JEE (Main)-2025 (Online) Phase-1 Past Paper PDF

22/01/2025 Morning Memory Based Answers & Solutions Time : 3 hrs. for M.M. : 300 JEE (Main)-2025 (Online) Phase-1...

22/01/2025 Morning Memory Based Answers & Solutions Time : 3 hrs. for M.M. : 300 JEE (Main)-2025 (Online) Phase-1 (Physics, Chemistry and Mathematics) IMPORTANT INSTRUCTIONS: (1) The test is of 3 hours duration. (2) This test paper consists of 75 questions. Each subject (PCM) has 25 questions. The maximum marks are 300. (3) This question paper contains Three Parts. Part-A is Physics, Part-B is Chemistry and Part-C is Mathematics. Each part has only two sections: Section-A and Section-B. (4) Section - A : Attempt all questions. (5) Section - B : Attempt all questions. (6) Section - A (01 – 20) contains 20 multiple choice questions which have only one correct answer. Each question carries +4 marks for correct answer and –1 mark for wrong answer. (7) Section - B (21 – 25) contains 5 Numerical value based questions. The answer to each question should be rounded off to the nearest integer. Each question carries +4 marks for correct answer and –1 mark for wrong answer. JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 1  JEE | NEET | Class 8 - 10 Download eSaral App PHYSICS SECTION - A 3. Find out equivalent capacitance for the situation show Multiple Choice Questions: This section contains 20 in figure. multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct. Choose the correct answer: B 1. Find the dimensions of 0 (1) [AL] (2) [AL–1] (3) [MAL] (4) [MALT–1] Answer (2) B i A0  K1K2 + K 2K3 + K3K1  Sol.  B  dl = 0i  0 =  [AL−1] (1) Ceq =   l d  K1 + K 2  2. Solid sphere of mass M, radius R exerts force F on a A0  2K1K 2 + K 2K3 + K3K1  (2) Ceq =   point mass. Now a concentric spherical mass M is d  2(K1 + K 2 )  7 removed. What is new force? A0  K1K2 + K 2K3 + K3K1  (3) Ceq =   d  2(K1 + K2 )  F 6 (1) (2) F 7 7 A0  K1K2 + K 2K3 + K3K1  (4) Ceq =   5F 3F 2d  (K1 + K 2 )  (3) (4) 7 7 Answer (2) Answer (2) Sol. Here, 2  A0K1 A0K1 C1 = = 2d d Sol. A0K 2 C2 = d A0K3 F= GMm C3 = 2d r2 Now C1 and C2 are in series  M GM − m 1 1 1 = + F =  7 So, 2 C1 C1 C2 r 1 d 1 1  d (K1 + K 2 ) 6  =  + =  F = F C1 A0K  K1 K 2  A0K1K 2 7 JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 2  JEE | NEET | Class 8 - 10 Download eSaral App A0K1K 2 5. Find the radius of curvature of the common surface of  C1 = d (K1 + K 2 ) two bubbles (R1 > R2) Now C1 is parallel to C3 A0  K1K 2 K   Ceq =  + 3 d  K1 + K 2 2  A0  2K1K 2 + K 2K3 + K3K1   Ceq =   d  2(K1 + K 2 )  R1R2 (1) R = R1 + R2 4. From a sphere of mass M and radius R, a cavity of 2R1R2 R (2) R = radius is created. Find the moment of inertia R1 – R2 2 R1R2 about an axis passing through the centre of sphere (3) R = R1 – R2 and cavity. R1R2 (4) R = (R1 – R2 ) Answer (3) 4S 4S Sol. P1 – P0 = ; P2 – P0 = R1 R2 4S  1 1 So, P2 – P1 = P = = 4S  –  31 R  R2 R1  (1) MR 2 48 1 R1 – R2 or = 31 R R1R2 (2) MR 2 80 6. From the given option, identify the diode connected in forward bias. 13 (3) MR 2 32 (1) 21 (4) MR 2 (2) 32 Answer (2) (3) 2 2 2  M  R  (4) Sol. I = MR 2 −    5 5  8  2  Answer (1) 31 Sol. Only in option (1), the p-side is connected at higher = MR 2 80 potential than the n-side of the diode. JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 3  JEE | NEET | Class 8 - 10 Download eSaral App 7. Radius of electron in ground state for hydrogen is a0, 9. A charge of value q is placed at the edge of a then radius of electron in He+ ion in 3rd excited state is imaginary cube of side a as shown in figure. Find the a0 net flux through the cube a. Then is a 1 (1) 2 1 (2) 4 1 (3) 16 1 (4) 8 Answer (4) q q (1) (2) 60 40 n2 Sol. r = r0  for H z q q (3) (4) 80 20 1 a0 = r 0 Answer (2) 1 q 42 Sol. 4 such cubes = a= r0 0 2 a0 1 q = 1 cube = a 8 40 8. Ice at –10°C is to be converted into steam at 110°C. 10. A closed organ pipe in 9th harmonic resonates with 4th Mass of ice is 10–3 kg. What amount of heat is harmonic of open organ pipe [lclosed = 10 cm]. Find required? length of open organ pipe. (1) Q = 730 cal (2) Q = 900 cal (1) L0 = 15 cm (3) Q = 1210 cal (4) Q = 870 cal 100 (2) L0 = cm Answer (1) 9 Sol. –10°C ice to 0°C ice → 0°C ice to 0°C water + 0°C (3) L0 = 110 cm water to 100°C water + 100°C water to 100°C 7 steam + 110°C steam. 80 (4) L0 = cm  1  9  Q =  1  10  + (1 80) + (1 1 100)  2  Answer (4)  1  9v 4v 8L +(1 540) +  1  10  = 730 cal Sol. =  L0 = c  2  4Lc 2L0 9 JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 4  JEE | NEET | Class 8 - 10 Download eSaral App 1. Now 13. If whole YDSE apparatus is immersed in a liquid of the battery is disconnected and dielectric slab of refractive index  then what is the effect on fringe dielectric constant K is inserted between the gaps of width? Q1 the plates. Now charge on capacitor is Q2. Find. Q2 (1) Fringe width increases 1 (1) 1 (2) (2) Fringe width decreases 2 2 (3) Fringe width remains unchanged (3) 2 (4) 3 (4) It may increase on one side and decrease on Answer (1) other side Q Sol. 1 = 1 (No further charge is supplied) Q2 Answer (2) 12. Which of the following graphs correctly represents D Sol.  = the variation of resistivity () with temperature (T).  So, for RI of  (1) D  = d 14. Two spherical black bodies of radius 0.8 m and 0.2 m are at temperatures of 400 K and 800 K respectively. (2) Find ratio of rate of heat loss. (1) 8 (2) 4 (3) (3) 2 (4) 1 Answer (4) (4) Sol. P1 = 4 (0.8)2 (400)4 P2 = 4 (0.2)2 (800)4 Answer (1) Sol. The resistivity of conductors increases with P1 4  4 = =1 increase in temperature non-linearly. P2 24 JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 5  JEE | NEET | Class 8 - 10 Download eSaral App 15. The equiconvex lens shown in figure is silvered on 17. The figure shows an electron entering the space one side. For what distance of object from the lens is between the plates of a parallel plate capacitor with an the image formed on the object itself? initial velocity, vx = 106 m/s parallel to the plates. If the length of plates is l = 10 cm and the electric field in the region E = 9.1 N/C, then the value of vy when the electron comes out of the plates is (Electronic mass = 9.1 × 10–31 kg) R (1) R (2)  R R (1) 1.6 × 104 m/s (2) 1.6 × 105 m/s (3) (4) (3) 1.6 × 107 m/s (4) 1.6 × 103 m/s 2 − 1 2 − 2 Answer (2) Answer (3) l Sol. Silvering of lens Sol. Time inside the electric field, t = vx 1 1 2 1  1  1  = − = ( − 1)  −   eE Feq fm f f  R  −R   Acceleration of electron along y-axis, a = m −2 4( − 1) 1 2( − 1) Velocity vy = at = − = R R f R eE l =  m vx −2(1 + 2 − 2) = R 1.6  10–19  9.1 10  10–2 = m/s −R 9.1 10–31  106 F= 2(2 − 1) = 1.6 × 105 m/s For object-image to coincide distance should be 2f 18. Find the equivalent power of the thin lens combination shown in the figure. |u| = 2|F| R = 2 − 1 16. Light of wavelength 550 nm is incident an surfaces of cerium and lithium. Work function are respectively 1.9 eV and 2.5 eV. Then electron will be ejected from (1) Cerium only (2) Lithium only (3) From both of them (4) None of them  R + R2   R + R2  Answer (1) (1) +  1  (2) −  1  1240 1240  R1 + R2   R1R2  Sol. E ( eV ) = = 2.25 (nm) 550  R + R2   R + R2  (3) −  1  (4) +  1  2.25 > 1.9 for cerium only  6R1R2   6R1R2  JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 6  JEE | NEET | Class 8 - 10 Download eSaral App Sol. v = 5gR Answer (3) Sol. Net power = P1 + P2 + P3 1 R KE A = mv 2 − mg 2 2 (1 − 1)  1 1  (3 − 1) = + (2 − 1)  + + R1  −R1 −R2  R2 KE A = 2mgR (1 − 2 ) (3 − 2 ) 1  3R  = + KEB = mv 2 − mg   R1 R2 2  2  = mgR 4 3 1 4 3 1 =  −  + −   3 2  R1  3 2  R2 KE A =2 KEB 1 1 1  = −  +  22. The current drawn from battery in the circuit shown 6  R1 R2  below is _____ A  R + R2  = − 1   6R1R2  19. 20. SECTION - B Numerical Value Type Questions: This section contains 5 Numerical based questions. The answer to each question should be rounded-off to the nearest integer. Answer (1) 21. The particle shown in figure is just able to complete 1 1 5 Sol. = +2= the vertical circular motion. Find the ratio of kinetic R1 2 2 energy at A to the kinetic energy at B. 2  R1 =  5 2 1 9 Now, R = + =  5 2 10 9  10 So, I = = 1A 10  9 23. 24. Answer (2) 25. JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 7  JEE | NEET | Class 8 - 10 Download eSaral App CHEMISTRY SECTION - A 4. Which of the following has maximum size out of – Multiple Choice Questions: This section contains 20 Al3+, Mg2+, F , Na+? multiple choice questions. Each question has 4 choices (1) Al3+ (2) Mg2+ – (1), (2), (3) and (4), out of which ONLY ONE is correct. (3) F (4) Na+ Choose the correct answer : Answer (3) 1. For complex ion [NiCl4]2– what is the charge on Sol. For isoelectronic species, more the negative metal and shape of complex respectively? charge more will be the size, also more the positive charge smaller will be the size. (1) +2, Tetrahedral (2) +2, Square planar The correct order of ionic size is : (3) +4, Tetrahedral (4) +4, Square Planar – Answer (1) Al3+ < Mg2+ < Na+ < F Sol. [NiCl4]2–  Ni2+ → 3d8 5. The IUPAC name of given specie is Cl– ligand is weak field ligand and hybridisation is sp3. Shape of complex is tetrahedral. 2. Compare boiling point of given solutions (1) 2, 3-dimethyl methyl carboxy butanoic acid (i) 10–4 M NaCl (ii) 10–3 M NaCl (2) 4-methoxy carbonyl-2, 3-dimethyl propanoic acid (iii) 10–2 M NaCl (iv) 10–4 M urea (3) 3-methoxycarbonyl-2-methyl butanoic acid (1) I > II > III > IV (2) III > II > I > IV (4) 1-carboxy-2, 3-dimethyl methyl butanoate (3) II > I > III > IV (4) III > I > II > IV Answer (3) Answer (2) Sol. Sol. Higher the elevation in boiling point, higher will be the boiling point Tb  i × m For urea i = 1 3-methoxycarbonyl-2-methyl butanoic acid For NaCl i = 2 6. Compare crystal field splitting energy() for given Boiling point order III > II > I > IV complexes 3. The correct decreasing order of electronegativity is (ii) Cu(NH3 )4  s +2 (i) K 4 [Fe(CN)6 ] (1) F > Cl > I > Br (2) Cl > F > Br > I (3) F > Cl > Br > I (4) Br > F > I > Cl (iii) K 4 Fe(SCN)6  (iv) Fe(en)3  Cl3 Answer (3) (1) I > II > III > IV (2) II > I > IV > III Sol. The correct order is (3) IV > I > III > II (4) IV > III > I > II F > Cl > Br > I Answer (2) JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 8  JEE | NEET | Class 8 - 10 Download eSaral App Sol. K 4 Fe(CN)6   d 6  SFL, (3) Statement-I is correct but statement-II is incorrect K2 Cu(NH3 )4   d 9  dsp2 (4) Statement-I is incorrect but statement-II is correct K 4 Fe(SCN)6   d 6  WFL Answer (1) Fe(en)3  Cl3  d 5  SFL Sol. CH3 − O − CH 2 stabilised by resonance. Splitting energy  Strength of ligand  Charge of 9. Which of the following acids is also known as vitamin C? CA. (1) Adipic acid (2) Ascorbic acid sp > o (3) Saccharic acid (4) Aspartic acid II > I > IV > III Answer (2) 7. Consider the given equilibrium reaction Sol. Ascorbic acid is also known as vitamin C. CO2 (g) + C(s) 2CO(g) 10. An electron of He+ is present in 3rd excited state. Find its de-Broglie wavelength. If initial pressure of CO2 is 0.6 atm and after (1) 6.64 Å (2) 1.66 Å equilibrium is established, total pressure is 0.8 atm. (3) 3.32 Å (4) 13.28 Å Then, find Kp. Answer (1) (1) 0.4 (2) 0.2 Sol. n = 2r (3) 0.6 (4) 0.8 For 3rd excited state, n = 4 Answer (1) n2 4 = 2 ×  × a z Sol. CO2 (g) + C(s) 2CO(g) 16 t=0 0.6 4 = 2 ×  × 0.529 Å 2 t = teq 0.6 – p 2p  = 2 × 3.14 × 0.529 × 2 Å = 6.64 Å Pt at equilibrium = 0.8 = 0.6 + p 11. Which of the following will show positive Fehling 0.2 = p test? (pCO )2 (2p)2   0.04 4  0.04 Kp = = = = = 0.4 (1) (pCO2 ) 0.6 − p 0.6 − 0.2 0.4 8. Statement-I: CH3 – O – CH2 – Cl will show nucleophilic substitution by SN1 mechanism in protic medium. (2) Statement-II: will not undergo (3) nucleophilic substitution via SN2 mechanism easily. (4) (1) Statement-I and statement-II both are correct (2) Statement-I and statement-II both are incorrect Answer (3) JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 9  JEE | NEET | Class 8 - 10 Download eSaral App Sol. Fehling test is given by Aldehydes except benzaldehyde 14. will give +ve Fehling test 12. 4f7 configuration is possible for (a) Eu3+, (b) Eu2+, (c) Gd3+, (d) Tb3+, (e) Sm2+ (1) (1) (a) and (c) (2) (b) and (c) (2) (3) (d) and (e) (4) Only (c) (3) Answer (2) Sol. Electronic configuration of: Eu3 +  4f 6 Tb3 +  4f 8 (4) Eu2+  4f 7 Sn2+  4f 5 Answer (1) Gd3 +  4f 7 Sol. 13. Given : NH2COONH4 (s) 2NH3 (g) + CO2 (g) If the partial pressure of CO2 gas at equilibrium is 0.4 atm and the total pressure is 1 atm, then the value of Kp at the same temperature is (1) 0.027 atm3 (2) 0.064 atm3 (3) 0.144 atm3 (4) 0.216 atm3 Answer (3) 15. CO2 gas is taken at 1 atm, 273K. Now it is allowed to pass through 0.1 M Ca(OH)2 aq. solution. Excess Sol. NH2COONH4 (s) 2NH3 (g) + CO2 (g) amount of Ca(OH)2 is neutralised with 40 mL of 0.1 Total pressure at equilibrium = 1.0 atm M HCl. Then find volume of Ca(OH)2 initially taken if 50% Ca(OH)2 is react with CO2 Partial pressure of CO2 at equilibrium = 0.4 atm (1) 40 mL  Partial pressure of NH3 at equilibrium = 0.6 atm (2) 20 mL K p = (pNH3 ) (pCO2 ) 2 (3) 80 mL = (0.6)2(0.4) (4) 50 mL = 0.144 atm3 Answer (1) JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 10  JEE | NEET | Class 8 - 10 Download eSaral App Sol. g meq of Ca(OH)2 = 2 × gm eq of HCl V 40 18. 0.1 mL  2 = 2  0.1 1 1000 1000 Identify C. VmL = 40 mL 16. In a closed insulated container, a liquid is stirred (1) with a paddle to increase the temperature, which of the following is true? (1) w = 0, E = q  0 (2) E = w  0, q = 0 (2) (3) E = w = 0, q  0 (4) E = 0, w = q  0 (3) Answer (2) Sol. In closed insulated container a liquid stirred with a paddle to increase the temperature, it behaves as (4) an adiabatic container, q = 0 Answer (1) From FLOT U = q + w; q = 0 Sol. E = w (but not zero) 17. Match the column and choose the correct option Column-I Column-II (Properties) (Order) 19. (A) Electronegativity (1) B Mg > AI Numerical Value Type Questions: This section (D) Electron affinity (4) Cl > F > Br > I contains 5 Numerical based questions. The answer to each question should be rounded-off to the nearest (1) A–1, B–2, C–3, D–4 integer. (2) A–4, B–3, C–2, D–1 (3) A–2, B–3, C–4, D–1 21. (4) A–3, B–2, C–4, D–1 Answer (1) Find molecular weight of (A) in g mol–1 Sol. Li+  Mg2+  Be2+   76 pm 72 pm 31 pm Answer (154) JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 11  JEE | NEET | Class 8 - 10 Download eSaral App 24. In Carius method 180 mg of organic compound Sol. gives 143.5 mg of AgCl. Find the percentage of Cl in the organic compound. (Nearest integer) Answer (20) Sol. Mass of organic compound = 180 mg Molecular weight of (A) = 154 g mol–1 Mass of AgCl = 143.5 mg 22. Calculate Number of stereoisomers of 143.5 Mass of Cl =  35.5 mg CH3 – CH = CH — CH —CH3 143.5 | OH = 35.5 mg Percentage of Cl in the organic compound Answer (4) 35.5  100 = Sol. Number of centres which can show 180 stereoisomerism in molecule = 2 = 19.72% 20% Number of isomers = 22 = 4 25. Two ampere current is allowed to pass through molten AlCl3 for 30 min. Find the mass (in mg) of 23. How many compounds have linear shape SO2, aluminium deposited at cathode. (Nearest integer) BeCl2, N3− , I3− , NO2+ , NO2 ? Answer (336) Answer (4) Sol. Total charge passed = 2 × 30 × 60 C Sol. 2  30  60 Number of Faradays passed = F 96500 36 Equivalents of Al deposited = 965 36  9 Mass of Al deposited = g 965 36  9  1000 = mg 965 = 335.75 mg  336 mg JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 12  JEE | NEET | Class 8 - 10 Download eSaral App MATHEMATICS SECTION - A Answer (2) 6 Multiple Choice Questions: This section contains 20 C2 Sol. P (E ) = multiple choice questions. Each question has 4 choices 10 C2 (1), (2), (3) and (4), out of which ONLY ONE is correct. 15 1 Choose the correct answer : = = 45 3 1. The shortest distance between the lines 3. Let A = {1, 2, 3} number of non-empty equivalence x − 1 y − 2 z − 1 and relations from A to A are = = 2 3 4 (1) 4 (2) 5 x + 2 y − 2 z + 1 is = = (3) 6 (4) 8 7 8 2 Answer (2) 88 78 (1) (2) Sol. The partitions far a set with 3 elements, {1, 2, 3} 1277 1277 {(1}, {2}, {3}} – Every element is in its own subset 66 55 (3) 1277 (4) 1277 {{1, 2}, {3}} – Two elements are together, one separate Answer (1) (a2 − a1 ) ⋅ (b1 × b2 ) {{1, 3}, {2}} – Two elements are together, one Sol. d = separate b1 × b2 {{2, 3}, {1}} – Two elements are together, one iˆ ˆj kˆ separate b1 × b2 =23 4 {{1, 2, 3}} – All elements are together in one subset 7 8 2 ∴ Therefore, total possible equivalence relation = 5 −26iˆ + 24 ˆj − 5kˆ , a2 − a1 = 3iˆ + 2kˆ = If f ( x ) 16(sec −1 x )2 + (cosec −1x )2. 4. = Then the maximum and minimum value of f(x) is (3iˆ + 2kˆ ) ⋅ ( −26iˆ + 24 ˆj − 5kˆ ) d= 262 + 242 + 52 1001π2 2π2 1105π2 4 π2 (1) and (2) and 33 9 68 17 −78 − 10 88 = = 1277 1277 1117π2 6 π2 1268π2 3 π2 (3) and (4) and 59 19 27 16 2. In a bag there are 6 white and 4 black balls two balls are drawn at random, then the probability that both Answer (2) ball are white are Sol. f ( x ) (4 sec −1 x )2 + (cosec −1x )2 = 1 1 (1) (2) =(4 sec −1 x + cosec −1x )2 − 8 sec −1 x cosec −1x 2 3 2 2 1  π π  (3) (4) =  3 sec −1 x +  − 8 sec −1 x  − sec −1 x  3 4  2 2  JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 13  JEE | NEET | Class 8 - 10 Download eSaral App −1 2 2 −1 −1 p π 4 = 9(sec x) + + 3π sec x − 4π sec x+ =8 3  + 4 3 4−p 8(sec −1 x )2 p 4= 2 4−p π = 17(sec −1 x )2 − π(sec −1 x ) + ⇒ 16 – 4p = p 4 ⇒ 5p = 16 2  π π2  π 17π2 16 ( ) 2 = 17  sec −1 x − (sec −1 x ) + 2  + − ⇒ p=  17 34  4 342 5   dx x 1  1  equals to  2 π   π2 π2 6. If + 2 = 3 , x(1) = 1. Then x   = 17  sec −1 x −  + − dy y y 2  34   4 68 (1) 2 – e (2) 3 – e 2 (3) 5 – e (4) 7 – e  π   4 π2 = 17  sec −1 x − + Answer (2)  34   17 ∫ y2 1 Sol. I.F = e 4 π2 − 1 Min = I.F = e y 17 1  1 Max if sec–1x = π − −   1   2 ∴ x ⋅ e y= ∫e y ⋅  3  dy y  π   4 π2 17  π −  + 1  1  34   17 − −   1  1  x ⋅ e y= ∫e y ⋅    2  dy  y  y  1089 2 4π2 1105π2 π + = 1 68 17 68 Put =t y 5. If 8 =3 + 1 4 4 1 ( ( 3 + p ) + 2 3 + p2 +... ∞ ) then the − 1 dy = dt value of p is y2 14 16 ∴ − ∫ e − t ⋅ t dt xe − t = (1) (2) 5 5   d (t )   − t e − t − ∫  xe − t = ⋅ ∫ e − t ⋅ dt  dt  3 4   dt   (3) (4) 5 5 xe − t =−  −t e − t − e − t  + c Answer (2) xe − t = te − t + e − t + c …(1)  3 3   p p2  Sol. 8=  3 + + +... + ∞  +  + +... + ∞  Given x(1) = 1  4 42   4 42   e −1 = e −1 + e −1 + c  1 1   p p2  −e −1 = c 8 3 1 + + = +... + ∞  +  + +... + ∞   4 4 2   4 4 2  ∴ from (1)    p x = t + 1 − (e − t ⋅ e t )  1  4 1 =8 3 + Put y =  1 − 1  1 − p 2  4 4 x= 3 − e JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 14  JEE | NEET | Class 8 - 10 Download eSaral App (2r − 1)(2r + 1)(2r + 3)(2r + 5) Sol. (1 – x) [(1 – x) (1 + x + x2)]2007 7. Let Tr = , then 64 = (1 – x) (1 – x3)2007 n = (1 – x3)2007 – x(1 – x3)2007 1 lim n →∞ ∑T is equal to [(1 – x3)2007 contains 3λ types of exponents while r =1 r x(1 – x3)2007 will have (3λ + 1) type while 2012 is 22 32 (3λ + 2) type] that is not possible ⇒ 0 (1) (2) 45 35 Coefficient of x2012 in (1 – x3)2007 = 0 27 32 Coefficient of x2011 in (1 – x3)2007 = 0 (3) (4) ⇒ Coefficient of x2012 in (1 – x)2008 (1 + x + x2)2007 = 0 45 45 9. If the images of the points A(1, 3), B(3, 1) and C(2, Answer (4) 4) in the line x + 2y = 4 are D, E and F respectively, (2r − 1)(2r + 1)(2r + 3)(2r + 5) then the centroid of the triangle DEF is Sol. Tr = 64  3 2 (1) (3, –1) (2)  − , −  1 64  5 5 ⇒ = Tr  1  1  3  5 16  r −  r +  r +  r +   2 1  1 2  2  2  2  2 (3)  , −  (4)  , −  5 5 5 5 4  5  1  Answer (3) r +  −  r −  1 3  2  2  Sol. Centroid of the ∆DEF is the mirror image of the ⇒ = Tr  1  1  3  5 centroid of the ∆ABC about the line x + 2y = 4. r − r + r + r +   2  2  2  2 G1 = Centroid of ∆ABC ≡ (2, 3), G2 ≡ Centroid of  ∆DEF. 1 4 1 ⇒  − Tr 3  1  1  3   r − r + r −  2   2   2   1    1  3  5  r + 2 r + 2 r +  2       n 1 4 1 1  y1 − 3 x +2 lim ∑ = − ⇒ 2, 1 = + ( y1 + 3 ) = 4   x1 − 2 2 n →∞ r = 1Tr 31 3 5 3 5 7 ⋅ ⋅ ⋅ ⋅ 2 2 2 2 2 2 2 1 ⇒ x1 =, y1 = − 1 1 5 5 − 3 5 7 5 7 9 2 1 ⋅ ⋅ ⋅ ⋅ ⇒ G2 =  , −  2 2 2 2 2 2 5 5 4  8  32 10. If A = {1, 2, 3, …., 10}. = = 3 15  45 m 8. Coefficient of x2012 in (1 – x)2008 (1 + x + x2)2007 B =  , m, n ∈ A and m < n and gcd of (m, n ) = 1}. n (1) 0 (2) 1 Then number of elements in set B is (3) 2 (4) 3 (1) 30 (2) 31 Answer (1) (3) 28 (4) 29 JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 15  JEE | NEET | Class 8 - 10 Download eSaral App Answer (2) 12. Let |Zi| = 1 for i = 1, 2, 3 satisfying 2 Sol. n = 1 m ∈ φ …0 Z1Z2 + Z2 Z3 + Z3 Z1 a + b 2 , where a, b an = m 1 π n = 2 m =1 ⇒ can be …1 rational numbers such that = arg(Z1 ) = , arg(Z2 ) 0 n 2 4 −π m 1 2 and arg(Z3 ) = , then find (a, b) n = 3 m = 1, 2 ⇒ can be , …2 4 n 3 3 (1) (5, 2) (2) (–5, –2) m 1 3 (3) (5, –2) (4) (–5, 2) n = 4 m = 1,3 ⇒ can be , …2 n 4 4 Answer (3) m 1 2 3 4 i π 1 1 n = 5 m = 1, 2, 3, 4 ⇒ = , , , …4 Sol. Z1= | 1| e 4= +i⋅ n 5 5 5 5 2 2 −(0) m 1 5 =Z2 | 1| e 1 + 0i n = 6 m = 1, 5 ⇒ = , …2 n 6 6 π −i 1 i = Z3 | 1| e = 4 − m 1 2 3 4 5 6 2 2 n = 7 m = 1, 2, 3, 4, 5, 6 ⇒ = , , , , , n 7 7 7 7 7 7  1 i  …6 Z= 1Z2  −  (1)  2 2 m 1 3 5 7  1 i  n = 8 m = 1, 3, 5,7 ⇒ = , , , …4 Z= 2 Z3 1 −  n 8 8 8 8  2 2 m 1 2 3 4 5 7 8  1 i  1 i  n = 9 m = 1, 2, 4, 5, 7, 8 ⇒ = , , , , , , Z3 Z1 =  +  +  n 9 9 9 9 9 9 9  2 2  2 2 …6  1 i   1 i  ⇒ Z1Z2 + Z2 Z3 + Z3 Z1 =  − + −  m 1 3 7 9  2 2  2 2 n = 10 m = 1, 3, 7, 9 ⇒ = , , , …4 n 10 10 10 10  1 1  1 +  −  + 2i   11. How many ways are there to pick 5 letters from 2 2 2 English alphabets such that M is the middle of the = 2 − 2i + i ( ) 2 letters (repetition not allowed). ⇒ Z1Z2 + Z2 Z3 + Z3 Z1 = 2 2 + i − 2 +1 (1) 26 C5.5! 2   ( ) ( ) 2 2 (2) 25 C4.4! =  2 + 1− 2    (3) 26 C4.4! = 5−2 2 (a, b) = (5, –2) (4) 25 C5.5! 13. Let a coin is tossed thrice. Let the random variable Answer (2) x is tail follows head. Let the mean of x is µ and variance is σ2. Find 64 (µ + σ2). Sol. A1 A 2 M A3 A 4 (1) 48 (2) 64 ↑ fixed (3) 132 (4) 128 25 C4 × 4! Answer (1) JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 16  JEE | NEET | Class 8 - 10 Download eSaral App Sol. f ′′( x ) xi Pi g ′′( x ) = + f ′′(3 − x ) 3 HHH 0 1 8 ⇒ g ′( x ) > 0 TTT 0 1 3 8 f ′   − f ′(3 − x ) > 0 3 HHT 1 1 8 f ′( x ) > 0 ⇒ f ′( x ) is increasing HTH 1 1  15. Let b = λiˆ + 4kˆ, λ > 0 and the projection vector of 8      THH 0 1 b on a = 2iˆ + 2 jˆ − kˆ is c. If | a + c | = 7 , then the 8  area of the parallelogram formed by vector b and TTH 0 1  c is (in square units) 8 THT 1 1 (1) 8 8 (2) 16 HTT 1 1 (3) 32 8 (4) 64 1 µ = ΣPi xi = 2 Answer (3) 2 σ =ΣPi xi2 –µ 2   2λ − 4  Sol. c =(b ⋅ aˆ )aˆ = a 1 1 1 6 = =– 2 4 4     2λ − 4   1 1 3  | a + c | = 7 ⇒ a 1 + = 7 64  +  = 64 × = 48  9  2 4 4 x 5 + 2λ 14. Let ( x ) 3f   + f (3 − x )∀x ∈ (0,3) g= and × 3= 7 ⇒| 5 + 2λ =| 21 3 9 f ′′( x ) > 0∀x ∈ (0,3) then g(x) decreases in interval  λ > 0 ⇒ λ =8 (0, α), then α is  4  7 2 ⇒ c= a and b= 4(2iˆ − kˆ ) (1) (2) 3 4 3 9 7 iˆ ˆj kˆ (3) (4)   16 16 4 3 ⇒ b × c= 2 0 1= ( −2iˆ + 4 jˆ + 4kˆ ) 3 3 Answer (3) 2 2 −1 x Sol. g ( x= ) 3f   + f (3 − x )   32 3 ⇒ b × c= | −iˆ + 2 ˆj + 2= kˆ | 32 3 1 3 g ′( x ) =⋅ 3 f ′   − f ′(3 − x )   3 3 ⇒ Area of parallelogram formed by b and c x   = f ′   − f ′(3 − x ) ⇒ b×c = 32 3 JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 17  JEE | NEET | Class 8 - 10 Download eSaral App 16. Let the parabola y = x2 + px – 3 cuts the coordinate 18. A hyperbola with foci (1, 14) and (1, –12) passes axes at P, Q and R. A circle with centre (–1, –1) through the point (1, 6). The length of the latus passes through P, Q and R, then the area of triangle PQR. rectum of the hyperbola is 5 3 (1) (2) 144 2 2 (1) (3) 3 (4) 5 5 Answer (2) (2) 50 Sol. Since at x = 0, y = –3, parabola cuts the coordinate y-axe at (0, –3) 288 ⇒ Equation of circle will be (3) 5 (x + 1)2 + (y + 1)2 = (–1 –0)2 + (–1 + 3)2 =1+4=5 (4) 100 x2 + 2x + y2 + 2y = – 3 = 0 Circle cuts x-axis at y = 0 Answer (3) ⇒ x2 + 2x – 3 = 0, (x + 3) (x – 1) = 0 ⇒ (–3, 0), (1, 0) Sol. |sp – s′p| = 2a, ss′ = 2ae ⇒ Area of ∆ s(1, 14), s′(1, –12), P(1, 6) −3 0 1 1 1 3 ⇒ 0 −3 0 = (3) = ⇒ 2a = |8 – 18| 2 1 0 0 2 2 ⇒ a = 5; 2ae = 26 (x − 2 3) 2 17. If the circle + y2 = 12 and parabola y 2 = 2 3 x intersects at P, Q and R. Then the area ⇒ ae = 13 of triangle PQR is (1) 10 sq. units (2) 12 sq. units (3) 14 sq. units (4) 16 sq. units Answer (2) Sol. Simply solving both we get x = 0, 2 3 2b 2 2a 2 (e 2 − 1) Length of latus rectum = 1 = a a 2(169 − 25) 288 = = 5 5 1 19. ∆PQR =× (4 3 )(2 3 ) 2 20. JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 18  JEE | NEET | Class 8 - 10 Download eSaral App SECTION - B 22. If a1, a2, a3..., an are in geometric progression such Numerical Value Type Questions: This section that a1a5 = 28, a2 + a4 = 29, then the value of a6 is contains 5 Numerical based questions. The answer to (1) 635 (2) 784 each question should be rounded-off to the nearest (3) 872 (4) 898 integer. Answer (2) 21. If A be a 3 × 3 square matrix such that det(A) = –2. Sol. a1a5 = 28 ⇒ a2r4 = 28 If det(3 adj(–6 adj(3A)) = 2n ⋅ 3m, where m ≥ n, then a2 + a4 = 29 ⇒ ar + ar3 = 29 4m + 2n is equal to ar, ar3 are roots of k2 – 29k + 28 = 0 Answer (104) ⇒ k = 1, k = 28 Sol. Concept: A. adj(A) = |A|I, det(λA) = λn det(A) ⇒ ar = 1, ar3 = 28 ⇒ det(A) = |A|n–1, where n is order 1 ⇒ det(3 adj(–6 adj(3A))) ⇒ r2 = 28, a 2 = 28 = 33 ⋅ det(adj(–6 adj(3A))) 1 = 33 ⋅ (–6 adj(3A)))2 a6 =ar 5 ⇒ a62 =a 2 r 10 = × (28)5 =(28)4 28 = 33 ⋅ (–6)6 |3A|4 ⇒ a6 = (28)2 = 784 = 39 ⋅ 26 ⋅ 312 ⋅ (–2)4 = 321 ⋅ 210 23. ∴ n = 10, m = 21 24. ∴ 4m + 2n = 104 25. JEE (Main)-2025 : Shift-1 (22-01-2025) www.esaral.com 19