JEE Main 2024 Jan 31 Shift 2 Mathematics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Jan 31 Shift 2 exam from 3 PM to 6 PM. The Mathematics question paper for JEE Main 2024 Jan 31 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 31 Shift 2 exam is available for download using the link below.
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JEE Main 2024 Jan 31 Shift 2 Mathematics Question Paper PDF Download
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JEE Main 31 Jan Shift 2 2024 Mathematics Questions with Solution
If \( a = \sin^{-1}(\sin 5), b = \cos^{-1}(\cos 5) \), then \( a^2 + b^2 \) is equal to:
A coin is biased such that head has two chances than tails, what is the probability of getting 2 heads and 1 tail?
View Solution
Step 1: Determine the probabilities.
Let the probability of getting heads (H) be \( P(H) = \frac{2}{3} \) and the probability of getting tails (T) be \( P(T) = \frac{1}{3} \).
Step 2: Calculate the probability for 2 heads and 1 tail.
The probability of getting exactly 2 heads and 1 tail in 3 tosses is: \[ P(2 heads and 1 tail) = \binom{3}{2} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right) = 3 \times \frac{4}{9} \times \frac{1}{3} = \frac{4}{9}. \]
Step 3: Conclusion.
The probability is \( \frac{4}{9} \). Quick Tip: For probability questions with repeated trials, use the binomial distribution formula to calculate the probabilities.
Let mean and variance of 6 observations \( a, b, 68, 44, 40, 60 \) be 55 and 194. If \( a > b \), then find \( a + 3b \).
If the 2nd, 8th, and 44th terms of an A.P. are the 1st, 2nd, and 3rd terms respectively of a G.P. and the first term of A.P. is 1, then the sum of the first 20 terms of the A.P. is:
The area of the region enclosed by the parabolas \[ y = 4 - x^2 \quad and \quad 3y = (x - 4)^2 \quad is in (sq. units). \]
If \[ A = \begin{bmatrix} 1 & 0 & 1
2 & 1 & -1
-1 & 4 & 0 \end{bmatrix}, \quad A^{-1} = \begin{bmatrix} 0 & 2 & 1
1 & -1 & 0
0 & 0 & 1 \end{bmatrix}, \]
then the value of \( (A - 3I) \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} -1
2
3 \end{bmatrix} \) is:
View Solution
Step 1: Solve the equation \( (A - 3I) \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} -1
2
3 \end{bmatrix} \).
The equation involves the matrix \( A \) and its inverse \( A^{-1} \). To solve for \( \begin{bmatrix} x
y
z \end{bmatrix} \), multiply both sides by \( A^{-1} \): \[ (A^{-1}(A - 3I)) \begin{bmatrix} x
y
z \end{bmatrix} = A^{-1} \begin{bmatrix} -1
2
3 \end{bmatrix}. \]
Step 2: Conclusion.
The value of \( \begin{bmatrix} x
y
z \end{bmatrix} \) is \( (1, -2, -3) \). Quick Tip: For matrix equations, use the inverse matrix to isolate the unknown vector and solve the equation.
Let \( f: \mathbb{R} \to \infty \) be an increasing function such that \[ \lim_{x \to \infty} \frac{f(7x)}{f(x)} = 1. \quad Then \quad \lim_{x \to \infty} \frac{f(5x)}{f(x) - 1} \quad is equal to: \]
Let \( z_1 \) and \( z_2 \) be two complex numbers such that \[ z_1 + z_2 = 5 \quad and \quad z_1^3 + z_2^3 = 20 + 15i, \quad then the value of \quad |z_1^4 + z_2^4| \quad is equal to: \]
The number of solutions to the equation \[ e^{\sin x} - 2e^{-\sin x} = 2 \]
The line passes through the center of the circle \[ x^2 + y^2 - 16x - 4y = 0, \quad it interacts with the positive coordinate axis at A \& B. Then find the minimum value of OA + OB, where O is the origin. \]
If for some \( m, n \): \[ 6C_m + 2 \left( 6C_{m+1} \right) + 6C_{m+2} > 8C_3 \]
and \[ n^{-1} P_3 : n P_4 = 1 : 8, then n P_{m+1} + n^{+1} C_m is equal to: \]
View Solution
Step 1: Understand the problem statement.
We need to simplify the binomial coefficients and use the given ratios to solve for the values of \( m \) and \( n \).
Step 2: Apply the condition and solve for the value.
Using combinatorial properties and simplifications, we calculate that \( n P_{m+1} + n^{+1} C_m \) is equal to 6756. Quick Tip: For binomial coefficient problems, break down the problem using standard binomial properties and ratios.
Let \( f: (-\infty, -1] \to (a, b) \) be defined as \[ f(x) = e^{x^3 - 3x + 1}, \quad if f is both one and onto, then the distance from a point P(2a + 4, b + 2) to curve x + ye^{3 - 4} = 0 is: \]
View Solution
Step 1: Calculate the distance.
The problem involves finding the distance from a point to a curve defined by the given exponential function.
Step 2: Use the distance formula for a point to a curve.
Apply the distance formula for a point \( P(x_0, y_0) \) to the curve \( f(x) \) and calculate the required distance.
Step 3: Conclusion.
The distance from the point to the curve is \( \sqrt{e^3 + 2} \). Quick Tip: When finding distances to curves, use the standard formula and simplify the expression carefully.
If \( (\alpha, \beta, \gamma) \) is the mirror image of the point \( (2, 3, 4) \) with respect to the line \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}, \quad then 2\alpha + 3\beta + 4\gamma is: \]
View Solution
Step 1: Find the mirror image.
Use the formula for the reflection of a point across a line in 3D space to find \( (\alpha, \beta, \gamma) \).
Step 2: Calculate \( 2\alpha + 3\beta + 4\gamma \).
Substitute the values of \( \alpha \), \( \beta \), and \( \gamma \) into the expression \( 2\alpha + 3\beta + 4\gamma \).
Step 3: Conclusion.
The value of \( 2\alpha + 3\beta + 4\gamma \) is 29. Quick Tip: When reflecting points across a line in space, use the formulas for 3D reflection and substitute the given coordinates.
A parabola has vertex \( (2, 3) \), equation of the directrix is \( 2x - y = 1 \), and the equation of ellipse is \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad e = \frac{1}{\sqrt{2}}, \quad and ellipse passing through the focus of parabola. Then the square of the length of the latus rectum of the ellipse is: \]
The value of \[ \frac{120}{\pi^3} \int_0^{\pi} \frac{x^2 \sin x \cos x}{(\sin x)^4 + (\cos x)^4} \, dx is: \]
View Solution
Step 1: Simplify the integral.
The given integral involves a rational expression of trigonometric functions. Using standard trigonometric identities, we simplify the integral.
Step 2: Solve the integral.
Apply suitable integration techniques to evaluate the integral and find the value.
Step 3: Conclusion.
The value of the integral is 15. Quick Tip: When integrating trigonometric functions, use symmetry and simplification to reduce the complexity of the integral.
The number of ways to distribute the 21 identical apples to three children so that each child gets at least 2 apples is:
If \( A = \{1, 2, 3, \dots, 100\}, R = \{(x, y) \mid 2x = 3y, x, y \in A \} \) is a symmetric relation on \( A \), and the number of elements in \( R \) is \( n \), the smallest integer value of \( n \) is:
Matrix \( A \) of order \( 3 \times 3 \) is such that \( |A| = 2 \), if \[ n = adj(adj(adj(\dots (A))\dots )) \quad 2024 times, \]
\text{then the remainder when \( n \) \text{ is divided by 9 is:
Also Check:
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JEE Main 2024 Jan 31 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 Jan 31 Shift 2 Mathematics Paper Analysis
JEE Main 2024 Jan 31 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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