ALL INDIA TEST SERIES

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1 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5 FIITJEE Students From All Programs have bagged 4 in Top, 66 in Top and 74 in Top 5 All India Ranks. FIITJEE Performance in JEE (Advanced), 4: 5 FIITJEE Students from Classroom / Integrated School Programs & 579 FIITJEE Students from All Programs have qualified in JEE (Advanced), 4. FIITJEE ALL IDIA TEST SERIES ASWERS, HITS & SLUTIS FULL TEST I (PAPER-) Q. o. PHYSICS CHEMISTRY MATHEMATICS. B B A. D B B. D A C 4. C C A 5. A, D A, B, D A, C 6. B, D A, B, C A, B, C, D 7. A, C B, D A, C, D 8. B, C, D B, C, D A, B, C, D 9. B C A. D D C. B A B.. (A) (q) (B) (p, q, s, t) (C) (p, q, s, t) (D) (r, s) (A) (t) (B) (q, s) (C) (p, q, r) (D) (t) JEE(Advanced)-5 (A s, t) (B p, q, r, t) (C p, s) (D q, s, t) (A q) (B p, t) (C p, q) (D s) (A) (s) (B) (q) (C) (s) (D) (p, r, t) (A) (p) (B) (q) (C) (p) (D) (r) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

2 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5 Physics. net dp F 5i ˆ tj ˆ 5kˆ dt PART I SECTI A = Freal Fpseudo and real F mg( k) ˆ 5kˆ 4. Q = dt KA d Q d A = K ( T)dT Q A = T K T by solving Q T K T A T Q () So, K A So, at = temperature T 45 K T 5. m = V = A A AA H m H A A AA A H F F B mg A A P = gh P = gh P > P F = mg F B = mg PA F = mg A P, F = mg A P 6. U GMm GM d y d GMm d y = Gm = U mv Gm d d y d FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

3 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5 7. v = f = v Particle h E E P p h P m 8. y 5e ( a b t) 9-. h = 6 m h 6 t 4 sec g Velocity of system after collision in vertical direction, v = m/s; Velocity of system after collision in horizontal direction, v ¼ 4 = m/s v t t = sec 4 sec v 5 m/s SECTI - B. Shape of sting at t = is given by. if vt L h(l vt) / L if vt L vt y(, t) h(l vt) / L if vt vt L if vt L SECTI C y y ((vt L), ) (vt, ) ((vt + L), ). gsin = 5g sin sin = 5 ( sin / cos /) cos / = 4. y = A sin t y = A sin t ( ) ( ) y r = A cos tsin t Resultant amplitude A r = A cos()t/ t ( ) t = s 4 t ( ) t = s 6 In one cycle of intensity of /s, the detector remain idle for s sec A r A A /6 s /4 s / s cycle / s /4 s 5/6s s FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

4 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5 4 In ½ sec cycle, active time is 6 = / sec / In sec interval, active time is = 8 sec (/ ) (C)V 6. V C 4 U C V 9 V = V V CV 9 V V U (C) C 9 9 CV 9 CV Initial total energy = Final total energy = (C)V CV 4 Charge flow through the battery CV CV CV 4 Wbattery CV 4CV CV Heat dissipated CV Chemistry PART II SECTI A. H C CH H C CH CH CaC CHCl C H C CH H C C H CH H C H C H C (A) CH CH H C CH (C) H CH i CHI ii Moist Ag H C CH CH CH H H H C CH CH (B) CH H FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

5 5 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5. CH H C C H C C Anhydride AlCl CH C CH CH C H CH (minor) i Zn ii SCl / Cl Hg /HCl C CH CH C H CH (major) i Zn Hg/HCl i SCl / Anhyd. AlCl / CS Cl CH Anhyd. AlCl / CS CH (minor) (major). C 8 H 7 Cl 87 m Molar weight of C 8 H 7 Cl = = 54.5 g Molar weight of = = 44 g According to Graham s Law of diffusion r MC 8H7Cl : r M 44 C8H7Cl.87 d 87 = 87 th row from side..87. dc8h 7Cl 87 = th row from weeping gas side..87 Therefore, the spectator from the side of in the 87 th. Row will be laughing and weeping simultaneously. Alternatively, the spectator from the side of weeping gas in th row will laugh and weep. 4. The given orbital diagram has two radial nodes, i.e. n The orbital, in which electron is present is s. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

6 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5 6 Cl Cl 5. (A) C H CH Cl H C CH C Cl Cl These occurs decrease in electron density and hence the ionization potential increases. Cl (B) Basic strength electron density over one -atom has lesser electron density over (C) one -atom than in Heat of combustion Stability stability o. of H and if no. of H of same stability ease of hyperconjugation. (D) Conjugate acid of weaker base is stronger acid. Basic strength H H H CH CH 6. B. pt. of CCl 4 < that of SiCl 4. The vapour must always be richer in more volatile component. Hence, the upper curve represents the composition of the vapour and the lower curve represents the composition of the solution at corresponding B. pt. 8. The correct basic order for (a) option CH > CH H > CH CH (-ve) charge on most electronegative atom (-ve) charge on least electronegative atom and it can also eplain by its conjugate rest all option are correct. Solution for the Q. o. 9 to. M PbC P Pb( ) B Pb C C D E F CaC G Ca(HC ) Q PbS 4 R a SECTI C H FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

7 7 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5. Z 4 B Z 4 A C Z A 4 B Z + y + z = 7. r = 5.49 o A r = 8.5 o A then r =. o A.59 n r Z.59 n. = Z n = 5 umber of maimum lines possible = CH CH ah CH Ca H 6 6 Int. o. of mole V V at t t sec V 6 V CHCa V = 5 7 < -6. Kw C H K = a =.6 8 < 6 so ph = log ( ) ph = 7 ph 5.4 Mn Mn e 4e H Eq of Mn + = Eq of (n = ) (n = 4) ½ mole of Mn + = Eq. of mole of Mn + = 4 Eq. of 6 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

8 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/ C s g C g Keq C g C s g Keq C s C g C g Keq X C s g C s C g Keq / C g g C g Keq Y Hence X : Y = : Mathematics PART III. Let z, z be the two roots with z = z z = c a c z a z SECTI A zz zz b z z and b = a a z + z = (z + z ) z z z z z z (z + z ) = z z b = ac. For [, ], f() f() ( f() is decreasing) f ' f ' f f f ' d f ' d f f d tan f f ' f d tan f f f ' tan < > d f f f '. P() = has n real roots P() = has n + real roots tan f f f ' f ' FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

9 9 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5 d P d P() + P() = has n real roots ( P() + P()) = has n + real roots d P' P has n real roots d i.e. P() + P() + P() = has n real roots (I) is true Let one of the roots of P() = is, and one root of Q() = is as given, P() Q() + P() Q() = d P Q' d Consider f() = P() Q() f() =, f() = Hence, by Rolles theorem d P Q' has at least one real root lying between and d 4. Case-I: a a If a [, ], the curves intersect at, and (a, a). The bounded region is contained in the triangle with vertices (, ),, and (, ) with area = 4 Hence, area can not eceed 4 Case-II: a a If a [, ]. In this case the bounded region is a quadrilateral with four vertices,, a,, a 4 a, and (, ). In this case area bounded = a a 4 a a a = 6 Case-III: If a [, 4]. This case is symmetric with case-i a, (, ) a, (, ) a, 5. Multiplying the first DE by gh, the second by fh and the third by fg, and adding the equations gives (fgh) = 6(fgh) + 6 Let f() g() h() = k() We have k() = 6(k()) + 6 Integrating and using k() = gives k() = tan 6 4 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

10 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5 ow st DE f ' k f k f ' tan 6 cot 6 f 4 4 Integrating and using f() = gives option A and C 6. By reflection property of ellipse PAF = QAF H is reflection of F QAH = QAF ow PH = PF, PH = PF and F H = F A + AF = a and F H = F B + BH = F B + F B = a Hence, PF H and PF H are congruent as all three sides are equal This leads to all the options H P A B F F H Q 7. Let f() = a( )( ) f() f( + ) = a ( )( + )( )( + ) = a ( + + )( + + ) f() f( + ) = a [( ( + ) + ) ] [( ( + ) + ) ] Hence, g() = a ( )( ) and h() = ( + ) + 8. Consider h() = ( + ln ), h: [, ) [, ) Clearly h() to increasing function in its domain f() denotes the solution of ( + ln ) = f: [, ) [, ), f is inverse function of h() As f ' lim h lim f f ' h' f ln f lnf f ln f ln ow consider L lim lim f ln f = lim / f lnf By L Hospitals Rule L lim lim lim f ' f f ' lnf lnf L lim lnf 9. (I) holds true obviously A B C a, a b c, b c For (II): = a (b + c ) + a (b + c ) + a (b + c ) + a (b + c ) = (a b + a b + a b + a b ) + (a c + a c + a c + a c ) = A B A C Similarly it can be checked that (III) also holds true. A B ab ab ab ab ln ln f FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

11 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5 a b a b A B A B a a b b A B a b a b A B a b a b a b B A. Hence, (I) holds true A B C a b c a b c a b c a b a b a b a c a c a c ab ab ab ac ac ac (II) Don t hold true Similarly (III) can be checked SECTI B. ormal y + t = at + at at + (a h)t k = t + t + t = t = (t + t ).. () t t + t t + t t = a h.. () a k Also, t t t... () a ABC is equilateral So, G H a a G tt tt tt, t t t t t t a G t t t t t t,... (4) a H a, t t t tt t... (5) Comparing (4) and (5), t t t = and from (), k = From equation (), we get a h a a a a h = a h = 5a circumcentre of ABC = ( a, ) Image of will lie on circle (5a, ) ne of the intersection point lie on verte ( k = ) So, t = at + (a 5a)t = t t = t(t ) = t =, Intersection points P a, a, Q (, ), R a, a A Q(t ) ( a, ) C P(t ) B (h, k) R(t ) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

12 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5. (A) (f()) + [f()] = h() h() = f() f() + f() f() = f() (f() + f()) = f() { g() f()} = (f()) g() So, h() < of > and h() > for < Then, = is a maima (B) f () + f() = g() f() f() + f() = g() and g() = f() {f() + } So, g() as g() is increasing and for f() f() + Hence, f (C) Let G f t dt.. () G' f f tdt and G' g and G() = = g() Also, g() is non-increasing Hence, g() (, ), g() [, ) i.e., G() for, G() for and G() = But G() (from equation ()) Hence, G() = R f t dt R i.e., f() = R Hence, f() = t (D) f e f t dt.. () t f e e f t dt t f e e f t dt t f f ' e e f e e f t dt t f ' f e f t dt = + f() + f() f() = + f() = + c and from equation () f() = Hence, c = i.e., f() = f, k k = h() G() FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

13 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5 SECTI C. Let f r ow, / / r sin d / r / r r / r cosd cos sin d r r cos d So, we have lim r ow, consider f r Also as sin f r r C r f r r f r / L r r d r, r / r f r d r r r r r f r r r r r f r f r r Hence, lim lim r r f r r r r f r C f r C ow, lim r r L lim r r L r f r r = For positive L we should have C = and L =. Let + y = and y = and y Given equation f() f() = ( f f ) f Hence, k (k is some real constant) f() = + k f() = k =, f() = f() = ( )( + ) on the interval, etreme value are and. The given equation can be re-written as quadratic in y y + (8 )y + ( + 6) = Its discriminant to equal to FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

14 AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5 4 = ( ) 4( + 6) = 4( 8 + 6) = 4( 6)( ) For solutions in integers this discriminant should by perfect square this happens if and only if z = ( 6) = ( ) 9 9 = ( ) z = ( + z )( z ) for name integer z Checking all factorisations 9 = ( 9) ( ) = ( ) ( ) = ( ) ( 9) = 9 = = 9 This gives all possible pairs (, y) as (, y) = (, ), (, 4), (, ), (8, 4), (8, 6), (6, ) Therefore eactly three pairs of natural number (, y) are possible 4. As no multiple of nor any odd multiple of 4 satisfies the equation we can multiply both sides by sin cos, which gives sin 8 cos 5 = sin 4 cos 4 cos 5 = 4 sin cos cos 4 sin 5 = (sin sin )(8 cos cos 4 cos 5) = sin cos sin + sin = sin sin sin = sin ( ) = n + ( ) n ( ) Which gives solutions,,,,,,,,, Let H i the event of getting a head on i th coin H getting odd number of heads on throwing n coins once Coin umber: 4.. m.. n P(H i ) m.. n.. () P H m n i () m n ow, P(H) = sum of series which has each term consisting of product of an odd number of terms from () and even number of terms from () such that total number of factors in each term is n 4 6 n Consider the product n n n LHS = PH PH n Also, PH PH n PH n =, =, =, + + = 4 6. (, ) and (, ) will lie on some line y = + Solving line and parabola ( + ) = a b b ( + a) + =.. () Again, y = a( y) b( y) by + y( + a b) + (b a) =.. () Both () and () has same roots which are and a Hence, + a b = ( + a) b ow, for, to eist disc of () > ( + a) 4b > (a )(a + ) > a (, ) (, ) (, ) (, ) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, ew Delhi -6, Ph 466, , Fa 6594

ALL INDIA TEST SERIES

ALL INDIA TEST SERIES From Classroom/Integrated School Programs 7 in Top 0, 3 in Top 00, 54 in Top 300, 06 in Top 500 All India Ranks & 34 Students from Classroom /Integrated School Programs & 373 Students from All Programs