JEE(Main + Advanced) 2024-2025 Past Paper PDF

(1001CJA101372240003) Test Pattern CLASSROOM CONTACT PROGRAMME JEE (Main)...

(1001CJA101372240003) Test Pattern CLASSROOM CONTACT PROGRAMME JEE (Main) UNIT TEST (Academic Session : 2024 - 2025) 05-05-2024 JEE(Main + Advanced) : ENTHUSIAST COURSE [PHASE : III(A)] ANSWER KEY PAPER-1 (OPTIONAL) PART-1 : PHYSICS Q. 1 2 3 4 5 6 7 8 9 10 A. B B A A C C A B D B SECTION-I Q. 11 12 13 14 15 16 17 18 19 20 A. B C D D D B B A C A Q. 1 2 3 4 5 6 7 8 9 10 SECTION-II A. 50 3 14 5 3000 6 45 53 200 6 PART-2 : CHEMISTRY Q. 1 2 3 4 5 6 7 8 9 10 A. A A A B B A B C A D SECTION-I Q. 11 12 13 14 15 16 17 18 19 20 A. C C B A C C A B A D Q. 1 2 3 4 5 6 7 8 9 10 SECTION-II A. 2 17 62 4 3 4 3 4 5 3 PART-3 : MATHEMATICS Q. 1 2 3 4 5 6 7 8 9 10 A. A B D C D D A C D C SECTION-I Q. 11 12 13 14 15 16 17 18 19 20 A. C B C A D B B D B A Q. 1 2 3 4 5 6 7 8 9 10 SECTION-II A. 99 3 2 5 0 0 2 2 1 14 HINT – SHEET PART-1 : PHYSICS 2. Ans ( B ) SECTION-I Place another identical hemisphere on given one, so that flat faces of two coincide The total flux 1. Ans ( B ) linked with cross- section of bottom half (lower → 2 = kλ ^i − kλ ^j E hemisphere) is Q as charge enclosed by given R R 2ε0 hemisphere is zero. So flux linked with flat face → 1 = − kλ ^i − kλ ^j E −Q R R of hemisphere is 2ε0 → 3 = kλ ^i + kλ ^j E 7. Ans ( A ) R R 12 : 00 – 3 : 25 = 8 : 35 → net = kλ ^i − kλ ^j E 8. Ans ( B ) R R f0 mθ = = 10 ; f0 = 10fe fe f0 + fe = 44 ; 11fe = 44 ; fe = 4 f0 = 40 1001CJA101372240003 HS-1/8  Target : JEE (Main + Advanced) 2025/05-05-2024/Paper-1 9. Ans ( D ) 15. Ans ( D ) 1 μL 1 1 Rate of change of momentum is proportional to =( − 1) ( − ) f μM R1 R2 external forces acting on the system. The total for plane surface, R → ∞ momentum of whole system remain constant when no external force is acted upon it. Internal 1 1 1 forces can change the kinetic energy of the P→ = × ⇒ f = 2R f 2 R system. Q → f = 4R 16. Ans ( B ) 2u⊥ 1 3 2 T= , which is same for both R→ = ( − 1) ( ) ⇒ f = −2R a⊥ f2 4 R 17. Ans ( B ) 1 3 3 2 S→ = ( × − 1) − ⇒ f = −4R dx f2 2 4 R = 5e−t dt 10. Ans ( B ) x ∞ Given u + v = 100 cm ∫ dx = ∫ 5e−t dt 0 0 ∞ Using Lens formula with sign convection 5e−t x=[ ] = −5 [e−∞ − e0 ] = 5 −1 0 1 1 1 1 1 1 + u = and + = 18. Ans ( A ) (100 − u) f (60 − u) (u + 40) f 2TVb = 3TVa equating them, u = 30 cm Va = 2 Vb 3 Now putting in any of the equations 19. Ans ( C ) 1 1 1 1 1 + = ⇒ = 70 30 f f 21 cm P = 5 m–1 12. Ans ( C ) F sin 53∘ = ma F cos 53∘ = mg µ1 θ 1 = µ2 θ 2 4 a 4g = ⇒a= 2µ = 3µ θ 3 g 3 2 20. Ans ( A ) θ= 3 Let at any time, their velocities are v1 and v2 14. Ans ( D ) respectively. Then v1 = v2 cos θ For tmin, →vM,w should be perpendicular to flow 336 a1 = a2 cos θ , if released from rest t= = 336 sec. 1 HS-2/8 1001CJA101372240003  Enthusiast Course/Phase-III(A)/05-05-2024/Paper-1 PART-1 : PHYSICS 7. Ans ( 45 ) SECTION-II T=8 1. Ans ( 50 ) 2uy =8 As image is virtual, so magnification is g uy = 40 positive. f ux(5 – 3) = 80 Thus, m = − (−u) − f ⇒ −u − f = −nf ux = 40 uy ⇒ u + f = nf tan θ = =1 ux ⇒ u = (n − 1)f θ = 45° u = (6 – 1)10 = 50 cm 8. Ans ( 53 ) 2. Ans ( 3 ) To see object at less than 100 cm, let man used π a lens of focal length of f. θ1 = − θ2 2 π θ1 + θ2 = Lens will form an image at 100 cm of an object 2 snell's law at 25cm to see the object clear. 3 sin θ 1 = 4 sin θ 2 v = – 100 cm and u = – 25 cm π 3 sin θ 1 = 4 sin ( − θ1 ) 2 1 1 1 − = tan θ 1 = 4 , θ 1 = 53° v u f 3 1 − 1 = 1 9. Ans ( 200 ) −100 (−25) f −1 + 4 1 1 3 = ; = 100 f f 100 100 Power of lens = = 3D f 6. Ans ( 6 ) Number of electric field lines are drawn in proportion to charge magnitudes. Resultant force on pulley = T √ 2 = mg√2 = 200 Newton 1001CJA101372240003 HS-3/8  Target : JEE (Main + Advanced) 2025/05-05-2024/Paper-1 10. Ans ( 6 ) 17. Ans ( A ) 18. Ans ( B ) (A) PCl5 + H2O → H3PO4 (B) 4XeF4 + 8H2O → 2Xe + 2XeO3 + 16HF + 1O2 (C) BCl3 + H2O → H3BO3 T = 6a1....(1) (D) B2H6 with N(CH3)3 gives symmetrical cleavage Mg – 2T = Ma2........(2) PART-2 : CHEMISTRY SECTION-II – 2Ta2 + Ta1 = 0 ⇒ a1 = 2a2.......(3) 1. Ans ( 2 ) Now 0.4 × 10 × 2 = 8N In truncated octahedron, hexagonal faces = 8 fr = 8N = 2a1 In truncated tetrahedron, hexagonal faces = 4 8 a1 = 4m/s2 ∴ Ratio = 4 =2 a1 T = 24 N & a2 = = 2m/s2 2. Ans ( 17 ) 2 Now Mg – 48 = M × 2 HCP ⇒ z = 6, TV = 12 M(g – 2) = 48 M = 48 g M = 6kg ⇒ 48 g of A has NA atoms 6 × 1023 PART-2 : CHEMISTRY ⇒ 24 × 10–6 g A = × 24 × 10–6 48 SECTION-I = 3 × 1017 atoms 1. Ans ( A ) Total THV in 24 × 10–6 g of A 32 × 5 × 5 = 250 × M × 1 M = 3.2 = 2 × 3 × 1017 = 6 × 1017 4. Ans ( B ) 3. Ans ( 62 ) P : 3,6; Q : 5; R : 4; S : 2, 5 1 1 5. Ans ( B ) i=α ( − 1) + 1 = 0.8 ( − 1) + 1 = 0.4 n 4 Solution exhibits ideal behaviour if it follows i × Kf × WB × 1000 Raoult’s law MB = ΔTf × WA PT = XA PA0 + XB PB0 0.4 × 1.86 × 2.5 × 1000 0.4(80) + 0.6 × 120 = 32 + 72 = 104 MB = = 62 0.3 × 100 PT = 104 mmHg 4. Ans ( 4 ) 100 < 104 ⇒ solution shows – ve deviation i, ii, iii, v 7. Ans ( B ) a r+ + r – = √ 3 = 2 HS-4/8 1001CJA101372240003  Enthusiast Course/Phase-III(A)/05-05-2024/Paper-1 9. Ans ( 5 ) 3. Ans ( D ) N2O5(s) = NO+2 NO−3 0 ≤ 3cos – 1a ≤ π (sp) (sp2 ) π 1 0 ⩽ cos−1 a ⩽ ⇒ ⩽a⩽1 3 2 PCl5(s) = P Cl+4 P Cl−6 3 (sp ) (sp3 d 2 ) 4. Ans ( C ) I2 Cl6 (l) = ICl−4 ICl+2 f'(x) = 3x2 + 2x + 3 + cosx (sp3 d 2 ) (sp3 ) D = 4 – 36 = – 32 XeF6 (s) = XeF5+ + F − minimum value of 3x2 + 2x + 3 = 32 3 2 (sp d ) 12 f'(x) > 0 function is.................... x = NO+2 , NO−3 , ICl−4 , ICl+2 = 4 so on to and one one. y = P Cl−6 = 1 5. Ans ( D ) 1 x+y=5 sin−1 (1 − ) (2 + x2 ) π PART-3 : MATHEMATICS y = sin – 1(1) when x → ∞ = 2 1 π SECTION-I When x = 0 ⇒ y = sin−1 ( ) = 2 6 1. Ans ( A ) 6. Ans ( D ) y = x2 – 7x + 10 f:R→R ⇒ x = y2 – 7y + 10 f(x) = y = 2x2 − 5x + 3 8x2 + 9x + 11 ⇒ y2 – 7y + 10 – x = 0 D = (9)2 – 4 × 8 × 11 < 0 7 ± √49 − 4(10 − x) 2x(x − 1) − 3(x − 1) ⇒y= 2x2 − 2x − 3x + 3 y= = 2 8x2 + 9x + 11 8x2 + 9x + 11 7 ± √9 + 4x (x − 1)(2x − 3) ⇒y= y= 2 8x2 + 9x + 11 –1 9 7 at x = 1, f(1) = 0 ƒ : [− , ∞ ) → [ , ∞ ) ⎫ 4 2 ⎬ many one at x = 32 , f ( 32 ) = 0 ⎭ 7 + √9 + 4x ⇒ f −1 (x) = 2x2 − 5x + 3 2 y= 2. Ans ( B ) 8x2 + 9x + 11 1 ⇒ 8yx2 + 9xy + 11y = 2x2 – 5x + 3 [ 3f(x) + f ( ) = x] × 3 x ⇒ (8y – 2)x2 + (9y + 5)x + 11y – 3 = 0 1 1 1 3f ( ) + f(x) = ⇒ D ≥ 0 coeff. of x2 ≠ 0 ⇒ 8y – 2 ≠ 0 ⇒ y ≠ x x 4 1 ⇒ (9y + 5)2 – 4(8y – 2)(11y – 3) ≥ 0 8f(x) = 3x − x ⇒ 81y2 + 25 + 90y – 4(86y2 – 24y – 22y + 6) ≥ 0 1 26 8f(3) = 9 − = ⇒ – 271y2 + 274y + 1 ≥ 0 3 3 13 ⇒ 271y2 – 274y – 1 ≤ 0 ⇒ f(3) = 12 y does not belong to R Range ≠ R (codomain) ⇒ Not onto 1001CJA101372240003 HS-5/8  Target : JEE (Main + Advanced) 2025/05-05-2024/Paper-1 7. Ans ( A ) 10. Ans ( C ) x−3 f (x) − 3 x+1 −3 fof (x) = = x−3 f (x) + 1 +1 x+1 x+3 fof (x) = − x−1 x+3 − x−1 − 3 fofof (x) = =x Df → (−∞, sec 2] ∪ [1, ∞) x+3 − x−1 + 1 11. Ans ( C ) 8. Ans ( C ) [x] 1 2 4 8 {x} = ⇒ {x} = 0, , ⇒ x = 0, , 3 3 3 3 3 9. Ans ( D ) 3x (A) Range of f (x) = 1 + 9x2 Put 3x = tan θ tan θ 1 ∴ f (x) = = sin 2θ 2 1 + tan θ 2 1 1 ∴ f (x) ∈ [− , ] 2 2 12. Ans ( B ) ∴ Number of integers in the range of f(x) = 1 cos ec−1 x f (x) = (B) Range of f (x) = √ x − a + √b − x √ {x} (where a < b) Domain ∈ (−∞, −1] ∪ [1, ∞) {x} ≠ 0 so x ≠ integers is [√ (b − a), √(b − a)√2] 13. Ans ( C ) ∴ Range of f (x) = √x − 1 + √x − 9 will be [√ 8, √8 ⋅ √2] = [2√2, 4] Let tan – 1x = θ ∴ Number of integers in the range of f(x) = 2 1 + x2 1 (C) f (x) = sin−1 ( ) sin(cot−1 ) 2x √ 1 + x2 ∣ 1 + x2 ∣ ∣ ⩽ 1 ∣ 1 ∣ 2x ∣ Let cot−1 =α ⇒ x2 + 1 ⩽ 2|x| √ 1 + x2 ⇒ (|x| – 1)2 < 0 x2 + 1 sin α = √ x2 + 2 ⇒ |x| – 1 = 0 ∴ x = ±1 π π ∴ Range of f(x) is {− , } 2 2 ∴ Number of integers in the range of f(x) = 0 (D) f(x) = [{sin x}] + {[cosx]} = 0 + 0 = 0 {As {x} ∈ [0,1) ∴ [{x}] = 0 and {n} = 0 ∀ x ∈ I} ∴ Number of integers in the range of f(x) = 1 HS-6/8 1001CJA101372240003  Enthusiast Course/Phase-III(A)/05-05-2024/Paper-1 14. Ans ( A ) 19. Ans ( B ) 2sin – 12x + cos – 13y + sec – 1z = 3p 1 + tan π4. tan(2cot−1 3) π ⇒ 2sin – 12x = p, cos – 13y = p, sec – 1z = π tan 4 − tan(2cot−1 3) 1 3 ⇒x= , y = − 1 , z = –1 1+ 4 =7 2 3 3 5 1− 4 ∴x+y+z=− 6 20. Ans ( A ) 15. Ans ( D ) α + β = – p, α β = q Statement-1 : is log10 [x] < 0 ( α – β )2 ⇒ p2 – 4q = 1 ∵ [x] = 1, 2, 3,........ ∴ p2 + 4q2 = 1 + 4q + 4q2 = (1 + 2q)2 ⇒ log10 [x] > 0 PART-3 : MATHEMATICS ⇒ Statement-1 is false. SECTION-II 1. Ans ( 99 ) and obviously Statement-2 is correct. 16. Ans ( B ) 2x2 + 6x – 5 = 0 x π sin−1 ( ) = − cos−1 x α + β = −3 2 6 ⎡ ⎢ x π ⎣ αβ = − 52 = sin( − cos−1 x) 2 6 α2 β2 ( α3 + β 3 ) x π π ( + ) = = sin cos(cos−1 x) − cos sin(cos−1 x) β α αβ 2 6 6 x x √ (α + β)3 − 3αβ (α + β) = − 1 − x2 = 2 2 αβ x ± 1 but x = – 1 is rejected. (−3)3 − 3 (− 52 ) (−3) ( −27 − 45 2 ) = = 5 (− ) ( − 52 ) 17. Ans ( B ) 2 tan – 1 tan(7) = 7 – 2 π 99 = = 19.80 cot – 1 cot(7) = 7 – 2 π 5 18. Ans ( D ) 18(tan – 1x)2 – 6 π tan – 1x – 3 π tan – 1x + π 2 = 0 6tan – 1x(3tan – 1x – π ) – π (3tan – 1x – π ) = 0 π π tan – 1x = and 6 3 1 x = √3 and √3 ∴ αβ = 1 4 Now, log √ 3 (8 + 1) = log √ 3( √ 3) = 4 1001CJA101372240003 HS-7/8  Target : JEE (Main + Advanced) 2025/05-05-2024/Paper-1 2. Ans ( 3 ) 8. Ans ( 2 ) x2 + x + 7 P (xy) + P ( yx ) cos(log 1 + cos(log y Let, y = xy ) x ) x+2 = P (x) P (y) 1 1 ⇒ xy + 2y = x2 + x + 7. cos(log x ) cos(log y ) ⇒ x2 + x(1 – y) + (7 – 2y) = 0 cos(log x + log y) + cos(log y − log x) = cos(log x) cos(log y) ∴∀x∈R⇒D≥0 ⇒ D = (1 – y)2 – 4(7 – 2y) ≥ 0 2 cos(log x) cos(log y) = =2 cos(log x) cos(log y) ⇒ y2 + 6y – 27 ≥ 0 ⇒ (y + 9) (y – 3) ≥ 0 9. Ans ( 1 ) g(x) = x2 + x – 1 g (f(x)) = 4x2 – 10x + 5 y ∈ ( – ∞ , – 9] ∪ [3, ∞ ) ∴ Smallest positive value of ƒ(x) is 3 Ans. = (2x – 2)2 + (2 – 2x) – 1 3. Ans ( 2 ) = (2 – 2x)2 + (2 – 2x) – 1 f(2) < 0 ⇒ f(x) = 2 – 2x 4. Ans ( 5 ) 5 −1 f( ) = Given equation ⇒ x2 – 3x – 4 = 0 4 2 ⇒ x = – 1, 4 ⇒ N = 5 10. Ans ( 14 ) 5. Ans ( 0 ) x3 – 6x2 + 2 = (x – α ) (x – β ) (x – γ ) (x – 2) (x – 5) > 0 Put x = 2 ⇒ (2 – α ) (2 – β ) (2 – γ ) = – 14 x ∈ ( – ∞ , 2) ∪ (5, ∞ ).... (I) (3 – x) (x – 4) > 0 ⇒ ( α – 2) ( β – 2) ( γ – 2) = 14 ⇒ x + (3, 4)....(II) I ∩ II ⇒ ϕ No solution 6. Ans ( 0 ) Draw f(x) and f – 1(x), they have zero point of intersection. 7. Ans ( 2 ) minimum value occure at x = 2 HS-8/8 1001CJA101372240003