AITS-2223-PT-I-JEEM-SOL.pdf

FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com FIITJEE ALL INDIA TEST SERIES PART TEST – I JEE (Main) - 202 3 TEST DATE: 12 - 11 - 2022 ANSWERS, HINTS & SOLUTIONS P P h h y y s s i i c c s s PART – A SECTION – A 1 A Sol. If the block has moved by distance x , Ma =   Mg 2 Mg 3 Mg 2 Mg l x 3 3 3l               smooth Mg F   2 A B C D l l l x   g 2 g 3 g a 2 g l x 3 3 3l                1 da 3 g dx 3l   [for 2 nd and 3 rd cells similarly] tan  1 : tan  2 : tan  3 = 3 : 2 : 1 2 A Sol. F I d B,         2 0 0 I B F I d       [MLT  2 ] = 2 0 Q T          0 = [MLQ  2 ] 3 A Sol. 2 L m r ( r ) m r ( r ) 2m r sin 2m sin                                    Torque = 2 2 dL J 2m sin(90 )sin 2m cos sin dt                 AITS - PT - I - PCM (So l.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 2 4 B Sol. 3 2 dm kt kt m 2 dt 3     2 dv dm dv m v mg m vkt mg dt dt dt      2 2 3 dv kt dv 3kt v g v g dt m dt kt 6       which is linear differential equation 3 3 v(kt 6) g(kt 6)dt c      , t = 0, v = 0  c = 0 4 3 g(kt 24t) v 4(kt 6)    Put 2 gm, k = 12 gm/s 3 and g = 100 cm/s 2 ) and t = 1 1000(12 24) v 500 4(12 6)     cm/sec 5 C Sol. 2 x 2 2 x ˆ F(x) e , x 0 2 x             Work done in a cycle is zero so conservative r F 0     , so angular momentum is conserved For equilibrium point F = 0  2 2 2 x F 0 2 x             x 2, 2   x > 0 so equilibrium point is 2 To check stable and unstable equilibrium point dF 0 dx   3 4 2x 0 2 x    At x 2,  dF 0 dx   3 4 2x 4 2 2 x 2 2      =  ve So, it is unstable point so particle moves away from point x = 2 So, it is obvious that i t will moves towards origin. 6 A Sol. Speed is never less than v 0 in x whereas in y it is less than v 0 first and then increases t v 0 so, y take more time. 7 D Sol. Friction will produce heat energy and so the total energy will not be PE + KE (or al l of the PE will not turn into KE). At a constant velocity there is no gain of KE. On a horizontal surface there is no change of PE. Falling freely, the PE turns solely into KE. 8 C Sol. 2 T sin m R    T cos mg   2 R tan g    gtan gtan R r sin         T cos  mg T sin  T R= r+l sin   AITS - PT - I - PCM (Sol.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 3 9 A Sol. T = m 1 g 1 kx 2T 2m g   1 2m g x k  Energy stored = 2 1 kx 2 = 2 2 1 2 4m g 1 k 2 k = 2 2 1 2m g k m 1 m 2 T T T T 2 T m 1 g T 1 10 B Sol. The rod will rotate about point A with angular acceleration: 2 2 Fx 3Fx I m m 3         3 Fx a 2 2 m      or a  x i.e., a - x graph is a straigth line passing through origin. 11 A Sol. 0 3v v cos60 cos30 2    , 0 3 3 v v 2  12 C Sol. 0 1 v t v 40  , t 2 0 1 30 v t S 40 2  0 2 0 v t v 3v 10   , 2 t 0 2 0 30 v t S 3v t 20   S 1 = S 2  t = 60 sec 13 B Sol. 2 2 min v g(y x y ) 30     m/s 2 v tan 3 gx    14 C Sol. 1 2 N f 2mg   ...(i) 2 1 N f  ...(ii) 1 1 f N   ...(iii) 2 2 f N   ...(iv) Taking moment about corner B of plate 2 1 (mg N )R N R   2 1 mg N N   ...(v) From (i), (ii) & (v), 1 2 f f mg   ...(vi) N 1 mg f 2 mg C N 2 B A f 1 AITS - PT - I - PCM (So l.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 4 From (iii), (iv) & (vi), 2 2 1 0      or 2 1    15 A Sol. Centre of mass of arc is 2R sin 3   m 1 = 2m, m 2 = m centre of mass 2m 2R sin 2R sin 3 6 m 3 2 3 6 3m                                                  2R( 3 1) 3   16 A Sol. f i impulse p p     p f  p i p f |p i | sin  mg |p i | cos   mg  sin  17 A Sol. t = 0 to t = 1 sec v = 4t m/s s = 2t 2 m s(t = 0) = 0 m, s(t= 1) = 2m average s v 2 t     m/s t = 1 to t = 2 sec v = 2 m/s s = 2t m s ( t = 1) = 2m, s(t = 2) = 4m v average = 2 m/s 18 C Sol. A cylinder of mass M For equilibrium N A cos 60  + N B cos 30  = Mg and A B N sin60 N sin30    On solving N B = A 3N ; A Mg N 2  60  30  N A N B Mg AITS - PT - I - PCM (Sol.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 5 19 C Sol. 0 x dM 1 dx a          0 a M 2    d(MO I) = dmx 2 2 2 Ma MOI dMx 6    y x dx 20 C Sol. Minimum stopping distance = s Force of friction mg   Work done against the friction W mgs   Initial kinetic energy of the toy cart   2 p / 2m     2 mgs p / 2m   2 1 2 2 1 s m s m        SECTION – B 21 3 Sol. Let the shell hit the plane at p(x, y), the range being AP = R, as shown in the figure. The equation for the projectile’s motion is 2 2 2 gx y x tan 2u cos     ...(i) Now y = R sin  ...(ii) 2R  A  B R P(x, y) x = R cos  ...(iii) Using equation (ii) and equation (iii) in equation (i) and simplifying 2 2 2u cos sin( ) R gcos       The maximum range is obtained by setting dR 0 d   , holding u,  and g constant. This gives cos(2 ) 0     or 2 2      22 5 Sol. for block A 2 A A T m g m R     For block B 2 A B A B T (m m )g m g m R        A B B A g(3m m ) R(m m )      AITS - PT - I - PCM (So l.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 6 23 120 Sol. As  p x = 0,  p y = 0 For x direction,  v cos cos v 2m mv 3m 2 2 2 3        cos 2mv mv 2    cos 1 cos 60 2 3     60 120 2        24 125 Sol. Kinetic energy 2 1 E mv 2   2 2 2 2 E v v 100 100 (1.5) 1 100 E v              E 100 125% E    25 16 Sol. 2 2 x y v v v   v x = 10  10 cos 5t, v y = +10 sin 5t 5t v 20 sin 2        t 0 5t S 20 sin dt 2         ,  t = 2  So, S = 16 m 26 30 Sol. A begins to climb B if F cos 60   Mg sin 60  F  3 Mg F  30 N 27 3 Sol. Power = K.E./ time = 2 2 1 n mv 3600 1 500 2 (0.2) t 60 2 3              = 3 500 10 3        watt 28 3 Sol. a = g(sin    cos  ) Also, a = R  , 2 Rz R( mgcos R) 2 gcos 1 I mR 2        g (sin   r cos  ) = 2  g cos  tan 3    AITS - PT - I - PCM (Sol.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 7 29 7 Sol. 2a 0 = N 1  10  2g 2kg 10N a 0 2g N 1 N 1 = 34 N 3a 0 = N 2  N 1  3g N 2 = 70 N 3kg N 1 a 0 N 2 3g 30 8 Sol. By the conservation of energy 2 2 0 1 1 mv mgh mv 2 2   ...(i) By conservation of angular momentum 0 mv R mvR 2  From equation (i) and equation (ii) 0 8gh v 3   k = 8 AITS - PT - I - PCM (So l.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 8 C C h h e e m m i i s s t t r r y y PART – B SECTION – A 31 B Sol. 1 Be B C N 2 2 3 2 2 2 2s p 2s p 2s 2s p Be and N have stable configurations, so do not accept electron easily, hence for these elements the process is Endothermic. S o, correct order is Be < N < B < C 32 D Sol. On moving down the group the 1 IE of Gr(1) elements decreases continuously hence reactivity increases. 33 D Sol. h v x 4 m      (Using uncertainty principle) 34 28 12 6.62 10 v 4 3.14 1.1 10 3 10             5 5 1 v 1.597 10 1.6 10 ms        34 A Sol. The ionic equation is         Acidic 4 2 medium CrO I HCl Cr I 3 1 6 0            4 2 2 2CrO 6I 16H 2Cr 3I 8H O            or 4 3 2 2 2 2BaCrO 6KI 16HCl 2CrCl 3I 8H O 6KCl 2BaCl         35 A Sol.       2 2 2HI g H g I g    (follows zero order kinetics)     o d R 1 k R 2 dt       o R R t 2k    1.50 0.30 t 2 0.06     1.20 t 10 s 0.12    36 A Sol. O F F 103 o O H H 104.5 o O Cl Cl 111 o Cl O O 1 1 8 o < < < AITS - PT - I - PCM (Sol.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 9 2 H O has larger bp – bp repulsion than 2 OF and 2 OCl has larger steric repulsion between two large Cl atoms while in 2 ClO the bond angle is   o o 118 close to 120 37 C Sol. 3 2 2 1 1 2NH N 3H K            ... (1)   2 2 2 N O 2NO K      ... (2)   3 2 2 2 3 3 3H O 3H O K 2      ... (3) Adding (1), (2) and (3)   3 2 2 eq 5 2NH O 2NO 3H O K 2       3 2 3 eq 1 K K K K   38 B Sol.       2 H O H aq OH aq         o 7 At 25 C, H OH 10 M              1 14 W K 10    o 6 At 35 C, H OH 10 M              2 12 W K 10    No w, 2 1 W 2 1 W 1 2 K T T H 2.303 log K R T T          12 14 10 H 308 298 2.303 log 2 308 298 10               H 10 2.303 2 2 308 298       1 H 84.551 Kcal mol     Since, 1 84.551 Kcal mol  is the enthalpy change for the reaction 39 D Sol.   4 1 1 1 KE PE mkr 2 2 2            4 2 1 1 KE mkr mv 4 2   4 2 4 mkr .2 1 v kr 4.m 2    ... (1) Since, 2 2 2 2 2 2 nh n h mvr v 2 4 m r      ... (2) From (1) and (2) 2 2 4 2 2 2 1 n h kr 2 4 m r   6 2 3 r n r n     AITS - PT - I - PCM (So l.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 10 40 D Sol.     2 2 2 H O g P 2 H g P K P  ... (1) 2 H T T 40 P P 0.4P 40 60     2 H O T T 60 P P 0.6P 40 60     2 2 H O T H T P 0.6P 3 1.5 P 0.4P 2      From Eq. (1)   2 P K 1.5 2.25   41 C Sol. (A) g n 0, in P shifts the equilibrium      (B)  in volume in (III) rd equilibrium shifts the equilibrium   g n 0    (C) At constant volume no effect of addition of He (D) At constant pressure, addition of He shifts the equilibrium   g n 0    42 B Sol. 2 atm 2 2 2 high temp. CaC N CaCN C Calcium cyanamide     43 B Sol. (A) On moving down the group, the size of metal i on increases, so stability of peroxides/superoxides increases. i.e. Large cation – Lagre anion interactions increases. (B) NaOH is hygroscopic in nature. (C), (D) Fact based on lattice H  and hyd H  44 A S ol. In solid state, BeCl 2 exists as a polymer. Be Cl Cl Be Cl Cl Be Cl Cl ( four electrons are shared between three atoms, so 3c - 4e type of bonds are present) 45 D Sol. I. – 1 corner shared II. – 2 corners shared III. – One unit shares 2 corner s while one unit shares 3 - corners. IV. – 2 corners shared V. – 4 corners shared VI – 3 corners shared AITS - PT - I - PCM (Sol.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 11 46 D Sol. Borazon  Diamond like structure (Non - planar) B orazole  Inorganic benzene (Planar) 3 6 B O  O B O B O B O O O   Planar 2 6 Al Cl Al Cl Cl Cl Cl Al Cl Cl (Non-planar) 47 B Sol. 2 NO  odd electron species  Para magnetic 2 NO  is linear while 2 NO  and 2 NO are bent. 2 NO   lp – bp repulsion 2 NO  bond angle is around 134 o Bond order of 3 4 NO is 1.33 3   Bond angles,   o 2 NO 115  ,     o o 2 2 NO 180 , NO 134  48 D Sol. LiH  Stable Na H  Unstable 2 3 Li CO Unstable  2 3 Na CO Stable  (Around 400 – 500 o C) 49 C Sol. I.  Bond length order is 2 2 2 H H H     II  Isoelectronic species, so identical bond order. III.  Fact. IV      3 3 NO Re sonance , BO no resonance   AITS - PT - I - PCM (So l.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 12 50 A Sol. B H H B H H H H Maximum 4 – H atoms lie in a plane. Maximum 6 – atoms (4H + 2B) lie in a plane. Bridge (B – H) is stronger than terminal (B – H) Terminal (H – B – H) bond angle > Bridge (H – B – H) bond angle. SECTION – B 51 12 Sol.       2 3 3 Ag CO s 2Ag aq CO aq     2 SP 3 K Ag CO            12 2 8.2 10 Ag 7.5 / 5                2 12 82 Ag 10 15                6 Ag 2.34 10 M          No w   SP K AgCl Ag Cl            6 SP 0.0026 K 2.34 10 35.5           10 SP K 1.71 10    x 1.71, y 10    x y 1.71 10 11.71 12       52 20 Sol. 2 1 KE mv 2    2 1 d KE d mv 2             d KE mv dv   ... (1) Since, h dv dx 4 m      h dv 4 m dx    ... (2) From (1) and (2)     h d KE mv. 4 m. dx   AITS - PT - I - PCM (Sol.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 13   8 34 12 1 3 10 6.62 10 d KE 6.62 60 4 3.14 10 3.14              14 16 1 d KE 10 1.25 10 J 80        x 1.25, y 16    xy 1.25 16 20     53 7 Sol. The ionic equation involved is           4 2 4 MnO aq SO g H aq Mn aq HSO aq          The balanced equation is             4 2 2 4 2MnO aq 5SO g 2H O H aq 2Mn aq 5HSO aq             Sum total of coefficient(s) = 2 + 5 = 7. 54 310 Sol. a E /RT k Ae   a E logK log A 2.303RT     13 0.693 98.6 1000 log log 4 10 693 2.303 8.314 T              or 13 3 4 10 98.6 1000 log 2.303 8.314 T 10        T 310.1 K 310K    55 64 Sol.     2 4 2 N O g 2NO g 0.28 1.1        2 2 4 2 2 NO P N O P 1.1 K 4.32 0.28 P    Now , volume is doubled, so pressure will become half and reaction proceeds in forward direction     2 4 2 N O g 2NO g 0.28 1.1 P 2P 2 2                    2 P 0.55 2P K 0.14 P           2 0.55 2 0.55 2P 4.32 0.14 P     0.3025 2.20P 0.14 P    ( Ignoring P 2 te rms) P 0.046 atm   2 NO P 0.55 2 0.046 0.642      AITS - PT - I - PCM (So l.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 14 2 64.2 10 atm    x 64.2 64    56 2 Sol.   3 2 2 4 XeO F AB L sp d Polar       2 2 2 SnCl AB L sp Polar       3 2 5 5 IF AB L sp d Polar       3 3 2 2 I AB L sp Polar      3 4 4 XeO AB sp Non Polar     (  )   3 3 7 7 XeF AB sp d Non Polar      (  )   2 2 2 SO AB L sp Polar     57 4 Sol. 1. Correct as EN of side atom increases, EN of carbon increases. 2. Correct, rd Mg 3 period   th K / S 4 period and isoelectronic    th Se 5 period   3. Correct as moving down the group cationic size increases so lattice energy decreases 4. Correct as 1 Hydration energy Size of cation  5. Incorrect as the bond energy order among halogens is 2 2 2 2 Cl Br F I    58 33 Sol.       2 H O 2 2 2 2 2 Ice cold Na O 2H O 2NaOH H O P Q        Room 2 2 2 2 temperature 2Na O 2H O 4NaOH O P R      Hence, 1 2 M M 34 32 33 2 2     59 80 Sol. At half neutralization S A  a S pH pK log A   a a pH pK pK 5.0     Now , at equivalence point,   W a 1 PH pK pK log C 2      1 9 14 5 log C 2    C 0.1 M   Let, V mL of NaOH is used AITS - PT - I - PCM (Sol.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 15   0.25 V Meq. NaOH 0.1 60 V     = Meq. (salt)  Volume (V) = 40 mL Meq. of acid = millimoles of acid = 0.25 × 40 = 10 Mass of acid (HX ) = 3 10 10 82    = 0.829 Mass% 0.82 100 80% 1.025    60 1 Sol. total H 25 0.1 50 0.2 12.5 milim oles           total OH 25 0.1 2.5 mi lim oles         final 12.5 2.5 H 25 50 25           final 10 1 H M M 100 10        final pH log H         1 pH log 1.00 10          AITS - PT - I - PCM (So l.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 16 M M a a t t h h e e m m a a t t i i c c s s PART – C SECTION – A 61 C Sol. f'(x) + f(x) = 12x 2 + 2 62 B Sol. Use expansion and collect the coefficient of x 2  P = 330 63 A Sol. For A 3 x x sin x x 6    and for B use limit as a sum  A = 1 2 , B = 2  64 D Sol. f'(x) =         2 2 2 1 sin2x 2 x a 0 sin x x a        x  (0, 1) and a  (2, 3) 65 C Sol. A  B =     f (1) 1 1 0 f (0) f x dx f x dx 0      , and also A = B. 66 D Sol. Domain is R. 67 D Sol.   /4 0 1 1 A tan x sin x dx ln2 1 2 2              68 A Sol. 2 2 1 2 3 I 2, I 2, I 1 2 2        69 D Sol. f'(x) = (lnx) 2 , g'(x) = (lnx), so area =     e 2 1 ln x ln x dx 3 e     70 D Sol. L 1 = L 2 = 2  71 B Sol.               n n 1 x 3 nx 3 1 1 T nx 3 n 1 x 3 n 1 x 3 nx 3              S 12 = V 1  V 2 + V 2  V 3 + ....... V 12  V 13 = V 1  V 13 AITS - PT - I - PCM (Sol.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 17 = 1 1 x 3 13x 3    x 1 1 1 3 lim 4 16 16    72 A Sol. tan  = 5 r 12 h   5h r 12  dv dt = 0.1 m 3 /min v = 3 2 1 25h r h 3 432     2 dv 25 h dh 25 dh 0.1 3 3 dt 144 dt 144 dt           16 dh 250 dt   h l  Now s =  rl = 2 2 2 65 h r h r 144     ds 65 dh 65 16 13 2h 2 3 dt 144 dt 144 250 75            m 2 /min 73 A Sol. x 2  4 = 2x + 11 x =  3, 5  A(  3, 5) 2x + 11 = 9 x = 1  B(  1, 9) 1 C ,9 2       and D(2, 0) A B C D y = 2x + 11 y = 12  6x y = 9 Required area         1 1/2 2 2 2 2 3 1 1/2 2x 11 x 4 dx 9 x 4 dx 12 6x x 4 dx                 =     1 1/2 2 2 2 2 3 1 1/2 2x x 15 dx 13 x dx 16 6x x dx              = 1 1/2 2 3 3 3 2 2 3 1 1/2 x x x x 15x 13x 16x 3x 3 3 3           = 40 153 81 511 3 8 8 12    74 B Sol.     2 t 5 2 4 2 e t f t t 2t 2        e f 1 25   AITS - PT - I - PCM (So l.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 18 Let x 2 = v  dv 2x dx     v 2 2 2 1 e v dv 2 v 2v 2       v 2 2 2 1 1 2v 2 e 2 v 2v 2 v 2v 2                  v 2 1 e 2 v 2v 2    2 x 4 2 1 e 2 x 2x 2       2 t x 4 2 0 1 e f t 2 x 2x 2    2 t 4 2 e 1 1 2 2 t 2t 2                e 1 f 1 10 4    f(1) + f  (1) = e e 1 7e 1 10 25 4 50 4     75 B Sol.         3 2 4 3 2 4 3 2 x k 2 x 3 n x m 3 x 2 p lim x kx 3x mx 2 x 2x nx 3x p                  k – 2 has to be 0 else limit wont be finite  k = 2 Now dividing Num. & Den by x 2     2 x 2 3 4 2 3 4 2 b m 3 3 n x x lim a 3 b 2 2 c 3 d 1 1 x x x x x x x x                3 n 2    3 n 4 2    n = 5. 76 A Sol. Let 2 x t   dt 2x dx   2 2022 2022 2022 0 1 t sin t dt I 2 cos t sin t        2022 2 2 2022 2022 2022 2022 2022 0 0 2 t sin t 1 sin t I 2I 2 cos t sin t cos t sin t              2022 2 2022 2022 0 sin t 2I 4 cos t sin t       2022 2 2022 2022 0 cos t 2I 4 cos t sin t       2 0 4I 4 dt      2 I 2   AITS - PT - I - PCM (Sol.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 19 7 7 A Sol.         x 2 y 2 dy dx x 2 y 2        put x  2 = h, y  2 = k dk h k dh h k    put k = vh  dv 1 v v h dh 1 v        1 2 2 1 v dh 1 dv tan v ln 1 v lnh c h 2 1 v                   2 1 2 y 2 y 2 1 tan ln 1 ln x 2 c x 2 2 x 2                          it passes through (3, 2)  ta n  1 0 = 1 2 ln1 + ln1 + c  c = 0  it also passes through (p + 2, 3) 1 2 1 1 1 tan ln 1 lnp p 2 p                    1 2 1 2 tan ln 1 p p          78 B Sol.   2 0 1 2 2 2 4 2 0 1 A x 6 x dx xdx x dx             = 8 2 7 35 6 3 3 3 3      2  4 1 79 C Sol.   4 e 4 10 1 log x f e dx 1 x    ...(1) Let x = 2 1 dx 1 v dv v     f(t) = 1/t 1/t 10 10 2 1 1 log (1/ v) log v 1 dv dv 1 v(v 1) v 1 v                  4 4 1/t e e 4 10 10 10 1 1 1 log v log v log x f t f e dv dx v v 1 v v 1 x x 1            ...(2) Add (1) and (2) f(e  4 ) + f(e 4 ) = 4 e 1 1 ln x 1 1 dx ln10 (1 x) x          AITS - PT - I - PCM (So l.) - JEE(Main)/ 20 2 3 FIITJEE Ltd., FIITJEE House, 29 - A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 20 =   4 4 e 2 e 1 1 ln x 1 ln x 1 ln10 x ln10 2   = 1 8 16 2ln10 ln10  80 C Sol. Let g(x) = x 3  2x 2 + 3x  9 g'(x) = 3x 2  4x + 3 > 0  x  R  g(x) is always increasing  g(2) = 8  8 + 6  9 =  3 is maximum value for x  2 If  3x + log 3 (k 2  5)   3 for x > 2 then we have maximum at x = 2   6 + log 3 (k 2  5)   3  log 3 (k 2  5)  3  k 2  5  27      k 4 2 k 4 2 0    k 4 2,4 2       and k 2  5 > 0      k , 5 5,       k    4 2, 5 5, 4 2        SECTION – B 81 1 So l. 2 2 x x 2 1 x x 1            82 1 Sol. A = 3 2 8 4    83 4 Sol. D f = [  2, 2] as     2 f x 3 4 x   R f = 0,2 3      84 2 Sol. f(x) = cos(x 2  3x) |x  2| sin|x  2| + (x  1) |x| |x  1| |x  7|  f(x) non differentiable at x = 0 and 7 only. 85 12 Sol. a = 1, b = 6, c = 5 86 96 Sol. Put x = 0 we get y = 1 ln(x + y) = 6xy Differentiate w.r.t. x 1 dy xdy 1 6 y x y dx dx                