
JEE Main 2024 Apr 9 Shift 1 Question Paper with Solution pdf is available for download here. Students found Physics easy and Mathematics hard. Physics carried the highest weightage and overall difficulty level was moderate.
| JEE Main 2024 Question Paper | Check Solution |
Let the line L intersect the lines:
x − 2 = −y = z − 1, 2(x + 1) = 2(y − 1) = z + 1, and be parallel to the line: (x − 2)/3 = (y − 1)/1 = (z − 2)/2. Then which of the following points lies on L?
The intersection of the lines and the parallel condition leads to the point (-1/3, 1, -1) lying on the line L.
The parabola y² = 4x divides the area of the circle x² + y² = 5 in two parts. The area of the smaller part is:
The smaller area is calculated using geometry formulas for the circle segment and the parabola sector, leading to the area 2/3 + 5 sin⁻¹(2√5).
The solution curve of the differential equation 2y (dy/dx) + 3 = 5 (dy/dx), passing through the point (0, 1), is a conic whose vertex lies on the line:
Solving the differential equation and using the initial condition (0, 1), the vertex of the parabola is found to lie on the line 2x + 3y = 9.
A ray of light coming from P(1, 2) reflects from Q on the x-axis and passes through R(4, 3). If S(h, k) makes PQRS a parallelogram, then hk² equals:
By using reflection laws and midpoint properties of parallelograms, the value of hk² is found to be 70.
For λ, μ ∈ R, if the system of equations 3x + 5y + λz = 3, 7x + 11y − 9z = 2, 97x + 155y − 189z = μ has infinitely many solutions, then μ + 2λ equals:
Using the condition for infinite solutions in the system of linear equations, μ + 2λ is found to equal 25.
The coefficient of x⁷⁰ in x²(1 + x)⁹⁸ + x³(1 + x)⁹⁷ + ... + x⁵⁴(1 + x)⁴⁶ is (99Cp) − (46Cq). A possible value of p + q is:
Summing the contributions of each term in the series expansion and simplifying, p + q is found to be 83.
If ∫ (2 − tanx)/(3 + tanx) dx = (1/2)(αx + loge|β sinx + γ cosx|) + C, then α + γ/β equals:
Using the method of partial fractions and substitution, α + γ/β is found to be 4.
A variable line passes through (3, 5) and intersects the positive axes at A and B. The minimum area of triangle OAB is:
By using calculus to minimize the area of triangle OAB, the minimum area is found to be 30.
If |cosθ cos(60° − θ) cos(60° + θ)| ≤ 1/8, θ ∈ [0, 2π], then the sum of all θ where cos3θ attains its maximum value is:
By solving the trigonometric inequality and analyzing the periodicity, the sum of θ is found to be 6π.
Let OA = 2a, OB = 6a + 5b, and OC = 3b. If the area of the parallelogram with sides OA and OC is 15 sq. units, the area of quadrilateral OABC is:
By calculating the areas of the parallelogram and using vector cross products for the diagonals, the area of quadrilateral OABC is 35.
If the domain of the function f(x) = sin⁻¹((x−1)/(2x+3)) is R−(α, β), then 12αβ is equal to:
The domain of the function is determined by solving the inequality for the argument of sin⁻¹, and the value of 12αβ is found to be 36.
If the sum of the series 1/(1 · (1 + d)) + 1/((1 + d)(1 + 2d)) + ... + 1/((1 + 9d)(1 + 10d)) is equal to 5, then 50d is equal to:
Using partial fraction decomposition and simplifying, we find that the value of 50d is 5.
Let f(x) = ax³ + bx² + cx + 41 be such that f(1) = 40, f'(1) = 2, and f''(1) = 4. Then a² + b² + c² is equal to:
By solving the system of equations for the coefficients a, b, and c using the given conditions, we find that a² + b² + c² = 51.
Let a circle passing through (2, 0) have its center at the point (h, k). Let (xc, yc) be the point of intersection of the lines 3x + 5y = 1 and (2 + c)x + 5c²y = 1. If h = lim c→1 xc and k = lim c→1 yc, then the equation of the circle is:
Using the center and radius of the circle determined from the intersection points, the equation of the circle is 25x² + 25y² − 20x + 2y − 60 = 0.
The shortest distance between the lines (x−3)/4 = (y+7)/−11 = (z−1)/5 and (x−5)/3 = (y−9)/−6 = (z+2)/1 is:
The shortest distance between skew lines is calculated using the perpendicular vector, and the result is 187√563.
The frequency distribution of the age of students in a class of 40 students is given below:
| Age | No. of Students |
|---|---|
| 15 | 5 |
| 16 | 8 |
| 17 | 5 |
| 18 | 12 |
| 19 | x |
| 20 | y |
If the mean deviation about the median is 1.25, then 4x + 5y is equal to:
By solving for x and y using the mean deviation equation, we find that 4x + 5y = 44.
The solution of the differential equation (x² + y²)dx − 5xy dy = 0, y(1) = 0, is:
By solving the differential equation, the solution is found to be |x² − 4y²|⁵ = x².
Let three vectors a = αî + 4ĵ + 2k̂, b = 5î + 3ĵ + 4k̂, c = xî + yĵ + zk̂, form a triangle such that c = a − b and the area of the triangle is 5√6. If α is a positive real number, then |c|² is:
By calculating the area of the triangle using the cross product and solving for α, the value of |c|² is 14.
Let α, β be the roots of the equation x² + 2√2x − 1 = 0. The quadratic equation whose roots are α⁴ + β⁴ and 1/10(α⁶ + β⁶) is:
By using identities for powers of roots and calculating the sums, the quadratic equation is found to be x² − 195x + 9506 = 0.
Let f(x) = x² + 9, g(x) = x/(x − 9), and a = f(g(10)), b = g(f(3)). If e and ℓ denote the eccentricity and the length of the latus rectum of the ellipse x²/a² + y²/b² = 1, then 8e² + ℓ² is equal to:
By calculating a and b and using the formulas for eccentricity and the latus rectum, 8e² + ℓ² is found to be 8.
Let a, b, c denote the outcomes of three independent rolls of a fair tetrahedral die, whose four faces are marked 1, 2, 3, 4. If the probability that ax² + bx + c = 0 has all real roots is m/n, where gcd(m, n) = 1, then m + n is equal to:
By analyzing the discriminant condition for real roots and counting favorable outcomes, the probability is m/n = 3/16, and m + n = 19.
The sum of the square of the modulus of the elements in the set: {z = a + ib : a, b ∈ Z, z ∈ C, |z − 1| ≤ 1, |z − 5| ≤ |z − 5i|} is:
By analyzing the geometric constraints of the complex number set and evaluating the square of the modulus, the sum is 9.
Let the set of all positive values of λ, for which the point of local minimum of the function:
f(x) = (1 + x(λ² − x²)) satisfies x² + x + 2 / x² + 5x + 6 < 0, be (α, β). Then α² + β² is equal to:
Solving for λ and checking conditions for the local minimum, α² + β² is found to be 39.
Let the following limit be equal to π/k, where k is an integer. Then k² is equal to:
Given:
lim n→∞ [n√(n⁴ + 1) − 2n(n² + 1)√(n⁴ + 1) + n√(n⁴ + 16) + ...]
The integral approximation and series expansions yield k² = 32.
The remainder when 4282024 is divided by 21 is:
Simplifying the large power modulo calculations, the remainder is found to be 1.
Let f : (0, π) → R be a function given by:
f(x) =
8/7 * tan(8x) / tan(7x), 0 < x < π/2 a − 8, x = π/2 (1 + |cot(x)|) * (b/a) * tan(|x|), π/2 < x < π
where a, b ∈ Z. If f is continuous at x = π/2, find a² + b².
To maintain continuity at x = π/2, solving for the values of a and b, a² + b² is found to be 81.
Let A be a non-singular matrix of order 3. If:
det(3adj(2adj((detA)A))) = 3⁻¹³ · 2⁻¹⁰, and:
det(3adj(2A)) = 2^m · 3^n, then |3m + 2n| is equal to:
By analyzing the determinant properties and solving the equation, |3m + 2n| is found to be 14.
Let the center of a circle, passing through the points (0, 0), (1, 0), and touching the circle x² + y² = 9, be (h, k). Then for all possible values of the coordinates of the center (h, k), 4(h² + k²) is equal to:
Using the geometric properties of the circle and applying the conditions of tangency, the value of 4(h² + k²) is found to be 9.
If a function f satisfies f(m + n) = f(m) + f(n) for all m, n ∈ N, and f(1) = 1, then the largest natural number λ such that:
Σk=12022 f(λ + k) ≤ (2022)², is equal to:
By solving for λ based on the functional equation and summation condition, λ is found to be 1010.
Let A = {2, 3, 6, 7} and B = {4, 5, 6, 8}. Let R be a relation defined on A × B by:
(a₁, b₁) R (a₂, b₂) ⇔ a₁ + a₂ = b₁ + b₂. Then the number of elements in R is:
By checking the pairs that satisfy the given condition for the relation, the total number of elements in R is 25.
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