IIT JEE-Advanced 2012 Solved papers by Triumph Academy
[1] IIT JEE – 2012 / Paper-I PHYSICS Part I: Physics Section – I (Single Correct Answer Type) This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R = 14 cm. For this bi-convex lens, for an object distance of 40 cm, the image distance will be (A) 280.0 cm (B) 40.0 cm Sol: [B] The focal length of each part g FGH IJK 1 F1 1I 1 b1.2 1g G J cm H 14 K 70 f b 1 1 1 1 15 . 1 cm1 f1 14 28 1 2 Hence equivalent focal length 1 1 1 1 1 1 cm-1 f f1 f 2 28 70 20 f eq 20 cm For u 40 cm 2 f v 2 f 40 cm (C) 215 . cm (D) 13.3 cm [2] IIT JEE – 2012 / Paper-I PHYSICS 2. A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed , as shown in the figure. At time t = 0, a small insect starts from O and moves with constant speed v with respect to the rod towards the other end. It reaches the end of the rod at t = T and stops. The angular speed of the system remains throughout. The mag nitude of the torque | | on the system about O, as a function of time is best represented by which plot? b g (A) (B) (C) (D) Sol: [B] The angular momentum of the system about the rotational axis is bg c h L LRod Mx2 LRod M vt or L LRod Mv2t 2 3. 2 dL 2 Mv 2t dt Hence t, till the particle reaches the end and stops, and then becomes zero Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is (A) FG IJ H K 65 2 1 4 FG IJ H K 97 (B) 4 T 1 4 FG IJ H K 97 (C) 2 T 1 4 T b g 1 (D) 97 4 T Sol: [C] The net heat received by the middle plate from the hotter plate is equal to the net heat given to the cooler plate hence eb g j e b gj A 3T T04 A T04 2T 4 4 [3] IIT JEE – 2012 / Paper-I PHYSICS 4. or 2T04 81T4 16T4 or 97 T4 T 2 4 0 F 97 I T G J H2K 0 1/ 4 T Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive surface charge density. The variation of the magnitude of the electric field E r and the electric potential V(r) bg with the distance r from the centre, is best represented by which graph ? (A) (B) (C) (D) Sol: [D] For a spherical shell E0 br R g br R g kQ r2 whereas the electric potential E V kQ (constant) for r < R R kQ for r R r The relevant graph is (D). V [4] IIT JEE – 2012 / Paper-I PHYSICS 5. FG H IJ K 4 MLg by using Searle's method, a wire of length L = 2m d 2 and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l 0.25 cm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. the number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement (A) due to the errors in the measurements of d and l are the same In the determination of Youngs modulus Y (B) due to the error in the measurement of d is twice that due to the error in the measurement of l (C) due to the error in the measurement of l is twice that due to the error in the measurement of d (D) due to the error in the measurement of d is four times that due to the error in the measurement of l Sol: [A] Y d 2 Y max max d max Here both screw gauge and micrometer used have same least count and hence maximum probable error in measurement of and d will be same i.e., max d max x Thus, c x max 0.25 mm and 2 d d max 2 x x 0.50 mm 0.25 and hence contributions due to the errors in the measurement of d and are the same 6. A small block is connected to one end of a massless spring of un -stretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency rad / s . Simultaneously at t = 0, a small pebble is projected with speed v from point P at an angle 3 of 450 as shown in the figure. Point P is at a horizontal distance of 10 m from O. If the pebble hits the block at t = 1s, the value of v is (take g = 10 m / s2) (A) 50 m / s (B) 51 m / s (C) 52 m / s (D) 53 m / s [5] IIT JEE – 2012 / Paper-I PHYSICS Sol: [A] The block starts its oscillation from its positive exrtreme and hence its position xblock from O as a function of time t can be written as xblock 4.9 0.2 cost b FG 1IJ 5.0 m H3 K g xblock t 1s 4.9 0.2 cos Hence range of the pebble R = 10 –5 = 5 m v 2 sin 2 5 v 50 m / s So g 7. Young's double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are G , R and B , respectively. Then, (A) G B R (B) B G R (C) R B G (D) R G B D and hence d green blue Sol: [D] Fringe width, Since red 8. So red green blue A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x - y plane with centre at O and constant angular speed . If the angular momentum of the system, calculated about O and P are denoted by LO and LP respectively, then (A) LO and LP do not vary with time (B) LO varies with time while LP remains constant (C) LO remains constant while LP varies with time (D) LO and LP both vary with time [6] IIT JEE – 2012 / Paper-I PHYSICS Sol: [C] The net force acting on the mass m is m 2r acting towards point O and hence net torque about point O is zero but about point P is non-zero. Hence L0 remains constant while LP varies with time. 9. A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the r m s speed F v bheliumgI is GH v barg ong JK rms rms (A) 0.32 (B) 0.45 (C) 2.24 Sol: [D] Both gases are at same temperature and thus vrms 10. vrmsb Heliumg vrmsb Argon g M Argon M Helium (D) 3.16 1 where M is molar mass of the gas M 40 316 . 4 Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage source of potential difference X. A Proton is released at rest midway between the two plates. It is found to move at 450 to the vertical JUST after release. Then X is nearly (A) 1 105 V (B) 1 107 V (C) 1 109 V Sol: [C] As the net force is at 450 to the vertical so 1 cm qE = 45 Mg qE mg X mg d or e or X mgd 16 . 1027 10 102 1 109 V e 16 . 1019 (D) 1 1010 V [7] IIT JEE – 2012 / Paper-I PHYSICS Section–II (Multiple Correct Answer Type) This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE or MORE THAN ONE is / are correct. 11. A cubical region of side a has its centre at the origin. It encloses three fixed point charges, q at 0, a / 4, 0 , 3q at (0, 0, 0) and q at 0, a / 4, O . Choose the correct option(s) b g b g (A) The net electric flux crossing the plane x = + a/2 is equal to the net electric flux crossing the plane x = – a/2. (B) The net electric flux crossing the plane y = +a/2 is more than the net electric flux crossing the plane y = – a/2. (C) q The net electric flux crossing the entire region is 0 (D) The net electric flux crossing the plane z = +a/2 is equal to the net electric flux crossing the plane x = + a/2. Sol: [A, C, D] The given distribution is symmetric about planes x a a and z . Further the distribution 2 2 a is separately symmetric about planes y . The total charge enclosed is q so accordingly all options 2 are correct. [8] IIT JEE – 2012 / Paper-I PHYSICS 12. For the resistance network shown in the figure, choose the correct option (s) (A) The current through PQ is zero (B) I1 3 A (C) The potential at S is less than that at Q (D) I 2 2 A Sol: [A, B, C, D] Due to the symmtry potentials VP VQ and VS VT So current through PQ is zero. P 6 S I2 I3 Q Further I1 b T 12 12V 3 A and I 2 12 V 2A 6 || 12 6 g Across each resistor potential decreases by 4V in the directin of current so VS VQ [9] IIT JEE – 2012 / Paper-I PHYSICS 13. A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle with the horizontal. A horizontal force of 1 N acts on the block through its center of mass as shown in the figure. The block remains stationary if (take g = 10 m/s2) 45 (B) 45 and a fricitional force acts on the block towards P. (A) (C) 45 and a frictional force acts on the block towards Q. (D) 45 and a fricitional force acts on the block towards Q. Sol: [A, C] 1N 1N The resultant of weight (1 N dowanrd) and applied force (1N) is b gb Fnet 1 sin cos g For 450 , Fnet is down the incline so friction acts towards Q where as for 450 , Fnet is up the incline so friction acts towards P. For 450 , the block remains at rest without friction [10] IIT JEE – 2012 / Paper-I PHYSICS 14. Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields E E0 j and B B0 j . At time t = 0, this charge has velocity v in the x - y plane, making an angle with the x axis. Which of the following option(s) is(are) correct for time t > 0. (A) If 00 , the charge moves in a circular path in the x-z plane (B) If 00 , the charge undergoes helical motion with constant pitch along the y- axis. (C) If 100 , the charge undergoes helical motion with its pitch increasing with time, along the y-axis. (D) If 90 , the charge undergoes linear but accelerated motion along the y-axis. y Sol: [C, D] B E v x For 00 900 , charge particle, undergoes helical motion with increasing pitch. For 900 , charge undergoes linear but accelerated motion along the y-axis. 15. A person blows into open-end of a long pipe. As a result, a high-pressure pulse of air travels down the pipe. When this pulse reaches the other end of the pipe (A) a high-pressure pulse starts traveling up the pipe, if the other end of the pipe is open. (B) a low-pressure pulse starts traveling up the pipe, if the other end of the pipe is open (C) a low-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed (D) a high-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed. Sol: [B, D] Pressure wave undergoes a phase charge of rad on refletion from open end and no phase change occurs on reflection from closed end. Accordingly B, D are correct. [11] IIT JEE – 2012 / Paper-I PHYSICS Section – III (Integer Answer Type) This section contains 5 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. (both enclusive) 16. An ninfinitely long solid cylinder of radius R has a uniform volume charge density . It has a spherical cavity of radius R/2 with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point P, which is at a distance 2R from the axis of the cylinder, is given by the expression 23R 16k 0 . The value of k is Sol. [6] Let us assume that the spherical cavity is filled with same charge density then E Ecomplete cylinder Espherical cavity E Q1 Q2 2 0 2 R 4 0 2 R b g b g 2 4 R3 R 2 3 8 2 0 2 R 4 0 4 R 2 b g LM N OP Q R 1 1 23 R 23 R 0 4 96 96 0 16 k 0 k6 [12] IIT JEE – 2012 / Paper-I PHYSICS 17. A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density J flows along the length. If the magnitude of the magnetic field at the point P is given by Sol: [5] Let us assume that cylindrical cavity is also carrying current with same current density Bnet Bcomplete cylinder Bcavity L a J OP M ( a J ) M 4 PQ B N 2 a F 3a I 2 G J H 2K L1 1 O Ja M P 5 Ja N 2 12 Q 12 2 2 or N 0aJ , then the value of N is 12 0 0 0 0 [13] IIT JEE – 2012 / Paper-I PHYSICS 18. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and IP P is I 0 and I P , respectively. Both these axes are perpendicular to the plane of the lamina. The ratio I 0 to the nearest integer is Sol: [2] Let us assume that the smaller disk is also filled with same surface mass density. Then as mass is propotional to area so if mbO m then m SD m 4 LM N b g 2 m 2R m R2 I0 R2 2 4 2 2 2 Lb2 Rg b2 Rg OP m L R 5 R O mR F 6 11I mM GH 8 JK MN 2 PQ 4 MN 2 PQ 2 I P' OP mR FG 2 3IJ 13 mR H 8K 8 Q 37 mR 2 8 I P 37 / 8 2.8 ~ 3 I O 13 / 8 2 2 2 2 [14] IIT JEE – 2012 / Paper-I PHYSICS 19. A circular wire loop of radius R is placed in the x-y plane centered at the origin O. A square loop of side b g a a R having two turns is placed with its center at z 3 R along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of 45 with respect to the z-axis. If the 0a 2 mutual inductance between the loops is given by p/ 2 , then the value of p is 2 R Sol: [7] The mutual inductance M b g N S BC A S cos 45 ie Here Ns = no of turns in square coil and BC be the magnetic field of circular coil at the position of square coil. bg c m 2 p=7 0 iC R2 2 R2 3a h 2 3/ 2 iC a2 1 2 2 a2 ~ 0 a ~ 07 / 2 8 2R 2 R bg [15] IIT JEE – 2012 / Paper-I PHYSICS 20. A proton is fired from very far away towards a nucleus with charge Q 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of b g the proton at its start is (take the proton mass, mp 5 / 3 1027 kg ; h / e 4.2 1015 J . s / C ; 1 9 109 m / F ; 1 fm 1015 m ) 4 0 Sol: [7] Q v e At closest approach U e K kQe p 2 h rmin 2 m 2 m2 h2rmin h rmin 2mkQe e 2mk 120 4.2 10 15 10 10 15 3 2 5 10 27 9 10 9 120 4.2 10 15 10 7 fm [16] IIT JEE – 2012 / Paper-I CHEMISTRY Part II Section–I : Straight Correct Answer Type This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 21. b g bNH g As per IUPAC nomenclature, the name of the complex Co H 2 O (A) Tetraaquadiaminecobalt (III) chloride (C) Diaminetetraaquacoblat (III) chloride 4 3 2 Cl 3 is (B) Tetraaquadiamminecobalt (III) chloride (D) Diamminetetraaquacobalt (III) chloride Ans. (D) Sol. Diamminetetraaquacobalt(III) chloride 22. b g In allene C 3 H 4 , the type(s) of hybridisationof the carbon atoms is(are) (A) sp and sp 3 (B) sp and sp 2 (C) only sp 2 (D) sp 2 and sp 3 Ans. (B) sp 2 sp sp 2 Sol. H 2 C C CH 2 23. For one mole of a vander Waals gas when b 0 and T 300 K , the PV vs. 1 / V plot is shown below. The value of the van der Waals constant a (atm. liter2 mol–2) is (A) 1.0 Ans. (C) (B) 4.5 (C) 1.5 (D) 3.0 [17] IIT JEE – 2012 / Paper-I CHEMISTRY Sol. Vander Walls equation for 1 mol of gas FG p a IJ bV bg RT H VK 2 given b = 0 FG p a IJ V RT H VK 2 pV a RT V pV RT pV vs a V 1 graph is a straight line have RT intercept and –a as slope V 201 . 21.6 15 . 3 2 . so a 15 Slope 24. The number of optically active products obtained from the complete ozonolysis of the given compound is (A) 0 (B) 1 (C) 2 (D) 4 Ans. (A) Sol. Ozonolysis products are b g CH 3CHO and OHC CH CH 3 CHO none is optically active 25. A compound M p X q has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empircial formula for the compound is (A) MX Ans. (B) (B) MX 2 (C) M 2 X (D) M 5 X14 [18] IIT JEE – 2012 / Paper-I CHEMISTRY Sol. 1 particle of M is located at bodycentre and 4 are on edge centre so total particle of M is 1 4 11 2 4 X is located at all corner and face centre 1 1 8 6 4 8 2 MX 2 26. The number of aldol reaction(s) that occurs in the given transformation is (A) 1 (B) 2 (C) 3 (D) 4 Ans. (C) Sol. CH 3CHO has 3 hydrogen atoms. First three steps are Alodol reactions. Last step is cannizaro reaction. 27. The colour of light abosrbed by an aqeuous solution of CuSO 4 is (A) oragne-red (B) blue-green (C) yellow (D) violet Ans. (A) Sol. The complementary colour of organe-red is blue 28. The carboxyl functional group (–COOH) is present in (A) picric acid (B) barabituric acid (C) ascorbic acid (D) aspirin Ans. (D) COOH OCOCH3 Sol. Aspirin 29. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [ a 0 is Bohr radius] h2 (A) 4 2 ma 20 Ans. (C) h2 (B) 16 2 ma 20 h2 (C) 32 2 ma 20 h2 (D) 64 2 ma 20 [19] IIT JEE – 2012 / Paper-I CHEMISTRY Sol. according to Bohr's postulate nh 2 squaring both side mvr m2 v 2 r 2 n2 h2 b2 g 2 1 n2 h2 mv 2 2 2 2 8 mr placing r 2 16a 20 for n = 2 b2g h 2 K.E. = 30. 2 8 2 m16a 20 h2 32 2 ma 20 Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen? (A) HNO 3 , NO, NH 4 Cl, N 2 (B) HNO 3 , NO, N 2 , NH 4 Cl (C) HNO 3 , NH 4 Cl, NO, N 2 (D) NO, HNO 3 , NH 4 Cl, N 2 Ans. (B) Sol. In HNO 3 oxidation state of N = + 5 NO oxidation state of N = + 2 NH 4 Cl oxidation state of N = – 3 N 2 oxidation state of N = 0 So order HNO 3 , NO, N 2 , NH 4 Cl [20] IIT JEE – 2012 / Paper-I CHEMISTRY Section–II Multiple Correct Answer(s) Type This section contains 5 multiple choice questions. Each question four 4 choices (A), (B), (C) and (D), out of which ONE or MORE are correct. 31. Identify the binarymixture(s) that can be separated into individual compound, bydifferential extraction, as shown in the given scehem. (A) C 6 H 5OH and C 6 H 5COOH (B) C 6 H 5COOH and C 6 H 5CH 2 OH (C) C 6 H 5CH 2 OH and C 6 H 5OH (D) C 6 H 5CH 2 OH and C 6 H 5CH 2 COOH Ans. (B, D) b g b g Sol. Both these mixtures can be separated by NaOH aq as well as NaHCO 3 aq 32. Choose the correct reason(s) for the stability of the lyophobic colloidal particles. (A) Preferential adsorption of ions on their surface from the solution (B) Preferential adsorptionof solvent on their surface from the solution (C) Attraction between different particles having opposite charges on their surface (D) Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles Ans. (A, D) Sol. Preferential adsorption of ions and potential difference betwen fixed and diffused layer of oppsite charges. 33. Which of the following molecules, in pure form, is(are) unstable at room temperature? (A) Ans. (B, C) Sol: Both are antiaromatic. (B) (C) (D) [21] 34. IIT JEE – 2012 / Paper-I CHEMISTRY b g Which of the following hydrogen halides react(s) with AgNO 3 aq to give a precipitate that dissolves in b g Na 2S2 O 3 aq ? (A) HCl (B) HF (C) HBr (D) HI Ans. (A, C, D) Sol. Bothe AgCl and AgBr precipitates dissolve in sodium thiosulphate solution. 35. For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown int he figure. Which of the following choice(s) is(are) correct? [taks S as change in entropy and w as workd done] (A) S x z S x y Sy z (B) w x z w x y w yz (C) w x y z w x y (D) S x y z Sx y Ans. (A, C) Sol. Entropy is state function S x z S x y Sy z so and Wx y z w x y because process y z is constant volume so work done is zero. [22] IIT JEE – 2012 / Paper-I CHEMISTRY Section–III Integer Answer Type This section contains 5 questions. The answer to each is a single-digit integer, ranging from 0 to 9 (both inclusive). 36. The substitutents R1 and R2 for nine peptides are listed in the table given below. How many of these peptides are positively charged at pH = 7.0? Peptide R1 I H II H III CH 2 COOH IV CH 2 CONH 2 V CH 2 CONH 2 VI CH 2 4 NH 2 VII CH 2 COOH VIII CH 2 OH IX CH 2 4 NH 2 b g b g R2 H CH 3 H CH 2 4 NH 2 CH 2 CONH 2 CH 2 4 NH 2 CH 2 CONH 2 CH 2 4 NH 2 CH 3 b g b g b g Ans. (4) 37. The periodic table consists of 18 groups.An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group, element X belongs in the periodic table? 63 29 Cu 11H 6 10 n 2 11H X Ans. (8) Sol. Cu 11H 6 10 n 2 11H X mass no of reactant side 63 1 64 mass no of product side 6 1 4 2 1 Mx 63 29 So M x 52 [23] IIT JEE – 2012 / Paper-I CHEMISTRY atomic no 29 1 2 2 Z x Z x 26 group no = 8 38. When the following aldohexose exits in its D-configuration, the total number of stereoisomers in its pyranose form is CHO | CH 2 | CHOH | CHOH | CHOH | CH 2 OH Ans. (8) Sol. Three chiral centres 2 3 8 39. . g mL1 . The molecular weight of HCl is 36.5 g mol 1. 29.2% (w/w) HCl stock solution has a densityof 125 The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is Ans. (8) Sol. 29.2% (w/w) of HCl solution means 29.2 g of HCl = 100 g of solution 100g mL of solution (given d 1.25 g / mL ) 125 . Now 200 mL of 0.4 M HCl is to be prepared means = 0.08 mol = 2.92 g of HCl = 8 mL of given solution [24] IIT JEE – 2012 / Paper-I CHEMISTRY 40. An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1 / 8 and 1 / 10 of its intial concentration are t 1/8 and t 1/10 respectively. What is the value of (take log10 2 0.3 ) Ans. (9) Sol. t 1/8 k 2.303 log 1 / 8a 0 a0 t 1/10 k 2.303 log t 1/8 0.9 t 1/10 t 1/8 10 9 t 1/10 1 / 10a 0 a0 t 1/8 10 ? t 1/10 [25] IIT JEE – 2012 / Paper-I MATHEMATICS PART III Section–I Single Correct Answer Type This section contains 10 multiple choice questions Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 41. The locus of the mid-point of the chord of contact of tangents drawn form points lying on the straight line 4 x 5y 20 to the circle x 2 y2 9 is c h (B) 20 x 2 y 2 36x 45y 0 c h (D) 36 x 2 y 2 20x 45y 0 (A) 20 x 2 y 2 36x 45y 0 (C) 36 x 2 y 2 20x 45y 0 b g c h c h Sol: [A] Consider a point on the line 5t 5, 4 t equation of the chord of contact of tangents drawn from the b g point 5t 5, 4 t : b g b g x 5t 5 y 4 t 9 Let . . .(1) b g mid -pt of the chord is h, k . Its equation is hx ky h 2 k 2 (1) and (2) represent the same line . . .(2) 5t 5 4 t 9 20t 20 20t 2 2 h k h k 4h 5k 42. b20t 20g b20tg 4 h 5k 20 4 h 5k 9 20 2 h k 4 h 5k 20 h 2 k 2 9 4 h 5k 0 20h 2 20k 2 36h 45k 0 Equation of locus is 20x 2 20y 2 36x 45y 0 2 c h b g The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is (A) 75 (B) 150 (C) 210 (D) 243 [26] IIT JEE – 2012 / Paper-I MATHEMATICS Sol: [B] Possible combination are 1, 1, 3 and 2, 2, 1 The number of required ways 43. 5! b1!g 3!b2 !g 2 3! 5! b2 !g 1! b2 !g 3 ! 60 90 150 2 R| bg S |T 2 Let f x x cos x , x 0 , x IR 0 , x0 then f is (A) differentiable both at x 0 and x 2 (B) differentiable at x 0 but not differentiable at x 2 (C) not differentiable at x 0 but differentiable at x 2 (D) differentiable neither at x 0 nor at x 2 Sol: [B] bg f ' 0 lim x 0 bg bg f x f 0 x0 x 2 cos lim x 0 0 x lim x cos x 0 x 0 x f is differentiable at x 0 R f ( 2 ) lim x 2 lim x 2 cos x 2 x2 2 x cos lim x 2 f ( x) f (2) lim x2 x 2 0 x x2 x 2 cos LM 0 formOP N0 Q x FG H x 2 sin x x x IJ FG IJ K H x K (Applying L' Hospital rule) 2 0 L f ( 2) lim x2 f ( x) f ( 2) lim x2 x2 0 x x2 x 2 cos [27] IIT JEE – 2012 / Paper-I MATHEMATICS lim x 2 cos x2 x 2 44. LM 0 formOP N0 Q x f is not differentiable at x 2 bg The function f : 0, 3 1, 29 , defined by f x 2 x 3 15x 2 36x 1, is (A) one-one and onto. (C) one-one but not onto (B) onto but not one-one. (D) neither one-one nor onto. Sol: [B] f ( x ) 2 x 3 15x 2 36 x 1 f ( x ) 6 x 2 30 x 36 6( x 2 ) ( x 3) f’ – + 2 f (2, 29) + 3 (3, 28) f ( 0) 1 , f ( 2 ) 29 , f ( 3) 28 2 3 f is not one-one but onto. FG x x 1 ax bIJ 4 then H x 1 K 2 45. If lim x (A) a 1, b 4 (B) a 1, b 4 (C) a 1, b 3 F x x 1 ax bI 4 GH x 1 JK 2 Sol: [B] lim x x x 1 ( ax 3) ( x 1) lim 4 x x 1 2 (1 a ) x 2 (1 a b ) x (1 b ) 4 x x 1 lim lim x (1 a ) x (1 a b ) 1 1 x 1 b x 4 1 a 0 AND 1 a b 4 a 1 AND b 4 (D) a 2, b 3 [28] IIT JEE – 2012 / Paper-I MATHEMATICS 46. Let z be a complex number such that the imaginary part of z is nonzero and a z 2 z 1 is real. Then a cannot take the value (A) 1 (B) 1 3 (C) 1 2 y0 a (D) 3 4 Sol: [D] Let z x iy , y 0 2 a z 2 2 1 = ( x iy ) x iy 1 ( x 2 y 2 x 1) i ( 2 y y ) 2yy 0 y ( 2 x 1) 0 2x 1 0 x 2 2 Now a x y x 1 47. 1 2 3 3 2 y 2 2 3 4 x 2 y2 1 is inscribed in a reactangle R whose sides are parallel to the coordinate axes. The ellipse E1 : 9 4 Another ellipse E 2 passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E 2 is (A) 2 2 (B) 3 2 (C) y (0, 4) Sol: [C] (3, 2) x 2 2 Let equation of ellipse E 2 is x y 1 2 16 a This passes through (3, 2) 9 4 1 a 2 16 This passes through (3, 2) 1 2 (D) 3 4 [29] IIT JEE – 2012 / Paper-I MATHEMATICS 9 4 1 a 2 16 48. a 2 12 e 1 12 3 1 1 16 4 2 Let P a ij be a 3 3 matrix and let Q b ij , where b ij 2 i j a ij for 1 i, j 3 . If the determinant of P is 2, then the determinant of the matrix Q is (A) 210 (B) 211 2 3 a 11 Q 2 3 a 21 2 4 a 31 Sol: [D] 2 3 a 12 2 4 a 22 2 5 a 32 (C) 212 (D) 213 2 4 a 13 2 5 a 23 2 6 a 33 a 11 2 3 4 ( 2 ) ( 2 ) ( 2 ) a 21 a 31 2 a 12 2 a 22 2 a 32 2 2 a 13 2 2 a 23 2 2 a 33 a 11 2 ( 2 ) ( 2 ) a 21 a 31 9 2 a 12 a 22 a 32 12 212 P 2 ( 2 ) 2 13 49. The integral (A) (B) RS T Sol: [C] UV W 1 1 1 (sec x tan x ) 2 K 11/ 2 11 7 (sec x tan x ) RS T UV W 1 1 1 (sec x tan x ) 2 K 11/ 2 11 7 (sec x tan x) (C) (D) z sec 2 x dx equals (for some arbitrary constant K) (sec x tan x ) 9 / 2 RS T UV W 1 1 1 (sec x tan x) 2 K 11/ 2 11 7 (sec x tan x) RS T UV W 1 1 1 (sec x tan x) 2 K 11/ 2 11 7 (sec x tan x) z sec 2 x (sec x tan x) I dx (sec x tan x ) 11/ 2 a 13 a 23 a 33 [30] IIT JEE – 2012 / Paper-I MATHEMATICS Let sec x tan x t sec x (sec x tan x ) dx dt 1 1 Also sec x tan x 2 sec x t t t I z z ( 2 sec x ) sec x (sec x tan x ) dx 2 (sec x tan x) 11/ 2 z FG t 1IJ dt H tK 2 t 11/ 2 1 ( t 9 / 2 t 13/ 2 ) dt 2 Ft 1I k t GH 7 11JK 1 FG 1 1 (sec x tan x) IJ k (sec x tan x) H 11 7 K FG H IJ K 1 t 7 / 2 t 11/ 2 k 2 7 / 2 11 / 2 1 2 11/ 2 2 50. The point P is the intersection of the striaght line joining the points Q( 2 , 3, 5) and R(1, 1, 4 ) with the plane 5x 4 y z 1 . If S is the foot of the perpendicular drawn from the point T( 2 , 1, 4 ) to QR , then the length of the line segment PS is (A) 1 2 (B) Sol: [A] Equation of the line QR (C) 2 2 (D) 2 2 x2 y3 z5 ( say ) 1 4 1 P ( 2 , 4 3, 5) P lies on the plane 5 ( 2 ) 4 ( 4 3) ( 5) 1 P 2 3 FG 4 , 1 , 13IJ H3 3 3K Dr's of QR = ( 2 1, 3 1, 5 4 ) (1, 4 , 1) FG H IJ FG K H 4 1 13 2 2 1 Dr's of PT = 3 2 , 3 1, 3 4 3 , 3 , 3 Angle between TP and QR = cos IJ K ( 2 ) (1) ( 2 ) ( 4 ) (1) (1) 1 45o 2 1 16 1 4 4 1 [31] IIT JEE – 2012 / Paper-I MATHEMATICS 4 4 1 1 9 9 9 Also TP o TS TP cos45 1 1 1 2 2 SECTION II : Multiple Correct Answer(s) Type The section contains 5 multiple choice questions. Each questions has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct 51. Let , [ 0, 2 ] be such that FG H IJ K 2 cos (1 sin ) sin 2 tan cot cos 1 , 2 2 tan( 2 ) 0 and 1 sin 3 2 Then cannot satisfy (A) 0 2 (B) b 4 2 3 g Sol: [A, C, D] 2 cos 1 sin sin 2 4 3 3 2 F 1 tan I 2 J cos 1 GG J GH tan 2 JK 2 2 cos 1 2 sin cos 1 sin 2 cos 1 = 2 sin cos 2 cos sin 2 sin( ) sin 2 (C) sin( ) cos Now 1 sin 1 2 1 2 3 AND tan 0 2 FG 3 5 IJ 3K H2 0 cos 1 sec( ) 1 2 (D) 3 2 2 [32] IIT JEE – 2012 / Paper-I MATHEMATICS 3 13 2 2 6 For 0 1 sin( ) For 4 2 3 1 2 (A) is an answer 2 3 (B) is Not an answer 52. 4 3 3 2 3 2 2 3 17 19 6 6 (C) is an answer 11 3 (D) is an answer 2 Let S be the area of the region enclosed by y e x , y 0 , x 0 , and x 1 . Then (A) S 1 e (B) S 1 F GH 1 1 (C) S 4 1 e I JK 1 e F GH 1 1 1 1 2 e 2 (D) S Sol: [A, B, D] S (1) e FG 1IJ H eK x2 S e z S 1 2 x 1 e for x [ 0, 1] z e x dx e x dx FG H 1 1 1 1 e 4 e S 1 IJ K S < area of OA 1 PB2 area of A 1A 2 QP' bg FG H 1 e 1 1 1 1 1 2 e 2 IJ K I JK [33] IIT JEE – 2012 / Paper-I MATHEMATICS 53. A ship is fitted with three engiens E 1 , E 2 and E 3 . The engines function independently of each other with respective probabilities 1 1 1 , and . For the ship to be operational at least two of its engines must 2 4 4 function. Let X denote the event that the ship is operational and let X1 , X 2 and X 3 denote respectively the events that the engines E 1 , E 2 and E 3 are functioning. Which of the following is (are) true? c (A) P X 1 | X 3 16 (B) P [ Exactly two engines of the ship are functioning | X]= (C) P X | X 2 5 16 (D) P X | X 1 Sol: [B,D] P( X 1 ) 7 8 7 16 1 1 1 , P ( X 2 ) , P( X 3 ) 2 4 4 P ( X ) P ( X 1 X 2 X 3 ) P ( X 1 X 2 X 3 ) P ( X1 X 2 X 3 ) P ( X1 X 2 X 3 ) 1 1 1 1 1 3 1 3 1 1 1 1 8 2 4 9 2 4 4 2 4 4 2 4 4 32 P ( X 1 / X) 1 P ( X 1 / X) c 1 P 1 ( X1 X) P ( X) P ( X1 X 2 X 3 ) P ( X1 X 2 X 3 ) P ( X1 X 2 X 3 ) P( X) 1 1 3 1 3 1 1 1 1 1 2 4 4 2 4 4 2 4 4 1 7 1 8 8 8 32 P (Exactly two functioning / X) = = P( X1 X2 X3 ) P( X1 X2 X3 ) P( X1 X2 X3 ) P( X) 1 1 3 1 3 1 1 1 1 7 2 4 4 2 4 4 2 4 4 8 8 32 x1 x2 x x3 [34] IIT JEE – 2012 / Paper-I MATHEMATICS P( X / X 2 ) P( X X 2 ) P ( X 1 X 2 X 3 ) P ( X 1 X 2 X 3 ) P ( X1 X 2 X 3 ) P( X 2 ) P( X 2 ) 1 1 3 1 1 1 1 1 1 5 2 4 4 2 4 4 2 4 4 1 8 4 P( X / X1 ) P ( X 1 X 2 X 3 ) P ( X 1 X 2 X 3 ) P ( X 1 X 2 X 3 ) P( X1 ) 1 3 1 1 1 1 3 1 1 1 7 2 4 4 2 2 4 4 2 4 4 1 16 2 2 54. 2 x y 1 , parallel to the straight line 2 x y 1 . The points of 9 4 contact of the tangents on the hyperbola are Tangents are drawn to the hyperbola (A) F 9 , 1I GH 2 2 2 JK F GH (B) Sol: [A,B] Equation of Tangent 9 2 2 , 1 2 I JK e (C) 3 3 , 2 2 j e (D) 3 3 , 2 2 j y 2 x 9(2) 4 2 y 2 x 4 2 . . . (1) tangent at point ( x 1 , y 1 ) 55. x 2 2 y 4 2 xx1 yy1 1 9 4 F GH x 1 9 2 9 I yF 1 I J GH 2 JK 2K 1 4 If y(x) satisfies the differential equation y y tan x 2 x sec x and y( 0) 0 , then FG IJ H K 2 y (A) 4 8 2 FG IJ H K 2 y (B) 4 18 Sol: [A,D] y y ( tan x ) 2 x sec x FG IJ H 3K 9 2 (C) y FG IJ H K 4 2 2 y (D) 3 3 3 3 [35] IIT JEE – 2012 / Paper-I MATHEMATICS Integration factor I ( x) e z tan xdx z cos x 2 Solution is given by y (cos x ) ( 2 x sec x ) cos xdx x K y x 2 sec x y ( 0) 0 K 0 SECTION III : Integer Answer Type This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive) 56. Let f : IR IR be defined as f ( x) x x 1 . The total number of points at which f attains either 2 a local maximum or a local minimum is Sol: [5] bg bg R| x c x 1h | x c1 x h S || x c1 x h T x x 1 f x x x2 1 x 1 2 1 x 0 2 0 x 1 1 x 2 2 R| x x 1 | x x 1 S || x x 1 T x x 1 x 1 1 x 0 0 x 1 2 2 2 1 x 2 57. The value of 6 log 3 2 Sol: [4] Let x 1 3 2 1 x 18 x 2 4 x 3 2 F 1 GG 3 2 H 4 4x 1 3 2 4 1 3 2 4 3 3 4 1 3 2 4 1 3 I ... J is 2 J K 4 .... 3 2 x 4x 18 x 2 x 4 0 [36] IIT JEE – 2012 / Paper-I MATHEMATICS b9 x 4gb2 x 1g 0 F 4I F 3I 6 log G J 6 log G J H 9K H 2K 3/ 2 x 4 9 x0 2 3/ 2 62 4 Let p ( x ) be a real polynomial of least degree which has a local maximum at x 1 and a local minimum at x 3 . If p(1) 6 and p( 3) 2 , then p ( 0) is Sol: [9] Least degree the polynomial having a local maximum and a local minimum is 4. 58. P ( x ) A ( x 1) ( x 3) A ( x 2 4 x 3) F x 2 x 3x I B GH 3 JK F1 I P (1) A G 2 3J B H3 K 3 P( x) A 2 4 AB6 3 P ( 3) 2 A ( 9 18 9 ) B 2 P ( B) 2 A ( 9 18 9 ) B 2 59. 4 A 3B 18 B2 B2 P ( 0) 3A 9 A3 2 2 If a , b and c are unit vectors satisfying a b b c c a Sol: [3] FG H IJ e K 9 , then 2a 5b 5c is 2 2 2 2 a b c 2 a . b b. c c . a 9 j 3 abbcca 2 2 2 2 2 2 a 5b 5c 4 a 25 b 25 c 20 a b 50 b c 50 c a 54 30 a b 50( a b b c c a ) FG IJ H K 3 54 30 a b 50 2 21 30 a b 2 2 a 5b 5c 21 30( 1) =9 2 2a 5b 5c 3 [37] IIT JEE – 2012 / Paper-I MATHEMATICS 60. Let S be the focus of the parabola y 2 8 x and let PQ be the common chord of the circle x 2 y 2 2 x 4 y 0 and the given parabola. The area of the triangle PQS is Sol: [4] x2 y2 2x 4y 0 x( x 2 ) y( y 4 ) 0 ( 0, 0) and (2, 4) are diametric points Q Also ( 2, 4 ) lies on the parabola P ( 0, 0) , Q ( 2 , 4 ) S ( 2 , 0) 1 Area of PQS = 2 4 4 2 P S [1] IIT JEE – 2012/ Paper - 2 PHYSICS Part I: Physics Section – I (Single Correct Answer Type) This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. Consider a disc rotating in the horizontal plane with a constant angular speed about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure.When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R. The velocity of projection is in the y-z plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed 1 8 rotation, (ii) their range is less than half the disc radius, and (iii) remains constant throughout. Then (A) P lands in the shaded region and Q in the unshaded region. (B) P lands in the unshaded region and Q in the shaded region. (C) Both P and Q land in the unshaded region. (D) Both P and Q land in the shaded region. Sol. [C] If the disk were stationary pebble thrown from point P would have landed say at P and from Q at Q . Q Q P Now considering rotation of disk, relative velocity of point P w.r.t. P is towards right at time of throw and hence it will now fall to the right of PP i.e. in unshaded region. Similarly relative velocity of Q w.r.t. Q is towards right and this will also fall in unshaded region. [2] IIT JEE – 2012/ Paper - 2 PHYSICS 2. In the given circuit, a charge of 80 C is given to the upper plate of the 4 F capacitor. Then in the steady state, the charge on the upper plate of the 3 F capacitor is (A) 32 C (B) 40 C (C) 48 C (D) 80 C Sol. [C] The net charge on upper plates both capacitor ( 2F and 3F ) will be 80C so charge on upper plate of 3C capacitor will be q 3. FG 3 IJ c80Ch 48C H 3 2K Two moles of ideal helium gas are in a rubber balloon at 30 0C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 350C. The amount of heat required in raising the temperature is nearly (take R = 8.31 J / mol K) (A) 62 J (B) 104 J Sol. [D] The process involved is isobaric so Q nC p T ( 2) 4. (C) 124 J (D) 208 J FG 5R IJ(5) 208 J H2K A loop carrying current I lies in the x - y plane as shown in the figure. The unit of vector k is coming out of the plane of the paper. The magnetic moment of the current loop is (A) a 2 Ik (B) FG 1IJ a Ik H2 K 2 (C) FG 1IJ a Ik H2 K 2 b g (D) 2 1 a 2 Ik [3] IIT JEE – 2012/ Paper - 2 PHYSICS Sol. [B] The system is equivalent to the figure shown so L F F aI I O m M4 G G J J a P I k MN H 2 H 2 K K PQ 2 2 5. FG 1IJ a I k H2 K 2 Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed . The discs are in the same horizontal plane. At time t= 0, the points P and Q are facing each other as shown in the figure. The relative speed between the two points P and Q is vr.. In one time period (T) of rotation of the discs, vr as a function of time is best represented by (A) (B) (C) (D) [4] IIT JEE – 2012/ Paper - 2 PHYSICS Sol. [A] From observation we can see FG H IJ K FG H IJ K T T 3T 0, vr t & 2v 2 4 4 so correct option is (A). vr t 0 & 6. A thin uniform cylindrical shell clossed at both ends is partially filled with water. It is floating vertically in water in half-submerged state. If c is the relative density of the material of the shell with respect to water, then the correct statement is that the shell is (A) more than half-filled if c is less than 0.5. (C) half-filled if c is more than 0.5. (B) more than half-filled if c is more than 1.0. (D) less than half-filled if c is less than 0.5. Sol. [A] If c 0.5, the cylinder will be half filled and if c 0.5, then the cylinder will be more than half filled. 7. A student is performing the experiment of Resonance Column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is 38°C in which the speed of sound is 336 m / s. The zero of the meter scale coincides with the top end of the Resonance Column tube. When the first resonance occurs, the reading of the water level in the column is (A) 14.0 cm (B) 15.2 cm (C) 16.4cm (D) 17.6 cm Sol. [B] For first resonance, 1 e or 8. 1 4 v 0.3d 15.2 cm 4f An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform current densityalong its length. The magnitude of the magnetic field, | B| as a function of the radial distance r from the axis is best represented by (A) (B) (C) (D) [5] IIT JEE – 2012/ Paper - 2 PHYSICS Sol. [D] The cylindrical symmetric current distribution, B 0ienclosed 2r For, r R , B0 2 R r R , B 0 2 F R GH i 2 R 2 4 I JK (r 2 R 2 / 4) 2r i and R r , B 0 2 r so correct option is D. Section–II: (Paragraph Type) This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each questions has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Paragraph for Questions 9 and 10 Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring sin 1 n2 material, then by Snell's law, sin n it is understood that the refracted ray bends towards the normal. But 2 1 it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation n FG c IJ H vK r r , where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, r and r the relative permittivity and permeability of the medium respectively.In normal materials both r and r are positive, implying positive n for the medium. When both r and r are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificiallyprepared and are called meta-materials. Theyexhibit significantlydifferent optical behavior, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials. [6] IIT JEE – 2012/ Paper - 2 PHYSICS 9. For light incident from air on a meta- material, the appropriate ray diagram is (A) (B) (C) (D) Sol. [C] sin 2 n1 sin 1 n2 as n2 0 so the ray will be refracted on the same side of the normal on entering the material. Thus, correct option is (C) 10. Choose the correct statement. (A) The speed of light in the meta- material is v c| n| (B) The speed of light in the meta-material is v c | n| (C) The speed of light in the meta-material is v c b g (D) The wavelength of the light in the meta-material m is given by m air | n|, where air is the wavelength of the light in air Sol. [B] Form the given expression, n so v c n c v [7] IIT JEE – 2012/ Paper - 2 PHYSICS Paragraph for Questions 11 and 12 The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal friction plane with angular speed the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed in this case. Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical and parallel to x-z plane; Case (b) the disc with its face making an angle of 45° with x-yplane and its horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed about the z-axis. 11. Which of the following statements regarding the angular speed about the instantaneous axis(passing through the centre of mass) is correct? (A) It is 2 for both the cases (B) It is for case (a): and (C) It is for case (a): and 2 for case (b) (D) It is for both cases. for case (b) 2 [8] IIT JEE – 2012/ Paper - 2 PHYSICS 12. Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct? (A) It is vertical for both the cases (a) and (b) (B) It is vertical for case (a); and is at 450 to the x-z plane and lies in the plane of the disc for case (b) (C) It is vertical for case (a); and is at 450 to the x-z plane and is normal to the plane of the disc for case (b) (D) It is vertical for case (a); and is at 450 to the x-z plane and is normal to the plane of the disc for case (b) Sol. [D, A] P A B A B Q o From figure, for case (a) when the complete system turns by 180 , the disk also turns by 180o about its vertical axis. P B A Q A B Q P From figure, for case (b) when the complete system turns by 180o, the disk also turns by 180o, about vertical axis. Hence angular speed is for both cases. Paragraph for Questions 13 and 14 The decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory, a c h proton (p) and an electron e are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energyhas a continuous spectrum. Considering a three-body decay process, i.e. n p e ve , around 1930, Pauli explained the bg observed electron energy spectrum. Assuming the anti-neutrino ve to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 106 eV . The kinetic energy carried by the proton is only the recoil energy. 13. If the anti-neutrino had a mass of 3 eV/c2 (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron? (A) 0 K 0.8 106 eV (B) 3.0 eV K 0.8 106 eV (C) 3.0 eV K 0.8 106 eV (D) 0 K 0.8 106 eV [9] IIT JEE – 2012/ Paper - 2 PHYSICS 14. What is the maximum energy of the anti-neutrino? (A) Zero (B) Much less than 0.8 106 eV (C) Nearly 0.8 106 eV (D) Much larger than 0.8 106 eV Sol. [D, C] Treating neutrino as mass particle, from momentum conservation, p p pe pve 0 It is possible that pe 0 & still momentum is conserved. Thus minimum kinetic energyof electron is zero. Total energy (Q - value) released will be lesser than 0.8 106 eV due to finite mass of neutrino. Hence, maximum KE of electron will be less than 0.8 106 eV . Maximum energyof antinutrino will be nearly 0.8 106 eV Section–III (Multiple Correct Answer(s) Type) This section contains 6 multiple choice questions. Each questions has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 15. A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it. The correct statement (s) is / (are) (A) The emf induced in the loop is zero if the current is constant (B) The emf induced in the loop is finite if the current is constant (C) The emf induced in the loop is zero if the current decreases at a steady rate (D) The emf induced in the loop is finite if the current decreases at a steady rate Sol. [A, C] The net flux passing through the loop due to wire is zero and hence no emf is induced in loop irrespective of the current being constant or varying. If zero is considered as finite also the options (A, B, C, D) can also be a possible answer. [10] IIT JEE – 2012/ Paper - 2 PHYSICS 16. In the given circuit, AC source has 100 rad / s. Considering the inductor and capacitor to be ideal, the correct choice (s) is (are) (A) The current through the circuit, I is 0.3 A (B) The current through the circuit, I is 0.3 2 A (C) The voltage across 100 resistor 10 2 V (D) The voltage across 50 resistor = 10 V 20 Sol. [C] In the upper branch, X C 100 , R 100 and than Z1 100 2 and current I1 .A 100 2 which leads by 450 w.r.t. voltage. While in lower branch X L 50 , R 50 and thus Z2 50 2 and current I 2 20 A which lags by 450 w.r.t. voltage. 50 2 Thus total current I is given by summation of phasors I1 and I2 which differ by 900 in phase and hence I I12 I 22 1 A 10 Also voltage across 100 resistor I1R Similarly across 50 resistor I 2 R 20 100 10 2V 100 2 20 50 V = 10 2V 50 2 [11] IIT JEE – 2012/ Paper - 2 PHYSICS 17. Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown in the figure. Given that K 1 q , which of the following statgement(s) is (are) correct 4 0 L2 (A) The electric field at O is 6K along OD (B) The potential at O is zero (C) The potential at all points on the line PR is same (D) The potential at all points on the line ST is same Sol. [A, B, C] From the symmetry of distribution, potential at O is zero and is same at all points on the line PR. Also electric field at O is 6 K along OD. 18. Two spherical planets P and Q have the same uniform density , masses M P and M Q , and surface areas d i A and 4A, respectively. A spherical plane R also has uniform density and its mass is M P M Q . The escape velocities from the planets P, Q and R, are VP , VQ and VR , respectively. Then (A) VQ VR VP (B) VR VQ VP (C) VR / VP 3 (D) VP / VQ 1 2 [12] IIT JEE – 2012/ Paper - 2 PHYSICS Sol. [B, D] For spherical planets having same density area and mass, M R 3 and as A R 2 Thus, so RQ Rp MQ 8 4 2 and MP 1 1 1 M R M P MQ 9 M P and hence RR 3R P Now, v esc 2 GM R R VP 1 V R VQ V P and V 2 Q 19. The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed / 2 . The ring and disc are separated by frictionless ball bearings. The system is in the x - z plane. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of 300 with the horizontal. Then with respect to the horizontal surface. (A) the point O has a linear velocity 3Ri (B) the point P has a linear velocity (C) the point P has a linear velocity 11 3 Ri Rk 4 4 13 3 Ri Rk 4 4 F GH (D) the point P has a linear velocity 3 I JK 3 1 Ri Rk 4 4 [13] IIT JEE – 2012/ Paper - 2 PHYSICS Sol. [A, B] Since outer ring is rolling without slipping Vcm Vo 3 R i Also, V P Vcm V P / cm 3Ri 20. FG R IJ ecos30k sin 30ij H2K VP/cm 300 P 300 O 11 3 R i R k 4 4 Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement (s) is (are) correct? (A) Both cylinders P and Q reach the grough at the same time (B) Cylinder P has larger linear acceleration than cylinder Q (C) Both cylinders reach the ground with same translational kinetic energy (D) Cylinder Q reaches the ground with larger angular speed Sol. [D] From the given information radius of gyration of P is greater than that Q i.e., k P k Q . Now for rolling cylinder down a fixed inclined plane g sin and thus QP QQ and hence Q reaches earlier. Also Q reaches the ground with larger 1 k 2 / R2 angular speed. a cm [14] IIT JEE – 2012/ Paper -2 CHEMISTRY Part II Section–I Single Correct Answer Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 21. The compound that undergoes decarboxylation must readily under mild condition is (A) (B) (C) (D) Sol: [B] It is a -keto acid 22. The shape of XeO 2 F2 molecule is (A) trigonal bipyramidal (C) tetrahedral F | Sol: [D] : Xe | F 23. (B) square planar (D) see-saw O O See saw structure For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2 C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take K b 0.76 K kg mol 1 ) (A) 724 (B) 740 (C) 736 (D) 718 Sol: [A] mass = 2.5 g moles of solute = so, 2.5 M molality of solute = Tb k b m 25 M [15] IIT JEE – 2012/ Paper -2 CHEMISTRY 2 0.76 25 M P X solute P 0.76 25 2 M P 0.26 as dilute solution P 5.55 M 9.5 Psolution 724 mm of Hg 24. b NiCl 2 P C2 H 5 g bC H g 2 6 5 2 exhibits temperature dependent magnetic behaviour (paramagnetic/ diamag- netic). The coodination geometries of Ni 2 in the paramagnetic and diamagnetic states are respectively (A) tetrahedral and tetrahedral (B) square planar and square planar (C) tetrahedral and square planar (D) square planar and tetrahedral Sol: [C] sp3 and dsp 2 hybridisation 25. The major product H of the given reaction sequence is – CN H 2SO 4 CH 3 CH 2 CO CH 3 G 95% H Heat (A) CH 3 CH C COOH | CH 3 (B) CH 3 CH C CN | CH 3 OH | (C) CH 3 CH 2 C COOH | CH 3 (D) CH 3 CH C CO NH 2 | CH 3 CN | CN H 2SO4 CH 3CH 2 C CH 3 95% Sol: [A] CH 3CH 2 C CH 3 || | O OH (a) OH | CH 3CH 2 C COOH CH 3CH 2 C COOH | | CH 3 CH 3 [16] IIT JEE – 2012/ Paper -2 CHEMISTRY 26. The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphorus containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other product are respectively (A) redox reaction; –3 and –5 (B) redox reaciton + 3 and + 5 (C) disproportionation reaction; –3 and +5 (D) disproportionation reaction; –3 and +3 Sol: [zero to all] 27. Using the data provided, calculate the multiple bond energy (kJ mol–1) of a C C bond in C 2 H 2 . That energy is (take the bond energy of C H bond as 350 kJ mol 1 ) 2C( s) H 2 ( g) C 2 H 2 ( g ) H 225kJ mol 1 2 C( s) 2C( g ) H 1410 kJ mol 1 H 2 ( g) 2 H ( g ) H 330 kJ mol 1 (A) 1165 (C) 865 (B) 837 (D) 815 2 C (g) 2 H (g) Sol: [D] 1410 2 × 330 – 2 DH–C – DCC 2 C( s) H 2 (g ) C 2 H 2 ( g) H 225 kJ / mol 1410 330 ( 2 350 D C C ) 225 1410 330 700 DCC 225 DCC 1410 330 925 = 815 28. In the cyanide process of silver from argentite ore, the oxidizing and reducing agents used are (A) O 2 and CO respectively (B) O 2 and Zn dust respectively (C) HNO 3 and Zn dust respectively (D) HNO 3 and CO dust respectively Sol: [B] [17] IIT JEE – 2012/ Paper -2 CHEMISTRY SECTION II : Paragraph Type This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct Paragraph for Questions 29 and 30 Bleaching powder and bleach solution are produced on a large scale and used in several house hold products. The effectiveness of bleach solution is often measured by iodometry. 29. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is (A) Cl 2 O 30. (B) Cl 2 O 7 (C) ClO 2 (D) Cl 2 O 6 25 mL of household bleach solution was mixed with 30 mL of 0.50 MKI and 10 mL of 4 N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N Na 2 S 2 O 3 was used to reach the end point. The molarity of the household bleach solutions is (A) 0.48 M (B) 0.96 M (C) 0.24 M (D) 0.024 M b g Sol: [A,C] Bleaching powder Ca OCl Cl , so oxoacid is HClO. Anhydride of HClO is Cl 2 O OCl 2I 2 H Cl I 2 H 2 O I3 2S2O42 3I S4O62 for 2 S 2 O 4 its normality is equal to its molarity 2 12mmol of S 2 O 3 So its M 1mol I 2 2 mol of S 2 O 32 1 mol of OCl 6 mm of OCl 1mol of I 2 6 mm of 0.24 25mL Paragraph for Questions 31 and 32 In the following reaction sequence, the compound J is an intermediate J ( C 9 H 8 O 2 ) gives effervescence on treatment with NaHCO3 and positive Baeyer's test [18] IIT JEE – 2012/ Paper -2 CHEMISTRY 31. The compound K is (A) Sol: [C] (B) O || CH (C) bCH CO g O 3 2 CH COONa 3 (D) CH CH C OH 2 , Pd / c || H O HO O SOCl 2 AlCl 3 Cl O O (K) 32. The compound I is (A) (B) (C) (D) Sol: [A] Paragraph for Questions 33 and 34 The electrochemical cell shown below is a concentration cell. c h 2 3 M | M 2 (saturated solution of a sparingly soluble salt,) MX2 M 0.001 mol dm M The emf of the cell depends on the difference in concentrations of M 2 ions at the two electrodes. The emf of the cell at 298 K is 0.059 V. 33. c h The value of G kJ mol 1 for the given cell is (take 1F 96500 C mol 1 ) (A) – 5.7 (B) 5.7 (C) 11.4 (D) – 114 . [19] IIT JEE – 2012/ Paper -2 CHEMISTRY 34. d i The solubility product K sp mol 3 dm9 of MX 2 at 298 K based on the information available for the given concentration cell is (take 2.303 R 298 / F 0.059 V ) (A) 1 1015 (B) 4 1015 (C) 1 1012 (D) 4 1012 Sol: [D, B] For the given cell c h 2 3 M | M 2 (saturated solution of a sparingly soluble salt,) MX2 M 0.001 mol dm M E cell 0.059 V G nFE 2 96500 0.059 11.40 kJ E cell cell FG Ksp IJ H4K 0.059 log 2 0.001 FG Ksp IJ H4K 0.059 0.059 log 2 3 10 10 2 F Ksp IJ G H4K 4 10 15 Ksp 1/ 3 1/ 3 0.001 1/ 3 [20] IIT JEE – 2012/ Paper -2 CHEMISTRY Section–III Section III : Muliple Corect Answer(s) Type This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 35. The given graphs / data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice (s) about I, II, III and IV is (are) correct? (A) I is physisorption and II is chemisorption (C) IV is chemisorption and II is chemisorption Sol: [A, C] (B) I is physisorption and III is chemisorption (D) IV is chemisorption and III is chemisorption [21] 36. IIT JEE – 2012/ Paper -2 CHEMISTRY The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statements(s) is (are) correct ? (A) T1 T2 (B) T3 T1 (C) w isothermal w adiabatic Sol: [A, C, D OR A, D] 37. (D) U isothermal Uadiabatic For the given aqueous reactions, which of the statements(s) is (are) true? b g excess KI K 3 Fe CN 6 H 2SO 4 dilute brownish - yellow solution ZnSO 4 yellow filtrate white precipitate + brownish Na 2S2 O 3 colourless solution (A) The first reaction is a redox reaction b g (B) White precipitate is Zn 2 Fe CN 6 2 (C) Addition of filtrate to starch solution gives blue colour. (D) White precipitate is soluble in NaOH solution Sol: [A, C, D] [22] IIT JEE – 2012/ Paper -2 CHEMISTRY 38. With reference to the scheme given, which of the given statement(s) about T, U, V and W is (are) correct? (A) T is soluble in hot aqueous NaOH (B) U is optically active (C) Molecular formula of W is C10 H 18O4 (D) V gives effervescence on treatment with aqueous NaHCO 3 Sol: [A, C, D] 39. Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct? (A) M and N are non-mirror image stereoisomers (B) M and O are identical (C) M and P are enantiomers (D) M and Q are identical Sol: [A, B, C] 40. With respect to graphite and diamond, which of the statement(s) given below is (are) correct? (A) Graphite is harder than diamond. (B) Graphite has higher electrical conductivity than diamond. (C) Graphite has higher thermal conductivitythan diamond. (D) Graphite has higher C-C bond order than diamond. Sol: [B, D] [23] IIT JEE – 2012/ Paper -2 MATHEMATICS Part III Section–I : Straight Correct Answer Type This section contains 8 multiple choice questions. Each question has fcour choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 41. Four fair dice D1 , D 2 , D 3 and D 4 , each having six faces numbered 1, 2, 3, 4, 5 and 6, are rolled simultaneously. The probability that D 4 shows a number appearing on one of D1 , D 2 and D 3 is 91 216 (A) (B) 108 216 (C) 125 216 (D) 127 216 Ans. [A] Sol. E1 : number of D 4 is shown by exactly one of D1 , D 2 , D 3 E 2 : number of D 4 is shown by exactly two of D1 , D 2 , D 3 E 3 : number of D 4 is shown by exactly all of D1 , D 2 , D 3 E : D 4 shows a number appearing on one of D1 , D 2 , D 3 bg b g b g b g F 1I F 5I F 1I F 5I F 1I 91 c C hG J G J c C hG J G J G J H 6K H 6K H 6K H 6K H 6K 216 P E P E E1 P E E 2 P E E 3 2 3 1 42. 2 3 3 2 If P is a 3 × 3 matrix such that P T 2 P I, where P T is the transpose of P and I is the 3 × 3 identity LMxOP LM0OP matrix, then there exists a column matrix X M y P M0P such that NM zPQ MN0PQ LM0OP (A) PX M0P (B) PX X (C) PX 2 X MN0PQ Ans. [D] Sol. Let LMa P a MMa N 11 21 31 a 12 a 22 a 32 PT 2P I a13 a 23 a 33 OP PP Q (D) PX X [24] IIT JEE – 2012/ Paper -2 MATHEMATICS LMa MMa Na 11 12 13 a 21 a 31 a 22 a 23 a 32 a 33 OP LM2a PP MM2a Q N2a 11 21 31 2a 12 2a 13 2a 22 2a 32 2a 23 2a 33 a 11 a 22 a 33 1 and a 12 a 21 a 31 a 13 a 23 a 32 0 OP LM1 PP MM0 Q N0 OP 0P 1PQ 0 0 1 0 LM1 0 0 OP P M 0 1 0 P so MN 0 0 1PQ LM1 0 0OP LMxOP LM xOP Hence PX M 0 1 0P M yP M yP X MN 0 0 1PQ MN zPQ MN zPQ 43. bg bg d 1 a 1ix d 1 a 1ix d 1 a 1i 0 where a 1 ba g and lim ba g are Then lim Let a and a be the roots of the equation 2 3 6 a 0 (A) a 0 5 and 1 2 (B) 1 and –1 2 (C) 7 and 2 2 Ans. [B] F 1 b1 ag F 1 a 1I G lim G lim G 2 J H 1 a 1K GH 13 b1 ag Sol. a 0 a 0 3 b g b g 1 1 a 1 a 1 6 lim lim 1 a0 a0 1 1 a 3 1 1 a 3 b g b g 1 6 3 1 , 2 2 1 , 1 2 5 6 2 3 1 2 1 2 2 3 I JJ 3 JK 2 (D) 9 and 3 2 [25] 44. IIT JEE – 2012/ Paper -2 MATHEMATICS The equation of a plane passing through the line of intersection of the planes x 2 y 3z 2 and 2 from the point (3, 1, –1) is 3 x y z 3 and at a distance (A) 5x 11y z 17 (B) 2 x y 3 2 1 (C) x y z 3 (D) x 2 y 1 2 Ans. [A] Sol. Let equation of plane through the line of intersectionof the planes x 2 y 3z 2 and x y z 3 is bx 2y 3z 2g bx y z 3g 0 . . . (i) b1 gx b2 g b3 gz 2 3 0 perpendicular distance from (3, 1, –1) b g b g b g b1 g b2 g b3 g 3 1 2 3 2 3 2 2 3 4 14 2 45. 2 2 3 2 2 3 2 3 3 32 4 14 7 2 32 32 4 14 Required plane 5x 11y z 17 0 5x 11y z 17 0 FG1 7 IJ x FG 2 7 IJ y FG 3 7 IJ z 2 3FG 7 IJ 0 H 2K H 2K H 2K H 2K Let a 1 , a 2 , a 3 ,... be in harmonic progression with a 1 5 and a 20 25 . The least positive integer n for which a n 0 is (A) 22 (B) 23 (C) 24 Ans. [D] 1 1 Sol. In corresponding AP, T1 , T20 5 25 1 1 (Let common difference of AP is d) T20 19d 5 25 d 40 25 19 (D) 25 [26] IIT JEE – 2012/ Paper -2 MATHEMATICS b gb g 4 1 0 Now Tn 0 n 1 5 25 19 46. 95 3 1 n 24 4 4 lest value of x = 25 n If a and b are vectors such that a b 29 and a 2 i 3j 4 k 2 i 3j 4 k b , then a pos sible value of a b 7 i 2 j 3k is e d ie (A) 0 j e j j (B) 3 (C) 4 (D) 8 Ans. [C] a 2 i 3j 4 k 2 i 3j 4 k b Sol. e j e j a e2 i 3j 4 k j e2 i 3j 4 k j b 0 da bi e2i 3j 4k j 0 but a b 0 da bi e2i 3j 4k j (say) a b 2 a 2 a 2 a 1 1 a b 2 i 3j 4 k j Now da bi e 7 i 2 j 3k j e2 i 3j 4 k j e7 i 2 j 3k j b 14 6 12g 4 47. e The value of the integral 2 z FGH 2 x 2 ln IJ K x cos x dx x is (A) 0 Ans. [B] 2 4 (B) 2 2 4 (C) 2 2 (D) 2 [27] IIT JEE – 2012/ Paper -2 MATHEMATICS 2 Sol. z FGH x 2 ln 2 IJ K x cos x dx x 2 z 2 z FG x IJ cos x dx H xK bodd functiong x 2 cos x dx ln 2 beven functiong 2 2 z 2 x 2 cos x dx 0 0 g b c h b LF I O 4 2 MG 2J b0gP NH 4 K Q 4 2 x sin x 2 x cos x 2 sin x 2 2 48. g 2 0 2 Let PQR be a triangle of area with a 2, b 5 7 and c , where a, b and c are the lengths of the 2 2 sides of the triangle copposite to the angles at P, Q and R respectively. Then 3 (A) 4 F 3I (C) GH JK 4 45 (B) 4 Ans. [C] 7 5 2 7 5 2 2 4 Sol. a 2, b , c , semiparimeter s 2 2 2 b b 2 sin P sin 2 P 2 sin P 1 cos P 2 sin P sin 2 P 2 sin P 1 cos P g g 1 cos P P tan 2 1 cos P 2 LM FG 4 7 IJ FG 4 5 IJ OP L bs bgbs cg OP M H 2 K H 2 K P FG 3 IJ M N Q MM PP H 4 K N Q 2 2 2 2 2 sin P sin 2 P equals 2 sin P sin 2 P F 45 I (D) GH JK 4 2 [28] IIT JEE – 2012/ Paper -2 MATHEMATICS Section–II : Paragrah Type This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has four choices (A), (B), (C) and (D), out of which ONE ONE is correct. Paragraph for Questions 49 and 50 Let a n denote the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let b n the number of such n-digit integers ending with digit 1 and c n the number of such n-digit integers ending with digit 0. 49. 50. The value of b 6 is (A) 7 (B) 8 Which of the following is correct? (A) a 17 a16 a 15 (B) c17 c16 c15 (C) 9 (D) 11 (C) b17 b16 c16 (D) a 17 c17 b16 Ans. [B, A] Sol. required number b 6 are 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 0 1 0 1 1 1 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 0 1 1 1 0 1 0 1 a17 1 16C1 15C 2 14 C 3 ... 9 C8 also a16 1 15C1 14 C 2 ... 8C8 a15 1 14 C1 13C 2 ... 8C 7 c h c a 16 a 15 1 15C1 14 C 2 ... 8 C8 1 14 C1 13C 2 ... 8C 7 2 15C1 15C 2 14 C 3 ... 9 C8 a17 h [29] IIT JEE – 2012/ Paper -2 MATHEMATICS Paragraph for Questions 51 and 52 A tangent PT is drawn to the circle x 2 y 2 4 at the point P b g d i 3 , 1 . A straight line L, perpendicular to PT is a tangent to the circle x 3 y 2 1 . 51. 2 A common tangent of the two circles is (A) x 4 52. (B) y 2 (C) x 3y 4 (D) x 2 2 y 6 (C) x 3 y 1 (D) x 3 y 5 A possible equation of L is (B) x 3 y 1 (A) x 3 y 1 Ans. [D, A] Sol. T divides A(0, 0), B(3, 0) in 2 : 1 y FG 2 3 0 , 0 0IJ b6, 0g H 2 1 2 1K Line through (6, 0) y 0 mb x 6g T A B 2 (3, 0) mx y 6m 0 . . . (i) it touches x 2 y 2 4 0 0 6m m2 1 (i) and 1 2 2 2 d 2 2 9 m, m 1 m xy xy 2 2 Tangent at P 2 6 2 2 6 2 2 1 2 2 0 x 2 2y 6 0 0 x 2 2y 6 0 x 2 2y 6 i 3 , 1 to the circle x 2 y 2 4 . . . (i) is x 3 y 1 4 . . . (ii) b g equation of line perpendicular to (i) which touches circle x 3 y 2 1 is L:y b g 1 1 x 3 1 1 3 3 3y x 3 2 x 3y 5 or x 3y 1 2 T [30] IIT JEE – 2012/ Paper -2 MATHEMATICS Paragraph for Questions 53 and 54 bg b g b g z FGH 2btt11g ln tIJK f btg dt for all x b1, g Let f x 1 x sin 2 x x 2 for all x IR , and let g x 2 x 1 53. Consider the statements: bg c h There exists some x IR such that 2 f b xg 1 2 xb1 xg P : There exists some x IR such that f x 2 x 2 1 x 2 Q: Then (A) both P and Q are true (C) P is false and Q is true 54. (B) P is true and Q is false (D) both P and Q are false Which of the following is true? b g g is decreasing on b1, g (A) g is increasing on 1, (B) b g g is decreasing on (1, 2) and increasing on b2, g (C) g is increasing on (1, 2) and decreasing on 2, (D) Ans. [C, B] bg c f x 2x 2 1 x2 Sol. h b1 xg sin x x 2x 2c1 x h b1 xg sin x x 2x 2 b1 xg sin x bx 1g 1 . . . (i) for for Let 2 2 2 2 2 2 2 2 2 2 2 x 1 sin x 1 1 b1 xg 2 which is false x = 1 equation (i) is NOT satisfied P is FALSE bg bg b g hb xg 2b1 xg sin x 2 x 1 hb0g 1 hb1g 1 h x 2f x 1 2x 1 x 2 2 [31] IIT JEE – 2012/ Paper -2 MATHEMATICS b g From IVT, at least one h 0, 1 bg h x 0 Q is TRUE b g z FGH 2btt11g ln tIJK f btg dt L 2bx 1g ln xOP f bxg 0 Differentiating: g' b xg z M N x 1 Q L 2bx 1g ln xOPeb1 xg sin x x j g' b xg M N bx 1g Q 2b x 1g 4 Now Let b xg ln x 2 ln x x 1 bx 1g x gx 1 x 1 2 b g b g 1 4x x 1 ' x 2 2 x x 1 x x 1 2 2 2 bg b g bx 1g ' b xg xb x 1g clearly ' b xg 0, x 1 is decreasing function b xg b1g x 1 b xg 0, x 1 g' b xg bxgeb1 xg sin x x j 0 x 1 gb xg is a decreasing function in b1, g 4 2 2 2 2 2 + + –1 – 0 – 1 [32] IIT JEE – 2012/ Paper -2 MATHEMATICS Section–III : Multiple Correct Answer(s) Type This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE or MORE are correct. 55. bg z x b gb g b g If f x e t t 2 t 3 dt for all x 0, , then 2 0 (A) f has a local maximum at x 2 (B) f is decreasing on (2, 3) bg b g (C) there exists some c 0, such that f " c 0 (D) f has a local minimum at x 3 Ans. [A, B, C, D] bg z x b gb g b g f x e t t 2 t 3 dt , x 0, Sol. 2 f’(x) 0 56. – 2 b g bx 2gbx 3g f ' b xg 0 have two roots f " b xg 0 will have one root. f' x e + x2 3 f(x) For every integer n, let a n and b n be real numbers. Let function f : IR IR be given by b g RSba T f x sin x, for for n cos x, n x 2 n, 2 n 1 , for all integers n. x 2 n 1, 2 n b g If f is continuoues, then which of the following hold(s) for all n? (A) a n 1 b n 1 0 (B) a n b n 1 (C) a n b n 1 1 Ans. [B, D] R|a sin x |b cos x f b xg S ||a sin x Tb cos x n 1 Sol. + n 1 n n x x x x 2 n 2, 2 n 1 2 n 3, 2 n 2 2 n, 2 n 1 2 n 1, 2 n continuity at x 2 n 2 b g b g cosb2 n 2g sinb2 n 2g 1 0 1 a n 1 sin 2 n 2 b n 1 cos 2 n 2 a n 1 b n 1 (D) a n 1 b n 1 [33] IIT JEE – 2012/ Paper -2 MATHEMATICS a n bn 1 continuity at x 2 n 1 b g b g a n 1 sin 2 n 1 b n cos 2 n 1 a n 1 b n 1 also a n b n 1 1 57. x 1 y 1 z x 1 y 1 z and are coplanar, then the the plane(s) contain2 k 2 5 2 k ing these two lines is(are) (A) y 2 z 1 (B) y z 1 (C) y z 1 (D) y 2 z 1 If the straight lines Ans. [B, C] Sol. x 1 y 1 z x 1 y 1 z and are coplanar 2 k 2 5 2 k b g b g 1 1 1 1 00 2 5 k 2 2 k 2 0 0 2 k 2 0 k 2 4 k 2 5 2 k If k 2 then required plane b g b g x 1 2 y 1 2 5 2 0 z0 2 0 2 bx 1gb4 4g by 1gb4 10g zb4 10g 0 6y 6 6z 0 y z 1 if k 2 then plane x 1 y 1 z 2 2 2 0 5 2 2 bx 1gb4 4g by 1gb4 10g zb4 10g 0 14b y 1g 14 z 0 y 1 z 0 y z 1 [34] IIT JEE – 2012/ Paper -2 MATHEMATICS 58. LM1 If the adjoint of a 3 × 3 matrix P is M2 MN1 (A) –2 OP PP Q 4 4 1 7 , then the possible value(s) of the determinant of P is(are) 1 3 (B) –1 (C) 1 (D) 2 Ans. [A, D] 1 4 4 Sol. adj P 2 1 7 4 1 1 3 59. P 4 P 2 2 b b g F 1I of f G J is(are) H 3K g Let f : 1, 1 IR be such that f cos 4 (A) 1 3 2 (B) 1 FG IJ FG H K H 3 2 (C) 1 Ans. [Zero marks to all] b g Sol. f cos 4 2 2 sec 2 2 cos2 1 cos 2 2 2 cos 1 cos 2 Now cos4 1 3 2 cos2 2 cos2 2 cos2 f cos 4 1 1 3 4 6 2 3 FG H 2 0, IJ FG K H , 2 2 b g 1 coscos22 cos12 1 3 F 1I fG J 1 H 3K 2 IJ K 2 for 0, . Then the value(s) , 2 2 sec 4 4 2 IJ K 2 3 (D) 1 2 3 [35] IIT JEE – 2012/ Paper -2 MATHEMATICS 60. b g Let X and Y be two events such that P X | Y b b g following is(are) correct? b g (A) P X Y 2 3 (B) X and Y are independent c Ans. [A, B] Sol. FG X IJ 1 , PFG Y IJ 1 , PbX Yg 1 H YK 2 H XK 3 6 F X I Pb X Y g 1 Pb Y g 1 PG J H Y K PbY g 2 3 F Y I PbY Xg 1 PbXg 1 PG J H X K PbXg 3 2 1 Now Pb X Yg Pb Xg PbYg 6 P X and Y are independent Also P X Y P X P Y P X Y b g bg bg b 1 1 1 2 2 3 6 3 c h bg b P Xc Y P Y P X Y 1 1 1 3 6 6 h c (D) P X Y (C) X and Y are not independent g g g 1 1 1 , P Y | X and P X Y . Which of the 2 6 3 1 3
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