JEE Main 2024 8 April Shift 2 Mathematics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 8 April Shift 2 exam from 3 PM to 6 PM. The Mathematics question paper for JEE Main 2024 8 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 8 April Shift 2 exam is available for download using the link below.
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JEE Main 2024 8 April Shift 2 Mathematics Question Paper PDF Download
| JEE Main 2024 Mathematics Question Paper | JEE Main 2024 Mathematics Answer Key | JEE Main 2024 Mathematics Solutions |
|---|---|---|
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JEE Main 8 Apr Shift 2 2024 Mathematics Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 1. If the image of the point (-4, 5) in the line x + 2y = 2 lies on the circle (x+4)2 + (y-3)2 = r2, then r is equal to: (1) 1 (2) 2 (3) 75 (4) 3 |
(2) 2 | The reflection formula for the image of a point in a line is used. Substituting the point (-4, 5) into the circle equation gives r = 2. |
| 2. Let a⃗ = î + 2ĵ + 3k̂, b⃗ = 2î + 3ĵ - 5k̂, and c⃗ = 3î - ĵ + λk̂ be three vectors. Let r⃗ be a unit vector along b⃗ + c⃗. If r⃗ · a⃗ = 3, then 3λ is equal to: (1) 27 (2) 25 (3) 21 (4) 15 |
(2) 25 | Using vector addition and normalization, we find r⃗ · a⃗ = 3, leading to λ = 25/3. Hence, 3λ = 25. |
| 3. If α ≠ a, β ≠ b, γ ≠ c, and |α b c; a β c; a b γ| = 0, then a/(α-a) + b/(β-b) + c/(γ-c) is equal to: (1) 2 (2) 3 (3) 0 (4) 1 |
(3) 0 | Expanding the determinant and analyzing the conditions, the expression simplifies to 0 due to linear dependence. |
| 4. In an increasing geometric progression of positive terms, the sum of the second and sixth terms is 70/3 and the product of the third and fifth terms is 49. Then the sum of the 4th, 6th, and 8th terms is: (1) 96 (2) 78 (3) 91 (4) 84 |
(3) 91 | Solving the equations for the geometric progression gives the common ratio r = √3. Substituting into the formula, the sum of the 4th, 6th, and 8th terms is 91. |
| 5. The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to: (1) 175 (2) 181 (3) 177 (4) 179 |
(4) 179 | Using combinations and considering cases of repetition, the total ways to select five letters is 179. |
| 6. The sum of all possible values of θ ∈ [−π, 2π], for which (1 + i cos θ) / (1 − 2i cos θ) is purely imaginary, is equal to: (1) 2π (2) 3π (3) 5π (4) 4π |
(2) 3π | Using the condition for the complex number to be purely imaginary and simplifying the given equation, the sum of all possible values of θ is 3π. |
| 7. If the system of equations x + 4y − z = λ, 7x + 9y + µz = −3, 5x + y + 2z = −1 has infinitely many solutions, then 2µ + 3λ is equal to: (1) 2 (2) −3 (3) 3 (4) −2 |
(2) −3 | Using the determinant method and applying the condition for infinitely many solutions, we find that 2µ + 3λ = −3. |
| 8. The shortest distance between the lines (x− λ)/2 = (y − 4)/3 = (z − 3)/4 and (x− 2)/4 = (y − 4)/6 = (z − 7)/8 is 13√29. Then a value of λ is: (1) −13/25 (2) 13/25 (3) 1 (4) −1 |
(3) 1 | Using the formula for the shortest distance between two skew lines and applying the given values, we find that λ = 1. |
| 9. If the value of (3 cos 36° + 5 sin 18°) / (5 cos 36° − 3 sin 18°) = a / √(5 − b/c), where a, b, c are natural numbers and gcd(a, c) = 1, then a + b + c is equal to: (1) 50 (2) 40 (3) 52 (4) 54 |
(3) 52 | By simplifying the given expression using trigonometric identities and rationalizing, we find that a + b + c = 52. |
| 10. Let y = y(x) be the solution curve of the differential equation sec y dy/dx + 2x sin y = x³ cos y, with the condition y(1) = 0. Then y(√3) is equal to: (1) π/3 (2) π/6 (3) π/4 (4) π/12 |
(3) π/4 | Solving the differential equation using the method of separation of variables and applying the initial condition, we get y(√3) = π/4. |
| 11. The area of the region in the first quadrant inside the circle x2 + y2 = 8 and outside the parabola y2 = 2x is equal to: (1) π/2 − 1/3 (2) π − 2/3 (3) π/2 − 2/3 (4) π − 1/3 |
(2) π − 2/3 | To find the area, first calculate the area inside the circle in the first quadrant using the equation of the circle. Then subtract the area under the parabola in the first quadrant. The area is given by the difference: π − 2/3. |
| 12. If the line segment joining the points (5, 2) and (2, a) subtends an angle π/4 at the origin, then the absolute value of the product of all possible values of a is: (1) 6 (2) 8 (3) 2 (4) 4 |
(4) 4 | The angle subtended by the line segment at the origin is given by the formula for the angle between two vectors. After solving the equation for the angle, we find the product of all possible values of a to be 4. |
| 13. Let ⃗a = 4ˆi − ˆj + ˆk, ⃗b = 11ˆi − ˆj + ˆk, and ⃗c be a vector such that (⃗a + ⃗b) × ⃗c = ⃗c × (−2⃗a + 3⃗b). If (2⃗a + 3⃗b) · ⃗c = 1670, then |⃗c|² is equal to: (1) 1627 (2) 1618 (3) 1600 (4) 1609 |
(2) 1618 | By using the given cross product equation and applying vector algebra, we determine that |⃗c|² = 1618. |
| 14. If the function f(x) = 2x³ − 9ax² + 12a²x + 1, a > 0, has a local maximum at x = α and a local minimum at x = α², then α and α² are the roots of the equation: (1) x² − 6x + 8 = 0 (2) 8x² + 6x − 8 = 0 (3) 8x² − 6x + 1 = 0 (4) x² + 6x + 8 = 0 |
(1) x² − 6x + 8 = 0 | By differentiating the given function and applying the conditions for local maxima and minima, we find that α and α² satisfy the equation x² − 6x + 8 = 0. |
| 15. There are three bags X, Y, and Z. Bag X contains 5 one-rupee coins and 4 five-rupee coins; Bag Y contains 4 one-rupee coins and 5 five-rupee coins; and Bag Z contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability that it came from bag Y is: (1) 1/3 (2) 1/2 (3) 1/4 (4) 5/12 |
(1) 1/3 | Using Bayes' Theorem, the probability that the coin came from Bag Y, given that it is a one-rupee coin, is calculated as follows: Let P(X), P(Y), and P(Z) be the probabilities of selecting Bag X, Y, and Z, respectively, and P(1|X), P(1|Y), and P(1|Z) be the probabilities of drawing a one-rupee coin from each bag. Applying Bayes' Theorem, we find that the probability is 1/3. |
| 16. Let ∫ loge 4 / sqrt(e^x - 1) dx = π / 6. Then eα and e−α are the roots of the equation: (1) 2x² − 5x + 2 = 0 (2) x² − 2x − 8 = 0 (3) 2x² − 5x − 2 = 0 (4) x² + 2x − 8 = 0 |
(1) 2x² − 5x + 2 = 0 | By solving the given integral and using appropriate transformations, we find that eα and e−α satisfy the quadratic equation 2x² − 5x + 2 = 0. |
| 17. Let f(x) = { −a if −a ≤ x ≤ 0 x + a if 0 < x ≤ a and g(x) = f(|x|) − |f(x)|². Then the function g : [−a, a] → [−a, a] is: (1) neither one-one nor onto (2) both one-one and onto (3) one-one (4) onto |
(1) neither one-one nor onto | The function g(x) is neither one-one (injective) nor onto (surjective) because it takes only two distinct values (−a and 0) over the entire domain [−a, a]. Thus, it does not cover the entire range and fails the criteria for being one-one. |
| 18. Let A = {2, 3, 6, 8, 9, 11} and B = {1, 4, 5, 10, 15}. Let R be a relation on A×B defined by (a, b)R(c, d) if and only if 3ad − 7bc is an even integer. Then the relation R is: (1) reflexive but not symmetric. (2) transitive but not symmetric. (3) reflexive and symmetric but not transitive. (4) an equivalence relation. |
(3) reflexive and symmetric but not transitive. | The relation R is reflexive because for any (a, b) in A×B, 3ab − 7ab is even. It is symmetric because if (a, b)R(c, d), then (c, d)R(a, b). However, it is not transitive as it fails to satisfy transitivity for certain combinations of elements. |
| 19. For a, b > 0, let f(x) = { tan((a + 1)x) + b tan x, x < 0 x³, x = 0 √(ax + b) / (2x² - √(ax) b√(ax√x)), x > 0 Then f(x) is continuous at x = 0. Then b / a is equal to: (1) 5 (2) 4 (3) 8 (4) 6 |
(4) 6 | For f(x) to be continuous at x = 0, the left-hand and right-hand limits must match the value of the function at x = 0. By solving the given limits and applying the continuity condition, we find that b/a = 6. |
| 20. If the term independent of x in the expansion of \( (\sqrt{a}x^2 + \frac{1}{2x^3})^{10} \) is 105, then \( a^2 \) is equal to: (1) 4 (2) 9 (3) 6 (4) 2 |
(1) 4 | By calculating the general term in the binomial expansion and solving for the term independent of \(x\), we find \( a^2 = 4 \). The solution involves equating powers and solving a cubic equation. |
| 21. Let A be the region enclosed by the parabola \( y^2 = 2x \) and the line \( x = 24 \). The maximum area of the rectangle inscribed in the region A is: | (1) 128 | The rectangle's area is maximized by solving an optimization problem using calculus, considering the boundary conditions of the parabola and the line. |
| 22. If α = limx→0+ (e√tan x − e√x) / (√tan x − √x) and β = limx→0 (1+sin x)1/2cot x are the roots of the quadratic equation ax² + bx − √e = 0, then 12 loge(a + b) is equal to: | (4) 6 | To solve for α and β, we evaluate the limits separately. α = 1 and β = e1/2. Substituting these into the quadratic equation ax² + bx − √e = 0, we find that a + b = −(1 + e1/2). Therefore, the value of 12 loge(a + b) is 6. |
| 23. Let S be the focus of the hyperbola x²/3 − y²/5 = 1, on the positive x-axis. Let C be the circle with its center at A(√6, √5) and passing through the point S. If O is the origin and SAB is a diameter of C, then the square of the area of the triangle OSB is equal to: | (4) 4 | First, calculate the coordinates of the focus S of the hyperbola. Using the geometry of the circle and the triangle formed by the points O, S, and B, we use the area formula for a triangle with known vertices. The square of the area is 4. |
| 24. Let P(α, β, γ) be the image of the point Q(1, 6, 4) in the line (x − 1) / 1 = (y − 1) / 2 = (z − 2) / 3. Then 2α + β + γ is equal to: (1) 5 (2) 6 (3) 10 (4) 11 |
(4) 11 | The image of point Q(1, 6, 4) with respect to the given line is found using the reflection formula in 3D space. After the calculation, we find that \( 2\alpha + \beta + \gamma = 11 \). |
| 24. Let P(α, β, γ) be the image of the point Q(1, 6, 4) in the line (x − 1) / 1 = (y − 1) / 2 = (z − 2) / 3. Then 2α + β + γ is equal to: | (4) 11 | The image of point Q(1, 6, 4) with respect to the given line is found using the reflection formula in 3D space. After the calculation, we find that \( 2\alpha + \beta + \gamma = 11 \). |
| 25. An arithmetic progression is written in the following way: The sum of all the terms of the 10th row is: | (2) 1505 | The sum of an arithmetic progression can be found using the formula \( S_n = \frac{n}{2} [2a + (n-1)d] \). After substituting the correct values, the sum of all terms in the 10th row is 1505. |
| 26. The number of distinct real roots of the equation \( |x+1||x+3| - 4|x+2| + 5 = 0 \) is: | (2) 2 | Solving the equation step by step, we analyze the behavior of the absolute values and find that there are 2 distinct real roots. |
| 27. Let a ray of light passing through the point (3, 10) reflect on the line 2x + y = 6 and the reflected ray pass through the point (7, 2). If the equation of the incident ray is \( ax + by + 1 = 0 \), then \( a^2 + b^2 + 3ab \) is equal to: | (4) 1 | Using the geometry of light reflection and the equation of the line of reflection, we calculate the values of \(a\) and \(b\), and find that \( a^2 + b^2 + 3ab = 1 \). |
| 28. Let \(a, b, c \in \mathbb{N}\) and \(a < b < c\). Let the mean, the mean deviation about the mean, and the variance of the 5 observations 9, 25, a, b, c be 18, 4, and 136/5, respectively. Then \( 2a + b - c \) is equal to: | (1) 33 | Using the given statistics (mean, mean deviation, and variance), we solve for \(a\), \(b\), and \(c\), and find that \( 2a + b - c = 33 \). |
| 29. Let α|x| = |y|exy − β, α, β ∈ N be the solution of the differential equation xdy − ydx + xy(xdy + ydx) = 0, with y(1) = 2. Then α + β is equal to: | (3) 4 | By solving the given differential equation using the method of integration and applying the initial condition y(1) = 2, we find that α + β = 4. |
| 30. If ∫1√5 (x−1)4 (x+3)6 dx = A(αx−1)/(βx+3) + B + C, where C is the constant of integration, then the value of α + β + 20AB is: | (2) 7 | After performing the integration and simplifying, we find that α + β + 20AB = 7. |
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 8 April Shift 2 Mathematics Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
|---|---|
| Aakash BYJUs | Download PDF |
| Vedantu | Download PDF |
| Reliable Institute | To be updated |
| Resonance | To be updated |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 2024 8 April Shift 2 Mathematics Paper Analysis
JEE Main 2024 8 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:
- JEE Main 2024 question paper pattern and marking scheme
- Most important chapters in JEE Mains 2024, Check chapter wise weightage here
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