ANSWER KEY
12 th ABCD (Date: 14-06-2009) Review Test-2
Code-B
Paper-1
CHEMISTRY PHYSICS MATHS
SECTION-2
SECTION-3
SECTION-1
PART-A
PART-A PART-A
Q.1 D
Q.1 D Q.2 D Q.1 B
Q.2 C Q.3 D Q.2 C
Q.3 C Q.4 A Q.3 A
Q.4 A Q.5 C Q.4 B
Q.5 C Q.6 B Q.5 D
Q.6 B Q.7 B Q.6 A
Q.7 A Q.8 C Q.7 C
Q.8 B Q.9 A Q.8 D
Q.9 C Q.10 B or D Q.9 B
Q.10 A Q.11 A Q.10 C
Q.11 D Q.12 C Q.11 A
Q.12 A Q.13 D Q.12 C
Q.13 B Q.14 A Q.13 A
Q.14 A Q.15 A, B, C, D Q.14 D
Q.15 C, D Q.16 A, B, C, D Q.15 A, B
Q.16 A, C Q.17 B, C, D Q.16 A, B, C, D
Q.17 C, D Q.18 C, D Q.17 B, C
Q.18 C Q.18 C, D
PART-B
Q.1 (A) Q, S, (B) P, R, S ; PART-B
PART-B
(C) P, R, S Q.1 (A) R ; (B) P ; (C) Q
Q.1 (A) P,Q,R,S (B) R,S
(C) Q PART-C
Q.1 0090 or 0270 PART-C
PART-C
Q.2 0002 Q.1 0025
Q.1 0033
Q.2 0005
Q.2 5280

PHYSICS
PART-A
Q.1
[Sol. After some time force on a become constant when relative motion starts but on B is going on increasing.]
Q.2
[Sol. Radius of curvature f ]
Q.3
[Sol. On the centre line ]
Q.4
1 2 1
[Sol. f = 120, fm = 30, = f + f ]
f  m
Q.5
[Sol. Wg + Wf = KE
2Fd  2Mgd
v= ]
M
Q.6
[Sol. If temperature constant speed does not change. ]
Q.7
[Sol. /2  

By symmetery  =
2

I = I0 cos2
4
I = I0/2 ]
Q.9
By conservation of energy mgh =  F dx
d
[Sol. ]
0
Q.10
[Sol.
H H
by using similar triangle the shaded length = 5H/6 ]
Q.11
[Sol. µ = 1 + 12 = 2
sin i 1
=2  sin r =  r = 30°
sin r 2
unit vector = cos 30ˆ  sin 30ˆ
i j ]
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PHYSICS
Q.12
[Sol. ]
Q.13
[Sol. No TIR take place in case of sphere ]
Q.14
[Sol. Molar heat capacity of given process is negative. ]
Q.15
O
[Sol. (A & B) F
I
2F
O
F
(C) F
O
F
(D) ]
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PHYSICS
Q.16
[Sol. All dimension will increase ]
Q.17
Sol. I < Iavg
if each source have equal intensity
4 I0 cos2 /2 < 2I0
1  3
cos2 /2 < i.e. << ]
2 2 2
Q.18
[Sol. Breaking force = breaking stress × cross-sectional area ]
PART-B
Q.1
[Sol. Use P-V graph.
P P P
C B A B B
(A) (B) (C) ]
A C A C
V V V
PART-C
Q.1
B
30°
30°
60°
[Sol. ]
A C
Ray
Q.2
[Sol. Q = 7J
Q = DU + W
7 = nCvT + PdV
5
=n RT + nRT
2
7
7= (nRT)  nRT = 2 J ]
2
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CHEMISTRY
PART-A
Q.1
[Sol. FeSO4 + HO – OH is Fenton's reagents
e 
HO — OH HO — OH
Fe2+  Fe3+ + e
OH is not formed. ]
Q.2
[Sol. AlF3 is ionic and during formation of ionic bond octet of Al3+ as well as that of F¯ have been completed.
AlCl3 and AlBr3 are coval;ent and hypovalent. ]
Q.3
[Sol. (a) Trans 1,2-dimethyl cyclohexane equatorial–equatorial position
(b) Trans 1,2-dimethyl cyclohexane equatorial–equatorial position
(c) Trans 1,2-dimethyl cyclohexane axial, axial position ]
Q.4
O O O
[Sol. H F F S
H F F H H
3 3 3 3
Hybridisation : sp Hybridisation : sp Hybridisation : sp Hybridisation : sp
Bond angle between H2O & OF2 is compared on the basis of Bent's rule.
FSF in SF2 >  HSH in H2S : From Drago's rule
FOF in OF2 >  FSF in SF2 : On descending in a group for same substituents bond
angle decreases. ]
Q.5
[Sol. Pvr = K
for diatomic gas
 V2 
  = 
r 7/5
P1
 
3
P2 =  V1 
  4
for monoatomic gas
 Vnew  3
5/3 21
 3  25
7/5
= 
P1
V  =  Vnew = Vinitial ×   ]
P2  initial  4 4
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CHEMISTRY
Q.6
O O
[Sol.  
CH3 – C – O CH 3 – C = O
O O
CH 3– C – OH CH 3– C = OH
+
r : pure double fuel
y : CH3 – OH pure single bond
x : Partial double bond character
p : Partial double bond character
q : Partial single bond character ]
Q.7
[Sol. The magnitude of negative charge developed at chlorine atoms in SiCl4 is more in comparision to negative
charge developed at chlorine atoms in CCl4, it is due to more electronegativity difference between Si
and Cl than that of between C and Cl. ]
Q.8
[Sol. U =q+w
= (40 × 200) + (–2 × 10 × 100)
= 6000 J Ans. ]
Q.9
O OH OH
OH
 
[Sol. MgHg 
 H

 
H 2O

O OH

–H
O
]
Q.10
Cl Cl Cl
I I
[Sol.
Cl Cl Cl
Hybridisation of each I : sp3 d2
Planar structure
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CHEMISTRY
2 bonds are of 3C – 4e–
4 bonds are of 2C – 2e–
Cl Cl Cl
Al Al
Cl Cl Cl
Hybridisation of each Al : sp3
Non planar structure
2 bonds are of 3C–4e–
4 bonds are of 2C–2e– ]
Q.12
F F
[Sol. (A) Xe
F F
Hybridisation of Xe : sp3 d2
All four Xe – F bonds are of equal lengths as they lie in same plane and are at 90° apart from each other.
F
F
F P
(B)
F
F
Hybridisation : sp3d
(P–F) axial > (P–F)equatorial
(1.58 Å) (1.53 Å)
F
F
S
(C)
F
F
Hybridisation : sp3d
(S–F)axial > (S–F)equatorial ]
Q.13
COOK
[Sol.   ]
It is optically
COOK inactive due to
plane of symmetry
Code-B
Q.14 Page # 3

CHEMISTRY
Q.15
 
dP P
[Sol.
dV V
 P

dP
dT   1 T

dV 1 T
(   1) V
]
dT
Q.16
N
[Sol. NH3 H H
H
Hybridisation of N : sp3
Bond angle HNH less than 109°
l.p. exists in hybrid orbitals.
P
H H
PH3
H
Hybridisation of P : N.A.
 HPH close towards 90°
l.p. exists in s-orbital.
dipole moment of pure orbital is zero.
As l.p. in NH3 exists in hybrid orbital while in PH3 it exists in s-orbital, hence NH3 is stronger Lewis base
than PH3.
D of PH3 is less than that of NH3 as bond moment of N–H is greater than that of P–H because of
nitrogen is more electronegative than phosphorus. ]
Q.18
Cl Cl
R R
[Sol. No plane of symmetry. Optically active but having C2 axis of symmetry.
.
R R
Cl Cl
Presence of AoS doesn't matter for O. Activity. ]
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CHEMISTRY
PART-C
Q.1
[Sol. 10 intermediate free radicals

Br


So 10 structural isomeric products are formed and total products including stereoisomers are 33.
Br Br Br
Br
(4) (2) (4) (2)
Br Br
(2) (1) Br (4) (8)
Br
Br
(2) Br (4) ]
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MATHEMATICS
PART-A
Q.1
[Sol. P' (x) = f (x) g ' (x) + g (x) f ' (x)
P' (2) = f (2) g ' (2) + g (2) f ' (2)
= (1) (2) + 4 (–1)
=–2
g ( x )f ' ( x )  f ( x ) g ' ( x )
Q ' (x) =
g 2 (x )
( 4)(1)  (1)( 2) 6 3
Q ' (2) = =– =–
16 16 8
C' (x) = f ' g( x )  g' ( x )
C ' (2) = f ' (4) · 2
=3·2=6 ]
Q.2
cos x  x sin x
[Sol.  x cos x
dx ; put x cos x = t  (cos x – x sin x) dx = 2t dt

2 t dt
= = 2t + C = 2 x cos x + C Ans.]
t
Q.3
a ( x1  y1 )  b ( x 2  y 2 )  c ( x 3  y 3 ) A(0, 0)
abc
[Sol. h+k=
3x+4y=0 4x+3y=0
 9  15
(0)  5  3    (3  4)
7 15/4 5

15 15
4  4 4 4 4 0
= =
5 
15 7
5
22 B C
(3, – 9/4) (3,– 4)
x=3
4 4 7/4
4
Alteantively: Equation of angle bisector of angle A
3x  4 y 4 x  3y
  x=±y
5 5
equation of internal bisector is x = – y
since h and k lie on the line x = – y
 h + k = 0 Ans.]
Q.4
 x 
[Sol. We have g(x) = f  f (x) 
 
on differentiating w.r.t. x, we get
 x   f ( x )  xf ' ( x )   1   f (1)  f ' (1) 
g'(x) = f ' 
 f (x)  ·
 
 ;
   f (1)  ×  f 2 (1) 
f '(1) = f '    
   f 2 (x)     
As f(1) = f '(1)
 g'(1) = 0 ]
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MATHEMATICS
Q.5
[Sol. Let g (x) = P(x) – x.
Hence g (x) vanishes at x = 1, 2, 3, 4, 5 and 6. Since it is a polynomial of degree six
so g (x) = P(x) – x = (x – 1)(x – 2)(x – 3)(x – 4)(x – 5)(x – 6)
 P(7) = 6 · 5 · 4 · 3 · 2 · 1 + 7 = 727 Ans.]
Q.6
2  2
– 1  3x – 1  1  0x  domain is 0, 
 3
[Sol.
3
2
when x = 0 then y = 1; x = , y = 4. Hence range is [1, 4] Ans. Ans. ]
3
Q.7
          
2 sin cos = sin( + ) 2 cos cos  cos  0
 
[Sol.
2 2 2 2 2
    
sin = 2 sin cos 2
2 2 2

–1=0
2
2 cos
2
cos( + ) = 0   +  = 90°
  = 90° = cot –1(0) Ans.]
Q.8
x100
[Sol. Lim = 0 (using L'Hospital's Rule)
x  ex
 2
x2
Lim  cos  (1)
x   x
 2   1 cos t 
Lim x 2  cos 1   Lim  ·4 2 2
= e x   x 
= e x 0  t 2 
; put =t  x=
x t
= e–2 Ans.]
Q.9
[Sol. Let 2k = A
(cot3A – cot A) sin4A
· 2 · sin 4 A
cos A cos 2A 1
 cot A(cot2A – 1) sin4A   sin 4A
sin A sin A 4
1  
 

sin 4A = 1  sin 2 k  2  = 1 [sin 22 + sin 23 + sin 24 + ....] =  sin 2 k  = 4 Ans.]
1 a
4 k 0 4 k 0 4 4 k 2

S
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MATHEMATICS
Q.10
[Sol.  f (x) is differentiable in (0, ) [TN, MOD] to be put
hence Lim f ( x ) must exist and is finite
x 
 y = f (x) must have a horizontal asymptote
as x  then only Lim f ( x ) will exist
x 
If f (x) has an inclined asymptotes as y = x 
1
then Lim f ( x ) 
x x 
 f (x) has a horizontal asymptote
hence Lim f ' ( x ) 
x 
 (C) (also see figure for f (x) = tan–1x)
e.g. Take the example given
1
(i) Let f (x) = x sin which is differentiable in (0, )
x
 cos
1 1 1
f ' (x) = sin
x x x
 1  1
f (x) + f ' (x) =  x sin    sin  cos 
1 1
    x x 
x    x  

Lim 1 Lim 0
x  x 
hence Lim f ( x )  L and Lim f ' ( x )  0 Ans.
x  x 
(ii) f (x) = tan–1x in (0, ) ]
Q.11
[Sol. (m2 – 1)x2 – (m + 2)x + 1 = 0
1
If m = 1, 3x = 1  x= which lies in (–1, 1)
3
hence m = 1 is possible
m = – 1, x = 1 which does not lie in (– 1, 1) hence m  – 1
For m  ± 1,
f (x) = (m2 – 1)x2 – (m + 2)x + 1 = 0
 f (1) · f (–1) < 0
(m2 – 1 + 1 – m – 2) (m2 – 1 + m + 2 + 1) < 0
(m2 – m – 2) (m2 + m + 2) < 0
but m2 + m + 2 > 0  m  R
hence, m2 – m – 2 < 0
(m – 2)(m + 1) < 0  m  (–1, 2)
if f (1) = 0 then m2 – m – 2 = 0; m = 2 or m = – 1
if m = 2 then equation is
3x2 – 4x + 1 = 0  3x2 – 3x – x + 1 = 0
 3x(x – 1) – (x – 1) = 0
(x – 1)(3x – 1) = 0  x = 1/3 or = 1
hence m = 2 is also permissible.
if f (–1) = 0  m2 + m + 2 = 0  no solution
hence m  (– 1, 2] Ans.]
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MATHEMATICS
Q.12
 2 dt
dx
[Sol. x =t 
x
 t   t 
I = 2  (e  e ) cos e  e    (e  e ) cos e  e   dt
t t t t t t
 4  4
d  t  d  t  t  
I = 2   sin  e  e    sin  e  e   dt
t
 dt  4  dt  4 
  t t   t  t     t  t 
= 2 sin  e  e    sin  e  e   = 4 sin  e   cos e   
  4  4    4 
sin(e )  cos(e ) =
 
t t
t
= 4 cos(e )· 2 2 cos(e  x
) sin(e x
)  cos(e x
)  C Ans. ]
2
Q.13
[Sol. Put [x] = x – {x}
f(x) = (a – 2b + c) x2 + 2(b – c) x{x} + c{x}2
will be periodic only if
a – 2b + c = 0 & b=c
 a=b=c ]
Q.14
[Sol. Statement-1 is false as it has a period equal to 2 and statement-2 is true
2  x  1

0
 N.D. 1  x  0
f (x) =  0 0  x 1  period 2 Ans.
 1 x  2

N.D.
0 2x3
]
Q.15
k 1 k 1

h 1 h 1
[Sol. =2
 ( h  1)  ( h  1) 
(k – 1)   =±2
 h2 1 
2(k  1)
= ± 2  k – 1 = ± (h2 – 1)
h2 1
(+) ve, k – 1 = h2 – 1  y = x2 Ans.  (A)
(–) ve, k – 1 = 1 – h 2  y = 2 – x2 Ans.  (B)]
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MATHEMATICS
Q.16
 1 
n
 x 
[Sol.  2 · x1 4 
 
 x    x
n 2
n 1
 n C1 x ·  n C2 2 ·  ......
n n 1 1
C0 34
2x 4 x
n n
nC
C1 C2
now 0, , are in A.P.
.
2 4
n n ( n  1) n ( n  1) n ( n  1)
1, , ; n=1+  (n – 1) =  n=8
2 8 8 8
 1 2 x 1 4 
8
hence we have  x 


 2  

r
8 r
·x 2
· r
Tr+1 = 8 C x 4
r
2
for integral powers of x
8r r
 must be an integer
2 4
16  2r  r 16  3r
 must be an integer  r = 0, 4, 8
4 4
 3 terms with integral power  (D)]
Q.17
[Sol. x + y = k; AB = 1
k
p=
2
; 2 12  p 2 = 1  4(1 – p2) = 1  4p2 = 3
k2
4·
3 3
  k=± =±
6
=3 k2 = Ans. ]
2 2 2 2
Q.18
[Sol. We have
exy + y cos x = 2 ....(1)
Put x = 0, we get
1+y=2
 y=1
 (0, 1) lies on the given curve.
On differentiating (1) w.r.t. x, we get
 dy 
 y  + y (–sin x) + cos x
dy
 exy  x
 dx 
=0 ....(2)
dx
As (0, 1) satisfying it, we get
dy
 0+1+ =0  y'(0) = –1 Ans.  (C)
dx
Code-B Page # 5

MATHEMATICS
Again differentiating (2), we get
 d 2 y dy dy   dy   y cos x  sin x dy 
2
   xy d2y
exy  x  x dx  y  – 
dy
  + cos x 2 – sin x
dx dx  + e   dx 
=0
 dx 
2
dx dx
As (0, 1) satisfy it, we get
 –2 + 1 – 1 + y''(0) = 0
 y''(0) = 2 Ans.  (D)]
PART-B
Q.1
e x
 1  e4x  1  e 4x dx
ex
Sol. Given I = dx and J =
e2x · e x
J=  1  e 4x dx
e x (1  e 2 x )
 I+J= 
1  e4x
dx
put ex = t  exdx = dt
1
1
1
1
1 t
I+J =  dt =    1 2 dt
2
t2
 t 2 dt =
1 t4 t2  2
1
t    2
t  t
 1
t y y =  t  2  dt  dy
1

 t 
put
t
t2 1
 y2  2 2 tan 1 tan 1
dy 1 y 1
= = =
2 2 2t
 e2x  1 
tan 1  
 2e x   C
1
J+I= ....(1)  (P)
2  
1  2
1
1
1
e (1  e ) 1 t  1 
     1  2 dt  t  t  y 
x 2x 2
J–I= t 2 dt = + t
dx = dt = +
 
|||ly
1 e 4x
1 t4 t2  2
1
t    2
t  t
y 2
 y2  2 =
dy 1
 J–I =+ ln
2 2 y 2
t  2
1
1 t2  2 t 1 1 e2x  2 ex  1
J–I=  t 1 · 2 · ....(2)  (Q)
2 2 t  2 t  1 2 2 e2x  2 ex  1
ln = =
2 2 t  2
1
t
Code-B Page # 6

MATHEMATICS
(1) – (2) gives
e2x  1 1 e2x  2 ex  1
tan 1 
1
2 ex 2 2 e2x  2 ex  1
2I =
2
1  1 e 2 x  1 1 e 2 x  2 e x  1 
 tan  ·  
2 e x 2 e 2 x  2 e x  1
I= (R) ]
2 2
PART-C
Q.1

1 1
1 5 1 1  5 1 y 5 · y1 5·y
1 1
· y · y1 – ·y · y1 ·  ·
1 1 y 1
[Sol. = 2; =2
5 5 5 y 5 y
 1 1  1 1
2 2

 y 5  y  5   100 y  y 5  y  5   4  100 y
1 1 2 2
y 
10 y
y5 5
     
   
2 2
y1 y1 y1
100 y 2
4(x2 – 1) = 2  y1 (x2 – 1) = 25y2
2  (x2 – 1)2y1y2 + y1 2x = 25 · 2 · yy1
2
y1
d2y dy
(x2 – 1)y2 + xy1 = 25y or (x2 – 1) 2 +x = 25y  k = 25 Ans.]
dx dx
 n  3n  2n 1  (n 1) + n  n  2n  3  (n 1)
Q.2
[Sol. Lim
= n  n 3 3 2 2
 
n n 3  3n 2  2n  1  (n  1)3  
n n 2  2n  3  (n  1) 2 
 
= Lim + Lim
n 
(n 3  3n 2  2n  1) 2 / 3  (n  1) n 3  3n 2  2n  1  (n  1) 2 n 
n  2n  3  (n  1)
1/ 3 2
n ( n ) n ( 2)
= Lim + Lim
n  (n 3  3n 2  2n  1) 2 / 3  (n  1) (n 3  3n 2  2n  1)1 / 3  (n  1) 2 n 
n 2  2n  3  ( n  1)
1 2 1 2
=– +1=
111 11
= + ]
3 3
Code-B Page # 7