JEE Main 2024 Jan 30 Shift 2 Mathematics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Jan 30 Shift 2 exam from 3 PM to 6 PM. The Mathematics question paper for JEE Main 2024 Jan 30 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 30 Shift 2 exam is available for download using the link below.
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JEE Main 2024 Jan 30 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 30 Jan Shift 2 2024 Mathematics Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 1: Consider the system of linear equations: x + y + z = 5, x + 2y + 2z = 9, x + 3y + λz = µ, where λ, µ ∈ R. Then, which of the following statements is NOT correct? (1) System has infinite number of solutions if λ = 1 and µ = 13 (2) System is inconsistent if λ = 1 and µ ≠ 13 (3) System is consistent if λ ≠ 1 and µ = 13 (4) System has unique solution if λ = 1 and µ ≠ 13 |
(4) System has unique solution if λ = 1 and µ ≠ 13 | Convert the system to matrix form and perform row reduction. For consistency, the determinant of the coefficient matrix must be non-zero. If λ = 1 and µ ≠ 13, the determinant vanishes, making the system inconsistent. |
| 2: For α, β ∈ (0, π/2), let 3 sin(α + β) = 2 sin(α − β), and a real number k be such that tanα = k tanβ. Then the value of k is equal to: (1) 2/3 (2) -5 (3) 3/2 (4) 5 |
(2) -5 | Using trigonometric identities, expand sin(α + β) and sin(α − β). Simplify and solve to find k = -5. |
| 3: Let A(10, 0) and B(0, β) be points on the line 5x + 7y = 50. Let P divide AB in the ratio 7:3. Let 3x−25 = 0 be a directrix of the ellipse x²/a² + y²/b² = 1, and focus S such that a perpendicular from S passes through P. Find the length of the latus rectum: (1) 25/3 (2) 32/9 (3) 25/9 (4) 32/5 |
(4) 32/5 | Substituting points and solving equations gives the ellipse parameters. Using LR = 2b²/a, calculate the latus rectum length as 32/5. |
| 4: Let a = i + αj + βk, where α, β ∈ R. Let b be a vector 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: (1) 90 (2) 75 (3) 95 (4) 85 |
(1) 90 |
We have |b| = 6 and the angle between a and b is θ = π/4. The magnitude of the cross product is given by |a × b| = |a||b|sinθ. |a × b| = 3√2, |b| = 6, and sin(π/4) = √2/2: We know |a × b|² = (3√2)² = 18. Thus, (α² + β²)|a × b|² = 5 × 18 = 90. Therefore, the correct answer is 90. |
| 5: Let f(x) = (x + 3)(x − 2)²(x + 1), x ∈ [-4, 4]. If M and m are the maximum and minimum values of f in [-4, 4], then the value of M - m is: (1) 600 (2) 392 (3) 608 (4) 108 |
(3) 608 | Find the critical points and evaluate f(x) at these points and boundaries. Maximum value M = 392, minimum value m = -216, so M - m = 608. |
| 6: Let a and b be distinct positive real numbers. The 11th term of a GP (first term a, third term b) equals the pth term of another GP (first term a, fifth term b). Find p: (1) 20 (2) 25 (3) 21 (4) 24 |
(3) 21 | Calculate the common ratios r₁ and r₂ for the respective GPs. Solving equations gives p = 21. |
| 7: If x² − y² + 2hxy + 2gx + 2fy + c = 0 is the locus of a point that moves such that it is always equidistant from the lines x + 2y + 7 = 0 and 2x − y + 8 = 0, then the value of g + c + h − f equals: (1) 14 (2) 6 (3) 8 (4) 29 |
(1) 14 | Since the point P (x, y) is equidistant from the lines x + 2y + 7 = 0 and 2x − y + 8 = 0, its locus is the angle bisector of these lines. Using the distance formula for equidistance and simplifying, we find g + c + h − f = 14. |
| 8: Let a and b be two vectors such that |b| = 1 and |b × a| = 2. Then |(b × a) − b|² is equal to: (1) 3 (2) 5 (3) 1 (4) 4 |
(2) 5 | Using the formula |u - v|² = |u|² + |v|² - 2(u · v), let u = (b × a) and v = b. We know:
= (2²) + (1²) - 2(0) = 4 + 1 = 5. |
| 9: Let y = f(x) be a thrice differentiable function in (−5, 5). Let the tangents to the curve y = f(x) at (1, f(1)) and (3, f(3)) make angles π/6 and π/4, respectively, with the positive x-axis. If 2∫₁¹/√3((f′(t))² + 1)f′′(t) dt = α + β√3, where α and β are integers, then the value of α + β equals: (1) -14 (2) 26 (3) -16 (4) 36 |
(2) 26 | From the tangents, f′(1) = 1/√3 and f′(3) = 1. Using the integral, solve to find α = 36 and β = -10, yielding α + β = 26. |
| 10: Let P be a point on the hyperbola H: x²/9 − y²/4 = 1, in the first quadrant such that the area of the triangle formed by P and the two foci of H is √13. Then, the square of the distance of P from the origin is: (1) 18 (2) 26 (3) 22 (4) 24 |
(3) 22 | The hyperbola parameters give a = 3, b = 2, and c = √13. Using parametric equations P = (3 sec θ, 2 tan θ) and the area condition, find θ = π/4. Substituting into the distance formula, x² + y² = 22. |
| 11: Bag A contains 3 white and 7 red balls, and bag B contains 3 white and 2 red balls. One bag is selected at random, and a ball is drawn. The probability of drawing the ball from bag A, given the ball drawn is white, is: (1) 1/4 (2) 1/9 (3) 1/3 (4) 3/10 |
(3) 1/3 | Using Bayes' theorem, calculate P(E1|E) where E1 is the event of selecting bag A and E is the event of drawing a white ball. P(E|E1) = 3/10, P(E|E2) = 3/5. Substituting into Bayes' formula, P(E1|E) = (3/10) / [(3/10) + (3/5)] = 1/3. |
| 12: Let f : R → R be defined by f(x) = ae²ˣ + beˣ + c. If f(0) = −1, f'(logₑ 2) = 21, and ∫[logₑ 4, 0] (f(x) − cx) dx = 39/2, then the value of |a + b + c| equals: (1) 16 (2) 10 (3) 12 (4) 8 |
(4) 8 | Using f(0) = −1, find a + b + c = −1. With f'(logₑ 2) = 21, solve 8a + 2b = 21. Evaluate the integral to find |a + b + c| = 8. |
| 13: Let L₁: r = (ˆi − ˆj + 2ˆk) + λ(ˆi − ˆj + 2ˆk), λ ∈ R, L₂: r = (ˆj − ˆk) + μ(3ˆi + ˆj + pˆk), μ ∈ R, and L₃: r = δ(ˆi + mˆj − ˆk), δ ∈ R be three lines such that L₁ is perpendicular to L₂, and L₃ is perpendicular to both L₁ and L₂. Then the point which lies on L₃ is: (1) (−1, 7, 4) (2) (−1, −7, 4) (3) (1, 7, 4) (4) (1, −7, 4) |
(1) (−1, 7, 4) | To find the point on L₃, solve the conditions for perpendicularity using dot products: L₁ · L₂ = 0 and L₃ · L₁ = L₃ · L₂ = 0. After simplification, the required point is (−1, 7, 4). |
| 14: Let a and b be real constants such that the function f defined by f(x) = { x² + 3x + a, x ≤ 1; bx + 2, x > 1 } is differentiable on R. Then, the value of ∫[2,−2] f(x) dx equals: (1) 15 (2) 19/6 (3) 21 (4) 17 |
(4) 17 | To ensure differentiability, equate f(1) and f'(1) for continuity and smoothness. Solve for a and b, and then compute the integral of f(x) in the given range to find the value as 17. |
| 15: Let f : R − {0} → R be a function satisfying f(x/y) = f(x)/f(y) for all x, y with f(y) ≠ 0. If f'(1) = 2024, then: (1) xf'(x) − 2024f(x) = 0 (2) xf(x) + 2024f(x) = 0 (3) f(x) + xf'(x) = 2024 (4) xf(x) − 2023f(x) = 0 |
(1) xf'(x) − 2024f(x) = 0 | Using the functional equation and differentiability, differentiate both sides to derive the relationship xf'(x) − 2024f(x) = 0. |
| 16: If z is a complex number, then the number of common roots of the equations z1985 + z100 + 1 = 0 and z2 + z + 1 = 0 is: (1) 1 (2) 2 (3) 0 (4) 3 |
(3) 0 | Solve z² + z + 1 = 0 to find the roots as ω and ω², where ω is a cube root of unity. Check if these roots satisfy z1985 + z100 + 1 = 0. Neither ω nor ω² satisfies the equation. Hence, there are no common roots. |
| 17: Suppose 2−p, p, 2−α, α are the coefficients of four consecutive terms in the expansion of (1 + x)n. Then the value of p² − α² + 6α + 2p equals: (1) 4 (2) 10 (3) 8 (4) 6 |
(2) 10 | Using binomial coefficients, derive p and α from Cr, Cr+1, Cr+2, and Cr+3. Simplify the expressions to find p² − α² + 6α + 2p = 10. |
| 18: If the domain of the function f(x) = logₑ(2x + 3 / 4x² + x − 3) + cos⁻¹(2x − 1 / x + 2) is (α, β], then the value of 5β − 4α is equal to: (1) 10 (2) 12 (3) 11 (4) 9 |
(2) 12 | Determine the domain for each component of f(x): the logarithmic term and the inverse cosine term. Intersect their domains to find α = 3/4 and β = 3. Compute 5β − 4α = 12. |
| 19: Let f : R → R be a function defined by f(x) = x / (1 + x⁴)1/4, and g(x) = f(f(f(x))). Then ∫[4√3, 8√3] x³g(x) dx equals: (1) 33 (2) 36 (3) 42 (4) 39 |
(4) 39 | Simplify g(x) iteratively using the function f(x). Apply substitution t = x⁴ and compute the integral using the limits. The result evaluates to 39. |
| 20: Let R = diag(sinθ, sin(θ + 2π/3), sin(θ + 4π/3)) be a diagonal 3×3 matrix. For a square matrix M, let trace(M) denote the sum of all the diagonal entries of M. Then among the statements: (I) Trace(R) = 0 (II) If trace(adj(adj(R))) = 0, then R has exactly one non-zero entry. Which of the following is true? (1) Both (I) and (II) are true (2) Neither (I) nor (II) is true (3) Only (II) is true (4) Only (I) is true |
(4) Only (I) is true | Trace(R) sums the diagonal entries: sinθ + sin(θ + 2π/3) + sin(θ + 4π/3), which equals 0. Statement (II) is false as trace(adj(adj(R))) = 0 does not imply one non-zero entry in R. |
| 21: Let Y = Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y − y = Y′(x)(X − x) and the coordinate axes, where (x, y) is any point on the curve, is always A = −y²/(2Y′(x)) + 1. If Y(1) = 1, then 12Y(2) equals: | 20 | Use the relationship given for the area and differentiate with respect to X. Solve the resulting differential equation with the condition Y(1) = 1 to find Y(X). Compute 12Y(2) = 20. |
| 22: Let a line passing through the point (−1, 2, 3) intersect the lines L₁: (x−1)/3 = (y−2)/2 = (z+1)/−2 and L₂: (x+2)/−3 = (y−2)/4 = (z−1)/−2 at points M(α, β, γ) and N(a, b, c), respectively. Then the value of (α + β + γ)² / (a + b + c) equals: | 196 | Find the parametric equations for the given lines and solve for the intersection points M and N. Calculate (α + β + γ) and (a + b + c) and evaluate the given expression to obtain 196. |
| 23: Consider two circles C₁: x² + y² = 25 and C₂: (x−α)² + y² = 16, where α ∈ (5, 9). Let the angle between the two radii (one to each circle) drawn from one of the intersection points of C₁ and C₂ be sin⁻¹(√63/8). If the length of the common chord of C₁ and C₂ is β, then the value of (αβ)² equals: | 1575 | Use the distance between centers and the radius lengths to find the equation of the common chord. Apply the given angle condition to calculate β. Evaluate (αβ)² to get 1575. |
| 24: Let α = Σ Cnk / (k + 1) and β = Σ (Cnk Cnk+1) / (k + 2). If 5α = 6, then n equals: | 10 | Simplify the summation expressions for α and β using binomial coefficient identities. Solve for n using the condition 5α = 6. The value of n is 10. |
| 25: Let Sn be the sum of the first n terms of an arithmetic progression 3, 7, 11, ... . If (n(n + 1)) / ΣSk < 42, then n equals: | 9 | Use the formula for the sum of an arithmetic progression to calculate Sn. Substitute into the inequality and solve for n to find the maximum integer satisfying the condition, which is 9. |
| 26: In an examination of Mathematics paper, there are 20 questions of equal marks, and the question paper is divided into three sections: A, B, and C. A student is required to attempt a total of 15 questions, taking at least 4 questions from each section. If section A has 8 questions, section B has 6 questions, and section C has 6 questions, then the total number of ways a student can select 15 questions is: | 11376 | Using the constraints xA + xB + xC = 15 with xA ≥ 4, xB ≥ 4, xC ≥ 4, transform variables to yA + yB + yC = 3. Compute the combinations using the stars and bars theorem and count selections from each section to get 11376 ways. |
| 27: The number of symmetric relations defined on the set {1, 2, 3, 4} which are not reflexive is: | 960 | Total symmetric relations = 2(n² + n)/2, reflexive symmetric relations = 2n(n − 1)/2. Subtract reflexive symmetric relations from total symmetric relations for n = 4. The result is 960. |
| 28: The number of real solutions of the equation x(x² + 3x) + |x−1| + 6|k−2| = 0 is: |
1 | Consider cases for |x − 1| and simplify the equation for each interval. Analyze the monotonicity of the cubic function and absolute terms. There is exactly one real solution. |
| 29: The area of the region enclosed by the parabola (y−2)² = x−1, the line x−2y+4 = 0, and the positive coordinate axes is: |
(1) 5 | Solve for the intersection points of the given parabola and line. Set up integrals for the enclosed area and compute using the limits of integration. The area evaluates to 5. |
| 30: The variance σ² of the data is: xi: 0, 1, 5, 6, 10, 12, 17 fi: 3, 2, 3, 2, 6, 3, 3 |
(1) 29.09 | Calculate the mean μ = Σ(fixi) / Σfi. Compute Σ(fixi²) and use the variance formula: σ² = [Σ(fixi²) / Σfi] − μ². The result is 29.09. |
Also Check:
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JEE Main 2024 Jan 30 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 Jan 30 Shift 2 Mathematics Paper Analysis
JEE Main 2024 Jan 30 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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