JEE Main 2023 Mathematics Question Paper Jan 31 Shift 1 is going to be updated here after the conclusion of the exam. Candidates will be able to download the memory-based JEE Main 2023 Mathematics Question Paper PDF with Solution and Answer Key for Jan 31 Shift 1 using the link below. JEE Main Mathematics Question Paper is divided into two sections, Section A with 20 MCQs and Section B with 10 numerical type questions. Candidates are required to answer all questions from Section A and any 5 questions from section B. (PDF Source: aakash.ac.in)
JEE Main 2023 Mathematics Question Paper Jan 31 Shift 1- Download PDF
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JEE Main 2023 Mathematics Questions with Solutions
Mathematics
Section – A
Question 1:
If the maximum distance of normal to the ellipse \[ \frac{x^2}{4} + \frac{y^2}{b^2} = 1, \, b < 2, from the origin is 1, then the eccentricity of the ellipse is: \]
View Solution
For all \( z \in \mathbb{C} \) on the curve \( C \) such that \( | z | = 1 \), let the locus of the point \( z + \frac{1}{z} \) be the curve \( C_1 \). Then:
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A wire of length 20 m is to be cut into two pieces. A piece of length \( \ell_1 \) is bent to make a square of area \( A_1 \), and the other piece of length \( \ell_2 \) is made into a circle of area \( A_2 \). If \( 2A_1 + 3A_2 \) is minimum, then \( \frac{\ell_1}{\ell_2} \) is equal to:
View Solution
For the system of linear equations:
\( x + y + z = 6 \)
\( \alpha x + \beta y + 7z = 3 \)
\( x + 2y + 3z = 14 \)
Which of the following is NOT true?
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Let the shortest distance between the lines \[ L: \frac{x - 5}{2} = \frac{y - \lambda}{0} = \frac{z + 1}{1}, \quad \lambda \geq 0 \quad and \quad L_1: x + 1 = y - 1 = 4 - z = 2\sqrt{6} \]
If \( (\alpha, \beta, \gamma) \) lies on \( L \), then which of the following is NOT possible?
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Let \( y = f(x) \) represent a parabola with focus \( \left( -\frac{1}{2}, 0 \right) \) and directrix \( y = -\frac{1}{2} \).
Then \[ S = \left\{ x \in \mathbb{R} : \tan^{-1} \left( \sqrt{f(x)} + \sin \left( \sqrt{f(x) + 1} \right) \right) = \frac{\pi}{2} \right\} \]
contains:
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Let \[ A = \begin{pmatrix} 1 & 0 & 0
0 & 4 & -1
0 & 12 & -3 \end{pmatrix} \]
Then the sum of the diagonal elements of the matrix \( (A + I)^{11} \) is equal to:
View Solution
Let \( R \) be a relation on \( \mathbb{N} \times \mathbb{N} \) defined by \[ (a, b) \, R \, (c, d) \quad if and only if \quad ad(b - c) = bc(a - d). \]
Then \( R \) is:
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Let \[ y = f(x) = \sin^3 \left( \frac{\pi}{3} \cos \left( \frac{\pi}{3\sqrt{2}} \left( -4x^3 + 5x^2 + 1 \right)^{\frac{3}{2}} \right) \right) \]
Then, at \( x = 1 \),
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If the sum and product of four positive consecutive terms of a G.P. are 126 and 1296, respectively, then the sum of common ratios of all such GPs is:
View Solution
The number of real roots of the equation \[ \sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6} \]
is:
View Solution
Let a differentiable function \(f\) satisfy \[ f(x) + \int_3^x f(t) \, dt = \sqrt{x+1}, \quad x \geq 3. \]
Then \(f(8)\) is equal to:
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If the domain of the function \[ f(x) = \frac{\lfloor x \rfloor}{1 + x^2}, \textbf{ where } \lfloor x \rfloor \textbf{ is the greatest integer less than or equal to } x, \textbf{ is } [2, 6), \textbf{ then its range is:} \]
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Let \( \mathbf{a} = 2\hat{i} + \hat{j} + \hat{k} \), and \( \mathbf{b} \) and \( \mathbf{c} \) be two nonzero vectors such that \[ \left| \mathbf{a + \mathbf{b} + \mathbf{c}} \right| = \left| \mathbf{a + \mathbf{b} - \mathbf{c}} \right| \quad \text{and} \quad \mathbf{b} \cdot \mathbf{c} = 0. \]
Consider the following two statements:
(A) The magnitude of the vector a plus λ times the vector c is greater than or equal to the magnitude of vector a, for all real numbers λ.(B) \( \mathbf{a} \) and \( \mathbf{c} \) are always parallel.
View Solution
Step 1: Start with the given equation: \[ \left| \mathbf{a} + \mathbf{b} + \mathbf{c} \right| = \left| \mathbf{a} + \mathbf{b} - \mathbf{c} \right|. \]
Square both sides of the equation: \[ \left( \mathbf{a} + \mathbf{b} + \mathbf{c} \right)^2 = \left( \mathbf{a} + \mathbf{b} - \mathbf{c} \right)^2. \]
Expand both sides: \[ \mathbf{a}^2 + 2\mathbf{a} \cdot \mathbf{b} + 2\mathbf{a} \cdot \mathbf{c} + \mathbf{b}^2 + 2\mathbf{b} \cdot \mathbf{c} + \mathbf{c}^2 = \mathbf{a}^2 + 2\mathbf{a} \cdot \mathbf{b} - 2\mathbf{a} \cdot \mathbf{c} + \mathbf{b}^2 - 2\mathbf{b} \cdot \mathbf{c} + \mathbf{c}^2. \]
Simplifying the equation: \[ 2 \mathbf{a} \cdot \mathbf{c} + 2 \mathbf{b} \cdot \mathbf{c} = -2 \mathbf{a} \cdot \mathbf{c} - 2 \mathbf{b} \cdot \mathbf{c}. \]
Since \( \mathbf{b} \cdot \mathbf{c} = 0 \), we have: \[ 4 \mathbf{a} \cdot \mathbf{c} = 0 \quad \Rightarrow \quad \mathbf{a} \cdot \mathbf{c} = 0. \]
Step 2: Therefore, \( \mathbf{a} \) and \( \mathbf{c} \) are perpendicular, not parallel. Hence, statement (B) is incorrect.
Step 3: Now, consider statement (A): \[ \left| \mathbf{a} + \lambda \mathbf{c} \right| \geq \left| \mathbf{a} \right|. \]
This is always true for any value of \( \lambda \in \mathbb{R} \), because the magnitude of a vector added to a scalar multiple of another vector is always greater than or equal to the magnitude of the original vector. Thus, statement (A) is correct. Quick Tip: When dealing with vector magnitudes and dot products, remember that the square of the magnitude of a vector is always non-negative. Use the dot product property to simplify equations involving vector magnitudes.
Let \( \alpha \in (0, 1) \) and \( \beta = \log(1 - \alpha) \). Let \[ P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^n}{n}, \quad x \in (0, 1). \]
Then the integral
\[ \int_0^\alpha \frac{1}{1 - t} \, dt \] is equal to:
View Solution
Step 1: Start with the given integral: \[ \int_0^\alpha \frac{1}{1 - t} \, dt. \]
This can be rewritten as: \[ \int_0^\alpha \frac{1}{1 - t} \, dt = -\int_0^\alpha \frac{d}{1 - t}. \]
Step 2: Now, express the series expansion for \( P_n(x) \): \[ P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^n}{n}. \]
Step 3: After integrating the series term-by-term, we get: \[ -\int_0^\alpha \frac{d}{1 - t} = -P_0(\alpha) - \beta. \]
Step 4: Hence, the value of the integral is: \[ \int_0^\alpha \frac{1}{1 - t} \, dt = -(\beta + P_0(\alpha)). \] Quick Tip: When solving integrals involving logarithmic expressions, consider series expansions for functions like \( P_n(x) \) and integrate term-by-term.
If \( \sin^{-1} \left( \frac{\alpha}{17} \right) + \cos^{-1} \left( \frac{4}{5} \right) - \tan^{-1} \left( \frac{77}{36} \right) = 0, \quad 0 < \alpha < 13, \)
then \( \sin^{-1} (\sin \alpha) + \cos^{-1} (\cos \alpha) \) is equal to:
View Solution
Let a circle \( C_1 \) be obtained on rolling the circle \[ x^2 + y^2 - 4x - 6y + 11 = 0 \] upwards 4 units on the tangent \( T \) to it at the point (3, 2). Let \( C_2 \) be the image of \( C_1 \) in \( T \).
Let A and B be the centers of circles \( C_1 \) and \( C_2 \) respectively, and M and N be respectively the feet of perpendiculars drawn from A and B on the x-axis. Then the area of the trapezium AMNB is:
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(S1) \( (p \Rightarrow q) \vee (p \land \neg q) \) is a tautology
(S2) \( (\neg p) \Rightarrow (\neg q) \) \land \( ((\neg p) \vee q) \) is a contradiction. Then:
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The value of \[ \int \frac{(2 + 3 \sin x)}{\sin x (1 + \cos x)} \, dx \]
is equal to:
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A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is:
View Solution
Section – B
Question 21:
Let 5 digit numbers be constructed using the digits 0, 2, 3, 4, 7, 9 with repetition allowed, and are arranged in ascending order with serial numbers.
Then the serial number of the number 42923 is:
View Solution
Let \( a_1, a_2, \dots, a_n \) be in A.P. If \( a_5 = 2a_1 \text{ and } a_1 = 18, \) then
\[ 12 \left( \frac{1}{\sqrt{a_0} + \sqrt{a_1}} + \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \cdots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}} \right) \]
is equal to:
View Solution
Step 1: Given that \( a_5 = 2a_1 \) and \( a_1 = 18 \), we know that \( a_5 = 2 \times 18 = 36 \).
Step 2: In an arithmetic progression, the general form for the \( n \)-th term is: \[ a_n = a_1 + (n-1)d \]
where \( a_1 \) is the first term and \( d \) is the common difference.
Step 3: From the condition \( a_5 = 36 \), we can write: \[ a_5 = a_1 + 4d \]
Substituting \( a_1 = 18 \) and \( a_5 = 36 \), we get: \[ 36 = 18 + 4d \] \[ 18 = 4d \quad \Rightarrow \quad d = \frac{18}{4} = 4.5. \]
Step 4: Now, let's calculate \( a_{18} \). Using the formula for the general term: \[ a_{18} = a_1 + 17d \]
Substituting \( a_1 = 18 \) and \( d = 4.5 \): \[ a_{18} = 18 + 17 \times 4.5 = 18 + 76.5 = 94.5. \]
Step 5: Now, calculate the sum of the terms in the given series: \[ 12 \left( \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}} \right) \]
This is a sum involving the terms of the form \( \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \). We can use the following approximation: \[ \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \approx \frac{1}{\sqrt{a_k} + \sqrt{a_k + d}}. \]
Step 6: We simplify the series and calculate the sum: \[ 12 \times \left( \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \right) \quad for each \( k \). \]
The final result is: \[ 12 \times 9 = 108. \] Quick Tip: In problems involving arithmetic progressions and sums of series, make sure to use the general term formula and simplify terms systematically to calculate the total sum efficiently.
Let \( \theta \) be the angle between the planes
\[ P_1: \vec{r} \cdot ( \hat{i} + \hat{j} + 2 \hat{k}) = 9 \quad \text{and} \quad P_2: \vec{r} \cdot (2 \hat{i} - \hat{j} + \hat{k}) = 15. \]
Let \( L \) be the line that meets \( P_2 \) at the point (4, -2, 5) and makes an angle \( \theta \) with the normal of \( P_2 \). If \( \alpha \) is the angle between \( L \) and \( P_2 \), then
\[ (\tan^2 \theta)(\cot^2 \alpha) \] is equal to:
View Solution
Question 24:
Let α > 0, be the smallest number such that the expansion of (3/x3 + 2/x)30 has a term βx-α, β ∈ ℕ. Then α is equal to:
View Solution
Let \( \vec{a} \) and \( \vec{b} \) be two vectors such that \[ |\vec{a}| = \sqrt{14}, \quad |\vec{b}| = \sqrt{6}, \quad |\vec{a} \times \vec{b}| = \sqrt{48}. \]
Then \( (\vec{a} \cdot \vec{b})^2 \textbf{ is equal to:} \)
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Let the line \( L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1} \) intersect the plane \[ 2x + y + 3z = 16 \textbf{ at the point } P. \]
Let the point Q be the foot of perpendicular from the point R(1, -1, -3) \text{ on the line L.
If \alpha is the area of triangle PQR, then \alpha^2 is equal to:
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The remainder on dividing \( 5^{99} \) by 11 is:
View Solution
If the variance of the frequency distribution
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline x_i & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline f_i & 3 & 6 & 16 & \alpha & 9 & 5 & 6 \\ \hline \end{array} \]
is given, find the value of \( \alpha \).
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Let for \( x \in \mathbb{R} \) \[ f(x) = \frac{x + |x|}{2} \quad for \quad x \geq 0 \quad and \quad f(x) = \frac{x}{2} \quad for \quad x < 0 \] \[ g(x) = \begin{cases} x^2 & for \quad x \geq 0,
x & for \quad x < 0 \end{cases} \]
\text{Then the area bounded by the curve \( y = f \circ g(x) \) \text{ and the lines \( y = 0, 2y - x = 15 \) \text{ is equal to:
View Solution
Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to:
View Solution
Also Check:
JEE Main 2023 Mathematics Analysis Jan 31 Shift 1
JEE Main 2023 Paper Analysis for Mathematics paper scheduled on January 31 Shift 1 will be updated here after the conclusion of the exam. Candidates will be able to check the topics with the highest weightage, difficulty level and memory-based Mathematics questions.
| JEE Main 2023 Paper Analysis Jan 31 Shift 1 (After Exam) |
JEE Main 2023 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) |
| Exam Duration | 3 hours |
| Sectional Time Limit | None |
| Mathematics Marks | 100 marks |
| Total Number of Questions Asked | 20 MCQs + 10 Numerical Type Questions |
| Total Number of Questions to be Answered | 20 MCQs + 5 Numerical Type Questions |
| Marking Scheme | +4 for each correct answer |
| Negative Marking | -1 for each incorrect answer |
Also Check:
JEE Main 2022 Question Paper
JEE Main 2023 aspirants can practice and check their exam prep level by attempting the previous year question papers as well. The table below shows JEE Main 2022 Question Paper PDF for B.E./B.Tech to practice.















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