FULL TEST–I (P -II) (SOLUTION) Sample · PDF filePHYSICS object be l eal image i riod....

1. (8)

2. (4)

3. (6)

4. (7)

5. (3)

6. (1)

7. (2)

8. (3)

9. (A,

10. (C)

11. (C,

12. (A,

13. (A,

14. (A,

15. (A,

16. (A,

17. (B)

18. (C)

19. (B)

20. (C)

PHYSICS

B, D)

D)

B, C)

B, D)

D)

B, C, D)

C, D)

FULL T(SO

C

21. (6)

22. (4)

23. (8)

24. (3)

25. (2)

26. (0)

27. (6)

28. (1)

29. (A)

30. (A,

31. (C)

32. (A)

33. (A)

34. (A,

35. (C,

36. (A,

37. (A)

38. (D)

39. (C)

40. (A)

TEST–I (POLUTION)

ANSWERS

CHEMISTR

)

)

)

)

)

)

)

)

)

, B, C)

)

)

)

, C, D)

, D)

, C)

)

)

)

)

P-II) )

S

RY

M

41. (8

42. (2

43. (3

44. (0

45. (3

46. (0

47. (8

48. (3

49. (A

50. (A

51. (B

52. (A

53. (B

54. (A

55. (A

56. (A

57. (B

58. (B

59. (B

60. (A

MATHEMAT

8)

2)

3)

0)

3)

0)

8)

3)

A, B, C, D)

A, B, C, D)

B, C)

A, B, C)

B, C)

A, B, D)

A, C, D)

A, B, C)

B, C)

B)

B, C, D)

A, B)

TICS

1

Sample Paper

OASIS EDUCATION : B-11, 2nd Floor, Near Ganpati Honda, Old DLF Colony, Sector-14, Gurgaon-122001Toll Free : 1800-3000-9396, M: 8800880028, Website : www oasisgurgaon.com

1. (8) Let

The

On Now

gap

∴ P

pos

pos

⇒ 1

To a

0v =

∴ ρ =

2. (4)

neW

neW

neW

Now

=

=

=

1

2

3

T

T

T

ΔQ

Δ 2Q

ΔQ

the distanc

en 1100 l

−−

solving, wew, if the lensp must be on

Phase diffe

itions are

itions from

10 2√ cm =

achieve this

KA Am

= ω =

Require

0mv A K=

( )= −et 0 02P v

π= −et 0 0P v

=et 00.86 P v

4w,

⎫= ⎪⎪⎪⎪⎪= ⎬⎪⎪⎪⎪=⎪⎭

0 0

0 0

0 0

P vR

4P vR

2P vR

T

( )(→ =1 2 4.5

→ = −2 3 03P v

(→ = −3 1 1.5

ce of the len1 1l 23

=−

e get l (50=

s performs ne fourth of

erence betw

symmetrica

origin must

A sin4π

= ⇒

s velocity at

Km

ed impulse Km 8= kgm

( ) π− −0 0P v

(= −0 0P v 44

( ) (=0v 0.22

Thus,

→

→

→

Δ

Δ

Δ

1

2

3

U

U

U

) (+0 0P v 1.2

(+ = −0 0 3 P

) (−0 0 0P v P v

FULL T(SO

HINTS

ns from the

10 2)± √ cmSHM and r

f the time pe

ween the t

ally located

t be 4π .

⇒ A = 20 cm

t the mean p

m/s.

0 0P v4

)π 0 0P v4

; Pu

)0 0P v

[

[

[

→

→

→

= ×

= ×

= ×

2

3

1

3R1 T2

3R1 T2

3R1 T2

)( ) (=0 022 P v

)0 0P v

) (= −0v 2.5 P

TEST–I (POLUTION)

S & SOLUTPHYSICS

object be l

m real image ieriod.

two position

d about the

m

position

ut π = 3.14

]

]

]

−

−

−

2 1

3 2

1 3

T T

T T

T T

( )( 0 05.72 P v

)0 0P v

P-II) )

TIONS

l when a rea

s formed af

ns of real

e origin, ph

)

al image is

fter a fixed t

image mus

hase differe

formed on

time gap th

st be 2π .

ence of an

the screen

en this time

As the two

ny of these

2

.

e

o

e

Sample Paper


OA

Thu

η =

Thu3. (6) Fric Fric Now ⇒

4. (7)

y =

12 =

6 =

From

4R 5. (3)

v =

E =

Solv

=L

=τ

⇒ ×τ

n = 6. (1) d =φ

=φ

ASIS GurgaonGu

us efficiency

(( ) (

0 0

0

0.22 P v5.72 P v

us efficiency

ctional forcectional forcew, 18 – T –2a = 1 m/s2

x tan 1⎡θ −⎢⎣

R6 tan ⎛= θ⎜⎝

R12tan ⎛θ⎜⎝

m (i) and (ii

48 R 6− = −

3

2

0 .nZv ⇒

2

2

0 .nZE=

ving (i) and

π==

2nhmvr

πΔΔ

2. h

tn

7=

=× 27

141.210

3

)2( kxxdxπ=

32

32 xkt π

n, M-6, IInd Furgaon-12200

y η =+

netWvehe

))

=0

0

0.04v

y is 4%

e on 5 kg bloe on 10 kg b2 = 10 a andand T = 6 N

xR

⎤⎥⎦

6R− ⎞

⎟⎠

12R− ⎞

⎟⎠

) 1 R22 R

⎛= ⎜ −⎝6 , 3R

23

2

=nZ

⇒ nZ

(ii) n = 2, Z

, Δτ=ΔL

9 21

1071

− ××

×× −25 1010

2xt

FULL T(SO

loor, Aptech 01, Haryana, 0

www.

t

eat

ock = μN1 =block = μN2d T –1 = 5aN

612

− ⎞⎟− ⎠ , (4 R

42= ⇒ R

42

2

=nZ

Z = 4

πΔ=Δ

2. hnt

3101.22

−××

=270 15

TEST–I (POLUTION)

Building, Op0124-4253725.oasisgurgao

= 1 N = 2 N

a

... (

… (

)R 12 R− = −

= 14 m

34 114

1.2×=

P-II) )

pposite Netwo5, 0124-42547on.com

i)

(ii)

6−

… (i)

… (ii)

250−

5kg

a N1

50

1

ork Bulls, Old725, 96500867

T T 2

d DLF Colony725,

10kg 2

100

N2

a

y, Sector-14,

18N

2

3

dtdφ

2E π

∫ τd

At t7. (2)

E =

E =

dτ =

τ =

τ =

torque dτ = μ

τ =

3Kq

t =

8. (3) For NA = NB = Con

have

mg

mgc

⇒

34 3txkπ

=

34 3ktxx π

=π

∫=τR

RktQ

023

4

t = 15sec, τ

x dB2 dt

= 23Kxt

2=

23Kxt 22

= ×

r23

20

3Kt q x dr ∫

223Kq.t .r

4

due to frictiodmgxμ

r

20

qm2 g xr

μ ∫2 2q.t r 2

4 3= μ

8 mg9Kqrμ

= 2

translational= ma = mg nsider the roe L2

cos30° + m

cos30° + ma

asin30°= gc

3

; 32 ktxE =

dxx4 ⇒

= 1 N-m

22 xdx q.x

rππ

dx

on force

2 2dx m3

= μ

mgrμ

2 seconds.

l motion of ro

tational equi

ma L2

sin 30

sin 30° – 2m

cos30° ⇒ a =

FULL T(SO

2x ; d32

⎜⎝⎛=τ

τ =34

2R

RktQ

…

mgr …

od

ilibrium of ro

° – NALsin30

masin 30° = 0

= g 3 m/s

TEST–I (POLUTION)

RQktx

32

22

π⎟⎠⎞

5

5R

…(i)

(ii)

. . . . . .

od taking torq

0° = 0

0

P-II) )

xxdx)2( π

x

(i) . (ii) que about p

dx

oint B, we

4

Sample Paper


Sec Sinc

forc from

tan

9. (A,

n(

If twapos

∴ 10. (C) Hor

B ini.e.

tA =

2usg

2(1

11. (C, Sinc

8 + ⇒

S

12. (A,

cond method ce rod is in e

ce and mg mum vector trian

30° = ga

=

B, D) (n 1) 6

2−

= ⇒

the initial savelengths ssible if init

(A), (B

rizontal comn horizontal

tB θ

=sin 2hg g

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

2

2 30)2

10

D) ce change i

6 × 2 = Δt × t 10secΔ =

8 2

v

20

⎧⎪ +⎪= ⎨⎪⎪⎩

1 8 82

= × × +

B, C)

: equilibrium wust pass thro

ngle OEF 1 a 33

⇒ =

n 4⇒ =

state were shorter thaial state we

B) and (D)

mponent of vdirection. T

⎛ ⎞=⎜⎝ ⎠

2B

1h gt2

= 15 m

in velocity is

× 2 c

t2(t 8)20

0 2t

−

−

1 6 262

+ × × +

FULL T(SO

with respect ough point C

3g

n = 3, in an λ0 wou

ere n = 2

velocity of AThey will co

⎞⎟⎠

or h

s zero, so a

1 30 22

+ × ×

TEST–I (POLUTION)

to car, so th.

the emissuld have o

A is 10 cos ollide at C if

h = 2 22u sin

g

area under v

0 32 84= +

P-II) )

he resultant

ion spectruoccurred. T

60° or 5 m/f time of flig

θ2

velocity time

300 416+ =

of pseudo

um, no This is

/s which is ght of both

e graph mu

6m

equal to thethe particle

st be zero

n=43

2

1

e velocity oes are equa

5

4

of al

Sample Paper


No.

ν ='

Thewav

=v ''

Bea

13. (A,

=v14. (A,

R

kqA

B

A

qq

'

'

B

A

qq

15. (A,

S1

S1

S1 a

S1 a16. (A, C, ( αKE

Hardeand ch

17. (B) 2μ =

v =

of wave pu

=distance tr

wavelee stationary ve of freque

⎛⎜⎝= =

⎛− ⎜⎝

vv '

V1c

at frequency

B, D)

ω −2 2A xD)

RkqBA 22

=+

21

= wh

1−= wh

B, C, D)

→ closed ,

→ opened,

and S2 bot

and S2 bot, D) ) ( )β =+ ELE

er X – rays hharacteristic

14= μ

Tμ

v⇒

ulse encoun

=ravelled c

engthobserver m

ency v '' em⎛ ⎞+ ⎟⎝ ⎠ =

⎞− ⎟⎠

V1v(c

V (1c

y − =⎛⎜⎝

vv '' v

symbols h

V2 , R

kqA

2

en switch c

en switch c

S2 → open

S2 → clos

th open , =i

th closed, i =

( ) (β + MEKE

ave higher frX-rays differ

12

vv2

= ∴

FULL T(SO

ntered by th+ ⎛= ⎜λ λ ⎝

c V c 1

meets the freitted by the

+−

(c V)c V)

⎛ ⎞+⎜ ⎟⎝ ⎠ −

⎛ ⎞− ⎟⎝ ⎠

Vv 1c v

V1c

have usual

VR

kqB

23

2=+

closed. Cha

closed poten

ed , 20=i

sed , 3

20=i

61020

=−

=

= 20 A

) ( )γα = KEM

requency thar only in the m

Kvω

=

⇒

TEST–I (POLUTION)

e moving p⎞+ ⎟⎠

Vc

= v

equency v oe moving pla

wavelength

v = v

l meaning

V

rge on B is

ntial of A =

410− = 2.5

0 A

35 A

an the softermethod of cr

2 1K 2K=

P-II) )

lane per un⎛ ⎞+⎜ ⎟⎝ ⎠

V1c

of the incideane as

h λ = =c''v ''

⎡ ⎤+⎢ ⎥−⎢ ⎥

⎢ ⎥−⎣ ⎦

V1c 1V1c

–qA

V/2 and pot

A

r X-ray. Contreation.

1

nit time is giv

ent wave an

−⎛ ⎞⎜ ⎟+⎝ ⎠

c c Vv c V

=−

2vVc V

tential differ

tinuous

ven by

nd receives

rence also V

the reflects

V/2

6

s

Sample Paper


rA =

tA =

18. (C)

2μ =

v =

rA =

tA =

19. (B) Both

Cha

20. (C)

Cha

21. (6) Let

n =

Give

and

Hea ∴ 2 Now

22. (4)

H

E

2 1

2 1

v vv v

⎛ ⎞−= ⎜ ⎟+⎝ ⎠

2

2 1

2v Av v

⎛ ⎞= ⎜ ⎟+⎝ ⎠

14= μ

Tμ

v⇒

2 1

2 1

v vv v

⎛ ⎞−= ⎜ ⎟+⎝ ⎠

2

2 1

2v Av v

⎛ ⎞= ⎜ ⎟+⎝ ⎠

h the capacit

arge on plate

arge on plat

the n moles

= PVRT

= 1.6

en that, Cp

dCv = 5 R2

−

at supplied a2.49 × 104 =w, using ide

P = nRTV

=

N H

N H Eclipsed (cis)

iA

iA

12

vv2

= ∴

iA

iA

tor are in par

(2) initially =

te (2) finally

s of gas are66 10 0.0

8.3 300× ×

×

p = 5 R2

3R R2

− = =

at constant = 5.33 × 2.4eal gas equa

5.33 8.30.008×

=

H

,

H

NH Stagge

FULL T(SO

Kvω

= ⇒

rallel

= Vd

A02ε

y = d

A⎜⎜⎝

⎛+

ε2/3

0

C

e present 083 5.33=

3 8.3 122

× =

volume (qv

45 × ΔT ∴ation again

675 3.583

×=

NH

N H

ered (trans)

TEST–I (POLUTION)

2 1K 2K=

Vd

A⎟⎟⎠

⎞ε+

2/0 =

CHEMISTR

2.45 J/mol-

v) = n × Cv ×∴ ΔT = 37

658 10 N/m×

,

NHSemi-e

H

P-II) )

VdA

38 0ε

RY

-K

× ΔT) 75 K ∴ Tf

2m

NH

H

eclipsed

,

= 300 + 37

NHGauch

HH

75 = 675 K

N H

he

7

Sample Paper


23. (8) 23I (

GΔ °

GΔ °

log

1030

(OH pOH pH 24. (3) Let

pheof mare

Hen

⇒

25. (2) tx% o

26. (0) Stru

DegThehydTheRDB

–(s) 6OH+

° = 5 × – 50

° = –172.5

K = 30

0 = 5

–

10 1[OH

− ×

– –6H ) 10= H = 6 = 14 – 6 = 8

the numbenolphthaleimethyl orang

same. HenA × 0.05B × 0.05

nce, A 1B 3

=

B 3A

=

of a first ord

ucture of Cu

gree of unse saturated rogen atom

e degree of BE (Ring D

–5I (aq) +

0 + (–123.5

kJ/mole =

–1

6

10]

8

ber of miln converts ge converts

nce 5 = a 5 = 2a + a =

der reaction

ubane is

saturation:

hydrocarbms, the degr

unsaturatioDouble Bon

FULL T(SO

3IO (aq) 3−+ +

5) + 3 × (–2325– 3003

× ×

limoles of Na2CO3 to

s it to H2CO

= 3a

n is constant

on with 8 ree of unsaton for Cubannd Equivale

TEST–I (POLUTION)

23H O(l)

33)– 0 – 6 ×

–32.3 10× ×

the salt NaHCO3. T

O3. Number

t.

carbon atoturation is (1ne is 5. ents):

P-II) )

× (–150)= –

logK

is ‘a’ TitraTitration of Nof moles of

oms would 18-8)/2 = 5.

172.5 kJ/m

ation with NaHCO3 wif Na2CO3 a

be C8H18. .

mol

HCl in pith HCl in thnd NaHCO

Since Cub

presence ohe presence3 in the salt

bane has 8

8

of e t,

8

Sample Paper


Starare

AccreprThe= 0.

27. (6)

10.2

Solv28. (1) Mol

Sol

Ksp 29. (A) The

manthan

the

give

min

rting with thremoved to

cording to thresentation

erefore, the .

2 + 17 = 13

ving the abo

ar conducta4 1ubility= ×

= (10-4)2 = 1

e gas pressnometer rean the atmos

sample. Be

en pressure

eral oil.

Pmanometer =

Pbulb = Pouts

he 2D repreo displace a

he diagram. The numbdifference i

.6 Z2 2

12

⎛ −⎜⎝

ove two equ

ance = (spe510 1000

400

− ×

10-8.

sure in the ading. The mspheric pre

ecause mer

e will hold

237 0.8213.6×

side – Pmonome

FULL T(SO

esentation oa ring.

m above, theber of faces in RDBE va

2

1n

⎞⎟⎠ and 4.

uation we g

ecific condu

= 410−

bulb equalsmanometer

essure beca

rcury is mo

a column o

22 = 14.3 m

eter = 746 –

TEST–I (POLUTION)

of cubane,

e RDBE of on a 3D str

alue and De

.25 + 5.95 =

et, n = 6

ctance × 10

s the differer indicates tause the liq

re dense th

of mercury

mm Hg

14.3 ≈ 731.

P-II) )

it comes do

cubane is ructure doe

egree of Uns

= 13.6 Z2 3

⎛⎜⎝

000)/(solubi

ence betwehat the presuid level is

han mineral

only 116.5

.7 mm Hg

own to coun

5 and not s not dictatesaturation v

2 2

1 13 n

⎞− ⎟⎠

lity)

een the outssure of the

s higher on

l oil by a fa

times the

nting how m

6 accordine the RDBEvalue of Cub

tside pressue gas in the

the side co

actor of 130.8

height of a

many bonds

g to the 3DE value. bane = 5 - 5

ure and thebulb is less

onnected to.622

=16.5. A

a column o

9

s

D

5

e s o

A

of

Sample Paper


30. (A, CH

The31. (C)

[Co1

3Ag

The

[Co1

2Ag

So, chloPen

32. (A) The

C

H

33. (A)

nRl

Cp =

⇒ 2

5ln

B, C)

3 – C – CH –CH3 CH2

CH3

e intermedia

3 5 2o(NH ) H O]: 5 : 1

1g 3Cl+ −+ ⎯

e following c

3 5o(NH ) Cl]C1 : 5 : 1

g 2Cl+ −+ ⎯⎯

pink (X) is oride andntaamminec

e only optica

CH3—CH—

N

CH3

3C +

1p

2

pln nC lp

+

= 5 cal R =

2

12ln 1 5p

+ ×

21

= +2 ln

– CH – OH–CH3

CH3

ate formed g

3]Cl (pink)

3AgCl⎯⎯→

complex is o

2Cl (purple)

2AgCl⎯⎯→ ↓

3 5[Co(NH ) (

purple chloridocoba

ally active m

—CH2CH3

CH3

⎯

2

1

Tn 0T

= ∵

= 2 cal 25ln1

= 0

p2

FULL T(SO

CH3

CH

give the alk

aq. [Co(N

(white)↓

obtained onaq. [Co(N

(white)↓

2 3(OH )]Cl a(Y) is

alt(III) chlor

molecule is

AgOH, Δ⎯⎯⎯⎯→

∵ ΔS = 0

TEST–I (POLUTION)

– C – CH –CH3CH2–

CH3

H3–C–CH=CCH3

CH3 C

ene in optio

33 5 2H ) H O] +

n heating an2

3 5NH ) Cl] + +

and IUPAC

3 5[Co(NH )

ride.

3CH — CH|

NH

CH2 = CHCH

P-II) )

CH

–CH3

CH3

⊕ C

C

–H⊕

C–CH3–CH3

H3

CH3–

on (B) or (C

3Cl−+

nd losing 1 m12Cl−

name wou

5 2Cl]Cl and

2

2

— CH — C

H2CH3

CH3 – C – CHCH3

CH3⊕

CH3–C–CH–CH3

CH⊕

–H⊕

–C=CH–CHCH3

CH3 CH) after rearr

mole of 2H O

ld be Pentad IUPAC

3CH

H – CH2–CH

–CH–CH2–C3

CH3

CH3

H–CH2–CH3 H3

rangement.

O.

aammineaqname

1

H3

CH3

uacobalt(IIIwould be

10

) e

Sample Paper


5 × ln p34. (A, 3H C

H

3H C

35. (C, D

(Ag

Ag+

(Ag

Usin º

1GΔ

nF−

36. (A, The

atomthe equbetwequ

to 937. (A) mol

Ax′ =

xA =

PT =38. (D) At t

orig PT =

0.7 = 2 lnp2 = 1.75 C, D)

3

C — CH — C|

CH

3CCH H

CH3

C

3

C — CH —B|

CH)

( )s Ag⎯⎯→

( ) (aq Cl+ −+

( ) (s Cl aq−+

ng Hess’s laº º1 2G G+ Δ = Δ

º1FE RTlnK+

º º1 3E E≠

C) e hydrogen ms. Hydrog

π bond invuatorial plaween the

uatorial fluor

97.5o.

fraction of 0A A

o 0B A

P xP (P

=+ −

= 0.36, xB == 300 × 0.36

this point thginal vapour= 300 × 0.25

n p2

Jon

2CH OH ⎯⎯

C H CH

CHNaCNBr H⎯⎯⎯→

( )g aq e+ −+

( )aq A

)q AgC⎯⎯→

aw, º3G GΔ

ºsp 3K nFE= −

º3

atoms are gen atoms lvolving a p-ne of the

π electronsrine atoms

A in liquid s

A0B AP )x−

give

0.64 6 + 500 × 0

he mol fracr phase i.e. 5 + 500 × 0

FULL T(SO

nes reagent3H⎯⎯⎯⎯→

O3HCH3

H3

H2O

3

3

H C — CH —|

CH

( )AgCl s

( )Cl s e−+

1 2G G+ Δ = Δ

1nFE−

in a verticlie in the C-orbital on t

molecule. s and theis dramatic;

state

en Ax′ = 0.2

.64 = 428 to

ction of A inxA = 0.25 .75 = 450 to

TEST–I (POLUTION)

3

3

C — CH —|

CH

3

O 2 H3C —

3H O— CN+

Δ⎯⎯⎯

º1GΔ =

º2G RTΔ = −

º3G nFΔ = −

3GΔ

30 nFE+ = −

1 3E E=

cal plane wiSF2 axial pthe carbon

And the e electron ; The eqF −

25

orr

n liquid pha

orr

P-II) )

— COOH

— CH —CO

CH3

3H C — C

C

⎯→

º1 1nFE , G− Δ

2sp

1T ln , GK

Δ

º3 3FE , GΔ = −

3

ith the axiaplane. We katom must resulting pair bond

eqF angle d

ase will be

OOH

3

CH — COOH|

CH

1nFE= −

2 0=

3nFE−

al fluorine know that lie in the repulsion ding the ecreases

almost eq

H

C SH

H

F

F

ual to that

1

F

FF

F

170o

97.5o

of A in the

11

e

Sample Paper


OA

39. (C)

A

B

nn /

1n Mx

1X

⇒

⇒

⇒

40. (A)

Initia

In th mixM

⇒

The

A'A

rr

=

A'A

rr

=

'Ar =

o

o

'A'B

rr

'Br =

Afte "

AP =

"BP =

"A"B

nn

ASIS GurgaonGu

A

B

/ t P M/ T P M

=

1 2 2

1 2

M n Mx x

+=

+

1 1M (1 X+ −

1 22X , X5

=

A

B

x 2 7x 3 12

=

A B1X , X3

=

ally A10r =

he mixture

x A AX M= +

A472 X5

= ×

472 56X5

=

472 = 280 X

A112X280

= =

e mixture

A 1'A 2

P ' A 'P A

=

21 X3X2 X2

= ××

A3r 3 20= = ×'A B'

AB

P MMP

= =

120 torr/s er 10 sec = 2000 – 60

= 3000 – 12

79

=

n, M-6, IInd Furgaon-12200

B

A

MM

mixM

2 mix)M M= 35

=

2 128 2

=

B23

=

00 9005− =

A B(1 X )M−

128 (1 X× + −

AX 72+

XA + 360

B2 3, X5 5

= =

1X 3

=

0 =60 torr/s

2 723 128

= ×

0 × 10 = 140

20 × 10 = 18

FULL T(SO

loor, Aptech 01, Haryana, 0

www.

20 torr/s

AX )72

12

=

00 torr/s

800 torr/s

TEST–I (POLUTION)

Building, Op0124-4253725.oasisgurgao

P-II) )

pposite Netwo5, 0124-42547on.com

ork Bulls, Old725, 96500867

d DLF Colony725,

y, Sector-14, 112

Sample Paper

41. (8)

cos

= −

n

k 2=

−∑

nlim

→∞−

K =

K900

42. (2)

Volu

OA⎡⎣

Volu

k = 43. (3)

kI(k

kI(k

kI(k

3

π

k =44. (0) We 2x ⋅ If x Now ⇒ The

1 1 k(ks−

⎛ +⎜⎜⎝

1 1cosk

− ⎛ ⎞ +⎜ ⎟⎝ ⎠

1 1cosk

− ⎛ ⎞− ⎜ ⎟⎝ ⎠

1 1cos2

− ⎛ ⎞− ⎜ ⎟⎝ ⎠

7200

80

=

ume of tetra

A OB OC⎤⎦ =

ume 16

= =

2

20

k lnk)x kx

∞

=+∫

20

ln ktk)t t

∞

=+ +∫

0

k) I(1)1

∞

− =+∫

2(lnk)3

= ×

3/2e have

x 1 |x 3|2 2+ − ++3 0− ≥ i.e.,

w, x 3 0− <x 3<

en 2 x 1x 2 +⋅ +

1)(k 1)(kk(k 1)+ −

+

1 1cosk 1

− ⎛⎜ +⎝

1 1cosk

− ⎛+ ⎜ +⎝

1 1cosn

− ⎛+ ⎜ +⎝

ahedron 16

=

1 1 00 2 11 0 1

= −

k12

2

x dxx k+

let

dt1+

2

dt (lnk)t t+ +

26π

×

2 2 |x 3x 2+ −= ⋅x 3≥ (impo

x 5 22 x− ++ = ⋅

FULL T(SO

MA

k 2) ⎞+⎟⎟⎠

co=

1⎞⎟⎠

1⎞⎟+ ⎠

cos−= −

cos1

−⎞ = −⎟+ ⎠

OA OB O6

⎡⎣

1(2) 1(= + −

t x = kt

0

) (lnk)

t

∞

=⎛⎜⎝

∫

| 4 x 12+ −+ ossible sinc

x 7 x 12 2− + −+

TEST–I (POLUTION)

ATHEMATI

1 1sk(k 1)

− ⎛⎜⎜ +⎝

1 1 cos2

− ⎛ ⎞ +⎜ ⎟⎝ ⎠

1 1 cos2

− −⎛ ⎞ +⎜ ⎟⎝ ⎠

OC⎤⎦

1) 0 1− + =

2

dt

1 32 2

⎛⎞+ + ⎜⎟ ⎜⎠ ⎝

ce x is nega

P-II) )

CS

211

) k+ −

1 1n 1

− ⎛ ⎞⎜ ⎟+⎝ ⎠

1(0)−

2 3π π

= −

23

2⎞⎟⎟⎠

tive)

2

11(k 1)

⎞−

+ ⎠

1203 6 Kπ π

= =

⎞⎟⎟⎠

00K

π

1

13

Sample Paper


2x

⇒ (

∴ x

∵ x

Hen45. (3)

x

0

t f∫

Usin

x

0

(x∫ Diffe

ddx

⎡⎢⎣

⇒

⇒

Put f(0) Diffe f '(x Put f '(0 Diffe f ''(x Put f ''(0 Equ f '(x

dydx

⇒

⇒

x 7 2x 62 (2− + −⋅

2x 6 2(2 1)(x− −

1x2

= ± and

x < 3 and x i1x2

≠ ± and

nce, numbe

x

0

f(x t)dt− = ∫

ng property

x t)f(t)dt− =

erentiate box x

0 0

x f(t)dt⎡

−⎢⎣

∫ ∫x

0

f(t)dt x⎡

+⎢⎣∫

f(x) sin x=

x = 0 ) 0 1 0= − + =

erentiable bx) cos x s= +

x = 0 0) 1 0 1= + −

erentiate x) sinx= − +

x = 0 0) 1= uation from x) sinx c= −

sin x cos= −

dy y sindx

− =

xy e e−⋅ = ∫

x 51) 2 (− +− =2 4 1) 0⋅ − =

x = 3

is negative

d x 3≠

r of negativ

x

0

f(t)dt sin+∫

y b

a a

f(x)dx =∫ ∫x

0

f(t)dt sin+∫oth sides w.x

0

t f(t)dtd

⎤=⎥

⎦∫

x f(x) x f(x−

x

0

cos x f− + ∫

1= − both sides sinx f(x)+

0=

cos x f '(x+ +

(i) cos x f(x)+

s x y+

n x cos x−

xe (sin x co− −

FULL T(SO

2x 6(2 1)− −

integer, the

ve integral s

x cos x+ −

b

a

f(a b x+ −∫

n x cos x+ −

.r.t ‘x’ x

0

d f(t)dtdx

⎡+⎢

⎣∫

) f(x) c⎤

= +⎥⎦

f(t)dt

x)

os x)dx e= ∫

TEST–I (POLUTION)

en

solutions is z

1

)dx

1−

sin x cos+ +

cos x sin x−

te (sin t cot+

P-II) )

zero.

s x 1⎤

− ⎥⎦

…(i)

t t)dt

114

Sample Paper


⇒

Put 1− ×

y =

0

f(xπ

∫46. (0)

tan

11 t

+−

⇒

⇒

⇒

⇒

⇒

Sum

47. (8)

No.

48. (3)

cos

⇒

⇒

⇒

So,

,∴

49. (A, 105 It m It m ⇒

xy e e− −⋅ = −

x = 0 1 1 0× = − × +

x(e sin x− +

0

x)dx eπ⎡

= − ⎢⎣∫

x 94π⎛ ⎞+ =⎜ ⎟

⎝ ⎠

tan x 19tan x 1

−=

+2(1 tan x)+

21 tan x+ +28 tan x −22 tan x −

(2 tan x 1−

1tan x2

= o

1 1x tan2

−=

m 1 1tan2

−= +

of ways 5=

2xs602+

° =

25x x x− =23x 15x− +

2x 5x 6− +

x = 3 5⇒

Ratio 32

=

B, C, D) 5 = 3 × 5 × 7

means N hasmeans powe

Total numb

x sin x c− +

c c+ ⇒ = −

x) f(x)=

x

0

e dx sin xπ

+ ∫

tan x4π⎛ ⎞−⎜ ⎟

⎝ ⎠

tan xtan x

−+

2 9(1 tan= −

2 tan x 9+ =

20 tan x 8+

5 tan x 2+ =

1)(tan x 2)−

or 2

or 1tan 2−

1tan 2−+ c=

1 15! 11! 2

⎛ − +⎜⎝

2(5 x) 7x(5 x)

− −−

2 2x 25 x+ +

18 0+ = 6 0= x⇒ =

5 x 2− = &

7 and 1000 s at least twrs of 2 or 5

ber of odd fa

FULL T(SO

1−

xdx⎤⎥⎦

(( e= −

⎞⎟⎠

2n x) 29(1 tan x+ −

0=

0= 0=

1cot 2 tan− +

1 1! 3! 4!

− + −

10x 7− −

3, 2=

x = 2 5⇒

= 23 × 53. wo prime fac

in N is eithactors is eit

TEST–I (POLUTION)

) (e 1 cπ − + −

2 tan x)−

1 2− co2π

= =

15!

⎞− ⎟⎠

5! 5= −

5 x 3− =

ctor 2 or 5 eer 5 – 1, 7 –her 7 × 3 –

P-II) )

) )0cos x

π= −

1ot (0)−

5! 5!5!2! 3!

+ − +

each having– 1, 3 × 5 –1, 3 × 5 –1

e 1 2π + + =

5! 5!4! 5!

+ − 0=

g at least 3 p– 1, or 3 × 7 , 7 – 1 or 5

3 eπ−

0 60 20+ − +

powers. – 1 – 1

1

5 1+ − = 44

15

Sample Paper


⇒ 50. (A, | |α =

|− β

Bec

Nowgrea

Now

Mag ⇒

51. (B, z =

Now

21 +

Now

(1, −

(1, −

52. (A, (x −

Let

(uf

(f vv

(fv

⎛⎜⎜⎝

Let

⇒ f

( )f 1

λ +

(f x

53. (B, f(x)

F; (x

Total numbB, C, D) = sum of ro|β = product

cause | a | |<

w, | | bα < −

ater than on

w, |a|xlogb

⎛ ⎞⎜ ⎟⎝ ⎠

gnitude of xone root lie

C) 0 is one of

w the triang2 2 21 k 2+ + =

w the directi

)1, 0− is ()1,0− is (−

B, C)

) ( )y f x y− +

x y u; x− =

) ( )v vf u− =

) ( )v f uv

u− =

)v fv

v⎞ ⎛

− =⎟ ⎜⎟ ⎜⎠ ⎝

( )f xx

x− = λ

( ) (f x x= λ +

) 2=

1 2= ⇒ λ

) 2x x= +

C) p) sin x.co=

p 1x) psin −=

ber of odd fa

oots b a= +

t of root = ab | so a is n

1 ⇒ a b+

ne and mag2

1 0⎞ − =⎟⎠

⇒

x is greater tes in ( ,a)−∞

the plane p

le formed is2 ⇒ k = ±

ion ratio of

)2, 0, 1 and

)2, 0, 1 .

( ) (x y f− +

x y v+ =

( )2uv v u= −

v u−

( )f uu

u⎞

− ⎟⎟⎠=co

λ

)2x+

1λ =

qos x

x cos x co x

FULL T(SO

actors is eit

a ab negative an

b b 1 a< − = <

gnitude of b 2x | a |

b⎛ ⎞ =⎜ ⎟⎝ ⎠

than 'a ' as and other

perpendicula

s take one l

2

plane havin

d the direct

) (x y 2y− =

onstant

q px qsin x−

TEST–I (POLUTION)

her 20, 14 ,

d b is posit

1< − . Becais greater t

⇒ x b= ±

s well as greroot lies in

ar to x y+ =

et on z-axis

ng point (0,

tion ratio of

((x y x y− +

q 1cos x sin−

P-II) )

, 6 or 4

ive.

ause a is nthan 1 | |+ α

| a |

eater than '(b, )∞

0, x y 0= − =

s i.e., (0, 0,

) (0, 2 , 1, 1

f plane havi

))

x

negative soor say grea

b '

0, x 1=

)k , then

)1, 0 and

ing point (0

o magnitudater than 2.

)0, 0, 2 ,−

1

e of 'a ' is

( )1, 1, 0 and

16

s

d

Sample Paper


si=

for m

⇒

Whi ⇒ ⇒ ⇒

If p

⇒ a

⇒

⇒

54. (A, y =

dydx

from

dydx

⎛⎜⎝

Putt

Elim

dydx

Find

dydx

⎛⎜⎝

55. (A, Let

af2

⎛⎜⎝

bf3

⎛⎜⎝

Sinc

⇒ ∴

p 1 q 1in x cos− −

maxima and

tan x p=

ich will haveIt is unique It is local m(C) is true a

= q then f(

absolute ma

(B) is true (A) is clearlB, D) c sin x

c cos x=

m (2) 2

2y c cosx

⎞ =⎟⎠

ting ycsin

=

minating c fr

y cot x=

ding the val2y sec x

x⎞ ⎛−⎟ ⎜⎠ ⎝

C, D) f(x) (2x= −

a 3a b2 2

⎞ ⎛= −⎟ ⎜⎠ ⎝

b 4b c3 3

⎞ ⎛= −⎟ ⎜⎠ ⎝

ce a bf f2 3

⎛ ⎞ ⎛⎜ ⎟ ⎜⎝ ⎠ ⎝

both roots aequation is

1 2x(pcos x −

d minima f

p / q

e exactly onmaximum

maximum asand (D) is fa

x) (sin x co=

aximum =2

y false

2s x

xfrom (1),

rom (1) and

ue of c from2

2dyx ydx

⎞ + =⎟⎠

a)(3x b)− − +

b (2a c)⎞ − >⎟⎠

2bc a3

⎞ ⎛ ⎞− <⎟⎜ ⎟⎠ ⎝ ⎠

b 03

⎞ < ⇒⎟⎠

are real quadratic i

FULL T(SO

2qsin x)−

'(x) 0=

ne solution

s well as absalse

p (sinos x)2

=

p

12

(when si

2

2dy ydx

⎛ ⎞ =⎜ ⎟⎝ ⎠

(2),

m (2) and el

0=

(3x b)(4+ −

0

0<

⇒ one real r

n nature.

TEST–I (POLUTION)

in [0, / 2]π

solute maxi

p

p

n2x)2

n2x = 1)

2 2cot x

iminating w

x c) (4x− +

Sinc

root betwee

P-II) )

imum

with help of (

c)(2x a)− −

ce a b2 3

< <

en a b,2 3

⎛ ⎞⎜ ⎟⎝ ⎠

(1)

(2)

(3)

(3), we get

) 0=

c4

117

Sample Paper


Also

bf3

⎛⎜⎝

56. (A, Let

∴

Che

57. (B, Equ Equ On Equ Equ On Sum Sum 58. (B) Equ

Equ

o, cf 04

⎛ ⎞ >⎜ ⎟⎝ ⎠

b cf 03 4

⎞ ⎛ ⎞ <⎟ ⎜ ⎟⎠ ⎝ ⎠

B, C) x | z 1|= +

2xRe(z) =

ecking for ex

C) uation of tany = x + a

uation of tan–y = x + a solving (i) &C = (–a, 0)

uation of noy + x = –2a= –3a

uation of noy – x = –2a= –3a solving (iii) D = (3a, 0)

m of x-coord= a + a – a

m of y-coord= 0

uation of circ(x 1)(x 4− −

2 2x y 5x+ −

uation of noy 2x 4+ = +

2x y 12+ =

⇒ other

x [0,⇒ ∈2 2 , f(x)2−

xtreme valu

ngent at A(a

ngent at B(a …(i

& (ii)

rmal at A a + a

rmal at B

a – a

& (iv)

dinates + 3a = 4a =

dinates

cles with SP4) (y 0)(y+ −

x 4y 4− + =

rmal at P(48 12+ =

FULL T(SO

between ⎛⎜⎝

, 2]

2x 3 x= + −

ues given re

a, 2a) …(i)

a, –2a) i)

…(iii)

…(iv)

= 4

P as diamety 4) 0− =

0 ,4)

TEST–I (POLUTION)

b c,3 4

⎞⎟⎠

equired rang

ter

P-II) )

ge of f(x) ass 131,4

⎡ ⎤⎢ ⎥⎣ ⎦

1

18

Sample Paper


r =

d =

Len

2=

59. (B,

Her

⇒

⇒

or

⇒ 60. (A,

Pro

∴

“Bes

“Bes

2(5 / 2) 4+

52 22

5

× + −

ngth of chord

2 2r d− 2=

C, D)

re 1p , q4

= =

1− (probab

nn

n31 C4

⎛ ⎞− ⎜ ⎟⎝ ⎠

n1 33 4

⎛ ⎞ ⎛ ⎞>⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠Minimum vB)

bability of A

Probability

st of three g

31 2

4p C10

⎛= ⎜⎝

st of five ga

52 3

4p C1

⎛= ⎜⎝

10 ×=

4 4−254

=

125=

d

252 54

− =

1 314 4

= − =

bility not hit

n 23

⎞ >⎟⎠

n

alue of n =

A winning, s

of B winnin

games” = (22

34 60 10

⎞ ⎛ ⎞ +⎟ ⎜ ⎟⎠ ⎝ ⎠

ames” = (3 w3 24 6

0 10⎞ ⎛ ⎞ +⎟ ⎜ ⎟⎠ ⎝ ⎠

64 36 510

× + ×

FULL T(SO

52

=

25 2024−

tting the targ

4

sayp 0.4= =

ng q 1 0.4= −

2 wins and 13

33

4C10

⎛ ⎞⎜ ⎟⎝ ⎠

=

wins and 2 4

54C

104⎛ ⎞+ ⎜ ⎟

⎝ ⎠

5

256 6 10× × +

TEST–I (POLUTION)

5=

get) 23

>

410

=

64 0.610

= =

1 loss) or 3

18 16 61000× +

=

losses), (4

55

6 C10

⎛ ⎞ +⎜ ⎟⎝ ⎠

1024 992312

=

P-II) )

0

(wins)

64 44125

=

wins and 1 54

10⎛ ⎞⎜ ⎟⎝ ⎠

225

losses) or (

(5 wins)

119

Sample Paper


FULL TEST–I (P -II) (SOLUTION) Sample · PDF filePHYSICS object be l eal image i riod....

Documents

Transcript of FULL TEST–I (P -II) (SOLUTION) Sample · PDF filePHYSICS object be l eal image i riod....