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JEE Main 2024 Question Paper with Answer Key PDF for Mathematics (January 29 - Shift 2)

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JEE Main 2024 Jan 29 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.

JEE Main 2024 Mathematics Question Paper with Answer Key PDF

JEE Main 2024 Question Paper PDF	JEE Main 2024 Answer Key PDF
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Question 1:

Let

A =

2	1	2
6	2	11
3	3	2

P =

1	2	0
5	0	2
7	1	5

The sum of the prime factors of |P⁻¹ A P - 2I| is:

Correct Answer: (1) 26 Solution:

We form P⁻¹ A P (similarity transform) and then subtract 2I from it. Taking the determinant: |P⁻¹ A P - 2I|=|P⁻¹ (A -2I) P|=|A -2I| (since determinant of P⁻¹ P=1). Next, compute A - 2I and find its determinant. Factor the result.

Those prime factors sum to 26.

Question 2:

The number of ways of arranging 8 identical books into 4 identical shelves (shelves can be empty) is:

Correct Answer: (4) 15 Solution:

Distributing n identical objects into k identical boxes is a "partition of n into at most k parts." We want partitions of 8 into ≤4 parts. Alternatively, use the "stars & bars" with identical boxes approach carefully.

The final count is 15 distinct distributions.

Question 3:

Let P(3,2,3), Q(4,6,2), R(7,3,2). The angle ∠QPR=?

π/6
cos⁻¹(7/18)
cos⁻¹(1/18)
π/3

Correct Answer: (4) π/3 Solution:

Compute vectors PQ=Q−P=(1,4,−1) and PR=R−P=(4,1,−1). Dot product PQ·PR = |PQ||PR| cos(∠QPR).

Evaluate: PQ·PR =1×4 +4×1 +(−1)(−1)=4+4+1=9. |PQ|=√(1²+4²+ (−1)²)=√(1+16+1)=√18=3√2, |PR|=√(4²+1²+ (−1)²)=√(16+1+1)=√18=3√2. cos(∠)=9/(3√2×3√2)=9/18=1/2 => ∠=π/3.

Question 4:

If the mean, variance of 5 observations are 24/5, 194/25, and the mean of the first 4 is 7/2, then the variance of first 4 is:

4/5
77/12
5/4
105/4

Correct Answer: (3) 5/4 Solution:

Let the 5 observations be x₁, x₂, x₃, x₄, x₅. Mean(5 obs)= (x₁+...+x₅)/5=24/5 => sum=24. Var(5 obs)=194/25 => use formula ∑xᵢ²−(∑xᵢ)²/5= 194/25×5= 194×(5/25)= 194/5 if needed. Also mean(4)= (x₁+ x₂+ x₃+ x₄)/4= 7/2 => sum(4)=14 => so x₅=24−14=10.

Then compute the variance of first 4 => 5/4.

Question 5:

f(x)=2x+3 x^(2/3), x in R. It has how many local maxima and minima?

One local min, no local max
One local max, no local min
One local max & one local min
Two local max & one local min

Correct Answer: (3) One local max & one local min Solution:

Differentiate: f'(x)=2+ (3× (2/3))×( x^(−1/3)) => etc. Solve f'(x)=0. Then check second derivative or sign changes to identify one maximum at x=−1 and one minimum at x=0.

Question 6:

z=2− i(2 tan(5π/8)), find modulus r and amplitude θ => (r,θ). The answer is (2 sec(3π/8), 3π/8).

(2 sec(3π/8), 3π/8)
(2 sec(3π/8), 5π/8)
(2 sec(5π/8), 3π/8)
(2 sec(11π/8), 11π/8)

Correct Answer: (1) Solution:

Real part=2, Imag part=−2 tan(5π/8). r=√(2² + [−2 tan(5π/8)]²)=2√(1+ tan²(5π/8))=2 sec(5π/8) or manipulated => 2 sec(3π/8).

Meanwhile angle is 3π/8 from quadrant analysis.

Question 7:

Equation 3 cos(2x) + cos³(2x)/(cos⁶x − sin⁶x)= x³−x²+6. Solutions sum=? => −1

0
1
−1
3

Correct Answer: (3) −1 Solution:

Detailed trigonometric simplifications lead to a cubic equation in x whose sum of roots (by Vieta’s) is −1.

Question 8:

OA=a, OB=12a+4b, OC=b, O is origin, S is parallelogram with sides OA,OC. The ratio area(OABC)/area(S)=?

Correct Answer: (4) 8 Solution:

S is parallelogram spanned by a,b => area(S)=|a×b|. Quadrilateral OABC includes vectors OA, AB=OB−OA= (12a+4b)−a=11a+4b, etc.

Carefully we find area=8|a×b| => ratio=8.

Question 9:

log a, log b, log c in A.P. and (log a−log 2b), (log 2b−log 3c), (log 3c−log a) in A.P. => ratio a:b:c=? => 9:6:4

9:6:4
16:4:1
25:10:4
6:3:2

Correct Answer: (1) 9:6:4 Solution:

"log a, log b, log c in A.P." => 2 log b=log a+ log c => b²= ac. Also "log a− log 2b, log 2b− log 3c, log 3c− log a in A.P." => more constraints. Solve => a:b:c=9:6:4.

Question 10:

∫ ( sin^(3/2)(x) + cos^(3/2)(x) ) / √( sin³(x) cos³(x) sin(x−θ )) dx = A cosθ sin x − B sinθ cos x + C => AB=? => 8 csc(2θ)

4 csc(2θ)
4 secθ
2 secθ
8 csc(2θ)

Correct Answer: (4) 8 csc(2θ) Solution:

By intricate trig transformations, the integral simplifies to a result of the form A cosθ sin x − B sinθ cos x. Then we identify AB=8 csc(2θ).

Question 11:

The distance from (2,3) to line 2x−3y+28=0 measured parallel to line √3x−y+1=0 => ? => 4 + 6√3

4√2
6√3
3+4√2
4+6√3

Correct Answer: (4) 4+6√3 Solution:

Standard distance formula is perpendicular, but we want "parallel to √3x−y+1=0." We effectively project the standard offset along that direction. The final numeric is 4+6√3.

Question 12:

If sin(y/x)= ln|x| + α/2 solves x cos(y/x) dy/dx= y cos(y/x)+ x, with y(1)=π/3 => α²=? => 3

Correct Answer: (1) 3 Solution:

By separation: rearr. The given solution sin(y/x)= ln|x| + α/2 => apply boundary (1, π/3). sin( (π/3)/1 )=sin(π/3)=√3/2 => so √3/2= ln(1)+ α/2 => α/2=√3/2 => α=√3 => α²=3.

Question 13:

A G.P. with a₁=1/8, a₂≠a₁, each term=arithmetic mean of next two => we get ratio r=−2. Then S₂₀−S₁₈=? => −2^15

2^15
−2^18
2^18
−2^15

Correct Answer: (4) −2^15 Solution:

Condition 2aₙ = aₙ₊₁ + aₙ₊₂ => implies r=−2 for the G.P. Then a₁=1/8 => Sₙ= a₁ (rⁿ−1)/(r−1). S₂₀−S₁₈= a₁(r¹⁹+ r²⁰ + ...?). Actually simpler: S₂₀−S₁₈= a₁₉+ a₂₀ => terms #19,#20 => final= −2^15.

Question 14:

Let A= intersection(3x+2y=14, 5x−y=6), B= intersection(4x+3y=8, 6x+y=5). The distance from P(5,−2) to line AB=? => 6

13/2
8
5/2
6

Correct Answer: (4) 6 Solution:

Solve for A,B => A(2,4), B(0.5,2). Then eqn(AB)=? Next use point-line distance formula from P(5,−2). Numeric => 6.

Question 15:

x=m/n is solution of cos(2 sin⁻¹ x)=1/9 => x=2/3. Then the quadratic m x²−n x−m+n=0 => roots α,β => (α,β) on line => 5x+8y=9

3x+2y=2
5x−8y=−9
3x−2y=−2
5x+8y=9

Correct Answer: (4) 5x+8y=9 Solution:

cos(2 sin⁻¹ x)=1/9 => 2 sin⁻¹ x=? => x=2/3. Then that x used in the given quadratic => find α,β. Summation => they satisfy 5x+8y=9.

Question 16:

f(x)= x/(x²−6x−16). Then f'(x) shows f is decreasing in entire domain except asymptotes => (−∞,−2),(−2,8),(8,∞). The correct statement is it decreases in all intervals => (2) .

decreases in (−2,8), increases in (−∞,−2)∪(8,∞)
decreases in all intervals of domain
decreases in (−∞,−2) & increases in (8,∞)
increases in entire domain

Correct Answer: (2) Solution:

Differentiate f(x)= x/(x²−6x−16). f'(x) <0 for x∈ R\{−2,8}. So the function is decreasing in each piece of domain around vertical asymptotes x=−2,8.

Question 17:

y= ln((1−x²)/(1+x²)), x in (−1,1). At x=1/2, compute 225(y'−y'') => 736

Correct Answer: (4) 736 Solution:

y= ln( (1−x²)/(1+x²) ) => y'= derivative, y''= second derivative. Evaluate difference at x=1/2, then multiply by 225 => 736.

Question 18:

The smallest equivalence relation R on {1,2,3,4} s.t. {(1,2),(1,3)} in R => total #elements in R=? => 10

Correct Answer: (1) 10 Solution:

An equivalence relation must be reflexive, symmetric, transitive. Including (1,2) & (1,3) => (2,1),(3,1). Then transitivity lumps 1,2,3 in the same class => also 4 alone. Counting all pairs => 10.

Question 19:

An integer from 1..50. Probability multiple of at least one of 4,6,7 => 21/50

8/25
21/50
9/50
14/25

Correct Answer: (2) 21/50 Solution:

Use inclusion-exclusion: #multiples(4)=⌊50/4⌋=12, #multiples(6)=8, #multiples(7)=7. Overlaps: multiples(4&6)=12?6? => LCM=12 => #=4, etc. Probability => 21/50.

Question 20:

u is unit vector with angles π/2, π/3, 2π/3 vs p₁=(1/√2,0,1/√2), p₂=(0,1/√2,1/√2), p₃=(1/√2,1/√2,0). v=(1/√2)(1,1,1). Then |u−v|²=? => 5/2

11/2
5/2
9
7

Correct Answer: (2) 5/2 Solution:

Dot products define angles => solve for u. Then subtract v => compute square of magnitude. Result=5/2.

Question 21:

Let α, β be roots of x² − √6 x + 3 = 0 with Im(α) > Im(β). Let a,b be integers not divisible by 3 and n a natural number such that αⁿ/β + α⁹⁹ + α⁹⁸ = 3ⁿ(a + i b). Then n + a + b=?

Correct Answer: 49 Solution:

α,β= √3 e^(± i π/4). Summing conditions yields n + a + b=49.

Question 22:

Three distinct consecutive terms a,b,c of an A.P. give lines ax+by+c=0 concurrent at P. Q(α,β) s.t. system x+y+z=6, 2x+5y+αz=β, x+2y+3z=4 has infinitely many solutions => (PQ)²=?

Correct Answer: 113 Solution:

Determinant=0 => β=8, P(1,−2), Q(8,6); (PQ)²=113.

Question 23:

Point P(α,β) on y²=4x also lies on chord of x²=8y with midpoint (1,5/4). Then (α−28)(β−8)=?

Correct Answer: 192 Solution:

Using midpoint formula & substituting in the two parabolas yields P => product=192.

Question 24:

∫ from π/3 to π/6 of √(1− sin2x) dx= α+β√2+γ√3 => then 3α+4β−γ=?

Correct Answer: 6 Solution:

The integral simplifies to −1 + 2√2 − √3 => so α=−1, β=2, γ=−1 => 3α+4β−γ=6.

Question 25:

The area of {(x,y):0≤x≤3, 0≤y≤min(x²+2,2x+2)}=A => 12A=?

Correct Answer: 164 Solution:

Split region at x²+2=2x+2 => x²−2x=0 => x=0,2. Integrate piecewise => total => 12A=164.

Question 26:

Lines: (x−5/4)/1=(y−4)/1=(z−5)/3 and (x+8)/12=(y+2)/5=(z+11)/9 => M,N are points s.t. MN=shortest distance, O=origin => OM·ON=?

Correct Answer: 9 Solution:

Parametric forms => find M,N => dot(OM,ON)=9.

Question 27:

f(x)= sqrt( lim(r→x)[2r²(f(r²)−f(x))f(r)/(r²−x²) − r² e^( f(r)/r )] ), with f(1)=1. If f(a)=0 => e^a=? => 2

Correct Answer: 2 Solution:

Simplifying limit yields a functional condition => we find a => e^a=2.

Question 28:

64^(3^3232) mod 9 => remainder=?

Correct Answer: 1 Solution:

64≡1 mod 9 => 1^(any)=1 => remainder=1.

Question 29:

Set C={(x,y): x²−2y=2023, x,y in N}. Then ∑(x,y in C)(x+y)=?

Correct Answer: 46 Solution:

x²−2y=2023 => check natural x => x=45 => y= (45²−2023)/2=1 => sum=46.

Question 30:

45x+5y+3=0 => slope=27r₁+9r₂². Then limit as x→3 of ∫ from x to 3 [8t²/(3r₂x²−r₂x²−r₁x³−3x)] dt=? =>12

Correct Answer: 12 Solution:

After simplifying integrand & applying L'Hopital if needed => final=12.

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*The article might have information for the previous academic years, please refer the official website of the exam.

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