MCP-06-02-2011 Paper-1 11th (PQRS & JK) Code A

8/6/2019 MCP-06-02-2011 Paper-1 11th (PQRS & JK) Code A

1/16


2/16

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.]


3/16

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.]


5/16

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.]


6/16

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


7/16

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.]


8/16

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


9/16

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 ]


10/16

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 lone pairs = 9 Number of lone pairs = 5

(D) SO3 SO O

O

BF3 BF F

F


Number of pairs = 6 Number of lone pair = 9 ]


11/16

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 ]


12/16

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 ]


13/16

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 ]


14/16

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)


15/16

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 ]


16/16

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 ]

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MCP-06-02-2011 Paper-1 11th (PQRS & JK) Code A

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