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IIT JEE Exam 2026 : Main

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Test - 01 22nd January-2025 (Morning) DURATION : 180 Minutes M.MARKS : 300 Answer Key Q1 (B) Q27 (B) Q2 (B) Q28 (A) Q3 (B) Q29 (B) Q4 (B) Q30 (C) Q5 (D) Q31 (B) Q6 (A) Q32 (B) Q7 (A) Q33 (C) Q8 (D) Q34 (A) Q9 (D) Q35 (B) Q10 (A) Q36 (C) Q11 (B) Q37 (A) Q12 (B) Q38 (D) Q13 (A) Q39 (C) Q14 (B) Q40 (B) Q15 (C) Q41 (B) Q16 (D) Q42 (D) Q17 (D) Q43 (A) Q18 (B) Q44 (B) Q19 (A) Q45 (C) Q20 (A) Q46 154 Q21 90 JEE MAIN Q47 20 Q22 4 Q48 45 Q23 40 Q49 69 Q24 8 Q50 6 Q25 5 Q51 (C) Q26 (C) Q52 (A) Android App | iOS App | PW Website JEE Q53 (D) Q65 (C) Q54 (D) Q66 (B) Q55 (D) Q67 (A) Q56 (B) Q68 (C) Q57 (C) Q69 (A) Q58 (D) Q70 (C) Q59 (A) Q71 34 Q60 (C) Q72 2035 Q61 (D) Q73 34 Q62 (C) Q74 216 Q63 (B) Q75 16 Q64 (A) Android App | iOS App | PW Website JEE Hints & Solutions Note: scan the QR code to watch video solution Q1 Text Solution: main scale. The vernier constant (least count) is actually (b) Pij = 2Pt + P 1 2 1 f eq 1 R R 1 R (4 4 + 2) (4 2) 1 = f eq Least Count = Value of One Main Scale Division- = Value of One Vernier Scale Division Thus, the correct option is: B 1 = fm 2 R given by: 1 f 4( 1) = = fQ R Video Solution: (4 2) R feq = 2 R = 2feq = 2 ( R 4 2 ) = R (2 1) Video Solution: Q4 Text Solution: 9 th harmonic of closed pipe = harmonic of open pipe = 9V1 4 1 4 d 2V2 2 Since, the frequency of the 9th harmonic of the closed pipe equals the frequency of the 4th harmonic of the open pipe, therefore: , Q2 Text Solution: 9V1 Using energy conservation, 1 2 2 mv A 1 2 = 1 2 2 mv B m(5g ) = 5mg 2 mg 2 4 1 + mgh 1 2 2 mv B + mg 2 = 9 4 1 2V2 2 B 1 2 == = 20 9 2 2 B 2 2 1 = 8 9 1 2 cm Video Solution: = K EB K EB = 2mg 1 2 2 mv C = 1 2 K EC = K EB K EC 2 mv D 1 2 + mg 2 mg + mg 2 = mg = 2 Video Solution: Q5 Text Solution: The equivalent EMF of two non-ideal batteries connected in parallel is given by: Ecq = E1 r2 +E2 r1 r1 +r2 where E1 , E2 are the EMFs of the two batteries and r1 , r2 are their internal resistances. Video Solution: Q3 Text Solution: A vernier scale is designed such that its divisions are slightly smaller than the main scale divisions. This difference allows precise measurements by aligning a particular vernier division with the Android App | iOS App | PW Website JEE The time t taken by the electron to travel through the plates is given by: t = L vx = 0.1 10 7 = 10 6 s Since the electron is initially moving horizontally, its initial vertical velocity is 0 . The vertical Q6 Text Solution: In an optically denser medium, the speed of light velocity is given by: decreases, which leads to a decrease in the vy = at wavelength (since vy = (1. 6 10 = = , where frequency c remains unchanged). 7 ) (10 ) 6 vy = 16 10 m/s As decreases, the fringe width w also decreases (since fringe width, w 14 = D d ), meaning the Video Solution: fringes come closer. The reason is also true because when light enters a denser medium, its speed decreases, but the frequency remains constant (as frequency is determined by the source). Option 1: Both (A) and (R) are true, and (R) is the correct explanation of (A). Q8 Text Solution: E = Video Solution: hc (6.626 10 34 550 10 E = 1.9878 10 550 10 8 )(3.0 10 ) E = 9 25 9 19 E = 3. 61 10 J = 2. 26eV Cesium (Cs): Work function = 1.9 eV Since 2.26 eV (incident photon energy) > 1.9 eV , Q7 Text Solution: photoelectric emission is possible. Given Data: Lithium (Li): Work function = 2.5 eV Length of the plates: Since 2.26 eV (incident photon energy) < 2.5 eV, L = 10cm = 0. 1m photoelectric emission is not possible. Initial horizontal velocity: vx Electric field: E Video Solution: 6 = 10 m/s = 9. 1V /cm = 910V /m Charge of an electron: e Mass of an electron: m 19 = 1. 6 10 31 = 9. 1 10 C kg The force on the electron due to the electric field is given by: 19 F = eE = (1. 6 10 16 10 a = F = Q9 Text Solution: since the line is at the center of the edge BC, the N m ) (910) = 1. 456 1.456 10 9.1 10 14 a = 1. 6 10 part of this line enclosed within the cube is only 16 one-fourth of the total length. Thus, the length of 31 m/s 2 the line charge enclosed in the cube is: (a/2).(1/4) = a/8 Android App | iOS App | PW Website JEE Charge enclosed = = qin 0 = a 2 4 = = a 8 h mv nh mvr = a 8 0 Video Solution: 1 4 = 2 2 rh , = nh r n = r1 n4 5.3 10 = n1 r4 11 4 1 84.8 10 11 1 4 Video Solution: Q10 Text Solution: Resistivity is independent of temperature for wire bound resistors. Video Solution: Q13 Text Solution: q = (K C C )V 6 100 = 4mC = 40 10 1 U = = = Q11 Text Solution: 1 2 1 2 2 CV C V 2 2 1 2 CV 2 = 1 2 (K 1)C V (2 1) 6 40 10 10000 = 0. 2J Video Solution: The circuit Q14 Text Solution: M 2 R eq = 0. 5 + = 45 50 i = 0.5 2 2+0.5 = 0. 9 0.9 0.9 = 1A = ( 5 10 + 10 25 ) I = I = MR 2 13 32 [ MR 4 ( R 2 2 ) 2 2 Video Solution: Video Solution: Q15 Text Solution: Q12 Text Solution: c is correct. Video Solution: Android App | iOS App | PW Website + M 4 ( R 2 2 ) ] 2 JEE 4 p1 = ( 1 b = ( p2 = ( P GM m (2R) G( F2 = F2 = p3 = ( . . . . . . (i) 2 GM m (2R) 11 48 ( GM m 4R 3 2 ) 1 ] = [ni] = L 2 ( 3 = 1 3 1 1 6 ( 1 |R1 | 1 )( 1 |R2 | 1 1 |R2 | 1 ) ) ) |R1 | |R1 | |R2 | 1 1 ( |R1 | |R1 | ) |R2 | |R1 | 1 A E A T 1 ) = 1 |R2 | 1 3|R2 | ) ) 1 ) |R2 | 4 2 Esmaller Video Solution: 1 1 Q19 Text Solution: Q17 Text Solution: 0 ( 1 Video Solution: Video Solution: [ 1 R F1 : F2 = 12 : 11 from (1) and (2) B )( ) . . . . . . (i) 2 2 p eq = )m 27 ( 2 M 1 2 Q16 Text Solution: F1 = 3|R1 | 1 = 1) ( 3 Elarger 4 (0.2) 800 = 2 = 4 (0.8) 400 1 16 16 = 1 Elager = Esmaller = E Video Solution: Q18 Text Solution: Q20 Text Solution: 3 Q1 = m S1 T = 10 2100 10 = 21J 3 Q2 = m L1 10 5 3. 35 10 = 335J 3 Q3 = m Sw T = 10 4180 100 = 418J 3 Q4 = m Lv = 10 6 2. 25 10 = 2250J 3 Q5 = m Sv T = 10 = 19. 2J Qne = 3043. 2J Android App | iOS App | PW Website 1920 10 JEE Video Solution: Q23 Text Solution: Q21 Text Solution: [90]VA = (2t i + 3n j + 2k) V = (2 i 2t j + 4pk) B V A V 4 6n + 8p = 0, 3n = 2 + 4p V A = V 4 + 9n p = 2 1 2L K 2A , R3 = 2 1 1 = Req 1 3 ( r AB V A ) + ( 3 3 ) k1 A = 2L R eqq = 1 + R1 1 r A/B = ( 1 1 ) i + ( 2 2 ) j S2 + k2 A 2L 2L (k1 +k2 )A From steady-state heat conduction: 100 = Req 0 R3 = (1 2) i + (1 + 1) j + 3k i j Substituting values: k L = 1 2 3 = i + 8 j 5k 2 1 2 100 = 2L (k 1 +k 2 0 L 2k )A 3 A After plugging in given values: L = 1 + 64 + 25 = 90 = 40 C Video Solution: Video Solution: Q24 Text Solution: v = Q22 Text Solution: r = r1 r2 r2 r1 = 2 4 4 2 L K 3A and R2 are in parallel. R1 Req L = mA , R2 = is: + 4 = 4 + 4 + 16p K 1A The equivalent resistance of parallel resistances B n = 4 2L R1 = = 0 B = 4cm m = uf u f v u = 24 1 24 1 = = 24 25( 24) Video Solution: Android App | iOS App | PW Website 24 25 = 1 25 JEE 1 1 v 1 v = 0[ +2 v u 2 ( 2 du = vi ; dt (vI ) 3 1 ) + dt dv , vt = m f dv ( 2 1 = u 1 v dt v0 = B<C<O<N 1 25 Metallic character increases down the group and ) decreases across a period B < Al < Mg < K = v0 ] dt du 2 at 2 2 2 u 3 (v0 ) + 1 u 2 a0 = 0 aI v 2 a1 = = = 2 = v 2 24 2 1 25 u 2v I 2 I v v 2 25 2 v 3 u 1 25 3 size decreases Si < P < S < Cl 2 v 0 2 Electron negativity increases across a period as Video Solution: 2 v 0 3 2 3 (24) 24 25 24 25 25 25 2 25 100a1 = 2 25 100 = 8 Video Solution: Q27 Text Solution: Concentrared Sulphuric acid can be used to obtain H2 S2 O8 by the process of electrolysis. At cathode : 2 HSO 4 H2 S 2 O8 + 2e Video Solution: Q25 Text Solution: uy = 60 sin 30 = 30m/s Height attained in the firest second, 2 h0 = uy (1) 1 2 g(1) h0 = 30 5 = 25m h1 = 30 + ( 10 2 ) (2 3 1) = 5 Q28 Text Solution: h1 h6 = 25 5 CH3 C CH + Na CH3 C = 5 + Video Solution: Na + 1 2 H2 1 mole of propyne gives 0. 5 mole of hydrogen gas CH3 C CH + NaNH2 CH3 C n Q26 Text Solution: < 4 g 40 g = 0. 1 mole So, 0. 1 mole occupies 2240 ml Ionic radii :Cations are smaller than anions Al = NH3 1 mole occupies 22400 ml Correct Match: 3+ + C Na + 2+ Mg < Na + < F Video Solution: Ionisation energy Increases across the period due to decrease in size but, N has higher Ionisation energy than O due to stable electronic configuration Android App | iOS App | PW Website C JEE Q29 Text Solution: CO2 (g) + C(s) 2 CO (g) 0. 5 Mg Al Si P S Cl 0.9 1.2 1.5 1.8 2.1 2.5 3.0 Correct order : Mg < Al < B < N 0. 5 x Na 2x Video Solution: PT = 0. 5 + x = 0. 8 x = 0. 8 0. 5 = 0. 3 KP = (0.6) 2 = 1. 8 0.2 Video Solution: Q33 Text Solution: Tb = iKb m Tb i x m For NaCl " i "= 2 and Urea " i " Q30 Text Solution: = 1 Higher the product of vant Hoff factor and concentration higher the Video Solution: boiling point. 4 i x C = (i) 2x 10 3 10 4 (ii) 1x 10 (iii) 2x 2 (iv) 2x 10 Hence the order for boiling point is : (ii) < (i) < (iii) < (iv) Video Solution: Q31 Text Solution: Given: Insulated thermally means q = 0 Acc. to 1st law, U = q + w U = 0 + w = w w > 0, U > 0 Q34 Text Solution: Video Solution: 6-methoxycarbonyl-2,5-dimethylhexanoic acid Video Solution: Q32 Text Solution: Electonegativity values acc. to Pauling Scale: Li Be B C N O F 1 1.5 2 2.5 3 3.5 4 Q35 Text Solution: Android App | iOS App | PW Website JEE (b) [N iC l4 ] 2 x 4 = 2 x = +2 Ni Ni 2+ 8 : [Ar] 3d 4s 2+ 0 : Since, C l is weak field ligand, therefore paring will not take place. 2 [N iC l 4 ] 3 : sp Q38 Text Solution: , Tetrahedral,2 unpaired electron, Paramagnetic [N i(C O)4 ] The carbon at position 2 is a chiral center, hence, x + 0 = 0 x = 0 10 N i : [Ar]3d 4s 3 [N i(C O)4 ] : sp contributes to 2 stereoisomers (R/S 0 , Tetrahedral, No unpaired electron, configurations). The C = C double bond at position 3 allows for cis-trans isomerism. Diamagnetic Therefore, it has total of 4 stereoisomers Video Solution: Video Solution: Q36 Text Solution: Radioactive decay follows first-order kinetics, Q39 Text Solution: where the decay constant k is independent of K3 [F e(S C N )6 ] temperature Fe Video Solution: 3+ : [Ar]3d 5 Since, S C N is weak field ligand, hence pairing will not take place. Q37 Text Solution: Eu Gd Eu 2+ : [Xe]4f 3+ : [Xe]4f 3+ 3+ Tb Sm C F S E = ( 0. 4t2g + 0. 6eg ) 0 : [Xe]4f : [Xe]4f 2+ 7 7 6 8 : [Xe]4f 6s 0 0 5d 6s 6s 6s 6 = ( 0. 4 3 + 0. 6 2) 0 0 = 0 0 Video Solution: 0 0 6s 0 Hence, only Eu2+ and Gd 3+ have 4f 7 configuration. Video Solution: Q40 Text Solution: Android App | iOS App | PW Website JEE Ascorbic acid is another name for vitamin C. Video Solution: Q43 Text Solution: As, r Z For Helium ion: Z Q41 Text Solution: C H3 O C H2 C l undergoes SN 1 reaction because of the presence of the OC H3 2 n = a0 group, which exerts a + M effect, that = 2 and n = 2 (given) 2 (2) r = a0 = 2a0 2 Video Solution: stabilizes the carbocation formed. Q44 Text Solution: Video Solution: Aliphatic aldehydes and alpha hydroxy ketones give positive results with Fehling s test. Hence, the compounds (C) which is alpha hydroxy ketone , compounds (D) & (E) which are aliphatic aldehydes give positive Fehling s test. Video Solution: Q42 Text Solution: A. Propene (C H3 C H = C H2 ) does not show geometrical isomerism as one of the double-bonded carbon atom has two identical hydrogen atoms. B. Correct statement Q45 Text Solution: C. Correct statement D. 2-Methylbut-2-ene (C H3 C (C H3 ) = C H C H3 ) does not show geometrical (c) Using Faraday's laws of electrolysis, W = Z I t isomerism because one of the doublebonded carbon atom has two identical methyl groups. Given: E. Trans-isomers have higher symmetry and I = 2A, t = 30min = 30 60sec better packing efficiency, leading to a higher W = melting point than cis-isomers. Video Solution: M nF ( Z = W = I t M nF ) 27 3 96500 2 30 60 W = 0. 336g Video Solution: Android App | iOS App | PW Website JEE Let the number of moles of C O2 be n Initial moles of C a(OH )2 be 2 n Excess moles of C a(OH )2 be n As, gm equivalent of C a(OH )2 = gm equivalent of HCl Thus, n 2 Q46 Text Solution: = 0. 1 (40/1000) 1 On solving, 3 n = 2 10 Volume of C O2 44. 8cm 3 = 2 10 22400 = 3 Video Solution: Video Solution: Q49 Text Solution: As, for first order reaction: 0.693 t1/2 = Also, k k = Ae Given: Q47 Text Solution: 20 Using Carius method for halogen estimation: A = 10 3 20 k = 10 = ( Mass of organic compound , Ea = 191. 48kJ mol = 191. 48 10 J mol Percentage of Cl Mass of C l in AgC l Ea /RT ) 100 = 10 10 20 10 191.48 10 e 10 = 10 1 1 , T = 1000K 3 8.314 1000 sec Therefore, Molar mass of 0.693 10 AgC l = 108 + 35. 5 = 143. 5g/mol i.e., 143.5 g AgCl contains = 35.5 g chlorine Given mass of AgCl = 143.5 mg ( 10 11 = 6. 93 10 12 = 69. 3 10 sec or 69. 3 picoseconds. Video Solution: Percentage of Cl : 35.5 180 ) 100 = 19. 72% = 20% Video Solution: Q50 Text Solution: Among the given molecules/ions, linear species are: Q48 Text Solution: Android App | iOS App | PW Website JEE Given, f (x + y) = f (x). f (y) then f (x) = e kx . . . . . . . . (i) f '(0) = 4a, Differentiate equation (i), we get f '(x) = ke kx At x = 0, f '(0) = ke 0 4a = k Now, f (x) = e Video Solution: 4ax f " (x) 3af '(x) f (x) = 0 k 2 3ak 1 = 0 Put the value of k, we get 2 16a 2 4a a = Q51 Text Solution: 2 12a 1 = 0 1 = 0 1 2 (a > 0) f (x) = e x Let R be the required relation 2 x For equivalence A = {(1, 1)(2, 2), (3, 3)} Area = e dx = e Case 1 : n(R) = 3, when R = A Video Solution: 0 2 1 Case 2 : n(R) = 5, R = A {(1, 2), (2, 1)} R = A {(1, 3), (3, 1)} R = A {(2, 3), (3, 2)} Number of Relations will be 3 Case 3 : R = {1, 2, 3} {1, 2, 3} Total number of equivalence relation Q53 Text Solution: = 1 + 3 + 1 = 5 Let 'G' be the centroid of triangle formed by (1,3), Video Solution: (3,1) and (2,4). G (2, 8 3 ) Image of 'G', with respect to x + 2y 2 2 1 = = 2 15 2 8 3 (2+ = 2 , = 16 3 2) 2 5 ( 16 3 ) 24 15 N ow, 15( ) = 15 ( Q52 Text Solution: = 1+4 = 0 Video Solution: Android App | iOS App | PW Website 2 15 + 24 15 ) = 22 JEE 16 ((sec 1 x) 2 + (cosec 2 1 x) ). . . . . . . . . . . . (i) 1 we know that, sec 1 and sec 1 x + cos ec x [0, ] { 2 x = 2 } now, put in (i), we get Q54 Text Solution: 16 ((sec 1 x) 2 + ( 2 Given, z1 = e i /4 , z2 = 1, z3 = e 1 let sec i /4 Now, 16 (2y 2 + z2 z 3 + z3 z 1 | |z 1 z = e i i = e 4 1 + 1 e 4 + e = 2 (cos i 4 4 + e i sin 4 4 + e 4 + 2 y + 2 4 ) = 0 4 e i 4 2 maximum value = 16 [2 ) + cos 2 + i sin 2 2 and minimum at y = (y = = 29 Video Solution: Q55 Text Solution: 2 + 2 ] 4 b 2a 4 for the quadratic in y) minimum value = 16 [ = 2 2 2 2 2 = 5, = 2 2 = 20 + (1 2 ) = 5 2 2 i 2 2 2 x = y Now, it is maximum aty = = 2 2 i + i = ( 2 ) 2 x) ) 2 i i 2 1 sec 2 16 2 4 + 2 4 ] 2 And sum of maximum & minimum value = 20 2 + 2 2 = 22 2 Video Solution: Q56 Text Solution: HH HH HT HT TH TH TT TT Outcomes number of times tail follows head H T H T H T H T 0 0 1 0 1 1 1 Probability distribution 0 xi P (x i ) 1 1/2 1/2 = xi p i = 2 = 2 1 2 = x pi 2 i 1 2 1 4 = 1 4 2 64 ( + ) = 64 ( Video Solution: Android App | iOS App | PW Website 1 2 + 1 4 ) = 48 0 JEE Q57 Text Solution: Now, direction ratio of P Q Given, < 3 2 + 1, 4 3 + 2, 5 4 + 2 a 1 a 5 = 28 4 a ar 2 a r 4 > P Q L1 = 28 , then (3 2 + 1)2 + (4 3 + 2)3 = 28 . . . . . . (i) + (5 4 + 2)4 = 0 a2 + a4 = 29 ar + ar 3 2 Similarly, P Q L2 ar (1 + r ) = 29 (3 2 + 1)3 + (4 3 + 2)4 2 2 2 2 2 a r (1 + r ) = (29) . . . . . . . . . . . + (5 4 + 2)5 = 0 = 2 r 2 (1+r ) 28 = 2 1 3 1 ; = 6 Now, 29 29 5 13 on comparing both sides: P( r = 28 Equation of Line P Q 2 a r 4 x = 28 (28) x 1 2 a6 = ar = 1 28 (28) 28 = 784 = 3 y 3 = 1 z = = y 3 2 z = 3 = 10 3 , 25 6 ) 13 3 3 1 Point ( 14 3 , 3, 22 3 ) lie on the line P Q y 2 3 = z 3 4 then P (2 + 1, 3 + 2, 4 + 3) x 2 , 13 Let point P lie on L1 L2 : 2 6 3 5 3 3 1 Q59 Text Solution: Q58 Text Solution: 2 ) & Q ( Video Solution: Video Solution: x 1 3 By observing the options: 28 5 , 3, 6 = 28 1 a = 3 5 1 2 2 . . . . . (ii) From equation (i) & (ii), we get from (i) & (ii), we get L1 : 50 38 + 21 = 0 . .(ii) a . . . . (i) 38 29 + 16 = 0 = 29 y 4 4 = z 5 5 and point Q lie on L2 then Q(3 + 2, 4 + 4, 5 + 5) Android App | iOS App | PW Website JEE 2 Given: e 5(ln x) +3 = x Q61 Text Solution: 8 Since 'M' is fixed in the middle position (3rd Taking log on both sides: 2 ln e position), 8 5(nnx) +3 = lnx we need to select 2 letters from the alphabet 2 5(ln x) that come before 'M' + 3 = 8 ln x and 2 letters that come after 'M'. (Let ln x = t) 5t 2 8t + 3 = 0 8 t1 + t2 = Therefore, the number of ways to choose 2 5 ln x1 + ln x2 = ln x1 x2 = x1 x2 = e letters from 12 is 8 5 C(12,2)= = 8 5 12! 2!(12 2)! = 12 11 2 1 = 66. Similarly, the number of ways to choose 2 letters 8/5 from 13 is Video Solution: C (13, 2) = 13! 2!(13 2)! = 13 12 2 1 = 78. Finally, since the selections of letters before and after 'M' are independent, we multiply the two results: 66 78 Video Solution: Q60 Text Solution: Tn = Sn Sn 1 = (2n 1)(2n+1)(2n+3)(2n+5) 64 (2n 3)(2n 1)(2n+1)(2n+3) Tn = 1 Tn 1 Tn = 8 8 (2n 1)(2n+1)(2n+3) = 2( n 1 (2n 1)(2n+1) r=1 1 1.3 n n r=1 n 1 1 3.5 ) + ( 1 3.5 1 5.7 ) + ] Tr 1 (2n 1)(2n+3) n lim n ) 1 lim 1 3 1 (2n 1)(2n+3) Tr = lim 2 [( = 2( 1 lim n Q62 Text Solution: 64 (2n 1)(2n+1)(2n+3) r=1 1 Tr = ) 2 3 Video Solution: Android App | iOS App | PW Website = 5148. JEE 1 2 Given, y dx + (x y Since at x = 0, y = 3, )dy = 0 parabola cuts the coordinate y-axis dx 1 + ( dy y2 1 )x = at R (0, 3) y3 comapre with general linear differential equation dx dy (x + 1) + Px = Q 1 we get, P = y Let equation of circle: y Now, I . F . = e P dy 3 = e 2 + (y + 1) = r 2 R (0, 3) will satisf y the circle 1 , Q = 2 2 2 1 y (0 + 1) dy 2 = e Then General solution : y r 2 = r 2 = 5 (x + 1) x. (IF) = Q. (IF) dy + C 2 + ( 3 + 1) 1 x 2 2 2 + (y + 1) + 2x + y 2 = 5 + 2y 3 = 0 N ow at y = 0, 1 x e 1 let, x e = (e y = t y 1 1 ) y 1 2 y 1 3 y dy x + 2x 3 = 0 (x + 3) (x 1) = 0 dy = dt x = 3, 1 t = t e dt y 2 P (1, 0) and Q ( 3, 0) Using integration by parts, we get x e x e 1 = te y 1 1 = y y e t + e t 1 y + e Area of PQR = + C e = 1 1 + e C = x e y e 2 = 2 e 2 + 1 3 0 1 0 3 + C 1 Video Solution: 1 e = 1 y 1 e 2 e 1 y + e 1 y 1 e 1 Now, at y = x 0 Area of PQR == 6 sq. units 1 y 2 1 + C e 1 At, x = 1, y = 1 1 1 2 x = 3 e Video Solution: 1 e Q64 Text Solution: The circle C1 has a radius of 2 and touches both coordinate axes in the second quadrant. Therefore, its center is (-2, 2). The equation of circle C1 is (x + 2) 2 2 + (y 2) = 2 2 Let the circle C2 with center (2, 5) and radius r. (x 2) Q63 Text Solution: 2 2 + (y 5) = r 2 The distance between the centers of circles C1 and C2 is: 2 d = (2 ( 2)) 2 + (5 2) = 5 For the circles to intersect at exactly two points, the following condition must hold: |r 2| < 5 < r + 2 |r 2| < 5 3 < r < 7 5 < r + 2 r > 3 On combining, we get 3 Android App | iOS App | PW Website < r < 7. JEE From the inequality 3 < r < 7 f (x + y) = f (x)f (y) + f (y)f (x) . . we can identify = 3 and = 7. . (i) 3 2 = 3(7) 2(3) = 15 put x = y = 0 in (i) Video Solution: f (0) = 1 2 Put y = 0 in (i), f (x) = f (x)f (0) + f (x) f (x) = f (x) = 7 tan 2 8 x + 7 tan 6 x 3 tan 4 x 2 n f (x) = 1 x + c 2 put f (0) = 0 c = 0 x f (x) = 7 tan 3 tan 2 On integrating, we get Q65 Text Solution: 3 tan f (x) 6 x (tan x (tan f (x) = (7 tan n f (x) = x + 1) 100 2 6 2 x + 1) n=1 x 3 tan 2 log e x 2 100 n n=1 2 f (n) = = 2525 Video Solution: 2 x)sec x On integrating both sides, we get I1 = /4 0 I1 = 1 0 (7 tan (7t 6 6 x 3 tan 2 3t )dt = [t 2 7 x) (sec 1 3 t ] 0 2 x)dx = 0 I2 == /4 0 x (7 tan = [x (tan /4 0 (tan 7 7 6 x 3 tan x tan x tan 3 3 x)] 2 x) (sec /4 0 2 x)dx Q67 Text Solution: for m = 1 x)dx for m = 2 = 0 /4 0 for m = 3 tan 3 x (tan 2 x 1) (1 + tan 2 x)dx for m = 4 = 1 0 (t 5 3 t )dt = [ 7I1 + 12I2 = 1 Video Solution: t 6 6 t 4 4 ] = 1 12 for m = 5 for m = 6 for m = 7 for m = 8 for m = 9 1 2 2 3 3 4 4 5 5 6 , , , , , 6 8 3 2 5 3 5 4 7 5 7 , , , , , 1 4 2 7 3 7 , , , 4 8 9 3 8 , 5 9 , 8 9 7 9 , 7 10 1 9 10 1 then n(B) = 31 Video Solution: Q66 Text Solution: Android App | iOS App | PW Website 1 10 4 , 3 10 5 3 9 5 2 , 4 1 7 7 1 , 3 9 JEE Q68 Text Solution: The probability P (A B) = Given equations are: (x 2 3)2 + y 2 9 . Here, m = 5 and n = 9, and gcd(5, 9) = 2 3 x 2 = 12 & y 5 = 1. point of intersection of both curves m + n = 5 + 9 = 14 (2 3 , 2 3 ) Video Solution: Radius of circle = 2 3 Q70 Text Solution: The distance between the foci is Area = 2[ 1 4 2 3 area of circle 0 2 3 x dx] 1 4 1 12 2 1 3 2 4 2 3 (x 3/2 The center of the hyperbola is the midpoint of which is ( 2 3 ) ] 0 2 2 .3 = 2 [3 4 3 . (2 2 . 3 . 3 1 4 1+1 2 , 14+( 12) 2 ) = (1, 1) Since the foci have the same x-coordinate, 1 = 14 ( 12) = 26 c = 13. the foci, = 2[ 2c )] the hyperbola has a vertical transverse axis. 2 (y 1) The equation is of the form 2 a 2 (x 1) b 2 = 1 The distance from the center (1, 1) to the point (1, = 2(3 8) = 6 16 sq. units 6) Video Solution: on the hyperbola is a = 6 1 = 5. We know that c 2 2 132 = a = 5 2 2 + b 2 + b b = 12 The length of the latus rectum is given by 2b a 2 = 2(144) 5 = 288 5 . Video Solution: Q69 Text Solution: The probability that the first ball is black is 10 6 After selecting a black ball, there are 5 black balls left out of 9 total balls. So, the probability that the second ball is black is 5 Q71 Text Solution: 9 P (A B) = Now, P (B) = 6 10 1 3 + 5 = 9 4 15 30 90 = = 5 15 1 3 + 4 = 15 9 15 = 3 5 By conditional probability, 1 P (A B) = P (A B) P (B) = 3 3 = 1 3 5 3 = 5 9 5 Android App | iOS App | PW Website JEE 11 Given, 5 Given, r=0 3ax f (x) = { 2 2 2, x < 1 = then LHL=RHL 2 3a 2 = a 11 5 f(x) is continuous, + b . . . . . . (i) = Also given, f(x) is differentiable, = then LHD = RHD m = n Now, + bx, x 1 a C2r+1 2r+2 C 2r+1 r=0 2r+2 1 5 12 1 2 r=0 [ 1 12 12 12 C (2 2 + 12 1 = C 5 1 12 r=0 12 11 2r+2 C 2r+1 2r+2 12 C 4 + 1) = 12 C 6 + 12 C 8 + 2047 12 compare with given condition, we get 6a = b . . . . . .(ii) m = 2047, n = 12 from (i) & (ii), we get then m n = 2047 12 = 2035 a = 2 (a > 1) & b = 12 Now, f (x) = { 6x 2 2 4 12x ; x < 1 1 2 1 Video Solution: ; x 1 Q73 Text Solution: Now, the required area = 3 ( 6x2 Given, |A|=-2 2 + 20)dx + (4 12x + 20)dx Now, 3adj( 6adj(3A)) = (16 + 12 3 ) + 6 3 = 3 adj( 6adj(3A)) = 22 + 12 3 Now, comparing with + 3 , we get = 22, = 12 4 6 3 = 3 ( 6) 3A = 3 then, + = 22 + 12 = 34 Video Solution: 12 = 3 21 21 4 6 2 ( 2) 2 10 = 2 m+n 3 mn So, m+n =10, mn =21 On simplifying, we get m = 7 & n = 3 (m > n) 4m + 2n = 4 7 + 2 3 = 34 Video Solution: Q72 Text Solution: Q74 Text Solution: Android App | iOS App | PW Website C 10 + 12 C 12 ] JEE L1 : x 1 3 = y 1 1 z+1 = we know that, projection vector = (let) 0 x = 3 + 1, y = + 1, z = 1 and L2 : x 2 2 = y = 0 z+4 c = ( b a = then x = 2 + 2, y = 0, z = 4 c = ( Now, given, L & L2 intersects each other 1 (3 + 1, + 1, 1) (2 + 2, 0, Now, 4) a +8 9 ) a a ) ( i + 2 j + 2k ) = 7 a + c i + 2 j + 2k + ( +8 9 )( i + 2 j + 2k ) = 7 On comparing, we get = 4 = 1, = 1, = 3 b = 4 i + 4k Point of intersection B(4, 0, 1) Now, Let point P is (2 + 2, 0, 3 4) point A is (1, 1, 1) Dr's of AP < 2 + 1, 1, 3 3 > 7 AP L2 = P( 40 13 , 0, 31 then, PB 13 2 4 3 ( i + 2 j + 2k ) Area of parallelogram = b c i j k = 4 13 ) = ( c = 0 4 = 16 sq. units 4 8 8 3 3 3 Video Solution: 144 169 + 324 169 ) = 468 169 2 Now, 26 ( PB) = 26 3 468 169 = 216 Video Solution: Q75 Text Solution: Android App | iOS App | PW Website JEE Android App | iOS App | PW Website

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