), i = √−1. Then n + a + b is equal to: |
49 | The roots α and β are determined as √3eiπ/4 and √3e−iπ/4. Using the equation properties and simplifications, n + a + b = 49. | ||||||||||||||||||
| 22: Let for any three distinct consecutive terms a, b, c of an A.P., the lines ax + by + c = 0 be concurrent at the point P, and Q(α, β) be a point such that the system of equations x + y + z = 6, 2x + 5y + αz = β, x + 2y + 3z = 4 has infinitely many solutions. Then (PQ)² is equal to: | 113 | The determinant of the system of equations is zero for infinite solutions. Solving, β = 8 and points P(1, −2) and Q(8, 6). The square of the distance (PQ)² = 113. | ||||||||||||||||||
| 23: Let P(α, β) be a point on the parabola y² = 4x. If P also lies on the chord of the parabola x² = 8y whose midpoint is (1, 5/4), then (α − 28)(β − 8) is equal to: | 192 | Using the midpoint formula and the equations of the parabola, the coordinates of P are determined, giving (α − 28)(β − 8) = 192. | ||||||||||||||||||
| 24: If ∫ π/3 π/6 √1 − sin 2x dx = α + β√2 + γ√3, where α, β, γ are rational numbers, then 3α + 4β − γ is equal to: | 6 | The integral simplifies using trigonometric identities as: ∫ π/3 π/6 |sinx − cosx| dx = ∫ π/3 π/6 (cosx − sinx) dx. By evaluating, we get: [sinx + cosx] from π/6 to π/3. Substituting limits: - At x = π/3: √3/2 + 1/2 = √2. - At x = π/6: 1/2 + √3/2 = (1 + √3)/2. Final result: √2 − (1/2 + √3/2) = −1 + 2√2 − √3. Calculating: α = −1, β = 2, γ = −1. Hence, 3α + 4β − γ = 3(−1) + 4(2) − (−1) = 6. | ||||||||||||||||||
| 25: Let the area of the region {(x, y) : 0 ≤ x ≤ 3, 0 ≤ y ≤ min(x² + 2, 2x + 2)} be A. Then 12A is equal to: | 164 | The region is divided into two parts, and integrals are calculated for each segment, resulting in total area A. Finally, 12A = 164. | ||||||||||||||||||
| 26: Let O be the origin, and M and N be the points on the lines: x−5/4 = y−4/1 = z−5/3 and x+8/12 = y+2/5 = z+11/9, respectively, such that MN is the shortest distance between the given lines. Then OM · ON is equal to: |
9 | Using the parametric equations of the lines and solving for the shortest distance, the dot product of OM and ON is calculated as 9. | ||||||||||||||||||
| 27: Let f(x) = √lim(r→x) [2r²(f(r²) − f(x))f(r)/(r²−x²) − r²e^(f(r)/r)] be differentiable in (−∞, 0) ∪ (0,∞) and f(1) = 1. Then the value of eᵃ, such that f(a) = 0, is equal to: | 2 | Simplifying the functional limit expression and substituting the given conditions, it is determined that eᵃ = 2 when f(a) = 0. | ||||||||||||||||||
| 28: Remainder when 64³²³² is divided by 9 is equal to: | 1 | Using modular arithmetic: 64 ≡ 1 (mod 9). Hence, 64³²³² ≡ 1³²³² ≡ 1 (mod 9). The remainder is 1. | ||||||||||||||||||
| 29: Let the set C = {(x, y) | x² − 2y = 2023, x, y ∈ N}. Then ∑(x, y)∈C(x + y) is equal to: | 46 | Solving the equation x² − 2y = 2023 under the constraints that x, y are natural numbers, the only solution is (x, y) = (45, 1). Thus, ∑(x, y)∈C(x + y) = 45 + 1 = 46. | ||||||||||||||||||
| 30: Let the slope of the line 45x + 5y + 3 = 0 be 27r₁ + 9r₂² for some r₁, r₂ ∈ R. Then: lim(x→3)∫(x to 3)(8t²/(3r₂x² − r₂x² − r₁x³ − 3x))dt is equal to: |
12 | Simplifying the integral and applying L’Hôpital’s Rule, the limit evaluates to 12. |
Also Check:
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JEE Main 2024 Jan 29 Shift 2 Mathematics Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
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| Aakash BYJUs | Download PDF |
| Reliable Institute | To be updated |
| Resonance | To be updated |
| Vedantu | To be updated |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 2024 Jan 29 Shift 2 Mathematics Paper Analysis
JEE Main 2024 Jan 29 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 Physics 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 |
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