ALL INDIA TEST SERIES

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1 Fom Classoom/Integated School Pogams 7 in Top 0, in Top 00, 54 in Top 00, 06 in Top 500 All India Ranks & 4 Students fom Classoom /Integated School Pogams & 7 Students fom All Pogams have been Awaded a Rank in JEE (Advanced), 0 FIITJEE ALL INDIA TEST SERIES ANSWERS, HINTS & SOLUTIONS FULL TEST II PAPER- Q.NO PHYSICS CHEMISTRY MATHEMATICS. A C A. B C A. C B B 4. B A A 5. C B A 6. A A A 7. D A B 8. B B A 9. D B A 0. A B A. A A B. B A C. C C C 4. C D B 5. B C C 6. A D C 7. B A D 8. A D B 9. A B A... (A) (q, ), (B) (q,, s) (C) (p), (D) (p) (A) (), (B) (p, q,, s) (C) (q), (D) (s) (A) (s) (B) () (C) (p) (D) (q) JEE(Advanced)-04 (A) (), (B) (p) (C) (s), (D) (q) (A) (p), (B) (q) (C) (), (D) (s) (A) (p, q,, s) (B) (p, q, s) (C) () (D) (p, ) (A) (p), (B) (q) (C) (), (D) (s) (A) (), (B) (s) (C) (q), (D) (p) (A) (s) (B) (p) (C) () (D) (q) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Saai, Savapiya Viha, New Delhi -006, Ph , , Fax 6594

2 AITS-FT-II(Pape-)-PCM(Sol)-JEE(Advanced)/4 Physics PART I. Accoding to the question g h g h R e R e h h 0 R R e e 5 R h = e 4. Fom figue it is clea 80 (i ) sini and sin = 0 = 90 sin60 sin i = 60 i = v v M m C M B 6. F = qvb sin = qvb ( = 90) So, B = F 0 qv = = T I 0 p Iq 7 Now, I = 4 amp. If the distance of point R fom thid cuent caying cuent is X, then B R = so x = m 9. P()d = dv dv (+d) d d dv P()d = d d v 0 P() = R v0 P = R FIITJEE Ltd., FIITJEE House, 9-A, Kalu Saai, Savapiya Viha, New Delhi -006, Ph , , Fax 6594

3 AITS-FT-II (Pape-)-PCM(Sol)-JEE(Advanced)/4 Chemisty PART II. H OH O O H O + OH H OH CH CH OH. (III) is most eactive (esonance activation) followed by N (inductive activation). (II) is moe deactivated (esonance deactivation) followed by (I) (inductive deactivation).. KE KE KE 0.99 KE 0.99 KE KE.0 KE % change is KE = % KE KE 00 KE 4. H 0, S 0 Reaction may be non-spontaneous at 5 o C G H T S = = 5. > 0 = Non-spontaneous To make it spontaneous G 0. We have to incease the tempeatue. H 80 0 T S 50 = 00 K = 97 o C.. Above citical tempeatue (T C ), gas can not be liquified on cooling, the aveage enegy of molecule deceases.. Positive chage on nitogen of diazo goup is stabilized by electon eleasing goup.. Within amino acid, poton is accepted fom COOH goup by NH goup to fom " COO R NH ". 4. Moe the numbe of alkyl substitute at double bond, geate its themodynamic stability. 5. C H bond is boken in non ate detemining step, theefoe, substitution of -H by deuteium doesn t affect the ate of eaction. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Saai, Savapiya Viha, New Delhi -006, Ph , , Fax 6594

4 4 AITS-FT-II(Pape-)-PCM(Sol)-JEE(Advanced)/4 6. Thee ae total fou types of -H and two types of cabonyl, hence a total of eight aldol would be fomed. 7. o Ecell =.7 V o E logkc logk c 8. Maximum wok = nfe = 6 0 kj FIITJEE Ltd., FIITJEE House, 9-A, Kalu Saai, Savapiya Viha, New Delhi -006, Ph , , Fax 6594

5 AITS-FT-II (Pape-)-PCM(Sol)-JEE(Advanced)/4 5 Mathematics PART III. Pefect squae = 00 = 9(excluding one) Pefect cubes = 00 / Pefect 4 th powes = / 4 00 Pefect 5 th powes = 00 / 5 Pefect 6 th powes = 00 / 6 Now, pefect 4 th powes have aleady been counted in pefect squaes and pefect 6 th powes have been counted with pefect squaes as well as with pefect cubes Hence the total ways = =. [sin x] > [cos x] x > 0 y / y = O - cos sin x Clealy [cos, x 0, cos x] = 0, x (cos, ] [sin 0, x [0, sin ) x] =, x sin, Hence [sin x] > [cos x] x [sin, ].. As oots ae of opposite sign, poduct of oots < 0 a + b < 0 a + b < a + ib < So, the point a + ib lies inside a cicle of cente (0, 0) and adius. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Saai, Savapiya Viha, New Delhi -006, Ph , , Fax 6594

6 6 AITS-FT-II(Pape-)-PCM(Sol)-JEE(Advanced)/4 4. z + i = 6 is a cicle with cente (, ) and adius 6 and z 4 i = z i is the pependicula bisecto of (4, ) and (, ) is a line paallel to axis imaginay. Now this line The line is tangent to cicle at complex numbe (8 i). Hence only one complex numbe satisfies the above equation. (8, ) (4, ) (, ) (, -) (8, -) 6. Coodinates of point T (a cos, 0) so distance fom focus of the point T is a (e cos ) 7. Thee ae 8 even and 9 odd numbes. So pobabilities of getting fist even numbe is pobabilities of getting second odd numbe = , so equied pobabilities = and 8. BI = cosec B = 4R sin A sin C A BI = sec B = 4R sin A cos C I II = (BI) (BI ) II = 4R = 4R sin A A sin R = 5 8 B 90 0 C I 9. y n =... n p log y = log lim log y n p k = log( x) dx = (k + ) log( + k) k 0. px + 4xy + qy + 4a(x + y + ) = 0 epesents pai of staight lines iff 4apq + 6a 4a p 4a q 6 a = 0 4 4a + ap + aq pq = 0. Fo eal, 6a 4.4(ap + aq pq) 0 (a p)(a q) 0 a p o a q. f(x) = ax bx + f(0) = f( ) = a + b + < 0 ( a + b + 4 < 0) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Saai, Savapiya Viha, New Delhi -006, Ph , , Fax 6594

7 AITS-FT-II (Pape-)-PCM(Sol)-JEE(Advanced)/4 7 f(0) f( ) < 0 One oot lie between (, 0) Nothing can be said about ab. Hee only one condition is given. So degee is. 4. Any point on the paabola is (t, t). Shift the oigin to ( 4, 0) so that the line becomes X + y = 0 and the point (t, t) becomes (4 + t, t) whee X = x + 4. If (X, y ) is the image of (4 + t, t) in X 0 4 t X + y = 0, then y 0 t X = t, y = (4 + t ) and in the oiginal coodinates x = 4 t, y = 4 t the equation of image is (x + 4) = (y + 4). 5. Shift the oigin to the point (0, ) so that any point (x, y ) on the eflected line is given by q(h ) x 0 p y 0 (since m = 0) h q(h ) x, y h px = qy + 6q p and hence the eflected line is px qy = 6q. 6. Equation of the tangent to the given cicle at (, ) is x + y = 0. Shift the oigin to the point (, 0) so that the two lines becomes y = X and X + y 7 = 0. Any X 0 h point on the line is (h, 7 h) and its eflection in y = X is given by y 0 y h X = 7 h, y = h X = 7 y x + = 7 y and hence the eflection of the tangent is y + x 5 = 0 o y = x 5 ( ) 4 which touch the paabola y = 5x. 7. In the new definition l = x x y z, etc. 8. d(o, P) = k x + y + z = k which epesents a set of 9 planes making intecepts of lengths k on positive as well negative sides of all thee axes. See the adjacent figue. (0, k, 0) Y Z (0, 0, k) X (k, 0, 0) (0, k, 0) Y X Z (0, 0, k) 9. The maximum value of d(o, P) = cicumadius of sphee = a. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Saai, Savapiya Viha, New Delhi -006, Ph , , Fax 6594

8 8 AITS-FT-II(Pape-)-PCM(Sol)-JEE(Advanced)/4 SECTION B. dx x xdx x a x a (A) I = x a x a x a x a x a x dx x dx = I a a x a x dx dx a a x a x x = c. a a x (B) Put x = a sec dx = a sec tan d a sec tand x a sin c c. I = a sec a tan a a x (C) Put x = a sin dx = a cos d acos I = acos d cot d a sin = cot + c a x x a x x = sin c cos c. x a x a (D) Put x = a sec dx = a sec tan d a tan asec tan d I = a tan d asec = a tan a + c = x x a asec c. a. (A) Requied aea = xdx / x = 4 / (B) sin x cos x dx / = sin xdx (C) Fo equation S + k = 0 to epesent pai of lines 0 k ( + k) ( + k + ) ( + 6) = 0 k = FIITJEE Ltd., FIITJEE House, 9-A, Kalu Saai, Savapiya Viha, New Delhi -006, Ph , , Fax 6594

9 AITS-FT-II (Pape-)-PCM(Sol)-JEE(Advanced)/4 9 (D) Let p.v. of given points be A ˆi ˆj k ˆ, Bi ˆ j ˆ kˆ and Ci ˆ kˆ, so that two vectos in the plane may be AB ˆi ˆj and AC i ˆ ˆj kˆ Thus, 0 0 ( ) + ( ) = 0 =. (A) The equation will have oots of opposite sign if it has eal oots and poduct of oots is negative 4 (b + ) (b b + ) 0 and < b < b b 0 (B) The pobability of poblem being solved is P A P B P C = =, (C) x = 5 (y + z) yz + x (y + z) = 8 yz + (y + z) (5 (y + z)) 8 = 0 y + y (z 5) + (z 5z + 8) = 0 Fo eal solution, (z 5) 4 (z 5z + 8) 0 7 (z ) z 0 z 7 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Saai, Savapiya Viha, New Delhi -006, Ph , , Fax 6594