AITS-FT-I-(Paper-)-PM(S)-JEE(Advanced)/ Frm Lng Term lassrm Prgrams and Medium / Shrt lassrm Prgram in Tp 0, 0 in Tp 0, in Tp 00, 75 in Tp 00, 59 in Tp 500 Ranks & 5 t t a l s e l e c t i n s i n I I T - J E E 0 FIITJEE JEE(Advanced)-0 ALL INDIA TEST SERIES ANSWERS, HINTS & SLUTINS FULL TEST I (Paper-) Q. N. PHYSIS HEMISTRY MATHEMATIS. A D. A A. A B B. A 5. D B D 6. B 7. A D B 8. A B 9., D A,, D A, 0. A, A, A, B. A, A, B, B,, D. A, B,, D A, B A,. A B B. A A 5. B D 6. B D B 7. D A B 8. A D A.. (A) (s) (B) (p) () (p) (D) (q, r) (A) (r) (B) (q) () (q) (D) (s) (A) (p, r, t) (B) (q) () (r, s) (D) (r, s) (A) (p, q, s) (B) (p, s) () (r, s) (D) (q, t) (A) (p, q, r, s) (B) (p) () (q, s) (D) (p, r) (A) (s) (B) (t) () (q) (D) (s) FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 606000, 65699, Fax 659
AITS-FT-I-(Paper-)-PM(S)-JEE(Advanced)/ Physics PART I SETIN A. E = V + (V V ) = V + 0 R L R As, E = V R V R = V = 0 V I = E = 0 =. amp R 00. Equivalent circuit can be drawn as Equivalent capacitance = 79 0. / 8/ 9/. This is simply discharging f a capacitr. t/r q = Q e where = πε 0 r 0 5. When the circuit will reach t steady state the inductr will given zer resistance. Hence the entire current will nly pass thrugh it. H+ H H 6. X = = 8. Here V /r r distance between Sun and Planet. 9. (y h) + x + h = dy x dx + = 0 dt x + h dt dy x dx = dt x + h dt A B h x=cm = m h h=cm = m y dy = va dt 5 u B = va (i) 5 h = v A dy dt (x + h ) 6 ab = v A (5) FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 606000, 65699, Fax 659
AITS-FT-I-(Paper-)-PM(S)-JEE(Advanced)/ a B 6 = va (ii) 5 0. Mg T = Ma (i) T = ma (ii) Slving (i) and (ii) Mg a = (M + m) FBD f man N Mg N = Ma N Mmg a = (M + m). πm T = =.s. qb Mg FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 606000, 65699, Fax 659
AITS-FT-I-(Paper-)-PM(S)-JEE(Advanced)/ hemistry PART II SETIN A. Et () H Et ( ) Ag ( ) /NH H H H + NaBH Et H Et. Pl 5 l l LiAlH H H H H P H H H H H. H H H H H H H H and H H H H H H H H 5 H 5 H 6H H 6H H The cnfiguratin differ nly at carbn n.. yclic frms f diasteremers that differ nly in the cnfiguratin f, are knwn as anmers.. Fr equimlar slutins, n = n x = 0.5 and 0.5 P = x P = 0.5 60 = 80 mm B B B P = x P = 0.5 60 = 0 mm T T T P = 80+ 0 = 0 mm Ttal Mle fractin f tluene in vapur phase = PT 0 = = 0.7 P 0 Ttal 5. Entrpy is a state functin, i.e. the change entrpy depends upn the initial and final states f the system and nt n hw that change is brught abut. FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 606000, 65699, Fax 659
5 AITS-FT-I-(Paper-)-PM(S)-JEE(Advanced)/ 7. 0.6 d rbitals in symmetrical field f ligands 0. FSE = (- 0.) + 0.6 8. Given ( K ) = 5 0 0. sp Ag K + AgN Ag + KN Given, [AgN ] = 0.05 M + Ag = N = 0.05M + ( sp ) high spin d cnfiguratin in an ctahedral field = -. + 0.6 = -0.6 Ag = K Ag 0.05 = 5 0 5 0 = = 0 M 0.05 = [ K] = 0 M 9 wt. f K 0 = 0 =.0 0 gm / litre H H Plane f symmetry. (A) anh.znl Ph H H Ph H cnc.hl l + Zn( H) l( immediate) ( white turbidity) Lucas reagent H H H H H l + Zn( H) l ( slw) (B) Lucas Ph H H PhH reagent l + Zn( H) l ( immediate) ( White turbidity) Ph H N reactin under nrmal cnditin Lucas reagent () Lucas ( H) H H ( H reagent ) H l + Zn( H) l ( Mderate rate) ( White turbidity) Lucas H H H H H H reagent H H H l + Zn( H) l ( Slw) ( White turbidity) FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 606000, 65699, Fax 659
AITS-FT-I-(Paper-)-PM(S)-JEE(Advanced)/ 6. HS + H + S+ PbS + PbS + S + S Snl + 6Hl + Snl + H. + + Zn + Fe Fe + Zn 0.059 E = E lgkeq n Given, E = 0.905 0.059 0.0 0.905 = E lg 0.00 0.0905 + 0.095 lg0 = 0.905 + 0.095 = 0. 0.059 E = lgkeq 0.059 0. = lgkeq K = 0 eq 0. 0.095. Fr mle f i.e. r ml f Al t change int Al+ ins. n = Thus, G = nef G + 87000 E = = =. vlt nf 96500 0.059. 5. E = lgk = 7.88 = lgk 0.059 7 K =.9 0 cell eq eq eq 6. pper sulphate when reacts with hypphsphrus acid, gets reduced t cuprus hydride. us + H P + 6H u H + H P + H S 7. ( ) + + ( ) + ( J) ( L) ( K) ( ) + + ( L) ( N) ( M) Ba H P 6H Ba H P PH Ba H P H S BaS H P gives apple green clur in the flame FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 606000, 65699, Fax 659
7 AITS-FT-I-(Paper-)-PM(S)-JEE(Advanced)/ 8. H P 7 (pyrphsphric acid) is a tetrabasic acid, i.e. -hydrxyl grups are present. It is prepared by remving ne water mlecule frm tw mlecules f rthphsphric acid. Each phsphrus atm lies in same tetrahedral envirnment, s called isplyacid. H P H H + H H P H H SETIN-B H P H P H H. (A) H (B) () H H H H () H ( ) H + H H H H H H () H + H H H H ( ) H H H H H H H H H (D) H H H H H H Reactins A,, and D shw S N reactins.. In case f (A), cmpund given in (A) exchanges H with D and becmes racemic. In case (B), cmpund in (B) reacts thrugh a carbcatin and gives racemic mixture. In case f (), n bnd attached t the sterecentre breaks in reactin f (), s retentin. In case f (D), cmpund in (D) gives Hffmann eliminatin and it is an example f antieliminatin. Prduct des nt have any steregenic centre. D Reactin( B) gives Reactin( ) gives H H H H H 5 D H 5 H H ( d, frm) Reactin( D) gives ( N sterecentre) FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 606000, 65699, Fax 659
AITS-FT-I-(Paper-)-PM(S)-JEE(Advanced)/ 8 Mathematics PART III. We must have f(a + a + 5) > f(a + ) a + a + 5 < a + a (, ) a can take values,, 0,. AB is equilateral a = b = c SETIN A. A = AB + B.. () BD = B + D.. () A = D + DA.. () DB = DA + AB.. () Equatin () + () and () + () gives A = BD S, equatin () and () give AB = D and equatin () and () give B = AD S s BD and ADB are cngruent BD = ADB = 90º ABD BD + ABD = BA DB lies in the plane f AB, s the pints are cplanar. Let a d = t; then bc = (d + zt)(d + t) = d + td + t Als a t t = +, s hk = d d t td bc d d + = + < d bc > hk 5. learly P (, ) Equatin f tangent at P is x() + y() = 8 A (, 0) and B (0, ) Similarly nrmal at P is x y = 6 A '(, 0) and B' 0, 0 5 Area BPB' = = and Area APA ' = 5 0 Rati = 6. As sin x + sin y cs α cs x x R π Let x = sin y sin y = + sin x cs α cs x cs α cs x sin x cs α+ = cs α = 0 sin y + cs α = + 0 = B P A B A FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 606000, 65699, Fax 659
9 AITS-FT-I-(Paper-)-PM(S)-JEE(Advanced)/ 7. sin x x I= dx < dx x = x 0 0 cs x / I < and J= x dx < = x J < 0 0 c 8. Equatin f tangent x+ t y = ct T (ct, 0) and T 0, t Equatin f nrmal ( t x ty = c t ) ; N c t, 0 ; N 0, c t t t c c NT = + ct; N'T' = ct + t t c c t c = + ct = and ' = ct + ct t t ( c + t ) t = + = ' ( c + t ) ' c π π 9. We have sin cs x = sin sinx cs x = sin x cs x(cs x sin x) = 0 tan x = 0 as cs x 0 tan x = and cs x = 5 0. learly, the pint lies n 7x y = 5 A Als, centre f the circle must lie n the bisectrs f the x + y + = 0 lines P(, ) x + y + = 0 and 7x y 5 = 0 given by (h, k) x+ y+ 7x y 5 7x y 5 = 0 =± x y = 5 and x + y + 5 = 0 B 50 Let (h, k) be the centre f the circle, then h k = 5.. () and h + k = 5.. () learly B is perpendicular BP k 7 = h + 7k 5 = 0.. () h n slving, we get centres as (9 ) and ( 6, ) r = 800 and r = 50 smaller circle has radius = 50 Therefre area f quadrilateral ABP = 50 00 sq. units. Let f(x) = ax + bx + c and f() = a + b + c < 0 f(x) < 0 fr all x R f( ) < 0 a b + c < 0 Again f() + f() < 0 a + 5b + c < 0 Als f( 5) < 0 5a 5b + c < 0 5b 5a c > 0 FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 606000, 65699, Fax 659
AITS-FT-I-(Paper-)-PM(S)-JEE(Advanced)/ 0. (x x) ( x) = 0 ; x < 0 f(x) = ( ) x+ x x = x ; x 0 f is cntinuus and differentiable 0 ; x < 0 Again f'(x) = x ; x 0 f (x) is cntinuus but nt differentiable at x = 0. Taking B as rigin, B as x axis and A as (h, k) and (, 0) h+ k We have D (, 0) and E, K + (h + )(h ) = 0.. () Als A = (h ) + k = 9.. () And area = K K = Slving () and (), we get h = and K = AB = y B (0, 0) A(h, k) D E (, 0) x AB BD. Let AD be angle bisectr = c = b A D Nw b + c > a b + c > 6 b + b > 6 b > Again b + b 6 < b < 6 b (, 6) and cnsequently c (, ) b 5. BE =, take BE = K and E = K E A must be (K, 6) 6 Nw equatin f line BA is y = x K Ky x = 0 Since perpendicular frm c n AB = 7 0 K = 7 K = K + Nw c = ( 6 0) + ( K 0) = 6 + 9K = 6 + c = 8 = x y B y A(K, 6) E E (7K, 0) x 6 8. Let yf(y)dy = a and ( ) 0 0 y f(y) dy + = b f(x) = ax + bx f(y) = ay + by ( ) ( ) 0 0 0 = f(x) = x y ay + by dy + 0x y ay + by dy + x a = a + b a = b.. () and a + 5b + = b a + b =.. () a =, b = f(x) = x + x a b a b x + + 0x + + x 5 FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 606000, 65699, Fax 659
AITS-FT-I-(Paper-)-PM(S)-JEE(Advanced)/ SETIN B ; x [, ) π. learly we have f(x) = ; x [, 0) and g(x) = sec x; x R (x + ) x ; x [0, ] ; secx [, ) We have fg( x) = ; sec x [, 0) secx ; secx [0, ] π π π π ; x,, { π, π} fg( x) = ; x = π, π π π secx ; x, Limit f fg exist at x = π, π,, pints f discntinuity f fg are π, π and pints f 5π differentiability f fg are, 6 { } π sec( ) ; x [, ) Again g(f(x)) = sec( ) ; x [, 0) π secx ; x [0, ] { } Limit f g(f(x)) des nt exist at x =. (A) Let E i ; i =,,, represents the events that the bag cntains I white balls: learly PE ( i ) =. Let W be the event that the ball drawn is white then, ( ) W 5 P W = P( E i ) P E = + + + = i 8 ( ) ( ) E PE PW/E / p Nw P = = = = p = 6 W P(W) 5/8 5 5 (B) Required number f ways = ( ) ( ) ( + + + + +... + + ) = + + +... + + + +... + ( ) ( ) 0 0 r r + r = + r= r= = = 0 = 0 m m = 5x 5y 5z x y z () 5 + + = + + x y z x y z x + y + z + = 5 + + = 5 + + + x y z x y z Nw apply A.M H.M n x, y and z FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 606000, 65699, Fax 659
AITS-FT-I-(Paper-)-PM(S)-JEE(Advanced)/ x + y + z 9 + + x y z 5 + + x y z 5x 5y 5z 9 Hence + + 5 + = x y z 5 Least value = n! K K K = K =!! (D) ( ) K= K= 9 7... = 6 p = 6! FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 606000, 65699, Fax 659