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PART-1 : PHYSICS
PART-3 : MATHEMATICS
PART-2 : CHEMISTRY
Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced
TEST DATE : 15 - 05 - 2016
Paper Code : 0000CT103115008
CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)
TARGET : JEE (ADVANCED) 2016
JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE
ANSWER KEY : PAPER-1
Q. 1 2 3 4 5 6 7 8 9 10
A. A,D C,D A,C,D A,D A,C,D B B A,B,C,D B,C A,C
A B C D A B C D
R R S P Q,S,T P,R,T P,R,T Q,S,T
Q. 1 2 3 4 5 6 7 8
A. 0 5 5 2 5 1 0 2
SECTION-I
SECTION-II Q.1 Q.2
SECTION-IV
Q. 1 2 3 4 5 6 7 8 9 10
A. C,D C,D A,C,D B,C,D A,B,C A,B B,D B,C,D B,C A,C,D
A B C D A B C D
P S R,T Q P,Q,S P,Q,S P,T Q,R
Q. 1 2 3 4 5 6 7 8
A. 8 8 4 2 4 3 2 4
SECTION-I
SECTION-II Q.1 Q.2
SECTION-IV
Q. 1 2 3 4 5 6 7 8 9 10
A. A,B,C,D A,B,D A,B,C B,C,D A,B,D A,D A,C,D A,B,C,D B,C,D A,C
A B C D A B C D
Q,R P,R Q,T Q,S P,Q,R T T S
Q. 1 2 3 4 5 6 7 8
A. 3 4 6 2 0 3 2 1
SECTION-I
SECTION-II Q.1 Q.2
SECTION-IV
PART-1 : PHYSICS
PART-3 : MATHEMATICS
PART-2 : CHEMISTRY
Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced
TEST DATE : 15 - 05 - 2016
Paper Code : 0000CT103115009
CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)
TARGET : JEE (ADVANCED) 2016
JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE
ANSWER KEY : PAPER-2
Q. 1 2 3 4 5 6 7 8 9 10
A. A,B,D A,B B,D A,D B,C A,C C A,B,C,D A A
Q. 11 12
A. C,D B,D
Q. 1 2 3 4 5 6 7 8
A. 8 5 3 3 4 2 3 3
SECTION-I
SECTION-IV
Q. 1 2 3 4 5 6 7 8 9 10
A. A,D A,B,D C,D B,C A,B A,C B A,B,D B,D A,D
Q. 11 12
A. D A
Q. 1 2 3 4 5 6 7 8
A. 5 5 4 9 4 4 9 6
SECTION-I
SECTION-IV
Q. 1 2 3 4 5 6 7 8 9 10
A. A,B,D B,D A,B,C,D B,C,D A,C A,B,C,D A,B,C,D B,C B A
Q. 11 12
A. D B
Q. 1 2 3 4 5 6 7 8
A. 0 2 9 3 1 4 2 6
SECTION-I
SECTION-IV
SOLUTION
HS-1/16Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005
+91-744-5156100 [email protected] www.allen.ac.in
SECTION–I
1. Ans. (A,D)
Sol. By conservation of linear momentum both will getsame velocity of centre of mass
2cm
1mv mgh
2
h1 = h
2
But by conservation of angular momentum aboutcentre of mass
System 2 have angular velocity also
K1 < K
2
[where k = 2 2cm
1 1mv I
2 2 ]
2. Ans. (C,D)
Sol. 3
2 0.085 4h 1mm
r 13.6 10 10 500
3. Ans. (A,C,D)
Sol. (A) Sx = –4 cos kx sin t
= –4 cos 2 2
0.05 sin 0.050.40 0.2
= 2 2
Q. 1 2 3 4 5 6 7 8 9 10
A. A,D C,D A,C,D A,D A,C,D B B A,B,C,D B,C A,C
A B C D A B C D
R R S P Q,S,T P,R,T P,R,T Q,S,T
Q. 1 2 3 4 5 6 7 8
A. 0 5 5 2 5 1 0 2
SECTION-I
SECTION-II Q.1 Q.2
SECTION-IV
(C) 0.4
v 2m / sT 0.2
p
2 1 2 2v 4cos cos 0.1
0.4 15 0.2 0.2
= +40 × 1
2 = 20
4. Ans. (A, D)
Sol.
20 1
2
2 12 0 2
1k E k2
1 kk E2
=
5
3
11
3
,
11
5
=
2 4
3 5
=
2
15
5. Ans. (A,C,D)Sol. Parallel ray will pass through focus after passing
through lens. Slab will shift the image by an amount
of = 1
t 1
=
23 1
3
= 1 cm
Final image will form at 11 cmIf converging rays meet at the point of image, imageis known as real. If diverging rays meet at thepointof image, image is known as virtual.
PAPER-1
Paper Code : 0000CT103115008
CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)
PART-1 : PHYSICS ANSWER KEY
Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced
TEST DATE : 15 - 05 - 2016
TARGET : JEE (ADVANCED) 2016
JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE
HS-2/16
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-1
0000CT103115008
6. Ans. (B)
Sol. As tension is always perpendicular to the velocity
its speed remain same.
7. Ans. (B)
8. Ans. (A,B,C,D)
Sol. x 2
2cxF
a
y 2
2cyma F
b
9. Ans. (B,C)
10. Ans. (A,C)
Sol.2 2 2
2 1 2
1 1 1kx m 4 m x
2 2 2
2
1
m 4 1
m 16 4
m1v
1 = m
2v
2
1
2
v 1
v 4
2 2 21 1 2 2
1 1 1kx m v m v
2 2 2
22 1
1 1 1
1 1 mm v 4v
2 2 4
2 21 1 1
1 1m 2 m 5v
2 2
1
4 2v
5 5 2
8v
5
2 mT
4 R
SECTION-II
1. Ans. (A)-(R); (B)-(R); (C)-(S); (D)-(P)
Sol. Path diffrence due to slab-1 = x = t(u –1)
= 1 5
5 m2 2
Phase diffrence 2 x
= 6
9
2 510 m
500 10 2
= 10
Path diffrence due to slab (2) = x = t[µ – 1]
= 3 3 9
m2 3 4
62 910 9
4
Path diffrence due to slab (3) = x = t[µ – 1]
= 1 1
[1] m4 4
6
9
2 110
500 10 4
(A) 10 fifth Maxima
(B) (10 – ) = 9 fifth minima
(C) (10 – (9 + ) = 0 Central Maxima
(D) [[10 + ] – 9] = 2 first Maxima
2. Ans. (A)-(Q,S,T); (B)-(P,R,T); (C)-(P,R,T);
(D)-(Q,S,T)
SECTION-IV
1. Ans. 0
2. Ans. 5
3. Ans. 5
Sol.2B a
i2R
21
0.1 4020
2 1
= 2 × 1
400
4. Ans. 2
Sol.dm / 0
0.08 0.081dt 0 1 r T
0.8 + 0.8 rT = 0.81
0.01r T
0.80 × 103
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-1
HS-3/160000CT103115008
1 100T 2
80 16
dmsT
dt
33 0.8 10 25
10 400010 4
= 2 × 103 W
5. Ans. 5
Sol. Potential at centre,
kQ kQ kQ kQ
12 12 9 9 =
1 2 1kQ kQ
6 9 18
Q –Q
Q–Q
12m9m
37°
15m
Potential at centre
=
9 69 10 0.01 10 90
V 5V18 18
6. Ans. 1
Sol.H
n T
QnC T W
n T
= 2R – R = R
7. Ans. 0
8. Ans. 2
Sol. Momentum of electron pe = 2meV
Momentum of particle, p
mp 2 eV
4
p p
e e
h / p m2
h / p m / 4
SOLUTION
PART-2 : CHEMISTRY ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10
A. C,D C,D A,C,D B,C,D A,B,C A,B B,D B,C,D B,C A,C,D
A B C D A B C D
P S R,T Q P,Q,S P,Q,S P,T Q,R
Q. 1 2 3 4 5 6 7 8
A. 8 8 4 2 4 3 2 4
SECTION-I
SECTION-II Q.1 Q.2
SECTION-IV
SECTION - I
1. Ans. (C, D)
All molecules move with different speeds and due
to molecular collision, kinetic energy of molecules
will change.
2. Ans. (C,D)
4 2 3
1 3P (s) Cl (g) PCl (g)32
4 2 ; H = 140
140 = P Cl
1 33 (B E)320 120
4 2
(B–E)P–Cl
= 120
40kJ / mole3
4 2 3
1 3P (s) H (g) PH (g)
4 2 ; H = 8
8 = P H
1 33 (B E)320 216
4 2
(B–E)P–H
= 132 kJ/mole
H2 2H ; H = 216 (H
f)
H =
216108
2
Cl2 2Cl ; H = 120 (H
f)
Cl =
12060
2
HS-4/16
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-1
0000CT103115008
3. Ans. (A,C,D)
Anode : 2Br– Br2 + 2e–
Cathode : 2H2O + 2e– H2 + 2OH–
K+ combines with OH– so KOH will also form.
4. Ans. (B,C,D)
5. Ans. (A,B,C)
6. Ans. (A,B)
7. Ans. (B,D)
8. Ans. (B,C,D)
9. Ans. (B,C)
10. Ans. (A,C,D)
SECTION - II
1. Ans (A)-(P) ; (B)-(S) ; (C)-(R,T); (D)-(Q)
(A) pH = 10 + log 0.1
100.1
(P)
(B) pOH = 6 + log 0.1
60.1
pH = 8 (S)
(C) pH = 1
2 [14 + 5 – 7] = 6 (R), (T)
(D) [H+] = 500 0.02
0.011000
,
[OH–] = 500 0.02
0.011000
so solution is neutral (Q).
2. Ans. (A)-(P,Q,S); (B)-(P,Q,S); (C)-(P,T);
(D)-(Q,R)
SECTION - IV
1. Ans. 8
10 H2O + SCN– SO
42–+CO
32– + NO
3–+2OH+ +16 e–
'n' factor or Ba(SCN)2 = 2 × 16 = 32
n 32
84 4
2. Ans. 125 [OMR Ans. 8]
2A(g) + B(g) 2C(g)
ng = 2 – (1 + 2) = – 1
Hr = 25.6 +
( 1) 2 30025kcal
1000
Sr = 2 × 500 – (2 × 200 + 100) = 500 cal.
G = 25 – 300 500
–125kcal1000
|G| = 125
Ans. 8
3. Ans. (4)
4. Ans. (2)
5. Ans. (4)
i, ii, iv, v
6. Ans. (3)
i, ii, vi
7. Ans. (2)
ii, iv
8. Ans. (4)
i, ii, iv, v
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-1
HS-5/160000CT103115008
SOLUTION
PART-3 : MATHEMATICS ANSWER KEY
SECTION-I
1. Ans. (A,B,C,D)
x 13 y 11 z 15L
2 2 1
A(13,–11,15)
L
P
A'( )
P 2 13, 2 11, 15
Putting P in plane – 6
F(1,1,9)
+ 13 = 2, –11 = 2, + 15 = 18
A' (–11,13,3)
Perpendicular distance 26 22 15 9
183
volume 5
3
1 9 9 39
6 2 2 2
2. Ans. (A,B,D)
z4 – z2 + 1 = 0 2z cos
3
2 5z cos
3
5z cos m z cos n
3 6
now Doyourself.
3. Ans. (A,B,C)
2
2 2
dy 2x 1 dy 1 x y 1
dx 1 x 1 x dx
2
xƒ x
x 1
12
–10
–1/2
1
4. Ans. (B,C,D)Do yourself
5. Ans. (A,B,D)
LR = 4a = 4
AB = 4 cosec2 = 16
1 2
2 1
t t 3
1
1
12 3 t
t
2
1 1t 2 3t 1
6. Ans. (A,D)
a and 1
b are the roots of the equation
6x2 + 20x + 15 = 0
1 10a
b 3 and
a 5
b 2
3
3 32
b 1
ab 9 ab 1 1 1a. 9 a
b b
1 6
5 1000 20159.2 27
7. Ans. (A,C,D)
1x
0
I ƒ(3 )dx
11 x
0
I ƒ(3 )dx
1
0
2I ƒ(3)dx Also ƒ(a.b) = ƒ(a) + ƒ(b) ƒ(a2) = 2ƒ(a)using this C & D also correct.
Q. 1 2 3 4 5 6 7 8 9 10
A. A,B,C,D A,B,D A,B,C B,C,D A,B,D A,D A,C,D A,B,C,D B,C,D A,C
A B C D A B C D
Q,R P,R Q,T Q,S P,Q,R T T S
Q. 1 2 3 4 5 6 7 8
A. 3 4 6 2 0 3 2 1
SECTION-I
SECTION-II Q.1 Q.2
SECTION-IV
HS-6/16
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-1
0000CT103115008
8. Ans. (A,B,C,D)
1
0
1 x 1
0 0 x
1
0
t x dt, x 0
ƒ x x t dt x t dt t x dt, 0 x 1
x t dt, x 1
2
1 2x, x 0
2
2x 2x 1ƒ x ,0 x 1
2
2x 1, x 1
2
9. Ans. (B,C,D)
33
3
1 1x x
x xy1 1 1x 2 x 3 x 2x x x
3
1 tx t y
x t 3t 2
2
1y
2t 3
t
2 21 1 2 1t 3 t 3 6 y
t t t 6
10. Ans. (A,C)ƒ'(x) < ƒ'(x) + xƒ"(x) ƒ'(x) < (xƒ'(x))1
x x
1
0 0
ƒ ' x dx xƒ ' x dx
ƒ x xƒ ' x ...(1)
Now at
2
ƒ x xƒ ' x ƒ xh x h ' x 0
x x
x (0,1) (from (1))
h Now g(x) > x {As ƒ is concave up in (0,1)}h(g(x)) > h(x)
ƒ g x ƒ x
g x x ƒ(x) g(x) < x2 x(0,1)
SECTION – II
1. Ans. (A)(Q,R); (B)(P,R); (C)(Q,T);
(D)(Q,S)
6 6 6
4 5 66
C C C 11 aP A
2 32 b
1 p
P B2 q
5 5 53 4 5
6
P AB C C CA 1 uP 2
B P B 2 2 v
P BAB 32 16 8 mP
A P A 11 64 11 n
2. Ans. (A)(P,Q,R); (B)(T); (C)(T);
(D)(S)
(A)
4991000
2k 1k 0
2k 1 C
100
499999
2kk 0
10 C
= 10.2998
(B) 4
3 1 I ƒ
4
3 1 ƒ '
4 24 4 4
0 2 4I ƒ ƒ ' 2 C 3 C 3 C
I + 1 = 2{9 + 18 + 1}
= 56
(C) Do yourself
(D) No. of 4 digit numbers
= Dearrangment of 4 objects
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-1
HS-7/160000CT103115008
SECTION – IV1. Ans. 3
Do yourself
2. Ans. 4
Do yourself3. Ans. 6
Use |PS – PS'| = 2b
4. Ans. 2
2 x 2x
x 2x
1 1log y 2 2 log y 2 2 0
2 2
2
x 2x 2x
x 2x 2x
1 1 1log y 2 2 2 2 2
2 2 2
2
x 2x
x 2x
0 0
1 1log y 2 2 2
2 2
x = 0; y = e–2
5. Ans. 0
c a c a b ...(1)
a.c 0
...(2)
Also a c a a c a a b
{Taking cross with a
}
2 2c a b a c a.b a a b
21 a c b a a b a 0
6. Ans. 3
x
1 x 1lim cot
x 1 2 2x 2
x
1 2 2lim cot
x 1 4 x 1
x
1
2 x 1 2lim
1 1tan
2 x 1
7. Ans. 2
Do yourself
8. Ans. 1
ƒ(1–) > ƒ(1)
2 > 1 + k
k < 1
SOLUTION
HS-8/16Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005
+91-744-5156100 [email protected] www.allen.ac.in
PART-1 : PHYSICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10
A. A,D A,B,D C,D B,C A,B A,C B A,B,D B,D A,D
Q. 11 12
A. D A
Q. 1 2 3 4 5 6 7 8
A. 5 5 4 9 4 4 9 6
SECTION-I
SECTION-IV
SECTION-I
1. Ans. (A,D)
Sol. max 1
V 1i Q CV
R LC
V C2V
R L
61 L 1 2R 10
2 C 2 8 = 250
2
0
1H V idt Li
2
Rt
L
0
VV e dt
R
– 21
Li2
0
Rt2L 2V L 1
e LiR R 2
22
2
V L 1Li
R 2
2 2 2
2 2 2
V L 1 V 1 V LL
R 2 R 2 R
2. Ans. (A,B,D)
Sol. CA B8RT3RT 2RT
M M M
CA B
8T3T 2T
C
280 3T 105
8
& TT
B = 420
B B
A A
P T 3
P T 2
C C
A A
V T 3
V T 8
B C C B
3R T T
2
3
R 420 1052
CA
= (280 – 105)R
3. Ans. (C,D)
Sol. (A)
00
RJJ 20
2 2
PAPER-2
Paper Code : 0000CT103115009
CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)
Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced
TEST DATE : 15 - 05 - 2016
TARGET : JEE (ADVANCED) 2016
JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-2
HS-9/160000CT103115009
0
22
R I
4 RR
4
02
R I
3 R4
4
(B) R/2
0
RJ
2 02
03
R I
3 R4
4
0I
3 R
4. Ans. (B, C)
Sol. (A)22 2
kQ kQ kQ0
a 9a 3a
5
(B) 22 2
4kQ kQ kQ0
a a a
5
(C)
2 2 2
kQ 34kQ kQ0
9a 25a 225a
(D) 2 2 2
22 2
kQ kQ kQ0
16a 4a 3a
15
5. Ans. (A,B)
Sol.2 21 1 5
MR MR '2 2 4
4'
5
2
2 2 21 5 4 1k MR MR
2 8 5 5
6. Ans. (A,C)
Sol. z
1C z 1
1
1
1C z 1
1 z
1
z 12
z 1
z1 = 2z – 1
2z 1 1
z 1 2
z 1z
2 2
7. Ans. (B)
Sol. 1 2
1 2
T T 10 30T 7.5days
T T 40
t = 15 days
eliminated = 1
5 12
Total driving till then = n2
15105e
=
3
21 55
2 2 2
µg
decayed = 1
5 12 2
Remaining in body = 5
2
decayed in body = 5 1
12 2
HS-10/16
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-2
0000CT103115009
8. Ans. (A,B,D)
Sol. Let magnetic field at centre of square due to
current in square is B1
2
2 213B B B
45°
1B 2B
0i
a4
2
[sin 45] × 2 × 4 2B
02 2 i2B
a
0
aBi
2
= M × B = ia2 B =
3 2
0
a B
2
(B) U = MB cos 90 = 0
(C) Bnet
= 2B B B 2 1
(D) Bnet
= 2B B B 2 1
9. Ans. (B,D)
Sol. A B,
mg(4) + 2 2B
1 1k 2 0 11 mv
2 2
B C,
mg(3) – 4 = 2 2C B
1m v v
2
10. Ans. (A,D)
Sol.2B
B
mvN
R
2C
C
mvN mg
R
11. Ans. (D)
Sol.h h h
p 2mE 2mqV where E = kinetic
energy
12. Ans. (A)
Sol.150 150
Å 1ÅV 150
SECTION-IV
1. Ans. 5
Sol. At equilibrium
8g = k × 0.2
8gk
0.2
When mass get turn then new equilibrium
shift by
1g 1g 0.2x
k 8g
x = 2.5 cm
If will be amplitude for SHM maximum height
= 2A = 5 cm
2. Ans. 5
Sol. 3
GM 4G
R 3
t3
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-2
HS-11/160000CT103115009
–3R——2
30°
30°
R
1t
43 3 G3
= 11
11250sec
4 203 10 8003 3
3. Ans. 4
Sol. T = c sxyrz
[T] = [s]x []y [r]z
= [MT–2]x [ML–3]y [L]z
[T] = Mx+y L–3y + z T–2x
x + y = 0
–2x = 1 x = –1
2, y =
1
2–3y + z = 0
3z
2
3rT c
s
T 1 3 r 1 s
T 2 2 r 2 s
1 13 4
2 2
4. Ans. 9
Sol.d
dT =
36810 m / K
9
Radius of wire r = 2mm
2
2
i
r
= e2r (T4 – TT
04)
2
2
i
r
e2r (T4)
2 3 42e r Ti
....(i)
As = 368
109
TT
Put = 368
109
T in equation (i) and after
solving we will get i = 0.18 A
5. Ans. 4
Sol. v Rg
2 1R 10
8
24R
5
2R 4T 4sec
2 2 5v
6. Ans. 4
Sol.1
330330 300
300 v
v1 = 30 m/s
2
330360 300
330 v
2
330330 v 10 275
12
v2 = 55 m/s
V = 25 m/s
100 = V × T
T = 4 sec
HS-12/16
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-2
0000CT103115009
7. Ans. 9
Sol. Using Kirchhoff’s first law at junctions a and
b, we have found the current in other wires of
the circuit on which currents were not shown.
4 F
1A
2A
13V
5
1
4
3
34V
2A
C
1A
2A a
2A
b
3A
6A
3A 2
Now, to calculate the energy stored in the
capacitor we will have to first find the potntial
difference Vab
across it.
Va – 3 × 5 – 3 × 1 + 3 × 2 = V
b
Va – V
b = V
ab = 12 volt
U = 1
2CV2
ab
= 1
2 × (0.125 × 10–6) (12)2 J
= 9 mJ
8. Ans. 6
Sol. From situation, it is clear that it is diverging
lens to left of C.
C B A
x 1/2 1
1
1x
2
+ 1
3x
2
= 1
f
1 1 1
1x fx
2
2 2 1
2x 1 2x 3 f
2
1 9x 2 4x 6 4
f 2x 3 2x 1 4x 8x 3
–f = 2 3
x 2x4
1 2 1
x 2x 1 f
2x 2x 1 1
2x 1 x f
–f = 2x2 + x = x2 + 2x + 3
4
2 3x x 0
4
1 1 3
2
=
3
2 or
1
2
–f = 2x2 + x
= 9 3
24 2
= 6cm
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-2
HS-13/160000CT103115009
SOLUTION
PART-2 : CHEMISTRY ANSWER KEY
SECTION–I
1. Ans. (A, B, D)
(a) Packing fraction of,
=
2
2
13 R
6 0.913 3 3
4R4
(b) C.N. = 6 ; it is a hexagonal 2D lattice
(c) r
0.155R so, diameter = 2r = 0.31R
(d)Distance between 2 layers is 2 2
R3
2. Ans. (A, B)
For the equal mix, Yben
= xtol
& ytol
= xBen
yben
– xBen
= minimum, which occurs at
Ptotal
= º ºBen TolP .P
Hence (a), (b) are correct
(C) At this, instant V.P. of equilibrium mixture
corresponds to V'
(D) If external pressure is increased then
condensation occurs and not vaporisation.
3. Ans. (B, D)
As2S
3 sol is negatively charged so the D.M.
moves to cathode while sol particles do not
move in either direction.
4. Ans. (A,D)5. Ans. (B,C)
6. Ans. (A,C)
7. Ans. (C)
8. Ans. (A,B,C,D)
9. Ans. (A)
H2O
2 decomposes by 1st order kinetics
K × 5 = ln 20
15
K = 0.06 min–1 ; t1/2
= ln 2
11.67 min0.06
10. Ans.(A)
N1V
1 = N
2V
2 ;
10
11.35/ 2 ×
11.35
2 = 0.1 × 5 × VV
11. Ans. (C,D)
12. Ans. (B,D)
SECTION–IV
1. Ans. (8)
0 0 01 1 2 2 3 3n E n E n E
(n1 = x – y , n
2 = y – z, n
3 = x – z )
(x – y) × 2 + (y – z) × 3 = (x – z) × 10
2x – 2y + 3y – 3z = 10 x – 10 zy – 8x + 7z = 0
y 7z8
x
2. Ans. (5)
n – l – 1 = 2 ; l = 3 n = 6
13.6 2 2
2 2
Z 313.6
6 3 Z = 6 ; oxidation
number = +5.
3. Ans. (3)4. Ans. (3)
5. Ans. (4)
6. Ans. (2)
7. Ans. (3)
8. Ans. (3)
Q. 1 2 3 4 5 6 7 8 9 10
A. A,B,D A,B B,D A,D B,C A,C C A,B,C,D A A
Q. 11 12
A. C,D B,D
Q. 1 2 3 4 5 6 7 8
A. 8 5 3 3 4 2 3 3
SECTION-I
SECTION-IV
HS-14/16
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-2
0000CT103115009
SOLUTION
PART-3 : MATHEMATICS ANSWER KEY
SECTION-I
1. Ans. (A,B,D)
2
h,k
dy 1 k
dx 3k
2 2
2k 0 1 k 3k1 k
h 2 3k h 2
...(1)
Also, 2
2 2h h1 k 1 k
3 9 ...(2)
From (1) and (2), we get
2h3 1
9 h 9 5h , k
h 2 3 2 2
9 5A ,
2 2
&
9 5B ,
2 2
2. Ans. (B,D)
We have
a aa b a b 1 1
b b
a
b lies on perpendicular bisector of (–1,0)
and (1,0)
so, a
b lies on imaginary axis.
aarg
b 2
arg a arg. b2
3. Ans. (A,B,C,D)
Let T(h,k) where h = t1t2, k = t
1+t
2.
Also t12 = 16t
22
Q. 1 2 3 4 5 6 7 8 9 10
A. A,B,D B,D A,B,C,D B,C,D A,C A,B,C,D A,B,C,D B,C B A
Q. 11 12
A. D B
Q. 1 2 3 4 5 6 7 8
A. 0 2 9 3 1 4 2 6
SECTION-I
SECTION-IV
On eliminating t1 & t
2, we get locus of T(h,k)
is 2 25y x
4
4. Ans. (B,C,D)
23a
a4r3as 2 32
,
where a is length of side of triangle
As, r Q a is irrational and multiple of 3 .
Also, R = 2r a
R Q3
.
Also 23
a Q4
and r1 = r
2 = r
3 =
3a Q
2 .
5. Ans. (A,C)
nxn 1 xn 2
n 1 nƒ x
ee
x 2x 2x 3x nxn 1 xn
1 2 2 3 n 1 nlim ....
e e e e ee
x
x nxn
1 nlim e
e e
6. Ans. (A,B,C,D)
we have
(sin2x + cos3y)2 +(cos3y + tan4z)2
+ (tan4z + sin2x)2 < 0
sin2x = cos3y = tan4z = 0
x ;y , ;z ,2 6 2 4 2
Now verify alternatives
7. Ans. (A,B,C,D)
2x
2
dy2x e dx
y
2x1e c
y , As ƒ(0) =
1
2
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-2
HS-15/160000CT103115009
c = 1
2x
1y
1 e
(0,1/2)
y
(1,1/1+e)
(1,0)O(0,0)x
1
0
1 11 0 ƒ x dx 1 0
1 e 2
8. Ans. (B,C)
xlim ƒ(x) 1
ƒ'(x) = 0 x = –1
ƒ(–1) =1
2
Also, x 1 x 1lim ƒ(x) , lim ƒ(x)
Range of ƒ(x) = 1
, (1, )2
+ +
x = –1 1
–
sign of ƒ'(x) x = –1 is point of local maximum of ƒ(x).
Paragraph for Question 9 to 10
C1 : x2 + y2 – 10x + 24 = 0 is circle whose centre
is (5,0) and radius is 1. Also , C2 : x2 + y2 –
10x + 21 = 0 is circle whose centre is (5,0) and
radius is 2.
9. Ans. (B)10. Ans. (A)
Paragraph for Question 14 to 16
1 2 3 4
1 2 3 4P B , P B , P B , P B
10 10 10 10
11. Ans. (D)
1
1 2 1 3 1 4 1P E 1
10 10 2 10 3 10 4
4 2
10 5
12. Ans. (B)
3 2
3 2
2
P B EP B E
P E
3 1
10 3
1 2 1 3 1 4 10
10 10 2 10 3 10 4
1110
3 310
SECTION – IV
1. Ans. 0
3 1 0
M 1 3 0
0 0 2
As, MMT = 4I 2MT + adj.M = 0
|2MT + ady.M| = 0
2. Ans. 2
Let P(x) = 12(x–1)(x– 3) (x – 5) + (2x + 1)
3. Ans. 9
n 6 61 1n 3 3
6P C . 3 . 4
6 n 61 1n 3 3
n 6Q C . 3 . 4
n 6
13
Q12 12 12
P
n 6
1 n 93
4. Ans. 3
we have
3 3 3
3 3 3
3 3 3
a a 1 a b a c
1ab b b 1 b c 11
a.b.cac bc c c 1
HS-16/16
ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-2
0000CT103115009
3 3 3
3 3 3
3 3 3
a 1 a a
b b 1 b 11
c c c 1
apply C1 C
1 + C
2 + C
3
& solving
a3 + b3 + c3 + 1 = 11
a3 + b3 + c3 = 10
possibilities are
(1,1,2), (1,2,1), (2,1,1)
Number of triplets = 3
5. Ans. 1
2 2x y1
9 4 ; P(3sec, 2tan)
Equation of chord of contact AB with respect
to P is T = 0
3x sec + 2y tan = 9 ...(1)
Also, equation of chord whose mid-point is
(h,k) is T = S1
hx + ky – 9 = h2 + k2 – 9 ...(2)
On comparing (1) and (2), we get
2 2
3sec 2 tan 9
h k h k
as, sec2 – tan2 = 1
locus of (h,k) is
22 2 2 2x y x y
19 4 9
= 1
6. Ans. 4
x2
– 6x + 12 = 0; Here ( – 6) =
8
24 24
8
62 1 2
= 224.
7. Ans. 2
Normal vector of required plane is parallel to
vector
ˆ ˆ ˆi j kˆ ˆ ˆ3 1 3 14i 27 j 5k
4 3 5
Equation of required plane is
14(x – 1) – 27(y – 2) + 5(z – 3) = 0
14x – 27y + 5z + 25 = 0
8. Ans. 6
dy 1 2y x
dx x x
(Linear differential
equation)
ƒ(x) = (x – 1)2 + 1
Required area
3
2
0
x 1 1 dx 3 3 6