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MATHEMATICS
Code-A Page # 1
PART-A
Q.1
[Sol. Image of (5, 0) in xy + 3 = 0 is (3, 8)
Hence the circle is (x + 3)2 + (y8)2 = 25
(5,0)
(3, 8)
x
y
y = x + 3
O
x2 + y2 + 6x 16y + 48 = 0
So, g = 6, f =
16, c = 48
Hence, (g + f + c) = 38 Ans.]
Q.2
[Sol. h(x) = )x(gf + 10 = | x |2 + 4 | x |1 + 10 = | x |2 + 4 | x | + 9 = 22|x| + 5
Hence range of h (x) is [9, )Minimum value occurs when x = 0 ]
Q.3
[Sol. Clearly, graph in option (A) best represent the graph of function y = 2|2x| .]
Q.4
[Sol. Put x2 = t 0at2 + bt + c = 0 ; b2 4ac
a
b> 0 and
a
c> 0
tO
f(t)
x
y
a
b< 0 and
a
c> 0 a > 0, c > 0, b < 0]
Q.5
[Sol. Using, tan( + ) =
tantan1
tantanand tan(arc tan a) = a a R, we have
y
1tanarc
x
1tanarctan = tan
10
1tanarc
xy11
y
1
x
1
=
10
1 (x10)(y10) = 101
The following four ordered pair of integer numbers are solutions of this equation:
(11, 111); (111, 11), (9,91), (91, 9) 4 ordered pairs Ans.]
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MATHEMATICS
Code-A Page # 2
Q.6
[Sol. We have 2a cos
2
CB= b + c
2 (2R sin A) cos
2
CB= 2R (sin B + sin C)
4sin2
Acos
2
Acos
2
CB= 2cos
2
Acos
2
CB
2cos2
A cos
2
CB
12
Asin2 = 0
Now, cos2
A= 0
2
A= 90 A = 180 (Not possible)
Also, cos
2
CB= 0
2
CB=
2
BC = (Not possible)
Hence sin
2
A=
2
1
2
A=
6
or A =
3
.
So, sec A = sec3
= 2 Ans.]
Paragraph for Question No. 7 to 9
[Sol. We have 1 + 1 + 1 =2
=2
3 + 3 + 3 = ( + + ) (2 + 2 + 2) + 3 3 + 3 + 3 = (2) (6 + 1) + 3 (2)Hence, 3 + 3 + 3 = 146 = 8 .......(4)
(i) Clearly, (3
+ 3
+ 3
) = 8 Ans.(ii) Let f(x) = x24ax12.
If both roots of f(x) = 0 lies on either side of + + , sof(1) < 0
(1)2 + 4a12 < 0 4a11 < 0 a 0 D2
= 4 [4] [42k] > 0
k >2 k < 2Hence k (2, 2) Ans.But k 0 because at k = 0, D1 = D2Hence only 2 lines are possible ]
Q.11
[Sol. Clearly,
kk100100
0kk
100
3)2x(C
= 100
3)2x( = (1 + x)100
Hence, coefficient of x50 is 100C50
. Now verify
Option (A), (D) equal to 100C50
.
(B) 99! (100C50
)
(C) 100C49
.]
Q.12
[Sol. (A)
A f gB C
gof: A C
(B)
A f gB C
gof: A C
(C)
A f gB C
gof: A C
(D)
A f gB C
gof: A C
Note: If f and g are both bijective then gof is also bijective but converse (in general) is not true. ]
Q.13
[Sol. Let equation of direct common tangent(s) be
(y
0) = m (x
3) mxy3m = 0
(3, 0) O
(1, 0)
R
Q
x
P
y
3,0
(3, 0)
3,0
As, p = r
2m1
|m30)1(m|
= 1
4m2 = 1 + m2 m2 =3
1 m =
3
1
Clearly,PQR is an equilateral triangle.Now, verify alternatives.]
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MATHEMATICS
Code-A Page # 4
PART-B
Q.1
[Sol.
(A) We have, x2 + y2 = y2 (1 + tan2t) = y2 sec2t = xyx
ysec2t =
ttan
t2sin sec2t =
tcostsin
tcostsin2=2 Ans.]
(B) Let E = (ab) + b +)ba(b
1
Using A.M. G.M. for the numbers (ab), b ,)ba(b
1
.
We have,3
1
)ba(b
1b)ba(
3
1
)ba(b
b)ba(
ab + b +)ba(b
1
3 [Note: Also minimum value of E = 3 occurs when a = 2, b = 1.]
Hence, a +)ba(b
1 3. Ans.]
(C) When | cos x | = 1 then | sin x | = 0
fmax = 40 + 5 = 9when | cos x | = 0 then | sin x | = 1
fmin
= 03 + 5 = 2 Ans.
Range of function f(x) is [2, 9]8 integral values.]
(D) As, tan1x + cot1x =2
x R
So, the given equation can be written as
(sin1x)3 + (cos1 x)3 = 7
8
3
(sin1 x + cos1 x)33 (sin1 x) (cos1 x) (sin1 x + cos1 x) = 73
2
3
2
3 (sin1x) (cos1 x)
2
= 7
3
2
(sin1 x) (cos1 x) =2
2........(1)
Now, as maximum value of cos1x is (cos1 x 0) and minimum value of sin1 x is2
and this
happens at the same x i.e., at x = 1.
Hence, x =1 is the only solution. Ans.]
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MATHEMATICS
Code-A Page # 5
PART-C
Q.1
[Sol. As 2x2x 39
7
5
19|2x||2x| 33E
Tr + 1 =n
Cr xnr
yr
(Put r = 5)
Now, T6
= 567 = 7C5 92x
2|2x| 33
92x2x2 3327
| x2 | = 4 x = 6 or 2Hence, sum of possible values of x = 62 = 4. Ans.]
Q.2
[Sol. Equation ofl1, l
2, l
3and l
4are y = mx + 1 ..............(1) l
1
y = mx + m ..............(2) l2
x + my = 0 ..............(3) l3
my + x = 1 ..............(4) l4distance between l1 and l2 = distance between l3 and l4
22 m1
1
m1
1m
| m1| = 1 m1 = | m1|
m = 2 ; m = 0 (rejected)
So,side of the square =5
1
l2
l1
l3
O(1,0) (1,0)
m
m
X
Y
m
1
(0, 1)
l4
m
1
Area =5
1 (p + q)least = 6. Ans.]
Q.3
[Sol. As
2sin
1cot2 = 2coseccot2 = 2cot1cot2 =
cot1 = cot1
Also,
,4
3, so
cot1 =(1 + cot )
we get
1tan no solution exist. Ans.]
Q.4
[Sol. Let f (x) = 5x22kx + 12
1
x
x
Now we want exactly one x for which f (x) < 0
Hence 1 < | x1x
2| < 2
| x1x
2| < 2 (x1x2)
2 < 4 (x1 + x2)24x
1x
2< 4
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MATHEMATICS
Code-A Page # 6
5
4
25
k42
< 4
k25 < 25 k2 < 30 k ......(1)
and | x1x
2| > 1 (x
1x
2)2 > 1 (x
1+ x
2)24x
1x
2> 1
5
4
25
k42
> 1
5
9
25
k42
k2 >4
45 k .....(2)
(1) and (2) positive integral value of k are 4 and 5.Hence sum = 9 Ans.
Alternatively: f (x) = 5x22kx + 12
1
x
x
Since k is positive
Hence vertex of f (x) lies on positive x-axis.
Also product of root is51 ,
so both roots of f (x) are positive and smaller root x1
has to be positive fraction
because if x1
> 1, then for x1x
2=
5
1, x
2has to be fraction which is not possible.
Hence for exactly one integral solution of f (x) < 0, f (1) < 0 and f (2) > 0
f (1) = 62k < 0 k > 3 .....(1)
and f (2) = 214k > 0 k 0, Sgn (x) = 1
f(x) = 0 +2
=
2
for x = 0 f(x) is not defined
for x < 0, f(x) = 2
=
2
M = 1N: Coefficient of t5 = coefficient of t2 in (1 + t2)5 coefficient of t3 in (1 + t3)8
= 5 8 = 40
Hence L = 36; M = 1 and N = 40
LM + N = 36 + 40 = 76 Ans.]
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CHEMISTRY
Code-A Page # 1
PART-A
Q.1
[Sol. Spherical node : 2 = 024 + 3 = 0 ; 23 + 3 = 0(3)1 (3) = 0 ; (1) (3) = 0
= 1, 3 ; = 3,1a
Zr2
0
r =Z
a
2
3 0,
Z
a
2
1 0]
Q.2
[Sol. Spin quantum number define the spin of electron. ]
Q.3
[Sol. (A) CH CNH3 2
||O
23 NHCCH|O
(B) NH2CNH2
NH
22 NHCNH|
NH
22 NHCNH|
NH
More basic due to high change density.
(C) 23 NHCCH||
NH
23 NHCCH|
NH
(D) Ph CNH2
O PhCNH2
O]
Q.4
[Sol. 118 % Oleum
100 gm+ 9 gm H O2
80 gm SO3 20 gm H SO2 4
H SO2 4 + SO3
69gmH SO2 4
40gm
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CHEMISTRY
Code-A Page # 2
(A) % H2SO
4by wt. = 100
109
69 = 63.30 %
(B) Mass % of combined SO3 100
109
180
98
69
= 51.67%
(C) with 109 gm New oleum water can react = 9 gm
with 100 gm New oleum water can react = 100109
9
new labelling = 100 +109
1009 108.25 %
(D) Mass % of free SO3 100
109
40 = 36.69% ]
Q.5
[Sol. In case of BaS size of cation and anion is greater than the other ions and lattice energy is irreversibly
proportional to size of ions. ]
Q.6
[Sol.
(IV)
CH OCH3 2..
> CH2
(I)
>)III(
CHCH 23
> (II)
Back bonding Sigma Resonance 3 H Bridge head unstableComplete octet Hyperconjugation ]
Q.7
[Sol. Double bond equivalent = Rings + Number of bonds
6 + 5 = 11 ]
Q.8
[Sol. 3 Functional groups = (i) Ester
(ii) Anhydride
(iii) 3 amine ]
Q.9
[Sol. CH bond energy radicalfreeofSatbility
1
IV > II > III > I ]
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CHEMISTRY
Code-A Page # 3
Q.10
[Sol. 1 mole Na2C
2O
4 1 mole CaC
2O
4[from reaction (i)]
1 mole CaC2O
4 1 mole H
2C
2O
4[from reaction (ii)]
1 mole H2C
2O
4 2 mole CO
2[from reaction (iii)]
2CO
n = 2
Urms
=nM
PV3
M
RT3
For CO2 gas
Urms
=22 COCO
M2
PV3
M
RT3 ]
Q.11
[Sol. (A) CO2
SO2
Number of atom = 3 Number of atom = 3
O = C = OS
O ONumber of lone pair = 4 Number of lone pair = 5
(B) XeF2
ICl2
Number of atom = 3 ClICl
FXe F Number of atom = 3
Number of lone pair = 9 Number of lone pairs = 9
(C) ICl2 ClICl
CNI 2 N CIC N
Number of atom = 3 Number of atom = 3
Number of lone pairs = 9 Number of lone pairs = 5
(D) SO3 SO O
O
BF3 BF F
F
Number of atom = 4 Number of atom = 4
Number of pairs = 6 Number of lone pair = 9 ]
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CHEMISTRY
Code-A Page # 4
Q.12
[Sol. (A) Non aromatic
(B) Aromatic
(C) Aromatic
(D) Aromatic ]
Q.13
[Sol. Stability order
III > IV > V > II > I
6 H 5 H 4 H 3 H 0 H
HOH AlkaneofSatbility
1
HOC Number of carbocation
Boiling point surface area V > IV > III > II > I
Rotational Barrier extent of double bond I > II > V > IV > III ]
PART-B
Q.1
[Sol. (A) ContainerI reacts with containerIIn factor of KI = 6
KMnO4
K2Cr
2O
7
1 5 = x 6 1 6 = x 6
x =6
5x = 1
(B) n factor of Cu2S = 8KMnO4
K2Cr
2O
7
1 5 = x = 8 1 6 = x 8
x =8
5x =
8
6=
4
3
(C) n factor of K2C
2O
4 3H
2C
2O
4 3H
2O = 8
KMnO4
K2Cr
2O
7
x =8
5x =
4
3
(D) n factor of NH4SCN = 6
KMnO4 K2Cr2O7
x =6
5x = 1 ]
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CHEMISTRY
Code-A Page # 5
PART-C
Q.1
[Sol. 3NaOH + H3PO
4 Na
3PO
4+ 3H
2O
Total m moles of H3PO
4= (10 0.5) = 5
Require NaOH = 15 m moles
for Molarity of NaOH required =300
15
20
1
21
21
VV
30
1V
10
1V
=
20
1
2V1
+3
2V
2= V
1+ V
2
V1
= V2
3
21
V1
=3
V2
V2
= 3V1
V1
+ V2
= 1000 ml
4V1
= 1000 ml V1 = 250 ml ]
Q.2
[Sol. A 30 W EA
(photon) = 13.6 eV
B 40 W EB
(photon) = 13.6 22 = 4 13.6 eV
For A (13.6 1.6 1019) nA
= 30
For B (4 13.6 1.6 1019) nB
= 40
4
3
n4
n
B
A
1
3
n
n
B
A ]
Q.3
[Sol. Naphthalene has 10 e's systemi.e. there are 5 bondsExpected (theoretical) heat of hydrogen = 120 5 = 600 kJ/molObserved (experimental) heat of hydrogen = 372 kJ/mol
R.E. =372(600)= 228 kJ / mol ]
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CHEMISTRY
Code-A Page # 6
Q.4
[Sol. O2
=32
320= 10 mole
P = 30 atm
T = 300 K
At constant volume :
constantT
P
400
'P
300
30
P' = 40 atm
At constant volume and temperature :[P n]
f
i
n
n
1
40
40 =fn
10
nf= mole
41
nescaped
= nin
f
= 104
1=
4
39
wt. of O2
escaped = gm324
39 = 312 gm Ans. ]
Q.5
[Sol.
C
C
H
CH
CH
HC
HC
NOO
:
14 : 4 ]
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PHYSICS
Code-A Page # 1
PART-A
Q.2
[Sol. Pgas
= Patm
+A
F]
Q.4
[Sol. F
is in direction P
45Pf
P
Pi
]
Q.5
[Sol. At t = 0, blade is at +ve extreme, so, x = R cost ]
Q.6
[Sol. In figure-1, energy at A is potential as well as kinetic. In figure-2 energy at A is only potential. ]
Q.7[Sol. See graph ]
Q.10
[Sol. As the motions of particles are simple harmonic, amplitude = 2cm.
(Agnular frequency) = 3 rad/sec.As the phase difference of particles separated by 1 cm is /8
=
28
= 16 cm
Hence, wave velocity =
2
= 24 cm/s.
y = (2) sin
)xt24(16
2cm. ]
Q.11
[Sol. For string fixed at both ends,
As= A sin
x2
Antinode of x =8
L3,
A = A sin
8
L3
2
4
L3= (2n + 1)
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PHYSICS
Code-A Page # 2
L =3
2(2n + 1)
L = 2, 6, 10 = 4 loop, 12 loop, 20 loop,..... ]
Q.12
[Sol. Ares
= cosAA2AA 22 = 3 3
y = y1
+ y2
=33
cm sin [4x700 t] ]
Q.13
[Sol. Static friction limit on 4kg block is more than 2g. ]
PART-B
Q.1
[Sol. f recorder
=
Ssound
Rsound
VV
VV165 Hz
recorder = 165
)VV( Ssound
Vsound
= 330 + Vwind
]
PART-C
Q.1
[Sol. f 440 = 4 (wrong)
440f = 4 (correct)
on heating up, tension decreases
440 =L2
1
T
]
Q.2
[Sol. 45(5 10 0.2 + 15 10 0.2) = 5+ve
T = 5 10 0.2 = 10 N ]
Q.3
[Sol. P = AT4 , T =1/ 4
P
T = 1000 K ]
Q.4
[Sol. I = I1 + I2 + 2 1I 2I cos ()
= 16 + 25 + 2 (4) (5) cos
1000
)548(
2= 41 + 20 = 61 ]
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PHYSICS
Code-A Page # 3
Q.5
[Sol. Let u be the speed of the person
Observed frequency of the wave coming from B1is
f1
=
V
UV
Observed frequency of the wave coming from B2is
f2 =
2BVV
UV
f2f1 = 4.5
f0
V
UV
VV
UV
2B= 4.5
350
u350
400
u350
=
500
5.4
500
5.4
= 2800
)v350(8)v350(7 pp
5
126= 15v
p350
15
3505
126
= vp
vp
=75
1876 25.01 ]