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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
PART-1 : PHYSICS ANSWER KEY
SECTION–I
1. Ans. (A,D)
Sol. By conservation of linear momentum both will get
same velocity of centre of mass
2
cm
1mv mgh
2
h1 = h
2
But by conservation of angular momentum about
centre of mass
System 2 have angular velocity also
K 1 < K
2
[where k =2 2
cm1 1mv I2 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 cos2 20.05 sin 0.05
0.40 0.2
= 2 2
Test Type : ALL INDIA OPEN TEST (MAJOR) Test Pattern : JEE-Advanced
TEST # 12 TEST DATE : 15 - 05 - 2016
TARGET : JEE (ADVANCED) 2016
LEADER TEST SERIES / JOINT PACKAGE COURSE
Paper Code : 0000CT103115008
DISTANCE LEARNING PROGRAMME(Academic Session : 2015 - 2016)
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.
2
0 12
2 12 0 2
1k E
k 21 k k E2
=
5
3
1
1 3
,
1
1 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 cm
If converging rays meet at the point of image, image
is known as real. If diverging rays meet at thepoint
of image, image is known as virtual.
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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 2
1 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 2
1 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 diffrence2 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
92 1 10
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 ai
2R
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 r T = 0.81
0.01r T
0.80 × 103
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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 90V 5V
18 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
m p 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
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3. Ans. (A,C,D)
Anode : 2Br – Br 2 + 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 + log0.1
100.1
(P)
(B) pOH = 6 + log0.1
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
4
2– +CO3
2– + NO3
– +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
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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 distance26 22 15 9
183
volume5
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 and1
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)1
x
0
I ƒ(3 )dx 1
1 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
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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 1x ,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 2
t 3t
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
' x dx xƒ ' x dx
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 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 6
6
C C C 11 a
P A 2 32 b
1 p
P B2 q
5 5 5
3 4 5
6
P AB C C CA 1 uP 2
B P B 2 2 v
P BAB 32 16 8 m
P A P A 11 64 11 n
2. Ans. (A)
(P,Q,R); (B)
(T); (C)
(T);
(D)
(S)
(A)
4991000
2k 1
k 0
2k 1 C
100
499999
2k
k 0
10 C
= 10.2998
(B) 4
3 1 I ƒ
4
3 1 ƒ '
4 2
4 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
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SECTION – IV
1. Ans. 3Do yourself
2. Ans. 4
Do yourself
3. Ans. 6Use |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 1tan2 x 1
7. Ans. 2
Do yourself
8. Ans. 1
ƒ(1 – ) > ƒ(1)
2 > 1 + k
k < 1
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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
Test Type : ALL INDIA OPEN TEST (MAJOR) Test Pattern : JEE-Advanced
TEST # 12 TEST DATE : 15 - 05 - 2016
TARGET : JEE (ADVANCED) 2016
LEADER TEST SERIES / JOINT PACKAGE COURSE
Paper Code : 0000CT103115009
DISTANCE LEARNING PROGRAMME(Academic Session : 2015 - 2016)
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 12A. 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 dtR
–
21 Li2
0
Rt2L 2V L 1
e LiR R 2
22
2
V L 1Li
R 2
2 2 2
2 2 2V L 1 V 1 V LLR 2 R 2 R
2. Ans. (A,B,D)
Sol.CA B8RT3RT 2RT
M M M
C
A 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 T2
3
R 420 1052
CA
= (280 – 105)R
3. Ans. (C,D)
Sol. (A)
00
R JJ 20
2 2
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0
22
R I
4 R R
4
02R I
3 R 4
4
(B)R/2
0
R J
2 02
03
R I
3 R 4
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 aa5
(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
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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
0ia
42
[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
mv N
R
2C
C
mv N 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
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– 3R
—— 2
30°
30°
R
1t
43 3 G3
=11
11250sec
4 203 10 8003 3
3. Ans. 4
Sol. T = c sxyr z
[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 = – 12
, y =12
–3y + z = 0
3z
2
3r T 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
2i
r
= e2r (T4 – TT04)
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
2
R 4T 4sec2 2 5v
6. Ans. 4
Sol.1
330330 300
300 v
v1 = 30 m/s
2330360 300330 v
2
330330 v 10 275
12
v2 = 55 m/s
V = 25 m/s
100 = V × T
T = 4 sec
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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 mJ8. 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 f x
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
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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
4R 4
(b) C.N. = 6 ; it is a hexagonal 2D lattice
(c)r
0.155R
so, diameter = 2r = 0.31R
(d)Distance between 2 layers is2 2
R 3
2. Ans. (A, B)
For the equal mix, Y ben
= xtol
& ytol
= xBen
y ben – 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 = ln20
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) × 102x – 2y + 3y – 3z = 10 x – 10 zy – 8x + 7z = 0
y 7z8
x
2. Ans. (5)
n – l – 1 = 2 ; l = 3 n = 6
13.62 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
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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 3k 1 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 = t1
t2
, k = t1
+t2
.
Also t12 = 16t2
2
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 & t2, we get locus of T(h,k)
is 2 25y x
4
4. Ans. (B,C,D)
23a a4
r 3as 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 23a Q
4 and r 1 = r 2 = r 3 =
3a Q
2 .
5. Ans. (A,C)
xn 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 alternatives7. Ans. (A,B,C,D)
2x
2
dy2x e dx
y
2x1e c
y , As ƒ(0) =
1
2
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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
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3 3 3
3 3 3
3 3 3
a 1 a a
b b 1 b 11
c c c 1
apply C1 C1 + C2 + C3
& 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 + k 2 – 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) is2
2 2 2 2x y x y1
9 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 k ˆ ˆ ˆ3 1 3 14i 27j 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
32
0
x 1 1 dx 3 3 6