FULL TEST – I Paper-2 [ANSWERS, HINTS & SOLUTIONS]
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- FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.comANSWERS, HINTS & SOLUTIONSFULL TEST –I(Paper-2)Q.No.PHYSICS CHEMISTRY MATHEMATICS1. C D B2. C A B3. D C A4. B B C5. B B D6. D C D7. B D C8. C B A9. A C C10. D B C11. B A D12. D C A13. B A C14. A D A15. A C A16. B B C17. B A B18. A B A19. B C A1.(A) (s), (B) (q),(C) (p), (D) (r)(A p, q, r), (B q),(C p, r, S), (D p)(A) (q), (B) (r),(C) (s), (D) (p)2.(A) (q); (B) (r);(C) (s); (D) (p)(A p, q, s), (B r, s),(C q), (D r, s)(A) (s), (B) (p),(C) (q), (D) (r)3.(A) (p), (B) (r),(C) (s), (D) (q)(A p, r), (B q, p),(C q), (D q, s)(A) (s), (B) (r),(C) (p), (D) (q)ALL INDIA TEST SERIESFIITJEEJEE(Advanced)-2014From Classroom/Integrated School Programs7 in Top 20, 23 in Top 100, 54 in Top 300, 106 in Top 500 All India Ranks & 2314 Studentsfrom Classroom /Integrated School Programs & 3723 Students from All Programs have been Awarded a Rank in JEE (Advanced), 2013
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- AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com2PPhhyyssiiccssPART – I2. T sin = mg/2T = T cos =mg mgcot2 2tan T sinT cosTT3.21mv pt2(P = const) v =2Ptma =dv 2P 1dt m2 tF = ma =mP 12t v5.1 2 3 4 5F F F F F F 2 5 2 4F F and F F 1 3 2 1F F 2F cos30 2F cos60 F3=22Gm4a; F2=22Gm3a; F1=22GmammmmmmF1F2F3F4F5F =22Gm 5 14a3 = m2a =3Gm 5 1a 43 T = 2 34 3aGm 5 3 46.VPPQRS7. a =v g a m g am = 20 m/s2t =2ha= 1 sec10. Time period becomes 2Rg. We can’t neglect the roundness of earth for the pendulum ofinfinite length.
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- AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com312. The ve sign isddt signifies the direction of induced emf.14. process AB U = constantP RTMand U t P = constProcess BC isochoricProcess CA isothermal15. Q = QAB+ QBC+ QCAQ = 5U0+ 3U0+010Uln2.5316. WAB= QAB UAB= 5U0 (3U0) = 2U017. For lens L1, ray must move parallel to the axis after refraction11 w1x R x = 10 cm18. For lens L2, image must form at centre of curvature of the curved surface after refraction throughplane part.22R x 0 x = 8 cm
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- AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com4CChheemmiissttrryy PART – II1.BrCH3HOMgEtherCH2MgBrCH3HOOHHRearrangment2.OH3COOO3CH OOH3COOOOCOOCH3OHOH3COOCOHOOCOOCH33.o o o oA A B A B B BP P X P P 1 X P X o o oA A BP P P P B ThusoAP 120 Torro o oA B BP P 75 P 45 Torr Hence C is correct answer.4. 4oBaSO1000Conc normality 45oBaSO1000 1000 8 10Normality400 = 2 × 10-4. 4NormalityMolarity 10 M Solubility2 Ksp = S2= 10–8M2.5.av1TK99.9 210 0.693 6.93T 10 tyK K Number of natural life times =6.693 1/K K = 6.93
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- AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com56.BOOBBOOBOHOHOHOHO7. 2 271145Ca OCl Cl.H O ClPercentage71100 49145 Hence D is correct answer.17.4 1Id c0.00331.32 10 m mindt 25 4 1IId c2.6 10 m mindt 3 1IIId c1.02 10 m mindt 18. On comparing rates order w.r.t A = 2, and w.r.t. B = 1. Thus rate law = K[A]2[B]19. 2dxK A Bdt 2dx / dtK 0.26A B
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- AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com6MMaatthheemmaattiiccss PART – III1. Total number of lines made =9C2= 36Now, these 36 lines are 9 sets each with 4 parallel members. With 5 vertices number of linesmade=5C2= 10Clearly, atleast 2 members belong to the same setSo, atleast one pair is parallel2.2 2 2OA OB ABcos2OA OB = 22 22 2OA OBOA OB3 OA OB 122OA OB 8 OA OB 4 For maximum cos ,2 23 OA OB 1 3 2 OA OB 18 OA OB 2 8 OA OB 4 =123 3. S1= a1S2= a1+ a2S3= a1+ a2+ a3Sn= a1+ a2+ ….. + anIf we divide all S1, S2, ….. Snby 23 we get remainders 1, 2, 3, 4, ….., 22So, two of these give same remainders Sp, Sq Sp– Sqwill be divisible by 23Sp– Sq= ap + 1+ap + 2+ ….. + aq4. Using Cauchy we get, 2 2 2 221 2 3 41 2 3 41 1 2 4z z z z 1 1 2 4 64z z z z 5. Point of intersection is1 bln2 a For C1,draed, now1dtan rdr =1ae a e 1 14 For C2,drbed 21tan be e 1b 234 Angle of intersection is2 134 4 2
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- AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com76. Let c, d, e be the three points where y = f(x) crosses x–axis. Then, f(c) = f(d) = f(e) = 0Assuming a < c < d < e < b. The function f satisfies Rolle’s theorem in two intervals (c, d) and(d, e). Since f and f are continuous and f(c) = f(d) = 0So, there exists, at least one point in the interval (c, d) and (d, e) such that derivative is zeroLet, C1 (c, d) such that f(C1) = 0 and C2 (d, e) such that f(C2) = 0. Now the function fsatisfies Rolle’s theorem since f, f are continuous and f(C1) = f(C2) = 0So, by Rolle’s theorem, there exists a number C3in between C1and C2such that f(C3) = 0 Minimum one root C3of the equation f(x) = 0 lies in the interval (a, b)7. Dividing the given differential equation by 3xy(y2– x2) 2 2 2 22 2 2 2y y 2x x 2y xdx dy 03xy y x 3xy y x 2 2 2 2dx xdx ydy dy0x yy x y x 2 22 2d y x1d ln xy 02y x 2 2 2 2d ln x y y x 0 2 2 2 2ln x y y x c x2y2(y2– x2) = c8. Let the circle be 22 2x y a . Let the point of intersection of tangents at P and Q be (h, k).Then equation of PQ, is2hx k y a 0 . As it passes througha,0, so,2ha k a 0 .2 2k a h a 0. D 0 k 4a h a 0 i.e.2y 4a x a .9. Consider 1 122 20 0x f x dx f x 2 xf x x f x dx = 2– 22+ 2= 0However f(x) assumes only positive values i.e. in (0, 1) ( – x)2(f(x)) > 0 integral can’t be zero10. Differential equation can be written as, (p – x)(p – 2 sin x)(2p + cos x) = 0 which has solution as(2y – x2– c)(y + 2 cos x – c)(2y + sin x – c) = 011. Put x = –1 we get (–1 + 1) p(–1) + 1 = (–1)n + 1(n + 1)! n 11n 1 ! So, n 11 x x 1 ..... x n xp xn 1 ! n 1 x 1 Clearly, 1, where 'n' is oddp n 1n, where 'n' is evenn 2
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- AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com813. If e is the eccentricity then,2 222e Now, we know,2 21 1a b 22 21ab h 2 22a bab h ,2 221ab h 222 22a b 4 ab hab h 222 21 1a b 4 ab h For an ellipse 2222a b a b 4 ab h So, 2222 22a b 4he a b a b 4h2 ab h 14. Put y = z = t = 0f(0)[f(x) + f(0)] = f(0)Put x = 02f2(0) = f(0) f(0) = 0,12If f(0) =12 f(x) +12= 1 f(x) =12If f(0) = 0, z = t = 0 f(x) f(y) = f(xy)Let, x = y = 1 f2(1) = f(1) f(1) = 0 or f(1) = 1We have f(0) = 0, f(1) = 0, y = 1f(x) = 0Also, f(0) = 0, f(1) = 1, x = 0, y = t = 1(f(0) + f(z)) (f(1) + f(1)) = f(–z) + f(z) 2f(z) = f(–z) + f(z) f(z) = f(–z)15. If y = x in f(x) f(y) = f(xy) f(x2) = f2(x) 0Put x = t, y = z[f(x) + f(y)]2= f(x2+ y2) f(x2+ y2) = f2(x) + f2(y) + 2f(x)f(y) f2(x) f(x2+ y2) f(x2) f is non decreasing for positive x16. Put y = z = t = 1 2(f(x) + 1) = f(x – 1) + f(x + 1)f(2) = 4, f(z) = 9, f(1) = 1, f(0) = 0f(n) = n2(Possible function), if f(n – 1) = (n – 1)22[f(n – 1) + 1] = f(n – 2) + f(n)
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- AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com9 f(n) = n2Now, forpxq(rational number) 2pf f q f pqq 2 2 2pf q p qq 2p pfq q (True for rational number)Now, if x R, lets prove for positive x since if it is proved the function is even and will follow fornegative xAssume for x > 0, f(x) < x2So, now a rational number ‘r’ between f xand x f x< r < x f(x) < r2< x2[f(r) = r2, f is non decreasing] f(r) = r2 f(x) [contradiction] f(x) < x2(impossible)Similarly we can prove contradiction f(x) > x2So, only possibility f(x) = x2substituting f(x) = x2we get(f(x) + f(z)) (f(y) + f(t)) = f(xy – zt) + f(xt + yz)(x2+ z2)(y2+ t2) = (xy – zt)2+ (xt + yz)2[Lagrange identity]So, f(x) =12, f(x) = 0, f(x) = x2are the required solutions17. am3+ m(2a – h) + k = 0 am3+ m(2a – x1) = 0 am2= x1– 2a {m = 0 (one possible value)}21x 2ama , if x (0, 2a) m2= (–) number, so non real roots18. Let P(h, k) be the point am3+ m(2a – h) + k = 0Since, m1, m1, m1are the possible roots 3m1= 0 m1= 0If m1= 0 is the root then k = 0am3+ m(2a – h) = 02h 2am 0a h = 2a(2a, 0) is the only point19. 8a3 / 212a4A x 2a dx27a = 8a5/ 22a2 2x 2a53 3a = 5/ 222 2 2 26a 36a 6a5 53 3a 3 3a =248 2a5 8a8a3 / 20102A 4axdx 2 a x3 =24 64a 8a 8a 2a3 3Area = 2(A2– A1) =24 32 16 2a3 5 =2211 352 2a32 2a15 15
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- AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com10SECTION – B1. (A) |A – xI| = 0There exists a non zero matrix X such that AX = xX [(Adj A)A]X = x(adj A)X |A|IX = x(adj A)X |A|X = x(adj A)X A Xadj A Xx A Xadj A X2(B) A' I A I ' A I A' I 0 if and only ifA I = 0A' I 0 if and only ifA I = 0oris the root ofA' xI 0 if and only if is the root of |A – yI| = 0 e–iis the required solution(C)11 12 1n22 2nnnA A .......... A0 A ..... A0 | |B I| | || | || | A = 0 (|A11| – )(|A22| – ) ….. (|Ann| – ) – 0Clearly the elements of principal diagonal become the roots(D) Let, AX = X [X is a non zero matrix]X'AX =X'X =X' AX'IX X'AXandX'IXare both realAlso,X'X 0, X 0X'AX/ XIX is real so can have real values2. (A) z = cos + i sin ,1z 2cosz 1z 2isinz pp1z 2cospz Now, 4 24 21 12isin 2cos z zz z =6 4 26 4 21 1 1z 2 z z 4z z z then, 2 cos 6 – 2.2. cos 4 – 2 cos 2 + 4 24i4sin4 22cos2 = 2(cos 6 – 2 cos 4 – cos 2 + 2) = 2
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