rt solutions-22!01!2012 xiii vxy paper ii code a sol

7/30/2019 Rt Solutions-22!01!2012 XIII VXY Paper II Code a Sol

1/16

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


2/16

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.]


3/16

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)


4/16

MATHEMATICS

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


5/16

MATHEMATICS

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.]


6/16

MATHEMATICS

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(


7/16

MATHEMATICS

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.]


8/16

MATHEMATICS

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.]


9/16

MATHEMATICS

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.]


10/16

PHYSICS

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 ]


11/16

PHYSICS

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]


12/16

CHEMISTRY

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. ]


13/16

CHEMISTRY

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


14/16

CHEMISTRY

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 :


(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]


[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 ]


15/16

CHEMISTRY

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)


16/16

CHEMISTRY

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)

]

rt solutions-22!01!2012 xiii vxy paper ii code a sol

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