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SOLUTION
PAPER-1
HS-1/16Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005
+91-744-5156100 [email protected] www.allen.ac.in
Paper Code : 0000CT103115002
CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)
PART-1 : PHYSICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10
A. B,D A,B B,C,D A,B,C A,C A,B,C,D B,C B,C A,B,C C,D
A B C D A B C D
S R S T T R S P
Q. 1 2 3 4 5 6 7 8
A. 6 1 4 6 2 6 2 6
Q.2
SECTION-IV
SECTION-I
SECTION-II Q.1
SECTION-I
1. Ans. (B,D)
2. Ans. (A,B)
3. Ans. (B,C,D)
Sol.0
QV d
2A
2Q Q
3Q—2
3Q—2
Q—2
–Q—2
If left plate is earthed,
0
QV d
A
–Q Q–––––––
+++++++
If right plate is earthed,
0
2QV d
A
2Q –2Q–––––––
+++++++
4. Ans. (A,B,C)
Sol. 0 = –nmv + m(v0 – v) 0v
vn 1
& vm = v
0 – v = v
0 –
0 0v nv
n 1 n 1
2 2
0 0v nv1 1KE nm m
2 n 1 2 n 1
20
1 nmv
2 n 1
5. Ans. (A,C)
Sol. E1 = 0 & V
1 = 4V
0
E2 = E; V
2 = 3V
0
E3 = 2E ; V
3 = 2 V
0
E4 = E; V
4 = V
0
6. Ans. (A,B, C, D)
Sol. In all cases
A P100 2 100
A P
7. Ans. (B,C)
Sol.
22 2 02 2
v1 1 2mg R R cos mv mR
2 2 2 5 R
Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced
TEST DATE : 14 - 02 - 2016
TARGET : JEE (ADVANCED) 2016
JEE (Main + Advanced) : ENTHUSIAST & LEADER COURSE
HS-2/16
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-1
0000CT103115002
= 22
7mv
10
2 2
10v gR 1 cos
7
22
2
mvmgcos
R 2
10g1 cos
7
17 cos 2 = 10
2
10cos
17
cos 2 < cos
1
2 >
1
mg (R – R cos 1) =
21
1mv
2
1 1v 2gR 1 cos
21
1 1
mvmgcos 2mg 1 cos
R
1
2cos
3
1
2 2gRv 2gR 1
3 3
2
10gR 7v
7 17
1
2gRv
3
8. Ans. (B,C)
Sol. Here NA = N
0e–t
NB = N
0(1 – e–t)
Now tB
A
Ne 1
N
If t << T; then
B
A
N1 t 1
N (ex = 1 + x for small x)
B
A
Nt
N
9. Ans. (A,B,C)
Sol. 2.5A (37 – T1) = 1 2 2
KAT T 2.5A T 7
Solving we getT
1 = 25 and T
2 = 19
Also heat lost by body with fur coat
= dH 5
1 37 25dt 2
= 30
10. Ans. (C,D)
Sol. It is clear that, I1 = I'
1 & I
2 = I'
2
But I1 > I
2 & I'
1 > I'
2
SECTION-II
1. Ans. (A)(S); (B)(R); (C)(S); (D)(T)
Sol. (A)
41
4 1 83 V1 3V 2 5
We know, 1
2
v 3m
u 5
Since v = –ve & |m| < 1
Hence image is virtual & diminished.
2. Ans. (A)(T); (B)(R); (C)(S); (D)(P)
Sol. (A) U = 3gh1 + 2gh
2
1 2dU dh dh3g 2g
dt dt dt
= –3gU + 2gU = –gU = 2g
t5
2 2
2
d U g
dt 5
(B)21
U Li2
= 2R
2 L0
1Li e
2
2R2 L0
dU 1 2RLi e
dt 2 L
2R2 22 L02 2
d U 1 4RLi e
dt 2 L
(C) U = mgy
y
dU dymg mg a 0
dt dt
2
2
d U
dt= mg a
y
(D)21
U kx2
= 2 20
1kx cos t
2
20
dU 1kx 2sin t cos t
dt 2
20
1kx sin 2 t
2
at t = 0 du
0dt
22 202
d U 1kx 2 cos 2 t
dt 2
at t = 0 2
2 202
d ukx
dt
2
2
d u0
dt
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-1
HS-3/160000CT103115002
SECTION-IV1. Ans. 6
Sol. 21
x 10 0.2 0.2m2
2 2D x
f 24cm4D
f
6cm4
2. Ans. 1
Sol. mvd
dx
= µqvB
m dv qB dx 15
6
0.2 1.5 10x
0.3 50 10 0.2
3. Ans. 4
Sol.
22 8
0
R 8 10 2 RT2
2
8
20
R Q8 10
4 t 4 R
Q = 8 2 38 10 64 R t
=
3
3
9
18 10 16
8 0.09 109 10
= 34 10
4. Ans. 6
Sol.0330 v320 320
53 330 10 3
v0 = 5 m/s
90
t15
= 6 sec
5. Ans. 2
Sol.8
e B v 3 510
= 12 Volt
q = CE (1 – e–t/)
24 = 6 × 12 (1 – e–t/) e–t/ = 2
3
i = t /E 12 2
e 2R 4 3
6. Ans. 6
Sol.2
max
muT mg
2
2min
mu mT mg u 4g
2
max
muT mg
2
min
muT 5mg
difference = 6 mg
7. Ans. 2
Sol. I0W
0 = IW
I0 × w
0 × q
0 = Iwq
2
2
mgm 402
m3 3
3
22
0
mmg Mg M
2 3
02
2
40 1 1.5 11.5 10
3 2 3
1 1.5 11.5 10 0.5 10 1 0.5 1
2 3
40 1550 23 4 2
25155 1
2
8. Ans. 6
Sol. C
C
q =12CV0
q =12CV0
3C
C
In second case
totalC
total
QV
C
C
24CVV 6Volt
4C
HS-4/16
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-1
0000CT103115002
SOLUTION
PART-2 : CHEMISTRY ANSWER KEY
SECTION - I
1. Ans. (A,B,C)
(A) For solution Z2 at P
1 pressure
XA = 0.25 X
B = 0.75
YA = 0.5 Y
B = 0.5
(B) For solution Z2 at P
3 pressure solution will
not vapourise so
XA = 0.4 ; X
B = 0.6
(C) For sulution Z1 at P
2 pressure solution will
not vapourise so
XA = 0.2 ; X
B = 0.8
2. Ans. (B)
For a reaction to occur Gº < 0
2MnO + 2C 2Mn + 2CO ; G = 50 kJ/mol
(not feasible)
2/3 Ca2O
3 + 2C 4/3 Cr + 2CO ;
G = – 50 kJ/mol (feasible)
3. Ans.(A,D)
(A) Fact
(B) Probability of finding an electron is nearly 90%
in an orbital
(C) No of angular nodes are l
(D) For 1s ||2 is maximum at nucleus
4. Ans. (A, B, C, D)
5. Ans. (A, B, C, D)
6. Ans. (A, C, D)
7. Ans. (A,B,C)
8. Ans. (A,C)
9. Ans. (A,D)
10. Ans. (A, B)
SECTION - II
1. Ans (A)(R,T); (B)(S,T); (C)(P, Q, T);
(D)(P, T)
(A) 10–6 =2x
1 xx = 10–3 ; pOH = 3; pH = 11
Q. 1 2 3 4 5 6 7 8 9 10
A. A,B,C B A,D A,B,C,D A,B,C,D A,C,D A,B,C A,C A,D A,B
A B C D A B C D
R,T S,T P,Q,T P,T P,R,T P,R P,Q,R,S,T P,Q,R,S
Q. 1 2 3 4 5 6 7 8
A. 2 4 4 9 1 2 6 5
Q.2
SECTION-IV
SECTION-I
SECTION-II Q.1
(B) 10–7=2x
0.1 xx=10–4 ; pOH = 4, pH = 10
(C) BOH + HCl BCl + H2O
20 × 0.5 2 × V 20 × 0.5
[BCl] = 20 0.5
0.420 5
2x
0.4 x
4 × 10–5 x = 4 × 10–3
pH = 2.4
(D) 3 14
10
(10 x)(x) 10
0.1 x 5 10
x = 10–3
[H+] = 10–3 + 10–3 = 2 × 10–3 pH = 2.7
2. Ans. (A) (P,R,T); (B) (P,R);
(C) (P,Q,R,S,T); (D) (P,Q,R,S)
SECTION-IV
1. Ans. 22
m
G. K.V
n n
l
2 11 4002Scm mol
100 2
2. Ans. 4
(i) It is enthalpy of hydration
(ii) For atomisation coefficient of NH3 should be
one
(iii) Products should be in gaseous ionic form
(iv) Hºr = 3/2
P–PHº
3. Ans. 4
4. Ans. 9
5. Ans. 1
6. Ans. 2
7. Ans. 6
8. Ans. 5
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-1
HS-5/160000CT103115002
SOLUTION
PART-3 : MATHEMATICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10
A. A,B A,C A,B,D A,C A,B,C A,B A,C A,B,C,D A,B A,C,D
A B C D A B C D
P,S P P,R T P,Q P R S,T
Q. 1 2 3 4 5 6 7 8
A. 9 6 4 4 6 3 3 4
SECTION-I
SECTION-II Q.1 Q.2
SECTION-IV
SECTION-I1. Ans. (A,B)
/ 2 / 2
n n / 2 nn
0 0
I x sin x cosx dx 2 x cos x dx4
Put x t4
/ 4n / 2 n
n
/ 4
I 2 t cos tdt4
/ 4n
1 n2
n
0
I 2 cos t dt
/ 4n / 41 n 1 n 2 22
n0
0
I 2 cos t sin t n 1 cos t 1 cos t dt
n
n 2
nI / 22 n 1
I
{After simplification}
2. Ans. (A,C)
Let g(x) = ax2 + bx + c
[ƒ(x)] = [g(x)] x R
–1 < ƒ(x) – g(x) < 1 x R
–1 < (1– a)x2 – x(1 + b) + (1 – c) < 1 x R
a = 1, b = –1
[x2 – x + 1] = [x2 – x + c] [1] = [c]
Let x2 – x + 1 = I + ƒ c = 1 + ƒ'
[I + ƒ] = [I – 1 + ƒ + c]
0 = [ƒ + c] – 1
1 = [c + ƒ]
1 = [1 + ƒ' + ƒ]
[ƒ' + ƒ] = 0 ƒ (0,1)
ƒ' = 0
c = 1
ƒ(x) = g(x) x R
3. Ans. (A,B,D)
ƒ(x) = (x – 1) |x – 3| – 4x + 12
2
2
9 x x 3ƒ x
x 8x 15 x 3
2x x 3
ƒ ' x2x 8, x 3
ƒ'(3–) = –6
ƒ'(3+) = –2
ƒ'(0) = 0
ƒ'(4) = 0
(1,ƒ(1))
4
5310
4. Ans. (A,C)z
P(x,y,z)
x
y
P'(x, ,0)
Obviously 'P' lies on plane y = zTo see the curve rotate y = z about x-axis tobecome z = 0. New co-ordiante of pointcorresponding to P in new plane is
4cos t,4 2 sin t,0
Locus is 2 2x y
1;z 016 32
5. Ans. (A,B,C)
P(22.52) = 22.521 1
1 1 2.5.42 5
no. of numbers multiple of 2 = 2.52
no. of numbers multiple of 5 = 22.5.no. of numbers multiple of both = 2.5 P(22.52) = 22.52 – 2.52 – 22.5+ 2 .5
HS-6/16
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-1
0000CT103115002
2 2 1 12 5 1 1
2 5
similarly check other options.
6. Ans. (A,B)
E : Both die shows face 6
A1 : Both are biased
A2 : Both are fair
A3 : One is biased & other is fair.
1
1 1 1A 6 8 8P
1 1 1 1 1 1 4 1 1E . . .6 8 8 6 6 6 6 6 8
9
73
2A 16P
E 73
7. Ans. (A,C)
2 3 1 1 1 1 1
P X3 5 3 2 5 2 5
6 1 7
15 15 15
Y 6 /15 6P
6 1X 7
15 15
8. Ans. (A,B,C,D)
No. of roots common to
1nz 1 & 2nz 1 is HCF(n1,n
2)
9. Ans. (A,B)
Do yourself
10. Ans. (A,C,D)
1003
1
100
I t dt
100
1
100
2I 1 dt
{king & odd}
I1 = –100
SECTION – II
1. Ans. (A)(P,S); (B)(P); (C)(P,R);(D)(T)
L1
S'(1,1)
y=x
L1
'
S
L2
'
(1,0)L2
y=0
Directrix
O
axis of parabola
M
In the figure ; line MO is parallel to axis of parabola.
Slope of axis 1
2 . L
1 & L
2 shown in figure
are parallel to axis of parabola and image of L1
& L2 in corresponding tangents pass through focus.
L1 x – 2y + 1 = 0
L1' –2x + y + 1 = 0
L2 x – 2y – 1 = 0
& L2' x + 2y – 1 = 0
3 1S ,
5 5
Now image of S lies on directrix
1 3
S' ,5 5
Equation of directrix 2x + y = 1now check options.
2. Ans. (A)(P,Q); (B)(P); (C)(R);(D)(S,T)
R
P
A( 0)A'(– ,0)
2 2x y1
2 1
Equation of AP a
y cot x ab 2
Equation of A'P a
y tan x ab 2
2 2 2 4
2 2 2
2 2 2 2
a y x ay x a
b a b b
2 2 4
2 2 2
x y a
b a b
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-1
HS-7/160000CT103115002
22 y
x 42
2 2x y1
4 8
Now check options.SECTION – IV
1. Ans. 9
(0,16)
Q( n3,24–18 n3)
(n3,0)0
P
A
Area of trap OAQP = 1
40 18 n3 n32
Req. area n3
2x
0
140 18 n3 e 18x 15 dx
2
5 n3 4 2. Ans. 6
ƒ(x) = x4 + px2 + qx + r
ƒ'(x) = 4x3 + 2px + q
2ƒ" x 12x 2p
ƒ"(x) = 0
2 p
6
Now
(1) 4 + p2 + q + r = 0
(2) 43 + 2p + q = 0
(1) - (2) ×
–34 – p2 + r = 02 23p p
r 036 6
p2 + 12r = 03. Ans. 4
16 3
8 0
1 82x u I ƒ u du I . ƒ u du 4
2 2
4. Ans. 4
ƒ(x) = 0 has 3 distinct real roots ƒ'(x) = 0 has atleast 2 distinct real roots. (ƒ(x).ƒ'(x))' = 0 has atleast 4 distinct real roots.
5. Ans. 6
8k 8k22 22 22
k 2k 7k
24k kk 1 k 1 k 1
z 1 z 1N 1 z z .... z
z 1 z 1
= 22 + (0–1)× 7 = 15
6. Ans. 3
ˆ ˆ ˆ ˆˆ ˆi j k . ai bj ck 3
3cos
2
/6 2
i–j+k
1
P
3.2cos 3 { a2 + b2 + c2 = 4}
locus of 'P' is circle with radius A = [A] = 3
7. Ans. 3D(origin)
A(a)C(c)
B(b)
a b c
2 2 2a b c 2a.b 0
2 2 2b c a 2b.c 0
2 2 2c a b 2a.c 0
2 2 2a b c 2a.b 2b.c 2c.a 0 ...(1)
Also 2
a b c 0
2 2 2a b c 2a.b 2a.c 2b.c 0 ....(2)
(1) & (2) similarly
a.b 0
b.c 0
c.a 0
equality cannot hold as points becomes collinear.8. Ans. 4
Let a b c d x
Consider
a c . b d c.d a.b c.b a.d
= 0
2xa.b c.d b.c a.d
2
SOLUTION
PAPER-2
HS-8/16Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005
+91-744-5156100 [email protected] www.allen.ac.in
Paper Code : 0000CT103115003
CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)
PART-1 : PHYSICS ANSWER KEY
SECTION-I1. Ans. (A,D)
Sol.21
nR 10 425
m 4
332 25
96 4m 15.36gm
25
nRTV
p
Vmax
at 3
16 40nR
400
1
1 300V
25 1
2
1 400V
25 2
2. Ans. (A,D)
Sol.
2 2R2h h
2g
2
2gh
R
2 21dK dm x
2
22 21 x
2 xdx h x2 2g
4 2 62 R R
K h2 2g 6
Q. 1 2 3 4 5 6 7 8 9 10
A. A,D A,D B,C A,B,C B,C A,D A,C C A,D C,D
Q. 11 12
A. A A,D
Q. 1 2 3 4 5 6 7 8
A. 6 8 5 4 2 7 4 2
SECTION-I
SECTION-IV
4 42 R 2gh R
h2 12g =
2 45 R h
6
3. Ans. (B,C)
Sol. 1
100
1 = 10 rad/s
2
400
1 = 20 rad/s
20t10t
30 t =
t = 30
sec
4. Ans. (A,B,C)
Sol. (A) t = 0
iL = 0
VV
2
(B) t =
2Vi
3R 2
Vi
3R A
VV
3
(C) after opening
A
Vi
3R
VL = V
A + V
R
V2R
3R
2V
3
A
VV
3
Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced
TEST DATE : 14 - 02 - 2016
TARGET : JEE (ADVANCED) 2016
JEE (Main + Advanced) : ENTHUSIAST & LEADER COURSE
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-2
HS-9/160000CT103115003
5. Ans. (B,C)
Sol. Pmax
at resonance 1
LC
61f 10 40 10
2
25Hz
Qfactor
= L
2 R
R 1005
2L 20
45f
2
6. Ans. (A, D)
Sol. mgˆ ˆ0.5g 1i 5i
Bˆ0.5 4i B
Option A Bˆ ˆ5i 6k those B
for which
has – ˆ5i are useful.
7. Ans. (A,C)
Sol. (B) unstable equilibrium
(C)
x
2 3
GMm GMmxF cos
R R so SHM.
(D) F proportional to x3
8. Ans. (C)
Sol. Pipe fixed
1 1S pdt m v
2 2 2S pdt m v1 2
2
1 2
m v Sv
S m
Pipe free same effect9. Ans. (A, D)
Sol.
100V200
50
100 – x
100
80V
0
100
i = 100
0.4A250
Vs = 1.5 V
no light
for current to flow initially
80 > 100 – xlight x > 20V
200
50
100 – x
100
y
0
100
1A
100
1 + y 0 y 100
0200 50
y y1
200 50
y 4y
1200
y = 4040 > 100 – x
x > 60V10. Ans. (C,D)
Sol.h
=
hC
2.5 =
12402 496nm
5
11. Ans. (A)
Sol. S
E
x1
L1
L2
x2
2 s3
GM
R
2s e
12 211
GM GMR x
xR x
s e s
12 2 311
M M MR x
x RR x
33 e1
s
M Rx
3M
s e s
22 2 322
GM GM GMR x
x RR x
s e22 2
2
M M3x
R R x
12. Ans. (A, D)
HS-10/16
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-2
0000CT103115003
SECTION-IV
1. Ans. 6
Sol. 14 10 47 5 2N B He
Q = (0.00307 – 0.01294 – 0.00260) 931
= 0.01247 × 931
2. Ans. 8
Sol.53°
37°
x
37° 53°
T2T1
T2 sin 37 = T
1 sin 53
3T2 = 4T
1
2
3xx cos53
5
1
4xx cos37
5
32
2
3x
5t 3 3 10T
3
1
3x 33 10
5 4T
31
11
4x4 45t x 80 105 T 5T
x
= 8 × 10–3 sec
3. Ans. 5
Sol.
120°
h
x
15°45°
2h
2 2 2 1.05 2x
3 1 3
2.1 3x
3 1
2.1 2.10x
1 0.577 0.42
4. Ans. 4
Sol.2 2
2 2
kQ kQ mv
r 4r r
22 3kQ
mv4r
2 221 kQ 2kQ
E 2 mv2 2r r
2 2 23kQ kQ 2kQ
4r 2r r
23kQ
4r
5. Ans. 2
Sol.
2r × S × 2 = 0.56 × 10–3
2r × 0.07 = 0.56 × 10–3
r = 2mm
6. Ans. 7
Sol.L
L g
g
1 Q 41 Q Q
5 Q 5
L
L
1 Q '1
10 Q
L L g
9 36Q ' Q Q
10 50
L
g
Q ' 36 141 1
Q 50 15
28%
7. Ans. 4
Sol. Rapprox
= 120
602
x v
120 15i i i 2 A
960 8
120R 64 R 4
15
8
8. Ans. 2
Sol.
R – x2 2
Rx
2 202 R x µ ni i = ma
= mvdv
dx
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-2
HS-11/160000CT103115003
R22 20
0
2 nivdv R x dx
m
2v
2 =
7 22 4 10 4000 i
m
R2 2
0
R x dx
x = R sin
dx = R cos d
/ 2 2
2 2
0
RR cos d 1 cos 2 d
2
= 2R 1
sin sin 02 2 2
= 2R
4
2 7 2v 8 10 4000 i2
2 0.02 4
2 7 22
3
v 8 10 4000 i1
2 4 10 4
v = 2m/s
SOLUTION
PART-2 : CHEMISTRY ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10
A. A,C,D D B,D C,D A,C,D A,B,D B A,C,D B,C D
Q. 11 12
A. A,B,D B
Q. 1 2 3 4 5 6 7 8
A. 2 1 2 0 6 4 6 3
SECTION-I
SECTION-IV
SECTION-I
1. Ans. (A,C,D)
2P(atm)
1 10 20
T=300K
V(Litre)
1
QP
(A) A real gas condense if external pressure is
just greater than vapour pressure
(B,C,D)
At point 'P' whole real gas is in gaseous phase
so density of gas = 1000
0.05gm / ml20 1000
At point 'Q' whole substance is in liquid phase
so density of liquid = 1000
1gm / ml1 1000
Note : during condensation at 300K density of
gas and liquid remains constant, only amount
varies.
2. Ans.(D)
In terms of edge length 'a'
X = 2
2a
Y = a
2
Z = 3a
4
3. Ans. (B, D)
4. Ans. (C, D)
5. Ans. (A,C,D)
6. Ans. (A,B,D)
7. Ans. (B)
8. Ans. (A,C,D)
9 Ans. (B, C)
If n/p ratio is smaller than required possibleemission is
53P135
54Q135 +
–1e0
HS-12/16
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-2
0000CT103115003
If n/p is smaller than required possible emission
are
54Q135
52R131 +
2He4
or
54Q135
53R135 +
+1e0
10. Ans.(D)
For 'Q' using steady state concept
Rate of appereance of Q = Rate of
disappeareance of Q
K1[P] = K
2[Q]
Number of nuclei of Q
= 1
2
k
k no. of nuclei of P
= 23 2010 / 60
6 10 101000
For 'R'
Since rate constant is very high for second step
than first step
Number of nuclei of R = Number of nuclei of
P disintegrated = 6 × 1023
11. Ans. (A,B,D)
12. Ans. (B)
SECTION-IV
1. Ans. 200 [OMR Ans. 2]
At freezing point vapour pressure of solid
= vapour pressure if liquid
10 – 3000 2000
5T T
5 = 1000
T
T = 200 K
2. Ans. (T5 = 100) OMR ANS (1)
Let heat involved in step 5 is Q5
qtotal
= – Wtotal
500 + 800 + Q5 = 700
Q5 = – 600 J
Since S total = 0
5
500 800 6000
200 200 T
2 + 4 – 6 = 0.
T5 = 100 K
3. Ans. 2
4. Ans. 0
Sol. FeO.Cr2O
3 + Na
2O
2 2NaFeO
2 + 4Na
2CrO
4
+ 2Na2O
(A) (B)
2NaFeO2 + H
2O NaOH + Fe
2O
3
(A) (C)
red brown ppt.
Na2CrO
4 + H
2O CrO
4–2 + 2Na+
(B) (D)
CrO4
–2 + 2H+ Cr2O
72– + H
2O
(D) (E)
(orange)
5. Ans. 6
6. Ans. 4
7. Ans. 6
8. Ans. 3
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-2
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SOLUTION
PART-3 : MATHEMATICS ANSWER KEY
SECTION-I1. Ans. (A,B,C)
1 x
0 0
ƒ(xt)dt 0 ƒ(t)dt 0 ƒ(x) 0
(A) y = 0 & y = x3 – 3x2 + P
intersect at 3 distinct points.
x3 – 3x2 + P = 0 P = 3x2 – x3
2
34
P (–4,0) P = –3,–2,–1
(B) ƒ(x) = 0 is a periodic function
(C) for minimum area ƒ(3) = 0
81 – 12 – a = 0 a = 69
(D) ƒ(x) = 0 is even as well as odd.
2. Ans. (B,D)
For k1 > k
2 k
1cos2x > k
2cos2x
–k1cos2x < – k
2cos2x
2 21 21 k cos x 1 k cos x
2 21 2
1 1
1 k cos x 1 k cos x
ƒ(k1) > ƒ(k
2)
Hence ƒ(x) is increasing function.
3. Ans. (A,B)
AB = BT
A = BTB–1
|A| = |BT|. |B–1| = 1 ...(1)
Also Adj A = Adj(BT B–1)
= Adj(B–1) Adj(BT)
AdjA = (AdjB)–1 Adj(BT)
Q. 1 2 3 4 5 6 7 8 9 10
A. A,B,C B,D A,B A,C B,C A,D A,C A,C C B
Q. 11 12
A. C C
Q. 1 2 3 4 5 6 7 8
A. 0 0 4 9 2 3 5 5
SECTION-I
SECTION-IV
(AdjB)T = (AdjB) AdjA
= (AdjB) B
(AdjB)T = |B|.I
AdjB = |B|.I
Adj(AdjA) = |AdjA| I
|A|n–2A = |A|n–1I (order n)
A = |A|I A = I (By (1))
B = I
4. Ans. (A,C)
If ƒ(x) touches x-axis at 2,0
it also touches at 2,0
Roots are 2, 2, 2, 2
5. Ans. (B,C)
1
n2
0
I n 2016 x cos x dx
2
0
1I 2016 x cos xdx
2
0
1I 2016 x cos xdx
2
0
12I 2016 cos xdx
40321I
4
1/ n2
0
n n
x 2016 cos xdx
lim nI n lim1/ n
2
2
2n
1 1 12016 cos
n n nlim 2016
1/ n
HS-14/16
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-2
0000CT103115003
6. Ans. (A,D)
z1 = , z
2 = 2
n ni2 i2n n 3 31 2z z e e
2n2cos
3
2016 20161 2z z 2
2015 20151 2
4030z z 2cos
3
2cos 13433
12 1
2
7. Ans. (A,C)
Q( )P(t)
A
B
Equation of chord of contact
x + y = 4 ...(1)
Also, Equation of tangent at P(t)
is ty = x + 2t2
Also t = + 2t2 ...(2)
from (1) & (2)
(t – 2t2)x + y = 4
(tx + y) – (2t2x + 4) = 0
tx + y = 0 & 2t2x + 4 = 0
yt
x
2
2
y2. x 4 0
x
y2 + 2x = 0
(–1/2,0)
Area bounded by S = 0 & the line 2x + 1 = 0
2 1 22
3 2 3
2x – y + 1 = 0 is a focal chord.
Hence angle is 90°.
8. Ans. (A,C)
Equation of family of planes
(2x – y + z – 2) + (x + 2y – z – 3) = 0
It must satisfy (3, 2, 1)
(6 – 2 + 1 – 2) + (3 + 4 – 1 – 3) = 0
3 + 3 = 0 = – 1
x – 3y + 2z + 1 = 0
Equation of acute angle bisector
2x y z 2 x 2y z 3
6 6
x – 3y + 2z + 1 = 0
Paragraph for Question 9 to 10
Solving x
y2
& 2
2xy 1
9 , we get.
6x
13 &
3y
13
(6/ 13, 3/ 13)
(3,0)Shaded region
3 2
6 / 13
1 6 3 x1 dx
2 913 13
3
2 1
6 / 13
9 1 x 9 x9 x sin
3 2 2 313
19 1 9 3 9 9 2. sin
3 2 2 213 13 13 13
13 3 2sin
4 2 13
13 3sin
2 13
similarly calculate the other area
as 1
2
3 1sin
2 1 9m
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-2
HS-15/160000CT103115003
9. Ans. (C)
10. Ans. (B)
162 = (a + b + c) (a + b – c) (a + c – b) (b + c – a)
Now, b = 2c & a = 3, we get
162 = (3c + 3) (3 + c) (3 – c) (3c – 3)
= 9(9 – c2) (c2 – 1)
Using A.M > G.M, we get
(9 – c2) + (c2 – 1) > 2 ((9 – c2) (c2 –1))1/2
Maximum value will be attained
for 9 – c2 = c2 – 1
c 5
162 = 9.4.4
Dmax
= 3
11. Ans. (C)
12. Ans. (C)
SECTION – IV
1. Ans. 0
A(2,2)
6 – 42 – 2,3
C
6 – 43
B
3y–2x= –2
6 – 23
3y+2x=6
(3,0)
(1,0)
+22
,6
6 – 4 – 4 + 4 = –2
3
2
Also 3.(12 2 )
( 2) 26
12 – 2 – 2 – 4 = –4
4 = 12 = 3
Slope of BC = 0
2. Ans. 0
(1 + x2 + x4 +......+x100 – x (1 + x2+......+x98))
(1 + x2+....+x100 – x(1 + x2 +....+x98))
= (1 + x2 + x4 +....+x100)2 – x2(1 + x2+....+x98)
which implies there is no term having oddpower of x.
3. Ans. 4
E
C
A F D B
/2
/2
/2–
/2–
Using sine law in BDE
BE BD
3cos sin
2 2
...(1)
Also sine law in BDC
BC BD
cos sin2 2
....(2)
Dividing we get
sinBE 2 EF3BC sin2
0
1lim EF
3 p + q = 4
4. Ans. 9
Using Baye's theorem
21
63 1 1 2
0 16 6 2 6
4
5
a + b = 9.
5. Ans. 2We have to calculate
2 2 2a 1 b 1 c 1
a b c
Now x3 – 2x2 + 3x – 4 = 0
a
b
c
abc = 4
HS-16/16
ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-2
0000CT103115003
Also x3 – 2x2 + 3x – 4 = (x – a) (x – b) (x –c)
Put x = i & –i and multiply, we get
(a2 + 1) (b2 + 1) (c2 + 1) = 8
2 2 2a 1 b 1 c 1
2abc
6. Ans. 3
VDABC 1 1
6 3 × (area of base) × height
Also 1 1 1
AC BC sin 45 AD6 3 2
AC
.BC.AD 12
Now, 3AC AC
BC AD 3. .BC.AD2 2
(using A.M.- G.M.)
AC
.BC.AD 12
Equality must hold
ACBC AD 1
2
AD must be perpendicular
CD2 = AD2 + AC2 = 3
7. Ans. 5
Given |z| = 1 &1
a z 1z
'a' moves on a circle with centre 1
zz
and radius 1.
1z
z can take any value in the interval
[–2,2], so the set S consists of the locus of unit
circles centered at a point on the line segment
with end points at –2 & 2 on the argand plane.
Therefore, the area S is the area of two semi
circles and area of rectangle which is + 8.
8. Ans. 5
0 2015 1 2015 2015 2015
1 1 1S ...
2 2 2 2 2 2
2015 2015 2014 2015 0 2015
1 1 1S ...
2 2 2 2 2 2
2015 2015
1 12S .......2016 times
2 2
2017
2016S
2
k = 5
