12 06 2011 xii abcd paper i code b
Post on 14-Apr-2018
220 Views
Preview:
TRANSCRIPT
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
1/15
12th ABCD (Date: 12-06-2011) Review Test-2
PAPER-1
Code-B
ANSWER KEY
CHEMISTRY
SECTION-2
PART-A
Q.1 C
Q.2 B
Q.3 B
Q.4 B
Q.5 C
Q.6 D
Q.7 C
Q.8 A
Q.9 C
Q.10 A
Q.11 B
Q.12 C
Q.13 A,B,C
Q.14 A,B,C,D
Q.15 B,C,D
Q.16 A,B,C,D
Q.17 B,C
PART-C
Q.1 0295
Q.2 0003
Q.3 0003Q.4 0029
[Marks will be awarded for
0025 or 0026 or 0027 or 0028
or 0030 also]
Q.5 0003
Q.6 0008
PHYSICS
SECTION-1
PART-A
Q.1 D
Q.2 C
Q.3 D
Q.4 A
Q.5 B
Q.6 B
Q.7 C
Q.8 B
Q.9 A
Q.10 A
Q.11 D
Q.12 C
Q.13 A,B
Q.14 A,C
Q.15 A
Q.16 C,D
Q.17 A,C
PART-C
Q.1 0012
Q.2 0005
Q.3 0100Q.4 0625
Q.5 0005
Q.6 0001
MATHS
SECTION-3
PART-A
Q.1 C
Q.2 C
Q.3 C
Q.4 D
Q.5 B
Q.6 D
Q.7 C
Q.8 A
Q.9 B
Q.10 B
Q.11 B
Q.12 A
Q.13 A,B,C,D
Q.14 A,B,C,D
Q.15 A,C,D
Q.16 A,B,C,D
Q.17 B,D
PART-C
Q.1 0005
Q.2 0003
Q.3 0001Q.4 0012
Q.5 0002
Q.6 0001
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
2/15
PHYSICS
Code B Page # 1
PART-A
Q.1
[Sol. Before q, E
should be in ()ve direction
between q1
& q2, E
should be in (+)ve direction
after q1E
should be in ()ve direction. ]
Q.2
[Sol.LineUSE
e
Line
Stable]
Q.4
[Sol. xmax
= 10 cm
dmin
= 0.9 m
d0I
= 0.9 2 = 1.8 m ]
Q.5
[Sol. 2nd OT = 3 funda = 660 Hz ]
Q.6
[Sol. Phase change of ]
Q.8
[Sol. TE series : A more +ve than B and less +ve than C
CAB : C+vely charged and A isvely charged. ]
Q.9
[Sol. VIV
M=(V
0V
M)
VI= 2V
MV
0= 2A
3 + A
2 max when at mean position.
=
k=
32
30
(2 + 3) = 5 rad/s
A2
+ A3
= 10 cm
2A23A
3= 0 2A
2= 3A
3 A
2=
2
3A
3
2
5A
3= 10 cm A
3= 4 cm A
2= 6cm
VI= 2 4 5 + 6 5 = 70 cm/s ]
Q.10
[Sol. We need a converging lens to converge diverging rays and form interference pattern. ]
Q.12
[Sol.
(11) S' S'' = 2 1 102
2cos
)2013sin(= 2 103 0.9
180
= 104 m
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
3/15
PHYSICS
Code B Page # 2
0
I
h
h=
fu
f
=
3040
30
=3
hI= 3 104 m
(12) v =fu
uf
=
3040
3040
= 120 cm
D = 200120 = 80 cm
= d
D= 4
9
103
8.010500
m = 3
4mm ]
Q.13
[Sol. m = 4 =0x
f
x0
=4
25= 6.25 cm ,
4
25
=6.25 cm
u =31.25 cm, u =18.75 cm ]
Q.14
[Sol.q
a 3a
Q
q isve
2a
kQ2= 2a
kQ+ 2
)a3(
kq q = 9Q
q is +ve
2akQ2 = 2a
kQ + 2a9kq q = 27Q ]
Q.16
[Sol. =2
1g sin t2
t =sing
2=
g
h2
sin
1]
PART-C
Q.1
[Sol.
S
30202mm
S'
50h1
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
4/15
PHYSICS
Code B Page # 3
1h
1.0=
50
50 h
1= 1mm
1.0
hL 1=
20
80 L + h
1= 4 mm L = 3 mm
B =d
D= 3
7
102
1105
= 2.5 104 m
N =B
L= 4
3
105.2
103
=
25
300= 12 ]
Q.2
[Sol.
2.5 tan
3.9x sin 37
x sin 3737
1.4m
3.9m
(3.9 x sin 37) tan C
= C
sin C
=54
2.5 tan C
+ (3.9x sin 37) tan C
= x cos 37 2.5 3
4+
3
4
5
3x9.3
= x
5
4
512
8.12 =
12
4.6x = 5.3 ]
Q.3
[Sol. u =80
v
1+
80
1=
40
1
v
1=
40
1
80
1 v =
3
80
u2
=3
8040 =
3
200
2v
1+
200
3=
40
1
2v
1=
200
5
200
3=
200
2
v2= 100 cm ]
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
5/15
PHYSICS
Code B Page # 4
Q.4
[Sol. f =L2
2
T=
21
23104.6
1600
=8.0
1
61064
16 =
6.1
1 103 =
16
10000= 625 Hz ]
Q.5
[Sol. 2t = n =
t = 2
= 25.12
105009
= 2 107
m2
A =t
v= 7102
1
= 5 106 m2 = 5 km2 ]
Q.6
[Sol.
mg cos 60 = 2
2
R
kqcos 30
q =3k
mgR2
=
3109
101010329
26
= 106 C = 1C ]
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
6/15
CHEMISTRY
Code-B Page # 1
PART-A
Q.1
[Sol. From (ii) + 3 (iii)(i) ; H = (700) + 3(158)(180) =994 kJ/mol]
Q.2
[Sol. P4O
10+ 6 H
2O 4H
3PO
3]
Q.3
[Sol. ]
Q.4
[Sol. Let x moles of each A and B are present in gm of solution
x 50 + x 100 = 75 x = 0.5
Now, for final solution , 2 =B
m
xm
B= 0.25 kg = 250 gm
Mass of B added = 250100 x = 200 gm]
Q.5
[Sol. Gerade and ungerade are consider by overlapping area, in this MO nodel plane is zero, so it is gerade.]
Q.6
[Sol.
At anode :
Cathode : H+ + e H H2
Na+ + OH NaOH pH ]
Q.7
[Sol.
OH
OH
H
OH O
]
Q.8
[Sol. TeF6SN2 mechanism
CCl4
, NF3
, SF6 SN1, mechanism, so not hydrolys at room temperature ]
Q.9
[Sol.24
OClS
OH2 2
342OSH
+ 2HCl ]
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
7/15
CHEMISTRY
Code-B Page # 2
Paragraph for question nos. 10 to 12
[Sol. (i) T =S.M
q=
2.4)1080(
105.103
6
= 31.25 K
(ii) m =vap
H
q
= 3
6
102
105.10
= 5250 gm = 5.25 kg
(iii) C12
H22
O11
+ 12 O2 12CO
2+ 11 H
2O
Moles of C12
H12
O11
burnt =3500
105.10 3 = 3
Water formed = 3 11 18 = 594 gm ]
Q.14
[Sol.
N Cl2
Ph
H O
2Cl ONO
Ph
Carbocation can be attact from either side. ]
Q.15
[Sol. Al2(CH
3)6Carbon does not have l.p. so it has 3C2e bond
BeCl2(solid), Al
2Cl
6, Fe
2Cl
6Chlorine have l.p. so they have 3C4e ]
Q.17 PhCH3
PhCH2Cl
This conversion can be done by :
(A) SOCl2
(B*) SO2Cl
2hv (C*) , (D) (i) AgOH, (ii) Cl
2/CCl
4
[Sol. PhCH3
PhCH2Cl PhCH
3PhCl
2Cl ]
PART-C
Q.1
[Sol. q = 0, w = 0U = 0
n Cv,m
(T2T
1)n2a
12V
1
V
1
or , 10 (T2300) = 5 2 100
10
1
20
1
T2
= 295 K]
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
8/15
CHEMISTRY
Code-B Page # 3
Q.2
[Sol. (B2, N
2, C
2)
Homodiatomic species upto nitrogen show sp mixing due to less energy difference between 2s and 2p
orbital.]
Q.3
[Sol.
3 H
5 H 6 H
]
Q.4
[Sol. 2HI H2
+ I2
4.65 103 M 0 0
(4.35 1032x) M xM xM
= 6.29 104 M
= 3.392 103 M
K = 23
44
)1039.3(
)1029.6)(1029.6(
=29
1]
Q.5
[Sol. Three oxygen atom shared per tetrahedron.]
Q.6
[Sol. ]
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
9/15
MATHEMATICS
Code B Page # 1
PART-A
Q.1
[Sol. l1
=
xsin0x
xsinLim
ln l1
=0x
Lim sin x ln(sin x) =0x
Lim x ln(sin x) = 0 l1 = 1.
l2
=
x1
x
0x
xsineLim
=
1xsinex
1Lim
x
0xe
=
x
xsin
x
1eLim
x
0xe = e2
l3
=
x
0x x
1Lim
= xnxLim
0xel
= e0 = 1
Hence, L = l1
+ l2
+ l3
= (1 + e2 + 1) = (e2 + 2). Ans.]
Q.2
[Sol. Let P(x) = x2 + ax + b, and let r be a common root.
P(r) = 0 and )r(PPP = 0 i.e. )0(PP = 0, P(0) = b
P(b) = 0 b2 + 4ab + b = 0 b (a + b + 1) = 0 P(0) P(1) = 0. ]
Q.3
[Sol. We have y = sin (5x) = sin x f ' (x) = cos x
f ' () =1 Ans.]
Q.4
[Sol. 2a = 7b a = b = 0 if a and b are integersIn case a and b are not integers then
7log22 = 7b a = log27 and b = 1
or49log22 = 7b a = log
249 and b = 2
or 2a =2log
77 a = 1 and b = log
72
Infinite solutions. Ans.]
Q.5
[Sol. The given equation can be re-written as
(2x1)2 + 1 = sin2 y, which is possible only when x =2
1, sin2 y = 1
y =2
,2
(as x2 + y2 3 )
Thus, there are only two pairs (x, y) satisfying the given equation. They are
2,
2
1and
2,
2
1.]
Q.6
[Sol.C C C C C C C1 2 3 4 5 6 7
All four must sit consecutive i.e., in 4 ways
Hence total = 4 4! = 96. Ans.]
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
10/15
MATHEMATICS
Code B Page # 2
Q.7
[Sol. dx)x(f = ln (sec x) + ln (sec 2x) + ln(sec 4x) + ln (sin 8x) + C
= ln
x4cosx2cosxcos
x8sin+ C = ln
x4cosx2cosxcosxsin8
x8sinxsin8
= ln
x8sin
x8sinxsin8= ln(8 sin x) = ln 8 + ln(sin x) + C = ln(sin x) + C. Ans.]
Paragraph for question nos. 8 and 9
[Sol. We havea2 + b2 + c22a + 4b2c + 6 = 0
(a1)2 + (b + 2)2 + (c1)2 = 0 a = c = 1, b =2 f(x) = x22x + 1 = (x 1)2
Let g(x) = px2 + qx + r
As, g(0) = r = 2, g'(0) = q =3 and g"(0) = 2p = 2, p = 1Hence, g(x) = x23x + 2 = (x1) (x2)
(i) Clearly, f (1) + g (1) = 0 + 0 = 0.
(ii) dx)x(g)x(f
=
dx2x3x
)1x(2
2
=
dx2x
1x
= dx2x1
1 = x + ln |x
2| + C. Ans.]
Paragraph for question nos 10 to 12
[Sol. Area of ABC
We have, =2
1ah = 12 ah = 24 h =
a
24=
AsinR2
24
Asin62
24h
h = 2 cosec AA
B C
A
h
aSo, y = f(x) = 2 cosec x
(i) Clearly, least value of f(x) is 2. Ans.
(ii) We have, g(x) = f(sin1x) = 2 cosec (sin1x) =x
2,
g'(x) = 2x
2
8
25
16
25.2
5
4'g
. Ans.
(iii) We have, h(x) = sec1
xeccos2
21 = sec1(cosec x) = x
2
Now,xcos)x(h
xsine2eLim
x2)x(h2
2x
=
xcosx2
xsine2eLim
x2
x2
2
2x
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
11/15
MATHEMATICS
Code B Page # 3
Put, h2
x
, we get
hsinh
)hcos1(1e2eLim
hh2
0h
=
h
hsinh
hcos1
h
1e
Lim22
2h
0h
=
2
1
2
11 . Ans. ]
Q.13
[Sol. Clearly, f(x) = 2x , so a = 8, b = 2 + 8 = 10, c = 82 = 6.
Clearly, =2
1(8) (6) = 24.
r =s
=
12
24= 2
A
B C
10
6
8
Also, tan A = 3
4
=
2
Atan1
2
Atan2
2 tan 2
A
= 2
1
. Ans.]
Q.14
[Sol. f(1) =
2x3x
2axxLim
2
23
1xexists and finite. [Note : f(1+) = )x(fLim
1x (1 form) = eln c = c. ]
a = 3, f(1) = 3 c = 3, b = 1,2. Ans.]
Q.15[Sol.
(A) As )x(fLimx
= c and f(x) be a continuous, so there will be
atleast one y for which we will have more than one x .
x-axisSo, f is not injective. Ans.
(B) Clearly, f(x) =f(x) but range = (9, 9)
f(x) is an odd and into function.(C) Given that f (2) = 4 and f (2) = 1.
Also f is continuous on [2, 2]
So by using intermediate value theorem, there exist a number c such that | c | < 2 and f (c) = e
(D) Clearly, f (x) = 5 ]
Q.16
[Sol. We have,
0xif,0
0xif,x
1sinx
)x(f ca
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
12/15
MATHEMATICS
Code B Page # 4
(A) As f(0) = 0, so f(x) is continuous at x = 0, if
0)x(fLim0x
=a
0xxLim
cx
1sin a > 0. Ans.
(B) f '(0) =h
)0(f)h(fLim
0h
=
h
0h
1sinh
Limc
a
0h
=
c
1a
0h h
1sinhLim = 0,
provided, a
1 > 0 a > 1. Ans.
(C) We have
0x,0
0x,x
1sinxa
x
c
x
1cosx
)x('f c1a
1cc
a
Clearly, f '(x) is continuous at x = 0, it a > 1 + c. Ans.
(D) We have f "(0) =h
)0('f)h('fLim
0h
=h
0h
1sinha
h
c
h
1cosh
Limc
1a
1cc
a
0h
= 0, provided, a > c + 2. Ans.]
Q.17
[Sol. Number of solution of equationxcos1
xsin
= 1 in [, 2] is zero.
(A) xx0x
1eLim
= 1
(B) derivative of esin xsin x1 with respect to x, when x = 0 is 0.
(C) Number of points of non-derivability of function y = | cos x | in (0, 2
) is 2.
(D) Clearly,
dxxsin = 0. Ans.]
PART-C
Q.1
[Sol. Using A.M. G.M. inc
1and
b
1,
a
9
3
1 c
1b1
a9 3
1
c1
b1
a9
or c
1
b
1
a
9
3 3
1
abc9
1 3 3
1
abc
9
or
abc
9271
abc 243 S = abc]
min.= 243 = 5
log3S = log
3243 = 5.
Characteristic of log of S to base 3 is 5. Ans.]
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
13/15
MATHEMATICS
Code B Page # 5
Q.2
[Sol. We have, f (0) = (1ah)1/h = e1/h(ah) = ea
Hence, f (0) = f (0) ea = b
Again f (0+) =11h
1)hc(Lim
31
0h
. For existence of limit c = 1 (think!)
So, f(0+
) = 11h
1)h1(Lim
31
0h
=
2
h
3
h
Lim0h = 3
2
Hence ea = b =3
2 a = ln
3
2; b =
3
2; c = 1
Hence, (ea + 2b + c) =3
2+
3
4+ 1 = 3 Ans.]
Q.3
[Sol. Vertex is at x = b. Also f(x) is decreasing in (, b) and increasing in (b,).
x = b is the point of minima.We have, f(x) = x22bx + 1
Case I: Let b 0So, f(1)f(0) = 4
(22b)1 = 4O
x
y
1
Case-I : y = f(x) 12b = 4 b =2
3(which is < 0)
Case II: Let b 1So, f(0)f(1) = 4
1(22b) = 4
Ox
y
1 b
y
Case-II : y f(x)
2b = 5 b =2
5(which is > 1)
Case III: If 0 < b < 1
Clearly,max. [f(0), f(1)]f(b) = 4
Here, f(1) > f(0)
Hence, f(1)f(b) = 4
22b(1b2) = 4O
x
y
1b
Case-III : f(x)
b22b + 1 = 4 b1 = 2 or2 b = 3 or b =1 (Rejected)
Here f(0) > f(1)So, f(0)f(b) = 4
1(1b2) = 4
Ox
y
1
Case-III : y = f(x)
b = 2 or2 (Rejected)
Hence b
2
5,
2
3
Hence, sum =2
5
2
3
= 1. Ans.]
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
14/15
MATHEMATICS
Code B Page # 6
Q.4
[Sol. dx)x1(
x1234
4
Take out x2 from the Dr , it will come out as x3
dx
xx
1
x
1x
232
2
3
Put 2x
1x2 = t2 2 dx
x
1x
3
= 2 t dt
dt
t
dtt3 = t
1+ C = 14
Cx1
x
; as f (0) = 0 C
1= 0.
Now
dxx1
x
4 = 221 Cxsin
2
1
but g (0) = 0 C2
= 0
g (x) =21
xsin2
1
Hence
2
1g =
12
k
k = 12 Ans
Alternatively : I =
dx
x1
x2x12/34
44
=
dx
x1
x2
x1
12/34
4
2/14
This is of the form f (x) + x f ' (x)
hence I = x f (x) =4x1
x
+ C ]
Q.5
[Sol. f(x) = x + cos x + 2
f '(x) = 1sin x and f '(0) = 1
f " (x) =
cos x and f " (0) =
1
Now,
dx
dy
1
dy
dx g '(y) =
)x('f
1
f
g(3) = 0
f(0) = 3
When y = 3 then x = 0
g '(3) =)0('f
1= 1 ......(1)
Again 2
2
dyxd =
dy
d
dx
dy1 =
dxd
dx
dy1
dydx =
3
dx
dy1
2
2
dxyd
g "(y) = 2)x(f
)x("for g "(3) =
3)0(f
)0("f=
1
1
g "(3) = 1 ......(2)
Hence, g '(3) + g "(3) = 1 + 1 = 2. Ans.]
-
7/30/2019 12 06 2011 Xii Abcd Paper i Code b
15/15
MATHEMATICS
Code B Page # 7
Alternatively: Clearly, f is a bijective mapping.
Also, f(0) = 3
g'(3) =)0('f
1.......(1)
and g"(3) = 3)0('f
)0("f......(2)
As, f(x) = x + cos x + 2 f '(x) = 1sin x f '(0) = 1and f " (x) =
cos x f '(1) =1
From (1) and (2), we get g'(3) =1
1= 1 and g"(3) = 3
)1(
)1(= 1.
Hence, )3("g)3('g = 1 + 1 = 2. Ans.]
Q.6
[Sol.
0
01
)1x(f)t(f
)t(f)1x()1x(ftLim
22
1xt
1)t('f
)t('f)1x()1x(ft2Lim
2
1xt
2(x + 1) f(x + 1)(x + 1)2 f '(x + 1) = f '(x + 1) 2(x + 1) f(x + 1) = [1 + (x + 1)2] f ' (x + 1)
2)1x(1
)1x(2
)1x(f
)1x('f
Integrating, we get
cn)1x(1ln)1x(fn 2 ll
Put x =
10 = 0 + ln c c = 1
f(x + 1) = 1 + (x + 1)2
So, f(x) = 1 + x2
Hence,
1x
2n)x(fnLim
1x
ll=
)1x(
2n)x1(nLim
2
1x
ll= 1. Ans.]
top related