DATE : 21/05/2017
CODE
6 Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.: 011-47623456 Fax : 011-47623472
Answers & Solutions
Time : 3 hrs.
Max. Marks: 183
for
JEE (Advanced)-2017 PAPER - 2 (Code - 6) INSTRUCTIONS QUESTION PAPER FORMAT AND MARKING SCHEME : 1.
The question paper has three parts : Physics, Chemistry and Mathematics.
2.
Each part has three sections as detailed in the following table : Section Question Number of Type Questions
Category-wise Marks for Each Question Full Marks
Partial Marks —
1
Single Correct Option
7
+3 If only the bubble corresponding to the correct option is darkened
2
One or more correct option(s)
7
+4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened
3
Comprehension
4
+3 If only the bubble corresponding to the correct answer is darkened
Zero Marks
Negative Marks
Maximum Marks of the Section
–1 0 If none of the In all other cases bubbles is darkened
21
–2 +1 0 For darkening a bubble If none of the In all other cases corresponding to each bubbles is correct option, provided darkened NO incorrect option is darkened
28
—
1
0 In all other cases
—
12
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
PHYSICS SECTION - 1 (Maximum Marks : 21) This section contains SEVEN 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
1.
In all other cases
Three vectors P , Q and R are shown in the figure. Let S be any point on the vector R . The distance between the points P and S is b R . The general relation among vectors P, Q and S is
Y P
(A) S 1 b P b 2Q
P
b|R |
S
S
Q
O
R QP Q X (B) S 1 b P bQ
(C) S 1 b2 P bQ
(D) S b 1 P bQ
Answer (B) Sol. S = P b | R | Rˆ R P b | R | = |R |
= P bR = P b Q – P
= (1 – b) P bQ 2.
A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is 3 × 105 times heavier than the Earth and is at a distance 2.5 × 104 times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is ve = 11.2 km s–1. The minimum initial velocity (vs) required for the rocket to be able to leave the Sun-Earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet) (A) vs = 72 km s–1
(B) vs = 22 km s–1
(C) vs = 62 km s–1
(D) vs = 42 km s–1
Answer (D) 2
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
Sol. vs
4
r = 2.5 × 10 R
m
E M1 = M
S 5 M2 = 3 × 10 M
loss in KE = Gain in PE
GM1m GM2m 1 mv s2 R r 2
1 2 GM G × 3 × 105M vs 2 R 2.5 × 104R
vs 2 ×
GM × 13 R
= 11.2 × 13 40.4 km/s
42 km/s 3.
A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is T = 0.01 seconds and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity g = 10 ms–2 and the velocity of sound is 300 ms–1. Then the fractional error in the measurement, L/L, is closest to (A) 1%
(B) 0.2%
(C) 5%
(D) 3%
Answer (A)
2L g
Sol. t1 = t2 =
L V
L
T = t1 + t2
2L L g V
T=
2 1 1 L L g 2 L V
T =
1 1 ⎞ ⎛ 1 ⎟ L 0.01 = ⎜ ⎝ 5 2 20 300 ⎠ 1 ⎞ ⎛ 1 0.01 = ⎜ ⎟ L ⎝ 20 300 ⎠
0.01 = L =
(15 1) L 300
0.01 300 16
L 3 100 = 100 = 1% L 16 20
3
t1
t2
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
4.
Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density remains uniform throughout the volume. The rate of fractional change in ⎛ 1 d ⎞ density ⎜ ⎟ is constant. The velocity v of any point on the surface of the expanding sphere is proportional ⎝ dt ⎠
to (B) R3
(A) R 1 R
(C)
(D) R2/3
Answer (A) Sol. M 0
4 R 3 3
4 ⎡ 2 dR d ⎤ ⎢3R R3 3 ⎣ dt dt ⎥⎦
Dividing by
0 3R 2
dR 1 d R3 dt dt
dR 3R 2 R 3K dt dR R dt 5.
hc ⎞ ⎛ A photoelectric material having work-function 0 is illuminated with light of wavelength ⎜ . The fastest 0 ⎟⎠ ⎝
photoelectron has a de-Broglie wavelength d. A change in wavelength of the incident light by result in a change d in d. Then the ratio d / is proportional to (A) d2 / 2
(B) d /
(C) 3d / 2
(D) 3d /
Answer (C) Sol. d
h ⎛ hc ⎞ ⎟ 2m ⎜ ⎝ ⎠
h ⎛ hc ⎞ ⎟ 2m ⎜ ⎝ ⎠ d 2 ⎛ hc ⎞ h f ⎟ 2 2m ⎜ ⎝ ⎠ d
h2 ⎛ 1 ⎞ 2m hc ⎜ 2 d ⎟ 3 d ⎝ ⎠ d d 3 k d2 d 4
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
6.
Consider regular polygons with number of sides n = 3, 4, 5.... as shown in the figure. The center of mass of all the polygons is at height h from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is . Then depends on n and h as
h
h
h 2⎛⎞ (B) h sin ⎜ ⎟ ⎝n⎠
1 ⎛ ⎞ 1⎟ (A) h ⎜ ⎜⎜ cos ⎛⎜ ⎞⎟ ⎟⎟ ⎝n⎠ ⎠ ⎝ 2⎛ ⎞ (C) h tan ⎜ ⎟ ⎝ 2n ⎠ Answer (A)
⎛ 2 ⎞ (D) h sin ⎜ ⎟ ⎝ n ⎠
/n Sol.
l
h
2 n
sin cos
n
h sin =l–h l
⎡ 1 ⎤ = h⎢ 1⎥ ⎣ sin ⎦
⎡ 1 ⎤ = h⎢ 1 ⎥ ⎢ cos ⎥ ⎣ ⎦ n 7.
A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at the center of the loop is I 4a
(A)
0I ⎡ 3 3 – 1⎤ ⎦ 4a ⎣
(B)
0I ⎡ 3 2 – 3⎤ ⎦ 4a ⎣
(C)
0I ⎡ 6 3 1⎤ ⎦ 4a ⎣
(D)
0I ⎡ 6 3 – 1⎤ ⎦ 4a ⎣
5
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
Answer (D) Sol. Considering one section out of symmetric star shaped conducting wire loop. From geometry :
O (Center of loop) 30 °
° 30
2a 0° 12
30° I
a
Magnetic field at the center of the loop due to all 12 identical sections is additive in nature. Bnet = 12 × =
0I cos30 cos120 4a
0I 6 ⎣⎡ 3 1⎦⎤ 4a
SECTION - 2 (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 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 all 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
Negative Marks
:
–2
If none of the bubbles is darkened 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. 8.
A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vertical is . Which of the following statements about its motion is/are correct? A L
B O
(A) When the bar makes an angle with the vertical, the displacement of its midpoint from the initial position is proportional to (1 – cos) (B) The midpoint of the bar will fall vertically downward (C) The trajectory of the point A is a parabola (D) Instantaneous torque about the point in contact with the floor is proportional to sin 6
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
Answer (A, B, D)
l Sol. Torque about O, at any instant is mg. sin . 2 Option (A)
mg
l 2
O
As no external force acts along x-axis, therefore centre of mass will fall vertically downward. Option (B) P (x,y)
l 2
l cos 2
l 2
O
Displacement of centre of mass along y-axis
l 1 cos 2
Option (D)
l x sin , y l cos 2 2
y2 ⎛ 2x ⎞ ⎜ ⎟ 2 1 ⎝ l ⎠ l
Trajectory is not parabola 9.
A source of constant voltage V is connected to a resistance R and two ideal inductors L1 and L2 through a switch S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At t = 0, the switch is closed and current begins to flow. Which of the following options is/are correct?
S R + V –
L1
L2
V L1 (A) After a long time, the current through L2 will be R L L 1 2 (B) The ratio of the currents through L1 and L2 is fixed at all times (t > 0)
V L2 (C) After a long time, the current through L1 will be R L L 1 2
(D) At t = 0, the current through the resistance R is
V R 7
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
Answer (A, B, C) Sol. Final current through battery =
V R
V ⎛ L2 ⎞ Current through L1 = R ⎜ L L ⎟ ⎝ 1 2⎠
V ⎛ L1 ⎞ Current through L2 = R ⎜ L L ⎟ ⎝ 1 2⎠ At t = 0 current through source = zero – tR ⎛ L1L2 ⎞ ⎞ ⎜⎝ L1 L2 ⎟⎠ e
At any time i = V ⎛ 0 – V R ⎜⎝ R ⎟⎠
L2 Current through L1 = i L L i1 1 2
iL1 Current through L2 = L L i 2 1 2 i1 L2 i 2 L1 10. A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct? Q
R
(A) Total flux through the curved and the flat surfaces is
Q 0
(B) The circumference of the flat surface is an equipotential (C) The component of the electric field normal to the flat surface is constant over the surface (D) The electric flux passing through the curved surface of the hemisphere is Answer (B, D) Sol. Net flux through curved surface and flat surface = 0 Q 45º R
R
Curved = – Plane
⎡ Q ⎤ ⎢ 1 cos ⎥ 2 ⎣ 0 ⎦ ⎡ Q ⎛ 1 ⎞⎤ ⎢ ⎜1 ⎟⎥ ⎝ 2 2⎠⎦ ⎣ 0 8
1 ⎞ Q ⎛ 1 ⎟ 2 0 ⎜⎝ 2⎠
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
The circumference points are equidistant from Q All points will be at the same potential. Option (B) and (D) are correct. 11. A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque about an axis normal to the plane of the paper passing through the point Q. Which of the following options is/are correct? S
Q
P R X
(A) If the force is applied normal to the circumference at point X then is constant (B) If the force is applied at point P tangentially then decreases continuously as the wheel climbs (C) If the force is applied tangentially at point S then 0 but the wheel never climbs the step (D) If the force is applied normal to the circumference at point P then is zero Answer (A, D) Sol. Correct options (A, D)
[ Treating magnitude of force constant]
For option (A): Applied force passes through point Q. So, its torque is zero. For option (D): Torque due to applied force at x remains constant. 3R (region 2 in the figure) pointing normally 2 into the plane of the paper. A particle with charge +Q and momentum p directed along x-axis enters region 2 from region 1 at point P1 (y = –R). Which of the following option(s) is/are correct?
12. A uniform magnetic field B exists in the region between x = 0 and x
Region 1
y
O +Q
P1
(y = –R)
× × × × × × × × ×
Region 2 × × B × × × × × × × 3R/2 9
Region 3 × × × × × × P2 x × × ×
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
(A) For B
8 p , the particle will enter region 3 through the point P2 on x-axis 13 QR
(B) When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its linear momentum between point P1 and the farthest point from y-axis is p / 2 (C) For a fixed B, particles of same charge Q and same velocity v, the distance between the point P1 and the point of re-entry into region 1 is inversely proportional to the mass of the particle (D) For B
2 p , the particle will re-enter region 1 3 QR
Answer (A, D) Sol. The particle will follow circular trajectory inside the magnetic field region. The magnetic field cannot change the magnitude of velocity and momentum.
For longest possible path, the radius of circular motion can be
3R . 2
×
O
P2
P1
At farthest point from y-axis, the momentum is directed upwards.
p 2p The radius and hence separation between p1 and re-entry point is proportional to m, if Q, v, B are same. The particle will return to region only if it completes the half circle.
r
3R 2
mV 3R B 2
p 3R QB 2 B
2p 3QR
If B
8p p 13R ;r 13QR QB 8 10
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
r
O
P2 R
r-r cos
P1
It passes through point P2 if r – r cos = R
R 12 sin 2 r 13 3
13R ⎛ 5 ⎞ 1– R 8 ⎜⎝ 13 ⎟⎠ R=R 13. Two coherent monochromatic point sources S1 and S2 of wavelength = 600 nm are placed symmetrically on either side of the centre of the circle as shown. The sources are separated by a distance d = 1.8 mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is . Which of the following options is/are correct? P1
P2
S2
S1 d
(A) A dark spot will be formed at the point P2 (B) At P2 the order of the fringe will be maximum (C) The angular separation between two consecutive bright spots decreases as we move from P1 to P2 along the first quadrant (D) The total number of fringes produced between P1 and P2 in the first quadrant is close to 3000 Answer (B, D) Sol. d = 1.8 × 10–3 m = 18 × 10–4 m
P1
P
S1
S2 d
11
P2
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
and = 6 × 10–7 m Path difference at point P (as shown) x = S1P – S2P = d sin , where angle is measured from vertical line as shown. For bright fringe d sin = m
...(i)
Point P1 is the point of central maxima. At point P2, path difference (x) = d If P2 is the point of bright fringe, then d m ⇒ m
d 3000
On differentiating equation (i) d cos () = (m) = constant for consecutive bright fringe cos as varies from 0 to
2
14. The instantaneous voltages at three terminals marked X, Y and Z are given by Vx V0 sin t ,
2 ⎞ ⎛ VY V0 sin ⎜ t and 3 ⎟⎠ ⎝ 4 ⎞ ⎛ VZ V0 sin ⎜ t 3 ⎟⎠ ⎝ An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be rms (A) VXY V0
3 2
(B) Independent of the choice of the two terminals
rms (C) VYZ V0
1 2
rms (D) VXY V0
Answer (A, B)
2 ⎞ ⎛ Sol. VXY = V0 sin t – V0 sin ⎜ t 3 ⎟⎠ ⎝ V0
60°
V0 3 V0 V0
rms VXY
3V0 2
2 ⎞ 4 ⎞ ⎛ ⎛ V0 sin ⎜ t VYZ = V0 sin ⎜ t ⎟ 3 ⎠ 3 ⎟⎠ ⎝ ⎝ 12
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
3 V0 V0
V0 60°
V0 rms VYZ
3V0 2
A, B
SECTION - 3 (Maximum Marks : 12) This section contains TWO Paragraphs. Based on each paragraph, there are TWO 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(s) 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
:
In all other cases
0
PARAGRAPH 1 Consider a simple RC circuit as shown in Figure 1. Process 1: In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V0 (i.e., charging continues for time T >> RC). In the process some dissipation (ED) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is EC. Process 2: In a different process the voltage is first set to the voltage is raised to
V0 and maintained for a charging time T >> RC. Then 3
2V0 without discharging the capacitor and again maintained for a time T >> RC. The 3
process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V0 as in Process 1. These two processes are depicted in Figure 2. v V0
S R
2V0/3
C
V + –
Process 1
Process 2
T >> RC
V0/3
T
Figure 1
2T
t
Figure 2
15. In Process 1, the energy stored in the capacitor EC and heat dissipated across resistance ED are related by : (A) EC = ED In 2
(B) EC = 2ED
(C) EC = ED
(D) EC = 13
1 ED 2
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
Answer (C) Sol. Final charge on capacitor = CV Wb = CV2 Ec =
1 CV 2 2
ED = Wb – Ec = CV2 –
=
1 CV 2 2
1 CV 2 2
EC ED
16. In Process 2, total energy dissipated across the resistance ED is : (A) ED
1 CV02 2
(B) ED 3CV02
(C) ED
1⎛ 1 ⎞ CV02 ⎟ 3 ⎜⎝ 2 ⎠
⎛1 2⎞ (D) ED 3 ⎜ CV0 ⎟ 2 ⎝ ⎠
Answer (C) Sol. ED = Wb – V =
CV0 ⎡V0 2V0 ⎤ 1 V0 ⎥ – CV02 ⎢ 3 ⎣3 3 ⎦ 2
=
CV0 ⎡V0 2V0 3V0 ⎤ 1 2 ⎥ – 2 CV0 3 ⎢⎣ 3 ⎦
=
CV0 1 2V0 – CV02 3 2
⎛ 2 1⎞ 2 = ⎜ – ⎟ CV0 ⎝3 2⎠ CV02 = 6
PARAGRAPH 2 One twirls a circular ring (of mass M and radius R) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity 0. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is and the acceleration due to gravity is g. 14
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
R r
Figure 1
R Figure 2
17. The total kinetic energy of the ring is
(A)
1 2 M 02 R r 2
(C) M 02 R r
(B)
2
3 2 M 02 R r 2
(D) M 02 R 2
Answer (D) Sol. k
1 1 Mv c2 Ic 2 2 2
1 1 M 02 (R r )2 MR 2 02 2 2
1 r ⎞ 1 ⎛ M 02R 2 ⎜ 1 ⎟ M 02R 2 2 2 ⎝ R⎠
VC = 0(R – r)
2
R– r
0
r << R
r 0 R
k
1 1 M 02R 2 M 02R 2 2 2
M02R 2 18. The minimum value of 0 below which the ring will drop down is
(A)
2g R r
(B)
3g 2 R r
(C)
g R r
(D)
g 2 R r
Answer (C) 15
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
f Sol. N
Mg N = M2 (R – r) f = Mg f N Mg M2 (R – r) 0
g R r
END OF PHYSICS
16
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
CHEMISTRY SECTION - 1 (Maximum Marks : 21) This section contains SEVEN 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 19. For the following cell, Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu(s) when the concentration of Zn2+ is 10 times the concentration of Cu2+, the expression for G (in J mol–1) is [F is Faraday constant; R is gas constant; T is temperature; E°(cell) = 1.1 V] (A) 2.303RT – 2.2F
(B) –2.2F
(C) 1.1F
(D) 2.303RT + 1.1F
Answer (A) Sol. Zn | ZnSO4 || CuSO4 | Cu (aq)
(aq)
G =G° + RT In Q
⎛ [Zn2 ] 10 ⎞ ⎜⎜ Q ⎟ 1 ⎟⎠ [Cu2 ] ⎝
G = G° + 2.303 RT log Q G°= – nF E°Cell = – 2F 1.1 = – 2.2 F G = – 2.2 F + 2.303 RT log
10 1
G 2.303 RT – 2.2 F
20. Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol–1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol–1] Among the following, the option representing change in the freezing point is 1
(B) Water + Et h
271
anol
Wate r Ice
V.P./bar
Ice
V.P./bar
(A)
1
Wate r
Water + Et h
273
270
T/K
T/K
1
+E ter Wa
Ice
270
ter Wa
nol tha
(D)
273
V.P./bar
V.P./bar
ter Wa
(C)
anol
273
1
271
T/K
17
+E ter Wa
Ice
273
T/K
nol tha
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
Answer (C)
⎡ W 1000 ⎤ Sol. Tf = iK f ⎢ 2 ⎥ ⎣ M2 W1 ⎦ ⎡ 34.5 1000 ⎤ = 1 2 ⎢ ⎣ 46 500 ⎥⎦ =3K
273 (K) – Tf = 3 (K) Tf = 270 K Also, with decrease in temperature, V.P. decreases. Graph (C) is correct. 21. The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are fG°[C(graphite)] = 0 kJ mol–1 fG°[C(diamond)] = 2.9 kJ mol–1 The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 × 10–6 m3 mol–1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is [Useful information : 1 J = 1 kg m2s–2; 1 Pa = 1 kg m–1 s–2; 1 bar = 105 Pa] (A) 29001 bar
(B) 58001 bar
(C) 14501 bar
(D) 1450 bar
Answer (C) Sol. Gº = V·P 2900 2 10 –6 P 2900 106 Pa 2 PF–1 = 14500 bar
P
PF = 14501bar 22. The order of basicity among the following compounds is
NH2
NH H3C
NH2
N
NH
HN
N
(A) II > I > IV > III
H2N NH III IV (B) IV > I > II > III
(C) I > IV > III > II
(D) IV > II > III > I
I
II
Answer (B)
NH2 Sol. H N 2
NH IV
Resonance with two
NH CH3
NH2
NH2 groups increases electron density on 'N' of
Lesser increase of electron density on = NH due to only one resonance with one –NH2 18
NH
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
sp2
N
–
This LPe is not available as it is involve in aromatic Sextet. 'N' is bonded to sp2 C on both sides.
N H III
N N H
This LPe– is not involve in aromaticity. So more available Also, 'N' is bonded to sp3 C on one side.
II IV > I > II > III 23. Which of the following combination will produce H2 gas? (A) Au metal and NaCN(aq) in the presence of air
(B) Fe Metal and conc. HNO3
(C) Zn metal and NaOH(aq)
(D) Cu metal and conc. HNO3
Answer (C) Sol. Zn + 2NaOH Na2ZnO2 + H2 Iron become passive with conc. HNO3. Copper liberate NO2 with HNO3 24. The order of the oxidation state of the phosphorus atom in H3PO2, H3PO4, H3PO3 and H4P2O6 is (A) H3PO4 > H3PO2 > H3PO3 > H4P2O6
(B) H3PO3 > H3PO2 > H3PO4 > H4P2O6
(C) H3PO4 > H4P2O6 > H3PO3 > H3PO2
(D) H3PO2 > H3PO3 > H4P2O6 > H3PO4
Answer (C) Sol. Oxidation state H3PO4
P=+5
H4P2O6
P=+4
H3PO3
P=+3
H3PO2
P=+1
H3PO4 > H4P2O6 > H3PO3 > H3PO2 25. The major product of the following reaction is
OH i) NaNO2, HCl, 0°C ii) aq.NaOH
NH2
OH
N=N
(A)
OH
(B)
Cl OH
–
O Na
(C)
(D) N=N
N2Cl
Answer (C) 19
+
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
OH NaNO2 + HCl 0°C
Sol.
–
OH
O aq.NaOH +
N+ NCl–
NH2
NN
OH
O
N=N H
N=N
SECTION - 2 (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 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 all 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
Negative Marks
:
–2
If none of the bubbles is darkened 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. 26. Among the following, the correct statement(s) is(are) (A) Al(CH3)3 has the three-centre two-electron bonds in its dimeric structure (B) AlCl3 has the three-centre two-electron bonds in its dimeric structure (C) BH3 has the three-centre two-electron bonds in its dimeric structure (D) The Lewis acidity of BCl3 is greater than that of AlCl3 Answer (A, C, D)
H Sol.
– e– H e
B H
–
e– e H
H It has two 3c-2e bonds.
B H
Cl 2AlCl3
CH3
Al Cl
No 3c-2e bond
Cl
CH3 Al
CH3
Cl Al
Cl
CH3 Al
CH3
Cl
CH3
Has two 3c-2e bonds.
Also BCl3 is stronger lewis acid than AlCl3. 20
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
27. For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by (A) With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases (B) With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative (C) With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases (D) With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive Answer (A, C, D) Sol. Whether reaction is endothermic or exothermic in forward direction increase in temperature cause intake of heat from surrounding to system in endothermic direction due to which entropy change in system is positive and S of surrounding is negative. 28. The option(s) with only amphoteric oxides is(are) (A) Cr2O3, BeO, SnO, SnO2 (B) ZnO, AI2O3, PbO, PbO2 (C) NO, B2O3, PbO, SnO2 (D) Cr2O3, CrO, SnO, PbO Answer (A, B) Sol. ZnO, Al2O3, PbO, PbO2, Cr2O3, BeO, SnO and SnO2 are amphoteric oxides. NO is neutral oxide CrO is basic oxide B2O3 is acidic oxide 29. In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. The correct option(s) among the following is(are) (A) The activation energy of the reaction is unaffected by the value of the steric factor (B) The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally (C) Since P = 4.5, the reaction will not proceed unless an effective catalyst is used (D) Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation Answer (A, B) Sol. Steric factor
A exp erimental A calculated
Steric factor = 4.5 It means Aexperimental > Acalculated [This seems that reaction occurs more quickly than particles collide, thus concept of steric factor was introduced] 30. For the following compounds, the correct statement(s) with respect to nucleophilic substitution reaction is(are)
CH3
CH3 Br
Br
I II (A) Compound IV undergoes inversion of configuration (B) The order of reactivity for I, III and IV is : IV > I > III (C) I and III follow SN1 mechanism (D) I and II follow SN2 mechanism 21
H3C – C – Br CH3 III
Br IV
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
Answer (A, C, D) Sol. When medium is highly polar and protic I & III will follow SN1. Option (A) is correct as Nu–
Br
Nu
SN2
Inversion in case of SN2. (B) is incorrect for both SN1 and SN2 conditions. Hence, (C) is correct. (D) I & II will follow SN2 when medium is polar aprotic and nucleophile is strong in high concentration. 31. Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is C8H8O. Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not Cannizzaro reaction (i)
i) O3/CH2CI2
P
Q (C8H8O)
ii) Zn/H2O
i) O3/CH2CI2
(ii) R
S (C8H8O)
ii) Zn/H2O
The option(s) with suitable combination of P and R, respectively, is(are) and H 3C
(A) H3C
CH3 (B)
CH3
H3C
and
H3C H3C
(C)
CH3
CH3
CH3 and
CH3
CH3
CH3 (D) H3C
CH3
and
CH3
Answer (B, D) (i) O3/CH2Cl2
Sol. CH3
(ii) Zn/H2O
O CH3
O
C–H+H–C–H Undergoes Cannizzaro's reaction
CH2
O
O3/CH2Cl2
C CH3
C
Zn/H2O
Undergoes Haloform reaction
22
O + H–C–H CH3
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
O
O3/CH2Cl2
CH3
C – H + CH3 – C – H
Zn/H2O
CH3
O
CH3
Undergoes Cannizzaro's reaction
CH3
CH3
O3/CH2Cl2 Zn/H2O
CH3
O
CH3
C
+ CH3
CH3
CH3 O
Undergoes Haloform reaction
32. The correct statement(s) about surface properties is(are) (A) The critical temperatures of ethane and nitrogen are 563 K and 126 K, respectively. The adsorption of ethane will be more than that of nitrogen on same amount of activated charcoal at a given temperature (B) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system (C) Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of the solution (D) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium Answer (A, B) Sol. Adsorption is an exothermic process and is accompanied by decrease is entropy,
H 0, S 0 sys
sys
More is critical temperature (Tc), more are intermolecular forces of attraction. More is extent of adsorption.
SECTION - 3 (Maximum Marks : 12) This section contains TWO Paragraphs. Based on each paragraph, there are TWO 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(s) 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
:
In all other cases
0
PARAGRAPH 1 Upon heating KCIO3 in the presence of catalytic amount of MnO2, a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure HNO3 gives Y and Z. 33. W and X are, respectively (A) O2 and P4O6
(B) O3 and P4O10
(C) O3 and P4O6
(D) O2 and P4O10
Answer (D) 23
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
34. Y and Z are, respectively (A) N2O4 and HPO3
(B) N2O3 and H3PO4
(C) N2O5 and HPO3
(D) N2O4 and H3PO3
Answer (C) Solutions of Q.No (33) & (34) [MnO ]
2 2KCl 3O2 Sol. 2KClO3
(W )
5O2 P4 P4O10 (X)
P4O10 4HNO3 2N2O5 4HPO3 (X)
(Z)
PARAGRAPH 2 The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q. The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 followed by treatment with H2O produces compound S. [Et in compound P is ethyl group]
(H3 C)3C
CO2Et
Q
R
S
P 35. The reactions, Q to R and R to S, are (A) Dehydration and Friedel-Crafts acylation (B) Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation (C) Aromatic sulfonation and Friedel-Crafts acylation (D) Friedel-Crafts alkylation and Friedel-Crafts acylation Answer (B) 36. The product S is
HO3S (H3C)3C
O
(H3C)3C
CH3
(A)
CH3
(B)
COCH3
COCH3
H3COC
COCH3 (C)
H3C
(H3 C)3C
CH3
(H3C)3C (D)
Answer (B) 24
H3C
CH3
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
Solutions of Q. 35 and 36 – +
O C(CH3)3
C – OEt
OMgBr CH3MgBr (excess)/(C2H5)2O
C(CH3)3
CH3–C–CH3
(P) H2O
H C(CH3)3
O
H
O–H
C(CH3)2
C(CH3)3
H2SO4/
CH3–C–CH3
0°C
(Q)
–H2O (dehydration)
C(CH3)3
C(CH3)2
CH3 CH3
C(CH3)3 (alkylation)
(R) O CH3–C–Cl/AlCl3 (Acylation)
C(CH3)3
CH3 CH3
COCH3 (S)
END OF CHEMISTRY
25
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
MATHEMATICS SECTION - 1 (Maximum Marks : 21) This section contains SEVEN 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
37. The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7, is (A) –14x + 2y + 15z = 3 (B) 14x + 2y + 15z = 31 (C) 14x + 2y – 15z = 1 (D) 14x – 2y + 15z = 27 Answer (B) Sol. Required equation of plane is
x 1 y 1 z 1 2 3
1 6
2 0 2
–14(x – 1) – 2(y – 1) + (–15)(y – 1) = 0
14 x 2y 15 y 31
38. Let S = {1, 2, 3, ..., 9}. For k = 1, 2, ..., 5, let Nk be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N1 + N2 + N3 + N4 + N5 = (A) 252 (B) 125 (C) 126 (D) 210 Answer (C) Sol. Required number of subsets = 5C1 × 4C4 + 5C2 × 4C3 + 5C3 × 4C2 + 5C4 × 4C1 + 5C5 × 4C0 = 5 + 40 + 60 + 20 + 1 = 126 Alternate method Coefficient of x5 in (1 + x)5(1 + x)4 = 9C5 = 126 26
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
⎛ 1⎞ 1 39. If f : » » is a twice differentiable function such that f (x) > 0 for all x » , and f ⎜ ⎟ , f (1) 1 , then ⎝ 2⎠ 2 (A) f (1) 1 (B) f (1) 0 (C)
1 f (1) 1 2
(D) 0 f (1)
1 2
Answer (A)
1 1 and f(1) = 1 Sol. f ( x ) 0 , f ⎛⎜ ⎞⎟ ⎝2⎠ 2 f ( x ) is always increasing
1 f (1) f ⎛⎜ ⎞⎟ ⎝2⎠ f (1) 1 1 2
B
1 1/2
A
1/2
1
f (1) 1
Slope of tangent at B > Slope of chord AB. 40. Let O be the origin and let PQR be an arbitrary triangle. The point S is such that OP OQ OR OS OR OP OQ OS OQ OR OP OS
Then the triangle PQR has S as its (A) Circumcentre (B) Incentre (C) Centroid (D) Orthocenter Answer (D) Sol. OP OQ OR OS OR OP OQ OS OQ OR OP OS
P
OP OQ OR OS OR OP OQ OS OP (OQ OR ) OS (OR OQ ) 0
S
RQ (OP OS ) 0
Q
RQ SP 0 RQ SP
and similarly from OR OP OQ OS OQ OR OP OS SR PQ
S is the orthocentre. 27
R
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
41. How many 3 × 3 matrices M with entries from {0, 1, 2} are there, for which the sum of the diagonal entries of MT M is 5? (A) 198 (B) 135 (C) 126 (D) 162 Answer (A)
⎡ a1 ⎢ Sol. Let M ⎢a2 ⎢⎣a3
b1 b2 b3
c1 ⎤ ⎡a1 a2 ⎥ T c2 ⎥ , M ⎢⎢b1 b2 ⎢⎣c1 c2 c3 ⎥⎦ 3
Sum of diagonal entries =
∑a i 1
2 i
a3 ⎤ b3 ⎥⎥ c3 ⎥⎦
bi2 ci2 5
Possible cases are (one 2, one 1 and seven zeros) or (five 1's and four 0's)
9! 9! 72 126 198 7! 5!4!
42. If y = y(x) satisfies the differential equation 8 x
9 x dy
4 9 x
1
dx, x 0 and y 0 7 , then
y(256) = (A) 80 (B) 9 (C) 16 (D) 3 Answer (D) Sol. As, dy
dx 8 x 9 x 4 9 x
Integrating, y 4 9 x c
at x = 0, y 7 c = 0 so, y 4 9 x at x = 256, y = 3 43. Three randomly chosen non-negative integers x, y and z are found to satisfy the equation x + y + z = 10. Then the probability that z is even, is (A)
5 11
(B)
1 2
(C)
6 11
(D)
36 55
28
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
Answer (C) Sol. x + y + z = 10 n(s) =
10+3–1C 3–1
=
12C 2
12 11 66 2
=
Let z = 2n, where n = 0, 1, 2, 3, 4, 5 x + y + 2n = 10 x + y = 10 – 2n Total such solution
5
∑ (11 2n ) 36
n 0
P(E) =
36 6 66 11
SECTION - 2 (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 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 all 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
Negative Marks
:
–2
If none of the bubbles is darkened 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. 44. If the line x = divides the area of region R (A) 0
x, y »2 : x 3 y x,0 x 1 into two equal parts, then
1 2
(B) 2 4 4 2 1 0 (C) 4 4 2 1 0 (D)
1 1 2
Answer (B, D)
1
0
3 3 Sol. ∫ ( x x ) dx ∫ ( x x ) dx
x2 x4 4
0
2 4 x x 2 4
1
29
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
2 x 4 ⎛ 1 1 ⎞ ⎛ 2 4 ⎞ ⎜ ⎟⎜ ⎟ 2 4 ⎝2 4⎠ ⎝ 2 4 ⎠
4 1 2 0 2 4
2 4 42 1 0
y
O (0, 0)
Let f ( ) 2 4 42 1
y = x3
(1, 0)
y=x
x
x=
1 1 f (0) 1 0 , f ⎛⎜ ⎞⎟ 0 ⎝2⎠ 8 f (1) 1 0
⎛1 ⎜ , ⎝2
⎞ 1⎟ ⎠
45. If f : » » is a differentiable function such that f(x) > 2f(x) for all x », and f(0) = 1, then (A) f(x) > e2x in (0, ) (B) f(x) < e2x in (0, ) (C) f(x) is decreasing in (0, ) (D) f(x) is increasing in (0, ) Answer (A, D) Sol. f ( x ) 2f ( x ) 0
e 2 x f ( x ) 2e 2 x f ( x ) 0 d e 2 x f ( x ) 0 e2x f ( x ) is increasing function. dx
e 2 x f ( x ) 1 for all x (0, )
f ( x ) e2 x ∵
f ( x ) 2f ( x ) e2 x 0
f(x) is increasing Also as, f ( x )
f ( x ) f (0) f (x) 1 f ( x ) x 0 x
i.e., f ( x ) e2 x x (0, 1)
e2 x x (1, ) 30
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
1 x(1 1 x ) ⎛ 1 ⎞ cos ⎜ for x 1. Then ⎝ 1 x ⎟⎠ 1 x
46. Let f ( x )
(A) limx 1 f ( x ) does not exist (B) limx 1 f ( x ) does not exist (C) limx 1 f ( x ) 0 (D) limx 1 f ( x ) 0 Answer (A, D) Sol. f ( x )
lim
x 1
1 x (1 | 1 x |) ⎛ 1 ⎞ cos ⎜ ⎟ | 1 x | ⎝ 1 x ⎠
1 x (1 x 1) ⎛ 1 ⎞ cos ⎜ ⎟ ( x 1) ⎝ 1 x ⎠
1 x2 ⎛ 1 ⎞ cos ⎜ ⎟ ⎝ 1 x ⎠ x 1 ( x 1) lim
⎛ 1 ⎞ lim – (1 x )cos ⎜ ⎟ = a number lying between –2 and 2 ⎝ 1 x ⎠ x 1
Hence, limit does not exist. lim
1 x (1 (1 x )) ⎛ 1 ⎞ cos ⎜ ⎟ (1 x ) ⎝ 1 x ⎠
lim
1 x (2 x ) ⎛ 1 ⎞ cos ⎜ ⎟ (1 x ) ⎝ 1 x ⎠
x 1
x 1
⎛ 1 ⎞ lim(1 x )cos ⎜ ⎟0 ⎝ 1 x ⎠
x 1
47. If g ( x ) ∫
sin(2 x )
sin x
sin1(t )dt, then
⎛ ⎞ (A) g ⎜ ⎟ 2 ⎝ 2⎠ ⎛ ⎞ (B) g ⎜⎝ ⎟⎠ 2 2 ⎛ ⎞ (C) g ⎜ ⎟ 2 ⎝ 2⎠ ⎛ ⎞ (D) g ⎜ ⎟ 2 ⎝ 2⎠ Answer (No options is correct) Sol. g ( x ) sin1 sin 2 x 2cos 2 x sin1 sin x cos x
⎛⎞ ⎛ ⎞ g⎜ ⎟ 0 , g⎜ ⎟ 0 ⎝2⎠ ⎝ 2⎠
None of the given options is correct. 31
JEE (ADVANCED)-2017 (PAPER-2) CODE-6 98
48. If I ∑ k 1 ∫ kk 1 (A) I
49 50
(B) I
49 50
k 1 dx, then x( x 1)
(C) I > loge99 (D) I < loge99 Answer (A, D)
⎛ k 1 dx ⎞ k ( 1) ∑ ⎜ ∫ x( x 1) ⎟ ⎝ k ⎠ k 1 98
Sol. I =
k 1
98
⎧ ⎛ x ⎞⎫ ∑ (k 1) ⎨log ⎜ ⎟⎬ k 1 ⎩ ⎝ 1 x ⎠ ⎭k
98
⎧
⎛ k 1⎞
98 ⎧ ⎛ k 1⎞ ⎛ k ⎞⎫ ∑ (k 1) ⎨log ⎜ ⎟⎠ log ⎜⎝ ⎟⎬ ⎝ k2 k 1⎠ ⎭ k 1 ⎩
⎛ k ⎞⎫
∑ ⎩⎨(k 1)log ⎜⎝ k 2 ⎟⎠ k log ⎜⎝ k 1⎟⎠ ⎭⎬ log(k 1) log k k 1
⎧ ⎛ 99 ⎞ ⎛ 1⎞ ⎫ ⎨99log ⎜ ⎟⎠ log ⎜⎝ ⎟⎠ ⎬ log99 log1 ⎝ 100 2 ⎭ ⎩ ⎛ 99 ⎞ 99log ⎜ ⎟ log2 loge (99) ⎝ 100 ⎠
cos(2 x ) cos(2 x ) sin(2 x ) cos x sin x , then 49. If f ( x ) cos x sin x sin x cos x
(A) f (x) = 0 at exactly three points in (–, ) (B) f (x) = 0 at more than three points in (–, ) (C) f(x) attains its maximum at x = 0 (D) f(x) attains its minimum at x = 0 Answer (B, C) Sol. C1 C1 C2
0 cos 2 x sin2 x f ( x ) 2cos x cos x sin x 0 sin x cos x f ( x ) 2cos x (cos 2 x cos x sin x sin 2 x ) f ( x ) 2cos x cos3 x ; (f(0) = 2 maximum at x = 0) 32
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
f ( x ) cos 4 x cos 2 x f ( x ) 2sin2 x (4cos 2 x 1)
sin 2 x 0 or cos 2 x
1 4
2x 0, , x 0,
1 , and cos 2 x gives 4 solutions in (–, ) 2 2 4
Total number of solutions = 7 50. Let and be non-zero real numbers such that 2(cos – cos) + cos cos = 1. Then which of the following is/are true?
⎛ ⎞ ⎛ ⎞ 3 tan ⎜ ⎟ tan ⎜ ⎟ 0 ⎝ 2⎠ ⎝ 2⎠
(A)
⎛ ⎞ ⎛ ⎞ (B) tan ⎜ ⎟ 3 tan ⎜ ⎟ 0 ⎝ 2⎠ ⎝ 2⎠ ⎛ ⎞ ⎛ ⎞ 3 tan ⎜ ⎟ tan ⎜ ⎟ 0 ⎝ 2⎠ ⎝ 2⎠
(C)
⎛ ⎞ ⎛ ⎞ (D) tan ⎜ ⎟ 3 tan ⎜ ⎟ 0 ⎝ 2⎠ ⎝ 2⎠ Answer (B, D) Sol. As 2(cos – cos) = 1 – coscos
⇒ cos
2cos 1 2 cos
Using componendo and dividendo ⇒
1 cos ⎛ 1 cos ⎞ 3⎜ ⎟ 1 cos ⎝ 1 cos ⎠
⇒ tan2
3 tan2 0 2 2
So, tan
3 tan 0 2 2
Or tan
3 tan 0 2 2
SECTION - 3 (Maximum Marks : 12) This section contains TWO Paragraphs. Based on each paragraph, there are TWO 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(s) 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
:
In all other cases
0
33
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
PARAGRAPH-1 Let O be the origin, and OX , OY , OZ be three unit vectors in the directions of the sides QR, RP, PQ , respectively, of a triangle PQR.
51.
| OX OY | (A) sin (Q + R) (B) sin 2R (C) sin (P + R) (D) sin (P + Q)
Answer (D) Sol. OX OY QR RP pq
pq sin R pq
sin(P Q )
52. If the triangle PQR varies, then the minimum value of cos(P Q ) cos(Q R ) cos(R P ) is (A)
5 3
(B)
3 2
(C)
5 3
(D)
3 2
Answer (B) Sol. cos(P + Q) + cos(Q + R) + cos(R + P) = –(cosP + cosQ + cosR) Maximum value of cosP + cosQ + cosR =
3 2
3 2 PARAGRAPH-2
Hence minimum of –(cosP + cosQ + cosR) =
Let p, q be integers and let , be the roots of the equation, x2 – x – 1 = 0, where . For n = 0, 1, 2, ... , let an = pn + qn. FACT : If a and b are rational numbers and a b 5 0 , then a = 0 = b. 53. If a4 = 28, then p + 2q = (A) 12 (B) 14 (C) 21 (D) 7 34
JEE (ADVANCED)-2017 (PAPER-2) CODE-6
Answer (A) Sol. a4 = 28 4
4
⎛ ⎞ ⎛ ⎞ p ⎜ 1 5 ⎟ q ⎜ 1 5 ⎟ 28 ⎝ 2 ⎠ ⎝ 2 ⎠
56( p q ) 24 5( p q ) 28 16 p=q=4 54. a12 = (A) 2a11 a10 (B) a11 a10 (C) a11 2a10 (D) a11 a10 Answer (D) Sol. 2 – – 1 = 0
12 = 11 + 10
...(i)
and
12 = 11 + 10
...(ii)
Multiplying (i) by p and (ii) by q and adding, a12 = a11 + a10
END OF MATHEMATICS
35
