JEE Main 2023 Question Paper with Solution Pdf for Mathematics (April 10 - Shift 1)
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The JEE Main 2023 Mathematics Question Paper with Solution PDF is available here for download. The exam was successfully conducted by NTA on April 10, 2023, in the first shift.
Students can download JEE Main previous year question papers PDFs with detailed solutions to practice and improve their performance. Solving these papers helps aspirants understand the JEE Main exam pattern, analyze the difficulty level, and prepare according to the latest JEE Main syllabus.
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JEE Main 2023 Mathematics Question Paper Apr 10 Shift 1 with Solution Pdf
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Question 1:
An arc PQ of a circle subtends a right angle at its centre O. The midpoint of the arc PQ is R. If →O P = →u, O→R = →v and →OQ = →αu+ →βv, then α, β2 are the roots of the equation:
View Solution
Step 1: Identify the vectors
Given →OP = →u and →OQ = ĵ, and O⃗R bisects the angle POQ. Thus, →OR = (1⁄√2) î + (1⁄√2) ĵ.
Step 2: Express O⃗Q in terms of u⃗ and v⃗
Using O⃗Q = αu⃗ + βv⃗, substitute O⃗R and solve for α and β.
Comparing coefficients, we get:
- α + (β⁄√2) = 0
- (β⁄√2) = 1 => β = √2
Step 3: Derive the quadratic equation
α = −1, and β2 = 2.
The sum of roots = α + β2 = −1 + 2 = 1.
The product of roots = α × β2 = (−1)(2) = −2.
Thus, the quadratic equation is x2 − x − 2 = 0.
Question 2:
A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in cm2) is equal to:
View Solution
Step 1: Define variables and volume
Let x be the side length of the square cut from each corner. The dimensions of the box are:
Length = 30 − 2x, Breadth = 30 − 2x, Height = x.
The volume V(x) = (30 − 2x)2 × x.
Step 2: Maximize the volume
Differentiate V(x):
V′(x) = 12x2 − 240x + 900. Solve V′(x) = 0 for critical points:
x = 5 or x = 15. Since x = 15 is not valid, x = 5.
Step 3: Calculate surface area
Surface Area S(x) = (30 − 2x)2 + 4x(30 − 2x).
Substitute x = 5: S(5) = 800 cm2.
Question 3:
Let O be the origin and the position vector of the point P be −̂i−2ĵ+3k̂. If the position vectors of A, B, and C are −2̂i+ ĵ−3k̂, 2̂i+4ĵ−2k̂, and −4̂i+2ĵ−k̂ respectively, then the projection of vector O⃗P on a vector perpendicular to vectors A⃗B and A⃗C is:
View Solution
Step 1: Find A⃗B and A⃗C
A⃗B = ⟨4, 3, 1⟩, A⃗C = ⟨−2, 1, 2⟩.
Step 2: Compute A⃗B × A⃗C
A⃗B × A⃗C = 5̂i − 10ĵ + 10k̂. Magnitude = 15.
Step 3: Compute projection
Projection = |(O⃗P ⋅ (A⃗B × A⃗C))|⁄|A⃗B × A⃗C| = 3.
Question 4:
If A is a 3×3 matrix and |A| = 2, then |3adj(|3A|A2)| is equal to:
View Solution
Step 1: Calculate |3A|
|3A| = 33 × |A| = 27 × 2.
Step 2: Adj of |3A|A2
Adj(|3A|A2) = (2 × 33)2 × |adj(A)|2.
Step 3: Final calculation
|3Adj(|3A|A2)| = 311 × 610.
Question 5:
Let two vertices of a triangle ABC be (2, 4, 6) and (0,−2,−5), and its centroid be (2, 1,−1). If the image of the third vertex in the plane x + 2y + 4z = 11 is (α, β, γ), then αβ + βγ + γα is equal to:
View Solution
Step 1: Find the third vertex
Let the third vertex be (x, y, z). Solve centroid equations to find C = (4, 1, −4).
Step 2: Find image coordinates
Use the plane equation x + 2y + 4z = 11 to compute the image of (4, 1, −4) as (6, 5, 4).
Step 3: Compute αβ + βγ + γα
αβ + βγ + γα = (6 × 5) + (5 × 4) + (4 × 6) = 74.
Question 6:
The negation of the statement (p ∨ q) ∧ (q ∨ (∼ r)) is:
View Solution
Step 1: Take the Negation
The given statement is (p ∨ q) ∧ (q ∨ (∼ r)). Taking its negation: ∼ [(p ∨ q) ∧ (q ∨ (∼ r))].
Step 2: Apply De Morgan's Laws
Using De Morgan's Laws: ∼ (p ∨ q) ∨ ∼ (q ∨ (∼ r)).
Step 3: Simplify Each Term
∼ (p ∨ q) = (∼ p ∧ ∼ q) and ∼ (q ∨ (∼ r)) = (∼ q ∧ r).
Rewriting: (∼ p ∧ ∼ q) ∨ (∼ q ∧ r).
Step 4: Use the Distributive Property
Distribute the terms: = (∼ p ∨ r) ∧ (∼ q).
Question 7:
The shortest distance between the lines x + 2 / 1 = y / −2 = z − 5 / 2 and x − 4 / 1 = y − 1 / 2 = z + 3 / 0 is:
View Solution
Step 1: Identify Given Lines
The direction vectors of the lines are l1 = ⟨1, −2, 2⟩ and l2 = ⟨1, 2, 0⟩. Position vectors are A = ⟨−2, 0, 5⟩ and B = ⟨4, 1, −3⟩.
Step 2: Use the Shortest Distance Formula
The formula is:
d = |(b − a) • (l1 × l2)| / |l1 × l2|.
Step 3: Compute Cross Product
l1 × l2 = ⟨−4, −2, 4⟩.
Step 4: Compute Distance
d = |6| = 9.
Question 8:
If the coefficient of x7 in (ax − 1 / bx2)13 and the coefficient of x−5 in (ax + 1 / bx2)13 are equal, then a4b4 is equal to:
View Solution
Step 1: Expand the General Term
Tr+1 = C13r * (ax)13-r (−1 / bx2)r.
Step 2: Equate Coefficients
For x7: r = 6. For x−5: r = 11.
Solve the equation to find a4b4 = 22.
Question 9:
A line segment AB of length λ moves such that the points A and B remain on the periphery of a circle of radius γ. Then the locus of the point, that divides the line segment AB in the ratio 2 : 3, is a circle of radius:
View Solution
Step 1: Use Geometry and Ratios
Determine that P divides AB in a given ratio, leading to locus equations.
Step 2: Apply Geometry Formulas
Radius = √19λ / 5.
Question 10:
For the system of linear equations 2x − y + 3z = 5, 3x + 2y − z = 7, 4x + 5y + αz = β, which of the following is NOT correct?
View Solution
Step 1: Analyze Determinant
Find Δ = 7(α + 5).
Step 2: Apply Conditions for Solutions
For α = −5, β = 8: Inconsistent.
For α = −6, β = 9: Incorrect.
Correct answers are evaluated based on determinant properties.
Question 11:
Let the first term a and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these terms is equal to:
View Solution
Step 1: Represent the terms of the GP
Let the terms be a, ar, and ar². The sum of squares is:
a² + (ar)² + (ar²)² = 33033
Step 2: Simplify the equation
a²(1 + r² + r⁴) = 33033. Factorize 33033: 33033 = 11² × 3 × 7 × 13.
This gives a² = 11² = 121, so a = 11.
Step 3: Solve for r
1 + r² + r⁴ = 273. Factorize: r²(r² + 1) = 272. Let r² = 16, so r = 4.
Step 4: Calculate the sum
Sum = a(1 + r + r²) = 11(1 + 4 + 16) = 231.
Question 12:
Let P be the point of intersection of the line x+3/3 = y+2/1 = 1−z/2 and the plane x + y + z = 2. If the distance of the point P from the plane 3x − 4y + 12z = 32 is q, then q and 2q are the roots of the equation:
View Solution
Step 1: Find the coordinates of P
Using the line equations, express P as:
x = 3λ − 3, y = λ − 2, z = 1 − 2λ. Substituting into the plane equation x + y + z = 2, solve for λ = 3.
Thus, P = (6, 1, −5).
Step 2: Distance from P to the plane
Use the distance formula: q = |3(6) − 4(1) + 12(−5) − 32| / √(3² + (−4)² + 12²), giving q = 6.
Step 3: Derive the quadratic equation
The roots are q and 2q. Sum of roots = 18, product = 72.
Equation: x² − 18x + 72 = 0.
Question 13:
Let f be a differentiable function such that x²f(x) − x = 4∫₀ˣ tf(t) dt, f(1) = 2/3. Then 18f(3) is equal to:
View Solution
Step 1: Differentiate the equation
x²f'(x) + 2xf(x) − 1 = 4xf(x). Rearrange to get:
x²f'(x) − 2xf(x) − 1 = 0.
Step 2: Solve the differential equation
Rewriting: f'(x) − (2/x)f(x) = 1/x². The integrating factor is 1/x². Solve to find:
f(x) = −1/(3x) + Cx².
Step 3: Apply initial condition
Substitute f(1) = 2/3 to find C = 1. Thus, f(x) = −1/(3x) + x².
Step 4: Find 18f(3)
f(3) = −1/9 + 9 = 80/9. Multiply: 18f(3) = 160.
Question 14:
Let N denote the sum of the numbers obtained when two dice are rolled. If the probability that 2N < N! is m/n, where m and n are coprime, then 4m − 3n is equal to:
View Solution
Step 1: Identify valid values of N
2N < N! is true for N ≥ 4.
Step 2: Calculate probabilities
P(N ≥ 4) = 1 − P(N < 4). Compute P(N < 4) using possible outcomes:
P(N < 4) = 3/36. Thus, P(N ≥ 4) = 33/36 = 11/12.
Step 3: Calculate 4m − 3n
m = 11, n = 12. Compute: 4m − 3n = 44 − 36 = 8.
Question 15:
If I(x) = ∫esin²xcosx(sin2x − sinx)dx and I(0) = 1, then I(π/3) is equal to:
View Solution
Step 1: Simplify the integral
Use substitution sin²x = t. Solve the integral to find:
I = esin²xcosx + C.
Step 2: Apply initial condition
I(0) = e0 + C = 1. Solve for C = 0. Thus, I = esin²xcosx.
Step 3: Evaluate I(π/3)
I(π/3) = e3/4 × 1/2 = 1/2e³⁄₄.
Question 16:
96 cos(π/33) cos(2π/33) cos(4π/33) cos(8π/33) cos(16π/33) is equal to:
View Solution
Step 1: Use the General Formula
The formula for cosA cos 2A ... cos 2n−1A is given by:
sin(2nA) / [2nsinA].
Step 2: Simplify the Expression
For A = π/33 and n = 5:
96 cos(π/33) cos(2π/33) ... cos(16π/33) = 96 sin(32π/33) / [32 sin(π/33)].
Using sin(π − x) = sin(x):
sin(32π/33) = sin(π/33).
Step 3: Final Calculation
= 96 sin(π/33) / [32 sin(π/33)] = 96 / 32 = 3.
Question 17:
Let the complex number z = x + iy be such that (2z − 3i) / (2z + i) is purely imaginary. If x + y2 = 0, then y4 + y2 − y is equal to:
View Solution
Step 1: Set the Real Part to Zero
The condition for purely imaginary numbers implies the real part of (2z − 3i) / (2z + i) must be zero.
Step 2: Substitute z = x + iy
Simplify (2z − 3i) / (2z + i) = [2x + 2yi − 3i] / [2x + i(2y + 1)]. Rationalize the denominator.
Step 3: Solve the Equation
Expand and simplify to find: x = −y2. Substitute into the given condition x + y2 = 0 and solve for y.
y satisfies y4 + y2 − y = 3/4.
Question 18:
If f(x) = [(tan 1°)x + loge(123)] / [x loge(1234) − (tan 1°)], x > 0, then the least value of f(f(x)) + f(f(4/x)) is:
View Solution
Step 1: Compute f(f(x))
Substitute f(x) into itself and simplify.
Step 2: Evaluate f(f(4/x))
Using the same method, find f(f(4/x)).
Step 3: Use AM-GM Inequality
f(f(x)) + f(f(4/x)) ≥ 4 by the AM-GM inequality.
Question 19:
The slope of the tangent at any point (x, y) on a curve y = y(x) is (x2 + y2) / (2xy), x > 0. If y(2) = 0, then a value of y(8) is:
View Solution
Step 1: Rewrite the Differential Equation
Substitute y = vx into dy/dx and separate variables.
Step 2: Solve the Equation
Integrate both sides to find y as a function of x. Apply the initial condition y(2) = 0 to find the constant of integration.
Step 3: Calculate y(8)
Substitute x = 8 into the solution and simplify to find y = 4√3.
Question 20:
Let the ellipse E: x2 + 9y2 = 9 intersect the positive x- and y-axes at the points A and B respectively. Let the major axis of E be a diameter of the circle C. Let the line passing through A and B meet the circle C at the point P. If the area of the triangle with vertices A, P, and the origin O is m/n, where m and n are coprime, then m − n is equal to:
View Solution
Step 1: Find the Line AB
The line AB is x/3 + y/1 = 1, or x + 3y = 3.
Step 2: Intersection of Line and Circle
Solve the system of equations for the line and circle x2 + y2 = 9 to find P.
Step 3: Area of the Triangle
Calculate the area of △AOP using the formula (1/2) × Base × Height. Simplify to find m − n = 17.
Question 21:
Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple is in a match, is 840, then the total number of persons who participated in the tournament is:
View Solution
Step 1: Represent the number of couples as n
The total number of ways matches can be formed is given by:
(n choose 2) × (n−2 choose 2) × 2 = 840.
Step 2: Simplify the equation
n(n − 1)(n − 2)(n − 3) = 3360. Solve for n by factorizing:
n = 8.
Step 3: Calculate the total number of persons
The total number of persons is 2n = 16.
Question 22:
The number of elements in the set {n ∈ Z : |n² − 10n + 19| < 6} is:
View Solution
Step 1: Rewrite the inequality
−6 < n² − 10n + 19 < 6.
Step 2: Solve the quadratic inequalities
Split the inequality into:
n² − 10n + 25 > 0 and n² − 10n + 13 < 0. Solve to find n.
Step 3: Determine the integer solutions
Combine the results to find the valid integers: {2, 3, 4, 5, 6, 8}.
Step 4: Count the elements
The total number of elements is 6.
Question 23:
The number of permutations of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is:
View Solution
Step 1: Total permutations
Total permutations = 7! = 5040.
Step 2: Apply inclusion-exclusion principle
Calculate permutations containing 153 and 2467, then subtract overlapping cases.
n(153 ∪ 2467) = 120 + 24 − 2 = 142.
Step 3: Calculate the required number
Required permutations = 5040 − 142 = 4898.
Question 24:
Let f(x) be defined as:
f(x) = x⌊x⌋, −2 < x < 0, and f(x) = (x − 1)⌊x⌋, 0 ≤ x < 2. If m and n are the number of points in (−2, 2) where y = |f(x)| is not continuous and not differentiable, then m + n is equal to:
View Solution
Step 1: Identify points of discontinuity
f(x) is discontinuous at x = −1.
Step 2: Identify points of non-differentiability
f(x) is non-differentiable at x = −1, 0, 1.
Step 3: Calculate m and n
m = 1 (discontinuities), n = 3 (non-differentiabilities).
m + n = 4.
Question 25:
Let a common tangent to the curves y² = 4x and (x − 4)² + y² = 16 touch the curves at the points P and Q. Then (PQ)² is equal to:
View Solution
Step 1: Find the equation of the tangent
The tangent to y² = 4x is y = mx + 1/m.
Step 2: Tangency condition for the circle
The perpendicular distance from the center (4, 0) to the tangent must equal the radius.
Step 3: Solve for m
Using the condition, find m² = 1/8.
Step 4: Calculate (PQ)²
Find the points P and Q and compute:
(PQ)² = 32.
Question 26:
If the mean of the frequency distribution is 28, then its variance is:
View Solution
Step 1: Calculate the Mean
The formula for the mean is:
x̄ = (Σfixi) / N
Substituting the given data:
28 = (10 + 45 + 25x + 175 + 180) / (14 + x).
Solve for x: x = 6.
Step 2: Calculate Variance
Variance = (Σfix2i) / N - (x̄)2.
Substitute the values to find Variance = 151.
Question 27:
The coefficient of x7 in (1 − x + 2x3)10 is:
View Solution
Step 1: Use the General Term
The general term in the expansion is:
Tn = (10! / a!b!c!) * (−x)b * (2x3)c, where a + b + c = 10 and b + 3c = 7.
Step 2: Solve for Coefficients
Find combinations of a, b, and c satisfying the equations. Compute coefficients for each combination.
Step 3: Final Calculation
Combine terms to find the coefficient of x7: 960.
Question 28:
If y = p(x) is the parabola passing through points (−1, 0), (0, 1), and (1, 0), and the area of the region {(x, y) : (x + 1)2 + (y − 1)2 ≤ 1, y ≤ p(x)} is A, then 12(π − 4A) is equal to:
View Solution
Step 1: Find the Equation of the Parabola
The parabola is x2 = −(y − 1).
Step 2: Calculate the Area
Find the area under the parabola using integration and subtract it from the sector area to find A.
Step 3: Final Calculation
Compute 12(π − 4A) = 16.
Question 29:
Let a, b, c be three distinct positive real numbers such that (2a)logea = (bc)logeb and b log2(ea) = a loge(c). Then 6a + 5bc is equal to:
View Solution
Step 1: Simplify the Equations
Using logarithmic properties, rewrite the given equations in terms of a, b, and c.
Step 2: Analyze Cases
Solve for relationships between a, b, and c, and determine conditions leading to multiple solutions.
Step 3: Conclude
Infinite solutions are possible; the question is a bonus.
Question 30:
The sum of all terms of the arithmetic progression 3, 8, 13, ..., 373, which are not divisible by 3, is:
View Solution
Step 1: Find Total Terms
The number of terms is n = 75.
Step 2: Calculate the Total Sum
Sum = n/2 * [a + l] = 14100.
Step 3: Subtract Terms Divisible by 3
Find and subtract the sum of terms divisible by 3 to get the result: 9525.
*The article might have information for the previous academic years, please refer the official website of the exam.