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JEE(Advanced)-2014 ANSWERS, HINTS & SOLUTIONS FULL TEST –I (Paper-2)

ALL INDIA TEST SERIES

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FIITJEE

Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

1.

C

D

B

2.

C

A

B

3.

D

C

A

4.

B

B

C

5.

B

B

D

6.

D

C

D

7.

B

D

C

8.

C

B

A

9.

A

C

C

10.

D

B

C

11.

B

A

D

12.

D

C

A

13.

B

A

C

14.

A

D

A

15.

A

C

A

16.

B

B

C

17.

B

A

B

18.

A

B

A

19.

B

C

A

1.

(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)

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AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/14

Physics 2.

3.

PART – I

T sin  = mg/2 T = T cos  mg mg = cot   2 2 tan 

T sin 

T

T cos 

T

1 mv 2  pt (P = const) 2 2Pt  v= m

a=

dv  dt

2P 1 m 2 t

F = ma =

5.

2

mP 1  2t v

      F  F1  F2  F3  F4  F5     F2  F5 and F2  F4   F1  F3  2F2 cos 30  2F1 cos 60

m

F3

F2

m

m

F4

F1

Gm2 Gm2 Gm2 F3 = ; F = ; F = 2 1 4a2 3a2 a2 Gm2  5 1  2 F=    = m a 2 a 4 3

m

F5

m

Gm  5 1     a3  4 3

=

T = 2

4 3a3



Gm 5 3  4

6.

m

P



S R

P

Q V

7.

a= t=

10.

v  g  a   m  g  a  m

= 20 m/s2

2h = 1 sec a

Time period becomes 2

R . We can’t neglect the roundness of earth for the pendulum of g

infinite length.


3


d signifies the direction of induced emf. dt

12.

The ve sign is 

14.

process AB U = constant P RT  and U  t  M  P = const Process BC isochoric Process CA  isothermal

15.

16.

Q = QAB + QBC + QCA 10U0 Q = 5U0 + 3U0 + ln2.5 3 W AB = QAB  UAB = 5U0  (3U0) = 2U0

17.

For lens L1, ray must move parallel to the axis after refraction

18.

For lens L2, image must form at centre of curvature of the curved surface after refraction through plane part. 2    0 R2 x  

1  w 1       x = 10 cm  x R1

x = 8 cm


4


Chemistry

PART – II

1.

CH2MgBr Br

OH H

H Mg  Ether

O H3C



H   Re arrangment



O

H3C

2.

O

O

O

H3CO

CH O 

3  

O

H3CO

O  O

O

H

O

 O

OH

COOCH3 H3COOC 3.

P

PAo X A

P

PAo

 PBo





PAo



PAo

 PBo

1  XB 

 O

O COOCH3

 PBo XB

B

Thus PAo  120 Torr

4.

PAo  PBo  75  PBo  45 Torr Hence C is correct answer. 1000 o BaSO   4 Conc  normality  Normality 

  1000 o BaSO 4



1000  8  10 5 = 2 × 10-4. 400

Normality  10 4 M  Solubility  2 Ksp = S2 = 10–8 M2. 1 Tav  K 10  0.693 6.93 T99.9  10  ty 2   K K  6.693   1  Number of natural life times =  /   K  K  = 6.93 Molarity 

5.


5


OH

6. O HO

B

O

O

B O

B B

OH

O

OH

7.

Ca  OCl Cl.H2 O  Cl2 145

71

71  100  49 145 Hence D is correct answer.  d  c   0.0033  1.32  104 m min1    dt 25  

Percentage 

17.

I

 d c   4 1  dt   2.6  10 m min  II

18. 19.

 dc   3 1    1.02  10 m min  dt III On comparing rates order w.r.t A = 2, and w.r.t. B = 1. Thus rate law = K[A]2[B] dx 2  K  A  B  dt  dx / dt  K  0.26  A 2 B



Mathematics

6

PART – III

1.

Total number of lines made = 9C2 = 36 Now, these 36 lines are 9 sets each with 4 parallel members. With 5 vertices number of lines made 5 = C2 = 10 Clearly, atleast 2 members belong to the same set So, atleast one pair is parallel

2.

cos  

OA 2  OB 2  AB2 2OA  OB 2

 OA  OB  OA 2  OB2    2 2  2   3  OA  OB   1 = 2OA  OB 8 OA  OB 4 For maximum cos , 1 3  OA 2  OB 2  1 3 2  OA  OB 1  =     8  OA  OB  2 8 OA  OB 4 2    3

3.

S1 = a1 S2 = a1 + a2 S3 = a1 + a2 + a3  Sn = a1 + a2 + ….. + an If we divide all S1, S2, ….. Sn by 23 we get remainders 1, 2, 3, 4, ….., 22 So, two of these give same remainders Sp, Sq  Sp – Sq will be divisible by 23 Sp – Sq = ap + 1 + ap + 2 + ….. + aq

4.

Using Cauchy we get,

5.

Point of intersection is  

 z1  z2

 12 12 22 42   2  z3  z 4        1  1  2  4   64 z z z z 2 3 4   1

1 b ln   2 a

For C1, dr d  ae , now tan 1  r =  ae  a 1e  1 d dr   1  4 For C2, dr  1  be  tan 2   be      e  1 d  b 3  2  4 3    Angle of intersection is 2  1    4 4 2


7


6.

Let c, d, e be the three points where y = f(x) crosses x–axis. Then, f(c) = f(d) = f(e) = 0 Assuming 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) = 0 So, there exists, at least one point in the interval (c, d) and (d, e) such that derivative is zero Let, C1  (c, d) such that f(C1) = 0 and C2  (d, e) such that f(C2) = 0. Now the function f satisfies Rolle’s theorem since f, f are continuous and f(C1) = f(C2) = 0 So, by Rolle’s theorem, there exists a number C3 in between C1 and C2 such that f(C3) = 0  Minimum one root C3 of the equation f(x) = 0 lies in the interval (a, b)

7.

Dividing the given differential equation by 3xy(y2 – x2) y  y2  2x2  3xy  y 2  x2  

dx 

x  2y 2  x2  3xy  y 2  x 2 

dy  0

dx xdx ydy dy    0 x y2  x 2 y2  x 2 y

 d  ln  xy   

2 2 1 d y  x  0 2 y2  x2

 d ln  x 2 y 2  y 2  x 2     0  ln  x 2 y 2  y 2  x2    c  x2y2(y2 – x2) = c 8.

2

Let the circle be x 2   y     a2 . Let the point of intersection of tangents at P and Q be (h, k). Then equation of PQ, is hx   k    y     a2  0 . As it passes through  a,0  , so, ha    k     a2  0 .   2  k  a  h  a   0. D  0  k 2  4a h  a   0

i.e. y 2  4a  x  a  . 1

9.

Consider

 0

1

   x 2 f  x dx 

   f  x   2xf  x   x f  x  dx =  2

2

2

– 22 + 2 = 0

0

However f(x) assumes only positive values i.e. in (0, 1) 2  ( – x) (f(x)) > 0  integral can’t be zero 10.

Differential equation can be written as, (p – x)(p – 2 sin x)(2p + cos x) = 0 which has solution as 2 (2y – x – c)(y + 2 cos x – c)(2y + sin x – c) = 0

11.

Put x = –1 we get (–1 + 1) p(–1) + 1 = (–1)  1n1   n  1 !

n+1

(n + 1)!

 1n1 x  x  1 .....  x  n  x   n  1 ! n  1 x 1 where 'n' is odd 1,  Clearly, p  n  1   n  n  2 , where 'n' is even So, p  x  


8


13.

If e is the eccentricity then, e2  1

Now, we know,   2  2  2

 ab

1



2

1 2

1 

2

 ab 

,  2 2 

2 1 2 2

 

 ab  h2

1 ab  h2

 a  b 2  4  ab  h2   ab  h 2

2    a  b   4  ab  h2 

For an ellipse So, e2 



ab  h2

   

2

2

 2  2

2 

2

2  a  b   a  b   4 ab  h2 

 a  b 2  4h2  2 2  a  b    a  b   4h  2  2 ab  h

14.

Put y = z = t = 0 f(0)[f(x) + f(0)] = f(0) Put x = 0 2f 2(0) = f(0) 1  f(0) = 0, 2 1 1 If f(0) =  f(x) + =1 2 2 1  f(x) = 2 If f(0) = 0, z = t = 0  f(x) f(y) = f(xy) Let, x = y = 1  f 2(1) = f(1)  f(1) = 0 or f(1) = 1 We have f(0) = 0, f(1) = 0, y = 1 f(x) = 0 Also, 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) 2 2  f(x ) = f (x)  0 Put x = t, y = z [f(x) + f(y)]2 = f(x2 + y2)  f(x2 + y2) = f2(x) + f 2(y) + 2f(x)f(y)  f2(x)  f(x2 + y2)  f(x2)  f is non decreasing for positive x

16.

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) = 0 f(n) = n2 (Possible function), if f(n – 1) = (n – 1)2 2[f(n – 1) + 1] = f(n – 2) + f(n)


9


 f(n) = n2

p (rational number) q p p f   f  q2   f  pq  f   q2  p2 q2 q   q

Now, for x 

2

p p  f      (True for rational number)  q  q Now, if x  R, lets prove for positive x since if it is proved the function is even and will follow for negative x Assume for x > 0, f(x) < x2 So, now a rational number ‘r’ between f  x  and x  f  x  < r < x 2 2  f(x) < r < x 2 [f(r) = r , f is non decreasing]  f(r) = r2  f(x) [contradiction] 2  f(x) < x (impossible) Similarly we can prove contradiction f(x) > x2 So, only possibility f(x) = x2 substituting f(x) = x2 we 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] 1 So, f(x) = , f(x) = 0, f(x) = x2 are the required solutions 2

17.

am3 + m(2a – h) + k = 0  am3 + m(2a – x1) = 0  am2 = x1 – 2a {m = 0 (one possible value)} x  2a  m2  1 , if x  (0, 2a) a  m2 = (–) number, so non real roots

18.

Let P(h, k) be the point am 3 + m(2a – h) + k = 0 Since, m1, m1, m1 are the possible roots  3m1 = 0  m1 = 0 If m1 = 0 is the root then k = 0 am3 + m(2a – h) = 0 h  2a  m2   0  h = 2a a (2a, 0) is the only point

19.

A1 

8a



2a

=

2 2  4  5 / 2 8a 3/2 x  2a  dx =  x  2a   2a 27a 3 3a 5

2  5 / 2 2 2 48 2a2 6a    36a2 6a = 5 3 3a 5 3 3a 5 2

8a

A1 

 0

4axdx  2 a 

4 64 2  3 / 2 8a a 8a 8a  2a2 x 0 = 3 3 3

11 352 2a2 4 3 Area = 2(A2 – A1) = 2  16 2a2    = 32 2a2   15 15 3 5


10


SECTION – B 1.

(A) |A – xI| = 0 There 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 AX    adj A  X x AX   adj A  X  2 (B) A ' I   A  I '  A  I  A ' I  0 if and only if A   I = 0  A ' I  0 if and only if A   I = 0 or  is the root of A ' xI  0 if and only if  is the root of |A – yI| = 0  e–i is the required solution  A11   A12 .......... A1n    A 22  ..... A 2n 0  0  | | (C) B  I    =0 | | |  |  | |   | A nn     |  (|A11| – )(|A22| – ) ….. (|Ann| – ) – 0 Clearly 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' AX and X'IX are both real Also, X' X  0 , X  0    X' AX / XIX is real so  can have real values 2.

(A) z = cos  + i sin , z 

1  2cos  z

1  2isin  z 1 zp  p  2cosp z z

4

2

1  1 1  1   4 2  Now,  2isin    2cos     z    z   =  z 6  6   2  z 4  4  z  z  z   z then, 2 cos 6 – 2.2. cos 4 – 2 cos 2  + 4 4 4 4 2 2  2 i sin  2 cos  = 2(cos 6 – 2 cos 4 – cos 2 + 2) =2

  2 1  z  2   4   z 


11

(B) Volume of tetrahedron =


C

1   ArABD   CQ 3

1  AB     DP  CQ 3 2  Let, CD be the largest side and AB = x  1 Let, T be closer to A than B x  BT  2 x2  CT 2  BC2  BT 2  1  4

=

 CT  1 

B

Q T

P A

x2 (same for DP, CQ) 4

1 x  x2  1  1   1  x 4  x 2   Vmax   3 2  4  24 8 (C) By AM  GM cos x cos3 x   cos2 x (x is an acute angle) 4 Now, setting x = A, B, C we get x1 + x3  cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 2x2 3  x1 + x2 + x3  3x 2  2 A B C Sa Sb Sc (D) Let, u  cot , v  cot , w  cot  u  , v , w 2 2 2 r r r S S  a   S  b  S  c   uvw r r 2 Now we can rewrite as 49[u + 4v2 + 9w2] = 36(u + v + w)2  (3u – 12v)2 + (4v – 9w)2 + (18w – 2u)2 = 0 1 1 u:v:w=1: : 4 9 S  a S  b S  c 2S  b  c Multiplying by r we get    36 9 4 94 2S  c  a 2S  a  b a b c =     4  36 36  9 13 40 45

So, V 

3.

1 3 4 2 and substitute, 8y + 4y + a – =0 2 2 3 2 Again, z = y we get, 8z2  4z  a   0 2 3 3 When a  there are 2 non real roots and two real and a  we have 4 non real roots 2 2 3  1, a  2  Sum =  2, a  3  2

(A) Put x  y 

2

xy (B) xy     1  0 < xy  1  2  3 3 3 3 x y (x + y ) = 2(xy)3((x + y)2 – 3xy) = 2(xy)3 (4 – 3xy)


D


12

Put z = xy 4

 4  3z  3z  z3  4  3z     1  4  3 3 3 3  x y (x + y )  2 (C) There are 9 possible numbers of the type 7775775, 7757575, 7575575 etc. (D) We have f(9) = f(4 + 5) = f(4 . 5) = f(20) = f(16 + 4) = f(16 . 4) = f(64)  f(64) = f(8 . 8) = f(8 + 8) = f(16) = f(4 . 4) = f(4 + 4) = f(8) = 9


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