12 06 2011 Xii Abcd Paper i Code A

7/30/2019 12 06 2011 Xii Abcd Paper i Code A

1/15

12th ABCD (Date: 12-06-2011) Review Test-2

PAPER-1

Code-A

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 A

Q.2 B

Q.3 B

Q.4 C

Q.5 D

Q.6 C

Q.7 D

Q.8 B

Q.9 A

Q.10 A

Q.11 D

Q.12 C

Q.13 A

Q.14 C,D

Q.15 A,C

Q.16 A,B

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 D

Q.2 B

Q.3 D

Q.4 C

Q.5 C

Q.6 C

Q.7 C

Q.8 A

Q.913

B

Q.10 B

Q.11 B

Q.12 A

Q.13 A,B,C,D

Q.14 B,D

Q.15 A,B,C,D

Q.16 A,B,C,D

Q.17 A,C,D

PART-C

Q.1 0005

Q.2 0003

Q.3 0001Q.4 0012

Q.5208

0002

Q.6 0001


2/15

PHYSICS

Code A Page # 1

PART-A

Q.1

[Sol. 2nd OT = 3 funda = 660 Hz ]

Q.2

[Sol. Phase change of ]

Q.4

[Sol. TE series : A more +ve than B and less +ve than CCAB : C+vely charged and A isvely charged. ]

Q.5

[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.6

[Sol.

Line

USEe

Line

Stable]

Q.8

[Sol. xmax

= 10 cm

dmin

= 0.9 m

d0I

= 0.9 2 = 1.8 m ]

Q.9

[Sol. VIV

M=(V

0V

M)

VI

= 2VM

V0

= 2A3

+ A2

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


3/15

PHYSICS

Code A 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.14

[Sol. =2

1g sin t2

t =sing

2=

g

h2

sin

1]

Q.16

[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.17

[Sol.q

a 3a

Q

q isve

2a

kQ2= 2a

kQ+ 2

)a3(

kq q = 9Q

q is +ve

2a

kQ2= 2a

kQ+ 2a9

kq q = 27Q ]

PART-C

Q.1

[Sol.

S

30202mm

S'

50h1


4/15

PHYSICS

Code A 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 ]


5/15

PHYSICS

Code A 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 ]


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 ]


7/15

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]


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. ]


9/15

MATHEMATICS

Code A Page # 1

PART-A

Q.1

[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 =2log77

a = 1 and b = log72 Infinite solutions. Ans.]

Q.2

[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.3

[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.]

Q.4

[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.]

Q.5

[Sol. l1 = xsin0x

xsinLim

ln l1 = 0xLim sin x ln(sin x) =

0xLim 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

=

xnxLim0xe

l

= e0 = 1

Hence, L = l1

+ l2

+ l3

= (1 + e2 + 1) = (e2 + 2). Ans.]


10/15

MATHEMATICS

Code A Page # 2

Q.6

[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.7

[Sol. We have y = sin (5x) = sin x f ' (x) = cos x

f ' () =1 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=

dx2x

11 = x + ln |x2| + 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

2

1= sec1(cosec x) = x

2

Now,xcos)x(h

xsine2eLim

x2)x(h2

2x

=

xcosx2

xsine2eLim

x2

x2

2

2x


11/15

MATHEMATICS

Code A 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. We have,

0xif,0

0xif,x

1sinx

)x(f ca

(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, a1 > 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.14

[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.]


12/15

MATHEMATICS

Code A Page # 4

Q.15

[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

A

tan2

2 tan

2

A=

2

1. Ans.]

Q.16

[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.17

[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 ]

PART-C

Q.1

[Sol. Using A.M. G.M. inc

1and

b

1,

a

9

3

1

c1b1a9

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.]


13/15

MATHEMATICS

Code A 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.]


14/15

MATHEMATICS

Code A 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) = 1

sin x and f '(0) = 1f " (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

dy

xd=

dy

d

dx

dy

1=

dx

d

dx

dy

1

dy

dx=

3

dx

dy

1

2

2

dx

yd

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.]


15/15

MATHEMATICS

Code A 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.]

Documents

12 06 2011 Xii Abcd Paper i Code A