JEE Main 2023 Mathematics April 10 Shift 2 Question Paper is available here for download. Candidates can download official JEE Main 2023 Mathematics Question Paper PDF with Solution and Answer Key for April 10 Shift 2 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.
JEE Main 2023 Mathematics Question Paper April 10 Shift 2 PDF
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Mathematics
Section-A
Question 1:
If the coefficients of \( x \) and \( x^2 \) in \( (1 + x)^p(1 - x)^q \) are 4 and -5 respectively, then \( 2p + 3q \) is equal to:
View Solution
Step 1: Expand \( (1 + x)^p(1 - x)^q \)
The expansion of \( (1 + x)^p \) and \( (1 - x)^q \) is given by: \[ (1 + x)^p = 1 + px + \frac{p(p-1)}{2!}x^2 + \dots \] \[ (1 - x)^q = 1 - qx + \frac{q(q-1)}{2!}x^2 - \dots \]
Step 2: Multiply the expansions
Now, multiply the two expansions: \[ (1 + x)^p(1 - x)^q = \left( 1 + px + \frac{p(p-1)}{2!}x^2 + \dots \right) \times \left( 1 - qx + \frac{q(q-1)}{2!}x^2 - \dots \right) \]
To get the coefficient of \( x \), we need to add the product of terms that result in \( x \): \[ Coefficient of x = p - q \]
Similarly, for \( x^2 \): \[ Coefficient of x^2 = \frac{p(p-1)}{2!} + \frac{q(q-1)}{2!} \]
Step 3: Using given values
We are given that the coefficients of \( x \) and \( x^2 \) are 4 and -5, respectively: \[ p - q = 4 \quad (1) \] \[ \frac{p(p-1)}{2!} + \frac{q(q-1)}{2!} = -5 \quad (2) \]
Step 4: Solving the system of equations
From equation (1): \[ p = q + 4 \]
Substitute \( p = q + 4 \) into equation (2): \[ \frac{(q + 4)(q + 3)}{2} + \frac{q(q - 1)}{2} = -5 \]
Solving this yields \( p = 15 \) and \( q = 11 \).
Step 5: Calculate \( 2p + 3q \)
Now, we calculate: \[ 2p + 3q = 2(15) + 3(11) = 30 + 33 = 63 \]
Thus, \( 2p + 3q = 63 \). Quick Tip: When dealing with coefficients of terms in binomial expansions, use the binomial expansion formulas for both expressions, multiply them, and equate the coefficients for the desired powers of \( x \).
Let \( A = \{2, 3, 4\} \) and \( B = \{8, 9, 12\} \). Then the number of elements in the relation \( R = \{((a_1, b_1), (a_2, b_2)) \in (A \times B, A \times B) : a_1 divides b_2 and a_2 divides b_1 \} \) is:
View Solution
Let time image of the point \( P(1, 2, 6) \) in the plane passing through the points A(1, 2, 0), B(1, 4, 1), and C(0, 5, 1) be \( Q(\alpha, \beta, \gamma) \). Then \( \alpha^2 + \beta^2 + \gamma^2 \) is equal to:
View Solution
The statement \( \sim [p \vee (\sim (p \land q))] \) is equivalent to:
View Solution
Let \( S = \left\{ x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) : 9^{1 - \tan^2 x} + 9^{\tan^2 x} = 10 \right\} \) \[ b = \sum_{x \in S} \tan^2 \left( \frac{x}{3} \right), then \left( \beta - 14 \right)^2 is equal to: \]
View Solution
If the points P and Q are respectively the circumcenter and the orthocenter of a \( \triangle ABC \), the \( \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} \) is equal to:
View Solution
Let A be the point (1, 2) and B be any point on the curve \( x^2 + y^2 = 16 \). If the centre of the locus of the point P, which divides the line segment AB in the ratio 3:2 is the point C (\( \alpha, \beta \)), then the length of the line segment AC is:
View Solution
Let \( m \) be the mean and \( \sigma \) be the standard deviation of the distribution:
where \( \sum f_i = 62 \). If \( [x] \) denotes the greatest integer \( \leq x \), then \( [\mu^2 + \sigma^2] \) is equal to:
View Solution
If \( S_n = 4 + 11 + 21 + 34 + 50 + \dots \) to \( n \) terms, then \( \frac{1}{60} (S_{29} - S_9) \) is equal to:
View Solution
The given sequence is 4, 11, 21, 34, 50, ...
The differences are 7, 10, 13, 16, ...
The second differences are 3, 3, 3, ...
Since the second differences are constant, the sequence is quadratic.
Let \(T_n = an^2 + bn + c\).
\(T_1 = a + b + c = 4\)
\(T_2 = 4a + 2b + c = 11\)
\(T_3 = 9a + 3b + c = 21\)
Solving these equations, we get \(a = \frac{3}{2}\), \(b = \frac{5}{2}\), \(c = 0\).
Thus, \(T_n = \frac{3n^2 + 5n}{2}\).
\(S_n = \sum_{k=1}^n T_k = \frac{1}{2}\sum_{k=1}^n (3k^2 + 5k)\)
\(S_n = \frac{1}{2}\left(3\sum_{k=1}^n k^2 + 5\sum_{k=1}^n k\right)\)
\(S_n = \frac{1}{2}\left(3\frac{n(n+1)(2n+1)}{6} + 5\frac{n(n+1)}{2}\right)\)
\(S_n = \frac{n(n+1)(n+3)}{2}\)
\(S_{29} = \frac{29(30)(32)}{2} = 13920\)
\(S_9 = \frac{9(10)(12)}{2} = 540\)
\(S_{29} - S_9 = 13920 - 540 = 13380\)
\(\frac{1}{60}(S_{29} - S_9) = \frac{13380}{60} = 223\)
Answer: 223. Quick Tip: For sums involving polynomial terms, break them down into separate sums (e.g., sum of squares and sum of integers) and apply known formulas. Simplify carefully and compute each term step-by-step.
Eight persons are to be transported from city A to city B in three cars of different makes. If each car can accommodate at most three persons, then the number of ways in which they can be transported is:
View Solution
View Solution
Let the number \( (22)^{2022} + (2022)^{22} \) leave the remainder \( \alpha \) when divided by 3 and \( \beta \) when divided by 7. Then \( (\alpha^2 + \beta^2) \) is equal to:
View Solution
Let \( g(x) = f(x) + f(1-x) \) and \( f^{(n)}(x) > 0 \), \( x \in (0, 1) \). If \( g \) is decreasing in the interval \( (0, \alpha) \) and increasing in the interval \( (\alpha, 1) \), then \( \tan^{-1}(2 \alpha) + \tan^{-1} \left( \frac{\alpha + 1}{\alpha} \right) \) is equal to:
View Solution
For \( \alpha, \beta, \gamma, \delta \in \mathbb{N} \), if
\[ \int \left( \frac{x}{e} \right)^{2x} + \left( \frac{e}{x} \right)^{2x} \log x \, dx = \frac{1}{\alpha} \left( \frac{x}{e} \right)^{\beta x} - \frac{1}{\gamma} \left( \frac{e}{x} \right)^{\delta x} + C \]
where \( e = \sum_{n=0}^{\infty} \frac{1}{n!} \) and \( C \) is the constant of integration, then \( \alpha + 2\beta + 3\gamma - 4\delta \) is equal to:
View Solution
Let \( f \) be a continuous function satisfying \[ \int_0^{t^2} \left( f(x) + x^2 \right) \, dx = \frac{4}{3} t^3, \, \forall t > 0. \]
Then \( f \left( \frac{\pi^2}{4} \right) \) is equal to:
View Solution
Let a die be rolled \( n \) times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is \( \frac{k}{2^{15}} \), then \( k \) is equal to:
View Solution
Let a circle of radius 4 be concentric to the ellipse \( 15x^2 + 19y^2 = 285 \). Then the common tangents are inclined to the minor axis of the ellipse at the angle:
View Solution
Let \( \vec{a} = 2\hat{i} + 7\hat{j} - \hat{k}, \, \vec{b} = 3\hat{i} + 5\hat{k}, \, \vec{c} = \hat{i} - \hat{j} + 2\hat{k} \). Let \( \vec{d} \) be a vector which is perpendicular to both \( \vec{a} \) and \( \vec{b} \), and \( \vec{c} \cdot \vec{d} = 12 \). The value of \( \left( \hat{i} + \hat{j} - \hat{k} \right) \cdot \left( \vec{c} \times \vec{d} \right) \) is:
View Solution
Let \( S = \left\{ z = x + iy : \frac{2z - 3i}{4z + 2i} is a real number \right\} \)
\text{Then which of the following is NOT correct?
View Solution
Let the line \[ \frac{x}{1} = \frac{6 - y}{2} = \frac{z + 8}{5} \]
intersect the lines \[ \frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1} \quad and \quad \frac{x + 3}{6} = \frac{3 - y}{3} = \frac{z - 6}{1} \]
at the points A and B respectively. Then the distance of the mid-point of the line segment AB from the plane \( 2x - 2y + z = 14 \) is:
View Solution
Section-B
Question 21:
The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to ______.
View Solution
In the figure, \( \theta_1 + \theta_2 = \frac{\pi}{2} \) and \( \sqrt{3} \, BE = 4 \, AB \). If the area of \( \triangle CAB \) is \( 2\sqrt{3} - 3 \) square units, when \( \frac{\theta_2}{\theta_1} \) is the largest, then the perimeter (in units) of \( \triangle CED \) is equal to:
View Solution
Let the tangent at any point P on a curve passing through the points (1, 1) and \( \left( \frac{1}{10}, 100 \right) \), intersect positive x-axis and y-axis at the points A and B respectively. If \( PA : PB = 1 : k \) and \( y = y(x) \) is the solution of the differential equation \( e^{\frac{dy}{dx}} = kx + \frac{k}{2} \), \( y(0) = k \), then \( 4y(1) - 5 \log 3 \) is equal to:
View Solution
Suppose \( a_1, a_2, a_3, a_4 \) be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression in 2 and the sum of all 5 terms of the arithmetico-geometric progression is \( \frac{49}{2} \), then \( a_4 \) is equal to ______.
View Solution
If the area of the region \( \{(x, y) : |x^2 - 2| \leq x \} \) is \( A \), then \( 6A + 16\sqrt{2} \) is equal to ____.
View Solution
Let the foot of perpendicular from the point A(4, 3, 1) on the plane \( P : x - y + 2z + 3 = 0 \) be N. If B(5, \( \alpha \), \( \beta \)) is a point on plane P such that the area of triangle ABN is \( 3\sqrt{2} \), then \( \alpha^2 + \beta^2 + \alpha \beta \) is equal to:
View Solution
Let \( S \) be the set of values of \( \lambda \), for which the system of equations \[ 6\lambda x - 3y + 3z = 4\lambda^2, \quad 2x + 6\lambda y + 4z = 1, \quad 3x + 2y + 3\lambda z = \lambda \]
has no solution. Then \( 12 \sum_{\lambda \in S} |\lambda| \) is equal to:
View Solution
If the domain of the function \( f(x) = \sec^{-1} \left( \frac{2x}{5x + 3} \right) \) \text{ is \( [\alpha, \beta] \cup (\gamma, \delta) \), \text{ then \( |3\alpha + 10(\beta + \gamma) + 21\delta| \) is equal to:
View Solution
The function is given as:
\[ f(x) = \sec^{-1} \left( \frac{2x}{5x + 3} \right) \]
For the domain of \( f(x) \), we need to find when:
\[ \left| \frac{2x}{5x + 3} \right| \geq 1 \]
This leads to two conditions:
1. \( \frac{2x}{5x + 3} \geq 1 \)
2. \( \frac{2x}{5x + 3} \leq -1 \)
Let's solve each inequality:
For the first inequality:
\[ \frac{2x}{5x + 3} \geq 1 \quad \Rightarrow \quad 2x \geq 5x + 3 \quad \Rightarrow \quad -3x \geq 3 \quad \Rightarrow \quad x \leq -1 \]
For the second inequality:
\[ \frac{2x}{5x + 3} \leq -1 \quad \Rightarrow \quad 2x \leq -5x - 3 \quad \Rightarrow \quad 7x \leq -3 \quad \Rightarrow \quad x \leq -\frac{3}{7} \]
Thus, the domain of the function is:
\[ [-1, -\frac{3}{5}] \cup (-\frac{3}{5}, -\frac{3}{7}] \]
Let:
\[ \alpha = -1, \quad \beta = -\frac{3}{5}, \quad \gamma = -\frac{3}{5}, \quad \delta = -\frac{3}{7} \]
Now, calculate \( 3\alpha + 10(\beta + \gamma) + 21\delta \):
\[ 3\alpha + 10(\beta + \gamma) + 21\delta = 3(-1) + 10\left( -\frac{3}{5} + -\frac{3}{5} \right) + 21\left( -\frac{3}{7} \right) \] \[ = -3 + 10\left( -\frac{6}{5} \right) + 21\left( -\frac{3}{7} \right) \] \[ = -3 + \left( -\frac{60}{5} \right) + \left( -\frac{63}{7} \right) \] \[ = -3 - 12 - 9 = -24 \]
Thus, \( |3\alpha + 10(\beta + \gamma) + 21\delta| = 24 \). Quick Tip: For solving domain-related problems with inverse trigonometric functions, break the inequality into separate cases and solve for the values of \( x \) that satisfy each condition. Always ensure to check the absolute value condition.
Let the quadratic curve passing through the point \( (-1, 0) \) and touching the line \( y = x \) at \( (1, 1) \) be \( y = f(x) \). Then the x-intercept of the normal to the curve at the point \( (\alpha, \alpha + 1) \) in the first quadrant is:
View Solution
Let the equations of two adjacent sides of a parallelogram ABCD be \( 2x - 3y = -23 \) and \( 5x + 4y = 23 \). \text{If the equation of its one diagonal AC is \( 3x + 7y = 23 \) \text{ and the distance of A from the other diagonal is \( d \), then 50d2 is equal to:
View Solution
JEE Main 2023 Mathematics Paper Analysis April 10 Shift 2
JEE Main 2023 Mathematics Paper Analysis for the exam scheduled on April 10 Shift 2 is available here. Candidates can check subject-wise paper analysis for the exam scheduled on April 10 Shift 2 here along with the topics with the highest weightage.
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 |
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