
JEE Main 2024 Jan 27 Shift 2 Mathematics Question Paper with Solution pdf is available for download here. Students found easy and hard. carried the highest weightage and overall difficulty level was moderate.
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Considering only the principal values of inverse trigonometric functions, the number of positive real values of x satisfying arctan(x) + arctan(2x) = π/4 is:
Applying the tangent addition formula transforms the equation to 2x = (1 − x)/(1 + x). Solving the quadratic equation reveals only one positive solution: x = (−3 + √17)/4.
Consider the function f: (0,2) → ℝ defined by f(x) = x² + 2/x, and the function g(x) by min{f(t)} for 0 < t ≤ x, 0 < x ≤ 1 and g(x) = (3/2 + x) for 1 < x < 2. Which statement about g is correct?
g(x) is continuous at x = 1 since the limits from both sides match g(1). However, the derivatives from the left and right do not coincide, making g(x) non-differentiable at x = 1.
Let R be the region between lines 3x − y + 1 = 0 and x + 2y − 5 = 0 containing the origin. Find values of a for which (a², a + 1) lie in R:
Solving the inequalities from the line equations for (a², a + 1) shows that a must be in the intervals (−3, 0) or (−1/3, 1).
The 20th term from the end of the progression 20, 19.25, 18.5, 17.75, ..., −129.25 is:
Recognizing the sequence as an arithmetic progression with a common difference of −0.75, calculating the 20th term from the end yields −115.
Let f: ℝ − {−1/2} → ℝ and g: ℝ − {−5/2} → ℝ be defined as f(x) = (2x + 3)/(2x + 1) and g(x) = |x| + 1/(2x + 5). What is the domain of f ∘ g?
The composite function f ∘ g is undefined at x = −5/2 because g(x) would result in a division by zero. Thus, the domain excludes x = −5/2.
For 0 < a < 1, find the value of the integral ∫₀^π dx / (1 − 2a cos(x) + a²):
By completing the square in the denominator and using a standard trigonometric integral, the result simplifies to π / (1 − a²).
Let g(x) = 3f(x/3) + f(3 − x) where fʺ(x) > 0 for all x in (0,3). If g is decreasing in (0, α) and increasing in (α,3), what is 8α?
Analyzing g(x) with fʺ(x) > 0 reveals that α = 9/4. Multiplying by 8 gives 8α = 18.
If lim(x→0) [(3 + α sin(x) + β cos(x) + log(1 − x)) / (3 tan²(x))] = 1/3, then 2α − β is equal to:
Expanding the numerator and denominator using Taylor series around x = 0 and equating coefficients results in 2α − β = 5.
If α and β are the roots of x² − x − 1 = 0, and Sₙ = 2023αⁿ + 2024βⁿ, then:
Utilizing the recurrence relation from the quadratic equation’s roots, it follows that S₁₂ equals S₁₁ plus S₁₀.
Let A and B be two finite sets with m and n elements, respectively. If the total subsets of set A are 56 more than B’s subsets, then the distance of the point P(m, n) from Q(−2, −3) is:
Solving 2^m = 2^n + 56 gives m = 6 and n = 3. The distance between P(6, 3) and Q(−2, −3) is calculated as 10.
The values of α for which the determinant |1 3 2; α+3 2 1; 1 1 3| = 0 lie in the interval:
Expanding the determinant and solving for α yields the interval (−3, 0).
An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability that the first draw gives all white balls and the second draw gives all black balls is:
The probability of drawing 4 white balls in the first draw is 15/1365. After removing 4 white balls, there are 9 black balls left.
The probability of drawing 4 black balls in the second draw is 126/330. The required probability is the product of these probabilities, giving 3/715.
The integral ∫(x⁸ − x²) dx / ((x¹² + 3x⁶ + 1) arctan(x³ + 1/x³)) is equal to:
Using the substitution u = arctan(x³ + 1/x³), we simplify the integral to du/u, which evaluates to log(u) + C.
The final answer is in the given form involving arctan(x³ + 1/x³)^(1/3).
If 2tan²(θ) − 5sec(θ) = 1 has exactly 7 solutions in the interval [0, nπ/2] for the least value of n ∈ ℕ, then Σ(k=1 to n) [k / 2k] is equal to:
Setting sec(θ) = t, we solve the quadratic equation to find that n = 13.
Using the formula for the sum, Σ = (1/213)*(214 − 15).
The position vectors of the vertices A, B, and C of a triangle are A = 2i − 3j + 3k, B = 2i + 2j + 3k, and C = −i + j + 3k respectively. If ℓ is the length of the angle bisector AD of ∠BAC, then 2ℓ² equals:
Using the position vectors and midpoint formula, we find the length of AD and calculate 2ℓ² as 45.
If y = y(x) is the solution curve of the differential equation (x² − 4) dy − (y² − 3y) dx = 0, x > 2, y(4) = 3/2, and the slope of the curve is never zero, then y(10) equals:
Separating variables and solving the differential equation, we integrate both sides and substitute x = 10 to find y(10) = 3/(1 + 8^(1/4)).
If e₁ is the eccentricity of the hyperbola x²/16 − y²/9 = 1 and e₂ is the eccentricity of the ellipse x²/a² + y²/b² = 1, which passes through the foci of the hyperbola and satisfies e₁e₂ = 1, then the length of the chord of the ellipse parallel to the x-axis and passing through (0,2) is:
For the ellipse x²/25 + y²/9 = 1, using the formula for the length of the chord parallel to the x-axis, we find the answer as 10√5/3.
Let α = (4!)! / (4!)³! and β = (5!)! / (5!)⁴!. Then:
Both α and β represent valid combinatorial expressions for arranging groups, making both natural numbers.
Let the position vectors of vertices A, B, and C of a triangle be 2i + 2j + k, i + 2j + 2k, and 2i + j + 2k respectively. Let ℓ₁, ℓ₂, and ℓ₃ be lengths of perpendiculars from the orthocenter to sides AB, BC, and CA. Then ℓ₁² + ℓ₂² + ℓ₃² equals:
Using midpoint formula and perpendicular distance calculations, we find ℓ₁² + ℓ₂² + ℓ₃² = 1/2.
The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking, it was found that an observation was read as 10 instead of 12. If μ and σ² denote the mean and variance of the correct observations, then 15(μ + μ² + σ²) is equal to:
Correcting the observation and recalculating mean and variance, we find 15(μ + μ² + σ²) = 2521.
The values of α for which lines 2x − y + 3 = 0, 6x + 3y + 1 = 0, and ax + 2y − 2 = 0 do not form a triangle are:
For lines not to form a triangle, they must be concurrent or parallel. Solving for α, we get (−3, 0).
If the area of the region {(x, y) : 0 ≤ y ≤ min(2x, 6x − x²)} is A, then 12A is equal to:
We calculate the area by integrating the expression and find A. Then, 12A is calculated to be 304.
Let Λ be a 2×2 real matrix and I be the identity matrix of order 2. If the roots of the equation |Λ − xI| = 0 are −1 and 3, then the sum of the diagonal elements of the matrix Λ² is:
Using the trace and determinant properties of Λ and calculating Λ², we find the sum of the diagonal elements as 10.
If the sum of squares of all real values of α, for which the lines 2x − y + 3 = 0, 6x + 3y + 1 = 0, and ax + 2y − 2 = 0 do not form a triangle is p, then the greatest integer less than or equal to p is:
Solving for the concurrent or parallel condition of these lines, we find p and the integer part is 32.
The coefficient of x^2012 in the expansion of (1 − x)^2008(1 + x + x²)^2007 is:
Analyzing the terms in the expansion, we see there is no term for x^2012, hence the coefficient is 0.
If the solution curve of the differential equation dy/dx = (x + y − 2)/(x − y) passes through the point (2, 1), the value of y(10) equals:
Solving the differential equation and substituting x = 10, we find y(10) as 3/(1 + (8)^(1/4)).
Let f(x) = ∫₀ˣ g(t) log((1 − t)/(1 + t)) dt, where g is a continuous odd function. If ∫ from −π/2 to π/2 of [f(x) x² cos(x)/(1 + e^x)] dx = π/2 − α, then α is equal to:
Using the properties of odd functions and simplifying the integral, we find that α = 2.
Consider a circle (x − α)² + (y − β)² = 50, where α, β > 0. If the circle touches the line y + x = 0 at point P, whose distance from the origin is 4√2, then (α + β)² is equal to:
Using the distance and tangency conditions, we find that (α + β)² = 100.
The lines (x − 2)/1 = (y − 1)/(−1) = (z − 7)/8 and (x + 3)/4 = (y + 2)/3 = (z + 2)/1 intersect at the point P. If the distance of P from the line (x + 1)/2 = (y − 1)/3 = (z − 1)/1 is ℓ, then 14ℓ² is equal to:
Calculating the intersection point and using the distance formula, we find 14ℓ² = 108.
Let the complex numbers α and 1/α lie on the circles |z − z₀| = 2 and |z − z₀| = 4 respectively, where z₀ = 1 + i. Then, the value of 100|α|² is:
Since α and 1/α lie on concentric circles with known radii, |α| = 2. Thus, 100|α|² = 20.
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