1-JEEM-2023-APRIL-06-FIRST-SHIFT-PAPER.pdf
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FIITJEE Solutions to JEE(Main) -2023 Test Date: 6th April 2023 (First Shift) PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours Maximum Marks: 300 Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. Important Instructions: 1. The test is...
FIITJEE Solutions to JEE(Main) -2023 Test Date: 6th April 2023 (First Shift) PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours Maximum Marks: 300 Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. Important Instructions: 1. The test is of 3 hours duration. 2. This test paper consists of 90 questions. Each subject (PCM) has 30 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 : Do any 5 questions out of 10 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 (1 – 10) contains 10 Numerical based questions. The answer to each question is rounded off to the nearest integer value. Each question carries +4 marks for correct answer and –1 mark for wrong answer. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-2 PART – A (PHYSICS) SECTION - A (One Options Correct Type) This section contains 20 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE option is correct. Q1. For the plane electromagnetic wave given by E Eo sin t kx and B BO sin t kx . the ratio of average electric energy density to average magnetic energy density is (A) 4 (B) 2 (C) 1/2 (D) 1 Q2. The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is (A) C (B) A (C) B (D) D A B n=4 C D n=3 n=2 n=1 Q3. The number of air molecules per cm 3 increased from 3 1019 to 12 1019. The ratio of collision frequency of air molecules before and after the increase in number respectively is: (A) 0.25 (B) 0.50 (C) 0.75 (D) 1.25 Q4. The induced emf can be produced in a coil by A. moving the coil with uniform speed inside uniform magnetic field B. moving the coil with non uniform speed inside uniform magnetic field C. rotating the coil inside the uniform magnetic field D. changing the area of the coil inside the uniform magnetic field Choose the correct answer from the options given below: (A) C and D only (B) B and D only (C) B and C only (D) A and C only Q5. Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : When a body is projected at an angle 45 , it’s range is maximum. Reason R : For maximum range, the value of sin 2 should be equal to one. In the light of the above statements, choose the correct answer from the options given below : (A) A is false but R is true (B) A is true but R is false (C) Both A and R are correct and R is the correct explanation of A (D) Both A and R are correct but R is NOT the correct explanation of A FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-3 Q6. For a uniformly charged thin spherical shell, the electric potential (V) radially away from the centre (O) of shell can be graphically represented as - O A R V V (A) (B) r=R r=R r r V V (C) (D) r=R r r=R r Q7. A planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing W on earth will weight on that planet: (A) 2W (B) 22/3 W 1/3 (C) 2 W (D) W Q8. Given below are statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Earth has atmosphere whereas moon doesn’t have any atmosphere. Reason R : The escape velocity on moon is very small as compared to that on earth. In the light of the above statements, choose the correct answer from the options given below. (A) Both A and R are correct but R is NOT the correct explanation of A (B) A is true but R is false (C) A is false but R is true (D) Both A and R are correct and R is the correct explanation of A Q9. A small ball of mass M and density is dropped in a viscous liquid of density 0. After some time, the ball falls with a constant velocity. What is the viscous force on the ball? (A) F Mg 1 0 (B) F Mg 1 0 (C) F Mg 1 0 Q10. (D) F Mg 1 0 Two resistances are given as R1 10 0.5 and R2 15 0.5 . The percentage error in the measurement of equivalent resistance when they are connected in parallel is – (A) 2.33 (B) 5.33 (C) 6.33 (D) 4.33 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-4 Q11. A monochromatic light wave with wavelength 1 and frequency v 1 in air enters another medium. If the angle of incidence and angle of refraction at the interface are 45 and 30 respectively, then the wavelength 2 and frequency v2 of the refracted wave are : (A) 2 1,v 2 2v1 (C) 2 1 2 1 (B) 2 1,v 2 1,v 2 v1 2 v1 (D) 2 2 1,v 2 v 1 Q12. A particle is moving with constant speed in a circular path. When the particle turns by an angle 90 , the ratio of instantaneous velocity to its average velocity is : x 2. The value of x will be. (A) 7 (B) 1 (C) 2 (D) 5 Q13. A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distributed across the cross-section. The magnetic field is (A) Inversely proportional to r in the region r < a and uniform through in the region r > a (B) uniform in the region r < a and inversely proportional to distance r from the axis, in the region r>a (C) zero in the region r < a and inversely proportional to r in the region r > a (D) directly proportional to r in the region r < a and inversely proportional to r in the region r > a Q14. Name the logic gate equivalent to the diagram attached (A) AND (B) OR (C) NAND (D) NOR +5V 1 o A 1 o B LED 1 Glow Y R Q15. A small block of mass 100g is tied to a spring of spring constant 7.5N/m and length 20 cm. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5 rad/s about point A. then tension in the spring is (A) 1.5 N (B) 0.50 N (C) 0.75 N (D) 0.25 N Q16. A mass m is attached to two strings as shown in figure. The spring constants of two springs are K1 and K2. For the frictionless surface, the time period of oscillation of mass m is Q17. K2 (A) 2 m K1 K 2 (B) 1 K1 K 2 m (C) 2 m K1 K 2 (D) 1 K1 K 2 m 2 2 m K1 The kinetic energy of an electron, particle and a proton are given as 4K, 2K and K are respectively. The de-Broglie wavelength associated with electron e , -particle ( and the proton p are as follows : (A) p e (C) p e (B) p e (D) p e FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-5 Q18. A source supplies heat to a system at the rate of 100W. If the system performs work at a rate of 200W. The rate at which internal energy of the system increases is (A) 600W (B) 500W (C) 1200W (D) 800W Q19. The receptivity of semiconductor varies with temperature. Which of the following curve represents the correct behaviour (A) (B) T T (C) (D) T Q20. T By what percentage will the transmission range of a TV tower be affected when the height of the tower in increased by 21% ? (A) 15% (B) 10% (C) 12% (D) 14% FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-6 SECTION - B (Numerical Answer Type) This section contains 10 Numerical based questions. The answer to each question is rounded off to the nearest integer value. Q1. A steel rod has a radius of 20mm and a length of 2.0m. A force of 62.8kN stretches it along its length. Young’s modulus of steel is 2.0 1011N / m2. The longitudinal strain produced in the wire is _________ 10 5 Q2. A particle of mass 10g moves in a straight line with retardation 2x, where x is the displacement in n 10 SI units. Its loss of kinetic energy for above displacement is J. The value of n will be _______ x Q3. A pole is vertically submerged in swimming pool, such that it gives a length of shadow 2.15m within water when sunlight is incident at an angle of 30 with the surface of water. If swimming pool is filled to a height of 1.5m, then the height of the pole above the water surface in centimetres is nW 4 / 3 __________ Q4. Two identical solid sphere each of mass 2kg and radii 10cm are fixed at the ends of a light rod. The separation between the centres of the spheres is 40cm. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is __________ 103 kg m2 Q5. An ideal transformer with purely resistive load operates at 12kV on the primary side. It supplies electrical energy to a number of nearby houses at 120V. The average rate of energy consumption in the houses served by the transformer is 60 kW. The value of resistive load (Rs) required in the secondary circuit will be _________ m . Q6. Q7. A parallel plate capacitor with plate area A and plate separation d is filled with a dielectric material of dielectric constant K = 4. The thickness of the dielectric material is x, where x < d. Let C1 and C2 be 1 the capacitance of the system for x d and 3 2d x. respectively. If C1 2F the value of C2 is 3 _________ F Plate - 1 d Dielectric x Plate - 2 Two identical circular wires of radius 20cm and carrying current 2 A are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires is _________ 10 8 T. Y I X I FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-7 Q8. The radius of fifth orbit of the Li is ________ 10 12 m. Take : radius of hydrogen atom = 0.51 A Q9. The length of a metallic wire is increased by 20% and its area of cross section is reduced by 4%. The percentage change in resistance of the metallic wire is ________ Q10. A person driving car at a constant speed of 15m/s is approaching a vertical wall. The person notices a change of 40Hz in the frequency of his horn upon reflection from the wall. The frequency of horn is _________Hz. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-8 PART – B (CHEMISTRY) SECTION - A (One Options Correct Type) This section contains 20 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE option is correct. Q1. Given below: are two statements one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Loss of electron from hydrogen atom results of 1.5 103 pm size. Reason R: Proton (H+) always exists in combined form. In the light of the above statements, choose the most appropriate answer from the options given below. (A) Both A and R are correct but R is NOT the correct explanation of A (B) Both A and R are correct and R is the correct explanation of A (C) A is not correct but R is correct (D) A is correct but R is not correct Q2. The possibility of photochemical smog formation is more at (A) Marshy lands (B) Industrial areas (C) Himalayan villages in winter (D) The places with healthy vegetation Q3. For a concentrated of a weak electrolyte (Keq=equilibrium constant) A2B2 of concentration ‘c’, the degree of dissociation ‘a’ is 1 1 K eq 5 (A) 5 6c K eq 5 (B) 2 25c 1 1 K eq 5 (C) 4 5c K eq 5 (D) 4 180c Q4. A compound is formed by two elements X and Y. The element Y forms cubic close packed arrangement and those of element X occupy one third of the tetrahedral voids. What is the formula of the compound? (A) XY3 (B) X3Y2 (C) X3Y (D) X2Y3 Q5. For the reaction Acetone RCH2Br I RCH2I Br major The correct statement is (A) The reaction can occur in acetic acid also. (B) The transition state formed in the above reaction is less polar than the localised anion. (C) The solvent used in the reaction solvates the ions in rate determining step. (D) Br can act as competing nucleophile. Q6. Polymer used in orlon is: (A) Polycarbonate (C) Polyacrylonitrile (B) Polyamide (D) Polyethene FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-9 Q7. The difference between electron gain enthalpies will be maximum between: (A) Ne and Cl (B) Ar and Cl (C) Ne and F (D) Ar and F Q8. The major products A and B from the following reactions are: H N LiAlH4 Br2 / AcOH B A O Br H N H N (A) A = , B= O Br OH Br Br H N NH2 (B) A = O Br B= , Br H N H N (C) A = O Br , B= Br H N H N , (D) A = O B= OH Br Q9. The standard electrode potential of M+ / M in aqueous solution does not depend on (A) Ionisation of a gaseous metal atom (B) Hydration of a gaseous metal ion (C) Ionisation of a solid metal atom (D) Sublimation of a solid metal Q10. Match List I with List II List IList II Enzymatic reaction Enzyme A. Sucrose Glucose and Fructose I. Zymase B. Glucose ethyl alcohol and CO2 II. Pepsin C. Starch Maltose III. Invertase D. Proteins Amino acids IV. Diastase Choose the corect answer from the option given below : (A) A-I, B- IV, C-III, D-II (B) A-I, B-II, C-IV, D-III (C) A-III, B-I, C-II, D-IV (D) A-III, B-I, C-IV, D-II Q11. Match List I with List II List IOxide A. N2O4 B. NO2 C. N2O5 D. N2O I. II. III. IV. List II Type of bond 1N = O bond 1N –O–N bond 1N–N bond 1N=N / NN bond FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-10 Choose the corect answer from the option given below : (A) A-II, B- IV, C-III, D-I (B) A-III, B-I, C-IV, D-II (C) A-II, B-I, C-III, D-IV (D) A-III, B-I, C-II, D-IV Q12. Match List I with List II List IList II Vitamin Deficiency disease A. Vitamin A I. Beri-Beri B. Thiamine II. Cheilosis C. Ascorbic acid III. Xeropthalmia D. Riboflavin IV. Scurvy Choose the corect answer from the option given below : (A) A-IV, B- I, C-III, D-II (B) A-III, B-II, C-IV, D-I (C) A-IV, B-II, C-III, D-I (D) A-III, B-I, C-IV, D-II Q13. The setting time of Cement is increased by adding (A) Silica (B) Gypsum (C) Limestone (D) Clay Q14. Which of the following options are correct for the reaction 2 Au CN 2 2 aq Zn(s) 2Au(s) Zn CN4 aq A. Redox reaction B. Displacement reaction C. Decomposition reaction D. Commination reaction Choose the correct answer from the options given below : (A) A only (B) A and B only (C) A and D only (D) C and D only Q15. Compound P HCl, Residue Q Filter Filtrate NaOH Oily Liquid R. Compound P is neutral, Q gives effervescence with NaHCO3 while R reacts with Hinsbergs reagent to give solid soluble in NaOH. Compound P is O O N (A) N (B) H CH3 H3C O O C N H (C) (D) H N CH3 H CH3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-11 Q16. Match List I with List II List IList II Element detected Reagent used / Product formed A. Nitrogen I. Na2[Fe(CN)5NO] B. Sulphur II. AgNO3 C. Phosphorous III. Fe4[Fe(CN)6]3 D. Halogen IV. (NH4)2MoO4 Choose the corect answer from the option given below : (A) A-II, B- I, C-IV, D-III (B) A-IV, B-II, C-I, D-III (C) A-III, B-I, C-IV, D-II (D) A-II, B-IV, C-I, D-III Q17. Strong reducing and oxidizing agents among the following respectively, are (A) Eu2+ and Ce4+ (B) Ce4+ and Tb4+ 4+ 2+ (C) Ce and Eu (D) Ce3+ and Ce4+ Q18. The major product formed in the following reaction is CONH2 COOCH3 Br2 / NaOH O O (B) (A) O O (C) NH O O (D) O O Q19. NH Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The spin only magnetic moment value for [Fe(CN)6 ]3 is 1.74 BM, whereas for Fe H2 O ]3 is 5.92 BM, 6 Reason R: In both complexes, Fe is present in +3 oxidation state. In the light of the above statements, choose the correct answer from the options given below: (A) Both A and R true and R is the correct explanation of A (B) A is false but R is true (C) Both A and R are true but R is NOT the correct explanation of A (D) A is true but R is false. Q20. Match List I with List II List IList II Name of reaction Reagent used A. Hell-Volhrd-Zelinsky reaction I. NaOH+I2 B. Iodoform reaction II. (i) CrO2Cl2,CS2 (ii) H2O C. Etard reaction III. (i) Br2 / red phosphorus (ii) H2O D. Gatterman-Koch reaction IV. CO, HCl anhyd. AlCl3 Choose the corect answer from the option given below : (A) A-III, B- I, C-IV, D-II (B) A-I, B-II, C-III, D-IV (C) A-III, B-I, C-II, D-IV (D) A-III, B-II, C-I, D-IV FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-12 SECTION - B (Numerical Answer Type) This section contains 10 Numerical based questions. The answer to each question is rounded off to the nearest integer value. Q1. Mass of Urea (NH2CONH2) required to be dissolved in 1000g of water in order reduce the vapour pressure of water by 25% is _______g. (Nearest integer) Given: Molar mass of N,C,O and and H are 14,12,16 and 1 g mol1respectively. Q2. Consider the graph of Gibbs free energy G vs Extent of reaction. The number of statement/s from the following which are true with respect to points (a), (b) and (c) is_____. A. Reaction is spontaneous at (a) and (b) B. Reaction is at equilibrium at point (b) and nonspontaneous at point (c) C. Reaction is spontaneous at (a) and non-spontaneous at (c) D. Reaction is non-spontaneous at (a) and (b) Gibbs energy a b c Extent of reaction Q3. Number of ambidentate ligands in a respective metal complex [M(en)(SCN)4] is_______. [En = ethylenediamine] Q4. The value of log K for the reaction A B at 298 K is__________. (Nearest integer) Given: Ho = 54.07 kJ mol1 So = 10 J K1 mol1 (Take 2.3038.314298 = 5705) Q5. For the adsorption of hydrogen on platinum, the activation energy is 30 kJ mol1 and for the adsorption of hydrogen on nickel, the activation is 41.4kJ mol1. The logarithm of the ratio of the rates of chemisorptions on equal areas of the metals at 300K is _________(Nearest integer) Given: ln 10 = 2.3 R = 8.3 JK1 mol1 Q6. The number of species from the following which have square pyramidal structure is_____ PF5 ,BrF4 ,IF5 ,BrF5 , XeOF4 ,ICl4 Q7. In ammonium – phosphomolybdate, the oxidation state of Mo is+______________. Q8. The wavelength of an electron of kinetic energy 4.50105m, (Nearest integer) Given: mass of electron is 91031kg, h= 6.61034Js. Q9. Number of bromo derivatives obtained on treating ethane with excess of Br2 in diffused is______________. Q10. If 5 moles of BaCl2 is mixed with 2 moles of Na3PO4, the maximum number of moles of Ba3(PO4)2formedis_________(Nearest integer) sunlight FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-13 PART – C (MATHEMATICS) SECTION - A (One Options Correct Type) This section contains 20 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE option is correct. Q1. ˆ ˆi 2jˆ 3k, ˆ 2iˆ ˆj 4kˆ and Let the position vectors of the points A, B, C and D be 5iˆ 5 ˆj 2k, 2 ˆi 5ˆj 6kˆ. Let the set S { : the points A, B, C and D are coplanar}. Then 2 is S equal to (A) 41 37 (C) 2 Q2. If 2x y 3y x 20 , then 2 loge 8 (A) 3 loge 4 3 loge 4 (C) 2 loge 8 (B) 25 (D) 13 dy at 2, 2 is equal to : dx 3 loge 16 (B) 4 loge 8 3 loge 8 (D) 2 loge 4 Q3. One vertex of a rectangular parallelepiped is at the origin O and the lengths of its edges along x, y and z axes are 3, 4 and 5 units respectively. Let P be the vertex 3, 4, 5 . Then the shortest distance between the diagonal OP and an edge parallel to z axis, not passing through O or P is : 12 (A) (B) 12 5 5 12 12 (C) (D) 5 5 5 Q4. The straight lines 1 and 2 pass through the origin and trisect the line segment of the line L : 9x 5y 45 between the axes. If m1 and m2 are the slopes of the lines 1 and 2 , then the point of intersection of the line y m1 m2 x with L lies on (A) y x 5 (C) 6x y 15 Q5. (B) y 2x 5 (D) 6x y 10 If the equation of the plane passing through the line of intersection of the planes x 1 y 3 z 2 2x y z 3, 4x 3y 5z 9 0 and parallel to the line is 2 4 5 ax by cz 6 0 , then a b c is equal to (A) 13 (B) 15 (C) 14 (D) 12 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-14 ˆ b ˆi 2ˆj 2kˆ and c ˆi 4ˆj 3kˆ. If d is a vector perpendicular to both b Let a 2iˆ 3ˆj 4k, 2 and c , and a d 18 , then a d is equal to Q6. (A) 640 (C) 720 (B) 760 (D) 680 Q7. A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the k probability of at least 4 successes is 11 , then k is equal to 3 (A) 164 (B) 75 (C) 123 (D) 82 Q8. Let A x R : x 3 x 4 3 , greatest integer function. Then, (A) A = B (C) B C, A B Let I x Q9. (A) loge (C) loge Q10. Q11. 4 x tan x 1 2 32 4 (B) A B, A B (D) A B dx. If I 0 0 , then I is equal to 4 2 4 4 (B) loge 2 4 4 (D) loge 2 16 2 t denotes 4 2 16 4 2 4 4 2 4 4 2 32 The sum of all the roots of the equation x 2 8x 15 2x 7 0 is : (A) 9 3 (B) 9 3 (C) 11 3 (D) 11 3 From the top A of a vertical wall AB of height 30 m, the angles of depression of the top P and bottom of Q of a vertical tower PQ are 150 and 600 respectively, B and Q are on the same horizontal level. If C is a point on AB such that CB = PQ, then the area (in m 2) of the quadrilateral BCPQ is equal to (C) 600 (A) 300 Q12. x 2 x sec 2 x tan x x3 3 B x R : 3 x r 3 3x , where r 1 10 3 1 3 1 (D) 300 3 1 (B) 200 3 3 If the system of equations x y az b 2x 5y 2z 6 x 2y 3z 3 has infinitely many solutions, then 2a 3b is equal to (A) 20 (B) 23 (C) 25 (D) 28 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-15 Q13. Statement P Q R Q is logically equivalent to (A) P R Q R (B) P R Q (C) P R Q R (D) P R Q Q14. Let a1, a 2 , a3 ,.......an be n positive consecutive terms of an arithmetic progression. If d > 0 is its common difference, then d 1 1 1 is lim ..... n n a1 a2 a2 a3 an1 an (A) d (B) 0 1 (C) 1 (D) d Q15. The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and 2 respectively. If the variance of all the 30 numbers in the two sets is 13, then 2 is equal to (A) 9 (B) 12 (C) 11 (D) 10 Q16. The sum of the first 20 terms of the series 5 11 19 29 41 ... is (A) 3450 (B) 3520 (C) 3420 (D) 3250 Q17. Let A aij , where aij 0 for all i, j and A 2 I. Let a be the sum of all diagonal elements of A 22 and b A. Then 3a2 4b2 is equal to (A) 7 (C) 4 Q18. Q19. (B) 14 (D) 3 2 1 1 Let 5f x 4f 3, x 0. Then 18 f x dx is equal to : x x 1 (A) 10loge 2 6 (B) 5loge 2 3 (C) 5loge 2 3 (D) 10loge 2 6 If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of n 4 1 2 4 is 3 6 : 1, then the third term from the beginning is : (A) 30 2 (C) 60 3 Q20. If 2n (B) 60 2 (D) 30 3 C3 : n C3 10 : 1 , then the ratio n2 3n : n2 3n 4 is (A) 27:11 (C) 35:16 (B) 65:37 (D) 2:1 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-16 SECTION - B (Numerical Answer Type) This section contains 10 Numerical based questions. The answer to each question is rounded off to the nearest integer value. Let y y x be a solution of the differential equation Q1. 0x Q2. Let R x cos x dy xy sin x y cos x 1 dx 0, . If y 3 , then y " 2y ' is equal to……. 6 6 2 3 3 6 A 1, 2, 3, 4,....,10 and a,b A A : 2 a b 2 B 0, 1, 2, 3, 4. The number of elements in the relation 3 a b B is……….. A circle passing through the point P , in the first quadrant touches the two coordinate axes at Q3. the points A and B. The point P is above the line AB. The point Q on the line segment AB is the foot of perpendicular from P on AB. If PQ is equal to 11 units, then the value of is……….. Let the point p, p 1 lie inside the region Q4. E 2 x, y : 3 x y 9 x 2 ,0 x 3. If the set of all values of p is the interval a, b , then 2 b b a is equal to………. Let the image of the point P 1, 2, 3 in the plane 2x y z 9 be Q. If the coordinates of the point Q5. R are 6,10,7 , then the square of the area of the triangle PQR is……….. Let the tangent to the curve x 2 2x 4y 9 0 at the point P 1, 3 on it meet the y-axis at A. Let Q6. the line passing through P and parallel to the line x 3y 6 meet the parabola y 2 4x at B. If B 2 lies on the line 2x 3y 8 , then AB is equal to………… Q7. If the area of the region S x, y : 2y y 2 x 2 2y, x y is equal to n2 , then the n 1 n 1 natural number n is equal to……….. Q8. Let a and t be the greatest integer t. Then the number of points, where the function f x a 13 sin x , x 0, is not differentiable, is………. Q9. The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is……….. 15 Q10. 1 The coefficient of x18 in the expansion of x 4 3 is………….. x FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-17 FIITJEE KEYS to JEE (Main)-2023 PART – A (PHYSICS) SECTION - A 1. D 2. D 3. A 4. A 5. C 6. A 7. C 8. D 9. B 10. D 11. C 12. C 13. D 14. D 15. C 16. A 17. B 18. D 19. B 20. B SECTION – B 1. 25 2. 2 3. 50 4. 176 5. 240 6. 3 7. 628 8. 425 9. 25 10. 420 PART – B (CHEMISTRY) SECTION - A 1. A 2. B 3. D 4. D 5. B 6. C 7. A 8. C 9. C 10. D 11. D 12. D 13. B 14. B 15. A 16. C 17. A 18 C 19. C 20. B SECTION - B 1. 1111 2. 2 3. 4 4. 10 5. 2 6. 3 7. 6 8. 7 9. 9 10. 1 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-18 PART – C (MATHEMATICS) SECTION - A 1. A 2. A 3. D 4. A 5. C 6. C 7. C 8. A 9. A 10. A 11. C 12. B 13. B 14. C 15. D 16. B 17. C 18. A 19. C 20. D SECTION - B 1. 2 2. 18 3. 121 4. 3 5. 594 6. 292 7. 5 8. 25 9. 3483638676 10. 5005 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-19 FIITJEE Solutions to JEE (Main)-2023 PART – A (PHYSICS) SECTION – A Sol1. 1 E2 UE 2 0 0 0 UB B2 20 2 1 E 2 2 c B c Sol2. As we know that hc 1 E E Now, for the shortest wavelength, the energy gap must be max. Thus, n 3 n 1 , is the correct option Sol3. Collision frequency, v f v or f 2d2 vnv 1 2 d2 nv f nv , where nv = no. density or f1 nv 1 f 3 1019 1 1 0.25 f2 nv 2 f2 12 1019 4 Sol4. In a coil, the induced emf is produced only if the flux through it changes w.r.t time in a magnetic field. Thus, for uniform field, flux can be changed either by rotating the coil or by changing the area of the coil. Sol5. For a projectile, range (R) is given as u2 sin2 R g Also, Range is max. when sin 2 is max ie 1 So, sin 2 1 or 2 90 or 45 So, both Assertion & Reason are true. Also reason is the correct explanation of assertion. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-20 Sol6. Sol7. For spherical shell (or hollow sphere), electric potential is given by kQ V ;r R R kQ V ; r R r V kQ R r r=R MP 2Me ; P e wt on earth w mge wt on planet w ' mgP 4 P GRP w ' mgP 3 4 w mge e GRe 3 1 w ' P RP 1 2 3 w e R e 1 w' 2 3w 4 4 MP 2Me P. RP3 & Me e Re 3 3 3 MP 2Me 4 4 P. RP 3 2e R e3 3 3 1 RP 2 3 Re Sol8. At moon, due to the low escape velocity, the rms velocity of molecules is greater than escape velocity. Hence, molecules escape and there is no atmosphere at moon. Thus both Assertion and Reason are correct and reason is correct explanation of Assertion. Sol9. When ball is falling with constant velocity : Fv + B = wt Fv wt B Mg 0 vg B Fv M Mg 0 g Fv Mg 1 0 wt Sol10. R1 10 0.5 ; R2 15 0.5 As we know that the equivalent resistance in parallel combination 1 1 1 R R1 R2 Differentiating both sides, we get R R1 R 2 R2 R12 R22 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-21 R R1 R2 2 2 R R R R 2 1 0.5 0.5 6 100 225 13 300 R 13 % 100 R 300 13 % 3 4.33% Sol11. Using snell’s law 1 sin 45 2 sin30 1 1 2 2 1 2 21 2 45 c c 2 v1 v 2 v1 2 v2 1 1, 1 30 1, 1 2 1 (as frequency remains unchanged) Also, v or v v 1 1 2 v 2 2 1 2 Sol12. 1 2 v inst v 2R R 2v 2 2 x 2 x2 Sol13. For solid infinite current carrying wire : Using ampere circuital law ir B 0 2 ; r R 2 R i 0 , r R 2r FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-22 Sol14. The truth table for the given circuit is : NOR Gate A 0 1 0 1 B 0 0 1 1 Y 1 0 0 0 Sol15. kx m 2 r Where r x So, kx m 2 x 7.5x 0.1 52 0.2 x 7.5x 0.5 2.5x 5x 0.5 x 0.1m Thus, tension in the spring is T = kx = 7.5 × 0.1 = 0.75N Sol16. T 2 m k eq Where k eq k1 k 2 T 2 m k1 k 2 Sol17. According to De-broglie, h h P 2mkE e P h 2me kEe h 2mP kEP h 2m kE h 2m e 4k h 8mek h 2mPk h 2m.4k h 8mP K Thus, P e Sol18. From 1st law of thermodynamics dQ dU dw dQ dU dw Also, dt dt dt d 1000w 200w dt dU 800 w dt FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-23 Sol19. A semiconductor starts conduction more as the temperature increases. It means resistance decreases with increase in temperature. So, if temperature increases, its resistivity decreases. m Also, 2 ne As temperature increases, decreases but n increases and n is dominant over . Sol20. Range of a TV tower is given as R1 2Rh New range, R2 2g h 0.21h 2gh 1.21 1.1 R1 It means new range increases by 10%. SECTION - B Sol1. y stress strain strain stress F y Ay 62.8 1000 62.8 1000 2 11 r 2 10 3.14 400 10 6 2 1011 200 10 5 25 105 8 Sol2. 1 sin60 sin r 4 sin r 3 3 3 3 3 4 2 8 - (i) y 27 37 cosr 1 0.75 64 8 tan r 60 air water r 1.5m 27 0.85 37 x 0.85 1.5 x 1.275m or 2.15m y y 2.15 1.275 0.875 1 y 0.875 y 0.50 1.732 3 0.875 So, length of pole above water surface is 0.50m or 50cm. tan30 Sol3. Isystem 2 Icom m d2 2 2 mr 2 md2 5 2 2 2 0.01 2 0.04 5 2 0.008 0.08 176 10 3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-24 Sol4. VS NS VP NP N 120 S 12000 NP N 1 S - (i) 100 NP For an ideal transformer, Pinput = Poutput iP vP iS v S 60000 w iP Now 60000 5 12000 v 12000 RP P 2400 iP 5 & RS Sol5. Sol6. vS 120 120 120 240m iS 60000 / 120 60000 A 0 A 4 0 2d d 3 3 4 A 0 C1 A 40 3 d A 0 2d d 3 3 4 A 0 According to question, 2 3 d A 0 3 - (i) d 2 A0 A 40 d 2d A 3 3 3 Now, C 2 2 0 2 3 A 4 d 2 A 0 0 d 2d 3 3 Bnet B1 B2 i i Bnet 0 ˆi 0 ˆj 2r 2r i or Bnet 2 0 2r 4 10 2 7 Sol7. rn 2 0.2 2 628 108 T n2 z or rn 0.51 n2 A z FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-25 For Li , z 3 So, r5 Sol8. R R' 0.51 52 1010 m 17 25 1012 m 425 1012 m 3 A 1.21 0.96 A 10 R 1.25R 8 Thus, resistance increases by 25% Sol9. Given fapp factual 40 330 15 fapp f0 f0 40 330 15 345 f0 f0 40 315 30 f0 40 315 4 f0 315 420H2 3 Sol10. a 2x vdv 2x dx On integrating, both sides, we get 0 x vdv 2 xdx v0 0 2 0 x v x2 2 2 v0 2 0 v 02 x2 2 v 20 2x 2 Thus, loss in KE KE i KEf 0 1 mv 02 0 2 1 10 2x 2 2 1000 x2 100 x 10 10 x Thus, n = 2 2 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-26 PART – B (CHEMISTRY) SECTION – A Sol1. Size of nucleus is of order 1.5103pm. and H+ always exists in combined form & there is no relation between these two statements. Sol2. Photochemical smog is a mixture of pollutants that are formed when nitrogen oxides and volatile organic compounds react in the presence of sunlight, creating a brown haze above cities. Sol3. 3 2 A 2B 3aq 2A a1 3B aq t0 At equilibrium C C C 2 0 2C 0 3C 3 A 3 B 2 4C2 2 27C3 3 Keq = C 1 A 2 B3 1 108C5 5 Keq 5 Keq 4 c 108C Sol4. ‘y’ form ccp lattice so number of ‘y’ atom in one unit cell = 4 8. 3 atom ‘x’ occupy one third of tetrahedral void, so number of ‘x’ atom in one unit cell = Therefore formula of compound = X8 Y4 or X2 Y3 Sol5. acetone R CH2 Br I R CH2 I + Br 2 SN R transition state I C H Br H Bond formation Bond Breaking From the above transition state we can say that for SN2 reaction, transition state formed is polar than the localized anion. less Sol6. Sol7. n CH2 CH CH2 CH CN Acrylonitrile CN n Polyacrylonitrile (PAN) or orlon Out of ‘F’ & ‘Cl’ , ‘Cl’ have more negative value of electrongain enthalpy. And out of Ne & Ar, have more positive value of electrongain enthalpy. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. Ne JEE-MAIN-2023 (6th April-First Shift)-PCM-27 Sol8. NH NH LiAlH4 O O Br (B) Sol9. NH Br2 AcOH (A) Because standard electrode potential is define for 1M concentration. Invertase Sol10. Sucrose glucose & fructose Zymase Glucose ethyl alcohol & CO2 Diastase Starch maltose Pep sin Proteins Amino acids. Sol11. NO 2 N2O 4 O O N N Sol12. Vitamine A Thiamine Ascorbic acid Riboflavin O : : : O O N O N2O N2O 5 O N O O O N N N O N N O O Xerophthalmia Beri-Beri Scurvy : Cheilosis Sol13. Calcium sulphate (CaSO4) is in the form of Gypsum and its function is to increase the initial setting time of cement. Sol14. (A) Reduction of Au & oxidation of Zn takes place so it is a redox reaction. (B) Zn is displacing Au, so it is a displacement reaction. Sol15. O O H2N OH + NH HCl H3C H3C (Q) Q Carboxylic acid, gives effervescence with NaHCO3 R1o amine, react with Hinsbergs reagent to give ppt (soluble in NaOH) (R) Sol16. Nitrogen detected by lassaigne’s method. Sulphur is detected by sodium nitropursside Phosphorous is detected by ammonium molybdate & Halogens are defected by AgNO3. Sol17. +4 oxidation state: oxidation agent. +2 oxidation state: Reducing agent. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-28 Sol18. CONH2 COOCH3 NH2 Br2 NH H NaOH C OCH3 C O OCH3 O NH C O Sol19. Explanation:3 3 Fe CN6 O.S. of Fe 3 Fe H2 O 6 O.S. of Fe 3 Fe3 Ar 3d5 4S0 Fe3 Ar 3d5 4S0 d5 field is strong Pairing of e takes place d5 field is weak Pairing of e not occurs. d 5 d n n 2 B.M. 5 n n 2 B.M. Put n =1 (unpaired e ) put n=5 1.74B.M. 5.92B.M. So both A & R are true but R is not the correct explanation of A. Sol20. HVZ reaction Br2 / Red P Iodoform reaction NaOH+I2 Etard reaction CrO2Cl2 / CS2 Gatterman-koch reaction CO, HCl & Anh. AlCl3. SECTION – B Sol1. As we know that P Ps XB 0 Ps For very dilute solution nsolute P Ps 0 nsolvent Ps Let the mass of urea required = x P 0.75P0 x 18 0 60 1000 0.75P0 x 1111g Sol2. at point a Slope = -ve dG = -ve reaction is spontaneous. at point b Slope = 0 dG = 0 reaction at equilibrium at point c Slope = +ve dG = +ve reaction is non-spontaneous. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-29 Sol3. SCN is a ambidentate ligand. So no of ambidentate ligand = 4 Sol4. As we know that G0 H0 TS0 G0 54070 298 10 57050 Jmol 1 Also G0 2.303RT logk 57050 2.303RT logk logk Sol5. k 2 Ea 1 Ea 2 k1 2.303RT log 57050 57050 10 2.303 8.314 298 5705 41.4 30 1000 2.3 8.3 300 1.99 2 Sol6. IF5, BrF5 & XeOF4 have square pyramidal structure. Sol7. Ammonium-phosphomolybdate (NH4)3PO4.12MoO3 Let the O.S of Mo is x x6=0 x =+6 Sol8. Sol9. h 6.6 10 34 2 9 1031 4.50 10 29 After solving above expression we get the value of 7.3 105. 2mKE hv CH3 CH3 Br2 Brominated product (excess) Types of Bromination Mono bromination No of different possible structure CH3 CH2 Br Dibromination CH3 CHBr2 & BrCH2 CH2 Br Tribromination CH3 CBr3 & Br2 CH CH2 Br Tetrabromintion BrCH2 CBr3 & Br2 CH CHBr2 Pentabromination Br2 CH CBr3 Br3 C CBr3 Hexabromination So total no of Bromo derivatives = 9 Sol10. 3BaCl2 2Na3 PO4 Ba3 PO4 2 6NaCl 5 2 Na3PO 4 limiting reagent So 2 mole of Na3PO4 will produce 1 mole of Ba3(PO4)2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-30 PART – C (MATHEMATICS) SECTION – A Sol1. C 2iˆ ˆj 4kˆ , D ˆi 5ˆj 6kˆ A 5iˆ 5ˆj 2kˆ , B ˆi 2ˆj 3kˆ AB 4iˆ 3 ˆj 3 2 kˆ AC 7iˆ 5 ˆj 4 2 kˆ AD 6iˆ 6 2 kˆ AB , AC , AD are coplanar. 4 3 3 2 7 5 4 2 0 2 5 6 0 6 0 6 2 2, 3 2 2 16 25 41 S Sol2. 2x y 3y x 20 dy 2yx y 1 3y x ny y dx 2x nx 3xy x 1 dy 23 3 22 n2 2 3n2 3 2 dx 2,2 2n2 3 2 n2 3 2 Sol3. y C 0,4,0 B x O 0,4,5 A 3,0,0 P 3,4,5 S R 0,0,5 Q Z FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-31 Equation of line OP r 3iˆ 4ˆj 5kˆ ……….(i) side parallel to z axis not passing O & P is CS Equation of CS r 4ˆj 5kˆ ………..(ii) shortest distance between lines OP and CS 3iˆ 4ˆj 5kˆ 5kˆ 4ˆj 3iˆ 4ˆj 5kˆ 5kˆ 4ˆj 15ˆj 20iˆ 15 Sol4. 2 20 2 60 12 25 5 y B(0,9) x y 1 5 9 AC : CB 1: 2 10 C , 3 3 and AD : DB 2 : 1 5 D ,6 3 L: y=m 2n(L2) D y=m 1x(L1) C A(5,0) L x 9 18 , m2 10 5 9 18 y x 10 5 m1 y 9 x 2 10 45 It cut the line L : 9x 5y 45 at , 7 7 Sol5. P1 2x y z 3 P2 4x 3y 5z 9 0 P1 P2 0 2x y z 3 4x 3y 5z 9 0 P3 2 4 x 1 3 y 1 5 z 3 9 0 x 1 y 3 z 2 2 4 5 2 2 4 4 1 3 5 1 5 0 P3 is parallel to 3 5 0 3 5 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-32 22x 14y 20z 12 0 5 5 5 5 P3 11x 7y 10z 6 0 a 11 b 7 c 10 a b c 11 7 10 14 P3 : Sol6. i d bc 1 j k 2 2 2iˆ ˆj 2kˆ 1 4 a d 18 4 3 8 18 2 d 4iˆ 2ˆj 4 kˆ i j k 2 ad 2 3 4 4 2 4 Sol7. 3 2 2 2 20i 83 16k 16 5iˆ 2ˆj 4kˆ 16 45 720 5 can occur in 4,1 3,2 2,3 1,4 ways 4 1 8 q 1 p 36 9. 9. P x 4 P x 4 P x 5 p 4 5 1 8 5 1 C4 C5 9 9 9 58 1 41 41 123 5 5 5 10 11 9 9 9 3 3 k 123 5 Sol8. x 3 x 4 3 2 x 7 3 x 2 x 1 A , 1 ………….(i) 3 (B) 3 x r r 1 10 x 3 33 x x 3 3 3 x 10 33 x 1 1 10 x x3 3 3 33x 33 3 3 x 3 3x x 1 x 1 B , 1 ………….(ii) A B FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-33 Sol9. I x x2 x sec x tan x x tan x 1 x 2 2 dx x2 2x dx x tan x 1 x tan x 1 x2 x cos x dx 2 x tan x 1 x sin x cos x I x x2 2n x sin x cos x C x tan x 1 I 0 0 c 0 2 1 I 2n 4 2 2 4 4 4 4 n 2 32 Sol10. 2 4 4 x 2 8x 15 2x 7 0 x 2 8x 15 2x 7 0 x 7 / 2 …………(i) x 2 8x 15 x 3 x 5 Case I x , 3 5, …………(ii) x 2 8x 15 2x 7 x 2 10x 22 0 10 100 88 5 3 2 x 5 3 x 3, 5 Case II x 2 8x 15 7 2x x 2 6x 8 0 x 4, 2 x 4 x sum of roots Sol11. 9 3 A 30 3; x 30 h 2 3 x 30 h 20 3 30 30 m x 10 3 P 60 B x 200 3 3 3 h 0 60 20 3 10 3 0 h h 60 20 3 area = hx 15 C FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. Q JEE-MAIN-2023 (6th April-First Shift)-PCM-34 600 3 1 3 is correct Sol12. x y az b 2x 5y 2z 6 x 2y 3z 3 System of equations can be written as 1 1 a x b 2 5 2 y 6 1 2 3 z 3 R1 R3 , R2 2R3 0 1 a 3 x b 3 4 y 0 0 1 1 2 3 z 3 0 0 a 7 x b 3 R1 R 2 0 1 4 y 0 1 2 3 z 3 For infinite solution a = 7 & b = 3 2a 3b 23 Sol13. Sol14. P Q R Q P Q R Q Q P R P R Q P R Q d 1 1 1 ..... n a1 a2 a2 a3 an 1 a2 a3 a2 an an 1 d a2 a1 lim .... n n d d d d an a1 lim n n d lim n lim n Sol15. d d a 1 a1 n 1 d 1 n n d n 1 xi 15 12 and xi2 122 14 15 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-35 yi2 142 2 15 And yi 15 14 and Now, 13 14 144 15 2 196 15 30 132 2 10 Sol16. S 5 11 19 29 .... Tn 1 Tn S 5 11 19 .... .... Tn1 Tn 0 5 6 8 10 ..... t n1 Tn n 1 12 n 2 2 2 n2 3n 1 Sn Tn n2 2n n Tn 5 n n 1 2n 1 S 20 3 6 10 7 41 6 n n 1 3 2 n 20 21 20 2 3520 Sol17. A aij aij 0 i, j 2 2 a Let A c 2 A I a2 bc ac dc b d a b c d ab bd 1 0 bc d2 0 1 a2 bc 1 a d c 0 ab bd 0 bc d2 1 Again A 1 ad bc 1 Solving a 0, b 1 3a2 4b2 4 Sol18. Sol19. 1 1 5f x 4f 3 ……….(i) x x 1 Replace x x 1 5f 4f x x 3 …………(ii) x Solving (i) and (ii) we get 9f x 5 4x 3 x E 21/ 4 1/ 4 2 2 1 5 f x dx 9 x 4x 3 dx 1 1 2 1 5log x 2x 2 3x 1 9 1 5log x 3 9 2 18 f x dx 10loge 2 6 1 n FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-36 n4 3 t 5 n C4 21/4 1/4 4 n C4 2 n4 4 ………..(i) 3 5th term from end 1 t '5 C 4 3 4 n 4 2 n Given 2 n4 4 3 1/4 2n Sol20. 10 n C4 3 n 4 4 2 …………(ii) t5 6 t '5 1 n4 4 66 3 2 n4 3 n 10 4 2 t3 4 8 3 C2 21/4 1/ 4 2 n4 4 63/ 2 45 4 3 1 2 60 3 C3 : n C3 10 : 1 2n ! 3! n 3 ! 10 3! 2n 3 ! n! 4 2n 1 10n 20 n8 n 2 Now n 2 3n 3n 4 64 24 88 2 64 24 4 44 SECTION – B Sol1. x cos x dy xy sin x y cos x 1 dx 0 2. dy x cos x y x sin x cos x 1 dx dy 1 1 y tan x dx x x cos x 0x 1 tan x x dx n sec x nx IF e e x sec x solution y x sec x sec 2 x dx yx sec x tan x C y 3 C 3 3 3 yx sec x tan x 3 y sin x cos x 3 x x FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-37 xy 2 sin x 3 yx x Sol2. dy 2cos x dx 3 d2 y dy 2 2sin x dx 2 dx 3 y" 2y ' 2 2 6 6 6 A 1, 2, 3,....,10 B 0, 1, 2, 4 a,b A A such that 2 2 a b 3 a b k 0 where k 0, 1, 2, 3, 4 we should have 9 4 2 k a perfect square for any possible (a, b) i.e., 9 + 8k is perfect square k 0 or k = 2 2 for k 0, 2 a b 3 a b 0 a b 0 a, b 1, 1 , 2, 2 .... 10, 10 . Total 10 elements belonging to R. 3 a b is not possible 2 for k = 0 2 a b 3 a b 2 0 1 (not possible) 2 a, b 1, 3 , 2, 4 ,..... 8, 10 8 element belonging to R Total = 18 a b 2 or a b Sol3. 2 Equation of circle is x a y a a 2 Passes P , a 2 P , 2 a a ……….(i) 2 Equation of AB x y a …………(ii) Let Q x1,y1 x1 y1 a ………..(iii) slope of PQ = 1 Equation of PQ x x1 y y1 r 1/ 2 1/ 2 For co-ordinate of Q , r 11 x1 1/ 2 y1 1/ 2 B 0,a Q A a,0 11 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-38 11 2 x1 & 11 2 y1 22 x1 y1 11 2 a 2 (i) 2 2 2a 2ab a2 0 2 2 2a a 2 0 2 2 2a 1/ 2 a a 0 11 121 11 2 2 2 1/ 2 a 2 Sol4. 2 y 3 x y 9 x2 ; 0 x 3 P, P 1 lies on y x 1 …………(i) Q Solving with x y 3 P 1, 2 P Again solving y x 1 & x 2 y 2 9 n 17 1 x x40 x 2 17 1 2 p 1, a 1 b b 4 2 2 2 b b a 3 2 Sol5. Let Q , , be the image of P, about the plane 2x y z 9 1 2 3 2 2 1 1 5, 0, 5 Then area of triangle PQR is 1 PQ PR 2 12iˆ 3ˆj 21kˆ 144 9 441 594 Square of area = 594 Sol6. Equation of tangent at P 1, 3 to the curve x 2 2x 4y 9 0 is y x 2 Then the point A is 0, 2 Equation of line passing through P and parallel to the line x 3y 6. The possible coordinate of B are 4, 4 or 16, 8 But 4, 4 does not satisfy 2x 3y 8 Thus the point B is 16, 8 2 Then AB 292 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (6th April-First Shift)-PCM-39 Sol7. x 2 2y 2y y 2 x 2 2y (i) x 2 y 2 2y 0 yx 2, 2 2 x y 1 1 2 (ii) x 2 2y 2 Required area = 2y 2y y 2 dy 1 7 6 4 Sol8. y 13 y 12 y2 y 1 at every line f(x) is discontinuous two times in y = 12 line it is discontinuous only once. total number of discontinuous points = 25 Sol9. 20 different oranges can be given to 3 children so that each gets at least once is 320 3 C1 220 3 C2 120 15 Sol10. 4 1 x 3 x Gen. term = r 1 15 15 15 r Cr x 4 1 3 x r Cr x 60 7r 60 7r 18 7r 42 r 6 coefficient of x18 15 C 6 91 55 5005 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
