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JEE (Adv) 2016 Question Paper with Solutions JEE(Advanced) 2016 Final Exam/Paper-1/Code-9
PHYSICS
JEE(Advanced) – 2016 TEST PAPER WITH ANSWER (HELD ON SUNDAY 22nd MAY, 2016)
PART-I : PHYSICS
1.
SECTION–1 : (Maximum Marks : 15) This section contains Five questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct For each question, darken the bubble corresponding to the correct option in the ORS For each question, marks will be awarded in one of the following categories : Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases A parallel beam of light is incident from air at an angle on the side PQ of a right angled triangular prism
of refractive index n 2 . Light undergoes total internal reflection in the prism at the face PR when has a minimum value of 45°. The angle of the prism is :
P
– n=2
Q
(A) 15° Ans. (A)
(B) 22.5°
R
(C) 30°
(D) 45°
P
45° r1=30°
r2 = 45°
Sol.
R
Q
1 sin 45° =
2 sin r1
r2 – r1 = = 45° – 30° = 15° CODE-9
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PHYSICS
2.
In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength () of incident light and the corresponding stopping potential (V0) are given below:
m V0 Volt 2.0 1.0 0.4
0.3 0.4 0.5 8
–1
–19
Given that c = 3 × 10 ms and e = 1.6 × 10 C, Planck's constant (in units of J s) found from such an experiment is : –34
–34
(A) 6.0 × 10
(B) 6.4 × 10
–34
(C) 6.6 × 10
(D) 6.8 × 10
–34
Ans. (B) hC
Sol. KE max
eVs
hC
1.6 10 19 2
h 3 108 3000 1010
... (i)
1.6 1019 1
h 3 108 4000 1010
... (ii)
h 3 108 1.6 10 19 From (ii) 10 4000 10 1.6 1019 2
1.6 1019
h 3 108 1 1 h 3 108 4 3 107 12 107 3 4
1.6 1019
h 3 108 1 12 107
1.6 4
1019 10 7 h 108
6.4 × 10 2/40
h 3 108 h 3 108 1.6 10 19 3000 1010 4000 1010
–34
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JEE (Adv) 2016 Question Paper with Solutions
3.
A uniform wooden stick of mass 1.6 kg and length rests in an inclined manner on a smooth, vertical wall of height h (<) such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of 30° with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio h/ and the frictional force f at the bottom of the stick are:
PHYSICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9
–2
(g = 10 ms ) (A)
3 16 3 h ,f N 16 3
(B)
16 3 h 3 ,f N 16 3
(C)
8 3 h 3 3 N ,f 3 16
(D)
16 3 h 3 3 N ,f 3 16
Ans. (D)
N1
Sol.
30°
h
y
x
N2
mg
60°
f
O
Force equation in x-direction, N1 cos 30° – f = 0
... (i)
Force equation in y-direction, N1 sin 30° + N2 – mg = 0
... (ii)
Torque equation about O, h 0 mg cos 60 N1 cos 30 2
... (iii)
Also, given N1 = N2
... (iv)
[Note taking reaction from floor as normal reaction only] solving (i), (ii), (iii) & (iv) we have h 3 3 16 3 & f 16 3
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PHYSICS
4.
A water cooler of storage capacity 120 litres can cool water at constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed 30°C and the entire stored 120 litres of water is initially cooled to 10°C. The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is :
Cooler
Device Hot
Cold
(Specific heat of water is 4.2 kJ kg–1 K–1 and the density of water is 1000 kg m–3) (A) 1600 Ans. (B)
(B) 2067
Sol. 3000 – P = (120 × 1)(4.2 × 103)
(C) 2533
(D) 3933
dT dt
20 dT dt 60 60 3
P = 2067 W 5.
An infinite line charge of uniform electric charge density lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity and electrical conductivity . The electrical conduction in the material follows Ohm's law. Which one of the following graphs best describes the subsequent variation of the magnitude of current density j(t) at any point in the material? j(t)
j(t)
(A)
j(t)
(B) (0,0)
t
j(t)
(C) (0,0)
t
(D) (0,0)
t
(0,0)
t
Ans. (A)
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JEE (Adv) 2016 Question Paper with Solutions
Sol. This is the problem of RC circuit where the product RC is a constant. So due to leakage current, charge & current density will exponentially decay & will become zero at infinite time. So correct answer is (A) for any small element
PHYSICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9
dr Resistance R = 2r Capacitance C =
2r dr
Product R × C =
= constant
q q0e
t
t
dq q 0 e I= dt
t q0 e I Current density = A 2r
je
t
SECTION–2 : (Maximum Marks : 32)
This section contains EIGHT questions.
Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct.
For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS
For each question, marks will be awarded in one of the following categories : Full Marks
: +4 If only the bubble(s) corresponding to the correct option(s) is (are) darkened.
Partial Marks : +1 For darkening a bubble corresponding to each correct option, Provided NO incorrect option is darkened. Zero Marks
:
0
If none of the bubbles is darkened.
Negative Marks : –2 In all other cases.
for example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result in –2 marks, as a wrong option is also darkened
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PHYSICS
6.
A transparent slab of thickness d has a refractive index n(z) that increases with z. Here z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices n1 and n2 (> n1), as shown in the figure. A ray of light is incident with angle i from medium 1 and emerges in medium 2 with refraction angle f with a lateral displacement . Which of the following statement(s) is(are) true ?
n1= constant z
i
1
n(z) d
n2= constant
(A) is independent of n2 (C) n1sini = (n2 – n1)sinf
f
2
(B) is dependent on n(z) (D) n1sini = n2sinf
Ans. (A,B,D)
Sol. For parallel slab
n1 sini = n2sinf
And depends on refractive angle in slab
depends on refractive index of slab and independent of n2 7.
A conducting loop in the shape of right angled isosceles triangle of height 10 cm is kept such that the 90° vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular –1 loop is in counterclockwise direction and increased at constant rate of 10 A s . Which of the following statement(s) is(are) true?
10cm
90°
(A) The induced current in the wire is in opposite direction to the current along the hypotenuse. (B) There is a repulsive force between the wire and the loop 0 volt
(C) If the loop is rotated at a constant angular speed about the wire, an additional emf of is induced in the wire
(D) The magnitude of induced emf in the wire is 0 volt. Ans. (B,D) 6/40
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E
x
Sol.
x
x
PHYSICS
induce electric field E
dx
by direction of induced electric field, current in wire is in same direction of current along the hypotenuse. 0.1
Flux through triangle if wire have current i
0i
0i
2x 2xdx 10 0
Mutual inductance = Induced emf in wire =
0 di 0 10 0 10 dt 10
A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement(s) is(are) true?
8.
0 10
(A) The refractive index of the lens is 2.5
(B) The radius of curvature of the convex surface is 45 cm (C) The faint image is erect and real
(D) The focal length of the lens is 20 cm. Ans. (A,D)
Sol. For lens
60cm
h O
I
30cm 2h
1 1 1 v u f
1 1 1 f 20cm 60 30 f Also
1 1 1 n 1 n 1 f R R
... (i) ... (ii)
For reflection from convex mirror (curved surface) CODE-9
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//// /
/// //// ////////////////////
PHYSICS
//// /
30cm
10cm
1 1 1 2 v u f R 1 1 1 2 10 30 f R
... (iii)
R= 30 cm from (i), (ii) & (iii) n = 2.5,
9.
faint image erect & virtual
An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true? (A) The temperature distribution over the filament is uniform
(B) The resistance over small sections of the filament decreases with time (C) The filament emits more light at higher band of frequencies before it breaks up (D) The filament consumes less electrical power towards the end of the life of the bulb Ans. (C,D)
Sol. Because of non-uniform evaporation at different section, area of cross-section would be different at different sections. Region of highest evaporation rate would have rapidly reduced area and would become break up crosssection. Resistance of the wire as whole increases with time. Overall resistance increases hence power decreases. At break up junction temperature would be highest, thus light of highest band frequency would be emitted at those cross-section. 10.
A length-scale () depends on the permittivity () of a dielectric material, Boltzmann constant kB, the absolute temperature T, the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles, Which of the following expressions(s) for is(are) dimensionally correct? nq 2 (A) k T B
k B T (B) nq 2
q2 (C) n 2 / 3 k T B
q2 (D) n1/3 k T B
Ans. (B,D) 8/40
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1 q2 F 4 0 r 2
PHYSICS
Sol. We know, q2 (Fr 2 )4 0
So, dimension q2 dim (Fr 2 ) = MLT–2× L2 = ML3T–2 0
3 2
Similarly; E K BT dim (KBT) = dim(Energy) = ML2T–2
(B)
(C)
(D)
L3 ML3 T 2 1 L ML2 T 2
(A)
nq 2 k B T
(E) vol ML2 T 2 L3 L MLT 2 L2 Fr 2
MLT 2 L2 L2 Fr 2 (vol)2 / 3 L3 L3 / 2 ML2 T 2 (K)
Fr 2 (vol)1/ 3 MLT 2 L2 L L Energy ML2 T 2 1 3 L ] vol
[ dimension n = dim 11.
Highly excited states for hydrogen like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n >> 1. Which of the following statement(s) is (are) true? (A) Relative change in the radii of two consecutive orbitals does not depend on Z (B) Relative change in the radii of two consecutive oribitals varies as 1/n (C) Relative change in the energy of two consecutive orbitals varies as 1/n3 (D) Relative change in the angular momenta of two consecutive orbitals varies as 1/n
Ans. (A, B, D)
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PHYSICS
n2 Sol. As radius r z 2
2
n 1 n 2 1 2n 1 r z z 2 2 n n r n n z 2 as energy E z 2 n
z2 z2 2 2 2 n 1 n 2 E n n 1 2 2 . n 1 2 2 z E n . n 1 2 n 1 E 2n 1 2n 1 2 n2 n n E
as angular momentum L =
12.
L L
nh 2
n 1 h nh 2 nh 2
2 1 1 n n
The position vector r of a particle of mass m is given by the following equation r(t) t 3 ˆi t 2 ˆj , where
10 3 ms , = 5 ms–2 and m = 0.1 kg. At t = 1 s, which of the following statement(s) is(are) 3
true about the particle? (A) The velocity v is given by v (10iˆ 10 ˆj)ms 1
(B) The angular momentum L with respect to the origin is given by L kˆ Nms 3
5
(C) The force F is given by F (iˆ 2ˆj)N 20 (D) The torque with respect to the origin is given by kˆ Nm 3
Ans. (A, B, D) 10/40
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r t ˆi t 2 ˆj
PHYSICS
Sol.
dr v 3t 2 ˆi 2 tjˆ dt
d2 r a 2 6tiˆ 2ˆj dt At t = 1 10 (A) v 3 1iˆ 2 J 1jˆ 3
= 10iˆ 10ˆj
(B) L r p
10 ˆ ˆ = 1i 5 1j ×0.1 10iˆ 10ˆj 3
5 = kˆ 3
10 ˆ ˆ F (C) m 6 1i 2 5 j = 2iˆ ˆj 3
(D) r F
10 ˆ ˆ ˆ ˆ = i 5j 2i j 3
=
10 ˆ k 10 kˆ 3
=
20 ˆ k 3
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PHYSICS
13.
Two loudspeakers M and N are located 20m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car is initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/hr along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Q. Let (t) represent the beat frequency measured by a person sitting in the car at time t. Let P, Q and R be the beat frequencies measured at locations P, Q and R, respectively. The speed of sound in air is 330 ms–1. Which of the following statement(s) is(are) true regarding the sound heard by the person ? (A) The plot below represents schematically the variation of beat frequency with time (t) P Q
Q
t
R
(B) The plot below represents schematically the variations of beat frequency with time (t) P
Q
Q
R
t
(C) The rate of change in beat frequency is maximum when the car passes through Q (D) P + R = 2 Q Ans. (A, C, D) Sol. fM =
C V cos f1 C
fN =
C V cos f2 C
(f1)M
–Q
P
(f2)N
f = fN – fM =
C V cos f 2 f1 C
d f V d f 2 f1 sin dt C dt &
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PHYSICS
[ C is correct] V P = 1 cos f C
Q = f V R = 1 cos f C
P + r = 2 Q
SECTION–3 : (Maximum Marks : 15) This section contains FIVE questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +3 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases.
Consider two solid spheres P and Q each of density 8 gm cm–3 and diameters 1 cm and 0.5 cm, respectively. Sphere P is dropped into a liquid of density 0.8 gm cm–3 and viscosity = 3 poiseulles. Sphere Q is dropped into a liquid of density 1.6 gm cm–3 and viscosity = 2 poiseulles. The ratio of the terminal velocities of P and Q is. Ans. 3
14.
Sol. VT
r 2 d m d L n 2
VTP rP n L2 d m d L1 VTQ rQ n L1 d m d L2 2
VTP 2 2 8 0.8 VTQ 1 3 8 1.6
VTP 3 VTQ 15.
Two inductors L1 (inductance 1 mH, internal resistance 3 ) and L2 (inductance 2mH, internal resistance 4), and a resistor R (resistance 12 ) are all connected in parallel across a 5V battery. The circuit is switched on at time t = 0. The ratio of the maximum to the minimum current (Imax/Imin) drawn from the battery is.
Ans. 8
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PHYSICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 L1=1mH
r1=3
L2=2mH
r2=4
R=12
Sol. = 5V
I max
5 A (Initially at t = 0) R 12
I min
1 1 1 (finally in steady state) R eq r1 r2 R
1
1
1
=
= 5 3 4 12 10 A 3
I max 8 I min
16.
12
12
12
The isotope 5 B having a mass 12.014 u undergoes -decay to 6 C. 6 C has an excited state of the nucleus (126 C*) at 4.041 MeV above its ground state. If 125 B decays to 12 C*, the maximum kinetic energy 6 of the -particle in units of MeV is (1u = 931.5 MeV/c2, where c is the speed of light in vacuum).
Ans. 9 Sol.
12 5
0 B 12 6 C 1e v
Mass defect = (12.014 – 12) u
Released energy = 13.041 MeV Energy used for excitation of
12 6
C = 4.041 MeV
Energy converted to KE of electron = 13.041 – 4.041 = 9 MeV 17.
A hydrogen atom in its ground state is irradiated by light of wavelength 970 Å. Taking hc/e = 1.237 × 10–6 eV m and the ground state energy of hydrogen atom as –13.6 eV, the number of lines present in the emission spectrum is
Ans. 6
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Sol.
PHYSICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 hc 12370 970
–13.6 + 12.7 =
13.6 n2
2
n = 16 n=4 Number of lines = nC2 = 6 18. A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by the metal. The sensor has a scale that displays log2(P/P0), where P0 is a constant. When the metal surface is at a temperature of 487 °C, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767 °C ? Ans. 9 Sol. P = eAT4 where T is in kelvin log 2
log 2
eA 487 273
4
P0
eA 2767 273 P0
1
...(i)
4
x
...(ii)
(ii) – (i)
4
3040 log 2 x 1 760
x=9
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CHEMISTRY
JEE(Advanced) – 2016 TEST PAPER WITH ANSWER
19.
PART-II : CHEMISTRY SECTION–1 : (Maximum Marks : 15) This section contains Five questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct For each question, darken the bubble corresponding to the correct option in the ORS For each question, marks will be awarded in one of the following categories : Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases P is the probability of finding the 1s electron of hydrogen atom in a spherical shell of infinitesimal 2 thickness, dr, at a distance r from the nucleus. The volume of this shell is 4r dr. The qualitative sketch of the dependence of P on r is -
(HELD ON SUNDAY 22nd MAY, 2016)
p
p
(B)
(A)
0
0
r
p
(C)
p
(D)
0
Ans. (B)
r
0
r
r
Sol. For 1s, radial part of wave function is 3
(r)
r
1 2 2 e a0 a0
probability of finding an e– in a spherical shell of thickness, 'dr' at distance 'r' from nucleus, P = 2(r) . 4r2dr 3
2r
1 a0 = 16r e dr a0 2
So P is zero at r = 0 and r =
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20.
One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings –1 (Ssurr) in J K is (1 L atm = 101.3 J) (A) 5.763
(B) 1.013
(C) –1.013
CHEMISTRY
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9
(D) –5.763
Ans. (C) Sol. From 1st law of thermodynamics qsys = U – w = 0 – [–Pext.V] = 3.0 atm × (2.0 L – 1.0 L) = 3.0 L-atm Ssurr =
q (q rev )surr sys T T 3.0 101.3J 300K
=–
= – 1.013 J/K
21.
The increasing order of atomic radii of the following group 13 elements is (A) Al < Ga < In < Tl
(B) Ga < Al < In < Tl
(C) Al < In < Ga < Tl
(D) Al < Ga < Tl < In
Ans. (B)
th
Sol. The order of radius of 13 group elements is Ga < Al < In < Tl.
Reason Due to poor shielding effect of d-orbital, radius of Ga is smallar than Al. 22.
On complete hydrogenation, natural rubber produces (A) ethylene-propylene copolymer
(B) vulcanised rubber
(C) polypropylene
(D) polybutylene
Ans. (A)
CH3
CH3 CH2 = C – CH = CH2
Sol.
Polymerisation
Isoprene
Natural rubber H2(excess) catalyst CH3
CH3 CH2 = CH2 + CH2 = CH
Ethylene
CODE-9
( CH2–C=CH–CH2 )n
Propylene
Copolymerisation
( CH2– CH – CH2 –CH2 )n Completely hydrogenated Natural rubber
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CHEMISTRY
23.
2–
Among [Ni(CO)4], [NiCl4] , [Co(NH3)4Cl2]Cl, Na3[CoF6], Na2O2 and CsO2, the total number of paramagnetic compounds is (A) 2
(B) 3
(C) 4
(D) 5
Ans. (B) Sol. Compound/Ion
Magnetic nature of compound
1.
[Ni(CO)4]
Diamagnetic
2.
[NiCl4]2–
Paramagnetic
3.
[Co(NH3)4Cl2]Cl
Diamagnetic
4.
Na3[CoF6]
Paramagnetic
5.
Na2O 2
Diamagnetic
6.
CsO2 Paramagnetic So total number of paramagnetic compounds is 3.
24.
SECTION–2 : (Maximum Marks : 32)
This section contains EIGHT questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble(s) corresponding to the correct option(s) is (are) darkened. Partial Marks : +1 For darkening a bubble corresponding to each correct option, Provided NO incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. for example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result in –2 marks, as a wrong option is also darkened A plot of the number of neutrons (N) against the number of protons (P) of stable nuclei exhibits upwards deviation from linearity for atomic number, Z > 20. For an unstable nucleus having N/P ratio less than 1, the possible mode(s) of decay is(are) (A) – decay ( emission)
(D) + decay (positron emission)
(C) Neutron emission Ans. (B, D) Sol. As
(B) orbital or K-electron capture
N N ratio is less than 1, for possible decay mode , ratio should increase. The possible modes P P
are -decay, K-capture and + -decay. Hence, correct option are (B), (D). (In – -decay or neutron decay , 18/40
N ratio will decrease further) P
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25.
CHEMISTRY
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 The correct statements(s) about of the following reaction sequence is(are) (i) O2
Cumene(C9H12)
(ii) H3O
+
P
CHCl3/NaOH
Q (major) + R (minor)
NaOH S PhCH2Br
Q (A) R is steam volatile
(B) Q gives dark violet coloration with 1% aqueous FeCl3 solution (C) S gives yellow precipitate with 2, 4,-dinitrophenylhydrazine (D) S gives dark violet coloration with 1% aqueous FeCl3 solution Ans. (B, C) CH3
CH3 O2
CH
Sol.
C–O–OH (Cumene hydroperoxide)
CH3
CH3
+
H /
OH
O
+ H3C–C–CH3 (Acetone)
(phenol) P
OH
OH
CH=O
CHCl3 +NaOH
(P)
OH
+
Q (major) Steam volatile (forms intramolecular hydrogen bonding)
OH CH=O
NaOH PhCH2Br
CH=O R (minor)
High boiling point due to inter molecular H-bonding So R is not steam volatile
OCH2Ph CH=O
S (does not give dark violet coloration with 1% FeCl3 solution) Q gives dark violet coloration with 1% aqueous FeCl3 solution because it has phenolic (–OH) group.|
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JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 26.
Positive Tollen's test is observed for H
O
OH
CHO (A)
H
(B)
H
O
Ph
(C) Ph
(D)
O
H
Ph
Ph
Ans. (A,B,C) Sol. Tollens's test is given by compounds having aldehyde group. Also -hydroxy carbonyl gives positive tollen's test.
H (A)
H
C =O
H
H
Tollen's reagent
–
H
CO2
H
H
+ Ag mirror
(+ve test)
Acraldehyde
CH=O
(B)
CO2–
Tollen's reagent
+ Ag mirror
(+ve test)
Benzaldehyde
(C) Ph–CH–C–Ph
Tollen's reagent
Ph–C–C–Ph
OH O Benzoin
+ Ag mirror
(+ve test)
O O
(D) PhCH=CH–C–Ph
Tollen's reagent
No reaction
(–ve test)
O
27.
The product(s) of the following reaction sequence is(are) NH2 (i) Acetic anhydride/pyridine (ii) KBrO3 / HBr +
(iii) H3O , heat (iv) NaNO2 / HCl, 273-278 K (v) Cu/HBr
Br
Br
Br
Br (A)
(B)
Br Br
Br
(C)
Br
(D)
Br Br Ans. (B) 20/40
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O NH2
O
Sol.
O
NH
O Py.
O
NH
NH2 +
KBrO3 + HBr Bromination
Acylation or acetylation
CHEMISTRY
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9
H3O Hydrolysis /
Br
Br NaNO2 + HCl 273-278 K
N2+ Cl
Br Cu HBr
Br
28.
Br
The compound(s) with TWO lone pairs of electrons on the central atom is(are) (B) ClF3
(C) XeF4
(A) BrF5
(D) SF4
Ans. (B,C)
Br
F
F
Sol. (A)
3 2
sp d hybridisation Number of lone pair on centre atom = 1 square pyramidal F (distorted octadehdral)
F
F
F
3
(B)
Cl
sp d hybridisation T-shape
F
Number of lone pair on centre atom = 2
F
F
F Xe
(C)
3 2
sp d hybridisation square planar
F
Number of lone pair on centre atom = 2
F
F F (D)
3
sp d hybridisation see-saw
S
Number of lone pair on centre atom = 1
F F Hence Ans. (B,C) CODE-9
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CHEMISTRY
29.
The crystalline form of borax has (A) Tetranuclear [B4O5(OH)4]2– unit (B) All boron atoms in the same plane 2
3
(C) Equal number of sp and sp hybridized boron atoms (D) One terminal hydroxide per boron atom Ans. (A,C,D) OH B
–
O
O
Sol. HO
OH
B
O
B
O
O
–
B
OH
(A) Having [B4O5(OH)4]2– tetranuclear (boron) unit (B) All boron atoms not in same plane 2
3
(C) Two boron are sp hybridised and two boron are sp hybridised (D) One terminal hydroxide per boron atom is present. 30.
The reagent(s) that can selectively precipiate S2– from a mixture of S2– and SO42– in aqueous soltuion is(are) : (B) BaCl2
(A) CuCl2
(C) Pb(OOCCH3)2
(D) Na2[Fe(CN)5NO]
Ans. (A OR A, C)
S2–
+
(A) CuCl2 (Soln)
CuS
(Soln)
(Soln)
(Black ppt.)
+ SO42– No ppt.
CuCl2 (Soln)
(Soln)
(B) BaCl2
+
S
n
2–
BaS
n
(Sol )
(Sol )
BaCl2
+ SO4
2–
n
2–
n
PbSO4
(Soln)
n
(Sol )
Na2[Fe(CN)5NO] n
(Sol )
–
+ 2CH3COO n
(Black ppt.)
Pb(OOCCH3)2 + SO42– (D) Na2[Fe(CN)5NO]
n
(Sol )
PbS
(Sol )
(Soln)
n
(Sol )
(White ppt.)
(C) Pb(OOCCH3)2 + S n
2Cl
BaSO4 + 2Cl–
(Sol )
(Sol )
–
+
(No ppt.)
n
(Sol )
22/40
2Cl–
+
(Sol )
+ 2CH3COO–
(White ppt.)
+ S2–
n
(Sol ) 2–
+ SO4
(Soln)
Na4[Fe(CN)5NOS] (Purple colour solution)
No ppt.
n
(Sol ) JEE(Advanced) 2016 Final Exam/Paper-1/Held on Sunday 22nd May, 2016
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–8
Note : PbSO4 Ksp = 2.5 × 10 PbS Ksp = 3 × 10–28
As in question selective precipitation is asked PbS will be precipitate much easier than PbSO4 though both are insoluble. Hence answer should be (C) also alongwith (A) 31.
CHEMISTRY
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9
According to the Arrhenius equation, (A) A high activation energy usually implies a fast reaction (B) Rate constant increase with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy (C) Higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant (D) The pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy.
Ans. (B,C,D) Sol. (A) k Ae E
a
/ RT
High Ea means less k, hence slower rate.
(B) e–Ea/RT = fraction of molecules having kinetic energy greater than activation energy which increase as temperature increases. k
E 1
k
1
2 a 2 (C) ln k R T T i.e., ln k E a 1 1 2 1
(D) Rate of reaction Total number of collisions × Fraction of collisions which can form product Rate of reaction
–Ea/RT
ZAB × (P × e
)
–Ea/RT
A e
SECTION–3 : (Maximum Marks : 15)
32.
This section contains FIVE questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +3 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as –3 its molality. Density of this solution at 298 K is 2.0 g cm . The ratio of the molecular weights of
MWsolute , is the solute and solvent, MW solvent Ans. (9) CODE-9
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Let
M1 = Molar mass solute M2 = Molar mass solvent
Molality,
0.1 m = 0.9 M 1000 2
....(1)
Molarity,
0.1 M = 0.1M + 0.9 M × 2 × 1000 1 2
.....(2)
m=M
200 0.1×1000 0.9 M 2 0.1M1 0.9 M 2
M1 9 M2
CHEMISTRY
Sol. 1 mole solution has 0.1 mole solute and 0.9 mole solvent
Alternate solution :
M=m
volume of solution = mass of solvent
Wsolute Wsolvent Wsolvent 2
Wsolute = Wsolvent
0.1 × Msolute = 0.9 × Msolvent
33.
M solute 9 M solvent In the following monobromination reaction, the number of possible chiral products is CH2CH2CH3 Br2(1.0 mole) Br H 300°C CH3 (1.0 mole)
(enantiomerically pure)
Ans. (5) Sol. CH2CH2CH3 Br
H CH3
(1.0 mole) (enantiomerically pure)
24/40
CH2CH2CH3 Br2(1.0 mole) 300°C
Br
H
CH2CH2CH3 +
CH2 – Br (Chiral)
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Br
Br CH3
(Achiral)
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CH3
H H
+
CH2CH3 Br Br
Br H
+
CH3 (Chiral)
34.
CH2CH3 H H H Br + H CH3 (Chiral)
CH3 Br H Br
CH3 (Achiral)
Br H + H
H H H + H Br H CH3 (Chiral)
CHEMISTRY
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9
CH2–Br H H Br CH3 (Chiral)
The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases x times. The value of x is
Ans. (4) U Avg
Rate of diffusion
Sol.
1
2 2 N*
UAvg
UAvg
2 2 N*
UAvg (kT)
Rate of diffusion
2 2 P 3 T2
P
3
rfinal (4) 2 = rinitial 2
rfinal 4 rinital 35.
The number of geometric isomers possible for the complex [CoL2Cl2]– (L = H2NCH2CH2O– ) is
Ans. (5) Sol. [CoL2Cl2]–
(L = H2NCH2CH2O–)
[Co(H2NCH2CH2O)2 Cl2]
It is
CODE-9
[M(AB)2C2]
type of complex
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CHEMISTRY
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9
A
C
M C
A
A
C
B
,
M C
B
B
A
B
C
B
M C
A
C
and
M
C
B
M
B
B
A B
A
A
A
A
B
C
C
Total geometrical isomers = 5 In neutral or faintly alkaline solution, 8 moles permanganate anion quantitatively oxidize thiosulphate anions to produce X moles of a sulphur containing product. the magnitude of X is
Ans. (6)
36.
7
2
4
6
Sol. MnO –4 S2 O2– MnO2 SO2– 3 4 X –
2–
Equivalents of MnO4 = equivalents of SO4 –
2–
Moles of MnO4 × n-factor = moles of SO4 × n-factor 8×3=X×4 X=6
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JEE(Advanced) – 2016 TEST PAPER WITH ANSWER (HELD ON SUNDAY 22nd MAY, 2016)
PART - III : MATHEMATICS
SECTION–1 : (Maximum Marks : 15) This section contains Five questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct For each question, darken the bubble corresponding to the correct option in the ORS For each question, marks will be awarded in one of the following categories : Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases
A debate club consists of 6 girls and 4 body. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 member) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is (A) 380 (B) 320 (C) 260 (D) 95 Ans. (A) Sol. (6C4 + 6C3.4C1). 4C1 = 380 38.
37.
The least value of R for which 4x 2
(A) Ans. (C)
1 64
1 32
(B)
1 1 , for all x > 0, is x
(C)
1 27
(D)
1 25
1 2 Sol. ƒ x 4x ; x 0 x ƒ ' x 8 x
1 x2
8x 3 1 x2 1/ 3
1 ƒ(x) attains its minimum at x 8
1 1/ 3 ƒ 1 8 1 4 8
2/3
31 / 3 1
CODE-9
1/ 3
8
1
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JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 39.
Let
. Suppose 1 and 1 are the roots of the equation x2 – 2xsec + 1 = 0 and 2 and 6 12 2
2 are the roots of the equation x + 2xtan – 1 = 0. If 1 > 1 and 2 > 2, then 1 + 2 equals (A) 2(sec – tan) (B) 2sec (C) –2tan (D) 0 Ans. (C) 2sec 4sec 2 4 2 1 sec | tan | { 1}
Sol. 1
2 tan 4 tan 2 4 2 2 tan sec
2
{ }
1 = sec – tan , 6 12 1 + 2 = – 2tan Let S x ( , ) : x 0, . The sum of all distinct solution of the equation 2
40.
3 sec x cosecx 2(tan x cot x) 0 in the set S is equal to -
(A) Ans. (C) Sol.
7 9
(B)
2 9
(C) 0
(D)
5 9
3 sin x cos x 2 cos 2x
cos 2x cos x 3 2x 2n x 3 x 6n 1
or 6n 1 3 9
7 5 x , , and in (–) 3 9 9 9 sum = 0 41. A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant T2 produces 80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that P(computer turns out to be defective given that is produced in plant T1) = 10P(computer turns out to be defective given that it is produced in plant T2) where P(E) denotes the probability of an event E. A computer produces in the factory is randomly selected and it does not turn out to be defective. Then the probabality that it is produced in plant T2 is 36 47 78 75 (A) (B) (C) (D) 73 79 93 83 Ans. (C) 28/40 CODE-9 JEE(Advanced) 2016 Final Exam/Paper-1/Held on Sunday 22nd May, 2016
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Sol. P T1
20 100
P(T2) =
MATHEMATICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 80 100
D Let P x T2 D P 10x T1 P D
7 (given) 100
7 D D P T1 P P T2 P T1 T2 100 20 80 7 10x x 100 100 100 1 40
x
D P T2
D 1 40 P T 2
39 40
D 30 D 10 P P T1 40 T1 40
80 39 78 T2 100 40 P D 20 30 80 39 93 100 40 100 40
SECTION–2 : (Maximum Marks : 32) This section contains EIGHT questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble(s) corresponding to the correct option(s) is (are) darkened. Partial Marks : +1 For darkening a bubble corresponding to each correct option, Provided NO incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. for example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result in –2 marks, as a wrong option is also darkened
CODE-9
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MATHEMATICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 42.
2
A solution curve of the differential equation (x + xy + 4x + 2y + 4)
dy y 2 0, x 0 , passes through dx
the point (1,3). The the solution curve(A) intersects y = x + 2 exactly at one point (B) intersects y = x + 2 exactly at two points (C) intersects y = (x + 2)
2
(D) does NOT intersect y = (x + 3)
2
Ans. (A,D) 2 Sol. (x xy 4x 2y 4)
((x 2) 2 y(x 2))
dy y2 0 dx
dy y2 dx
Let x + 2 = X, y = Y (X)(X Y)
dY Y2 dX
–X2dY = XYdY – Y2dX 2
–X dY = Y(XdY – YdX)
dY XdY YdX Y X2
Y n | Y | C X
– n | y |
y C x2
it is passing through (1, 3)
–n3 =1 + C
C = –1 – n3
curve
y n | y | 1 n3 0, x 0 x2
......(i)
put y = x + 2 in equation (i) then
x2 n | x 2 | 1 n3 0 x2
x = 1, –5(reject)
curve intersect y = x + 2 at point (1, 3)
for option (C), put y = (x + 2)2, we will get x + 2 + 2n(x + 2) = 1 + n3 30/40
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JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 Clearly left hand side is an increasing function Hence, it is always greater than 2 + 2n2 therefore no solution 2 for option (C) put y = (x + 3) in equation (i)
(x 3)2 n(x 3)2 1 n3 0 x2 (x 3)2 (x 3)2 n 1 0 3 x2
x>0
x+3>x+2 and x + 3 > 3
(x 3)2 (x 3)2 n 1 So 3 x2 (x 3)2 (x 3)2 1 0 has no solution n 3 x2
2
curve y = (x + 3) does not intersect 43.
Consider a pyramid OPQRS located in the first octant (x > 0,y > 0, z > 0) with O as origin, and OP and OR along the x-axis and the y-axis, respectively. The base OPQR of the pyramid is a square with OP= 3. The point S is directly above the mid-point T of diagonal OQ such that TS = 3. Then(A) the acute angle between OQ and OS is
3
(B) the equaiton of the plane containing the triangle OQS is x – y = 0 (C) the length of the perpendicular from P to the plane containing the triangle OQS is
(D) the perpendicular distance from O to the straight line containing RS is
3 2
15 2
Ans. (B,C,D) 3 3 S 2 , 2 ,3 R(0,3,0)
O
y
T
Sol. P(3,0,0)
Q(3,3,0)
x
CODE-9
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MATHEMATICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 Given OP = OR = 3 and OPQR is a square OQ = 3 2
using SOT, tan
OT
3
and ST = 3
2
ST 2 OT
tan 1 2
clearly, equation of plane containing triangle OQS is Y – X = 0 Also, length of perpendicular from P to the plane containing the triangle OQS is PT =
3 2
3 3 Also equation of RS is r 3ˆj t ˆi ˆj 3kˆ 2 2 O(0,0,0)
3t 3t , 3 ,3t 2 2
3t 3t Let co-ordinates of M , 3 ,3t 2 2
44.
OM.RS 0
M
R(0,3,0)
3 3t 9 t 3 9t 0 4 2 2
1 5 M , ,1 2 2
OM
9t 9 9t 2 2
t
3 3 S 2 , 2 ,3
1 3
1 25 30 15 1 4 4 4 2
In a triangle XYZ, let x,y,z be the lengths of sides opposite to the angles X,Y,Z, respectively and 2s = x + y + z. If
sx sy sz and area of incircle of the triangle XYZ is 3 2 4
8 , then3
(A) area of the triangle XYZ is 6 6 (B) the radius of circumcircle of the triangle XYZ is (C) sin
35 6 6
X Y Z 4 sin sin 2 2 2 35
3 2XY (D) sin 2 5 Ans. (A,C,D) 32/40
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Sol.
X
sx sy sz k Let 4 3 2 s – x = 4k y + z – x = 8k
s – y = 3k
x + z – y = 6k
s – z = 2k
x + y – z = 4k
MATHEMATICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9
y
z Y
x
Z
x = 5k, y = 6k, z = 7k 3s – (x + y + z) = 9k s = 9k Let r be inradius
r 2
8 3
2
8 3 s
8 .s 3
s(s x)(s y)(s z) 9k.4k.3k.2k 24.9k 2
8 .s 3
8 9k 3
8 .9k 3
k=1 x = 5, y = 6, z = 7
8 8 .9k .9 6 6 3 3
R = circumradius =
xyz 5.6.7 35 4 4.6 6 4 6
X Y Z r 2 36.6 4 sin sin sin s 2 2 2 4R xyz s.xyz 9.5.6.7 35
cos Z
x 2 y 2 z 2 25 36 49 1 2.5.6 5 2xy
1 Z 1 cos Z XY 5 3 cos2 sin 2 2 2 2 5 2 1
CODE-9
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MATHEMATICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 45.
Let RS be the diameter of the circle x2 + y2 = 1, where S is the point (1,0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. then the locus of E passes through the point(s)-
1 1 (A) , 3 3
1 1 (C) , 3 3
1 1 (B) , 4 2
1 1 (D) , 4 2
Ans. (A,C)
(cos, sin) y=1 Q
E(h,k) (–1,0)R
Sol.
S(1,0)
Tangent at P : x cos + y sin = 1
...(i)
Tangent at S : x = 1
...(ii)
1 cos By (i) & (ii) : Q 1, sin
Line through Q parallel to RS : y
1 cos y tan 2 sin
Normal at P : y
......(iii)
sin x y tan .x cos
......(iv)
1 tan 2 Point of intersection of equation (iii) and (iv), E : h
eliminating : h
2
2 ; k tan 2
1 k2 y 2 1 2x 2
Options (A) and (C) satisfies the locus. 34/40
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46.
2
2
2
The circle C1: x + y = 3, with centre at O, intersects the parabola x = 2y at the point P in the first quadrant. Let the tangent to the circle C1 at P touches other two circles C2 and C3 at R2 and R3, respectively. Suppose C2 and C3 have equal radii 2 3 and centres Q2 and Q3, respectively. If Q2 and Q3 lie on
MATHEMATICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9
the y-axis, then(A) Q2Q3 = 12
(B) R2R3 = 4 6
(C) area of the triangle OR2R3 is 6 2
(D) area of the triangle PQ2Q3 is 4 2
Ans. (A,B,C) Sol. On solving x2 + y2 = 3 and x2 = 2y we get point P( 2,1) Equation of tangent at P
2.x y 3 Q2(0,9) C 2
Let Q2 be (0,k) and radius is 2 3
R2
2y = x2
2 0 k 3 2 3 2 1
2
2
P(2,1)
x +y =3
R3
Q3(0,–3)
k = 9, – 3
C3
Q2(0,9) and Q3(0, –3) hence Q2Q3 = 12
R2R3 is internal common tangent of circle C2 and C3 R2R3
Q2Q3
2
2 3 2 3
2
122 48 96 4 6
Perpendicular distance of origin O from R2R3 is equal to radius of circle C1 = 3 1 1 Hence area of OR2R3 R2 R3 3 .4 6. 3 6 2 2 2
Perpendicular Distance of P from Q2Q3 2 Area of PQ2Q3 =
CODE-9
1 12 2 6 2 2
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MATHEMATICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 47.
3
Let ƒ : R R, g : R R and h : R R be differentiable functions such that ƒ(x) = x + 3x + 2, g(ƒ(x)) = x and h(g(g(x))) = x for all x R. Then(A) g ' 2
1 15
(B) h'(1) = 666
(C) h(0) = 16
(D) h(g(3)) = 36
Ans. (B,C) Sol. (A) ƒ ' x 3x 2 3
1 ƒ '0
so, g'(2) =
g '(2)
(Given g(x) = ƒ–1(x))
1 3
(B) h(g(g(x)) = x
h '(g(g(x))
1 g'(g(x)).g'(x)
Now, g(g(x)) = 1 g(x) = ƒ(1) = 6
x = ƒ(6) = 236
so h '(1)
1 1 h'(1) = 666 g '(6).g'(236) 1 . 1 6 111
(C) g(g(x)) = 0
g(x) = g–1(0) g(x) = ƒ(0) g(x) = 2 x = g–1(2) x = ƒ(2) x = 16
so h(0) = 16 –1 (D) g(x) = 3 x = g (3) x = ƒ(3) x = 38
so h(g(3)) = 38 48.
Let ƒ : (0, ) R be a differentiable function such that ƒ ' x 2
ƒ x for all x (0,) and x
ƒ(1) 1. Then
1 (A) lim ƒ ' 1 x x 0
1 (B) lim xƒ 2 x x 0
2 (C) lim x ƒ ' x 0 (D) |ƒ(x)| < 2 for all x (0,2) x 0
Ans. (A) 36/40
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JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 Sol. Let y = ƒ(x) dy y 2 (linear differential equation) dx x
y.e
dx x
2 e
dx x
2 e
dx x dx c
yx = 2 xdx c yx = x2 + c c ; As ƒ(1) 1 c 0 x
ƒ ' x 1
c ,c 0 x2
ƒ(x) = x +
1 2 lim ƒ ' lim 1 cx 1 x x 0 x 0
1 1 2 (B) lim xƒ lim x cx lim 1 cx 1 x x x 0 x 0 x 0
c 2 2 2 (C) lim x ƒ ' x lim x 1 2 lim x c c x x 0 x 0 x 0 c (D) ƒ x x , c 0 x
for c > 0
lim ƒ x function is not bounded in (0,2). x 0
49.
3 1 2 Let P 2 0 , where R, Suppose Q = [qij] is a matrix such that PQ = kI, where k R, 3 5 0 k k2 k 0 and I is the identity matrix of order 3. If q23 = – and det(Q) = , then8 2
(A) = 0, k = 8
(B) 4 – k + 8 = 0 9
(C) det (Padj(Q)) = 2
13
(D) det(Qadj(P)) = 2
Ans. (B,C) CODE-9
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JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 Sol. PQ = kI 3 |P|.|Q| = k |P| =2k 0 P is an invertible matrix PQ = kI –1 Q = kP I Q
adj.P 2
q23 =
k 8
k 3 4 k4 2 8 |P| = 2k k = 10 + 6 ...(i) Put value of k in (i).. we get =–1 4 – k + 8 = 0
2
50.
k2 k5 29 & det (P(adj.Q)) = |P| |adj.Q| = 2k. 2 2 SECTION–3 : (Maximum Marks : 15) This section contains FIVE questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories : : +3 If only the bubble corresponding to the correct answer is darkened. Full Marks Zero Marks : 0 In all other cases. 2
Let m be the smallest positive integer such that the coefficient of x in the expansion of 2 3 49 50 51 (1 + x) + (1 + x) + ....... + (1 + x) + (1 + mx) is (3n + 1) C3 for some positive integer n. Then the value of n is
Ans. 5
Sol. Coefficient of x2 in the expansion of (1 + x)2 + (1 + x)3 + ....... (1 + x)49 + (1 + mx)50 is 2
C2 + 3C2 + ...... 49C2 +
3
C3 + C2 + ....... C2 + C2m = (3n + 1) C3
3
50
50
49
C2m2 = (3n + 1) 51C3
50
C3 + 50C2m2 = (3n + 1)
2
51
51
C3
50.49.48 50.49 2 51.50.49 m = (3n + 1) 2 6 6
m2 = 51n + 1 must be a perfect quuared n = 5 and
m = 16
Ans. 5 38/40
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51.
The total number of distinct x R for which
x x2 1 x3 2x 4x 2 1 8x 3 3x 9x 2 1 27x 3
MATHEMATICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9
= 10 is
Ans. 2
1 1 1 x3 3 3 Sol. x 2 4 1 8x 10 3 9 1 27x 3
1 1 1 1 1 1 3 3 x 1 2 4 x .x 2 4 8 0 3 9 27 1 3 9 3
x3(25 – 23) + 6x6.2 = 10 6x6 + x3 – 5 = 0
5 x3 , 1 6
two real solutions
52.
–1 3i Let z = , where i = 2
(z)r 1 , and r, s {1, 2, 3}. Let P = z 2s
z 2s and I be the identity zr
matrix of order 2. Then the total number of ordered pairs (r, s) for which P2 = –I is Ans. 1
Sol. z =
()r P 2s
2s 2 , P = –I r
2r 4s r 2s (( 1) r 1) P 2 r 2s I r 4s 2r ((1) 1) r
(–1) + 1 = 0 r is odd r = 1,3 2r 4s also + = –1 r 3 2 4s by r = 1 + = –1 s=1 (r, s) = (1, 1) only 1 pair CODE-9
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MATHEMATICS
JEE(Advanced) 2016 Final Exam/Paper-1/Code-9 53.
x
t2 The total number of distinct x [0, 1] for which 1 t 4 dt = 2x – 1 is 0
Ans. 1 x
t2 dt 2x 1 Sol. Let ƒ(x) 4 0 1 t x2 2 ƒ'(x) = 1 x4 1 x4 2 as x2 3 2
ƒ(x) is continuous and decreasing
ƒ'(x)
x2 1 4 2 1 x
1
t2 3 dt 2 4 2 0 1 t
ƒ(0) = 1 and ƒ(1)
by IVT ƒ(x) = 0 possesses exactly one solution in [0, 1] 54.
Let R be such that lim x 0
Ans. 7
Sol. If 1, then lim
x 0
x 2 sin(x) = 1. Then 6( + ) equals x sin x
x sin x 0 x sin x x
sin x x lim x 0 3 x sin x 1/ 6 x 3 x x3
1
6 = 1
1 6
6( ) = 7
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