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FIITJEE Solutions to AIEEE – 2009
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AIEEE–2009, BOOKLET CODE(A)
Note: (i) The test is of 3 hours duration.
(ii) The test consists of 90 questions. The maximum marks are 432.
(iii) There are three parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct
response.
Part A − Physics (144 marks) − Question No. 1 to 2 and 9 to 30 consists FOUR (4) marks each and Question No. 3
to 8 consists EIGHT (8) marks each for each correct response.
Part B − Chemistry (144 marks) − Question No. 31 to 39 and 46 to 60 consists FOUR (4) marks each and Question
No. 40 to 45 consists EIGHT (8) marks each for each correct response.
Part C − Mathematics(144 marks) − Question No. 61 to 82 and 89 to 90 consists FOUR (4) marks each and Question
No. 83 to 88 consists EIGHT (8) marks each for each correct response.
(iv) Candidates will be awarded marks as stated above for correct response of each question. 1/4th marks will be deducted for
indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an
item in the answer sheet.
(v) * marked questions are from syllabus of class XI CBSE.
Physics
PART − A
1. This question contains Statement1 and Statement2. Of the four choices given after the statements,
choose the one that best describes the two statements.
Statement – 1: For a charged particle moving from point P to point Q, the net work done by an
electrostatic field on the particle is independent of the path connecting point P to point Q.
Statement2: The net work done by a conservative force on an object moving along a closed loop is
zero
(1) Statement1 is true, Statement2 is false
(2) Statement1 is true, Statement2 is true; Statement2 is the correct explanation of Statement1.
(3) Statement1 is true, Statement2 is true; Statement2 is not the correct explanation of Statement1.
(4) Statement1 is false, Statement2 is true
Sol: (2)
Work done by conservative force does not
depend on the path. Electrostatic force is a
conservative force.
2. The above is a plot of binding energy per nucleon
Eb, against the nuclear mass M; A, B, C, D, E, F
correspond to different nuclei. Consider four
reactions:
(i) A + B → C + ε (ii) C → A + B + ε
(iii) D + E → F + ε and (iv) F → D + E + ε
where ε is the energy released? In which reactions
is ε positive?
(1) (i) and (iv) (2) (i) and (iii)
(3) (ii) and (iv) (4) (ii) and (iii)
Sol: (1)
1st reaction is fusion and 4th reaction is fission.
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3. A pn junction (D) shown in the figure can act
as a rectifier. An alternating current source
(V) is connected in the circuit.
(1)
(2)
(3)
(4)
Sol: (3)
Given figure is half wave rectifier
4. The logic circuit shown below has the input waveforms ‘A’ and ‘B’ as shown. Pick out the correct
output waveform.
(1)
(2)
(3) (4)
Sol: (1)
Truth Table
A B Y
1 1 1
1 0 0
0 1 0
0 0 0
*5. If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple
harmonic motion of time period T, then, which of the following does not change with time?
(1) a2T2 + 4π2v2 (2) aT
x
(3) aT + 2πv (4) aT
v
Sol: (2)
2 2 2
2
aT xT 4 T 4
x x T T
ω π π
= = × = = constant.
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6. In an optics experiment, with the position of the object fixed, a student varies the position of a convex
lens and for each position, the screen is adjusted to get a clear image of the object. A graph between
the object distance u and the image distance v, from the lens, is plotted using the same scale for the
two axes. A straight line passing through the origin and making an angle of 45o with the xaxis meets
the experimental curve at P. The coordinates of P will be
(1) (2f, 2f) (2) f , f
2 2
(3) (f, f) (4) (4f, 4f)
Sol: (1)
It is possible when object kept at centre of
curvature.
u = v
u = 2f, v = 2f.
*7. A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through
its end. Its maximum angular speed is ω. Its centre of mass rises to a maximum height of
(1)
1 2 2
3 g
l ω
(2) 1
6 g
lω
(3)
1 2 2
2 g
l ω
(4)
1 2 2
6 g
l ω
Sol: (4)
T.Ei = T.Ef
2 1I mgh
2
ω =
2 2 1 1m mgh
2 3
× l ω = ⇒
1 2 2 h
6 g
ω
= l
h
•
8. Let 4
P(r) Q r
R
=
π
be the charge density distribution for a solid sphere of radius R and total charge Q.
for a point ‘p’ inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric
field is
(1) 0 (2) 2
o 1
Q
4πε r
(3)
2
1
4
o
Qr
4πε R
(4)
2
1
4
o
Qr
3πε R
Sol: (3)
r1
2
4 2
2 0 1
1 4
0 0
Q r4 r dr
R Qr E4 r E
4 R
π
π
π = ⇒ =
ε πε
∫
. r
P
R
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9. The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation.
Infrared radiation will be obtained in the transition from
(1) 2 → 1 (2) 3 → 2
(3) 4 → 2 (4) 5→ 4
Sol: (4)
IR corresponds to least value of 2 2
1 2
1 1
n n
−
i.e. from Paschen, Bracket and Pfund series. Thus the transition corresponds to 5 → 3.
*10. One kg of a diatomic gas is at a pressure of 8 × 104 N/m2. The density of the gas is 4 kg/m3. What is
the energy of the gas due to its thermal motion?
(1) 3 × 104 J (2) 5 × 104 J
(3) 6 × 104 J (4) 7 × 104 J
Sol: (2)
Thermal energy corresponds to internal energy
Mass = 1 kg
density = 8 kg/m3
⇒ Volume = 3 mass 1m
density 8
=
Pressure = 8 × 104 N/m2
∴ Internal Energy =
5
2
P × V = 5 × 104 J
11. This question contains Statement1 and Statement2. Of the four choices given after the statements,
choose the one that best describes the two statements.
Statement1: The temperature dependence of resistance is usually given as R = Ro(1 + αΔt). The
resistance of a wire changes from 100 Ω to 150 Ω when its temperature is increased from 27oC to
227oC. This implies that α = 2.5 ×10−3 /o C.
Statement 2: R = Ri (1 + αΔT) is valid only when the change in the temperature ΔT is small and ΔR =
(R – Ro) << Ro.
(1) Statement1 is true, Statement2 is false
(2) Statement1 is true, Statement2 is true; Statement2 is the correct explanation of Statement1.
(3) Statement1 is true, Statement2 is true; Statement2 is not the correct explanation of Statement1.
(4) Statement1 is false, Statement2 is true
Sol: (1)
Directions: Question numbers 12 and 13 are based on the following paragraph.
A current loop ABCD is held fixed on the plane of the paper as shown
in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop
are joined by two straight wires AB and CD. A steady current I is
flowing in the loop. Angle made by AB and CD at the origin O is 30o.
Another straight thin wire with steady current I1 flowing out of the plane
of the paper is kept at the origin.
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12. The magnitude of the magnetic field (B) due to loop ABCD at the origin (O) is
(1) zero (2)
( ) o b a
24ab
μ −
(3) oI b a
4 ab
μ −
π
(4) o ( ) ( ) I
2 b a a b
4 3
μ π − + + π
Sol: (2)
Net magnetic field due to loop ABCD at O is
B = BAB + BBC + BCD + BDA
= o o I I
0 0
4 a 6 4 b 6
μ π μ π
+ × + − ×
π π
= o o I I
24a 24b
μ μ
− = oI ( )
b a
24ab
μ
−
13. Due to the presence of the current I1 at the origin
(1) The forces on AB and DC are zero
(2) The forces on AD and BC are zero
(3) The magnitude of the net force on the loop is given by o 1 ( ) ( ) II
2 b a a b
4 3
μ π − + + π
(4) The magnitude of the net force on the loop is given by o 1 ( ) II
b a
24ab
μ
−
Sol: (2)
The forces on AD and BC are zero because magnetic field due to a straight wire on AD and BC is
parallel to elementary length of the loop.
14. A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young’s
double slit and gives rise to two overlapping interference patterns on the screen. The central
maximum of both lights coincide. Further, it is observed that the third bright fringe of known light
coincides with the 4th bright fringe of the unknown light. From this data, the wavelength of the
unknown light is
(1) 393.4 nm (2) 885.0 nm
(3) 442.5 nm (4) 776.8 nm
Sol: (3)
3λ1 = 4λ2
⇒ 2 1
3 3 590
4 4
λ = λ = × =
1770
4
= 442.5 nm
15. Two points P and Q are maintained at the potentials of 10V and 4V respectively. The work done in
moving 100 electrons from P to Q is
(1) −19 ×10−17 J (2) 9.60 ×10−17 J
(3) −2.24 ×10−16 J (4) 2.24 ×10−16 J
Sol: (4)
W = QdV = Q(Vq  VP) = 100 × (1.6 × 1019) × (– 4 – 10)
= + 100 × 1.6 × 1019 × 14 = +2.24 1016 J.
16. The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected
photoelectrons was found to be 1.68 eV. The work function of the metal is (hc = 1240 eV nm)
(1) 3.09 eV (2) 1.41 eV
(3) 151 eV (4) 1.68 eV
Sol: (2)
2
o
1mv eV 1.68eV
2
= = ⇒
h hc 1240evnm
400nm
ν = =
λ
= 3.1 eV ⇒ 3.1 eV = Wo + 1.6 eV
∴ Wo = 1.42 eV
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*17. A particle has an initial velocity 3ˆi + 4ˆj and an acceleration of 0.4ˆi + 0.3ˆj . Its speed after 10 s is
(1) 10 units (2) 7 2 units
(3) 7 units (4) 8.5 units
Sol: (2)
u = 3ˆi + 4ˆj
r
; a = 0.4ˆi + 0.3ˆj
r
u = u + at
r r r
= 3ˆi + 4ˆj + (0.4ˆi + 0.3ˆj)10 = 3ˆi + ˆj + 4ˆi + 3ˆj = 7ˆi + 7ˆj
Speed is 72 + 72 = 7 2 units
*18. A motor cycle starts from rest and accelerates along a straight path at 2 m/s2. At the starting point of
the motor cycle there is a stationary electric sire. How far has the motor cycle gone when the driver
hears the frequency of the siren at 94% of its value when the motor cycle was at rest? (speed of
sound = 330 ms1).
(1) 49 m (2) 98 m
(3) 147 m (4) 196 m
Sol: (2)
Motor cycle, u = 0, a = 2 m/s2
Observer is in motion and source is at rest.
⇒ O
S
v v
n n
v v
−
′ =
+
⇒ O 94 330 v n n
100 330
−
= ⇒ 330 – vO =
330 94
100
×
⇒ O
v 330 94 33
10
×
= − =
33 6 m/s
10
×
s =
v2 u2 9 33 33 9 1089
2a 100 100
− × × ×
= = ≈ 98 m.
*19. Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume
that the duration of collision is negligible and the collision with the plate is totally elastic. Then the
velocity as a function of time the height as function of time will be
(1) (2)
(3) (4)
Sol: (3)
2 h 1 gt
2
= , v =  gt and after the collision, v = gt.
(parabolic) (straight line)
Collision is perfectly elastic then ball reaches to same height again and again with same velocity.
t1 2t2 3t1 t
v1
+v1
v
t
h
y
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20. A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the
other two corners. If the net electrical force on Q is zero, then the Q/q equals
(1) −2 2 (2) 1
(3) 1 (4) 1
2
−
Sol: (1)
Three forces F41, F42 and f43 acting on Q are shown
Resultant of F41 + F43
= 2 Feach
= 2
o
2 1 Qq
4πε d
Resultant on Q becomes zero only when ‘q’ charges are of
negative nature.
( ) 4,2 2
o
F 1 Q Q
4 2d
×
=
πε
⇒ 2 2
2 dQ Q Q
d 2d
×
=
⇒
2 q Q Q
2
×
× =
∴ q Q
2 2
= − or Q 2 2
q
= −
Q q
Q
F42
F43
F41
q
*21. A long metallic bar is carrying heat from one of its ends to the other end under steadystate. The
variation of temperature θ along the length x of the bar from its hot end is best described by which of
the following figure.
(1)
(2)
(3)
(4)
Sol: (2)
We know that
dQ KA d
dt dx
θ
=
In steady state flow of heat
d dQ. 1 .dx
dt kA
θ =
⇒ θH  θ = k x ′ ⇒ θ = θH  k x ′
Equation θ = θH  k′ x represents a straight line.
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22. A transparent solid cylindrical rod has a refractive index of
2
3
. It is surrounded by air. A light ray is incident at the mid
point of one end of the rod as shown in the figure.
θ
The incident angle θ for which the light ray grazes along the wall of the rod is
(1) sin 1 1
2
−
(2) sin 1 3
2
−
(3) sin 1 2
3
−
(4) sin 1 1
3
−
Sol: (4)
SinC = 3
2
….. (1)
Sin r = sin (90 – C) = cosC =
1
2
2
1
sin
sinr
θ μ
=
μ
sin 2 1
3 2
θ = ×
sin 1 1
3
−
θ =
θ
C
*23. Three sound waves of equal amplitudes have frequencies (v – 1), v, (v + 1). They superpose to give
beats. The number of beats produced per second will be
(1) 4 (2) 3
(3) 2 (4) 1
Sol: (3)
Maximum number of beats = ν + 1− (ν −1) = 2
*24. The height at which the acceleration due to gravity becomes
g
9
(where g = the acceleration due to
gravity on the surface of the earth) in terms of R, the radius of the earth is
(1) 2R (2)
R
2
(3) R
2
(4) 2 R
Sol: (1)
( )2
g GM
R h
′ =
+
, acceleration due to gravity at height h
⇒
( )
2
2 2
g GM. R
9 R R h
=
+
=
2 g R
R h
+
⇒
2 1 R
9 R h
= +
⇒
R 1
R h 3
=
+
⇒ 3R = R + h ⇒ 2R = h
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*25. Two wires are made of the same material and have the same volume. However wire 1 has crosssectional
area A and wire2 has crosssectional area 3A. If the length of wire 1 increases by Δx on
applying force F, how much force is needed to stretch wire 2 by the same amount?
(1) F (2) 4F
(3) 6F (4) 9F
Sol: (4)
A1l1 = A2l2 ⇒ 1 1 1 1
2
2
A A
A 3A 3
×
= = = l l l
l ⇒ 1
2
= 3 l
l
Δx1 = 1
1
F
A
×
γ
l ….. (i)
2
2 2
F
x
3A
Δ =
γ
l …..(ii)
Here Δx1 = Δx2
2 1
2 1
F F
3A A
=
γ γ
l l
F2 = 1
1 1
2
3F × = 3F × 3 l
l
= 9F
*26. In an experiment the angles are required to be measured using an instrument. 29 divisions of the
main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the
main scale is halfadegree(=0.5o), then the least count of the instrument is
(1) one minute (2) half minute
(3) one degree (4) half degree
Sol: (1)
Least count =
value of main scale division
No of divisions on vernier scale
=
1 MSD
30
=
1 1o 1 o
30 2 60
× = = 1 minute
27. An inductor of inductance L = 400 mH and resistors of resistances R1
= 2Ω and R2 = 2Ω are connected to a battery of emf 12V as shown in
the figure. The internal resistance of the battery is negligible. The
switch S is closed at t = 0. The potential drop across L as a function
of time is
(1) 6e−5tV (2) 3t 12e V
t
−
(3) 6(1− e−t / 0.2 )V (4) 12e−5tV
Sol: (4)
1
1
F
I
R
= =
12 6A
2
=
2
2 2
dI
E L R I
dt
= + ×
( t / tc )
I2 Io 1 e= − − ⇒ o
2
I E 12 6A
R 2
= = =
3
c
t L 400 10 0.2
R 2
× −
= = =
( t / 0.2 )
I2 6 1 e= − −
Potential drop across L = E – R2I2 = 12 – 2 × 6 (1 – ebt) = 12 e5t
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Directions: Question numbers 28, 29 and 30 are based on the following paragraph.
Two moles of helium gas are taken over the cycle ABCDA, as shown in the P – T diagram.
*28. Assuming the gas to be ideal the work done on the gas in taking it from A to B is
(1) 200 R (2) 300 R
(3) 400 R (4) 500 R
Sol: (3)
WAB = ΔQ  ΔU = nCpdT – nCvdT (at
constant pressure)
= n(Cp – Cv)dt
= nRdT = 2 × R × (500 – 300) = 400 R
A B
D C
P
n = 2, γ = 1.67
300 K 500 K T
2 × 105 Pa
1 × 105 Pa
*29. The work done on the gas in taking it from D to A is
(1) – 414 R (2) + 414 R
(3) – 690 R (4) + 690 R
Sol: (1)
At constant temperature (isothermal process)
WDA = 1
2
P
nRTln
P
=
5
5
2.303 2R 300log 10
2 10
× ×
×
=
2.303 600Rlog 1
2
×
= 0.693 × 600 R =  414 R.
*30. The net work done on the gas in the cycle ABCDA is
(1) Zero (2) 276 R
(3) 1076 R (4) 1904 R
Sol: (2)
Net work done in a cycle = WAB + WBC + WCB + WBA
= 400 R + 2 × 2.303 × 500 R ln 2 – 400R – 414 R
= 1000R x ln 2 – 600R x ln 2 = 400R x ln 2 = 276R
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CHEMISTRY
PART − B
31. Knowing that the Chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the
following statements in incorrect ?
(1) Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic in
character.
(2) The ionic sizes of Ln (III) decrease in general with increasing atomic number.
(3) Ln (III) compounds are generally colourless.
(4) Ln (III) hydroxides are mainly basic in character.
Sol: (3)
Ln+3 compounds are mostly coloured.
32. A liquid was mixed with ethanol and a drop of concentrated 2 4 H SO was added. A compound with a
fruity smell was formed. The liquid was :
(1) 3 CH OH (2) HCHO
(3) 3 3 CH COCH (4) 3 CH COOH
Sol: (4)
Esterification reaction is involved
3 ( ) CH COOH l + 2 5 ( )
C H OH H
+
→ l 3 2 5( ) 2 ( ) CH COOC H +H O l l
*33. Arrange the carbanions, ( )3 3 3 CH C, CCl , ( )3 2 6 5 2 CH CH, C H CH , in order of their decreasing stability :
(1) ( ) ( ) 6 5 2 3 3 3 3 2 C H CH > CCl > CH C > CH CH (2) ( ) ( ) 3 2 3 6 5 2 3 3 CH CH > CCl > C H CH > CH C
(3) ( ) ( ) 3 6 5 2 3 2 3 3 CCl > C H CH > CH CH > CH C (4) ( ) ( ) 3 3 3 2 6 5 2 3 CH C > CH CH > C H CH > CCl
Sol: (3)
2o carbanion is more stable than 3o and Cl is –I effect group.
*34. The alkene that exhibits geometrical isomerism is :
(1) propene (2) 2methyl propene
(3) 2butene (4) 2 methyl 2 butene
Sol: (3)
C=C
H
CH3 CH3
H
C=C
H
H3C H
CH3
cis Trans
*35. In which of the following arrangements, the sequence is not strictly according to the property written
against it ?
(1) 2 2 2 2 CO < SiO < SnO < PbO : increasing oxidising power
(2) HF< HCl < HBr < HI : increasing acid strength
(3) 3 3 3 3 NH < PH < AsH < SbH : increasing basic strength
(4) B < C < O < N : increasing first ionization enthalpy.
Sol: (3)
Correct basic strength is 3 3 3 3 NH > PH > AsH > BiH
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36. The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is :
(1) benzoic acid (2) salicylaldehyde
(3) salicylic acid (4) phthalic acid
Sol: (3)
Kolbe – Schmidt reaction is
OH ONa OH OH
NaOH→ 2
o
CO
6atm, 140 C
→
COONa
3 H O+
→
COOH
Salicylic Acid
37. Which of the following statements is incorrect regarding physissorptions ?
(1) It occurs because of vander Waal’s forces.
(2) More easily liquefiable gases are adsorbed readily.
(3) Under high pressure it results into multi molecular layer on adsorbent surface.
(4) Enthalpy of adsorption ( ) adsorption ΔH is low and positive.
Sol: (4)
Enthalpy of adsorption regarding physissorption is not positive and it is negative.
38. Which of the following on heating with aqueous KOH, produces acetaldehyde ?
(1) 3 CH COCl (2) 3 2 CH CH Cl
(3) 2 2 CH Cl CH Cl (4) 3 2 CH CHCl
Sol: (4)
3 2
CH CHCl aq.KOH→ 3
/OH
CH CH
\OH
3
2
CH CHO
H O
→
−
*39. In an atom, an electron is moving with a speed of 600m/s with an accuracy of 0.005%. Certainity
with which the position of the electron can be located is (h = 6.6 ×10−34 kg m2s−1 ,
mass of electron, 31
me = 9.1×10− kg )
(1) 1.52 ×10−4m (2) 5.10 ×10−3m
(3) 1.92 ×10−3m (4) 3.84 ×10−3m
Sol: (3)
x.m v h
4
Δ Δ =
π
Δx = h
4π mΔv
Δv = 600 0.005 0.03
100
× =
⇒
34
3
31
x 6.625 10 1.92 10 m
4 3.14 9.1 10 0.03
−
−
−
×
Δ = = ×
× × × ×
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40. In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
3 2 2 2
CH OH( ) 3 O (g) CO (g) 2H O( )
2
l + → + l At 298K standard Gibb’s energies of formation for
3 2 CH OH(l), H O(l) and 2 CO (g) are 166.2, 237.2 and 394.4 kJ mol−1 respectively. If standard
enthalpy of combustion of methanol is 726kJ mol−1 , efficiency of the fuel cell will be
(1) 80 % (2) 87%
(3) 90% (4) 97%
Sol: (4)
3 CH OH(l) + 2 2 2
3O (g) CO (g) 2H O( )
2
→ + l ΔH = −726kJ mol−1
Also o
f 3 ΔG CH OH(l) = 166.2 kJ mol1
o
f 2 ΔG H O(l) = 237.2 kJ mol1
o
f 2 ΔG CO (l) = 394.4 kJ mol1
Q G Δ = of
G ΣΔ products of
−ΣΔG reactants.
= 394.4 2 (237.2) + 166.2
= −702.6 kJ mol1
now Efficiency of fuel cell = G 100
H
Δ
×
Δ
= 702.6 100
726
×
= 97%
41. Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol
of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this
solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X
and Y in their pure states will be, respectively :
(1) 200 and 300 (2) 300 and 400
(3) 400 and 600 (4) 500 and 600
Sol: (3)
o o
T X X Y Y P = P x + P x
X x = mol fraction of X
Y x = mol fraction of Y
∴ 550 = o o
x Y
P 1 P 3
1 3 1 3
+ + +
=
o o
X Y P 3P
4 4
+
∴ 550 (4) = o o
X Y P + 3P ………….. (1)
Further 1 mol of Y is added and total pressure increases by 10 mm Hg.
∴ 550 + 10 = o o
X Y
P 1 P 4
1 4 1 4
+ + +
∴ 560 (5) = o o
X Y P + 4P ………….(2)
By solving (1) and (2)
We get, o
X P = 400 mm Hg
o
Y P = 600 mm Hg
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42. The half life period of a first order chemical reaction is 6.93 minutes. The time required for the
completion of 99% of the chemical reaction will be (log 2=0.301) :
(1) 230.3 minutes (2) 23.03 minutes
(3) 46.06 minutes (4) 460.6 minutes
Sol: (3)
1
1/ 2
0.6932 0.6932 min
t 6.93
Qλ = = −
Also t =
[ ]
[ ]
o 2.303 A log
λ A
[ ] o A = initial concentration (amount)
[A] = final concentration (amount)
∴ t = 2.303 6.93 log100
0.6932 1
×
= 46.06 minutes
43. Given : 3
o
Fe /Fe E 0.036V, + = − 2
o
Fe / Fe E + = 0.439V. The value of standard electrode potential for the
change, 3 2
(aq) Fe + + e− → Fe + (aq) will be :
(1) 0.072 V (2) 0.385 V
(3) 0.770 V (4) 0.270
Sol: (3)
Q Fe3+ + 3e− →Fe; Eo = −0.036V
∴ O o
1 ΔG = −nFE = −3F(−0.036)
= +0.108 F
Also Fe2+ + 2e− →Fe; Eo = 0.439 V
∴ O2
ΔG = nFEo
= 2 F( 0.439)
= 0.878 F
To find Eo for 3 2
(aq) Fe + + e− →Fe + (aq)
ΔGO = nFE o
= 1FE o
Q o o o
1 2 G = G −G
∴ Go = 0.108F – 0.878F
∴ FEo = +0.108F – 0.878F
∴ EO = 0.878 – 0.108
= 0.77v
*44. On the basis of the following thermochemical data : ( o
(aq) ΔfG H+ = 0)
2 H O(l) →H+ (aq) + OH− (aq); ΔH = 57.32kJ
2 2 2
H (g) 1O (g) H O( )
2
+ → l ; ΔH = −286.20kJ
The value of enthalpy of formation of OH− ion at 25oC is :
(1) 22.88 kJ (2) 228.88 kJ
(3) +228.88 kJ (4) 343.52 kJ
Sol: (2)
By adding the two given equations, we have
2(g) H + 2(g) (aq) (aq)
1O H OH
2
→ + + − ; ΔH =228.88 Kj
Here of
ΔH of (aq) H+ = 0
∴ of
ΔH of OH− = 228.88 kJ
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45. Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom ?
(1) 108 pm (2) 127 pm
(3) 157 pm (4) 181 pm
Sol: (2)
For FCC,
2a = 4r (the atoms touches each other along the face diagonal)
r = 2a 2 361
4 4
×
=
= 127 pm
46. Which of the following has an optical isomer ?
(1) ( ) 3 3 CO NH Cl + (2) ( )( ) 2
3 2 CO en NH +
(3) ( ) ( ) 3
2 4 CO H O en + (4) ( ) ( ) 3
2 3 2 CO en NH +
Sol: (4)
It is an octahedral complex of the type ( ) 2 2 M AA X
Where AA is bidentate ligand.
*47. Solid Ba ( ) 3 2 NO is gradually dissolved in a 1.0 ×10−4M 2 3 Na CO solution. At what concentration of
Ba2+ will a precipitate begin to form ?(Ksp for Ba CO3 = 5.1 × 10 9 − ).
(1) 4.1 ×10−5M (2) 5.1×10−5M
(3) 8.1×10−8M (4) 8.1×10−7M
Sol: (2)
( )3 2 3 3 3 Ba NO + CaCO → BaCO + 2NaNO
Here 2 [ ] 4
3 2 3 CO− = Na CO = 10− M
2 2
sp 3 K = Ba+ CO− ⇒ 5.1×10−9 = Ba2+ (10−4 )⇒ Ba+2 = 5.1×10−5
At this value, just precipitation starts.
48. Which one of the following reactions of Xenon compounds is not feasible ?
(1) 3 XeO + 6HF → Xe 6 2 F + 3H O
(2) 4 2 3 3Xe F + 6H O → 2 Xe + XeO + 12 HF + 1.5 2 O
(3) 2 2 2 2XeF + 2H O → 2Xe + 4HF + O
(4) 6 7 XeF + RbF → Rb(XeF ]
Sol: (1)
Remaining are feasible
*49. Using MO theory predict which of the following species has the shortest bond length ?
(1) 22
O + (2) 2 O+
(3) 2 O− (4) 22
O −
Sol: (1)
Bond length 1
bond order
α
Bond order = no..of bonding e no.of antibonding e
2
−
Bond orders of 2
2 2 2 O+ , O− , O− and 2
2 O+ are respectively
2.5, 1.5, 1 and 3.
50. In context with the transition elements, which of the following statements is incorrect ?
(1) In addition to the normal oxidation states, the zero oxidation state is also shown by these elements
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in complexes.
(2) In the highest oxidation states, the transition metal show basic character and form cationic
complexes.
(3) In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d
electrons are used for bonding.
(4) Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in bonding
decreases.
Sol: (2)
In higher Oxidation states transition elements show acidic nature
*51. Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 ×103ms−1
(Mass of proton = 1.67 ×10−27 kg and h = 6.63 ×10−34 Js ) :
(1) 0.032 nm (2) 0.40 nm
(3) 2.5 nm (4) 14.0 nm
Sol: (2)
h
mv
λ =
34
27 3
6.63 10 0.40
1.67 10 10
−
−
×
= ≡
× ×
nm
52. A binary liquid solution is prepared by mixing nheptane and ethanol. Which one of the following
statements is correct regarding the behaviour of the solution ?
(1) The solution formed is an ideal solution
(2) The solution is nonideal, showing +ve deviation from Raoult’s law.
(3) The solution is nonideal, showing –ve deviation from Raoult’s law.
(4) nheptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law.
Sol: (2)
The interactions between n –heptane and ethanol are weaker than that in pure components.
*53. The number of stereoisomers possible for a compound of the molecular formula
( ) 3 CH − CH = CH − CH OH −Me is :
(1) 3 (2) 2
(3) 4 (4) 6
Sol: (3)
About the double bond, two geometrical isomers are possible and the compound is having one chiral
carbon.
*54. The IUPAC name of neopentane is
(1) 2methylbutane (2) 2, 2dimethylpropane
(3) 2methylpropane (4) 2,2dimethylbutane
Sol: (2)
Neopentane is
3
3 3
3
CH

HC C CH

CH
− −
*55. The set representing the correct order of ionic radius is :
(1) Li+ > Be2+ > Na+ > Mg2+ (2) Na+ > Li+ > Mg2+ > Be2+
(3) Li+ > Na+ > Mg2+ > Be2+ (4) Mg2+ > Be2+ > Li+ > Na+
Sol: (2)
Follow the periodic trends
56. The two functional groups present in a typical carbohydrate are :
(1) OH and COOH (2) CHO and COOH
(3) > C = O and – OH (4) – OH and CHO
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Sol: (3)
Carbohydrates are polyhydroxy carbonyl compounds.
*57. The bond dissociation energy of B – F in BF3 is 646 kJ mol−1 whereas that of CF in 4 CF is 515kJ
mol−1 . The correct reason for higher BF bond dissociation energy as compared to that of C F is :
(1) smaller size of Batom as compared to that of C atom
(2) stronger σ bond between B and F in 3 BF as compared to that between C and F in 4 CF
(3) significant pπ – pπ interaction between B and F in 3 BF whereas there is no possibility of such
interaction between C and F in 4 CF .
(4) lower degree of pπ – pπ interaction between B and F in 3 BF than that between C and F in 4 CF .
Sol: (3)
option itself is the reason
58. In Cannizzaro reaction given below
2 Ph CHO ( )
2 2
( )
: OH Ph CH OH PhCO −
−
→ + && the slowest step is :
(1) the attack of :
( )
OH
−
at the carboxyl group
(2) the transfer of hydride to the carbonyl group
(3) the abstraction of proton from the carboxylic group
(4) the deprotonation of Ph 2 CH OH
Sol: (2)
Hydride transfer is the slowest step.
59. Which of the following pairs represents linkage isomers ?
(1) ( ) [ ] 3 4 4 Cu NH Pt Cl and ( ) [ ] 3 4 4 Pt NH CuCl
(2) ( ) ( ) 3 2 2 Pd P Ph NCS and ( ) ( ) 3 2 2 Pd P Ph SCN
(3) ( ) 3 5 3 4 CO NH NO SO and ( ) 3 5 4 3 CO NH SO NO
(4) ( ) 2 3 4 2 PtCl NH Br and ( ) 2 3 4 2 PtBr NH Cl
Sol: (2)
NCS is ambidentate ligand and it can be linked through N (or) S
60. BunaN synthetic rubber is a copolymer of :
(1) 2 2
Cl

H C = CH − C = CH and 2 2 H C = CH − CH = CH
(2) 2 2 H C = CH − CH = CH and 5 6 2 H C − CH = CH
(3) 2 H C = CH − CN and 2 2 H C = CH − CH = CH
(4) 2 H C = CH − CN and 2 2
3
H C CH C CH

CH
= − =
Sol: (3)
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Mathematics
PART − C
61. Let a, b, c be such that b(a + c) ≠ 0 . If
( )n 2 ( )n 1 ( )n
a a 1 a 1 a 1 b 1 c 1
b b 1 b 1 a 1 b 1 c 1 0,
c c 1 c 1 1 a 1 b 1 c + +
+ − + + −
− + − + − − + =
− + − − −
then the
value of ‘n’ is
(1) zero (2) any even integer
(3) any odd integer (4) any integer
Sol: (3)
( )n
a a 1 a 1 a 1 b 1 c 1
b b 1 b 1 1 a 1 b 1 c 1
c c 1 c 1 a b c
+ − + + −
− + − + − − − +
− + −
( )n
a a 1 a 1 a 1 a 1 a
b b 1 b 1 1 b 1 b 1 b
c c 1 c 1 c 1 c 1 c
+ − + −
= − + − + − + − −
− + − +
( )n 1
a a 1 a 1 a 1 a a 1
b b 1 b 1 1 b 1 b b 1
c c 1 c 1 c 1 c c 1
+
+ − + −
= − + − + − + − −
− + − +
( )n 2
a a 1 a 1 a a 1 a 1
b b 1 b 1 1 b b 1 b 1
c c 1 c 1 c c 1 c 1
+
+ − + −
= − + − + − − + −
− + − +
This is equal to zero only if n + 2 is odd i.e. n is odd integer.
62. If the mean deviation of number 1, 1 + d, 1 + 2d, ….. , 1 + 100d from their mean is 255, then the d is
equal to
(1) 10.0 (2) 20.0
(3) 10.1 (4) 20.2
Sol: (3)
( )
n
Mean x sum of quantities
n
= =
(a l)
2
n
+
1[1 1 100d] 1 50d
2
= + + = +
[ ] i
M.D. 1 x x 255 1 50d 49d 48d …. d 0 d …… 50d 2
n 101
= Σ − ⇒ = + + + + + + + + = d 50 x 51
101 2
d 255 x 101 10.1
50 x 51
⇒ = =
*63. If the roots of the equation bx2 + cx + a = 0 be imaginary, then for all real values of x, the expression
3b2x2 + 6bcx + 2c2 is
(1) greater than 4ab (2) less than 4ab
(3) greater than – 4ab (4) less than – 4ab
Sol: (3)
bx2 + cx + a = 0
Roots are imaginary ⇒ c2 − 4ab < 0⇒ c2 < 4ab⇒ −c2 > −4ab
3b2x2 + 6bcx + 2c2
since 3b2 > 0
Given expression has minimum value
Minimum value =
( )( )
( )
2 2 22 2 2
2
2 2
4 3b 2c 36b c 12b c c 4ab
4 3b 12b
−
= − = − > − .
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*64. Let A and B denote the statements
A: cosα + cosβ + cos γ = 0
B: sinα + sinβ + sinγ = 0
If cos( ) cos( ) cos( ) 3
2
β − γ + γ − α + α − β = − , then
(1) A is true and B is false (2) A is false and B is true
(3) both A and B are true (4) both A and B are false
Sol: (3)
cos( ) cos( ) cos( ) 3
2
β − γ + γ − α + α − β = −
⇒ 2cos(β − γ) + cos(γ − α) + cos(α − β) + 3 = 0
⇒ 2cos(β − γ) + cos(γ − α) + cos(α − β) + sin2 α + cos2 α + sin2 β + cos2 β + sin2 γ + cos2 γ = 0
⇒ ( )2 ( )2 sinα + sinβ + sinγ + cosα + cosβ + cos γ = 0
*65. The lines p(p2 +1)x − y + q = 0 and ( ) ( ) p2 + 1 2 x + p2 +1 y + 2q = 0 are perpendicular to a common line
for
(1) no value of p (2) exactly one value of p
(3) exactly two values of p (4) more than two values of p
Sol: (2)
Lines must be parallel, therefore slopes are equal ⇒ p(p2 + 1) = −(p2 +1) ⇒ p = – 1
66. If A, B and C are three sets such that A ∩B = A ∩C and A ∪B = A ∪C , then
(1) A = B (2) A = C
(3) B = C (4) A ∩B = φ
Sol: (3)
67. If u, v, w
r r uur
are noncoplanar vectors and p, q are real numbers, then the equality
3u pv pw − pv w qu − 2w qv qu = 0
r r uur r uur r uur r r
holds for
(1) exactly one value of (p, q) (2) exactly two values of (p, q)
(3) more than two but not all values of (p , q) (4) all values of (p, q)
Sol: (1)
(3p2 − pq + 2q2 )u v w = 0
r r uur
But u v w ≠ 0
r r uur
3p2 − pq + 2q2 = 0
2 2 2
2p2 p2 pq q 7q 0 2p2 p q 7 q2 0
2 4 2 4
+ − + + = ⇒ + − + =
⇒ p = 0, q = 0, p q
2
=
This possible only when p = 0, q = 0 exactly one value of (p, q)
68. Let the line x 2 y 1 z 2
3 5 2
− − +
= =
−
lies in the plane x + 3y − αz + β = 0 . Then (α, β) equals
(1) (6, – 17) (2) ( – 6, 7)
(3) (5, – 15) (4) ( – 5, 15)
Sol: (2)
Dr’s of line = (3, − 5, 2)
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Dr’s of normal to the plane = (1, 3, − α)
Line is perpendicular to normal ⇒ 3(1) − 5(3) + 2(−α) = 0 ⇒ 3 −15 − 2α = 0 ⇒ 2α = −12⇒ α = −6
Also (2, 1, − 2) lies on the plane
2 + 3 + 6(−2) + β = 0⇒β = 7
∴ (α, β) = (−6, 7)
*69. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and
arranged in a row on the shelf so that the dictionary is always in the middle. Then the number of such
arrangements is
(1) less than 500 (2) at least 500 but less than 750
(3) at least 750 but less than 1000 (4) at least 1000
Sol: (4)
4 novels can be selected from 6 novels in 6
4 C ways. 1 dictionary can be selected from 3 dictionaries
in 3
1 C ways. As the dictionary selected is fixed in the middle, the remaining 4 novels can be arranged
in 4! ways.
∴ The required number of ways of arrangement = 6 3
4 1 C x C x 4! = 1080
70. [ ]
0
cot x dx
π∫
, [•] denotes the greatest integer function, is equal to
(1)
2
π
(2) 1
(3) – 1 (4)
2
π
−
Sol: (4)
Let [ ]
0
I cotx dx
π
= ∫ …(1)
( )
0
cot x dx
π
= ∫ π − [ ]
0
cot x dx
π
= ∫ − …(2)
Adding (1) and (2)
2I = [ ]
0
cot x dx
π∫
+ [ ]
0
cot x dx
π
∫ − = ( )
0
1 dx
π
= ∫ −
[x] [ x] 1if x Z
0 if x Z
+ − = − ∉
= ∈
Q
[ ]0 x π = − = −π
∴ I
2
π
= −
71. For real x, let f (x) = x3 + 5x + 1, then
(1) f is oneone but not onto R (2) f is onto R but not oneone
(3) f is oneone and onto R (4) f is neither oneone nor onto R
Sol: (3)
Given f (x) = x3 + 5x + 1
Now f ‘(x) = 3×2 + 5 > 0, ∀x ∈R
∴ f(x) is strictly increasing function
∴ It is oneone
Clearly, f(x) is a continuous function and also increasing on R,
Lt
x → −∞
f (x) = −∞ and Lt
x → ∞
f (x) = ∞
∴ f(x) takes every value between −∞ and ∞ .
Thus, f(x) is onto function.
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72. In a binomial distribution B n, p 1
4
=
, if the probability of at least one success is greater than or
equal to 9
10
, then n is greater than
(1) 4 3
10 10
1
log − log
(2) 4 3
10 10
1
log + log
(3) 4 3
10 10
9
log − log
(4) 4 3
10 10
4
log − log
Sol: (1)
n
n
3
4
1 q 9 3 1 n log 10
10 4 10
− ≥ ⇒ ≤ ⇒ ≥ −
4 3
10 10
n 1
log log
⇒ ≥
−
*73. If P and Q are the points of intersection of the circles x2 + y2 + 3x + 7y + 2p − 5 = 0 and
x2 + y2 + 2x + 2y − p2 = 0 , then there is a circle passing through P, Q and (1, 1) for
(1) all values of p (2) all except one value of p
(3) all except two values of p (4) exactly one value of p
Sol: (1)
Given circles S = x2 + y2 + 3x + 7y + 2p − 5 = 0
S’ = x2 + y2 + 2x + 2y − p2 = 0
Equation of required circle is S + λS’ = 0
As it passes through (1, 1) the value of
( )
( 2 )
7 2p
6 p
− +
λ =
−
If 7 + 2p = 0, it becomes the second circle
∴it is true for all values of p
74. The projections of a vector on the three coordinate axis are 6, – 3, 2 respectively. The direction
cosines of the vector are
(1) 6, − 3, 2 (2) 6, 3, 2
5 5 5
−
(3) 6, 3, 2
7 7 7
− (4) 6, 3, 2
7 7 7
− −
Sol: (3)
Projection of a vector on coordinate axis are 2 1 2 1 2 1 x − x , y − y , z − z
2 1 2 1 2 1 x − x = 6, y − y = −3, z − z = 2
( )2 ( )2 ( )2
2 1 2 1 2 1 x − x + y − y + z − z = 36 + 9 + 4 = 7
The D.C’s of the vector are 6, 3, 2
7 7 7
−
*75. If Z 4 2
z
− = , then the maximum value of Z is equal to
(1) 3 +1 (2) 5 +1
(3) 2 (4) 2 + 2
Sol: (2)
Z Z 4 4
Z Z
= − +
Z Z 4 4
Z Z
⇒ = − +
Z Z 4 4
Z Z
⇒ ≤ − + Z 2 4
Z
⇒ ≤ +
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⇒ Z 2 − 2 Z − 4 ≤ 0
( Z − ( 5 + 1))( Z − (1− 5 )) ≤ 0⇒1− 5 ≤ Z ≤ 5 + 1
*76. Three distinct points A, B and C are given in the 2 – dimensional coordinate plane such that the ratio
of the distance of any one of them from the point (1, 0) to the distance from the point ( – 1, 0) is equal
to 1
3
. Then the circumcentre of the triangle ABC is at the point
(1) (0, 0) (2) 5 , 0
4
(3) 5, 0
2
(4) 5, 0
3
Sol: (2)
P = (1, 0) ; Q(−1, 0)
Let A = (x, y)
AP BP CP 1
AQ BQ CQ 3
= = = ..(1)
⇒ 3AP = AQ⇒ 9AP2 = AQ2 ⇒ 9(x −1)2 + 9y2 = (x +1)2 + y2
⇒ 9×2 −18x + 9 + 9y2 = x2 + 2x + 1+ y2 ⇒ 8×2 − 20x + 8y2 + 8 = 0
x2 y2 5 x 1 0
2
⇒ + − + = …(2)
∴ A lies on the circle
Similarly B, C are also lies on the same circle
∴ Circumcentre of ABC = Centre of Circle (1) = 5, 0
4
*77. The remainder left out when ( )2n 1 2n 8 62+ − is divided by 9 is
(1) 0 (2) 2
(3) 7 (4) 8
Sol: (2)
( )2n 1 2n 8 62+ − = ( )n ( )2n 1 1 63 63 1 + + − −
( )n ( )2n 1 ( n n ( )2 ( )n ) ( (2n 1) (2n 1) ( )2 ( )( )(2n 1) )
1 2 1 2 1 63 1 63 1 c 63 c 63 …. 63 1 c 63 c 63 …. 1 63 = + + − + = + + + + + − + + + + + − +
(n n ( ) ( )n 1 (2n 1) (2n 1) ( ) ( )(2n) )
1 2 1 2 2 63 c c 63 …. 63 c c 63 …. 63 = + + + + − − + + + + −
∴ Reminder is 2
*78. The ellipse x2 + 4y2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in
inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is
(1) x2 +16y2 = 16 (2) x2 +12y2 = 16
(3) 4×2 + 48y2 = 48 (4) 4×2 + 64y2 = 48
Sol: (2)
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( )
2 2
x2 4y2 4 x y 1 a 2, b 1 P 2, 1
4 1
+ = ⇒ + = ⇒ = = ⇒ =
Required Ellipse is
2 2 2 2
2 2 2 2
x y 1 x y 1
a b 4 b
+ = ⇒ + =
(2, 1) lies on it
2
2 2
4 1 1 1 1 1 3 b 4
16 b b 4 4 3
⇒ + = ⇒ = − = ⇒ =
∴
2 2 2 2
x y 1 x 3y 1 x2 12y2 16
16 4 16 4
3
+ = ⇒ + = ⇒ + =
1
A’ 2 A 2
V’ V
P (2, 1)
(4, 0)
*79. The sum to the infinity of the series 2 3 4
1 2 6 10 14 ……
3 3 3 3
+ + + + + is
(1) 2 (2) 3
(3) 4 (4) 6
Sol: (2)
Let 2 3 4
S 1 2 6 10 14 ….
3 3 3 3
= + + + + + …(1)
2 3 4
1S 1 2 6 10 ….
3 3 3 3 3
= + + + + …(2)
Dividing (1) & (2)
2 3 4
S 1 1 1 1 4 4 4 ….
3 3 3 3 3
− = + + + + +
2 2
2S 4 4 1 1 1 ……
3 3 3 3 3
= + + + +
2 2
2S 4 4 1 4 4 3 4 2 6
3 3 3 1 1 3 3 2 3 3 2
3
⇒ = + = + = + =
−
2S 6 S 3
3 3
⇒ = ⇒ =
80. The differential equation which represents the family of curves c2x
1 y = c e , where 1 c and 2 c are
arbitrary constants is
(1) y ‘ = y2 (2) y ” = y ‘ y
(3) yy” = y’ (4) ( )2 yy” = y’
Sol: (4)
c2x
1 y = c e …(1)
c2x
2 1 y ‘ = c c e
2 y’ = c y …(2)
2 y ” = c y ‘
From (2)
2
c y’
y
=
So,
( ) ( )
2
y ‘ 2
y” yy” y’
y
= ⇒ =
81. One ticket is selected at random from 50 tickets numbered 00, 01, 02, …., 49. Then the probability
that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero,
equals
(1) 1
14
(2) 1
7
(3) 5
14
(4) 1
50
Sol: (1)
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S = { 00, 01, 02, …., 49 }
Let A be the even that sum of the digits on the selected ticket is 8 then
A = { 08, 17, 26, 35, 44 }
Let B be the event that the product of the digits is zero
B = { 00, 01, 02, 03, …., 09, 10, 20, 30, 40 }
A ∩B = {8}
Required probability = ( ) ( )
( )
1
P A B 50 1 P A/B
P B 14 14
50
∩
= = =
82. Let y be an implicit function of x defined by x2x − 2xx cot y −1= 0 . Then y ‘(1) equals
(1) – 1 (2) 1
(3) log 2 (4) – log 2
Sol: (1)
x2x − 2xx cot y −1= 0 …(1)
Now x = 1,
1 – 2 coty – 1 = 0 ⇒ coty = 0 ⇒ y =
2
π
Now differentiating eq. (1) w.r.t. ‘x’
2x2x (1 logx) 2 xx ( c osec2y) dy cot y xx (1 logx) 0
dx
+ − − + + =
Now at 1,
2
π
( ) ( )
1,
2
2 1 log1 2 1 1 dy 0 0
dx π
+ − − + =
1, 1,
2 2
2 2 dy 0 dy 1
dx π dx π
⇒ + = ⇒ = −
83. The area of the region bounded by the parabola ( )2 y − 2 = x −1, the tangent to the parabola at the
point (2, 3) and the xaxis is
(1) 3 (2) 6
(3) 9 (4) 12
Sol: (3)
Equation of tangent at (2, 3) to
( )2 y − 2 = x −1 is 1 S = 0
⇒ x − 2y + 4 = 0
Required Area = Area of ΔOCB + Area of
OAPD – Area of ΔPCD
( ) ( ) ( )
3
2
0
1 4 x 2 y 4y 5 dy 1 1 x 2
2 2
= + ∫ − + −
3 3
2
0
4 y 2y 5y 1 4 9 18 15 1
3
= + − + − = − − + −
= 28 −19 = 9sq. units
0
D (0, 3)
P (2, 3)
2y = x + 4
A (1, 2)
B (4, 0)
C (0, 2) A (1, 2)
(or)
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Area = ( ) ( ) 3 3
2 2
0 0
∫ 2y − 4 − y + 4y − 5 dy = ∫ −y + 6y − 5 dy ( ) ( ) 3 3 3
2
0 0
y 3 27 3 y dy 9 sq.units
3 3
−
= − − = = =
∫
84. Given P(x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of P'(x) = 0 . If P(−1) < P(1) ,
then in the interval [−1, 1]
(1) P(−1) is the minimum and P(1) is the maximum of P
(2) P(−1) is not minimum but P(1) is the maximum of P
(3) P(−1) is the minimum and P(1) is not the maximum of P
(4) neither P(−1) is the minimum nor P(1) is the maximum of P
Sol: (2)
P(x) = x4 + ax3 + bx2 + cx + d
P'(x) = 4x3 + 3ax2 + 2bx + c
Q x = 0 is a solution for P'(x) = 0 , ⇒ c = 0
∴ P(x) = x4 + ax3 + bx2 + d …(1)
Also, we have P(−1) < P(1)
⇒1− a + b + d < 1+ a + b + d⇒ a > 0
Q P'(x) = 0 , only when x = 0 and P(x) is differentiable in ( – 1, 1), we should have the maximum and
minimum at the points x = – 1, 0 and 1 only
Also, we have P(−1) < P(1)
∴ Max. of P(x) = Max. { P(0), P(1) } & Min. of P(x) = Min. { P(1), P(0) }
In the interval [ 0 , 1 ],
P'(x) = 4x3 + 3ax2 + 2bx = x(4x2 + 3ax + 2b)
QP'(x) has only one root x = 0, 4x2 + 3ax + 2b = 0 has no real roots.
∴ ( )
2
3a 2 32b 0 3a b
32
− < ⇒ <
∴ b > 0
Thus, we have a > 0 and b > 0
∴ P'(x) = 4×3 + 3ax2 + 2bx > 0, ∀x ∈(0, 1)
Hence P(x) is increasing in [ 0, 1 ]
∴ Max. of P(x) = P(1)
Similarly, P(x) is decreasing in [1 , 0]
Therefore Min. P(x) does not occur at x = – 1
85. The shortest distance between the line y − x = 1 and the curve x = y2 is
(1) 3 2
8
(2) 2 3
8
(3) 3 2
5
(4) 3
4
Sol: (1)
x − y + 1= 0 …(1)
x = y2
1 2y dy
dx
= dy 1
dx 2y
⇒ = = Slope of given line (1)
1 1 y 1
2y 2
= ⇒ = ⇒
2 y 1 x 1 1
2 2 4
= ⇒ = =
⇒ (x, y) 1, 1
4 2
=
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∴ The shortest distance is
1 1 1
4 2 3 3 2
1 1 4 2 8
− +
= =
+
Directions: Question number 86 to 90 are Assertion – Reason type questions. Each of these questions
contains two statements
Statement1 (Assertion) and Statement2 (Reason).
Each of these questions also have four alternative choices, only one of which is the correct answer. You have
to select the correct choice
86. Let ( ) ( )2 f x = x +1 −1, x ≥ −1
Statement1 : The set {x : f (x) = f −1 (x)} = {0, −1}
Statement2 : f is a bijection.
(1) Statement1 is true, Statement2 is true; Statement2 is a correct explanation for Statement1
(2) Statement1 is true, Statement2 is true; Statement2 is not a correct explanation for Statement1
(3) Statement1 is true, Statement2 is false
(4) Statement1 is false, Statement2 is true
Sol: (3)
There is no information about codomain therefore f(x) is not necessarily onto.
87. Let f (x) = x x and g(x) = sinx .
Statement1 : gof is differentiable at x = 0 and its derivative is continuous at that point.
Statement2 : gof is twice differentiable at x = 0.
(1) Statement1 is true, Statement2 is true; Statement2 is a correct explanation for Statement1
(2) Statement1 is true, Statement2 is true; Statement2 is not a correct explanation for Statement1
(3) Statement1 is true, Statement2 is false
(4) Statement1 is false, Statement2 is true
Sol: (3)
f (x) = x x and g(x) = sinx
gof (x) = sin(x x )
2
2
sin x ,x 0
sin x ,x 0
− <
=
≥
∴ ( ) ( )
2
2
gof ' x 2x cos x ,x 0
2xcos x ,x 0
− <
=
≥
Clearly, L(gof )'(0) = 0 = R(gof )'(0)
∴ gof is differentiable at x = 0 and also its derivative is continuous at x = 0
Now, ( ) ( )
2 2 2
2 2 2
gof " x 2cos x 4x sinx ,x 0
2cosx 4x sinx ,x 0
− + <
=
− ≥
∴L(gof )"(0) = −2 and R(gof )"(0) = 2
∴L(gof )"(0) ≠ R(gof )"(0)
∴ gof(x) is not twice differentiable at x = 0.
*88. Statement1 : The variance of first n even natural numbers is
n2 1
4
−
Statement2 : The sum of first n natural numbers is
n(n 1)
2
+
and the sum of squares of first n natural
numbers is
n(n 1)(2n 1)
6
+ +
(1) Statement1 is true, Statement2 is true; Statement2 is a correct explanation for Statement1
(2) Statement1 is true, Statement2 is true; Statement2 is not a correct explanation for Statement1
(3) Statement1 is true, Statement2 is false
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(4) Statement1 is false, Statement2 is true
Sol: (4)
Statement2 is true
Statement1:
Sum of n even natural numbers = n (n + 1)
( ) n(n 1)
Mean x n 1
n
+
= = +
( ) ( ) ( ) ( ) 2 2 2 2 2 2
i
Variance 1 x x 1 2 4 ..... 2n n 1
n n
= − = + + + − + Σ
2 2 2 2 ( )2 ( )( ) ( )2 1 4 n n 1 2n 1 2 1 2 ..... n n 1 n 1
n n 6
+ +
= + + + − + = − +
(n 1) 2(2n 1) 3(n 1) (n 1)[4n 2 3n 3] (n 1)(n 1) n2 1
3 3 3 3
+ + − + + + − − + − − = = = =
∴ Statement 1 is false.
89. Statement1 : ~ (p ↔~ q) is equivalent to p ↔ q.
Statement2 : ~ (p ↔~ q) is a tautology.
(1) Statement1 is true, Statement2 is true; Statement2 is a correct explanation for Statement1
(2) Statement1 is true, Statement2 is true; Statement2 is not a correct explanation for Statement1
(3) Statement1 is true, Statement2 is false
(4) Statement1 is false, Statement2 is true
Sol: (3)
p q p ↔ q ~q p ↔~ q ~ (p ↔~ q)
T T T F F T
T F F T T F
F T F F T F
F F T T F T
90. Let A be a 2 x 2 matrix
Statement1 : adj(adj A) = A
Statement2 : adj A = A
(1) Statement1 is true, Statement2 is true; Statement2 is a correct explanation for Statement1
(2) Statement1 is true, Statement2 is true; Statement2 is not a correct explanation for Statement1
(3) Statement1 is true, Statement2 is false
(4) Statement1 is false, Statement2 is true
Sol: (2)
n 1 2 1 adj A A A A − − = = =
( ) n 2 0 adj adj A A A A A A − = = =
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AIEEE–2009, ANSWER KEY
Test Booklet CodeA Test Booklet CodeB Test Booklet CodeC Test Booklet CodeD
PHY CHE MAT CHE MAT PHY MAT PHY CHE CHE PHY MAT
1. (2)
2. (1)
3. (3)
4. (1)
5. (2)
6. (1)
7. (4)
8. (3)
9. (4)
10. (2)
11. (1)
12. (2)
13. (2)
14. (3)
15. (4)
16. (2)
17. (2)
18. (2)
19. (3)
20. (1)
21. (2)
22. (4)
23. (3)
24. (1)
25. (4)
26. (1)
27. (4)
28. (3)
29. (1)
30. (2)
31. (3)
32. (4)
33. (3)
34. (3)
35. (3)
36. (3)
37. (4)
38. (4)
39. (3)
40. (4)
41. (3)
42. (3)
43. (3)
44. (2)
45. (2)
46. (4)
47. (2)
48. (1)
49. (1)
50. (2)
51. (2)
52. (2)
53. (3)
54. (2)
55. (2)
56. (3)
57. (3)
58. (2)
59. (2)
60. (3)
61. (3)
62. (3)
63. (3)
64. (3)
65. (2)
66. (3)
67. (1)
68. (2)
69. (4)
70. (4)
71. (3)
72. (1)
73. (1)
74. (3)
75. (2)
76. (2)
77. (2)
78. (2)
79. (2)
80. (4)
81. (1)
82. (1)
83. (3)
84. (2)
85. (1)
86. (3)
87. (3)
88. (4)
89. (3)
90. (2)
1. (1)
2. (4)
3. (1)
4. (3)
5. (3)
6. (1)
7. (1)
8. (1)
9. (2)
10. (1)
11. (2)
12. (4)
13. (3)
14. (3)
15. (2)
16. (2)
17. (2)
18. (1)
19. (1)
20. (2)
21. (2)
22. (2)
23. (2)
24. (1)
25. (1)
26. (1)
27. (3)
28. (2)
29. (2)
30. (2)
31. (2)
32. (1)
33. (2)
34. (3)
35. (2)
36. (2)
37. (1)
38. (4)
39. (1)
40. (3)
41. (4)
42. (1)
43. (2)
44. (2)
45. (3)
46. (2)
47. (1)
48. (4)
49. (2)
50. (1)
51. (2)
52. (2)
53. (4)
54. (1)
55. (2)
56. (1)
57. (1)
58. (4)
59. (4)
60. (3)
61. (2)
62. (2)
63. (1)
64. (3)
65. (1)
66. (1)
67. (2)
68. (4)
69. (1)
70. (4)
71. (4)
72. (1)
73. (3)
74. (4)
75. (3)
76. (2)
77. (1)
78. (1)
79. (2)
80. (1)
81. (4)
82. (3)
83. (3)
84. (1)
85. (4)
86. (1)
87. (3)
88. (4)
89. (2)
90. (2)
1. (4)
2. (4)
3. (4)
4. (3)
5. (2)
6. (1)
7. (4)
8. (4)
9. (4)
10. (3)
11. (1)
12. (3)
13. (4)
14. (3)
15. (2)
16. (4)
17. (3)
18. (3)
19. (4)
20. (1)
21. (2)
22. (3)
23. (3)
24. (3)
25. (4)
26. (1)
27. (2)
28. (3)
29. (2)
30. (2)
31. (4)
32. (2)
33. (3)
34. (4)
35. (2)
36. (3)
37. (1)
38. (4)
39. (2)
40. (2)
41. (1)
42. (4)
43. (2)
44. (1)
45. (4)
46. (3)
47. (1)
48. (2)
49. (3)
50. (3)
51. (3)
52. (4)
53. (1)
54. (3)
55. (1)
56. (2)
57. (1)
58. (3)
59. (3)
60. (2)
61. (4)
62. (3)
63. (4)
64. (1)
65. (3)
66. (4)
67. (3)
68. (4)
69. (4)
70. (4)
71. (3)
72. (4)
73. (1)
74. (4)
75. (4)
76. (3)
77. (4)
78. (4)
79. (3)
80. (2)
81. (2)
82. (3)
83. (3)
84. (4)
85. (1)
86. (1)
87. (2)
88. (3)
89. (4)
90. (1)
1. (1)
2. (1)
3. (1)
4. (1)
5. (4)
6. (2)
7. (1)
8. (1)
9. (1)
10. (4)
11. (4)
12. (4)
13. (4)
14. (1)
15. (3)
16. (3)
17. (1)
18. (1)
19. (1)
20. (4)
21. (4)
22. (1)
23. (2)
24. (4)
25. (2)
26. (4)
27. (2)
28. (1)
29. (2)
30. (4)
31. (4)
32. (3)
33. (4)
34. (4)
35. (1)
36. (3)
37. (3)
38. (4)
39. (3)
40. (1)
41. (4)
42. (4)
43. (2)
44. (4)
45. (2)
46. (2)
47. (2)
48. (2)
49. (1)
50. (1)
51. (3)
52. (3)
53. (1)
54. (3)
55. (4)
56. (4)
57. (1)
58. (2)
59. (3)
60. (2)
61. (4)
62. (4)
63. (1)
64. (1)
65. (2)
66. (3)
67. (2)
68. (4)
69. (4)
70. (2)
71. (3)
72. (3)
73. (1)
74. (4)
75. (4)
76. (1)
77. (2)
78. (1)
79. (1)
80. (3)
81. (1)
82. (4)
83. (4)
84. (1)
85. (1)
86. (3)
87. (4)
88. (3)
89. (1)
90. (1)