Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi Ph.: Fax : Answers & Solutions

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1 DATE : 5/05/04 CDE Regd. ffice : Aakash Tower, Plot No.-4, Sec-, MLU, Dwarka, New Delhi-0075 Ph.: Fax : Answers & Solutions Time : hrs. Max. Marks: 80 for JEE (Advanced)-04 PAPER - (Code - ) INSTRUCTINS Question Paper Format The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of three sections.. Section contains 0 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which NE is correct.. Section contains paragraphs each describing theory, experiment and data etc. Six questions relate to three paragraphs with two questions on each paragraph. Each question pertaining to a particular passage should have only one correct answer among the four given choices (A), (B), (C) and (D).. Section contains 4 multiple choice questions. Each question has two lists (List-, P, Q, R and S; List- :,, and 4). The options for the correct match are provided as (A), (B), (C) and (D) out of which NLY NE is correct. Marking Scheme 4. For each question in Section,, and you will be awarded marks if you darken the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one ( ) mark will be awarded.

2 PART I : PHYSICS SECTIN - : (nly ne ption Correct Type) This section contains 0 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which NLY NE ption is correct.. A glass capillary tube is of the shape of a truncated cone with an apex angle so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height h, where the radius of its cross section is b. If the surface tension of water is S, its density is, and its contact angle with glass is. the value of h will be (g is the acceleration due to gravity) (A) (B) S cos( ) bg S cos( ) bg h (C) S cos( /) bg (D) S cos( /) bg Answer (D) b h Sb cos b h S cos( ) bg h. A planet of radius R (radius of Earth) has the same mass density as Earth. Scientists dig a well of 0 depth 5 R on it and lower a wire of the same length and of linear mass density 0 kgm into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth = m and the acceleration due to gravity on Earth is 0 ms ) (A) 96 N (C) 0 N (B) 08 N (D) 50 N ()

3 Answer (B) R P Re m P = E GM G 4 g R R R x dx 4 g GR R P gp RP ge gp m/s g R 0 e E Acceleration due to gravity at a depth x g x x g R P Force on a small segment of wire at a depth x from surface df df ( dx) gx x gdx R R /5 x F 0 gdx R x x R R/5 0 g R R 9 9 g 5 0 R g N 5. Charges Q, Q and 4Q are uniformly distributed in three dielectric solid spheres, and of radii R/, R and R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of sphere, and are E, E and E respectively, then P P P Q R Q R 4Q R R/ R Sphere Sphere Sphere (A) E > E > E (B) E > E > E (C) E > E > E (D) E > E > E ()

4 Answer (C) E KQ R E K ( Q ) KQ R R Q E K(4 QR ) KQ 8R R E E E 4. If Cu is the wavelength of K X-ray line of copper (atomic number 9) and Mo is the wavelength of the K X-ray line of molybdenum (atomic number 4), then the ratio Cu / Mo is close to (A).99 (B).4 (C) 0.50 (D) 0.48 Answer (B) Cu ( ZMo ) 4 Mo ( Z 8 Cu ).4 5. A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is A 90 B (A) Always radially outwards (B) Always radially inwards (C) Radially outwards initially and radially inwards later (D) Radially inwards initially and radially outwards later Answer (D) 6. A metal surface is illuminated by light of two different wavelengths 48 nm and 0 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u and u, respectively. If the ratio u : u = : and hc = 40 ev nm, the work function of the metal is nearly (A).7 ev (B). ev (C).8 ev (D).5 ev Answer (A) hc mu W mu hc W (4)

5 hc W u u hc W 4hc hc 4 W W ( u ) u 4hc hc W W 6 5 = W 0 48 = W W =.7 ev 7. A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The figures are only illustrative and not to the scale. K K (A) (B) t t K K (C) (D) Answer (B) v t v t t t K v t v t K.E. t t (5)

6 8. During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90, as shown in the figure. The least count of the scale used in the metre bridge is mm. The unknown resistance is R 90 (A) 60 ± 0.5 (C) 60 ± 0.5 Answer (C) R R x (00 x) 40.0 cm (Balanced Wheatstone) (B) 5 ± 0.56 (D) 5 ± 0. For R, R R R x x For R, ln R ln x ln(00 x) R x (00 x) R x (00 x) R x x R x (00 x) R R = 0.5 R = 60 ± Parallel rays of light of intensity I = 9 Wm are incident on a spherical black body kept in surroundings of temperature 00 K. Take Stefan-Boltzmann constant = Wm K 4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to (A) 0 K (C) 990 K (B) 660 K (D) 550 K (6)

7 Answer (A) 4 R.( T T ) 9R T T T (00) (40 8) 0 T = 0 K 0. A point source S is placed at the bottom of a transparent block of height 0 mm and refractive index.7. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter.54 mm on the top of the block. The refractive index of the liquid is Liquid S (A). (B).0 (C).6 (D).4 Answer (C) Block 0 mm 0 mm = 0 S sin.7 sin0.7.6 =.7 SECTIN - : Comprehension Type (nly ne ption Correct) This section contains paragraphs, each describing theory, experiments, data etc. 6 questions related to the three paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given options (A), (B), (C) and (D). Paragraph For Questions and The figure shows circular loop of radius a with two long parallel wires (numbered and ) all in the plane of the paper. The distance of each wire from the centre of the loop is d. The loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if seen from above. Q d d S Wire a Wire P (7) R

8 . When d a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case (A) Current in wire and wire is the direction PQ and RS, respectively and h a (B) Current in wire and wire is the direction PQ and SR, respectively and h a (C) Current in wire and wire is the direction PQ and SR, respectively and h.a (D) Current in wire and wire is the direction PQ and RS, respectively and h.a Answer (C). 0ia 0ia a h ( a h ) ( ) a a h a 4 a 4a a h 0a 4a 4h 6a 4h a h h.a. Consider d a, and the loop is rotated about its diameter parallel to the wires by 0 from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop) (A) 0I a d (B) 0I a d (C) 0I a d (D) 0I a d Answer (B) 0I B d 0I I a d 0 I a d (8)

9 Paragraph For Questions and 4 In the figure, a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulation material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with moles of an ideal monatomic gas at 700 K and the upper compartment is filled with moles of an ideal diatomic gas 5 at 400 K. The heat capacities per mole of an ideal monatomic gas are CV = R, CP = R, and those for an 5 7 ideal diatomic gas are CV = R, CP = R.. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be (A) 550 K (B) 55 K (C) 5 K (D) 490 K Answer (D) Q = Q R 7R ( T 700) ( T 400) 6T T = 0 T T = 490 K 4. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. The total work done by the gases till the time they achieve equilibrium will be (A) 50 R (B) 00 R (C) 00 R (D) 00 R Answer (D) nc ( T700) nc (400 T) P P 5R 7R ( T 700) (400 T) 600 5T T T T 55 W = U = nc (55 700) nc (55 400) V V R 5R 75 5 W = U = 00 R = 55R + 65R (9)

10 Paragraph For Questions 5 and 6 A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 0 mm and mm respectively. The upper end of the container is open to the atmosphere. 5. If the piston is pushed at a speed of 5 mms, the air comes out of the nozzle with a speed of (A) 0. ms (B) ms (C) ms (D) 8 ms Answer (C) A V = A V (0.) VL VL VL V L = 6. If the density of air is a and that of the liquid l, then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to (A) a l (B) a l (C) l a (D) l Answer (A) P P v a a P P v l l P = P lvl ava vl a l v a Volume flow rate a l (0)

11 SECTIN - : Matching List Type (nly ne ption Correct) This section contains 4 questions, each having two matching lists. Choice for the correct combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which one is correct. 7. A person in lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of. m from the person. In the following, state of the lift's motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II and select the correct answer using the code given below the lists. List-I List-II P. Lift is accelerating vertically up. d =. m Q. Lift is accelerating vertically down. d >. m with an acceleration less than the gravitational acceleration R. Lift is moving vertically up. d <. m with constant speed S. Lift is falling freely 4. No water leaks out of the jar Code : (A) P-, Q-, R-, S-4 (B) P-, Q-, R-, S-4 (C) P-, Q-, R-, S-4 (D) P-, Q-, R-, S- Answer (C) a R gh h g H d v t x H( g a) h g a h H( g+a) d Independent of acceleration of lift. d 8. Four charges Q, Q, Q and Q 4 of same magnitude are fixed along the x axis at x = a, a, +a and +a, respectively. A positive charge q is placed on the positive y axis at a distance b > 0. Four options of the signs of these charges are given in List I. The direction of the forces on the charge q is given in List II. Match List I with List II and select the correct answer using the code given below the lists. q (0, b) List-I Q ( a,0) Q ( a,0) Q (+ a,0) List-II Q4 (+ a,0) P. Q, Q, Q, Q 4 all positive. +x Q. Q, Q positive; Q, Q 4 negative. x R. Q, Q 4 positive; Q, Q negative. +y S. Q, Q positive; Q, Q 4 negative 4. y ()

12 Code : (A) P-, Q-, R-4, S- (B) P-4, Q-, R-, S- (C) P-, Q-, R-, S-4 (D) P-4, Q-, R-, S- Answer (A) Hint: P. Not along +y Q. Not along +x Q Q Q Q 4 Q Q Q Q 4 R. Not along y S. Not along x Q Q Q Q 4 Q Q Q Q 4 9. Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is.5. Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists. List-I List-II P.. r Q.. r R.. r S. 4. r Code : (A) P-, Q-, R-, S-4 (B) P-, Q-4, R-, S- (C) P-4, Q-, R-, S- (D) P-, Q-, R-, S-4 ()

13 Answer (B) (P) ( u ) (.5 ) f r r r r r f ' f' f f P f r (Q) 0.5 (.5 ) f r r f r f f r f f f f r r Q 4 (R) 0.5 (.5 ) f r f r r f r f r r f f f r r f r R (S) (.5 ) 0.5 f r f r r r r 0.5 (.5 ) f r f r f r f f f r r r S f r ()

14 0. A block of mass m = kg another mass m = kg, are placed together (see figure) on an inclined plane with angle of inclination. Various values of are given in List I. The coefficient of friction between the block m and the plane is always zero. The coefficient of static and dynamic friction between the block m and the plane are equal to = 0.. In List II expressions for the friction on block m are given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by g. [Useful information : tan(5.5 ) 0.; tan(.5 ) 0.; tan(6.5 ) 0.] m m List-I List-II P. = 5. m g sin Q. = 0. (m + m )g sin R. = 5. m g cos S. = 0 4. (m + m )g cos Code : (A) P-, Q-, R-, S- (B) P-, Q-, R-, S- (C) P-, Q-, R-, S-4 (D) P-, Q-, R-, S- Answer (D) m m mgcos ( m m) gsin tan m m m 0. So block m starts sliding at >.5. When = 5 and = 0, then block will be rest, so friction will be ( m m) gsin For = 5 and = 0, block starts sliding. So friction will be mg cos. So option (D) is correct. (4)

15 PART II : CHEMISTRY SECTIN - : (nly ne ption Correct Type) This section contains 0 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which NLY NE ption is correct.. Assuming s-p mixing is NT operative, the paramagnetic species among the following is (A) Be (B) B (C) C (D) N Answer (C) Be = s *s s *s B = s *s s *s p z C = s *s s *s p z p x p y N = s *s s *s p z p x p y. For the process H (I) H (g) at T = 00 ºC and atmosphere pressure, the correct choice is (A) Ssystem > 0 and S surroundings > 0 (B) Ssystem > 0 and S surroundings < 0 (C) Ssystem < 0 and S surroundings > 0 (D) Ssystem < 0 and S surroundings < 0 Answer (B) Given conditions are boiling conditions for water due to which S total = 0 S system + S surroundings = 0 S system = S surroundings For process, S system > 0 S surroundings < 0. For the elementary reaction M N, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is (A) 4 (B) (C) (D) Answer (B) r [M] r [M] 8 = () = So, order of reaction is. (5)

16 4. For the identification of -naphthol using dye test, it is necessary to use (A) Dichloromethane solution of -naphthol (B) Acidic solution of -naphthol (C) Neutral solution of -naphthol (D) Alkaline solution of -naphthol Answer (D) + NCl H H NaH N N 5. Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure. [Figure] and and (I) (II) (III) The correct order of their boiling point is (A) I > II > III (B) III > II > I (C) II > III > I (D) III > I > II Answer (B) As branching increases boiling point decreases, so order of boiling point is III > II > I. 6. The acidic hydrolysis of ether (X) shown below is fastest when [Figure] R Acid H + RH (A) ne phenyl group is replaced by a methyl group (B) ne phenyl group is replaced by a para-methoxyphenyl group (C) Two phenyl groups are replaced by two para-methoxyphenyl groups (D) No structural change is made to X Answer (C) CH group has +R effect. It increases the stability of the carbocation. 7. Hydrogen peroxide in its reaction with KI 4 and NH H respectively, is acting as a (A) Reducing agent, oxidising agent (B) Reducing agent, reducing agent (C) xidising agent, oxidising agent (D) xidising agent, reducing agent (6)

17 Answer (A) H acting as a reducing agent with KI 4 and H acting as an oxidising agent with NH H. 8. The major product in the following reaction is Cl. CH MgBr, dry ether, 0 C. aq. acid CH (A) HC CH (B) HC H CH CH CH (C) CH (D) CH Answer (D) Cl CH MgI Cl CH CH CH 9. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is XeF 6 Complete Hydrolysis P + ther product Q H /H Slow disproportionation in H /H Products (A) 0 (B) (C) (D) Answer (C) XeF 6 Complete Hydrolysis Xe + H F H /H HXe 4 Slow disproportionation in H /H 4 Xe 6 + Xe(g) + H + (g) (7)

18 0. The product formed in the reaction of SCl with white phosphorous is (A) PCl (B) S Cl (C) SCl (D) PCl Answer (A) P 8SCl 4PCl 4S S Cl 4 (White) SECTIN - : Comprehension Type (nly ne ption Correct) This section contains paragraphs, each describing theory, experiments, data etc. 6 questions related to the three paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given options (A), (B), (C) and (D). Paragraph For Questions and X and Y are two volatile liquids with molar weights of 0 g mol and 40 g mol respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 4 cm, as shown in the figure. The tube is filled with an inert gas at atmosphere pressure and a temperature of 00 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours. L = 4 cm Cotton wool soaked in X d Initial formation of the product Cotton wool soaked in Y. The value of d in cm (shown in the figure), as estimated from Graham's law, is (A) 8 (B) (C) 6 (D) 0 Answer (C) x 40 4 x 0 x 4 x x = 6 (8)

19 . The experimental value of d is found to be smaller than the estimate obtained using Graham's law. This is due to (A) Larger mean free path for X as compared to that of Y (B) Larger mean free path for Y as compared to that of X (C) Increased collision frequency of Y with the inert gas as compared to that of X with the inert gas (D) Increased collision frequency of X with the inert gas as compared to that of Y with the inert gas Answer (D) Increased collision frequency of X with the inert gas as compared to that of y with the inert gas. Therefore, the experimental value of d is found to be smaller than the estimate obtained using Graham's law. Paragraph For Questions and 4 Schemes and describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both the schemes. H M H. NaNH (excess). CH CH I ( equivalent). CH I ( equivalent) 4. H, Lindlar s catalyst X Scheme- N H. NaNH ( equivalent). H Br. H, (mild) 4. H, Pd/C 5. Cr Y Scheme-. The product X is HC H (A) H H (B) HC H CH C H H (C) H H (D) CH C H H Answer (A) H NaNH + CH CH I CNa ( eq) H Na + Na + C CH I CH H H H Lindlar's catalyst CH (9)

20 4. The correct statement with respect to product Y is (A) It gives a positive Tollens test and is a functional isomer of X (B) It gives a positive Tollens test and is a geometrical isomer of X (C) It gives a positive iodoform test and is a functional isomer of X (D) It gives a positive iodoform test and is a geometrical isomer of X Answer (C) H. NaNH ( eq). H Br +. H (mild) 4. H, Pd/C 5. Cr Paragraph For Questions 5 and 6 An aqueous solution of metal ion M reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M always forms tetrahedral complexes with these reagents. Aqueous solution of M on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarized in the scheme given below : Scheme : Tetrahedral Q Excess M R Excess Square planar Tetrahedral Q Excess M R Excess Tetrahedral S, stoichiometric amount 5. M, Q and R, respectively are White precipitate R Excess Precipitate dissolves (A) Zn +, KCN and HCl (B) Ni +, HCl and KCN (C) Cd +, KCN and HCl (D) Co +, HCl and KCN Answer (B) Ni + + HCl [NiCl 4 ] Ni + + KCN [Ni(CN) 4 ] 6. Reagent S is (A) K 4 [Fe(CN) 6 ] (B) Na HP 4 (C) K Cr 4 (D) KH (0)

21 Answer (D) H 4 white ppt Soluble Zn H Zn(H) [Zn(H) ] SECTIN - : Matching List Type (nly ne ption Correct) This section contains 4 questions, each having two matching lists. Choice for the correct combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which one is correct. 7. Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given below the lists. {en = H NCH CH NH ; atomic numbers : Ti =, Cr = 4; Co = 7; Pt = 78} List-I List-II P. [Cr(NH ) 4 Cl ]Cl. Paramagnetic and exhibits ionisation isomerism Q. [Ti(H ) 5 Cl](N ). Diamagnetic and exhibits cis-trans isomerism R. [Pt(en)(NH )Cl]N. Paramagnetic and exhibits cis-trans isomerism S. [Co(NH ) 4 (N ) ]N 4. Diamagnetic and exhibits ionisation isomerism Code : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 Answer (B) Magnetic character Isomerism P. [Cr(NH ) 4 Cl ]Cl Paramagnetic cis/trans Q. [Ti(H ) 5 Cl](N ) Paramagnetic ionization R. [Pt(en)(NH )Cl]N Diamagnetic ionization S. [Co(NH ) 4 (N ) ]N Diamagnetic cis/trans 8. Match the orbital overlap figures shown in List-I with the description given in List-II and select the correct answer using the code given below the lists. List-I List-II P.. p d antibonding Q.. d d bonding ()

22 R.. p d bonding S. 4. d d antibonding Code : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 Answer (C) P. d d ( bonding) Q. p d ( bonding) R. p d ( antibonding) S. d d ( antibonding) 9. Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from List I with an appropriate structure from List II and select the correct answer using the code given below the lists. P R+ R C R R Q C R+ R R+ X+ carbonyl compound (Peroxyester) R RC + R C R+ X + carbonyl compound S RC + R C R+ R ()

23 List-I P. Pathway P. List-II CHCH 6 5 CH Q. Pathway Q. CH 6 5 CH R. Pathway R. C H CH 6 5 CH CH CHC6H 5 S. Pathway S 4. C H 6 5 CH CH CH 6 5 Code : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 Answer (A) () C6H5CH C CH C H CH + C + CH 6 5 () C H CH C 6 5 CH CH CH C C + C6H5CH + CH C CH C H 6 5 CH C H 6 5 C (4) CH 6 5 C CH CH CH C H C + C CH 6 5 CH CH () CH C 6 5 CH CH 6 5 C + CH ()

24 40. Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction schemes (I, II, III, IV) provided in List II and select the correct answer using the code given below the lists. List-I List-II P. H H. Scheme I (i) KMn 4, H, heat (ii) H, H? (iii) SCl (iv) NH CHN 7 6 H Scheme II Q. H. (i) Sn/HCl (ii) CHCCl (iii) conc. HS4 (iv) HN (v) dil. HS 4, heat (vi) H? CHN 6 6 N Scheme III R.. (i) red hot iron, 87 K (ii) fuming HN, HS 4, heat (iii) HS.NH (iv) NaN, HS 4 (v) hydrolysis? CHN 6 5 N Scheme IV S. CH 4. (i) conc. HS 4, 60ºC (ii) conc. HN, conc. HS 4 (iii) dil. HS 4, heat? CHN Code : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 Answer (C) N Scheme I (S) CH N Scheme II (R) Scheme III H C C H (P) H Scheme IV H (4)

25 PART III : MATHEMATICS SECTIN - : (nly ne ption Correct Type) This section contains 0 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which NLY NE ption is correct. 4. The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has (A) nly purely imaginary roots (B) All real roots (C) Two real and two purely imaginary roots (D) Neither real nor purely imaginary roots Answer (D) Let p(x) = x + a (a > 0) ( roots are purely imaginary) p(p(x)) = (x + a) + a (a R) x 4 + a(x ) + a + a = 0 x = a 4a 4a 4a = a ai x = a ai x iy 4. Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is (A) (B) (C) (D) 4 Answer (A) n(s) = 5! Case I : G 4! = 48 Case II : G! = n(e) = PE ( ) 5! 4. Six cards and six envelopes are numbered,,, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered is always placed in envelope numbered. Then the number of ways it can be done is (A) 64 (B) 65 (C) 5 (D) 67 (5)

26 Answer (C) 6!!!! 4! 5! 6! In a triangle the sum of two sides is x and the product of the same two sides is y. If x c = y, where c is the third side of the triangle, then the ratio of the in radius to the circum-radius of the triangle is (A) y x( x c) (B) y ( cxc) (C) y 4 x( x c) (D) y 4( cxc) Answer (B) a + b = x, ab = y, x c = y Now, (a + b) c = ab a + b c = ab a b c ab cosc c = 0 c c R sinc R c c r ( sc) tan x c r c ( ) r x c r ( x c ) ( ) x c R c c ( x c)( x c) cx ( c) ( x c ) cx ( c) y cx ( c) 45. The common tangents to the circle x + y = and the parabola y = 8x touch the circle at the points P, Q and the parabola at the poitns R, S. Then the area of the quadrilateral PQRS is (A) (B) 6 (C) 9 (D) 5 (6)

27 Answer (D) x + y = Equation of tangent to circle x + y = is y mx m Equation of tangent to y = 8x is y mx m ( m ) m 4 ( m ) m m ( + m ) = m 4 + m = 0 (m + )(m ) = 0 m = y = x + y = x (, ) (, ) (, 4) (, 4) Area = = 5 sq. unit 46. The function y = f(x) is the solution of the differential equation f(0) = 0. Then f ( x ) dx is (A) (B) 4 dy xy x x dx x x 4 in (, ) satisfying (C) 6 4 Answer (B) dy x x x y dx x x 4 (D) 6 I.F. x dx x x e So 5 4 x y x ( x x) dx x c 5 Given f(0) = 0 c = 0 So y 5 x x 5 x f( x) (7)

28 5 x x ( ) 5 x dx I f x dx dx x Let x = sin I sin d 0 ( cos ) d sin x 47. Let f : [0, ] R be a function which is continuous on [0, ] and is differentiable on (0, ) with f(0) =. Let x ( ) ( ) 0 F x f t dt for x [0, ]. If F(x) = f(x) for all x (0, ), then F() equals (A) e (B) e 4 (C) e (D) e 4 Answer (B) F(x) = f(x).x = f(x) f'( x) x f( x) f( x) ke Given f(0) = So x x x F( x) e dx 0 x e So F() = e 4 f( x) e x 48. Coefficient of x in the expansion of ( + x ) 4 ( + x ) 7 ( + x 4 ) is (A) 05 (B) 06 (C) (D) 0 Answer (C) Power of Coefficient of x x x x C C C C C C 4 7 C C C C C C For x (0, ), the equation sinx + sinx sinx = has (A) Infinitely many solutions (B) Three solutions (C) ne solution (D) No solution (8)

29 Answer (D) sinx + sinx sinx = cosx( sinx) + 4sinxcosx = sinx[cosx cosx] = sinx(cosx (cos x )) = sinx( + cosx cos x) = sinx cos x Possible only when sinx = (i) and cos x 0 cos x From (i) and (ii) No solution. (ii) 50. The following integral 7 (cosec x) dx is equal to 4 (A) log( ) u u 6 ( e e ) du (B) 0 log( ) u u 7 ( e e ) du 0 (C) log( ) u u 7 ( ) 0 e e du (D) log( ) u u 6 ( ) 0 e e du Answer (A) 7 I (cosec x) dx 4 cosec x cot x cosec cosec x cot x xdx 4 Let cosecx + cotx = e u cosecxdx = du 6 0 u u 6 I ( e e ) du ln( ) ln( ) 0 u u 6 ( e e ) du (9)

30 SECTIN - : Comprehension Type (nly ne ption Correct) This section contains paragraphs, each describing theory, experiments, data etc. six questions related to the three paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given options (A), (B), (C) and (D). Paragraph For Questions 5 and 5 Box contains three cards bearing numbers,, ; box contains five cards bearing numbers,,, 4, 5; and box contains seven cards bearing numbers,,, 4, 5, 6, 7. A card is drawn from each of the boxes. Let x i be number on the card drawn from the i th box, i =,,. 5. The probability that x + x + x is odd, is (A) 9 05 (B) 5 05 (C) (D) Answer (B) B B 4 5 x + x + x = odd B Case I : ne odd ( ) + (4 ) + ( ) Two even = = 9 Case II : odd 4 = 4 = = 5 Total cases = 5 7 = 05 Probability = The probability that x, x, x are in an arithmetic progression, is 7 (A) 9 05 (C) 05 Answer (C) (B) (D) x = x + x x and x must be even number. both x {, } x and x odd x {,, 5, 7} also x + x {, 4, 6, 8, 0} (0)

31 x x (,, 5, 7) (,, 5, 7) 8 cases x and x both even x = x = (, 4, 6) cases Total = cases Required probability = 05 Paragraph For Questions 5 and 54 Let a, r, s, t be nonzero real numbers. Let P(at, at), Q, R(ar, ar) and S(as, as) be distinct points on the parabola y = 4ax. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (a, 0). 5. The value of r is (A) t (B) t + t (C) t Answer (D) (D) t t (, ) P at at S a t, a t ( ar, ar) R ( a, 0) K(a, 0) a t, Q a t ()

32 Let co-ordinates of Q be (at, at ) Since PQ is focal chord t t = t = t a a Co-ordinates of Q, t t at 0 t Slope of PK = = at a t Slope of RQ a ar + r + t t = = a ar r t t = r t Slope of PK = Slope of RQ t = t r t t t r = r = t t t 54. If st =, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is (A) ( t + ) t (C) at ( + ) t Answer (B) Since st = (B) (D) at ( + ) t at ( + ) t s = t a a Co-ordinates of S =, t t Equation of tangent at P y = t (x + at ) x = yt at...(i) Slope of normal at S s = t ()

33 Equation of normal a a y = x t t t a ty + a = x t a x = ty+ a+...(ii) t From (i) and (ii) a y.t at = ty + a + t a ty = at + t + a ty = a t + t at ( + ) y = t y = Given that for each a (0, ), h a + h 0 h a lim t ( t) dt at + t t Paragraph For Questions 55 and 56 exists. Let this limit be g(a). In addition, it is given that the function g(a) is differentiable on (0, ). 55. The value of g is (A) (B) π (C) Answer (A) g(a) = g ( h) a + h 0 h ( h) / + h 0 h ( h) a lim t ( t) dt = lim ( ) = lim + h 0 h / t t dt dt t( t) t lim sin = + h 0 h 0 h h = lim [sin (( h) )) sin (h )] + π π = + = π (D) () π 4

34 56. The value of g' is (A) π (B) (C) π (D) 0 Answer (D) g(a) = h 0 h a a lim t ( t) dt + h h a a t g '( a) lim t ( t) ln dt h0 t h h t h0 h t g ' lim t ( t) ln dt g ' 0 SECTIN - : Matching List Type(nly ne ption Correct) This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which one is correct. 57. List-I List-II P. The number of polynomials f(x) with non-negative integer. 8 coefficients of degree, satisfying f(0) = 0 and 0 fxdx ( ), is Q. The number of points in the interval [, ]at. which f(x) = sin(x ) + cos(x ) attains its maximum value, is R. x dx ( x e ) equals. 4 S. 0 x cos x log dx x x cos x log dx x equals 4. 0 P Q R S (A) 4 (B) 4 (C) 4 (D) 4 (4)

35 Answer (D) (P) Let f(x) = a 0 x + a x + a ; a 0, a, a > 0 Now, a = 0 and f ( x ) dx 0 ax 0 ax a0 a 0 a 0 + a = 6 i.e., for integral solution either a0, a 0 or a0 0, a i.e., f(x) = x or f(x) = x P () (Q) f( x ) at max 9 x, 4 4 x, So, number of points = 4 Q () (R) I x e x dx again, I x e x dx Adding, I x dx { x } 6 I = 8 R () (S) As x cosx log x is an odd function So, x cosxlog 0 dx x S (4) (5)

36 58. List-I List-II P. Let y(x) = cos( cos x), x [,], x. Then. d y( x) dy( x) ( x ) x equals yx ( ) dx dx Q. Let A, A,..., A n (n > ) be the vertices of a regular. polygon of n sides with its centre at the origin. Let a k be the position vector of the point A k, k =,,..., n. If ( a a ) ( a. a ), n n k k k k k k then the minimum value of n is R. If the normal from the point P(h, ) on the ellipse. 8 x y 6 is perpendicular to the line x + y = 8, then the value of h is S. Number of positive solutions satisfying the equation 4. 9 tan tan tan x 4x x is P Q R S (A) 4 (B) 4 (C) 4 (D) 4 Answer (A) (P) dy sin(cos x) dx x dy ( x ) 9( cos ( cos x)) 9( y ) dx dy d y dy dy ( x ) ( ) 8 x y dx dx dx dx d y ( ) dy x x 9 y dx dx (Q) ( n) R sin ( n) R cos n n tan n n = 8 (6)

37 (R) As, equation of normal is 6x y cos sin Given its slope = tan Also (i) passes through (h, ) (i) So, 6h h = (S) tan 4 x x tan x ( )(4 ) x x 6x x 8x 6x x 4x x x 7x 6 = 0 x, But x + > 0 and 4x + > 0 So, solution are x = (where x 0) 59. Let f : R R, f : [0, ) R, f : R R and f 4 : R [0, ) be defined by x if x 0, f ( x) x e if x 0; f (x) = x ; and f (x) = sin x if x 0, x if x 0 f( f( x)) if x 0, f4 ( x) f( f( x)) if x 0. List-I List-II P. f 4 is. nto but not one-one Q. f is. Neither continuous nor one-one R. f of is. Differentiable but not one-one S. f is 4. Continuous and one-one P Q R S (A) 4 (B) 4 (C) 4 (D) 4 (7)

38 Answer (D) (P) f 4 (x) = f[ f( x)], x < 0 f[ f( x)], x 0 nto but not one-one = x x < e x, 0, x 0 Now f(x) is not differentiable at x = 0, not one-one but continuous. (P) (Q) Now, f = sin x, x < 0 x, x 0 Differentiable but not one-one. (Q) (R) f of = x x < e x, 0, x 0 (R) (S) f : [0, ] R, f (x) = x (S) k k Let zk cos i sin ; k =,,..., List-I List-II P. For each z k there exists a z j such that z k. z j =. True Q. There exists a k {,,..., 9} such that z.z = z k has. False no solution z in the set of complex numbers R. z z... z9 equals. 0 9 k S. k cos equals 4. 0 (8)

39 P Q R S (A) 4 (B) 4 (C) 4 (D) 4 Answer (C) z k = π π cos + isin = e 0 0 k iπk 0 (P) z k.z j = e π i ( k + j ) 0 = π ( k+ j) π ( k+ j) cos = and sin = k + j = 0n and k + j = 5m; so true z k (Q) z = = e z π i ( k ) 0 So, if k {,,...9}, z has solution; i.e., false (Q) (R) z 0 = 0 (z z )(z z )...(z z 9 ) = + z + z z 9 So, z z... z 9 = 0 i.e., (R) (S) π 4π 8π cos + cos cos π π + 8 cos 0 0 = = (9)

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