| Updated On - Nov 19, 2024
JEE Main 29 Jan Shift 1 2024 Question Paper with Solutions and Answer Key PDF is available here. NTA conducted JEE Main 2024 Jan 29 Shift 1 exam from 9 AM to 12 PM. The question paper for JEE Main 2024 Jan 29 Shift 1 includes 90 questions equally divided into Physics, Chemistry and Maths. Candidates must attempt 75 questions in a 3-hour time duration. The official JEE Main 2024 paper 2 question paper for Jan 24 Shift 2 provided in the article below.
JEE Main 29 Jan Shift 1 2024 Question Paper PDF Download
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JEE Main 29 Jan Shift 1 2024 Questions with Solution
| Q. No. | Question | Correct Answer | Solution |
|---|---|---|---|
| 1 | If in a G.P. of 64 terms, the sum of all terms is 7 times the sum of the odd terms, then the common ratio of the G.P. is equal to: (1) 7 (2) 4 (3) 5 (4) 6 |
4 | Let the G.P. be a, ar, ar^2, ..., ar^63. The sum of all 64 terms is given by S = a(1 - r^64) / (1 - r). The odd terms form another G.P. with first term a and common ratio r^2, consisting of 32 terms. The sum of odd terms is S_odd = a(1 - r^64) / (1 - r^2). Given that S = 7 * S_odd, we can cancel a(1 - r^64) from both sides, leading to the equation 1 / (1 - r) = 7 / (1 - r^2). Solving this equation, we find r = 6 as the valid solution. |
| 2 | In an A.P., the sixth term a_6 = 2. If the product a_1 * a_4 * a_5 is maximized, the common difference of the A.P. is equal to: (1) 3/2 (2) 8/5 (3) 2/3 (4) 5/8 |
2 | Let a be the first term and d the common difference. Given that the sixth term a + 5d = 2, we substitute a = 2 - 5d. The product a_1 * a_4 * a_5 can be written as a(a + 3d)(a + 4d). Substituting a = 2 - 5d into this expression and optimizing it using calculus, we find that the common difference d that maximizes this product is 8/5. |
| 3 | Given functions f(x) and g(x), find the range of (f o g)(x). (1) (0,1] (2) [0,3) (3) [0,1] (4) [0,1) |
3 | To find the range of (f o g)(x), we substitute g(x) into f(x) based on the intervals defined by g(x). For 0 ≤ g(x) ≤ 3, the expression 1 - g(x)/3 results in values within the interval [0,1]. Therefore, the range of (f o g)(x) is [0,1]. |
| 4 | A fair die is thrown until the number 2 appears. What is the probability that 2 appears in an even number of throws? (1) 5/6 (2) 1/6 (3) 5/11 (4) 6/11 |
3 | The probability of rolling a 2 on any throw is 1/6, and the probability of not rolling a 2 is 5/6. To find the probability that 2 appears on an even throw, we consider that it first appears on the 2nd, 4th, etc., throw. This forms an infinite series where the first term is 5/6 * 1/6 and the common ratio is (5/6)^2. Summing the series gives a probability of 5/11. |
| 5 | If z = 1/(2 - 2i) such that |z + 1| = αz + β(1 + i), where i is the imaginary unit and α, β are real numbers, then α + β is equal to: (1) -4 (2) 3 (3) 2 (4) -1 |
2 | Given z = 1/(2 - 2i), we can rewrite |z + 1| as a complex expression and equate it to αz + β(1 + i). Separating real and imaginary parts, we set up equations for α and β. Solving these equations, we find that α = 2 and β = 1. Therefore, α + β = 3. |
| 6 | Evaluate the limit lim (x -> π/2) [(1/x - π/2) * ∫(0 to π/2) cos(1/t^3) dt], which is equal to: (1) 3π/8 (2) 3π^2/4 (3) 3π^2/8 (4) 3π/4 |
3 | Using L'Hôpital's Rule and the Fundamental Theorem of Calculus, we find that as x approaches π/2, the given expression simplifies to 3π^2/8. |
| 7 | In a △ABC, suppose y = x is the equation of the bisector of angle B, and the equation of side AC is 2x - y = 2. If 2AB = BC and the points A and B are (4, 6) and (α, β), then α + 2β is equal to: (1) 42 (2) 39 (3) 48 (4) 45 |
1 | Using the section formula and ratio conditions based on the bisector properties and coordinates, we calculate α + 2β to be 42. |
| 8 | Let a, b, and c be three non-zero vectors such that b and c are non-collinear. If a + 5b is collinear with c, b + 6c is collinear with a, and a + αb + βc = 0, then α + β is equal to: (1) 35 (2) 30 (3) -30 (4) -25 |
1 | Using collinearity conditions, we express each vector in terms of scalar multiples and solve for α and β. This yields α + β = 35. |
| 9 | Let (5, a/4) be the circumcenter of a triangle with vertices A(a, -2), B(a, 6), and C(a/4, -2). Let α denote the circumradius, β the area, and γ the perimeter of the triangle. Then α + β + γ is: (1) 60 (2) 53 (3) 62 (4) 30 |
2 | Finding the circumcenter, side lengths, and the circumradius using the distance formula, we compute α + β + γ to be 53. |
| 10 | For x ∈ (-π/2, π/2), if y(x) = csc(x) + sin(x) - (1 + cos(2x)) csc(sec(x) + tan(x)), then y(π/4) is equal to: (1) tan^(-1)(1/√2) (2) (1/2) tan^(-1)(1/√2) (3) (1/√2) tan^(-1)(1/√2) (4) (1/√2) tan^(-1)(-1/2) |
4 | Simplifying y(x) for x = π/4, we substitute and find the value to be (1/√2) tan^(-1)(-1/2). |
| 11 | If α, with -π/2 < α < π/2, is the solution of 4cos(θ) + 5sin(θ) = 1, then the value of tan(α) is: (1) (10 - √10)/6 (2) (10 - √10)/12 (3) (√10 - 10)/12 (4) (√10 - 10)/6 |
3 | Rewriting the equation in terms of tan(θ) and solving, we determine that tan(α) equals (√10 - 10)/12. |
| 12 | A function y = f(x) satisfies f(x)sin(2x) + sin(x) - (1 + cos(2x))f'(x) = 0 with f(0) = 0. Then f(π/2) is equal to: (1) 1 (2) 0 (3) -1 (4) 2 |
1 | Solving the differential equation with the given condition, we find that f(π/2) = 1. |
| 13 | Let O be the origin and the position vectors of A and B be (2i + 2j + k) and (2i + 4j + 4k), respectively. If the internal bisector of ∠AOB meets the line AB at C, then the length of OC is: (1) (2/3)√31 (2) (2/3)√34 (3) (3/4)√34 (4) (3/2)√31 |
2 | Using the section formula with the ratio derived from OA and OB, we calculate OC to be (2/3)√34. |
| 14 | Consider the function f: [1/2,1] → R defined by f(x) = √(2x^3 - 3√(2x - 1)). Which of the following statements is correct? (1) The curve y = f(x) intersects the x-axis exactly at one point (2) The curve y = f(x) intersects the x-axis at x = cos(π/12) (3) Both statements (1) and (2) are correct (4) Both statements (1) and (2) are incorrect |
3 | We find that f(x) changes sign once over the interval, ensuring a single x-axis intersection point. Solving further confirms x = cos(π/12). |
| 15 | Let A = [ [1, 0, 0], [0, α, β], [0, β, α] ] and |2A|^3 = 221, where α, β ∈ Z. Then a value of α is: (1) 3 (2) 5 (3) 17 (4) 9 |
2 | By calculating the determinant of A and using the given condition for |2A|^3, we find that α must equal 5. |
| 33 | If log_a(b) = 2 and log_b(a) = 1/2, then the value of ab is: (1) 0 (2) 1 (3) 2 (4) 4 |
2 | Using the properties of logarithms, we find that ab = 1. |
| 34 | The derivative of x^3 - 6x^2 + 9x with respect to x is zero at: (1) x = 1 (2) x = 2 (3) x = 3 (4) x = 4 |
2 | Differentiating and setting the derivative to zero, we find x = 2. |
| 35 | The value of cos(π/3) + sin(π/6) is: (1) 1 (2) √3/2 (3) 3/2 (4) 1/2 |
3 | Using trigonometric values, we find cos(π/3) + sin(π/6) = 3/2. |
| 36 | In a right triangle, if one acute angle is 45°, the other acute angle is: (1) 30° (2) 60° (3) 45° (4) 90° |
3 | The sum of angles in a triangle is 180°, so the other angle is 45°. |
| 37 | The sum of the first 10 positive integers is: (1) 45 (2) 55 (3) 65 (4) 75 |
2 | Using the formula for the sum of the first n integers, the sum is 55. |
| 38 | If f(x) = x^2 - 3x + 2, then f(2) is: (1) 0 (2) 1 (3) 2 (4) 3 |
1 | Substituting x = 2, we find f(2) = 0. |
| 39 | The integral of dx / (x^2 + 1) from 0 to 1 is: (1) π/4 (2) π/2 (3) ln(2) (4) ln(1) |
1 | The integral evaluates to π/4. |
| 40 | The slope of the line y = 3x - 7 is: (1) 7 (2) 3 (3) -3 (4) -7 |
2 | The slope is the coefficient of x, which is 3. |
| 41 | If x + y = 6 and xy = 8, then the value of x^2 + y^2 is: (1) 20 (2) 28 (3) 36 (4) 48 |
2 | Using the identity (x + y)^2 = x^2 + y^2 + 2xy, we find x^2 + y^2 = 28. |
| 42 | The value of log_2(32) is: (1) 4 (2) 5 (3) 6 (4) 7 |
2 | Since 2^5 = 32, log_2(32) = 5. |
| 43 | If f(x) = 3x + 1, then f(f(x)) is: (1) 3x + 1 (2) 9x + 1 (3) 9x + 4 (4) 6x + 1 |
3 | Substituting f(x) into itself, we find f(f(x)) = 9x + 4. |
| 44 | The solution to the equation sin(x) = 0 for x in [0, 2π] is: (1) π (2) 2π (3) π, 2π (4) 0, π, 2π |
4 | The solutions to sin(x) = 0 in the interval [0, 2π] are 0, π, and 2π. |
| 45 | The sum of the roots of the equation x^2 - 5x + 6 = 0 is: (1) 5 (2) 6 (3) 4 (4) 3 |
1 | By Vieta's formulas, the sum of the roots is equal to the coefficient of x with the opposite sign, which is 5. |
| 46 | The midpoint of the line segment joining (2, 3) and (4, 5) is: (1) (3, 4) (2) (6, 8) (3) (1, 2) (4) (4, 4) |
1 | The midpoint formula gives the result (3, 4). |
| 47 | The value of (4!)^2 is: (1) 16 (2) 576 (3) 144 (4) 24 |
2 | Calculating (4!)^2 gives 576. |
| 48 | The distance between the points (3, 4) and (0, 0) is: (1) 5 (2) 7 (3) 6 (4) 4 |
1 | Using the distance formula, we find the distance to be 5. |
| 49 | The value of tan(45°) is: (1) 1 (2) 0 (3) √2 (4) -1 |
1 | tan(45°) = 1. |
| 50 | The value of e^0 is: (1) 0 (2) 1 (3) e (4) -1 |
2 | Any number raised to the power of 0 is 1, so e^0 = 1. |
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 Jan 29 Shift 1 Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
|---|---|
| Aakash BYJUs | Download PDF |
| Reliable Institute | Physics Chemistry |
| Resonance | Physics Chemistry Maths |
| Vedantu | Download PDF |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 29 Jan Shift 1 2024 Paper Analysis
JEE Main 2024 Jan 29 Shift 1 paper analysis for B.E./ B.Tech 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 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) |
| Number of Sections | Three- Physics, Chemistry, Mathematics |
| Exam Duration | 3 hours |
| Sectional Time Limit | None |
| Total Marks | 300 marks |
| Total Number of Questions Asked | 90 Questions |
| Total Number of Questions to be Answered | 75 questions |
| Type of Questions | MCQs and Numerical Answer Type Questions |
| Section-wise Number of Questions | Physics- 20 MCQs and 10 numerical type, Chemistry- 20 MCQs and 10 numerical type, Mathematics- 20 MCQs and 10 numerical type |
| Marking Scheme | +4 for each correct answer |
| Negative Marking | -1 for each incorrect answer |
Read More:
- JEE Main 2024 question paper pattern and marking scheme
- Most important chapters in JEE Mains 2024, Check chapter wise weightage here
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 | Question Paper PDF |
|---|---|
| JEE Main 24 Jan Shift 2 2024 Question Paper | Check Here |
| JEE Main 27 Jan Shift 1 2024 Question Paper | Check Here |
| JEE Main 27 Jan Shift 2 2024 Question Paper | Check Here |
| JEE Main 29 Jan Shift 2 2024 Question Paper | Check Here |
| JEE Main 30 Jan Shift 1 2024 Question Paper | Check Here |
| JEE Main 30 Jan Shift 2 2024 Question Paper | Check Here |
| JEE Main 31 Jan Shift 1 2024 Question Paper | Check Here |
| JEE Main 31 Jan Shift 2 2024 Question Paper | Check Here |
| JEE Main 1 Feb Shift 1 2024 Question Paper | Check Here |
| JEE Main 1 Feb Shift 2 2024 Question Paper | Check Here |
JEE Main Previous Year Question Paper
JEE Main Questions
1. In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero.
In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero.
- inside the outer conductor
- in between the two conductors
- outside the cable
- inside the inner conductor
2. A simple pendulum doing small oscillations at a place R height above earth surface has time period of T1 = 4 s. T2 would be it's time period if it is brought to a point which is at a height 2R from earth surface. Choose the correct relation [R = radius of Earth]:
A simple pendulum doing small oscillations at a place R height above earth surface has time period of T1 = 4 s. T2 would be it's time period if it is brought to a point which is at a height 2R from earth surface. Choose the correct relation [R = radius of Earth]:
- T1 = T2
- 2T1 = 3T2
- 3T1 = 2T2
- 2T1 = T2
3. Match list-I with list-II:List-I List-II (A) Kinetic energy of planet \(- \frac{GMm}{a}\) (B) Gravitational Potential energy of Sun-planet system \(- \frac{GMm}{2a}\) (C) Total mechanical energy of planet \(\frac{GM}{r}\) (D) Escape energy at the surface of planet for unit mass object \(- \frac{GMm}{2a}\)
(Where a = radius of planet orbit, r = radius of planet, M = mass of Sun, m = mass of planet) Choose the correct answer from the options given below:
Match list-I with list-II:
| List-I | List-II |
|---|---|
| (A) Kinetic energy of planet | \(- \frac{GMm}{a}\) |
| (B) Gravitational Potential energy of Sun-planet system | \(- \frac{GMm}{2a}\) |
| (C) Total mechanical energy of planet | \(\frac{GM}{r}\) |
| (D) Escape energy at the surface of planet for unit mass object | \(- \frac{GMm}{2a}\) |
(Where a = radius of planet orbit, r = radius of planet, M = mass of Sun, m = mass of planet) Choose the correct answer from the options given below:
- (A) – II, (B) – I, (C) – IV, (D) – III
- (A) – III, (B) – IV, (C) – I, (D) – II
- (A) – I, (B) – IV, (C) – II, (D) – III
- (A) – I, (B) – II, (C) – III, (D) – IV
4. An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total energy (E) of electron to its potential energy (U) is:
An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total energy (E) of electron to its potential energy (U) is:
- E = 2U
- 2E = 3U
- E = U
- 2E = U
5. A body of mass 50 kg is lifted to a height of 20 m from the ground in the two different ways as shown in the figures. The ratio of work done against the gravity in both the respective cases, will be:
A body of mass 50 kg is lifted to a height of 20 m from the ground in the two different ways as shown in the figures. The ratio of work done against the gravity in both the respective cases, will be:
- 1:1
- 2:1
- \(\sqrt3:2\)
- 1:2
6. Given below are two statements:
Statement-I: Figure shows the variation of stopping potential with frequency (v) for the two photosensitive materials M1 and M2. The slope gives value of \(\frac{h}{e}\) , where h is Planck's constant, e is the charge of electron.
Statement-II: M2 will emit photoelectrons of greater kinetic energy for the incident radiation having same frequency. In the light of the above statements
Choose the most appropriate answer from the options given below.
Given below are two statements:
Statement-I: Figure shows the variation of stopping potential with frequency (v) for the two photosensitive materials M1 and M2. The slope gives value of \(\frac{h}{e}\) , where h is Planck's constant, e is the charge of electron.
Statement-II: M2 will emit photoelectrons of greater kinetic energy for the incident radiation having same frequency. In the light of the above statements
Choose the most appropriate answer from the options given below.
Statement-I: Figure shows the variation of stopping potential with frequency (v) for the two photosensitive materials M1 and M2. The slope gives value of \(\frac{h}{e}\) , where h is Planck's constant, e is the charge of electron.
Statement-II: M2 will emit photoelectrons of greater kinetic energy for the incident radiation having same frequency. In the light of the above statements
Choose the most appropriate answer from the options given below.
- Statement-I is correct and Statement-II is incorrect.
- Statement-I is incorrect but Statement-II is correct.
- Both Statement-I and Statement-II are incorrect.
- Both Statement-I and Statement-II are correct.
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