NARAYANA IIT/PMT ACADEMY JEE ADVANCE

Transcription

1 JEE ADVANCE - This booklet contains printed pages XYZ PAPER - : PHYSIC, CHEMISTY & MATHEMATICS Please Read the Instructions carefully You are allotted 5 minutes specifically for this purpose Important Instructions : Test Booklet Code INSTRUCTINS A General : This booklet is your Questions Paper Do not break the seals of this booklet before being instructed to do so by the invigilators The question paper CDE is printed on the right hand top corner of this sheet and on the back page (Page No 44) of this booklet Blank spaces and blank pages are provided in the question paper for your rough work No additional sheets will be provided for rough work Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic gadgets are NT allowed inside the examination hall Write your name and roll number in the space provided on the back cover of this booklet Answers to the questions and personal details are to be filled on a two part carbon less paper, which is provided separately These parts should only be separated at the end of the examination when instructed by the invigilator The upper sheet is machine gradable objective Response Sheet (RS) which will be retained by the invigilator You will be allowed to take away the bottom sheet at the end of the examination Using a black ball point pen darken the bubbles on the upper original sheet Apply sufficient pressure so that the impression is created on the bottom duplicate sheet D NT TAMPER WITH/MUTILATE THE RS R THE BKLET n breaking the seals of the booklet check that it contains 44 pages and all the 6 questions and corresponding answers choices are legible Read carefully the instruction printed at the beginning of each section and TTAL MARKS 8. B, Filling the right part of the RS The RS also has a CDE printed on its left and right parts Check that the CDE printed on the RS (on both sheets) is the same as that on this booklet and put your signature affirming that you have verified this IF THE CDES D NT MATCH, ASK FR A CHANGE F THE BKLET

2 JEE ADVANCE - PART I : PHYSICS SECTIN -: (nly ne option correct Type) This section contains multiple choice questions Each question has four choices (A), (B), (C) and (D) out of which NLY NE is correct x y The work done on a particle of mass m by a force, K ˆi ˆj / / x y (x y ) (K being a constant of appropriate dimensions), when the particle is taken from the point (a,) to the point (, a) along a circular path of radius a about the origin in the x-y plane is (A) K a (B) K a (C) K a (D) Ans (D) Sol Direction of Force tan y x ds F Force is radially outwards and displacement is tangential Hence work done = Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure ne of the blocks has thermal conductivity k and the other k The temperature difference between the ends along the x-axis is the same in both the configurations It takes 9s to transport a certain amount of heat from the hot end to the cold end in the configuration I The time to transport the same amount of heat in the configuration II is (A) s (B) s (C) 45 s (D) 6 s Ans Sol (A) In series, equivalent thermal conductivity

3 JEE ADVANCE - In parallel Heat Supplied K K K eq eq K K K K eq 4K K (A A ) K A K A eq K Keq Q Q K A K A T T 4K A K A. 9 t 6 = t t = sec t t ( Q Q ) Two non-reactive monatomic ideal gases have their atomic masses in the ratio : The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : The ratio of their densities is (A) : 4 (B) : (C) 6 : 9 (D) 8 : 9 Ans (D) Sol P M = d RT P M = d RT PM d d 4 8 P M d d 9 4 A particle of mass m is projected from the ground with an initial speed u at an angle with the horizontal At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed u The angle that the composite system makes with the horizontal immediately after the collision is (A) 4 (B) 4 (C) 4 (D) Ans (A)

4 JEE ADVANCE - u o Sol H max u cos o u sin o u sin g u sin V u g g V u cos v u cos tan 4 5 A pulse of light of duration ns is absorbed completely by a small object initially at rest 8 Power of the pulse is mw and the speed of light is ms - The final momentum of the object is 7 (A). kg ms - 7 (B). kg ms - 7 (C). kg ms - 7 (D) 9. kg ms - Ans (B) Sol E = mw ns = - J = -9 J P = E/C 9 8 = -7 kg m/s 6 In the Young s double slit experiment using a monochromatic light of wavelength, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is (A) (n ) (B) (n ) 4 (C) (n ) 8 (D) (n ) 6 Ans (B) Sol I I cos cos (n ) / 4

5 JEE ADVANCE -. x x (n ) / 4 7 The image of an object, formed by a plano-convex lens at a distance of 8m behind the lens, is real and is one-third the size of the object The wavelength of light inside the lens is times the wavelength in free space The radius of the curved surface of the lens is (A) m (B) m (C) m (D) 6 m Ans (C) Sol / / m hi v h o u v 8 u u u = 4 ( ) f R v u R R = m 8 ne end of a horizontal thick copper wire of length L and radius R is welded to an end of another horizontal thin copper wire of length L and radius R When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is (A) 5 (B) 5 (C) (D) 4 Ans (C) Sol Y F (R) L L R R L F L R Y 5

6 JEE ADVANCE - Y L L F R L L L F L R Y 9 A ray of light travelling in the direction ˆi j ˆ is incident on a plane mirror After reflection, it travels along the direction ˆi j ˆ The angle of incidence is (A) (B) 45 (C) 6 (D) 75 Ans (A) Sol Direction of Normal is (i j) (i )j j ˆ Angle between incident ray and normal (i j).( j) cos o ˆ ˆ ˆ (i j j The diameter of a cylinder is measured using a Vernier calipers with no zero error It is found that the zero of the Vernier scale lies between 5 cm and 55 cm of the main scale The Vernier scale has 5 divisions equivalent to 45 cm The 4 th division of the Vernier scale exactly coincides with one of the main scale divisions The diameter of the cylinder is (A) 5 cm (B) 54 cm (C) 56 cm (D) 548 cm Ans (B) Sol MSD = 5 cm.45 VSD.49cm 5 LC = 5 49 = cm Reading = cm = cm = 54 cm 6

7 JEE ADVANCE - SECTIN - : (ne or more options correct Type) The section contains 5 multiple choice questions Each question has four choices (A), (B), (C) and (D) out of which NE and MRE are correct In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C The switch S is pressed first to fully charge the capacitor C and then released The switch S is then pressed to charge the capacitor C After some time, S is released and then S is pressed After some time, (A) the charge on the upper plate of C is CV (B) the charge on the upper plate of C is CV (C) the charge on the upper plate of C is (D) the charge on the upper plate of C is CV S S S V C C V Ans B, D Sol: When s is closed and then released the charge on upper plate of C is +CV o When S is closed and then released charge will be distributed equally in C and C So the charge on the upper plate of C is +CV o When S is pressed the charge on upper plate of C will become ( CV o ) A particle of mass M and positive charge Q, moving with a constant velocity u ˆ 4ims, enters a region of uniform static magnetic field normal to the x-y plane The region of the magnetic field extends form x = to x = L for all values of y After passing through this region, the particle emerges on the other side after milliseconds with a velocity u i ˆ ˆ j ms. The correct Ans Sol: statements (s) is (are) (A) The direction of the magnetic field is z direction (B) The direction of the magnetic field is +z direction (C) The magnitude of the magnetic field 5 M Q units (D) The magnitude of the magnetic field is M Q units A, C 7

8 y NARAYANA IIT/PMT ACADEMY JEE ADVANCE - x x x x x x x x x x x x x x x x o x x x x x x x x o x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x = x x x x x x x x x = L x x x x x x x x x x x x x x x x x x x x x x x x x As particle turns toward +ve y axis the direction of field must be along z axis (as) [use : F q v B ] Since u ˆ ˆ i j tan o So particle emerges at an angle of o with initial direction Hence Time to turn by o is T T 6 t m qb B 5 M Q A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = ( m) sin[(68 m ) x] cos [(68 s )t] Assuming = 4, the correct statement (s) is (are) (A) The number of nodes is 5 (B) The length of the string is 5 m (C) The maximum displacement of the midpoint of the string, from its equilibrium position is m (D) The fundamental frequency is Hz Ans B, C Sol: y x, t.m sin 6.8m x cos 68s t Here, k 6.8.m And 68 s v ms k Since string is vibrating in 5 th harmonic so 5 5. L.5 m 8

9 JEE ADVANCE - and fundamental frequency v f Hz L.5 The midpoint is the antinode hence A = m 4 A solid sphere of radius R and density is attached to one end of a mass-less spring of force constant k The other end of the spring is connected to another solid sphere of radius R and density The complete arrangement is placed in a liquid of density and is allowed to reach equilibrium The correct statement (s) is (are) 4 R g (A) the net elongation of the spring is k R g (B) the net elongation of the spring is k (C) the light sphere is partially submerged (D) the light sphere is completely submerged Ans A, D Sol: A B 4 R g kx B 4 R g If whole system is submerged then upthrust is equal to the weight of the system hence the sphere A must be completely submerged n sphere B net force must be zero as weight of sphere B is greater than upthrust, hence there must be elongation in spring 4 4 So for sphere B kx R g R g. 4 R g x k 5 Two non-conducting solid spheres of radii R and R, having uniform volume charge densities and respectively, touch each other The net electric field at a distance R and from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero The ratio be (A) 4 (B) 5 can 9

10 Ans Sol: NARAYANA IIT/PMT ACADEMY JEE ADVANCE - (C) (D) 4 5 B, D If and both are both positive or both negative then field will be zero at a point in between the line joining the centre R ie, R R o o 4 Case II: If one of the charge density is negative and other is positive the point will lies on the left side of smaller sphere R R o R o 5R 5 SECTIN-: (Integer value correct Type) The section contains 5 questions The answer to each question is a single digit integer, ranging from to 9 (both inclusive) Integer 6 A bob of mass m, suspended by a string of length l, is given a minimum velocity reqired to complete a full circle in the vertical plane At the highest point, it collides elastically with another bob of mass m suspended by a string of length l, which is initially at rest Both the strings are mass-less and inextensible If the second bob, after collision acquires the minimum speed l required to complete a full circle in the vertical plane, the ratio l is Ans 5 Sol: The speed of the mass attached to string of length l at the highest point will be gl which should be equal to the critical speed of mass attached to the string of length l at the bottom-most point ie gl 5gl l or 5 l 7 A particle of mass kg is moving in one dimension under a force that delivers a constant power 5 W to the particle If the initial speed (in ms ) of the particle is zero, the speed (in ms ) after 5 s is Ans 5

11 Sol: w k pt mv v 5ms NARAYANA IIT/PMT ACADEMY JEE ADVANCE - 8 The work functions of Silver and Sodium are 46 and ev, respectively The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is Ans Sol: kemax hf W and kemax ev s slope of either graph is h e which is constant ratio of slopes = v s f 9 A freshly prepared sample of a radioisotope of half-life 86 s has activity disintegrations per second Given that ln = 69, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 8 s after preparation of the sample is Ans 4 Sol: t No e Required fraction = No e.4 % fraction 4% A uniform circular disc of mass 5 kg and radius 4 m is rotating with an angular velocity of rad s about its own axis, which is vertical Two uniform circular rings, each of mass 65 kg and radius m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis The new angular velocity (in rad s ) of the system is Ans 8 Sol: From conservation of angular momentum 5 R 5 R R R 8 rad / s

12 JEE ADVANCE - PART II : CHEMISTRY SECTIN (nly ne option correct Type) This section contains multiple choice questions Each question has four choices (A) (B), (C) and (D) out of the which NLY NE is correct Consider the following complex ions, P, Q and R P = [FeF 6 ] -, Q = [V(H ) 6 ] + and R = [Fe(H ) 6 ] + The correct order of the complex ions, according to their spin only magnetic moment values (in BM) is (A) R < Q < P (B) Q < R < P (C) R < P < Q (D) Q < P < R Ans (B) [FeF 6 ], Fe + has 5 unpaired electron = 5 BM [V(H ) 6 ] +, V + has unpaired electrons = 5 BM [Fe(H ) 6 ] +, Fe + has 4 unpaired electrons = 4 BM P > R > Q The arrangement of X ions around A + ion in solid AX is given in the figure (not drawn to scale) If the radius of X is 5 pm, the radius of A + is (A) 4 pm (B) 5 pm (C) 8 pm (D) 57 pm Ans (A) cation in octahedral voids r r.44 (without distortion) r 5.44 r + = 44 5 = 5 4 (approx) Sulfide ores are common for the metals (A) Ag, Cu and Pb (B) Ag, Cu and Sn (C) Ag, Mg and Pb (D) Al, Cu and Pb Ans (A) Common sulphide ore are given by Ag Ag S : Silver glance Cu CuS : Copper glance Pb PbS : Galena 4 The standard enthalpies of formation of C (g), H (l) and glucose (s) at 5 C are -4 kj/mol, - kj/mol and - kj/mol, respectively The standard enthalpy of combustion per gram of

13 Ans NARAYANA IIT/PMT ACADEMY JEE ADVANCE - glucoses at 5 C is (A) +9 kj (B) -9 kj (C) -6 kj (D) +6 kj (C) I C + C H = 4 kj/mol II H + H H = kj/mol III 6C + 6H + C 6 H 6 H = kj/mol C 6 H C + 6H H = ( 4 6) + ( 6) ( ) H = 9 kj / mole H of per gram of Glucose = kJ / gram 5 Upon treatment with ammoniacal H S the metal ion that precipitates as a sulfide is (A) Fe(III) (B) Al (III) (C) Mg(II) (D) Zn (II) Ans (D) n reaction with H S in presence of NH 4 Cl/NH 4 H only Zn reacts and gives ZnS (white) Zn + + H S ZnS (white ppt) (IV group) 6 Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 5 C For this process, the correct statement is (A) The adsorption requires activation at 5 C (B) The adsorption is accompanied by a decreases in enthalpy (C) The adsorption increases with increases of temperate (D) The adsorption is irreversible Ans (B) Adsorption of phenolphthalein is exothermic at 5 C 7 KI in acetone, undergoes S N reaction with each of P, Q, R and S The rate of the reaction vary as H C - Cl Cl P Q R (A) P > Q > R > S (B) S > P > R > Q (C) P > R > Q > S (D) R > P > S > Q Cl S Cl Ans (B) S > P > R > Q S N

14 JEE ADVANCE - Note : Methyl halide has lower rate of S N then allyl halide according to the data given in Jerry March (4 th Edition, Page No 9, Book Name Advance rganic Chemistry) Average relative S N rates for some alkyl substrates Reactant Relative rate Methyl Isopropyl 5 Allyl 4 But considering the different options given in question the most appropriate order S > P > R > Q S N 8 In the reaction P + Q R + S the time taken for 75% reaction of P is twice the time taken for 5% reaction of P the concentration of Q varies with reaction time as shown in the figure The overall order of the reaction is [Q] [Q] Ans time (A) (B) (C) (D) (D) In the reaction P + Q R + S time taken for 75% of composition is two times of the decomposition of 5% of P then the order of reaction with respect to P is for t t P C C / C /4 according to graph concentration of Q completely decomposed in given time then order of reaction with respect to Q is overall order of reaction = + = 9 Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of (A) N (B) N (C) N (D) N 4 Ans (B) n long standing concentrated HN gives brown colored gas HN (con) N (g) + H + (brown coloured) 4

15 JEE ADVANCE - The compound that does NT liberate C, on treatment with aqueous sodium bicarbonate solution is (A) Benzoic acid (B) Benzenesulphonic acid (C) Salicylic acid (D) Carbolic acid (Phenol) Ans (D) Phenol does not reacts with NaHC Because phenol is less acidic then HC Thus according to acid-base concept Ph can not abstract the hydrogen from HC and do not give C gas CH S H CH H H SECTIN : (ne or more options correct Type) This section contains 5 multiple choice questions Each question has four choices (A), (B), (C) and (D) out of which NE or MRE are correct The initial rate of hydrolysis of methyl acetate ( M) by a weak acid (HA,M) is / th of that of a strong acid (HX, M), at 5 o C The K a of HA is (A) -4 (B) -5 (C) -6 (D) - Ans (A) r H r H r r H H H H K C K a C = -4 K a = -4 a The hyperconjugative stabilities of tert-butyl cation and -butene, respectively, are due to (A) p (empty) and * electron delocalisations (B) * and electron delocalisations (C) p (filled) and electron delocalisations (D) p (filled) * and * electron delocalisations 5

16 Ans NARAYANA IIT/PMT ACADEMY JEE ADVANCE - (A) t-butyl carbocation is stabilized by -butene is stabilized by * overlapping p overlapping The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are) Cr NH Cl Cl and Cr NH Cl Cl Ans (A) 5 (B) 4 (C) Co NH Cl and Pt NH H Cl CoBr Cl and (D) PtBrCl 4 Pt NH N Cl and Pt NH Cl Br (B), (D) Cr NH Cl Cl does not exhibit isomerism, Cr NH Cl Cl exhibits geometrical (A) 5 isomerism (B) 4 Co NH Cl and Pt NH H Cl Both can show geometrical isomerism (C) CoBr Cl - tetrahedral so it does not exhibit isomerism PtBr Cl - square planar so it exhibits geometrical isomerism (D) Pt NH N Cl and Pt NH Cl Br Both can show isonisation isomerism 4 Among P, Q, R and S, the aromatic compound(s) is/are (A) P (B) Q (C) R (D) S Cl 4 AlCl P NaH Q NH4 C o R 5 C HCl S Ans (A), (B), (C), (D) 6

17 Cl NARAYANA IIT/PMT ACADEMY JEE ADVANCE - H AlCl NaH electron aromatic Na AlCl 4 H 6 electron aromatic H H H base H enolization 6 electron aromatic H H H H 6 electron aromatic 5 Benzene and naphthalene form an ideal solution at room temperature For this process, the true statement(s) is(are) (A) G is positive (B) S system is positive Ans (C) Ssurroundings (D) H (B), (C), (D) G H T S system system system H T S system surrounding So as solution is ideal So Ssurrounding As mixing is spontaneous so H T S So as system system H system =, system Hsystem Gsystem S must be greater than zero 7

18 JEE ADVANCE - SECTIN : (Integer value correct Type) This section contains 5 questions The answer to each question is a single digit integer, ranging from to 9 (both inclusive) 6 The atomic masses of He and Ne are 4 and amu, respectively The value of the de Broglie wavelength of He gas at 7 o C is M times that of the de Broglie wavelength of Ne at 77 o C M is Ans 5 He : m = 4, T = K Ne : m =, T = K 7 Ans 8 So He Ne m He Ne h m Ne h h m Ne He m T Ne Ne He Ne Ne Ne He He m T He He v T M T m T m EDTA is ethylenediaminetetraacetate ion The total number of N Co bond angles in Co EDTA complex ion is 8 The total number of carboxylic acid groups in the product P is 8

19 NARAYANA IIT/PMT ACADEMY JEE ADVANCE - Ans. H, P.. H H H H keto acid / H H H 9 A tetrapeptide has CH group on alanine This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis For this tetrapeptide, the number of possible sequences (primary structures) with NH group attached to a chiral center is Ans 4 CH CH CH CH CH Valine (V) NH CH CH CH NH Alanine (A) HC CH CH Phenylalanine (P) NH HN CH CH Glycine (G) As (A) is at the end so options possible with NH at beginning are (i) VPGA, (ii) VGPA, (iii) PVGA and (iv) PGVA 4 The total number of lone-pairs of electrons in melamine is Ans 6 9

20 H N N NARAYANA IIT/PMT ACADEMY JEE ADVANCE - NH N N NH Melamine Total number of lone pairs = 6 PART III : MATHEMATICS SECTIN (nly ne option correct Type) This section contains multiple choice questions Each question has four choices (A) (B), (C) and (D) out of the which NLY NE is correct 4 For a b c, the distance between, and the point of intersection of the lines ax by c and bx ay c is less than. then (A) a b c (B) a b c (C) a b c (D) a b c 4 (A) c c Point of intersection, a b a b Distance of (, ) from point of intersection c c a b a b a b c a b Acc to question a b c a b a b c 4 The area enclosed by the curves y sin x cos x and y cos x sin x over the interval, is (A) 4 (B) (C) (D) 4 (B) Required Area /4 / sin x dx cos /4 x dx /4 / cos x sin x /4

21 JEE ADVANCE The number of points in, for which x xsin x cos x, is (A) 6 (B) 4 (C) (D) 4 (C) Let f x x xsin x cos x f ' x x sin x xcos x sin x f ' x x cos x When x f ' x f x is si function When x f ' x f x is sd function At x = f ' x f When x f x f x will be zero at exactly two points 44 The value of cot (A) 5 44 (B) n cot cot n n k k k n cot cot k n k (B) 5 n. n n k k n n cot n n n T r r r cot tan tan r r r r r r k is (C) 4 (D) 4

22 Tr tan r tan r n T r tan 4 NARAYANA IIT/PMT ACADEMY JEE ADVANCE - 4 cot tan A curve passes through the point,. 6 y y sec, x. x x y (A) sin log x x y (C) sec log x x 45 (A) Given dy y y sec (i) dx x x Let y vx dy dv v x (ii) dx dx Substituting (ii) in (i) dv v x v sec v dx dx cos v dx x At x, y v at x 6 6 v x dx cos v dv x /6 v sin v log x /6 sin v sin log x 6 y sin log x x Then the equation of the curve is x Let the slope of the curve at each point x, y be y (B) cos ec log x x y (D) cos log x x

23 46 Let NARAYANA IIT/PMT ACADEMY JEE ADVANCE - f :, (the set of all real numbers) be a positive, non-constant and differentiable function such that f ' x f x and interval (A) e, e (B) e, e (C) 46 (D) f ' x f x r x x e f x e f x ' d e x f x dx x e f x is sd x e f x e f x e x n f in x f x dx e, / / / f x dx e f. Then the value of / e, e (D) f, x dx lies in the 47 Let PR iˆ ˆj k ˆ and SQ iˆ ˆj 4k ˆ determine diagonals of a parallelogram PQRS and PT iˆ ˆj k ˆ be another vector Then the volume of the parallelepiped determined by the vector PT, PQ and PS is (A) 5 (B) (C) (D) 47 (C) S R e b P Q a a b iˆ ˆj k ˆ c iˆ ˆj k ˆ a b iˆ ˆj 4k ˆ a b a b iˆ ˆj k ˆ

24 JEE ADVANCE - a b iˆ ˆj k ˆ a b 5 iˆ ˆj k ˆ Now volume of parallelepiped c a b c. a b 5 48 Perpendiculars are drawn from points on the line feet of perpendiculars lie on the line x y z (A) 5 8 x y z (C) (D) x y z The given line is General point on the line is P r, r, r Now Now the direction ratio of PQ are r, r, r which are proportional to (,, ) r r r k r k k r r Point,, lies on pane x y z so k r k r k r k 6 4r r r 6 4r r r and 7 6 4r 6 r r 5 6 x y z r k 6 4r x y z to the plane x y z. The x y z (B) 5 x y z (D) 7 5 (-,-, ) A P Q 4

25 JEE ADVANCE - 49 Four person independently solve a certain problem correctly with probabilities,,, Then the probability that the problem is solved correctly by at least one of them is (A) 5 (B) (C) (D) (A) Prob that the problem is solved correctly = prob that problem is not solved correctly by none of them Let complex numbers and lie on circles 5 56 x x y y r and x x y y 4 r, respectively If z x iy satisfies the equation z r, then (A) 5 (c) (B) z r (i) z r (C) 7 (D) z r (ii) (ii)/(i) gives z z r r z z 4 z z z z z 4 z z z. (A) 4 z 4 4 z z z 4 (B) From (i) z z z r r z z r 5

26 From (B) & (C) NARAYANA IIT/PMT ACADEMY JEE ADVANCE - r (C) r 4 r 4 4 r 4 7 SECTIN : (ne or more options correct Type) This section contains 5 multiple choice questions Each question has four choices (A), (B), (C) and (D) out of which NE or MRE are correct 5 A line l passing through the origin is perpendicular to the lines ˆ ˆ ˆ : t i t j 4 t k, t : s ˆi s ˆj s k, ˆ s Then, the coordinate(s) of the point(s) on at a distance of of (A) and is (are) 7 7 5,, Key (B, D) Sol Given equation of line and are x y z 4 () and x y z () dr of line are,, Equation of line is x y z () Any point on () is k,k, k To find point of intersection of () and () We have (B),, (C),, (D) 7 from the point of intersection 7 7 8,,

27 JEE ADVANCE - k k k 4 4k 6 k k 4 7k 7 k Point of intersection of () and () is P,, Any point on line is Q S, S, S PQ 7 S 6 S S 7 9S 8S 9S S 8S 9S S S,S 9 Q point is 7 7 8,, or,, 5 Let f x x sin x, x Then for all natural numbers n,f ' x vanishes at Key (B, C) Sol (A) a unique point in the interval (B) a unique point in the interval n,n n,n (C) a unique point in the interval n,n (D) two points in the interval n,n 7

28 JEE ADVANCE - y y x,,, x f ' x x cos x. sin x, x > f ' x x cos sin x Now f ' x xcos x sin x x tan x Now taking y tan x, and y x From graph, A unique point in the interval n,n n N 5 Let S k Then S n can take value(s) n 4n k k k x x x 5 and a unique in the internal n,n (A) 56 (B) 88 (C) (D) Key (A, D) Sol 4n k k S k n k S n terms n S n n n n n n n n n n n S 6 r 4 r 9 6 r 6 r 4 n r r r r r r 8

29 JEE ADVANCE - n n n 6 r 8 r r r r n n r r r 6 r n n n = 4n 4n Therefore the required solutions are 56, 54 For matrices M and N, which of the following statement(s) is (are) NT correct? T (A) N M N is symmetric or skew symmetric, according as M is symmetric or skew symmetric (B) MN NM is skew symmetric for all symmetric matrices M and N (C) MN is symmetric for all symmetric matrices M and N (D) adjm adjn adj MN for all invertible matrices M and N Sol (C, D) (A) T T T T T T T N MN N MN MN N So option A is correct (B) T T T MN NM MN NM T T N M N T N MN T N MN T T T T T N N N M M N M T M T (If M is symmetric M M) T (If M is skew-symmetric M M) NM MN MN NM So option B is correct T T T (C) MN N M NM (M and N are symmetric ) So option (C) is wrong (D) adjm adjm adj MN Given So option D is wrong 55 A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 5 is converted into an open rectangular box by folding after removing squares of equal area from all four corners If the total area of removed squares is, the resulting box has maximum volume Then the lengths of the sides of the rectangular sheet are (A) 4 (B) (C) 45 (D) 6 Key (A, C) Sol 9

30 a a NARAYANA IIT/PMT ACADEMY JEE ADVANCE - a a 8x a 8x a 5x a Volume of the box V 8x a 5x a.a V 4a 46a x x.a For max volume dv dv and da da dv a 9ax ax da dv 5x 4a 9x at a da Now given that sum of Area 4a a 5 5x So from a x SECTIN (Integer Value correct Type) This section contains 5 questions The answer to each question is a single digit integer, ranging from to 9 (both inclusive) 56 Consider the set of eight vectors ˆ ˆ ˆ V ai bj ck : a,b,c, There non-coplanar vectors can be chosen from V in P ways Then p is Ans 5 Sol Consider a cube of side length unit, whose centre is at origin (assumed sides parallel to coordinate axis) 4 5 Required no of Triplets of non-coplanar vectors C... P 5 57 f the three independent events E, E and E, the probability that only E occurs is, only E occurs is and only E occurs is Let the probability p that none of events E, E or E occurs satisfy the equations p and p All the given probabilities are assumed to lie in the interval (, )

31 JEE ADVANCE - Probability of occurrence of E Then Probability of occurrence of E Ans 6 Sol Let P E x,p E y,p E z x y z y x z z x y and p x y z Given equations are p and p () p p () x y and () y z x 6z x z 6 P E P E 6 () 58 The coefficients of three consecutive terms of Ans 6 Sol C C C 5::4 n 5 n 5 n 5 r r r n 5! n 5! n 5! : : 5: :4 r! n r 6! r! n r 5! r : n r 4! n n r 5 :: 5::4 n r 6 r r r n r 6 n r 6 n r 6 (I) n r 5 7 r 5 5n 5r 5 7r 7 n 5 x are in the ratio 5 : : 4 Then n = 5n r 8 (II) 4n r 4 (III) n = 6 59 A pack contains n cards numbered from to n Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 4 If the smaller of the numbers on the removed cards is k, then k Ans 5

32 Sol n n NARAYANA IIT/PMT ACADEMY JEE ADVANCE - k 4 n n 4k 45 n n 45 k 4 But k n 4 n n 45 4n 4 k n n 49, n N But for n 49,5, k N So n = 5, n n n n 5 K = 5 x y 6 A vertical line passing through the point h, intersects the ellipse at the points P 4 and Q Let the tangents to the ellipse at P and Q meet at the point R If h area of the triangle Ans 9 Sol PQR, 8 h and h, then 8 5 max min / h / h p cos, sin (h, ) R sec, Q cos, sin Area of PQR sin sec cos sin cos Which is decreasing function of cos

33 JEE ADVANCE - h cos 8 5 cos So

34 JEE ADVANCE - ANSWER KEY PHYSICS CHEMISTRY MATHEMATICS Q.No. Answer Q.No. Answer Q.No. Answer D B 4 A A A 4 B D A 4 C 4 A 4 C 44 B 5 B 5 D 45 A 6 B 6 B 46 D 7 C 7 B 47 C 8 C 8 D 48 D 9 A 9 B 49 A B D 5 C B,D A 5 B, D A,C A 5 B, C B,C B,D 5 A, D 4 A,D 4 A,B,C,D 54 C, D 5 B,D 5 B,C,D 55 A, C