JEE Main 2024 Feb 1 Shift 2 Mathematics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Feb 1 Shift 2 exam from 3 PM to 6 PM. The Mathematics question paper for JEE Main 2024 Feb 1 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 Feb 1 Shift 2 exam is available for download using the link below.
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JEE Main 2024 Feb 1 Shift 2 Mathematics Question Paper PDF Download
| JEE Main 2024 Mathematics Question Paper | JEE Main 2024 Mathematics Answer Key | JEE Main 2024 Mathematics Solution |
|---|---|---|
| Download PDF | Download PDF | Download PDF |
JEE Main 1 Feb Shift 2 2024 Mathematics Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 1: Let f(x) = |2x² + 5||x| − 3, x ∈ R. If m and n denote the number of points where f is not continuous and not differentiable respectively, then m+n is equal to: (1) 5 (2) 2 (3) 0 (4) 3 |
(4) 3 | The function f(x) = |2x² + 5||x| − 3 is discontinuous and non-differentiable at x = 0. Total points = 3. |
| 2: Let α and β be the roots of the equation px² + qx − r = 0, where p ≠ 0. If p, q, r are consecutive terms of a non-constant G.P. and 1/α + 1/β = 3/4, then the value of (α − β)² is: (1) 80/9 (2) 9 (3) 20/3 (4) 8 |
(1) 80/9 | Using roots and G.P. properties, simplify to find (α − β)² = 80/9. |
| 3: The number of solutions of the equation 4 sin²x − 4 cos³x + 9 − 4 cosx = 0, for x ∈ [−2π, 2π], is: (1) 1 (2) 3 (3) 2 (4) 0 |
(4) 0 | On solving, no real roots are found within the valid range. Therefore, the solution count is 0. |
| 4: The value of ∫₀¹ (2x³ − 3x² − x + 1)¹/³ dx is equal to: (1) 0 (2) 1 (3) 2 (4) -1 |
(1) 0 | Using symmetry properties of definite integrals, the integral evaluates to 0. |
| 5: Let P be a point on the ellipse x²/9 + y²/4 = 1. Let the line passing through P and parallel to the y-axis meet the circle x² + y² = 9 at point Q such that P and Q are on the same side of the x-axis. Then, the eccentricity of the locus of the point R on PQ such that PR : RQ = 4 : 3 as P moves on the ellipse, is: (1) 11/19 (2) 13/21 (3) √139/23 (4) √13/7 |
(4) √13/7 | Substitute points and solve to determine the ellipse parameters. The eccentricity is √13/7. |
| 6: Let m and n be the coefficients of the seventh and thirteenth terms respectively in the expansion of (1/3x¹/3 + 1/2x²/3 + 1)¹⁸. Then n/m³ is: (1) 4/9 (2) 1/9 (3) 1/4 (4) 9/4 |
(4) 9/4 | Using the general term in binomial expansion, compute coefficients m and n. Simplify to find n/m³ = 9/4. |
| 7: Let α be a non-zero real number. Suppose f : R → R is a differentiable function such that f(0) = 2 and limx→∞ f(x) = 1. If f'(x) = αf(x) + 3, then f(-log 2) is equal to: (1) 3 (2) 5 (3) 9 (4) 7 |
(3) 9 | Solve the first-order differential equation using initial conditions and limit properties to find f(-log 2) = 9. |
| 8: Let P and Q be the points on the line (x + 3)/8 = (y − 4)/2 = (z + 1)/2 which are at a distance of 6 units from the point R(1, 2, 3). If the centroid of the triangle PQR is (α, β, γ), then α² + β² + γ² is: (1) 26 (2) 36 (3) 18 (4) 24 |
(3) 18 | Calculate coordinates of P and Q using the line equation. Compute the centroid and find α² + β² + γ² = 18. |
| 9: Consider a triangle ABC where A(1,2,3), B(-2,8,0), and C(3,6,7). If the angle bisector of BAC meets the line BC at D, then the length of the projection of the vector AD on the vector AC is: (1) 37/2√38 (2) √38/2 (3) 39/2√38 (4) √19 |
(1) 37/2√38 | Use the angle bisector theorem to locate D. Apply the formula for projection to find the length as 37/2√38. |
| 10: Let Sn denote the sum of the first n terms of an arithmetic progression. If S10 = 390 and the ratio of the tenth and fifth terms is 15:7, then S15 − S5 is equal to: (1) 800 (2) 890 (3) 790 (4) 690 |
(3) 790 | Using the sum formula and given ratio, compute S15 − S5 to find the result as 790. |
| 11: If ∫₀π/3 cos⁴x dx = aπ + b√3, where a and b are rational numbers, then 9a + 8b is equal to: (1) 2 (2) 1 (3) 3 (4) 3/2 |
(1) 2 | Using reduction formulas for powers of cosines, evaluate the integral to find a = 1/3 and b = 1/2. Substitute to get 9a + 8b = 2. |
| 12: If z is a complex number such that |z| ≥ 1, then the minimum value of |z + 1/(2(3 + 4i))| is: (1) 5/2 (2) 2 (3) 3 (4) 3/2 |
(1) 5/2 | Geometrically interpret the problem as finding the minimum distance of z from the point -1/(2(3 + 4i)). The calculated distance is 5/2. |
| 13: If the domain of the function f(x) = √(x² − 25)/(4 − x²) + log₁₀(x² + 2x − 15) is (-∞, α) ∪ [β,∞), then α² + β³ is: (1) 140 (2) 175 (3) 150 (4) 125 |
(3) 150 | Solve inequalities for the domain of square root and logarithm. α = -5, β = 5. Calculate α² + β³ = 150. |
| 14: Consider the relations R₁ and R₂ defined as aR₁b ⇔ a² + b² = 1 for all a, b ∈ R, and (a, b)R₂(c, d) ⇔ a + d = b + c for all (a, b), (c, d) ∈ N × N. Then: (1) Only R₁ is an equivalence relation (2) Only R₂ is an equivalence relation (3) Both R₁ and R₂ are equivalence relations (4) Neither R₁ nor R₂ is an equivalence relation |
(2) Only R₂ is an equivalence relation | Check reflexivity, symmetry, and transitivity. R₁ fails reflexivity and transitivity. R₂ satisfies all equivalence properties. |
| 15: If the mirror image of the point P(3, 4, 9) in the line (x − 1)/3 = (y + 1)/2 = (z − 2)/1 is (α, β, γ), then 14(α + β + γ) is: (1) 102 (2) 138 (3) 108 (4) 132 |
(3) 108 | Use the reflection formula for 3D geometry to find (α, β, γ). Substitute into 14(α + β + γ) to get 108. |
| 16: Let f(x) = { x−1, x is even; 2x, x is odd }, x ∈ N. If for some a ∈ N, f(f(f(a))) = 21, then: (1) 121 (2) 144 (3) 169 (4) 225 |
(2) 144 | Analyze the function step-by-step for even and odd cases. Solve for a such that f(f(f(a))) = 21 and compute the required limit. |
| 17: Let the system of equations x + 2y + 3z = 5, 2x + 3y + z = 9, 4x + 3y + λz = µ have an infinite number of solutions. Then λ + 2µ is equal to: (1) 28 (2) 17 (3) 22 (4) 15 |
(2) 17 | Set determinant of coefficient matrix to zero for consistency. Solve equations to find λ and µ. Calculate λ + 2µ = 17. |
| 18: Consider 10 observations x₁, x₂, ..., x₁₀ such that ∑(xᵢ−α) = 2 and ∑(xᵢ−β)² = 40, where α, β are positive integers. Let the mean and variance of the observations be 6/5 and 84/25 respectively. The ratio β/α is equal to: (1) 2 (2) 3/2 (3) 5/2 (4) 1 |
(1) 2 | Use mean and variance formulas along with the given conditions to solve for α and β. The ratio β/α = 2. |
| 19: Let Ajay not appear in the JEE exam with probability p = 2/7, while both Ajay and Vijay will appear with probability q = 1/5. Then the probability that Ajay will appear and Vijay will not appear is: (1) 9/35 (2) 18/35 (3) 24/35 (4) 3/35 |
(2) 18/35 | Using complementary probabilities and conditional probabilities, compute the required probability as 18/35. |
| 20: Let the locus of the midpoints of the chords of circle x² + (y−1)² = 1 drawn from the origin intersect the line x + y = 1 at P and Q. Then, the length of PQ is: (1) 1/√2 (2) √2 (3) 1/2 (4) 1 |
(1) 1/√2 | Find the locus of midpoints as a circle. Solve for intersections with the line x + y = 1. Use distance formula to find PQ = 1/√2. |
| 21: Three successive terms of a G.P. with common ratio r (r > 1) are the lengths of the sides of a triangle. If [r] denotes the greatest integer less than or equal to r, then 3[r] + ⌊-r⌋ is equal to: | 6 | Using the triangle inequality conditions for the sides of a triangle, determine r and calculate 3[r] + ⌊-r⌋ = 6. |
| 22: Let A = I₂ − MMᵀ, where M is a real matrix of order 2 × 1 such that MᵀM = I₁. If λ is a real number such that AX = λX holds for some non-zero real matrix X of order 2 × 1, then the sum of squares of all possible values of λ is equal to: | 2 | The matrix A is a projection matrix. Eigenvalues are 0 and 1. The sum of squares of all eigenvalues is 0² + 1² = 2. |
| 23: Let f : (0, ∞) → R and F(x) = ∫₀ˣ tf(t) dt. If F(x²) = x⁴ + x⁵, then ∑₁² f(r²) is equal to: | 219 | Differentiate F(x²) to find f(x²). Substitute r = 1, 2, ..., 12 to compute the sum of f(r²) = 219. |
| 24: If y = √((x + 1)(x² − √x)) / (x√x + x + √x) + 1/15(3cos²x − 5)cos³x, then 96y'(π/6) is equal to: | 105 | Differentiate y with respect to x. Substitute x = π/6 and simplify to find 96y'(π/6) = 105. |
| 25: Let a = î + αĵ + βk̂, α, β ∈ R. Let a vector b be such that the angle between a and b is π/4 and |b| = 6. If |a × b| = 3√2, then the value of (α² + β²)|a × b|² is equal to: | 90 | Using |a⃗ × b⃗| = |a⃗||b⃗|sinθ and |b⃗| = 6, solve for |a⃗|. Compute α² + β² and substitute to get (α² + β²)|a⃗ × b⃗|² = 90. |
| 26: The lines L₁, L₂, ..., L₂₀ are distinct. For n = 1, 2, 3, ..., 10, all the lines L₂ₙ₋₁ are parallel to each other, and all the lines L₂ₙ pass through a given point P. The maximum number of points of intersection of pairs of lines from the set {L₁, L₂, ..., L₂₀} is equal to: | 101 | The odd-numbered lines intersect each even-numbered line at a distinct point. Additionally, all even-numbered lines meet at P, resulting in a total of 101 points of intersection. |
| 27: Three points O(0, 0), P(a, a²), Q(−b, b²), where a > 0 and b > 0, are on the parabola y = x². Let S₁ be the area of the region bounded by the line PQ and the parabola, and S₂ be the area of the triangle OPQ. If the minimum value of S₁/S₂ is m/n, where gcd(m, n) = 1, then m + n is: | 7 | Using integration for S₁ and geometry for S₂, calculate the ratio S₁/S₂ and minimize it. The minimum value is 4/3, so m + n = 7. |
| 28: The sum of squares of all possible values of k, for which the area of the region bounded by the parabolas 2y² = kx and ky² = 2(y − x) is maximum, is equal to: | 8 | Solve for k by maximizing the bounded area using calculus. The sum of squares of the possible values of k is 8. |
| 29: If dx/dy = 1 + x − y² and x(1) = 1, then 5x(2) is equal to: | 5 | Solve the first-order differential equation with the initial condition x(1) = 1. Evaluate x(2) and calculate 5x(2) = 5. |
| 30: Let △ABC be an isosceles triangle where A = (−1, 0), AB = AC, and BC = 4. If the line BC intersects the line y = x + 3 at (α, β), then β⁴ is equal to: | 36 | Find the points B and C using symmetry and the triangle's properties. Solve for the intersection with y = x + 3. Calculate β⁴ = 36. |
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 Feb 1 Shift 2 Mathematics Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
|---|---|
| Aakash BYJUs | To be updated |
| 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 Feb 1 Shift 2 Mathematics Paper Analysis
JEE Main 2024 Feb 1 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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