2017-Jee-Advanced
Question Paper-1_Key & Solutions
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bubble.
2017-Jee-Advanced
Question Paper-1_Key & Solutions PART-1:PHYSICS SECTION -1 (Maximum Marks: 28)
This Section Contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are) Correct. For each question, darken the bubble(s) corresponding to all the correct options(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 all the correct options(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 get +4 marks; darkening only (A) and (D) will get +2 marks; and darkening (A) and (B) will get -2 marks; as a wrong option is also darkened. **************************************************************************************************
Q1.
In the circuit shown,
L 1 H , C 1 Fand R 1k . They are connected in series with an a.c. source
V V0 sin t as shown. Which of the following options is/are correct? L 1 H
C 1 F R 1 K
V0 sin t
A) At
0 the current flowing through the circuit becomes nearly zero
B) The frequency at which the current will be phase with the voltage is independent of R C) The current will be in phase with the voltage if D) At
104 rad . s 1
106 rad .s 1 , the circuit behaves like a capacitor
Key:
A, B
Sol:
~ O V0 sin A so no current and option A is correct (A) current in phase with voltage implies resonance so, L
1 or 0 C
1 frequency depends on LC
capacitance and inductance not on resistance. (B) is correct
1 6
10 10
6
106 X L X C
106 rad01 so (C) is not correct. or circuit behaves as inductive so (D) is not correct.
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0
2017-Jee-Advanced Q2.
Question Paper-1_Key & Solutions
For an isosceles prism of angle A and refractive index
, it is found that the angle of minimum deviation
m A . Which of the following options is/are correct? A) For the angle of incidence i1
A , the ray inside the prism is parallel to the base of the prism
B) At minimum deviation, the incident angle i1 and the refracting angle r1 at the first refracting surface are related by
r1 i1 / 2
C) For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is i1 sin
D) For this prism the refractive index
Sol:
2 A sin A 4cos 1 cos A 2
1 and the angle of prism A are related as A cos 1 2 2
A, B, C
A m sin 2 sin A m A & sin A / 2 sin A / 2 or or
2sin A / 2cos A / 2 2cos A / 2 sin A / 2
cos A / 2 / 2 and A / 2 cos1 / 2
or A 2cos
1
/ 2 (D) is not correct.
m i1 i2 A i1 i2 2 Aand i1 i2 A So option A is correct.
r1 r2 and 2cos A / 2
sin i1 sin r1
sin A 2sin A / 2cos A / 2 sin r1 sin r1 2cos A / 2
r1 A / 2 ri / 2 option (B) is correct. Emergent ray tangential to the surface means
i2 90 r2 C (critical angle)
r1 A C
sin i1 sin r1
sin i1 sin r1 2cos A / 2 sin A C
2cos A / 2 sin A cos C cos Asin C
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Key:
1
2017-Jee-Advanced
Question Paper-1_Key & Solutions
1 2cos A / 2 sin A 1 sin 2 C cos A
1 1 2cos A / 2 sin A 1 2 cos A 1 1 2cos A / 2 sin A 1 cos A 4cos 2 A / 2 2cos A / 2 sin A 4cos 2 A / 2 1 1 2cos A / 2 cos A 2cos A / 2 2cos A / 2
sin A 4cos 2 A / 2 1 cos A
i1 sin 1 sin 1 A 4cos2 A / 2 1 cos A option C is correct A circular insulated copper wire loop is twisted to form two loops of area A and 2A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field B points into the plane of the paper. At common diameter as axis with a constant angular velocity
t 0 , the loop starts rotating about the
in the magnetic field. Which of the following
options is/are correct?
B Area A
Area 2A
A) The emf induced in the loop is proportional to the sum of the areas of the two loops B) The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper C) The net emf induced due to both loops is proportional to cos t D) The amplitude of the maximum net emf induced due to the both loops is equal to the amplitude of maximum
emf induced in the smaller loop alone
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Q3.
2017-Jee-Advanced
Question Paper-1_Key & Solutions
Key:
B,D
Sol:
emf induced in the loop 1
d BA cos t dt
emf induced in the loop 2
d B2 A cos t dt
But these two are in opposite in sense as Area vectors are in opposite direction.
net emf induced difference of area so option A is wrong Rate of change of flux Will be max when Q4.
d d BA cos t or B 2 A cos t dt dt
t 90 so option B is correct and option D is currect
A flat plate is moving normal to its plane through as a gas under the action of a constant force F . The gas is kept at a very low pressure. The speed of the plate
v is much less than the average speed u of the gas molecules.
Which of the following options is/are true? A) At a later time the external force F balance the resistance force B) The plate will continue to the move with constant non-zero acceleration, at all times C) The resistive force experienced by the plate is proportional to v D) The pressure difference between the leading and trailing faces of the plate is proportional to Key:
ACD
Sol:
S area of plate
uv
F
V
2
1
u v P f 1
p1 p2 p1 p2
2
2
p2
2
2
u v
2
2
u v u v 2
2
4Uv
p1 p2 2uv Resistance form p1
p2 s
2uvS
2US V
KV
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K 2US (constant)
2017-Jee-Advanced Resistance force
Question Paper-1_Key & Solutions
v
After long time, resistance can balance external force. Q5.
A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at x 0 , in a co-ordinate system fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is x and the velocity is
v . At that instant, which of the
following options is/are correct?
R y
m R
x
M
x0 A) The velocity of the point mass m is: v
2 gR m 1 M
B) The x component of displacement of the centre of mass of the block M is: C) The position of the point mass is x 2 D) The velocity of the block M is: V Key:
A,B
Sol:
1) mV MV 1
v and v
1
mR M m
mR M m
m M
2 gR
are velocities of m and M
2 1 1 2) mgR mV 2 M V 1 2 2
1 1 m mgR mV 2 M V 2 2 M
2 gR V m 1 M
Option (A)
Option (B)
x
2
m R x Mx 0
mR M m
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But in negative x direction hence B is correct.
2017-Jee-Advanced Q6.
Question Paper-1_Key & Solutions
A block M hangs vertically at the bottom end of a uniform rope of constant mass per unit length. The top end of the rope is attached to a fixed rigid support at at point O on the rope. The pulse takes time
O . A transverse wave pulse (Pulse 1) of wavelength 0 is produced
TOA to reach point A. If wave pulse of wavelength 0 is produced at
point A (Pulse 2) without disturbing the position of M it takes time TAO to reach point
O . Which of the
following options is/are correct?
O
Pulse1
Pulse 2 A M A) The time
TAO TOA
B) The wavelength of pulse 1 becomes longer when it reaches point A C) The velocity of any pulse along the rope is independent of its frequency and wavelength D) The velocity of the two pulses (Pulse 1 and pulse 2) are the same at the midpoint of rope Key:
ACD
Sol:
option A is correct as variation of velocity is same w.r.t “y”
T
Velocity Q7.
so “C” & “D” are correct. 2
A human body has a surface area of approximately 1m . The normal body temperature is 10 K above the surrounding room temperature
T0 . Take the room temperature to be T0 300 K . For T0 300 K , the value of
T04 460Wm2 ( where is the Stefan Boltzmann constant). Which of the following options is/are correct? A) If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths B) If the surrounding temperature reduces by a small amount
T0 T0 , then to maintain the same body
temperature the same (living) human being needs to radiate W
4 T03T0 more energy per unit time
C) The amount of energy radiated by the body in 1 second is close to 60 joules D) Reducing the exposed surface area of the body (e.g. by curling up) allows humans to maintain the same body temperature while reducing the energy lost by radiation Key:
BCD or D
Sol:
(A) is clearly wrong from win’s displacement law.
(B) correct:- W1 A T T0
4
W2 A T 4 T0 T0
4
4
W2 W1 4 AT03T0
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(C) net heat radiated is 60J per second. (D) True, by reducing are a net power emitted is reduced while his body will same heat inside so will be easy to maintain temperature.
2017-Jee-Advanced
Question Paper-1_Key & Solutions SECTION -2 (Maximum Marks: 15)
Q8.
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. Negative Marks : -2 In all other cases. **************************************************************************************************
An electron in a hydrogen atom undergoes a transition from an orbit with quantum number
ni to another with
quantum number n f . Vi and V f are respectively the initial and final potential energies of the electrons. If
Vi 6.25 , then the smallest possible n f is Vf Key:
5
Sol:
Potential energy
Vi
27.2 ineV n2
27.2 27.2 and V f ni2 n 2f
2 Vi n f 6.25 (given) V f ni2
nf ni
2.5
for ni
2
n f 5 (smallest possible value) 2
A drop of liquid of radius R 10 m having surface tension S
0.1 Nm1 divides itself into K identical drops. 4
3
In this process the total change the surface energy U 10 J . If K 10 then the value of Key:
6
Sol:
Surface energy U1
is
4 R 2 S (for big drop)
U 2 K 4 r 2 S ( for small drop) R3 Kr 3 r
R K 1/3
U K 4 r 2 S 4 R2 S
4 R 2 S K 4 R 2 S 4 R 2 S K 1/3 1 2/3 K 4 R 2 SK 1/3 4 102
2
0.1 1/3 K 103 4
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Q9.
2017-Jee-Advanced
Question Paper-1_Key & Solutions
105 K 1/3 103 K 1/3 102 10 /3 102
6 Q10.
A stationary surface emits sound of frequency the source with a speed of
f0 492 Hz . The sound is reflected by a large car approaching
2ms 1 . The reflected signal is received by the source and superposed with the
original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is
330ms 1 and the car reflects the sound at the frequency it has received). Key:
6
Sol:
frequency of sound received by car is
f1
V 2 492 V
V 330us 1
Reflected sound from the car will have a frequency f 2
V f1 V 2
V V 2 492 V 2 V
V 2 492 V 2 Beat frequency
f
f1 f
492
V 2 492 492 V 2
V 2 V 2 492
V 2
4 492 6 328
Q11.
I is an isotope of iodine that decays to an isotope of Xenon with a half-life of 8 days A small amount of a
131
131
serum laelled with I is injected into the blood of a person. The activity of the amount of
131
I injected was
2.4 105 Becquerel (Bq). It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After 11.5 hours 2.5 ml of blood is drawn from the person’s body, and gives an activity of 115 Bq. The total volume of blood in the person’s body, in liters is approximately (you may use e 1 x for x
Key:
5
Sol:
131
x 1 and ln 2 0.7 ).
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I Xe
2017-Jee-Advanced dN N dt
Question Paper-1_Key & Solutions
2.4 105
i.e. initially.
N0 N
2.4 105
2.4 105
e
t
2.4 105
e
11.50.693 824
For 2.5 ml we get 115 Bq So for “V” vol. we get
115 V N 2.5
115V 2.4 105 t e 2.5 V 0.0521105 0.95
5lit In litres. Q12.
A monochromatic light is travelling in a medium of refractive index n=1.6. It enters a stack of glass layers from the bottom side at an angle
300 . The interfaces of the glass layers are parallel to each other. The refractive
indices of different glass layers are monotonically decreasing as th
index of the m slab and
m 1 m m 1
th
and m
th
nm n mn, where nm is the refractive
n 0.1 (see the figure). The ray is refracted out parallel to the interface between the
slabs from the right side of the stack. What is the value of m?
n mn n m 1 n
n3 n n2 n n n n
3 2 1
Key: 8
n sin 300 n mn sin 900
" m 8" ( but n for a material i.e n mn can’t be less than “1” ).
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Sol:
2017-Jee-Advanced
Question Paper-1_Key & Solutions SECTION -3 (Maximum Marks: 18)
This section Contains SIX questions of matching type. This section Contains TWO tables (each having 3 columns and 4 rows) Based on each table, there are THREE 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 charged particle (electron or proton) is introduced at the origin (x=0,y=0,z=0) with a given initial velocity . A uniform electric field E and a uniform magnetic field B exist everywhere. The velocity , electric field E and magnetic field B are given columns 1,2 and 3, respectively. The quantities
E0 , B0 are positive in magnitude.
Column 1
Column 2
(I) Electron with 2 (II) Electron with (III) Proton with
E0 x B0
E0 y B0
0
(IV) Proton with 2
E0 x B0
Column 3
(i) E E0 z
(P) B B0 x
(ii) E E0 y
(Q) B B0 x
(iii) E E0 x
(R) B B0 y
(iv) E E0 x
(S) B B0 z
Q13. In which case would the particle move in a straight line along the negative direction of y-axis (i.e., move along- y )? A) (IV) (ii) (S)
B) (II) (iii) (Q)
C) (III) (ii) (R)
D) (III) (ii) (P)
Key: C Sol:
Y By ase ele r ton pro
X Ey
Proton released from rest at origin
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F magnetic q(V y B y )
2017-Jee-Advanced Vy
Question Paper-1_Key & Solutions
qE y t j m
V y is antiparallel to B y F magnetic 0 Because of Electrostatic force, if the proton moves along –ve y-direction. Q14.
Key:
In which case will the particle move in a straight line with constant velocity? A) (II) (iii) (S)
B) (III) (iii) (P)
C) (IV) (i) (S)
D) (III) (ii) (R)
A
Sol: Y
V
FM
E
When
X
FE
F m F E 0, the electron moves with constant velocity along y-axis. qVy B0 qE0
q
E0 B0 qE0 B0
Fnet 0
V y remains constant. 15.
Key:
In which case will the particle describe a helical path with axis along the positive z direction? A) (II) (ii) (R)
B) (III) (iii) (P)
C) (IV) (i) (S)
D) (IV) (ii) (R)
C
Y
E
V
X
B
The proton describes helical path with increasing pitch, with axis of helix along z-axis and plane of helix is
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Sol:
Z
2017-Jee-Advanced
Question Paper-1_Key & Solutions
An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding P-V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamic processes. Here
is the ration of heat capacities at constant pressure and constant volume. The number of moles
in the gas is n. Column 1 (I) W12
1 PV PV 1 2 2 1 1
Column 2
Column 3
(i) Isothermal
(P)
P
1
2
V (II)
W12 PV 2 2 PV 1 1
(ii) Isochoric
(Q) P 1
2 V
(III)
W12 0
(iii) Isobaric
P
1 2
V
(R) (IV) W12 nRT ln(
V2 ) V1
(iv) Adiabatic
P
1
2
(S)
V
Q16.
Which one of the following options correctly represents a thermodynamic process that is used as a correction
`
in the determination of the speed of sound in an ideal gas? A) (IV) (ii) (R)
B) (I) (ii) (Q)
C) (I) (iv) (Q)
D) (III) (iv) (R)
Key:
C
Sol:
Vsound
RT M0
as the sound wave propagates, the air in a chamber undergoes compression & rare fraction, &
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Undergo a adiabatic process. So curves are steeper than isothermal
2017-Jee-Advanced
Question Paper-1_Key & Solutions
dP P 1 dV Adi V dP P 2 dV Iso V Graph Q17.
' Q ' satisfies eqn(1)
Which of the following options is the only correct representation of a process in which U Q PV ? A) (II) (iii) (S)
B) (II) (iii) (P)
C) (III) (iii) (P)
D) (II) (iv) (R)
Key:
B
Sol:
U Q PV
U PV Q As U 0 W 0 Q 0 The Process represents, Isobaric process
Wgas P V P v2 v1
Pv2 Pv1 graph ‘p’ satisfies isobaricprocess Which one of the following options is the correct combination? A) (II) (iv) (P)
B) (III) (ii) (S)
C) (II) (iv) (R)
D) (IV) (ii) (S)
Key:
B
Sol:
Work done in isochoric process is zero
W12 0 as V 0 Graph ‘S’ represents isochoric processes
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Q18.
2017-Jee-Advanced
Question Paper-1_Key & Solutions PART-2:CHEMISTRY SECTION -1 (Maximum Marks: 28)
This Section Contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are) Correct. For each question, darken the bubble(s) corresponding to all the correct options(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 all the correct options(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 get +4 marks; darkening only (A) and (D) will get +2 marks; and darkening (A) and (B) will get -2 marks; as a wrong option is also darkened.
**************************************************************************************************
Q19.
The colour of the
X 2 molecules of group 17 elements changes gradually from yellow to violet down the group.
This is due to A) The physical state of
X 2 at room temperature changes from gas to solid down the group
B) Decrease in HOMO-LUMO gap down the group C) Decrease in
* O * gap down the group
D) Decrease ionization energy down the group Key:
BC
Sol:
IF Consider,
X 2 as F2
F2 MoT config is / s 2 * /s 2 2s 2 * 2s 2 * 2 pz 2 2 px 2 = 2 p y 2
* 2 px2 * 2 py2 HOMO * 2 pz0 LuMo Q20.
if gap between HOMO & LUMO Due to absorbed energy decreases complimentary colour is seen light energy is seen both ‘B’ & ‘C’ are thus applicable
Addition of excess aqueous ammonia to a pink coloured aqueous solution of
MCl2 .6H 2O (X) and NH 4Cl
gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1: 3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y. 2
A) The hybridization of the central metal ion in Y is d sp B) When X and Z are in equilibrium at
3
0C , the colour of the solution is pink
C) Z is a tetrahedral complex D) Addition of silver nitrate to Y gives only two equivalent of silver chloride ABC
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Key:
2017-Jee-Advanced Sol:
Question Paper-1_Key & Solutions
X Co H 2O 6 Cl2 pink sp3d 2
Y Co NH 3 6 Cl3 d 2 sp3
Z CoCl4 blue sp3 Option (B) is correct CBSE book equilibrium chapter. Hence ‘A’ is correct Z is tetrahedral , ‘C’ is correct Q21.
An ideal gas is expanded from
p1,V1,T1 to p2 ,V2 , T2 under different conditions. The correct statement(s)
among the following is (are) A) If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic B) The work done by the gas is less when it is expanded reversibly from V1 to V2 under adiabatic conditions as compared to that when expanded reversibly from V1 to
V2 under isothermal conditions
C) The work done on the gas is maximum when it is compressed irreversibly from constant pressure
p2 ,V2 to p1,V1 against
p1
D) The change in internal energy of the gas is (i) zero, if it is expanded reversibly with if it is expanded reversibly under adiabatic conditions with Key:
BC
Sol:
in free expansion the heat supplied in zero (z=0)
T1 T2 , and (ii) positive,
T1 T2
And work done is zero in both isothermal and adiabatic process But
T1 T2 isothermal
P
Adiabatic process
V In adiabatic process
PV cons tan t
is always greater than 1 so, opp. Pressure is less for same exp is
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Case of adiabatic process compare to isothermal so work done is less.
2017-Jee-Advanced
Question Paper-1_Key & Solutions
compression
PV 1 1
P2 ,V2
P
V Expansion
PV 1 1 P
PV 2 2
V Work done in compression is more than expansion or (Area under curve is more) wirr wrev comp
Exp
opp., pressure is more in compression Pext V In case of adiabatic expansion E And temperature also . Q22.
For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here
xL and xM represent mole fractions of L and M, respectively,
in the solution. The correct statement(s) applicable to this system is (are)
Z
pL
1
xM
0
A) Attractive intermolecular interactions between L-L in pure liquid L and M-M in pure liquid M are stronger than those between L-M when mixed in solution
xL 0
C) The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed from
xL 0 to xL 1
D) The point Z represents vapour pressure of pure liquid L and Raoult’s law is obeyed when
xL 1
AD
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Key:
B) The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed when
2017-Jee-Advanced
Question Paper-1_Key & Solutions
Sol:
actual given behaviour
Z
Q
pL
Xm 1 XL 0
expected ideal behaviour
XM 0 XL 1
xM
As actual is above expected ideal
+ve deviation L – L attn > L – M attn Also m –mattn > L – M attm Pt. z X L 1 only ‘L’ is present PL PL , PM 0
@ Z PL PL0 From pt Q to Z the graph (actual) is very similar to expected ideal behaviour where X L 1 it mm ideal behaviour . Q23.
The IUPAC name(s) of the following compound is (are)
Cl
H 3C
A) 1-chloro-4-methybenzene B) 4-chlorotoluene C) 1-methy-4-chlorobenzene D) 4-methylchlorobenzene AB
Cl 1
6
2
5
3 4
Sol:
CH 3
Alphabetical order is the priority order “C” of chloro has greater priority over “W” of methyl
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Key:
2017-Jee-Advanced
Question Paper-1_Key & Solutions
1-chloro -4-methyl benzene A
4- Chlorotoluene B According to NCERT BOOK Q24.
The correct statement(s) for the following addition reactions is (are)
H 3C H (i)
H 3C (ii)
H
H
Br2 CHCl3
M and N
CH 3 CH 3
Br2 CHCl3
O and P
H
A) O and P are identical molecules B) Bromination proceeds through trans-addition in both the reactions C) (M and O) and (N and P) are two pairs of enantiomers D) (M and O) and (N and P) are two pairs of diastereomers Key:
BD
Sol:
CH 3
CH 3 H
OH
H
OH
Br2 / CHCl3
(i )
HO
H
HO
H
CH 3
CH 3 M
N CH 3
CH 3
Br2
(ii )
CHCl3
OH
H
H
HO
CH 3 O
HO
H
H
OH CH 3
P
Bromination is anti addition . M & N are identical
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(M& O) & (N& P) two pairs of diastereomere
2017-Jee-Advanced Q25.
Question Paper-1_Key & Solutions
The correct statement(s) about the oxoacids, A) The conjugate base of
HClO4 and HClO , is(are)
HClO4 is weaker base than H 2O
B) The central atom in both
HClO4 and HClO is sp 3 hybridized
C)
HClO4 is formed in the reaction between Cl2 and H 2O
D)
HClO4 is more acidic than HClO because of the resonance stabilization of its anion
Key:
ABD
Sol:
HClO4 and HClO O O
Cl
O
O O
is resonance stabilized, so charge density is less, it is a weak base than mo
sp
Cl
3
Cl
O
OH
O
sp
O
3
so HClO4 is strong acid due to resonance stabilization of anion
ClO4 SECTION -2 (Maximum Marks: 15)
26.
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. Negative Marks : -2 In all other cases. **************************************************************************************************
The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrode is 120 cm with an area 7
of cross section of 1cm . The conductance of this solution was found to be 5 10 S . The pH of the solution is 2
4. The value of limiting molar conductivity m of this weak monobasic acid in aqueous solution is 0
Z 102 S cm1 mol 1 . The value of Z is Key:
6
Sol:
120 cm 7 1 K G cell constant k 5 107 S 2 600 10 S cm 1 cm
m
600 107 1000 600 107 1000 40 C 0.0015M
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pH 4
2017-Jee-Advanced
Question Paper-1_Key & Solutions
HT C 104
1 15
M 0m
0m
m
40 600 1/ 15
600S cm2 mol 1 6 102 2 2 10 27.
The sum of the number of lone pairs of electrons on each central atom in the following species is
TeBr6 , BrF2 2
, SNF3 , and XeF3
(Atomic numbers: N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54) Key:
6
F F
F 3 3 F sp d
Te F Sol:
F
Br
2 lonepair F
F
N
1 lonepair
F
S
F
0 zero-lone pair
F
Xe
3 lonepair
Total 6 lone pair Among the following, the number of aromatic compounds(s) is
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28.
2017-Jee-Advanced
Key:
Question Paper-1_Key & Solutions
5
Sol:
Aromatic
Aromatic
ions
Aromatic
Aromatic
Compounds
Aromatic
* It is clearly mentioned in the question word “compounds” then answer is “2” only 29.
A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is
8 g cm3 , then the number of atoms present in 256 g of the crystal is N 1024 .
The value of N is
2 FCC: Z=4
d
Z M a 3 .N O
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Key: Sol:
2017-Jee-Advanced 8
Question Paper-1_Key & Solutions
4 ( Mol.Wt )
(400)3 1012 NO 3
mol.mass 2 400 1012 N0 3
2 400 1012 N 0 3
no.of moles
256
2 400 1012 N 0 3
No.of atoms
256
2 N0 30.
Among
H 2 , He2 , Li2 , Be2 , B2 , C2 , N 2 , O2 , and F2 the number of diamagnetic species is
(Among numbers: H 1, Key:
5
Sol:
H 2 diamagnetic
He 2 , Li 3 , Be 4 , B 5 , C 6 , N 7 , O 8 , F 9 )
He2 paramagnetic
Li2 Diamagnetic Be2 Does not exist B2 Paramagnetic due to s p mixing C2 diamagnetic due to s p mixing N2 diamagnetic(14e ) O2 paramagnetic(17e )
F2 diamagnetic SECTION -3 (Maximum Marks: 18) This section Contains SIX questions of matching type. This section Contains TWO tables (each having 3 columns and 4 rows) Based on each table, there are THREE 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. **************************************************************************************************
Answer Q.31,Q.32 Q33 by appropriately matching the information given in the three columns of the following table. The wave function n,l ,ml is a mathematical function whose value depends upon spherical polar coordinates (r,
, ) of the electron and characterized by the quantum numbers n, l and ml . Here r is distance from nucleus, is colatitude and is azimuth. In the mathematical functions given in the Table, Z is atomic number and a0 is Bohr radius. Column 1
Column 2
Column3
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2017-Jee-Advanced
Question Paper-1_Key & Solutions
(I) Is orbital
3
(i) n,l ,ml
(P)
Zr
Z 2 e a0 a0
n,l ,m r 1
0
r / a0
(II) 2S orbital
(ii) one radial node
(Q) Probability density at nucleus
(III) 2 Pz orbital
(iii) n,l ,ml
(R) Probability density is maximum at
5 2
Z re a0
Zr 2 a0
nucleus
COS
(iv) xy-plane is a nodal plane
(IV) 3d z 2 orbital
1 a03
(S) Energy needed to excite electron from n=2 state to n= 4 state is
27 times the 32
energy needed to excite electron from n=2 state to n=6 state For He ion, the only INCORRECT combination is A) (I) (i) (R)
B) (II) (ii) (Q)
C) (I) (i) (S)
Key:
D
Sol:
1s 0 radial nodes , no angular component ( no cos )
D) (I) (iii) (R)
2s 1 radial node, , no angular component ( no cos )
2 pz 0 radial node, angular component is present ( cos ) 3d z2 0 radial node, angular component is present ( cos ) a)1s
(i) ‘R’ 1s orbital ’0’ radial node probability max@nucleas
correct b)2s 1 radial node of correct c) 1s (i) ’s’ correct
1s (i ) ' s ' correct
(2 4)
3 16
(2 6)
8 36
3 27 16 8 32 36
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Q31.
2017-Jee-Advanced
Question Paper-1_Key & Solutions
d ) 1s no angular component cos should not be present wrong. Q32.
For the given orbital in column I, the only CORRECT combination for any hydrogen-like species is A) (I) (ii) (S)
Key: Sol:
B) (IV) (iv) (R)
C) (III) (iii) (P)
D) (II) (ii) (P)
D
1s 0 radial nodes , no angular component ( no cos ) 2s 1 radial node, , no angular component ( no cos )
2 pz 0 radial node, angular component is present ( cos ) 3d z2 0 radial node, angular component is present ( cos ) a) 1s 0 radial node wrong as (ii),(i) 1 radial node b) 3d z2 dz 2 ' R ' is wrong po int c)for 2 pz graph ' p ' iswrong d) correct 133.
For hydrogen atom, the only CORRECT combination is A) (II) (i) (Q)
Key: Sol:
B) (I) (iv) (R)
C) (I) (i) (P)
D) (I) (i) (S)
D
1s 0 radial nodes , no angular component ( no cos ) 2s 1 radial node, , no angular component ( no cos )
2 pz 0 radial node, angular component is present ( cos ) 3d z2 0 radial node, angular component is present ( cos ) a) 2s 1radial node / coloumn(2) (i) no radial node wrong
b) 1s no node plane wrong c) 1s no radial node coloumn (3) ‘p’ 1 radial node wrong d) correct Answer Q.34,Q.35 and Q.36 by appropriately matching the information given in the three columns of the following table.
Columns 1, 2 and 3, contain starting materials, reaction conditions, and type of reactions, respectively. Column-1
Column-3
(I) Toluene
(i) NaOH / Br2
(P) Condensation
(II) Acetophenone
(ii) Br2 / hv
(Q) Carboxylation
(III) Benzaldehyde
(iii) CH 3CO 2 O / CH 3COOK
(R) Substitution
(IV) Phenol
(iv) NaOH / CO2
(S) Haloform
The only CORRECT combination in which the reaction proceeds through radical mechanism is A) (II) (iii) (R)
B) (III) (ii) (P)
C) (IV) ii) (Q)
D) (I) (ii) (R)
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34.
Column-2
2017-Jee-Advanced Key:
Question Paper-1_Key & Solutions
D
CH 3
CH 2 Br
Br2 / h
Sol:
Free radical substitution Ans: (I) (II) (R) 35.
For the synthesis of benzoic acid, the only CORRECT combination is A) (III) (iv) (R)
Key:
B) (IV) (ii) (P)
C) (II) (i) (S)
D) (I) (iv) (Q)
C
Sol:
COOH
0
NaoH / Br
C CH 3
+
CHBr3
afterH
Acetophenone
Haloform r n 36.
The only CORRECT combination that gives two different carboxylic acids is A) (IV) (iii) (Q)
Key:
S
B) (I) (i) (S)
C) (III) (iii) (P)
D) (II) (iv) (R)
C
Sol:
AC2O
CH O
CH 3COOK
CH CH COOH ACOH
Cis & trans After acidification
Condensation R n P (III)(iii)(P)
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(III)
2017-Jee-Advanced
Question Paper-1_Key & Solutions PART-3:MATHS SECTION -1 (Maximum Marks: 28)
This Section Contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are) Correct. For each question, darken the bubble(s) corresponding to all the correct options(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 all the correct options(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 get +4 marks; darkening only (A) and (D) will get +2 marks; and darkening (A) and (B) will get -2 marks; as a wrong option is also darkened. **************************************************************************************************
Q37.
Let a, b, x and y be real numbers such that a b 1 and y
0 . If the complex number z x iy satisfies
az b Im y , then which of the following is (are) possible value z 1 (S) of x? A) 1 1 y C) 1 1 y
B) 1 1 y
2
2
D) 1 1 y
2
Key:
AB
Sol:
az b az b zz z 1 z 1
2
a zz az bz b a zz bz az b z z z 1 z 1
az bz bz az z z z 1 z 1
a b z z
z 1 z 1
zz
z z 0 or z 1 z 1 1
x 1
2
y2 1
x 1 z y2
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x 1 1 y 2
2017-Jee-Advanced 38.
Lef
Question Paper-1_Key & Solutions
0,1 be a continuous function. Then which of the following functions (S) has (have) the value
f:
zero at some point in the interval 0,1 ?
A) f x
2
B) x f x
f t sin t dt
9
0
C) x
2
x
x
f t cos t dt
D) e x
f t sin t dt 0
0
Key: BC Sol: f x is ve ;
2 0
f t sin dt is ve
f x 02 f t sin dt is ve B) f x x f x f 0 0 f 0 ve 9
f 1 1 f 1 ve f 0 0 for atleast one x 0,1 C) f x x
2 0
x
f t cot dt
f 0 0 02 f t cost dt is - ve
1
f 1 1 f 02 f t cos t dt is ve f 0 0 for atleast one x 0,1 D) f 0 e f 0 f t sec t dt i 0
0
f 1 e f t sec t dt ve 1
0
f 1 x e x f x sin x 0 for x 0,1 39.
x2 y2 If 2 x y 1 0 is tangent to the hyperbola 2 1 , then which of the following CANNOT be sides of a a 16 right angled triangle?
Sol:
2a,4,1
B)
a,4,1
C)
a,4,2
D)
2a,8,1
BCD
2x y 1 0
x2 y2 1 a 2 16
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Key:
A)
2017-Jee-Advanced Tangent with slope 2
Question Paper-1_Key & Solutions y 2 x a 2 4 16
2 x y 4a 2 16 Comparing 4a 17 2
17 2
a
2a 40.
2
17 2a, 4,1
is Pythagorean triplet
Let X and Y be two events such that P X A) P X Y
'
C) P X / Y
'
1 1 2 , P X / Y and P Y / X . Then; 3 2 5
1 5
B) P Y
12
4 15
D) P X Y
Key:
BC
Sol:
1 1 2 P X , P X / Y and P Y / X 3 2 5
2 5
1 2 2 P X Y P X .P Y / X 3 5 15 P Y
P X Y 2 / 15 4 P X / Y 1 / 2 15
P X / Y '
P Y
P Y P X Y P Y
4 / 15 2 / 15 2 / 15 1 4 / 15 4 / 15 2
P X Y
1 4 2 9 2 7 3 15 15 15 15 15
Which of the following is (are) NOT the square of a 3 3 matrix with real entries?
1 0 0 A) 0 1 0 0 0 1
1 0 0 B) 0 1 0 0 0 1
1 0 0 C) 0 1 0 0 0 1
1 0 0 D) 0 1 0 0 0 1
Key:
AC
Sol:
Determent value should not be negative A , C
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41.
P X ' Y
2017-Jee-Advanced
Question Paper-1_Key & Solutions
1 0 0 For options B I 0 1 0 and for option (D) 0 0 1 2
42.
1 0 0 1 0 0 1 0 0 0 1 1 0 1 1 0 1 1 0 2 1 0 2 1 0 0 1
Let x be the greatest integer less than or equals to x .Then, at which of the following points(S) the functions
f x x cos x x is discontinuous? A) x 0
B) x 1
C) x 2
D) x 1
Key: BCD Sol: f x x cos
x x
f (0 ) f (0 ) f (0) f (1 ) 1cos x 1 dis continuous f (1 ) 1cos(2 x ) 1 f (2 ) 2cos3 x 2 f (2 ) 2cos 4 x 2 dis continuous f (2) 2cos 4 x 2 f ( 1 ) 1cos 2 x 1 dis continuous f ( 1 ) 1cos3 x 1 If chord, which is not a tangent, of the parabola y 16 x has the equation 2x 2
the which of the following is (are) possible values (S) of
y p , and midpoint h, k ,
p, h and k ?
A)
p 2, h 3, k 4
B)
p 5, h 4, k 3
C)
p 1, h 1, k 3
D)
p 2, h 2, k 4
Key: A Sol: S1
S11
ky 8 x h k 2 16h ky 8 x 8h k 2 0 ky 8x k 2 8h 0 Comparing with
y 2x p 0 k k 2 8h 4 1 p k 4 4 p k 2 8h
8h 4 p 16 2h p 4
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43.
2017-Jee-Advanced
Question Paper-1_Key & Solutions SECTION -2 (Maximum Marks: 15)
Q44.
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. Negative Marks : -2 In all other cases. **************************************************************************************************
The sides of a right angled triangle are in the Arithmetic progression. If the Triangle has area 24, then what is the length of its smallest side?
Key:
6
Sol:
Let the sides a-d, a, a+d
(a d )2 a 2 (a d )2 4ad a 2 a 4d Sides are 3d, 4d, 5d
Area : 6d 2 24 d=2
Smallest side = 6
Q45.
1 2 x 1 y = For a real number if the system 2 1 z
1 1 1
Of linear equations, has infinitely many solutions, then 1 2
Key:
1
Sol:
1 2 1 = 2 1 14 4 4 2 2 0 4 2 2 1 0 2 1
1 But gives no solution 1
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1 2 1
2017-Jee-Advanced Q46.
Question Paper-1_Key & Solutions
Words of length 10 are formed using the letters A, B, C D,E,F,G,H,I,J Let ‘x’ be the number of such words where no letter is repeated ; and let y be the number of such words Where exactly one letter is repeated twice and no other letter is repeated. Then
Key:
5
Sol:
x 10!
y = 9x
5
10! 10 9 10! y 10 C2 2 2 45 10! 2! 2! 2
y 45 10! 5 9 x 9 10! Q47.
For how many Values of p, the circle x 2 y 2 2 x 4 y p 0 and the coordinate axes have exactly three common points?
Key:
2
Sol: 48
Let f :
be a differentiable function such that f(0)=0, f(
)=3 and f 1 (0)=1. If g(x)= 2
2
f
'
(t )cos ect cot t cos ectf (t ) dt for x(0,
x
Key:
2
Sol:
g x f cos ec x f x 2 = 3
2
] then
Limx0 g ( x)
f x sin x
Let g x Lt 3 Lt x0
x0
x0
f x sin x
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f ' 0 3 Lt 2 x0 cos n
2017-Jee-Advanced
Question Paper-1_Key & Solutions SECTION -3 (Maximum Marks: 18)
Q49.
This section Contains SIX questions of matching type. This section Contains TWO tables (each having 3 columns and 4 rows) Based on each table, there are THREE 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. **************************************************************************************************
Answer Q.49,Q.50 and Q.51 by appropriately matching the information given in the three columns of the following tables Column 1
Column 2
Column 3
I. x 2 y 2 a 2
(i) my m2 x a
II. x 2 a 2 y 2 a 2
(ii) y mx a m 1
a 2a , 2 m m
p.
ma
2
Q.
2 m 1
III. y 2 4ax
(iii) y= mx
a 2 m2 1
R.
a 2 m
2 2 a m 1
IV. x 2 a 2 y 2 a 2
(iv) y= mx
a 2 m2 1
S.
a 2 m
2 2 a m 1
Q49.
1 2
The Tangent to a suitable conic (Column) at 3, is found to be
m2 1 a
,
,
a 2 m2 1
,
a 2m2 1
1
1
3 x+2y=4, then Which of the following
option is the only CORRECT combination?
Sol:
B) II iii R
C) IV iv S
D) II iv R
D
3x 2 y 4
y
3 x2 2
1 2
If 3, lieson II ,3
y
a2 a2 a2 4 4
3 3 x 4. 1 Satisfies D is correct 2 4
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Key:
A) IV iii S
2017-Jee-Advanced Q50.
Question Paper-1_Key & Solutions
If a tangent to a suitable conic (Column) is found to be
y x 8 and it’s point of contact is 8,16 then which
of the following option is the only CORRECT combination? A) III i P Key:
A
Sol:
y 1. x 8 y mx
Q51.
For a
B) I ii Q
C) II iv R
D) III ii Q
8,16 lieson162 4a.8 a 8
a matches A is correct m
2 , If a Tangent is drawn to a suitable conic (Column 1) at the point of contact
1,1 , then which of the following options is the only CORRECT combination for obtaining its equation? A) II ii Q
B) I i P
C) I ii Q
D) III i P
Key:
C
Sol:
I : x 2 y 2 2 tangent at 1,1 is x y 2 y x 2 1 1
I , ii, Q is C is correct Let f x x loge x x loge x, x 0, '
x and
f '' x .
Column 2 contains information about the limiting behavior of f x , f
'
x and
f '' x at infinity.
Column 3 contains information about increasing/decreasing nature of f x and f Column 1
(I) f x 0 for some x 1, e (II) f
'
x 0 for some '
x 0 for some x 0,1
(IV) f
''
x 0 for some x 1, e
x .
Column 2
Column 3
(I) lim x f x 0
(P) f is increasing in 0,1
(II) lim x f x
x 1, e (III) f
2
'
(III) lim x f
'
(IV) lim x f
x ''
x 0
(Q) f is decreasing in e, e
'
'
(S) f I decreasing in e, e
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(R) f is increasing in 0,1
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Column 1 contains information about zeros of f x , f
2017-Jee-Advanced Q52.
Which of the following options is the only CORRECT combination? A) I ii R
Key: Sol:
Question Paper-1_Key & Solutions B) IV i S
C) III iv P
D) II iii S
D
f x x log x x lg x
f ' x 1
f '' x
2 1 1 log x logx x x
1 1 x2 x
f 1 1 0, f e2 e2 2 2e2 0
f ' 1 1 0, f ' e
1 1 0 e
Let f x x
Let f ' x x
Let f '' x 0 x
f '' x 0, for x 0,1 f i s increa sin g f ' x 0, for x e, e2 f i s decrea sin g
f '' x 0, for x 0,1 f ' i s decrea sin g f '' x 0, for x e, e2 f ' i s decrea sin g Which of the following options I the only CORRECT combination? A) I i P Key: Sol:
B) II ii Q
C) III iii R
D) IV iv S
B
f x x log x x lg x
f ' x 1 f '' x
2 1 1 log x logx x x
1 1 x2 x
f 1 1 0, f e2 e2 2 2e2 0
f ' 1 1 0, f ' e
1 1 0 e
Let f x x
Let f ' x x
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53.
2017-Jee-Advanced
Question Paper-1_Key & Solutions
Let f '' x 0 x
Let f x x
Let f ' x x
Let f '' x 0 x
f '' x 0, for x 0,1 f i s increa sin g f ' x 0, for x e, e2 f i s decrea sin g
f '' x 0, for x 0,1 f ' i s decrea sin g f '' x 0, for x e, e2 f ' i s decrea sin g Which of the following options is the only INCORRECT combinations? A) II iii P Key: Sol:
B) I iii P
C) III i R
D) II iv Q
C
f x x log x x lg x
f ' x 1 f '' x
2 1 1 log x logx x x
1 1 x2 x
f 1 1 0, f e2 e2 2 2e2 0
f ' 1 1 0, f ' e Let f x
1 1 0 e
x
Let f ' x x
Let f '' x 0 x
Let f x x
Let f ' x x
Let f '' x 0 x
f '' x 0, for x 0,1 f i s increa sin g f ' x 0, for x e, e2 f i s decrea sin g
f '' x 0, for x 0,1 f ' i s decrea sin g f '' x 0, for x e, e2 f ' i s decrea sin g
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Q54.
2017-Jee-Advanced
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Question Paper-1_Key & Solutions
