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13th VXY (Date: 22-01-2012) Review Test-6
PAPER-2
Code-A
ANSWER KEY
PHYSICS
SECTION-2PART-A
Q.1 D
Q.2 C
Q.3 B
Q.4 B
Q.5 D
Q.6 A
Q.7 B
Q.8 B
Q.9 B
Q.10 A
Q.11 C
Q.12 C
Q.13 B
Q.14 C
Q.15 A
Q.16 A
PART-B
Q.1 (A) S
(B) S
(C) Q
PART-C
Q.1 0002
Q.2 0006
Q.3 2375
Q.4 0006
MATHS
SECTION-1PART-A
Q.1 A
Q.2 B
Q.3 D
Q.4 B
Q.5 A
Q.6 B
Q.7 C
Q.8 A
Q.9 A
Q.10 B
Q.11 D
Q.12 D
Q.13 C
Q.14 C
Q.15 B
Q.16 D
PART-B
Q.1 (A) R
(B) S
(C) Q
PART-C
Q.1 0002
Q.2 0075
Q.3 0021
Q.4 0439
CHEMISTRY
SECTION-3PART-A
Q.1 D
Q.2 A
Q.3 B
Q.4 C
Q.5 C
Q.6 D
Q.7 C
Q.8 C
ONLY V-GROUP :Q.9 D
Q.10 B
Q.11 D
ONLY XY-GROUP :
Q.9 B
Q.10 A
Q.11 D
Q.12 B
Q.13 D
Q.14 CQ.15 D
Q.16 D
PART-B
Q.1 (A) Q,R
(B) R
(C) R,S
PART-C
Q.1 0810
Q.2 0004
Q.3 9306
Q.4 3421
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MATHEMATICS
Code-A Page # 1
PART-A
Q.1
[Sol. Put tan x = t sec2x dx = dt. So I = dx1xtan3
3xtan12
0
2
2
= dt
)t1)(1t3(
3t32
0
22
2
= dt1t3
2
t1
132
022
=
32
02
2
32
02
3
1t
dt
3
2
t1
dt
= 12
+ 3
1
ln
32
13
]
Q.2
[Sol. We must have f(p) < 0, where f(x) = 2x22(2p + 1)x + p(p + 1) px-axis
x = x = So, f(p) < 0 p2 + p > 0 p > 0 or p < 1. ]
Q.3
[Sol. If y = ax3 + bx2 + cx + d has only one critical point on R, sodx
dy= 0 has both roots real and equal.
Putdx
dy= 0 3ax2 + 2bx + c = 0. So, D = 0 4b2 12ac = 0 b2 = 3ac
Given, ac = 2 b2 = 6
| b | = 6 . Ans.]
Q.4
[Sol. Given, 1 + x2 + 2x sin (cos1y) = 0
As x = 0 will not satisfy it, so x +x
1= 2 sin (cos1y).
As, cos1y [0, ], so 0 sin (cos1y) 1.So, above equation is possible when x = 1 and y = 0.
Only one ordered pair (1, 0) is possible. ]
Q.5
[Sol. Use L'Hospital rule, we have
x
a
t
ax)0a(dttsinet
ax
1Lim
form
0
0=
x2
1xsinexLim x
ax= asine
2
1 a. Ans.]
Q.6
[Sol. GG GG GG | R | R | R | R | R |
6C2
gaps for GG and GG
& selection of 2 more gaps from the remaining 4 in 4C2
ways for G and G.
Total ways = 6C2 4C
2=
!2!2
!4
!4!2
!6=
!2!2!2
!6=
8
720= 90. Ans.]
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MATHEMATICS
Code-A Page # 2
Q.7
[Sol. As, trace A = (x 2) + (x2x + 3) + (x 7) = x2 + x 6
Given, trace A = 0 x2 + x 6 = 0 = (x + 3) (x 2)
x = 3 or 2. Ans.]
Q.8
[Sol. Given, C1
(3, 2) and r1
= 1249 = 5
Also, C2 (2, 3) and r2 = 1294 = 5
Also, C1C
2= 11 = 2
As, | r1 r2 | < C1C2 < r1 + r2
S1
and S2
intersect each other at two distinct points S1
and S2
have 2 direct common tangents.
Also, (C1C
2)2 2
22
1rr
S1
and S2
are not orthogonal circles.
Also, equation of radical axis of S1
and S2
is S1S2 = 0 i.e. x y = 0.
As, S1
(1, 2) < 0 and S2 (1, 2) > 0 (1, 2) lies inside S1 but outside S2.]
Q.9, 10, 11
[Sol. Any tangent to 14
y
25
x22
, is
y = mx 4252 .........(1)
If above line in equation (1) is also tangent to x2 + y2 = 16, so
4m1
4m25)0(m0
2
2
25 m2 + 4 = 16(1 + m2) 9m2 = 12
m2 =3
4or m =
3
2.
On putting m = 3
2in equation (1), we get
T1: 074y3x2 .........(i)
T2
: 074y3x2 .........(ii)
T3
: 074y3x2 ........(iii)
T4
: 074y3x2 ........(iv)
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Code-A Page # 3
(i) The equation of common tangent between C1 and C2 having negative gradient in the first quadrant is
74y3x2 .
So, x-intercept = 72 (A) is correct.
(ii) As quadrilateral formed by common tangents
between C1
and C2
is a rhombus, so area of rhombus
=
3
78
742
1
dd2
121
O(0,0)
(2 7, 0) (2 7, 0)
3
74,0
3
74,0
x
y
= 33
112
3
716
(B) is correct. Ans.
(iii) Director circle of C1 is x2 + y2 = 32. So, any point on it is sin24,cos24 .
Also, auxiliary circle of C2, is x2 + y2 = 25.
So, equation of chord of contact, is
sin24ycos24x = 25 ..........(1)Let, mid point of chord of contact be (h, k). So, also equation of chord of contact is
hx + ky = h2 + k2 [Using T = S1] ..........(2)As, equation (1) and equation (2) represents same line, so on comparing, we get
h
cos24 =
k
sin24 = 22 kh
25
cos = 22 kh24
h25
, sin =
22 kh24k25
Now, on squaring and adding, we get
1kh32k625
kh32
h625222
2
222
2
625 (h2
+ k2
) = 32(h2
+ k2
)2
h2
+ k2
= 32
625
Locus of (h, k) is x2 + y2 =32
625(D) is correct.Ans.]
Q.12, 13, 14
[Sol. We have,
0x,x
0x1),1x(
1x,1x
)x(F and
2x,x2 2x1,2x
1x0,1x
0x,1x
)x(G
Now, H(x) = F(x) + G(x) =
2x,2x2x
2x1,2x2)2x(x
1x0,1x2)1x(x
0x1,21x)1x(
1x,x21x1x
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Code-A Page # 4
y
(0,3) (1,3)
(2,2)
y=2(0,2)
y=2(1,2)
y = 2x
(2,4)
12 1 2x
y=2x+1
A1 A2 A3 A4
y=2x2
(0,1)
O
(0,4)
(i) From above graph, the function H(x) has local maximum at x = 1 (D) is correct.(ii) From above graph, the function H(x) is discontinuous at two points viz. x = 0
and x = 1 (C) is correct.
(iii)
2
2
dx)x(H = AA1
+ A2
+ A3
+ A4
=2
1(4 + 2) 1 + 21 + 131
2
1 +
21
2
1
= 3 + 2 + 2 + 1 = 8. Ans.]
Q.15
[Sol. We have, f(x) = (x2 + x 2) (x2 + 2x 3), x R
or f(x) = (x + 2) (x 1)2 (x + 3)
S-1 is correct because f(1) > f(1) < f (1+) .
S-2 is correct as f(x) has a repeated root at x = 1.
But S-2 is not correct explanation of S-1 as f '(c) = 0 does not imply that f has an extrema at x = c. ]
e.g., f(x) = (x 1)3 is differentiable x R and f '(1) = 0 but f(x) does not have extrema at x = 1. (B) is correct.
Q.16
[Sol. For two non-zero vectors to be perpendicular, their dot product must be zero.
i.e., CBAACCBBA
= 3 CBA
0
Statement-1 is false.
Obviously, Statement-2 is true.
(D) is correct.]
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Code-A Page # 5
PART-B
Q.1
Sol.
(A) Equation of given tangent to y2 = 8x, is y = x + 2 ... (1)
So, equation of any line perpendicular to it is
y = x + c. If this line is tangent to y2 = 8x, so
x=-2
yy=x+2
900
(-2, 0)
xc = m
agives c = 1
2
c = 2.
Equation of other tangent is y = x 2 ... (2)Now, on solving (1) and (2), we get M(2, 0).
So distance of M(2, 0) from
N(1, 2 2 ) =22 )220()12( = 981 = 3 Ans.
(B) The equation of ellipse x2 + 2y2 = 6 ... (1)
or 13
y
6
x22
Differentiate (1) on both sides with respect to x, we get(0, 0)
(x, y)x
y
x+y=7
2x + 4ydx
dy= 0
dx
dy=
y2
x= 1. (Given)
(As, slope of line x + y 7 = 0 is 1)
x = 2y ... (2) Solving (1) and (2), we get
4y2 + 2y2 = 6 y = 1So, points are (2, 1) and (2, 1).
But, the point lies in first quadrant. So, P(2, 1).
Also, Distance between P(2, 1) and Q(2, 1) =16)11()22(
22 = 4. Ans.
(C) The co-ordinates of P are
2
33,
2
5.
Also, e =25
91 =
5
4.
So, co-ordinates of foci are S(4, 0) and S'(4, 0) and SS' = 8.y
xS'(-4, 0) 8 S(4, 0)
I(x , y )11
2
33,
2
5P
7 3Also, SP = a ex1 = 5
5
4
2
5= 5 2 = 3.
and S'P = a + ex1
= 5
5
4
2
5= 5 + 2 = 7
The co-ordinates of incentre I(x1, y
1) are
x1
= 2837
2
58)4(3)47(
and y1
=3
2
837
2
338)03()07(
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Code-A Page # 6
I
3
2,2 .
Now, distance from I
3
2,2 from
3
2,0 =
22
3
2
3
2)02(
= 24 . Ans.
PART-C
Q.1
Sol. Given, f(x) =
x2cos (x + 3) + 2
1
sin (x +3) =
2xcos x
2
1
sin x
Now, f '(x) =
2x( sin x) +
1cos x 2
1
( cos x)
f '(x) = (2 x) sin (x)+
0 1 2 3 4
+x-axis
Clearly, x = 1 is the point of local maximum. Also, x = 3 is the point of local minimum.
Note : x = 2 is inflection point of function. ]
Q.2
[Sol. Line AB : kj2i3kir
As, P(3 + 1, 2, + 1) which lies on x + y + z = 6 2 + 2 = 6 = 2Hence P is (7, 4, 3)
Now, line of intersection of the planes x + y + z = 1 and x + z = 0
is parallel to vector =
101
111
kji= )10(k)11(j)1(i = ki = kj0i
Now, the line is1
z
0
1y
1
x
= t (say)
k)t3(j5i)7t(NP
P(7, 4, 3)
N(t, 1, t)
VNP
= 0 1(t 7) + 0 + 3 + t = 0 2t = 4 t = 2
So, N (2, 1, 2)
k5j5i5NP
Hence, 75NP d d = 75 Ans.]
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Code-A Page # 7
Q.3
[Sol. Let x = 5 and y = 100.
Now, x, A1, A
2, A
3, ........., A
n, y are in A.P. ......(1)
and x, H1, H
2, H
2, H
3, ...., H
n, y are in H.P. .......(2)
Given that, P = Ar 1
and Q = Hr 1
From (1), y = x + (n + 1) d d =1n
xy
P = Ar 1 = rth term = x + (r 1) d = x + (r1)
1n
xy
)1n(x
xy1r1
x
P
..........(3)
Now, from (2),
y
1,
H
1,,.........
H
1,
H
1,
x
1
n21
are in A.P..
y
1
= (n + 2)th
term = x
1
+ (n + 1) d' d' =
1n
x
1
y
1
1rH
1
Q
1
= rth term = 'd1rx
1 =
1n
x
1
y
1
1rx
1
1nx
yx1r
x
y
Q
y
... (4)
Hence,
Q
y
x
P= 1 +
x
y(independent of n and r)
[Put x = 5 and y = 100]
Here,
Q
100
5
P= 1 +
5
100= 21. Ans.
Objective approach: Take P as first arithmetic mean of 5 and 100 i.e. P =2
105.
Hence5P =
221 and Q as first harmonic mean of 5 and 100 hence Q =
10510052
HenceQ
100=
2
1
Q
100
5
P= 21 Ans.]
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Code-A Page # 8
Q.4
[Sol. Clearly, =p21
312
111
= 1 (p 6) 1(2p 3) + 1 (4 1) (Expanding along R1)
= pCase-I : If p 0, then system of equations has unique solution.Case-II: If p = 0, put z = k, we get x + y = 4 k and 2x + y = 6 3k
On solving, we getx = 2 2k, y = 2 + k
Now, substituting these values of x, y and z in equation x + 2y + p z = q, we get
(2 2k) + 2 (2 + k) + p k = q 6 + 0k = q i.e, q = 6Thus for q 6, there is no solution and for q = 6, there are infinite solution.Hence, for unique solution p 0, q R L = 20 21 = 420
for no solution we must have p = 0, q 6 M = 1 20 = 20for infinite solution p = 0 and q = 6 N = 1 1 = 1
L + M N = 420 + 20 1 = 440 1 = 439. Ans.Alternatively: x + y + z = 4 ... (1)
2x + y + 3z = 6 ... (2)
x + 2y + pz = q ... (3)Solving (1) and (2) x = 2 2z and y = 2 + zPut in equation (3), we get
pz = q 6
Hence, for unique solution p 0, q R L = 20 21 = 420for no solution we must have p = 0, q 6 M = 1 20 = 20for infinite solution p = 0 and q = 6 N = 1 1 = 1
L + M N = 420 + 20 1 = 439 Ans.]
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Code-A Page # 1
PART-A
Q.1
[Sol.20
1+
20
1+
15
1=
60
46= 6 ]
Q.4
[Sol. p =
sA
]cosv[dt
dm
=
secAcosAv
2
= v2 cos2 ]
Q.11
[Sol. = NiABI= NqABI = C
2
2=
I2
C 2
=I
C
NABQ = I I
C
Q = K ]
PART-C
Q.1
[Sol. di =T
dq=
2
dq=
2
dxx2
dB =
x2
di0
B =2
0
R
0x
dxx=
2
R0
]
Q.2
[Sol. i2 =2
Rr
=
41
5
= 1A
Pd = i2R2 = 4V
Pd. A = 2V
Q = 3 2 = 6 C ]
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Code-A Page # 2
Q.3
[Sol. A1v
1= A
2v
2
10 5 = 5 v2
v2= 10 m/s
pg
p1
+g2
v21
=pg
v2
+g2
v22
4
1
10
p
+ 20
25
= 4
5
10
102+ 20
100
41
10
p= 251.25 = 23.75
p1
= 2375 102 Pa ]
Q.4
[Sol. 1
=3
1200= 20
2
=27
1200=
3
20
1t = (2x + 1)
2
2t = (2m + 1)
2
1
3=
1m2
1n2
2n + 1 = 3
1t =
2
3 t =
80
6]
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Code-A Page # 1
PART-A
Q.2
[Sol.
P O P
O
P
O
O
O O
O O
P
O
O
P O P
O
P
O
O O
P:
::
:
O
pd present NO - bond6 POP linkages 6 POP linkages
P = sp3 hybridised P = sp3 hybridised
= 4 = 3 + 1 lp ]
Q.3
[Sol. During intramolecular cannizaro more reaction CHO is oxidised and less reactive in reduced.]
Q.4
[Sol. N2
+ 1/2 O2 N = N = O
100 = 100)]600BE()250950[( NN
200 = 1200BEN=N
600
BEN=N
= 400 kJmol1 ]
Q.6
[Sol. t1/2
=.avg
k
2ln=
)2(ln2
2ln 20 = 10 min
A B
1x x/2
A C
1x x/2
(1x) 60 =2
x 40 + 80
x = 0.5
Time in which reaction will be optically inactive = 10 min. ]
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Code-A Page # 2
Q.7
[Sol. LAH can rduce NHC||
O
intoCH2NHand C = O into CH2
While SHB can not reduce NHCR||
O
, it reduce C = O ,CHO and ClC||
O
into alcohol. ]
Q.8
[Sol. (A) Ecell
= 2/1H
]P[
]H[log
1
059.0
2
2HP ( Ecell )
(B) its a fact
(C) for spontaneous process H =ve and S = +ve(D) From structure
OSOOSOO
O
O
O
+6 +6
]
ONLY V-GROUP :
Paragraph for question nos. 9 to 11
[Sol.(9)NH Cl + K Cr O + conc. H SO4 2 2 7 2 4
Salt solid
NaOH
(E)
HCl
White fumes(A)
(B)Orange red fumes
NaOH
(C)
yellow solution
CH COOH3(CH COO) Pb3 2
(D)yellow ppt
gas
NaCl + NH + H O3 2
NH Cl4
CrO Cl2 2
Na CrO2 4
PbCrO4
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Code-A Page # 3
(10)
O
Cr
ClOCl
O
Cr
O
OO
d s hybridisation3
BaCl2 + CrO42
colourYellowBaCrO4
(11) NH4Cl + AgNO
3
pptWhiteAgCl
OHNH 23 [Ag(NH3)2]Cl 3
HNO AgCl
Z = [Ag(NH3)2]+Cl
C.N. = 2 , hybridisation = sp ]
ONLY XY-GROUP :
Paragraph for question nos. 9 to 11
(9) In polling process impurity of metal oxide i reduced3Cu2O + CH
4 6Cu + CO
2+ 2H
2O
(10) In Bessemerisation Cu2O is reduced by Cu
2S
2Cu2O + Cu
2S 6Cu + SO
2]
Paragraph for question nos. 12 to 14
[Sol. C=N C=N
C=N
OH OH2Ph Ph
Ph
H H
H
H+
(i) LAH(i) MeMgX
(ii) H /H O+
2
CHCl /OH3
(ii) H O2(A) (B)(D)
(C)
PhC NH
PhC N
H O2
PhCH NH2 2
PhCH NC2
PhCCH3
O
Note : PhCN czn be prepared. By heating
O||
NHCPh 2 into Ph CN in presence of P2O5
(Dehydrating agent)2NHCPh
||O
OH
OP
2
52
PhCN ]
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Code-A Page # 4
Q.15
[Sol.
Na/NaH/OHEE
22]
Q.16
[Sol. Resonating structure with complete octet is more stable. ]
PART-B
Q.1
[Sol. (A)Zn/OH)ii(
O)i(
2
3 O + O
H
(B)Zn/OH)ii(
O)i(
2
3 O
(C)
H
Zn/OH)ii(
O)i(
2
3
OCCH
|H
3 +
O
]
PART-C
Q.1
[Sol. Equal volume of AgNO3(0.2M) and KCN (1M) are mixed
[Ag+] = 0.1 M [CN ] = 0.5 MAg+ + 2CN Ag(CN)
2 ...(i)
Initial conc. 0.1 0.5 0
at eqm. conc. 106 0.3 0.1
k1
= 26)3.0(10
1.0
Similarly equal volume of Zn(NO3)2(0.2M) and KCN(1M) are mixed.
[Zn+2] = 0.1 M [CN] = 0.5 MZn+2 + 4CN Zn(CN)
42 ...(ii)
Initial conc. 0.1 0.5 0
at eqm.conc. 1012 0.1 0.1
k2
= 412)1.0(10
1.0
2[Ag(CN)4](aq) + Zn+2(aq) [Zn(CN)
4]2(aq) + 2Ag+(aq)
Kc= 2
1
2
k
k(By reaction (i) and (ii)
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Code-A Page # 5
=
46
2
124
)3.0(10
)1.0(
10)1.0(
1.0
= 810 ]
Q.2
[Sol. spin = 2.8 BMso unpaired electron = 2
Li2+ B
2/C
2+2 B
2 C
2N
2O
2 N
22/O
2
unpaired e 1 2 1 0 0 1 2
Ans. 4 ]
Q.3
[Sol.(a) Ksp
= [Mg2+][OH]2
[OH]2 = 1010
[OH] = 105
[H+] = 109
pH = 9
(b) HG = PdV + VdPHG = VdP
= 1 [7040]
= 30 bar hit = 3000 J = 3 kJ
(c) For zero order t1/2
a
(d) PbU 20682
23692
8 '' and 6 '' particule are emitted.]
Q.4
[Sol.O
OH
OH
HCC
OH
(Phenol salt with Na, NaOH)
(Carboxylic acid form saltwith Na, NaOH, NaHCO )3
(Alcohol form salt with Na)
(Form
saltw
ithNa)
]