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Advanced (Paper – II) May, 2016
Test Paper Code - 8
KEYS
Q. No.
PHYSICS
Q. No.
CHEMISTRY
Q. No.
MATHEMATICS
1.
(A)
19.
(A)
37.
(B)
2.
(C)
20.
(A)
38.
(C)
3.
(C)
21.
(D)
39.
(C)
4.
(C)
22.
(A)
40.
(C)
5.
(B)
23.
(A)
41.
(B)
6.
(A)
24.
(D)
42.
(A)
7.
(A), (B), (C) and (D)
25.
(A) and (C)
43.
(B) and (C)
8.
(A) and (C)
26.
(C) / (C) and (D)
44.
(B) and (C)
9.
(C) and (D)
27.
(A), (B) and (C)
45.
(A), (C) and (D)
10.
(A), (B) and (D)
28.
(B) and (C)
46.
(A), (C) and (D)
11.
(A) and (D)
29.
(B), (C) and (D)
47.
(B) and (C)
12.
(A), (B) and (D)
30.
(B) and (D)
48.
(A) and (B)
13.
(B) and (C)
31.
(A) and (B)
49.
(A) and (D)
14.
(B)
32.
(B), (C) and (D)
50.
(B), (C) and (D)
15.
(C)
33.
(B)
51.
(B)
16.
(A)
34.
(C)
52.
(C)
17.
(B)
35.
(A)
53.
(A)
18.
(D)
36.
(B)
54.
(C)
II IIT (Advanced Paper II) May, 2016
2
Advanced (Paper – II) May, 2016
Test Paper Code - 8
Questions with Solutions
PART I : PHYSICS
Multiple choice questions with one correct alternative
1. There are two Vernier calipers both of which have 1 cm
divided into 10 equal divisions on the main scale. The
Vernier scale of one of the calipers (C1) has 10 equal
divisions that correspond to 9 main scale divisions. The
Vernier scale of the other caliper (C2) has 10 equal
divisions that correspond to 11 main scale divisions. The
readings of the two calipers are shown in the figure. The
measured values (in cm) by calipers C1 and C2,
respectively, are
(A) 2.87 and 2.83
(B) 2.87 and 2.86
(C) 2.85 and 2.82
(D) 2.87 and 2.87
Ans (A)
For vernier C1:
10 VSD = 9 MSD = 9 mm
1 VSD = 0.9 mm
 LC = 1 MSD – 1 VSD = 1 mm – 0.9 mm = 0.1 mm
Reading of C1 = MSR + (VSR)(L.C.) = 28 mm + (7) (0.1)
Reaing of C1 28.7 mm = 2.87 cm
For vernier C2: the vernier C2 is abnormal. Here 1 MSD < 1 VSD
So we have to find the reading from basics.
The point where both of the marks are matching:
distance measured from main scale = distance measured from vernier scale
28 mm + (l mm)(8) = (28 mm + x) + (1.1 mm) (7)[(28 mm + x) represents reading of C2]
solving x = 0.3 mm
So reading of C2 = 28 mm + 0.3 mm = 2.83 cm
2. The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is
given by
2
3 Z  Z  1 e
E
5 4 0 R
15
The measured masses of the neutron, 11 H, 15
7 N and 8 O are 1.008665 u, 1.007825 u, 15.000109 u and
15.003065 u, respectively. Given that the radii of both the
2
15
7
N and
15
8
O nuclei are same, 1 u = 931.5
2
MeV/c (c is the speed of light) and e /(40) = 1.44 MeV fm. Assuming that the difference between the
15
binding energies of 15
7 N and 8 O is purely due to the electrostatic energy, the radius of either of the
nuclei is (l fm = 10–15 m)
(A) 2.85 fm
II IIT (Advanced Paper II) May, 2016
(B) 3.03 fm
(C) 3.42 fm
3
(D) 3.80 fm
Ans (C)
2
3 Z  Z  1 e
E
5 4 0 R
15
n 15
7 N 11 H
8 O 


Q  M n  M O15  m N15  M H1 C 2
8
1
1
= 0.003796 × 931.5
= 3.5359 MeV
3 e2
1
E  
 8  7  7  6 
5 4 0 R
3
1
  1.44 MeV fm    14  3.5359 MeV
5
R
R = 3.42 fm
3. A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at
pressure Pi = 105 Pa and volume Vi = 10–3 m3 changes to a final state at Pf = (1/32) × 105 Pa and
Vf = 8 × 10–3 m3 in an adiabatic quasi-static process, such that P3V5 = constant. Consider another
thermodynamic process that brings the system from the same initial state to the same final state in two
steps: an isobaric expansion at Pi followed by an isochoric (isovolumetric) process at volume Vf. The
amount of heat supplied to the system in the two-step process is approximately
(A) 112 J
(B) 294 J
(C) 588 J
(D) 813 J
Ans (C)
5
PV 3  C   
5
3
Q1 = nCPT
5
 
R
PV
5
35
3
n
T 
   105  7  103    7  102   102 J
 1
 1  5 
2
2
  1
3 
 1

 1 105  8  103
P  V  32 

nR
31 8  102
93
Q2  nC V T 
T 



3 
 102 J
5
 1
 1
32
2
8
1
3
 35 93 
Q  Q1  Q2      102
8
 2
140  93
4700
Q 
 100 
 587.5 J
8
8
II IIT (Advanced Paper II) May, 2016
4
4. An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of
half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the
permissible level required for safe operation of the laboratory. What is the minimum number of days
after which the laboratory can be considered safe for use?
(A) 64
(B) 90
(C) 108
(D) 120
Ans (C)
A  A 0 2 t /TH
A
t
 0  A 0 2 t /TH  6 
 t  6TH  108 days
64
TH
5. A small object is placed 50 cm to the left of a thin convex lens of focal length 30 cm. A convex spherical
mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror
is tilted such that the axis of the mirror is at an angle  30° to the axis of the lens, as shown in the
figure.
If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of
the point (x, y) at which the image is formed are

(A) 50  25 3,25


(B) 25,25 3

(C) (0, 0)
Ans (B)
For I1
 u = –50 cm f = +30 cm
v = + 75 cm
Final image is I2
y
tan 60  3
y  50 3  3x
50  x
y  3x  50 3
A and B both options satisfy this, size of image > size of object
2
y 2   50  x  sin 30  25sin 30
y2 + (50 – x)2 > 625
625 625
for option (A)

 625
3
9
for option (B) 625(3) + 625 > 625
So (B) is correct.
II IIT (Advanced Paper II) May, 2016
5

(D) 125 / 3,25 / 3

6. The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the
wires has a length of 1 m at 10 °C. Now the end P is maintained at 10 °C, while the end S is heated and
maintained at 400 °C. The system is thermally insulated from its surroundings. If the thermal
conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of
PQ is 1.2 × 10–5 K–1, the change in length of the wire PQ is
(A) 0.78 mm
(B) 0.90 mm
(C) 1.56 mm
(D) 2.34 mm
Ans (A)
dl  dx1    10 
l   dl
   10  130

   10  130x

 x  1
1
l   130x  1dx
l  130 1
0
l  130  1.2  105 
x2
2
1
 78  105  0.78 mm
2
Multiple choice questions with one or more than one correct alternative/s
7. In the circuit shown below, the key is pressed at time t = 0. Which of
the following statement(s) is(are) true?
(A) The voltmeter displays –5 V as soon as the key is pressed, and
displays + 5 V after a long time
(B) The voltmeter will display 0 V at time t = ln 2 seconds
1
(C) The current in the ammeter becomes of the initial value after 1
e
second
(D) The current in the ammeter becomes zero after a long time
Ans (A), (B), (C) & (D)
t
t
 
 


q1   200  103  1  e 1  ;q 2  100  10 3  1  e 1 




q1
dq
  50  103  2
C
dt
3
 200 10 1  e t   50 103 100 103 e t 


 
40  106
1  e t 
106
 50  103  e  t 
20
1
 e t
t = ln2
2
I = I1 + I2 = (200 × 10-3) (e-t) + (100 × 10–3) e–t
= 100 × 10–3 [2e–t + e–t]
= (300 × 10–3) e–t
 300  103 


e


At t = , I = 0
II IIT (Advanced Paper II) May, 2016
6
8. Consider two identical galvanometers and two identical resistors with resistance R. If the internal
resistance of the galvanometers Rc < R/2, which of the following statement(s) about any one of the
galvanometers is(are) true?
(A) The maximum voltage range is obtained when all the components are connected in series
(B) The maximum voltage range is obtained when the two resistors and one galvanometer are connected
in series, and the second galvanometer is connected in parallel to the first galvanometer
(C) The maximum current range is obtained when all the components are connected in parallel
(D) The maximum current range is obtained when the two galvanometers are connected in series and the
combination is connected in parallel with both the resistors
Ans (A) & (C)
For max. Voltage vmax = I[Rg + Seq]
for max current
Here Seq will be parallel combination of two shunt and one galvanometer
9. A rigid wire loop of square shape having side of length L and resistance R is moving along the x-axis
with a constant velocity v0 in the plane of the paper. At t = 0, the right edge of the loop enters a region of
length 3L where there is a uniform magnetic field B0 into the plane of the paper, as shown in the figure.
For sufficiently large v0, the loop eventually crosses the region. Let x be the location of the right edge of
the loop. Let v(x), I(x) and F(x) represent the velocity of the loop, current in the loop, and force on the
loop, respectively, as a function of x. Counter-clockwise current is taken as positive.
Which of the following schematic plot(s) is(are) correct? (Ignore gravity)
II IIT (Advanced Paper II) May, 2016
7
(A)
(B)
(C)
(D)
`
Ans (C) & (D)
While entering i.e. x < L
B2 l 2 v
R
2 2
f Bl v
vdv
a2 
 Kv  
m
mR
dx
F  ilB 
v
x
 dv  K  dx  v  v0  kx
v0
0
2 2
f
Bl
R
 V0  kx      x
Bl
 i0  x
R
For 3L > x > L
f=0
4L > x > 3L
i   v 0  kx 
II IIT (Advanced Paper II) May, 2016
i=0
v = constant.
8
f  ilB 
B2 l 2 v
R
B2 l 2
vdv
v  kv  
mR
dx
v  v'0  kx
f      x
a
i = i0 –  x
10. A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and
moves without friction on a horizontal surface. The block oscillates with small amplitude A about an
equilibrium position x0. Consider two cases: (i) when the block is at x0; and (ii) when the block is at
x = x0 + A. In both the cases, a particle with mass m(< M) is softly placed on the block after which they
stick to each other. Which of the following statement(s) is(are) true about the motion after the mass m is
placed on the mass M?
M
(A) The amplitude of oscillation in the first case changes by a factor of
, whereas in the second
mM
case it remains unchanged
(B) The final time period of oscillation in both the cases is same
(C) The total energy decreases in both the cases
(D) The instantaneous speed at x0 of the combined masses decreases in both the cases
Ans (A), (B) & (D)
Case-1
X0
II IIT (Advanced Paper II) May, 2016
9
+
In case-1
MV1 = (M + m) V2
 M 
V2  
 V1
Mm
k
 M  k
A2  
A1

Mm
Mm M
A2 
M
A1
Mm
In case-2
A2 = A1
T  2
Mm
in both case.
K
Total energy decreases in first case where as remains same in 2nd case.
Instantaneous speed at x0 decreases in both cases.
11. While conducting the Young’s double slit experiment, a student replaced the two slits with a large
opaque plate in the x-y plane containing two small holes that act as two coherent point sources (S1, S2)
emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane
(for z > 0) at a distance D = 3 m from the
mid-point of S1S2, as shown schematically in
the figure. The distance between the sources
d = 0.6003 mm. The origin O is at the
intersection of the screen and the line
joining S1S2. Which of the following is(are)
true of the intensity pattern on the screen?
(A) The region very close to the point O will be dark
(B) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction
(C) Straight bright and dark bands parallel to the x-axis
(D) Semi circular bright and dark bands centered at point O
Ans (A) & (D)
from theory fringes will be semi circular
d
1
and  1000 

2

at 0, x  1000 
2
so at 0 it will be dark
12. In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a
7(R  r)
periodic motion is T  2
. The values of R and r are measured to be (60 ± 1) mm and
5g
(10 ± 1) mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s,
0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s.
Which of the following statement(s) is(are) true?
(A) The error in the measurement of r is 10%
(B) The error in the measurement of T is 3.57%
(C) The error in the measurement of T is 2%
(D) The error in the determined value of g is 11%
II IIT (Advanced Paper II) May, 2016
10
Ans (A), (B) & (D)
Sl.No. T
(1)
0.52
absolute error = |T – Tmean|
0.04
(2)
0.56
0.00
(3)
0.57
0.01
(4)
(5)
0.54
0.59
0.02
0.03
2.78
5
 0.56
Tmean 
Tmean
(T)mean = 0.02
0.02
 100  3.57%
0.56
According to the question
T2
g
Rr
dg 2dT dR  dr


g
T
R r
dg
11
 2  3.57%  
 100%
g
60  10
% error in T 
dg
 11%
g
13. Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a
massless, rigid rod of length l = 24 a through their centers. This assembly is laid on a firm and flat
surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod

is . The angular momentum of the entire assembly about the point ‘O’ is L (see the figure). Which of
the following statement(s) is(are) true?
(A) The magnitude of angular momentum of center of mass of the assembly about the point O is 81 ma2
(B) The magnitude of angular momentum of the assembly about its center of mass is l7ma2 /2
(C) The center of mass of the assembly rotates about the z-axis with an angular speed of /5

(D) The magnitude of the z-component of L is 55 ma2
II IIT (Advanced Paper II) May, 2016
11
Ans (B) & (C)
(D) Velocity of point P: a = l then
a

 Angular velocity of C.M. w.r.t. point O.
l
Angular velocity of C.M. w.r.t. z axis =  cos 
a 24
a 24
C.M.z 

l 5
24a 5

CM  z 
5
2
4m  2a 
ma 2
17ma 2 
(B) LD CM 


2
2
2
2
9
l
9
l
81m
l

81ml 2 a


(C) LCM  O   5m     


5
5
l
5  5
2
81mla 81 24a m
LCM  O 

5
5
(A) L Z  LCM O cos   LD CM sin 

81 24 2
24 17ma 2  1
81 24ma 2  17ma 2 
a m



5
5
2
25
24
2 24
14. Light of wavelength ph falls on a cathode plate inside a vacuum tube as shown in the figure. The work
function of the cathode surface is  and the anode is a wire mesh of conducting material kept at a
distance d from the cathode. A potential difference V is maintained between the electrodes. If the
minimum de Brogue wavelength of the electrons passing through the anode is e, which of the following
statement(s) is(are) true?
(A) e is approximately halved, if d is doubled
(B) For large potential difference (V >> /e), e is approximately halved if V is made four times
(C) e decreases with increase in  and ph
(D) e increases at the same rate as ph for ph < hc/)
Ans (B)
e = Min. de Broglie wave length
II IIT (Advanced Paper II) May, 2016
12
hc
hc
    eV
 ph
e
 1

 1

hc  2 d pn   hc   2 d e 


 e

 pn

2
d ph  ph
 2
(A wrong)
d e
e

V 
e
Energy of electron = eV
P2
h
1
E
 
2m
p
2mE
P  2mE
Read the passage given below and answer questions 15 and 16 by choosing the correct
alternative
A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial
frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant

angular velocity  is an example of a non-inertial frame of reference. The relationship between the force Frot

experienced by a particle of mass m moving on the rotating disc and the force Fin experienced by the particle



  


in an inertial frame of reference is Frot  Fin  2m  v rot    m   r    , where v rot is the velocity of the

particle in the rotating frame of reference and r is the position vector of the particle with respect to the
centre of the disc.
Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant
angular speed  about its vertical axis through its center. We assign a coordinate system with the origin at
the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the


rotation axis   k̂ . A small block of mass m is gently placed in the slot at r   R / 2  ˆi at t = 0 and is


constrained to move only along the slot.
15. The distance r of the block at time t is
R
R
(A) cos t
(B)  e 2 t  e 2 t 
2
4
Ans (C)
mr2 = ma
a = r2
II IIT (Advanced Paper II) May, 2016
(C)
13
R t
 e  et 
4
(D)
R
cos 2t
2
vdv
 r2
dr
v
r
2
 vdv    rdr
R
2
0
v   r2 
R

R
2
dr
R2
r 
4
2
Assume r 
dr 
R2
4
t
  dt
…(1)
0
R
sec 
2
R
sec  tan d
2
R
sec  tan d t
2
  dt
 R2
0
2
tan 
4
 2r
4r 2  R 2 
t  ln  

R
 R

R
r  e t  e t 
4
16. The net reaction of the disc on the block is
1
(A) m2 R  e t  et  ˆj  mgkˆ
2
1
(C) m2 R  e 2 t  e 2 t  ˆj  mgkˆ
2
Ans (A)


Frot  Fin  2m v rot ˆi  kˆ  m kˆ  riˆ  kˆ

mr2 ˆi  Fin  2mvrot  ˆj  m2 riˆ

Fin  2mv r ˆj
R
r  e t  e t 
4
dr
R
 v r  e t  e t 
dt
4
 

(B) m2 R cos tjˆ  mgkˆ
(D) m2 R sin t ˆj  mgkˆ

 
II IIT (Advanced Paper II) May, 2016
…(1)
14

R  t
e  e t  ˆj
Fin  2m
4

mR2 t
 e  e t  ˆj
Fin 
2 
Also reaction is due to disc surface then

mR2 t
e  e t  ˆj  mgkˆ
Freaction 
2
Read the passage given below and answer questions 17 and 18 by choosing the correct
alternative
Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an
insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft
material and coated with a conducting material are placed on the bottom plate. The balls have a radius r <<
h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at
+V0 and the top plate at –V0. Due to their conducting surface, the
balls will get charged, will become equipotential with the plate
and are repelled by it. The balls will eventually collide with the
top plate, where the coefficient of restitution can be taken to be
zero due to the soft nature of the material of the balls. The
electric field in the chamber can be considered to be that of a
parallel plate capacitor. Assume that there are no collisions
between the balls and the interaction between them is negligible.
(Ignore gravity)
17. Which one of the following statements is correct?
(A) The balls will bounce back to the bottom plate carrying the same charge they went up with
(B) The balls will bounce back to the bottom plate carrying the opposite charge they went up with
(C) The balls will stick to the top plate and remain there
(D) The balls will execute simple harmonic motion between the two plates
Ans (B)
The balls will bounce back to the bottom plate carrying the opposite charge they went up with.
18. The average current in the steady state registered by the ammeter in the circuit will be
1
(A) proportional to the potential V0
(B) proportional to V02
(C) zero
(D) proportional to V02
Ans (D)
II IIT (Advanced Paper II) May, 2016
15
Vr
Kq
 V0  q  0
r
K
1  qE  2

t  h
2 m 
1 V0 r 2V0 2
t h
2 K hm
h 2 mk
t2  2
V0 r
t
h
V0
mk
r
During every collision 2q charge will flow from circuit
2q 2V02 r r
Average current Iavg 

t
h mk k
2
Iavg  V0
II IIT (Advanced Paper II) May, 2016
16
PART II: CHEMISTRY
Multiple choice questions with one correct alternative
19. The geometries of the ammonia complexes of Ni2+ , Pt2+ and Zn2+, respectively, are
(A) octahedral, square planar and tetrahedral
(B) square planar, octahedral and tetrahedral
(C) tetrahedral, square planar and octahedral
(D) octahedral, tetrahedral and square planar
Ans (A)
[Ni(NH3)6]2+
octahedral
2+
[Pt(NH3)4]
square planar
2+
[Zn(NH3)4]
tetrahedral
20. In the following reaction sequence in aqueous solution, the species X, Y and Z respectively, are
(A) [Ag(S2O3)2]3, Ag2S2O3, Ag2S
(C) [Ag(SO3)2]3, Ag2S2O3, Ag
Ans (A)
2S2 O32   Ag   [Ag(S2O 3 ) 2 ]3 soluble complex
(B) [Ag(S2O3)3]5, Ag2SO3, Ag2S
(D) [Ag(SO3)3]3, Ag2SO4, Ag
[Ag(S2O 3 ) 2 ]3  Ag   Ag 2 (S2O 3 ) 2   S2O3
white ppt
The precipitate is unstable, turning dark on standing, when silver sulphide is formed.
Ag 2S2O 3   H 2O  Ag 2S   2H   SO42 
X  [Ag(S2O3)2]3
Y  [Ag2(S2O3)]
Z  Ag2S
21. For the following electrochemical cell at 298 K,
Pt(s) | H2(g, 1 bar)] H+(aq, 1 M) || M4+ (aq), M2+ (aq) | Pt(s)
[M 2  (aq)]
E cell  0.092 V when
 10x .
4
[M (aq)]
RT
Given: E oM4  /M 2  0.151 V; 2.303
 0.059 V
F
The value of x is
(A) 2
(B) 1
(C) 1
Ans (D)
Emf  Ecathode  E anode  E M  E H 2  0.092
Emf  E oM  E oH 2 
0.0591
[M 2 ]
log10
2
[M 4 ]
2
[M 2 ]  0.151  0.059 
log 4  
 2
[M ] 
0.059

[M 2 ]
 102
4
[M ]
II IIT (Advanced Paper II) May, 2016
17
(D) 2
22. The correct order of acidity for the following compound is
(A) I > II > III > IV
(B) III > I > II > IV
(C) III > IV > II > I
(D) I > III > IV > II
Ans (A)
The conjugate base of compounds I and II are stabilized by intramolecular hydrogen bond.
The conjugate base is destabilized by +R effect. The order of acidity is I > II > III > IV.
23. The major product of the following reaction sequence is
(A)
(B)
(C)
(D)
Ans (A)
O
..
H
HO
H2O
O
OH
O
..
OH
(i) HCHO
Cannizaro's
reaction with HCHO
(ii) H2O
HCO2H
OH
HCHO/H+
O
O
O
OH
OH
H2O
24. The qualitative sketches I, II and III given below show the variation of surface tension with molar
concentration of three different aqueous solutions of KCl, CH3OH and CH 3 (CH 2 )11 OSO 3 Na  at room
temperature. The correct assignment of the sketches is
II IIT (Advanced Paper II) May, 2016
18
(A)
II : CH3OH
III : CH 3 (CH 2 )11 OSO 3 Na 
III : KCl
(B)
I : KCl
I : CH 3 (CH 2 )11 OSO 3 Na 
(C)
I : KCl
II : CH3OH
II : CH 3 (CH 2 )11 OSO 3 Na 
(D)
I : CH3OH
II : KCl
III : CH3OH
III : CH 3 (CH 2 )11 OSO 3 Na 
Ans (D)
Strong electrolyte KCl
CH3OH
surfactent
Multiple choice questions with one or more than one correct alternative/s
25. According to Molecular Orbital Theory
(A) C22  is expected to be diamagnetic
(B) O22  is expected to have a longer bond length than O2
(C) N 2 and N 2 have the same bond order
(D) He 2 has the same energy as two isolated He atoms
Ans (A) and (C)
2
Total 14 electron C22  1s2 1s* 2 2s2 *2s2 ( 2p 2  2p2 ) 2p
all electrons are paired.
2
x
y
Nb  Na 9  4 5

  2.5
2
2
2
10  5 5
N 2 bond order 
  2.5
2
2
(B) O22  has shorter bond length than O2
N 2 bond order 
(D) He 2 has less energy than isolated He atoms
2 1
1
1s2 1s* 1 bond order

2
2
26. Reagent(s) which can be used to bring about the following transformation is(are)
(A) LiAlH4 in (C2H5)2O
(B) BH3 in THF
(C) NaBH4 in C2H5OH
(D) Raney Ni/H2 in THF
Ans (C) / (C) and (D)
(C) Aldehyde is reduced by NaBH4 in C2H5OH
(D) Raney Ni/H2 in THF  in some specified conditions may bring the above conversion.
II IIT (Advanced Paper II) May, 2016
19
27. Extraction of copper from copper pyrite (CuFeS2) involves
(A) crushing followed by concentration of the ore by froth-flotation
(B) removal of iron as slag
(C) self-reduction step to produce ‘blister copper’ following evolution of SO2
(D) refining of ‘blister copper’ by carbon reduction
Ans (A), (B) and (C)
A : Sulphide ores are concentrated by forth floatation method
B : 2CuFeS2 + O2  Cu2S + 2FeS + SO2
2FeS + O2  2FeO + 2SO2
FeO + SiO2  FeSiO3 (slag)
C : 2Cu2S + 3O2  2Cu2O + 2SO2
2CuO + Cu2S  6Cu + SO2
28. For ‘invert sugar’, the correct statement(s) is(are)
(Given: specific rotations of (+)-sucrose, (+)-maltose, L-()-glucose and L-(+)-fructose in aqueous
solution are +66, +140, 52 and +92, respectively)
(A) ‘invert sugar’ is prepared by acid catalyzed hydrolysis of maltose
(B) ‘invert sugar’ is an equimolar mixture of D-(+)-glucose and D-()-fructose
(C) specific rotation of ‘invert sugar’ is 20
(D) on reaction with Br2 water, ‘invert sugar’ forms saccharic acid as one of the products
Ans (B) and (C)
(C) 1 sugar  ½ dextrose + ½ fructose
52
92
(50  52)  (50  92)
40
Specific rotation of the mixture 
   20
100
2
29. Among the following, reaction(s) which gives (give) tert-butyl benzene as the major product is(are)
(A)
(B)
(C)
(D)
Ans (B), (C) and (D)
The electrophiles generated in reaction (B) and (C)
(B)
Cl
AlCl3
AlCl4
+
+
+
+
II IIT (Advanced Paper II) May, 2016
20
+ H2SO4
(C)
+
+
+
(D) BF3.OEt2 +
+
OH
O
..
BF3
H
+
1,2-shift
..
+ +
BF3(OH)
+
+
30. The nitrogen containing compound produced in the reaction of HNO3 with P4O10
(A) can also be prepared by reaction of P4 and HNO3
(B) is diamagnetic
(C) contains one N-N bond
(D) reacts with Na metal producing a brown gas
Ans (B) and (D)
N2O5 + Na  NaNO3 + NO2
HNO3 + P4O10  4HPO3 + 2N2O5
O
O
N
O
O
N
O
It does not contain NN
31. Mixture(s) showing positive deviation from Raoult’s law at 35C is(are)
(A) carbon tetrachloride + methanol
(B) carbon disulphide + acetone
(C) benzene + toluene
(D) phenol + aniline
Ans (A) and (B)
32. The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is(are)
(A) The number of the nearest neighbours of an atom present in the topmost layer is 12
(B) The efficiency of atom packing is 74 %
(C) The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively
(D) The unit cell edge length is 2 2 times the radius of the atom
Ans (B), (C) and (D)
II IIT (Advanced Paper II) May, 2016
21
Read the passage given below and answer questions 33 and 34 by choosing the correct
alternative
Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following
equation:
The standard reaction Gibbs energy, rG, of this reaction is positive. At the start of the reaction, there is
one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by . Thus,
equilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total
pressure of 2 bar. Consider the gases to behave ideally. (Given: R = 0.083 L bar K1 mol1)
33. The equilibrium constant KP for this reaction at 298 K, in terms of equilibrium , is
2
2
8equilibrium
8equilibrium
42equilibrium
42equilibrium
(A)
(B)
(C)
(D)
2
2
2  equilibrium
4  equilibrium
2  equilibrium
4  equilibrium
Ans (B)
 2X(g)
X 2 (g) 
[X2]
[X]
0
decomposed/moles
1

2
moles remaining at
equil.
1
Initial no moles

2

Total number of moles at equilibrium = 1 
pX 2
pX 2


   1
2
2
 
1  
  
  2   2 pX  
2


1  
1



 2
 2
2(2  )
4

; pX 
(2  )
(2  )
2
 4 
 (2  ) 
16 2
 

2(2  )
2(2  )(2  )
(2  )
kp 
p2X
p X2
Kp 
82
4  2
34. The INCORECT statement among the following for this reaction, is
(A) Decrease in the total pressure will result in formation of more moles of gaseous X
(B) At the start of the reaction, dissociation of gaseous X2 takes place spontaneously
(C) equilibrium = 0.7
(D) KC < 1
Ans (C)
8 2
8  (0.7) 2
C is incorrect.  upon substituting  = 0.7 in k p 

 1.1168  1
2
4 
4  (0.7) 2
But for G to be positive, kp has to be less than 1.
So, option ‘C’ is incorrect.
II IIT (Advanced Paper II) May, 2016
22
Read the passage given below and answer questions 35 and 36 by choosing the correct
alternative
Treatment of compound O with KMnO4/H+ gave P, which no heating with ammonia gave Q. The compound
Q on treatment with Br2/NaOH produced R. On strong heating. Q gave S, which on further treatment with
ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T.
35. The compound R is
(A)
36. The compound T is
(A) glycine
(B)
(C)
(D)
(B) alanine
(C) valine
(D) serine
Solutions to passage questions
O
O
O
(O)
NH2
O
(P)
O
alanine (T)
+
O
O
OH
H3O+
heat
HO
HBr
O
OH
OH
O
35. (A)
36. (B)
23
(R)
O
N
..
O
O
II IIT (Advanced Paper II) May, 2016
(Q)
O
O
N
NH2
NH3
Br
NH2
NH2
NH2
OH
OH
O
H
PART III: MATHEMATICS
Multiple choice questions with one correct alternative
 1 0 0
37. Let P   4 1 0  and I be the identity matrix of order 3. If Q = [qij] is a matrix such that P50 Q = I,
16 4 1 
q  q 32
then 31
equals
q 21
(A) 52
(B) 103
Ans (B)
0
0
 1 0 0
 1



1
P   4 1 0   4 1
1
0
16 4 1 
 4  4 4  1 1 
0
0
 1 0 0
 1



2
P   8 1 0  4  2
1
0 
 48 8 1 
 8  6 4  2 1 
(C) 201
(D) 205
0
0
 1 0 0
 1



P  12 1 0    4  3
1
0 
96 12 1 
12  8 4  3 1 
0 0
0
0
 1
 1



4
P   16 1 0    4  4
1
0
160 16 1 
16  10 4  4 1 
0 0
0
0
 1
 1



5
P   20 1 0    4  5
1
0 
240 20 1 
20  12 4  5 1 
3
 1
P   4n
 4n  a n
n
0
0
1 0
4n 1 
an  nth term of A.P. 4, 6, 8, 10, 12, . . .
an = 4 + (n  1) × 2
0 0
 1

 P   200
1 0
 20400 200 1 
50
a50 = 4 + 49 × 2 = 4 + 98 = 102
Given P50  Q = I  Q = P50  I
0 0
 0

 Q   200
0 0
20400 200 0
q31  q 32 20400  200 20600


 103
q 21
200
200
38. Let P be the image of the point (3, 1, 7) with respect to the plane x  y + z = 3. Then the equation of the
x y z
plane passing through P and containing the straight line   is
1 2 1
(A) x + y  3z = 0
(B) 3x + z = 0
(C) x  4y + 7z = 0
(D) 2x  y = 0
Ans (C)
II IIT (Advanced Paper II) May, 2016
24
Let A = (3, 1, 7)
A
E:xy+z=3
Putting A in E  3  1 + 7  3 = 6  0
E

‘A’ lies outside E
For the image of A(x1, y1, z1) w.r.t. E : ax + by + cz + d0 = 0
P
x  x1 y  y1 z  z1 2(ax1  by1  cz1  d)



a
b
c
a 2  b 2  c2
x  3 y  1 z  7 2(3  1  7  3)




1
1
1
11 1
x  3 y 1 z  7



 4
1
1
1
x  3 =  4,
y  1 = 4,
z  7 =  4  x = 1, y = 5, z = 3
P(1, 5, 3)
Equation of a plane  through the points origin, P and parallel to the vector (1, 2, 1) direction of the
x y z
given line can be taken as 1 5 3  0
1

a
2 1
c
 x(5  6)  y(1  3) + z (2  5) = 0
  x + 4y  72 = 0
 x  4y + 7z = 0


P
 E
 R(r)
c
(1, 2, 1)

39. Area of the region (x, y)  R 2 : y  | x  3 | , 5y  x  9  15 is equal to
1
6
Ans (C)
(A)
(B)
3
A1 

 x  3 dx 
4
1
A2 

3
x  3 dx 
4
3
(C)
3
2
(D)
2
3
16
3
1
15
Area of the trapezium  (1  2)5 
2
2
15
 A 3   A1  A 2
2
15 2 16 3
   
2 3 3 2
y2 = x + 3
(1, 2)
(6, 3)
y2 = x  3
(4, 1)
A3
A1
4 3
II IIT (Advanced Paper II) May, 2016
A2
1
25
x  5y + 9 = 0
5
3
13
40. The value of
1
is equal to
  (k  1)  
  k 
k 1
sin  
sin




6 
4
4 6 

(B) 2  3  3 
(A) 3  3
(C) 2  3  1
(D) 2  2  3 
Ans (C)
13

1
  k 1 
  k 
sin i  
  sin   
4
6


4 6 
1
1
1
1



 . . .
 sin 5
5
7
7
3
 9 
 29 
sin .
sin sin
sin . sin
sin   sin 

4 12
12
12
12
4
 4 
 12 
k 1


1
1
1
1


2
2
2
2

2 


... 
 9 
 29  
 sin  . sin 5 sin 5 .sin 7  sin 7  .sin 3
sin   sin 


4
12
12
12
12
4
 4 
 12  
  5  
 7  5 
 3 7 
 29 9  
sin 
  
 sin  12  4  sin  12  12  sin  4  12 
4 




  ... 
 12

 2 

5

5

7

7

3

9

29
 



 sin sin
sin .sin
sin .sin
sin   sin 


4
12
12
12
12
4
 4   12  

5


5
7
5
29
9
29
 
 5
sin
cos  cos
.sin 
 sin 12 cos 4  cos 12 sin 4 sin 12 cos 12  cos 12 sin 12
12
4
12
4
 2

... 


5
5
7
9
29


sin . sin
sin .sin
sin .sin

4
12
12
12
4
12

5
5
7
9
29 
 
 2 cot  cot
 cot
 cot
 . . .  cot
 cot
12
12
12
4
12 
 4
29 
 
 2  cot  cot
 2 1   2  3    2 1  2  3   2  3  1
12 
 4
41. Let bi > 1 for i = 1, 2, . . ., 101. Suppose loge b1, loge b2, . . . , loge b101 are in Arithmetic Progression
(A.P.) with the common difference loge 2. Suppose a1, a2, . . ., a101 are in A.P. such that a1 = b1 and
a51 = b51. If t = b1 + b2 + . . . + b51 and s = a1 + a2 + . . . + a51, then
(A) s > t and a101 > b101
(B) s > t and a101 < b101 (C) s < t and a101 > b101 (D) s < t and a101 < b101
Ans (B)
loge b1, loge b2, . . . loge b101
A.P. (Common difference = loge 2)
 b1, b2, . . . , b101
G.P. (common ratio = 2)
b
 Common ratio r  2  2
b1
 S = a1 + a2 + . . . + a51
51
  a1  a 51 
2
51
  b1  b51 
2
51
 b1  b1 .250 

2 
II IIT (Advanced Paper II) May, 2016
t = b1+ b2 + . . . + b51
t = b1 . (251  1)
From (1) and (2)
26
…(2)
51
.b1 (250  1)
2
51
S  .b1 (251  2)
4
Next, a101 = a1+ 100 d
= b1 + 100 d
S
 b1  100.
S>t
…(1)
a51 = b51
 a1 + 50 d = b1(250)
b1 (250  1)
50
 b1 + 50 d = b1.250
= b1 + b1.251  2b1
 d
a101 = b1.251  b1 = b1 (251  1)
i.e., a101 = b1 (251  1)
b101 = b1.2100
From (3) and (4), a101 < b101
b1 (250  1)
50
…(3)
…(4)

2
42. The value of
x 2 cos x
 1  e x dx is equal to

2
2

2
4
Ans (A)
(A)
(B)
2
2
4


(C) 2  e 2
(D) 2  e 2

2
I
x 2 cos x
 1  ex dx

2

2
x 2 cos x
I 
dx 
x
 1 e

2

2
x 2 cos x
 1 dx
1  x

2
e

2
I
e x .x 2 cos x
 1  ex dx

2

2
2I 
(1  e x )x 2 cos x
 1  ex dx

2

2
2I 
x

2
cos xdx

2

2
2I  2  x 2 cos xdx
0

2
0
I   x 2 ( sin x)  

2
 2x( sin x)dx 
0

2
 2  x(  cos x)   ( cos x)dx  2

0
4
 2


2
 2(x cos x)0 /2  2(sin x)0 /2 
 2(0)  2    2 
4
4
 4

2

II IIT (Advanced Paper II) May, 2016
27
Multiple choice questions with one or more than one correct alternative/s
43. Let
 1
f :  ,
 2

 1 
2  R and g :   , 2  R

 2 
be functions defined by f(x) = [x2  3] and
g(x) = | x | f(x) + | 4x  7 | f(x), where [y] denotes the greatest integer less than or equal to y for y  R.
Then
 1 
(A) f is discontinuous exactly at three points in   , 2
 2 
 1 
(B) f is discontinuous exactly at four points in   , 2
 2 
 1 
(C) g is NOT differentiable exactly at four points in   , 2 
 2 
 1 
(D) g is NOT differentiable exactly at five points in   , 2 
 2 
Ans (B), (C)
 1 
f :  , 2   R
 2 
f(x) = [x2  3]
1
 x2
2
1
 x2  2
4
11
  x2  3  1
4
2
[x  3] is discontinuous at integral points i.e.,  2,  1, 0, 1
 1 
 f is discontinuous exactly at four points in   , 2
 2 
g(x) = | x | f(x) + | 4x  7 | f(x)
= | x | [x2  3] + | 4x  7 | [x2  3]
1

  x 1
 3 | x | 3 | 4x  7 |
2

 2 | x | 2 | 4x  7 | 1  x  2

  | x |  | 4x  7 |
2 x 3


0
3  x2

 3x  3(4x  7)

 3x  3(4x  7)

2x  2(4x  7)
   x  (4x  7)


0



0

II IIT (Advanced Paper II) May, 2016
1
 x0
2
0  x 1
1x  2
2  x 3
3 x
7
4
7
 x2
4
28
1

x0
15x  21,
2

0  x 1
 9x  21,

1 x  2
 6x  14
g(x)   3x  7
2 x  3

7

0
3x

4

7

0
 x2

4
1

x0
15,
2

0  x 1
 9,

1x  2
 6,
g(x)   3,
2 x 3

7
 0,
3 x

4

7
 0,
 x2

4
Hence, g is not differentiable exactly at four points  0, 1,
2, 3  .
1 ˆ ˆ
ˆ
 i  j  2kˆ  . Given that there exists a vector v
44. Let û  u1ˆi  u 2 ˆj  u 3 kˆ be a unit vector in R3 and 
6


3
ˆ . uˆ  v   1. Which of the following statement(s) is/are correct?
in R such that | uˆ  v |  1 and w

(A) There is exactly one choice for such v

(B) There are infinitely many choices for such v
(C) If û lies in the xy-plane then | u1 | = | u2 |
(D) If û lies in the xz-plane then 2 | u1 | = | u3 |
Ans (B)

(A) u  (v1 , v 2 , v3 )

1
w
(1, 1, 2)
6
 
  
| u  v | 1
w. (u  v)  1
 


| u | | v | sin   1, | w |  1, | u |  1

| v | sin   1
  
. (u  v)  1
  
 | w | | u  v | cos   1

 
 | w | | u | | v | sin   cos 


Since | u |  | w |  1
 
(u  v)
…(1)
u
…(2)
From (1) we have cos  = 1   = 0

 
 w is along  u  v 
  
(B) Since   u  v is normal to the plane of u, v,
‘v’ can be taken as any arbitrary vector

 there exists infinitely many choices for v
II IIT (Advanced Paper II) May, 2016

w

29

v

1 ˆ 1 ˆ


(C) If we can take u  
i
j   u1    u 2 
2
2

1 ˆ 2 ˆ


If we can take u  
i
j
We can have | u 3 |  2 | u1 |
3
3



But in general | u |  1  doesnot imply | u1 | | u 2 |
 3 ˆ 1 ˆ


If v is in xy planes : v   
i  j
 2
2 


(D) If u is z-x plane
u  iu1  ku 3
 1 ˆ 2 
| u |  1 for   iˆ
j

3
 3


But in general | u |  | iv1  kv3 |  1 does not imiply

(u 3 )  2 | u1 |
1


45. Let a, b  R and a2 + b2  0. Suppose S   z  C: z 
, t  R , t  0  , where i  1 . If z = x + iy
a  ibt


and z  S, then (x, y) lies on
1
 1

(A) the circle with radius
and centre  , 0  for a > 0, b  0
2a
 2a 
1
 1

(B) the circle with radius 
and centre   , 0  for a < 0, b  0
2a
 2a

(C) the x-axis for a  0, b = 0
(D) the y-axis for a = 0, b  0
Ans (A), (C), (D)
a  ibt
z  x  iy  2
a  b2 t 2
Comparing real and imaginary parts
a
bt
x 2
, y 2
2 2
a b t
a  b2 t 2
2
1 
1

(A) For a > 0,  x    (y  0) 2  2
2a 
4a

x
 x 2  y2   0
a
x
1
1
 x 2  y2   2
 2
2 2
a (a  b t ) a  b 2 t 2
1
1
 1 
 x  2x,
 2  y2   
2a 4a
 2a 
2
2
2
2
1 

 1 
  x    y2   
2a 

 2a 
1
 1

Circle of radius
and centre at  , 0 
2a
2a


(B) Let a < 0,
Let a =  a1 (a1 > 0)
x
2x
1
1
x 2  y2   0  x 2 
 y2  2  2
a1
2a1
4a
4a
II IIT (Advanced Paper II) May, 2016
(a > 0)
30
2
1 

 1 
  x  2   y2   2 
2a 

 2a 
2
2
2
1 

 1 
  x    y2    
2a 

 2a 
 1

 1 
Circle with centre  , 0  and radius    when a < 0
 2a 
 2a 
(C) a  0, b = 0
we have from I,
y = 0  x-axis
(D) a = 0, b  0,
we have from I,
x = 0  y-axis
2
46. Let P be the point on the parabola y = 4x which is at the shortest distance from the center S of the circle
x2 + y2  4x  16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally.
Then
(A) SP  2 5
(B) SQ : QP   5  1 : 2
(C) the x-intercept of the normal to the parabola at P is 6
1
(D) the slope of the tangent to the circle at Q is
2
Ans (A), (C), (D)
y2 = 4x
x2 + y2  4x  16y + 64 = 0
S  (2, 8) r  4  64  64  2
Point on parabola P   t 2 , 2t 
8  2t
2  t2
2t  t 3  8  2t
Slope of normal  t 
t=2
P  (4, 4)
SP  (4  2)2  (4  8) 2  20
Slope of normal at P(4, 4) m =  2
Equation of normal 2x + y = 12
x-intercept = 6
Slope of SQ =  2
1
Slope of tangent at Q = Q 
2
x

n
n
n


n n (x  n)  x   . . .  x  


2
n


 for all x > 0. Then
47. Let f (x)  lim 
2
2 
n  
 2 n 
n 
2
2  2
 n! (x  n )  x   . . .  x  2  
4
n 




f (3) f (2)
1
1
2
(A) f    f (1)
(B) f    f  
(C) f  (2)  0
(D)

f (3)
f (2)
2
 3
3
Ans (B) & (C)
II IIT (Advanced Paper II) May, 2016
31
x

n
x x 1
x 1
n 2n   1    . . .   


n  n 2
n n

f (x)  lt 
2
n 
  x2 1 
 x2 1  

2n  x
 n! n  n 2  1  n 2  22  . . .  n 2  n 2  

 






x  x 1 x 1
 1    . . .   



x
n  n 2 n n

ln f (x)  lt
ln 
2
n  n
  x2 1 
 x2 2  
 x
 n!  n 2  1  n 2  22  . . .  n 2  n 2  
 



 
  x   2x 
 nx  
 1 . 
 1 . . . 
 1 


x
n   n
n



 

 lt
ln
2
2 2
2
n  n
 2 x

 nx

 x
  n 2  1  n 2  1 . . .  n 2  1  
 




 lt
n 
 r2 x2

x  n
 rx  n
 l n   1   l n  2  1 
n  r 1  n
 r 1  n

1
 x  l n (xt  1)  l n x 2 t 2  1)  dt
0
1
 xt  1 
 x l n  2 2
 dt
 x t 1 
0
x
x
l n(f (x))   l n(1  x) du   l n(1  x 2 )dx
0
xt = u, xdt = du
u=0
LL
u=xL
0
f (x)
 1 x 
 l n
2 
f (x)
1 x 
f (2)
 3
 l n    0  f (2)  0
f (2)
5
f (3)
 4
2
 l n   l n  
f (3)
 10 
5
f (3) f (2)


f (3) f (2)
f (x)
f (x)
Also,
 0 in (0, 1) and
 0 in (1, )
f (s)
f (x)
 f(x) is increasing in (0, 1) and decreasing in [1, )
( f(x) is positive)
1
1
2
 f    f (1) and f    f  
2
3
 3
48. Let a, b  R and f : R  R be defined by f(x) = a cos (| x3  x |) + b | x | sin (| x3 + x | ). Then f is
(A) differentiable at x = 0 if a = 0 and b = 1
(B) differentiable at x = 1 if a = 1 and b = 0
(C) NOT differentiable at x = 0 if a = 1 and b = 0
(D) NOT differentiable at x = 1 if a = 1 and b = 1
Ans (A), (D)
f(x) = a cos (x3  x) + bx sin (x3 + x)
At x = 0
II IIT (Advanced Paper II) May, 2016
32
f (0  h)  f (0)
h
a cos( h 3  h)  bh sin(  h 3  h)  a
a cos(h 3  h)  bh sin(h 3  h)  a
 lt
 lt
h 0
h 0
h
h
2
 a cos(h(h  1))
a
 lt 
 bsin(h 3  h)  
h 0 
h
h
f (0  h)  f (0)
R.H.D.  lt
h 0
h
a cos(h 3  h)  bh sin(h 3  h)  a
 lt
h 0
h
3
 a cos(h  h)
a
 lt 
 bsin(h 3  h)  
h 0
h
h

L.H.D.  lt
h 0
 f(x) is differentiable at x = 0 if a = 0 and b = 1
At x = 1
f (1  h)  f (1)
L.H.D.  lt
h 0
h
a cos((1  h)3  (1  h))  b(1  h)sin((1  h)3  (1  h))  (a  bsin 2)
 lt
h 0
h
 a cos((1  h)3  (1  h) b(1  h) sin((1  h)3  (1  h) a  bsin 2 
 lt 



h 0
h
h
h


f (1  h)  f (1)
h
a cos ((1  h)3  (1  h))  b(1  h)sin ((1  h)3  (1  h))  (a  bsin 2)
 lt
h 0
h
3
 a cos(1  h)  (1  h) bsin((1  h)3  1  h)
(a  bsin 2) 
 lt 

 bsin((1  h)3  (1  h) 

h 0
h
h
h


R.H.D.  lt
h 0
 f(x) is not differentiable at x = 1 if a = 1 and b = 1
49. Let f : R  (0, ) and g : R  R be twice differentiable functions such that f  and g are continuous
f (x) g(x)
functions on R. Suppose f (2) = g(2) = 0, f (2)  0 and g(2)  0. If lim
 1, then
x  2 f (x) g (x)
(A) f has a local minimum at x = 2
(B) f has a local maximum at x = 2
(C) f (2) > f(2)
(D) f(x)  f (x) = 0 for atleast one x  R
Ans (A), (D)
f (x)g(x)
f (x)g (x)  g(x)f (x)
lim
 lim
x  2 f (x)g (x)
x  2 f (x)g (x)  f (x)g (x)
f (2)g(2)  g(2)f (2)
f (2) g(2) f (2)



g(2)  0 
f (2)g(2)  f (2)g(2) f (2)g(2) f (2)
f (2)

 1  f (2)  f (2)  0 and f (2)  0
f (2)
 f has a local minimum at x = 2.
Hence (A) is correct and (B) is wrong.
Since f (2)  f (2),
(C) is wrong
f(x)  f (x) = 0 for x = 2
II IIT (Advanced Paper II) May, 2016
 (D) is correct.
33
50. Let a, ,   R. Consider the system of linear equations
ax + 2y = 
3x  2y = 
Which of the following statement(s) is/are correct?
(A) If a = 3, then the system has infinitely many solutions for all values of  and 
(B) If a  3, then the system has a unique solution for all values of  and 
(C) If  +  = 0, then the system has infinitely many solutions for a =  3
(D) If  +   0, then the system has no solution for a = 3
Ans (B), (C), (D)
When a =  3, then 3x + 2y = 
3x  2y = 
If   , the system cannot have infinite solutions.
So, (B) is correct.
a 2
For unique solution
 0  a  3
 (C) is correct.
3 2
If  +   0, then the equations become inconsistent when a = 3.
Hence, no solution exists.
 (D) is correct.
Read the passage given below and answer questions 51 and 52 by choosing the correct
alternative
Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes of the
two games are independent. The probabilities of T1 winning, drawing and losing a game against T2 are
1 1
1
, and , respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in
2 6
3
game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two games.
51. P(X > Y) is
1
5
1
7
(A)
(B)
(C)
(D)
4
12
2
12
Ans (B)
Probability that the score of T1 > probability that the score of T2.
= (T1 wins in the 1st game) and (T1 wins in the 2nd game)
+ (draw in the 1st game) and (T1 wins in the 2nd game)
+ (T1 wins in 1st game) and (draw in the 2nd game)
1 1 1 1 1 1 5
 .  .  . 
2 2 6 2 2 6 12
52. P(X = Y) is
11
1
13
1
(A)
(B)
(C)
(D)
36
3
36
2
Ans (C)
Similarly P(X = Y) = (draw in both games) + (T1 wins in 1st game and T2 wins in 2nd game)
+ (T1loses in 1st game and T1 wins in the 1st game)
1 1 1 1 1 1 13
 .  .  . 
6 6 2 3 3 2 36
II IIT (Advanced Paper II) May, 2016
34
Read the passage given below and answer questions 53 and 54 by choosing the correct
alternative
x 2 y2

 1 . Suppose a parabola
9
8
having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N
in the fourth quadrant.
53. The orthocentre of the triangle F1MN is
 9 
2 
 9 
2

(A)   , 0 
(B)  , 0 
(C)  , 0 
(D)  , 6 
 10 
3 
 10 
3

Ans (A)
x 2 y2
1

 1, e 
9
8
3
F1  (1, 0), F2  (1, 0)
Equation of parabola y2 = 4x
Point of intersection of parabola and ellipse
x 2 4x

1
9
8
2x2 + 9x  18 = 0
3
x  or x  6
2
3
x
2
3

3

y   6  M   , 6 , N   ,  6 
2

2

2 6
Slope of F1 N 
5
5 
3
Equation of normal through M y  6 
x  
2
2 6
15
2 6y  12  5x 
2
9
On x-axis y = 0, x  
10
54. If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis
at Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2 is
(A) 3 : 4
(B) 4 : 5
(C) 5 : 8
(D) 2 : 3
Ans (C)
1
Area of MF1F2 
6 6 0  6
2
Area of quadrilateral MF1 NF2  2 6
Let F1(x1, 0) and F2(x2, 0), for x1< 0 and x2 > 0, be the foci of the ellipse
y2 = 4x
Slope of normal  
6
2
Equation of normal 2y  2 6   x 6 
II IIT (Advanced Paper II) May, 2016
3 6
2
3

tangent at  , 6 
2

3
x
2  2 6y  1
9
8
35
x 6  2y 
7
6
2
7 
Q  , 0
2 
x
6y

 1 on x-axis
6
4
x=6
R  (6, 0)
1 7
6  6 6 
2 2
15
 5
 
6
6
2 2
 4
Area of MQR 
5
6
Area of MQR
5
4

Area of MF1 NF2 2 6 8
***
II IIT (Advanced Paper II) May, 2016
36