Solution_for_ISAT_2012

SOLUTION & ANSWER FOR ISAT-2012 SET – E [PHYSICS, CHEMISTRY & MATHEMATICS]  A + Dmin  sin  2   n= A sin  2

PART A − PHYSICS 1.

In a closed container filled with air at a pressure p0 there is an air bubble -----1 Ans : p0R 24 Sol:

4T ––(i) R piVi = pi’Vi’ (θ is constant) pV p ⇒ pi’ = i i = i ( Q R ⇒ 2R) Vi ' 8 p0 4T p p 4T pi’ − = ⇒ i− 0 = ––(ii) 16 2R 8 16 2R 28T p R Solving (i) and (ii) ⇒ pi = ⇒T= 0 R 24

A solid hemisphere of radius R of some material is attached on top of a solid cylinder of ------

4 1 2 πR3ρ × = πR3ρ 3 2 3 2 2 M2 = πR2. Rρ = πR3ρ 3 3 M1 = M2 and M1 + M2 = M M ⇒ M1 = M 2 = 2 M1 =

4.

Two physicists `A’ and `B’ calculate the efficiency of a Carnot engine running between two heat reservoirs by measuring-----Ans :

5 5 and 9 3 For A

η = 1−

T2 (T2 constant, T1 varies) T1

=

Ι2 =

1  M  2 MR2 . R = 2 2 4

For B

MR MR 9 + = MR 2 5 4 20

 3 + 1  Ans :  2 A = 90° (from figure) Dmin = 60° (Data)

 T  dT  =  2  1   T1  T1 

dη 10 5 = × 100 = % for A η 3 × 600 9

η = 1− 2

T12

300 × 10 10 =  600  3 × 600 900 × 900 ×    900 

⇒%

2

dT1

 T2  dT1   300 10     × dη  T1  T1   900 900  ⇒ = = η   T2   300  1 −   1 − 900    T   1 

2 M 2 MR 2 ; . .R = 5 2 5

For the prism shown in the figure, the angle of incidence is adjusted such that------

Sol:

 3 + 1    2   

Ι1 =

Ι = Ι1 + Ι2 =

3.

=

⇒ dη = + T2.

9 Ans : MR2 20 Sol:

sin 45° cos 30° + cos 35° sin 30° sin 45°

pi − p0 =

Sol: 2.

=

T2 (T1 constant, T2 variable) T1

⇒ dη = −

1 .dT2 T1

1 − .dT2 dη 10 T1 ∴ = = − η  300   T2  900 × 1 − 1 −    T1   900   10 = 600 dη 10 5 % = × 100 = % for B η 600 3

5.

Sol:

Two positive and two negative charges of magnitude q are kept on the x-y plane as shown ------

Distributions symmetric about tst are more accurate. Distributions with smaller widths about tst are more precise. ⇒ Statement (A) is correct.

Ans : X

−q

Sol:

+2q

9.

−q

Velocity v (m/s) versus time graph of a cyclist moving along the -----Ans :

3 ˆ 15 i, 4 4

Sol:

4+6 S1 =   × 5 = 25 m  2 

Y

Resultant field of the given configuration is zero along any point on Z-axis. In configuration (B), the dipole moments of two dipoles (−q, +q), (−q, +q) are in opposite directions ⇒ field along Z-direction is zero.

 4+8 S2 =   × (− 5 ) = −30 m  2  2+6 S3 =   × 5 = 20 m  2 

∴ S = S1 + S2 + S3 = 75 m S = S1 + S2 + S3 = 15 m

6.

Two hemispheres made of glass (µ = 1.5) are kept as shown in the figure. The radius -----Ans :

Sol:

2R = 3 7.

v Av

2R 3

 µ − 1 Normal shift = t    µ   1.5 − 1 = (R + R )   1.5 

A right-angled prism ABC (∠C < ∠B) made of a material of refractive index µ0 is immersed------

10. A sphere of mass M and radius R is surrounded by a shell of the same mass and radius 2R. A small hole ------

Ans :

3GM R

Sol:

Ushell = −

 µ  Ans : sin−1   µ0  Sol:

8.

r1 + r2 = φ for prism r1 = 0 (Data) ⇒ r2 = φ, should be the critical angle  µ  µ  ⇒ sinφ = ⇒ φ = sin−1  µ0  µ0 

Four screw gauges are to be calibrated to the standard thickness `tst’ of a wire. Series of measurements -----Ans : Screw gauge 1 is less precise but more accurate than screw gauge 4.

S 15 3 ˆ = = i t 20 4 S 75 15 = = = 20 20 4

v Av =

GMm 2R GMm Usolid sphere = − R GMm GMm 3GMm ∴ Ui = − − =− 2R R 2R 1 3GMm 2 i.e. mv e = 2 2R

⇒ ve =

3GM R

11. A particle of mass m is projected in the vertical plane (taken to be the x-y plane) with speed v at an angle -----ˆ Ans : −0.5mv cosθgt2 k

Sol:

J = F dt = −mgdt j t

L =r×J =

∫ (u cos θt )ˆi × −mgdtˆj 0

∫

= − mg u cos θ tdtkˆ

Sol:

2 = −0.5mgu cosθt kˆ

12. Consider a damped simple harmonic oscillator given by the equation of motion -----Ans : x

t

v

Sol:

π rad 2 and since starting is from extreme position, x and v must be in opposite directions. x and v have phase difference of

τ constant ⇒ FR constant ⇒ F constant Impulse = Ft = 10 F But impulse = ∆p = mv mv ⇒ 10 F = mv ⇒ F = 10 Fcentripetal = friction = µF = mvω µmv ⇒ = mv ω 10 µ 1 ⇒ω= = 0.02 rad s− 10

16. A model potential between two molecules A and B in a solid is shown in the figure, where x gives the distance of B with respect-----Ans :

Sol:

13. A metal sphere is kept in a uniform electric field as shown. What is the correct-----Ans :

1 T 1 2 kx 2 KE = constant × kBT 1 2 ⇒ kx 2 ∝ kBT ⇒ x ∝ T 2 dx 1 ⇒ 2xdx ∝ dT ⇒ ∝ dT x dx 1 1 ∝ 2 ∝ xdT T x PE = KE =

––(i) ––(ii)

17. The mass density of a dusty planet of radius R is seen to vary from its center as -----Sol:

No electric field inside sphere and field lines are normal to surface.

14. An electron travelling with velocity v = 3ˆi + 5ˆj in an electric field ------

Ans :

Sol:

45 16 r   dm = 4πr 2ρ0 1 − dr   R  R/2

⇒ M1 =

Ans : 5ˆi − 3ˆj + 18kˆ Sol:

(

)

(

i

j

k

= −3 5

0

)

R

M= E1 =

= 5ˆi − 3ˆj + 18kˆ 15. A small block is kept on a frictionless horizontal table. A wooden plank pivoted at O, but otherwise free to rotate, pushes the block by applying a constant ------

E2 =

GM1 R   2

=

2

GM 2

3   R 2  E 45 ⇒ 1 = E2 16

5πρ0R3 48

πρ0R3 3

∫ dm = 0

6 4 −1

Ans : 0.2 rad/s

dm =

0

qE + q v × B = 0

⇒ E = − v×B

∫

5 πρ0GR ; 12

=

4 πρ0GR 27

18. A particle is moving in a force field given by F = y 2 ˆi − r 2ˆj . Starting from A the particle has to

R

∴ UB =

∫ 0

reach ------

µ λ2Q2Lπ r 4 dUB = 0 .

4 4A 2

R

0

Ans : (−1, 1)

µ λ2Q2LπR 4  2 A = 0  πR = A, R 2 =  π 16 A 2 

Sol:

=

dW = F.d r

∫ dW ; For AB, dr = dxˆi ;

W=

(Q LA = volume) µ λ2Q2 × volume ∴ UB = 0 16 πA σ Q Electric field E = = ε0 Aε0

path

For BC, dr = dyˆj , for AD, dr = dyˆj and for DC, dr = dx ˆi

∫ F.dr + ∫ F.dr

⇒ WABC =

AB

(1, 0 ) =

∫

(1, 1)

(0, 0 )

(1, 0 ) AD

(0, 1)

∫

=

∫ F.dr + ∫ F.dr

W ADC =

=

∴ Energy density =

BC

∫ F.dyˆj = −1

F.dx ˆi +

(1, 1)

F.dyˆj +

(0, 0 )

∫ F.dxˆi = +1

( 0, 1)

∴

( )

1 ε 0µ 0 Aλ2 8 π t

i(r) = πr 2. j =

= =

µ0 .πr 2i µ0ir = A.2πr 2A =

µ02λ2Q2r 2

2 1 B(r ) .(2πrLdr ) 2 µ0

1 µ02λ2Q2r 2 1 . . .2πrLdr 2 µ0 4A 2 µ0λ2Q2 4A 2

=

Sol:

(Q i = −λQ) 4A 2 4A 2 Consider a cylindrical shell of radius r, thickness dr and length L (distance between plates of capacitor).

dUB (r ) =

(

)

ε0µ0λ2 A 8π

––(i)

πr 2i A

µ02i2r 2

 UB  µ0λ2Q2 2 A 2ε 0 = × volume  × 2 UE  16 πA  Q × volume

Ans :

Br.2πr = µ0i(r) ⇒ B(r) =

1 Q2 × (volume ) 2 A 2ε 0

20. Three sinusoidal oscillations A sin(21t), and A sin(19t) are superposed. Which of ------

Q = Q0e−λ dQ t = −λQ0e−λ = −λQ i= dt i j= A

B2(r ) =

1 Q2 2 A 2ε 0

UE =

19. A capacitor made of two parallel circular plates of area A holds a charge Q0 initially. Suppose that it discharges as ------

Sol:

1 1 Q2 ε0E2 = ε0 . 2 2 2 A ε0

∴ Energy in electric field,

DC

∴ (−1, 1) is the answer.

Ans :

µ0λ2Q2 (LA ) A µ0λ2Q2 (LA ) = π 16 πA 16 A 2

3

.Lπr dr

y1 = A sin19t y2 = A sin20t y3 = A sin21t y = y1 + y2 + y3 C+D C−D  . cos Q sin C + sin D = 2 sin  2 2   ⇒ y = A[2 cos 2πt + 1] sin20t ⇒ There will be intermediate maxima with smaller amplitude.

21. A vertical resonance pipe is filled with water and resonates with a tuning fork at minimum air column length of 30 cm -----Ans : 4 : 1

Sol:

γRT = M

Vair =

1.4 28 × 10

−3

∴ Minimum repulsion energy between

× RT

electrons (when they are diametrically

λair = 30 × 4 = 120 cm = 0.12 m V f of tuning fork = air λ air =

1 .4 28 × 10 −3

×

RT 0.12

23. If the Hydrogen atom ionization temperature is T, the temperature at which He atoms ------

––(i)

λmixture = 42 × 4 = 168 cm = 0.168 vmixutre = λmixture × f ] = 0.168 ×

1.4 28 × 10

But vmixture =

−3

×

RT 0.12

γ mixtureRT Mmixture

Ans : 6T

––(ii)

0.168 = × 50 0.12 5 γmixture = (Q both monoatomic) 3

⇒ Mmixture = =

Sol:

24. The coefficient of viscosity of a fluid is known to vary with temperature -----Ans :

γ mixture × (0.12)2

(0.168 )2 × 50 5 (0.12 )2 = 0.017 kg × 3 (0.168 )2 × 50

= 17 gram n1M1 + n2M2 = 17 (n1 + n2 )

log(T v1]

T

Sol:

⇒ v T .CTe

For singly ionized He atom, ground state Ze2 = −54.4 eV energy is − 2R

⇒ −

2e 2 = −54.4 eV 2R 2

⇒

e − 54.4 = = 27.2 eV 2R −2

−

⇒ vTT ∝ e

T T0

= constant

T T0

T , where k is a constant.

10 ⇒ graph of ln vTT vs T will be a straight line with positive slope.

ln vTT = k

≅4:1

Ans : 27.2 eV

vT =

⇒ vTη = constant

4V1 + 20V2 = 17V1 + 17V2 3V2 = 13V1 V 13 ∴ 1 = V2 3

22. The minimum repulsive energy between the two electrons would ------

2 r g(ρ − σ ) 9 η 2

n1 ∝ V1, n2 ∝ V2 4V1 + 20 V2 = 17 (V1 + V2 )

Sol:

E = Ionization energy of He-atom = 2 × 54.4 eV − repulsion energy = 108.8 eV − 27.2 eV 13.6 eV ∝ T ⇒ E ∝ 8T − 2T ∝ 6T ∴ AT 6T, He atom ionizes completely.

––(iii)

γmixture 0.168 1.4 = × Mmixture 0.12 28 × 10 −3

⇒

e2 = 27.2 eV 2R

opposite) =

25. A particle moves in a force field of the form r×L F = k 2 , where r is the position vector -----r Ans: Magnitude of angular momentum decreases exponentially, but its direction remains unchanged.

Sol:

F is ⊥ to r , ⊥ to L and ⊥ to plane containing r and L r

⇒ F is in the plane of motion and opposing the motion. τ is not zero ⇒ L is decreasing F is also not constant as L is changing

= − 5.8 × 10− 32. Lithium nitrate when heated gives-----5

26. The approximate formation-----

standard

enthalpies

of

Ans : ∆H (octane) is more negative than ∆H (methanol) Sol:

Combustion of C8H8 involves more number of carbons and hydrogens compared to CH4O

Ans : Li2O, NO2 and O2 Sol:

4LiNO3 → 2Li2O + 2NO2 + O2

33. For a fixed mass of an ideal gas the correct-----Ans :

P3 > P2 > P1

27. The Boyle temperatures of three gases are------

0

Ans : I-hydrogen, II-oxygen, III-ethene Sol:

More the compressibility factor greater is the negative deviation from ideal behaviour.

28. The reduction potentials of M trend------

2+

/ M follow the

Sol:

P3

0 Temperature, T (K)

Plot of V α T is a straight line and the slope of the isobar decreases with increase of pressure

34. Match each one with the correct method ------

Sol:

o EM 2+ / M for V = − 1.18 V

Fe = − 0.44 V Ni = − 0.25 V Cu = 0.34 V 29. The total number of isomers expected for------

Cr2O3 − Al reduction Fe2O3 → CO reduction Cu2S → self reduction ZnS → Roasted to ZnO and then CO reduction

35. The intermediate formed in the following ------

Ans : 9 Sol:

P2

Ans : (a) → (ii), b → (iv), c → (iii), (d) → (i)

Ans : V < Fe < Ni < Cu Sol:

P1

Volume, V

⇒ Variation is exponential.

PART B − CHEMISTRY

Cl

HO

NO2

± cis [Pt(en)2(CNS)2] ± cis [Pt(en)2(NCS)2] ± cis [Pt(en)2(NCS) (CNS)] trans [Pt(en)2(CNS)2] trans [Pt(en)2(NCS)2] trans [Pt(en)2(NCS) (CNS)]

Ans :

NO2

Cl 30. In the conversion of dinitrogen to hydrazine, -----

NO2 Sol:

Ans : 4 and 4 Sol:

N2 + 4H + 4e− → N2H4 +

Sol:

E = − 4 × 10− T − 9 × 10− 5 + 3.6 × 10− T dE 5 7 = − 0.4 × 10− − 9 × 10− × 2 T dT 5 5 = − 0.4 × 10− − 5.4 × 10− 5

NO2 OH slow NO 2

NO2

31. The temperature of dependence of the e.m.f -----Ans : − 5.8 × 10−5 7

× T

OH

Cl

OH NO2

fast

+ Cl

2

NO 2

36. The crystal field splitting energy (∆0), of------

CH 3

Ans : I < II < III < IV Sol:

Sol:

The arrangement of ligands spectrochemical series is CN− > NCS− > F− > Br−

in

the

CH 2

respectively

37. The compounds that form stable hydrates are----Ans : II and IV Sol:

and

Alkenes (A) and (B) are

CH3

CH3 OH

II is Indane-1,2,3-trione. It forms a stable hydrate known as ninhydrin which is stabilized by intramolecular hydrogen bonding. IV is chloral which also forms stable chloral hydrate

Hydroboration Oxidation (A)

(X)

CH2

CH2OH

0

38. The symbols F, H, S, Vm and E denote -----Hydroboration

Ans : F, H, S, are extensive; Vm and E0 intensive Sol:

Oxidation (B)

F, H and S are extensive properties, as they depend on the quantity of the system 0 Vm and E are intensive properties

CH3

st

39. For a 1 order reaction of the form------

CH2 oxymercuration

or

Ans : I and IV Sol:

(Y)

(A)

reduction (B)

OH

H 3C

ln A A 0 = − kt i.e., A A 0 decreases with increase of kt A 1 = A 0 ekt A decreases with increase of kt A0

40. Consider the reaction 2A Ans : 0.05 Sol: 2A K=

(x 2)

(Z) 42. The major product of the following reaction is-----

B.------

Ans : C6H5CH=CHCH=NNHCONH2 Sol:

B

H2N − NH − CO − NH2 → ( Semicarbazide )

(a − x )2

C6H5−CH=CH−CH=N−NH−CO−NH2 O

on solving, x = 0.15 and 0.1 x = 0.15 is not possible ∴ Amount of B at equilibrium = 0.05

The − NH2 group away from the

41. Two isomeric alkenes A and B on hydrogenation in the presence -----X=

Y= Me

Ans :

Z= CH2 OH

C group of semicarbazide reacts with aldehydes and ketones 43. At 100 K, a reaction is 30% complete in 10 minutes, while at 200 K------

Me OH

OH

C6H5−CH=CH−CHO +

Ans : 1150 J

Sol:

log

K2 Ea T2 − T1 = K1 2.303 R T1T2

50. In the oxidation of sulphite using permanganate, the number of protons ------

Ea × 100 2.303 × 8.314 × 100 × 200 Ea = 1150 J

log 2 =

Ans : 3 Sol:

2MnO 4− + 5SO32 − + 6H → 2Mn +

5SO 24 −

44. The stability order of the following carbocation is----

+

+ 3H2O

PART C − MATHEMATICS

Ans : II > III > I > IV Sol:

2+

II is the most stable carbocation because positive charge is at allylic position with respect to two double bonds.

IV is the least stable carbocation as it is antiaromatic

51. Let the line segment joining the centers of the 2 2 circles x − 2x + y = 0 -----2

Sol:

45. For the following Newman projection------

H

2

Ans : 5x + 5y + 2x + 16y + 8 = 0

Cl

Centre of x2 + y2 − 2x = 0 (1, 0) radius = 1 centre and radius of 2 2 x + y + 4x + 8y + 16 = 0 is (−2, −4) radius = 2

Cl H

Ans : H

Sol:

P

H

(b)

Distance between the centers = 5 ∴ PQ = 5 − 1 − 2 = 2 Clearly centre of the required circle lies the third quadrant and radius of the required circle, is 1 which is 5x2 + 5y2 + 2x + 16y + 8 = 0

46. For bromoalkanes-----Ans : I and III Sol:

Statements I and III are correct

47. The number of unpaired electrons present ------

[Co(C2O4)3] − is a spin paired complex 2 3 3 with d sp hybridisation where as [CoF6] − 3 2 is a spin free complex with sp d hybridisation 3

48. The correct statement regarding the functioning of a catalyst is that it-----Ans : II and IV Sol:

52. If the angle between the vectors a and b is

Statements II & IV are correct

49. The relationship among the following pairs of isomers is-----Ans : I − A, II − A, III − B, IV − B

Ans : 2 3

Sol:

1 a×b = 3 2 If adjacent sides are represented by Ι a and Ι b Area of the triangle =

⇒ a×b = 6 ⇒ ab sinθ = 6 π ⇒ ab sin =6 3 12 ⇒ ab = 3

∴ a • b = ab cos =

12 3

Sol:

π 3

------

Ans : 0 and 4 Sol:

Q

•

Geometrical and optical isomers known as configurational isomers

are

×

π 3

1 6 ⇒ =2 3 2 3

π   cos 9 53. Let P =  − sin π 9 

π 9  -and α, β, γ ----π cos  9 

Ans : 1 Sol:

3π 5π <x< 4 4

sin

αp6 + βp3 + γΙ = 0  1 3  −  2  +β ⇒α  2 − 3 −1   2   2 1 0 0 +γ   =  0 1 0

56. The number of distinct real values of λ for which the vectors -----Ans : 1

 1   2 − 3   2

3  2  1   2 

Sol:

0  0

−α β 3 3 + + γ = 0 and α+ β =0 2 2 2 2 ⇒ −α + β +2γ = 0 and α + β = 0 ⇒ α = −β ⇒ ∴ 2β +2γ = 0 ⇒ β + γ = 0 ⇒ β = −γ ∴ α = −β = r ∴α−γ=0 2 2 2 ( )( )( ) ∴ (α + β + γ ) α − β β − γ γ−α 2 2 2 0 = (α + β +γ ) = 1 ⇒

54. A random variable X takes values −1, 0, 1, 2 with probabilities-----Ans :

Sol:

0 1 − λ3 0 =0 2π − sin λ − λ

λ3 1 λ4

0 1 − λ3 0 2π − sin λ0

2λ − sinλ + λ = 0 7 X +2λ = sinλ 7 6 f(x) = λ −2λ f’(x) = 7λ +2 f’(x) > 0 ∴ f(x) is increasing So it intersects with sinλ only once 7

57. -The minimum value of |z1 − z2| as z1 and z2 vary over the curves-----

Ans :

−1 5 and 16 4

λ3 1 1

5 7 2 3

−1 0 1 2 1 + 3p 1 − p 1 + 2p 1 − 4p P(x) = 4 4 4 4 ∴ x = Σxp(x) 2 − 9p = ; But p ∈R and from the given 4 probabilities since 0< p(x) <1  −1 1  we get p∈  ,  .  3 4 x:

 −1 5  Hence x ∈  ,   16 4   −π π  55. Let f(x) = log(sinx + cosΙ), x∈  −  ----- 4 4

Sol:

z1 lies on the circle z − (i.e) |z − z0| =

f(x) = log

π  2 sin  + x  4 

π  f’(x) =- cot  x +  4  π π π 3π 0 < x+ < or π< x + < 4 2 4 2

i 3

=

7 3

2i 3

= z −9−

18i 3

(i.e) z2 lies on the perpendicular bisector   18  −2  and  9,  of  − 1,  3  3     8   = 8z0 ⇒ z2 passes through  4, 3  

∴ Minimum of |z1 − z2| = |z2 − z0| − |z1 − z0|

= 7 |z0| − Sol:

−

1 7 i   where z0 =  + 2 3 3  

z2 lies on z + 1 +

= |8z0 − z0| −

 −π π  Ans :  ,   4 4

1 2

=

7 3

7 3

5 7 2 3

1

10

10

(1+tanθ) +(2+tanθ) + tan9 θ 10 -----+(20+tanθ) ) − 20 tanθ-----

58. Let f(θ) =

Ans : 2100 Sol:

+

Put t = tanθ

t9

=

(1 + t )

−t 1+ t − t

10

+

(2 + t )10 − t10 + − − − + 2+t−t

(

1 t9

 (2 + t )10 − t10  1  (1 + t )10 − t10    + 2  --9 t  1 + t − t   (2 + t ) − t  1 9 9 = 9 [10. t + 2 × 10 . t + ----+20] t = 10 [ 1+ 2 + -------+20] [20 × 21] = 2100 = 10 2

=

59. Let r > 1 and n > 2 be integers. Suppose L and M are coefficients ------

dx

∫ x2 + 2

−

2 3

dx

∫ x+2

(By partial fractions) 1 1 1 = − log 3 + log 2 − log 2 4 4 2 2 2 1  x  2  + log x 2 + 2 0 + tan−1  3 3 2  2 2 − log(x + 2) 3

(1 + t )10 + (2 + t )10 + − − +(20 + t )10 − 20t10 = 10

2 3

)

 2 2  1 1  =  − +  log 3 + tan−1  3  3  4 3   5 − log 2 2

=

2 1 5 tan −1 2 + log 3 − log 2 3 12 12

61. The age distribution of 400 persons in a colony having median age 32 is given below------

Ans : n = 2r+1 Ans : −10 Sol:

Let l = 3r m = r +2 th th Given l term = L & m term is M L = (2n −1)Cl −1 & M = (2n −1)Cm−1

∴

m (2n − 1)! 2n (l − 1)! (2n − l )!

=

l (2n − 1)! 2n (m − 1)! (2n − m)!

2n ! l ! (2n − l )!

=

2n ! m ! (2n − m)!

2n

=

Cl

⇒l=m OR ⇒ 3r = r +2 OR

2

(

log x 2 + 2

(x + 2)2

0

∫ log(x 2

=

(

2 log x 2

+2

(x + 2)2

) dr is------

2 1 5 tan −1 2 + log 3 − log 2 3 12 12

Ans : =

∫

Cm

l +m = 2n 2r = 4r+1 ∴ n = 2r+1

∫0

60. The value of the integral

Sol:

2n

2

) dx = )

Sol:

0

(

)

−

2

2 x  −1 

∫ x2 + 2  x + 2 d x

1 1  2 = −  log(3 × 2) − log 2  + 2 4  3

xdx

∫ x2 + 2

below 25 110 below 30------110 + x below 35------185 + x below 40------240 + x below 45------240 + x + y below 50-----240 + x + y ∴ 270 + × + y = 400 ⇒ x + y = 130 -----(1) N   − c.f h 2  Medians = l +  f 200 − (110 + x ) = 30 + × 5 = 32 7.5 ⇒ x = 60 ∴ y = 70 ∴ x − y = −10

62. The probability that a randomly selected calculator from a store is of brand r is proportional to r, -----Ans :

+ 2 (x + 2)−2 dx

 −1  = log x 2 + 2    x+2

Sol:

8 63 Let n be the total no of calculators.

Since p (r) α r and Σp(r) = 1 k 2k 6k ⇒ + +−−−−+ =1 n n n ⇒ n = 21 k r ⇒ p(r) = , r= 1, 2, -------, 6 21 6 5 1 Again p(Dr) = when , ,− − − − −, 21 21 21 r =1, 2, -----,6

π  cos x +  6 

6

P (Defective calculator)

∑ p(r ) × p(Dr ) r =1

 π π  π    cos x +  − 2 cos +  cos(x −   = 0  6 6  6    

1 6 2 5 = × + × +−−− 21 21 21 21 6 1 56 8 + × = = 21 21 212 63

π  π π  ⇒ cos x +   2 cos x cos − 2 cos  6  6 6  =0 π π  ⇒ 2 cos x +  cos (cosx − 1) = 0 6 6 

63. There are 10 girls and 8 boys in a class room including Mr. Ravi, Ms. Rani and Ms. Radha.-----

π  ⇒ cos x +  = 0 or cosx − 1 = 0 6  π π ⇒ x + = ± or cosx = 1 6 2 π −2π  π π ⇒x= or or x = 0 ∈  − ,  3 3  2 2

Ans : 308 Sol:

10g 8g

8b 6b

Ravi is in : 9 C8 ×7 C5 = 9 ×

7×6 = 189 2

Rani is in : 7 C6 ×8 C7 = 7 × 8 = 56

∴ But

Both Ravi & Rani are out: 9

64. Let f: [0, 4] → R be a continuous functions such that |f(x)| ≤ 2 for all x ∈ [0, 4]-----Ans : [−6+2x.,10 −2x] Sol:

∴ Number of solutions = 2

C8 ×7 C6 = 9 × 7 = 63

Total = 189 +56+ 63 = 308

2=

x

4

x

∫x 2 dt = 8 − 2x

∫0 f (t)dt − 4∫x f (t) dt ∫0

− 2 f ( t ) dt ≤ −

66. The equation of the circle which cuts each of the 2 2 three circles x +y = 4,-----Ans : No correct option (None of the equations represents a circle) Sol:

4

x

2x − 8 ≤ 2

∫0 f (t) dt ≤ 8 − 2x

⇒ 2x − 6 ≤

∫0 f (t) dt ≤ 10 − 2x

Radical axis of S1 and S2 is 2x − 1 = 0 and that of S2 and S3 is y − 1 = 0 1  Radical center of the three circle is  ,1 . 2  which is in the interior of all the three circles. ∴ No circle orthogonal to all the three.

x

65. The number of solutions of the equation------

−2π  −π π  ∉ ,  3  2 2

67. Suppose an ellipse and a hyperbola have the same pair of f------

Ans : 2 Sol:

π π π   cos  x +  − 2 cos x +  cos 6 6 6   π = sin2 − cos2 x 6 π  π π   cos  x +   cos x +  − 2 cos  6  6 6  

7 3

Ans :

2

π  = −  cos2 x − sin2  6  

Sol:

1 for ellipse 2

e= i=

1 b2 = 1− 2 2 a

1 b2 = 1− 2 4 a

π  π π   = cos  x +   cos x +  − 2 cos  6 6 6      

∴

 π π    = −  cos x +  cos x −   6 6     

∴ b = 7 and a =

b2 a2 2

= 1−

1 3 = 4 4 2

28 3

1 − f (x )2 = f (x) f ' (x )

 7  ∴ Foci = ae =  ± , 0   3  Equation of the hyperbola that passes through (2, 2) is x2 a2

y2

−

7 − a2 3

= 1⇒ a = 1

7 Hence a e = ⇒e= 3 2 2

x+

Ans : | α2 − β2| Let α’, β’ be the roots of a (x +5) − 25a (x +5) + 50 = 0 ∴ |α’ − β’| will remain the same for 3 2 a y = 25ay +50 = 0 3

 π 2 71. In the interval 0,  , the equation cos x − cosx  2

−x = 0 has ------

2

Ans : Exactly one solution Sol:

 ay   ay  ⇒ a   − 5  + 2 = 0  5   5  2

1− y 2 ⇒ C ⇒ C = 1

7 1 ∴f  =± 4 4  

2

⇒ ay and β

1− y2

3  1 Qf  = 2 2  

7 3

68. -Let a be on a on − zero real number and α, β be the root of the equation ax2 +5x+2 = 0-----

Sol:

y dy

∴ dx =

− 5y +2 = 0. Whose roots are α

−aα' −aβ' and β = ∴α= 5 5 −5 | α − β| = | α2 − β2| |α’ − β’| = a −5 since α + β = a

f(x) = cos x − cosx − x = 0 f’(x) = −sin2x +sinx −1  π < 0 in  0,  which is decreasing  2 2

x = 0 is the only solution 72. The points with position vectors -----Ans : ⇒ (1 − α) (β +1) = 0 Sol:

Vectors are denoted by A, B, C, D then AB = (α −1) i +2j +2k

69. The set of all 2 ×2 matrices which commute with 1 1 the matrix   -----1 0

AC = (α −1) i − j +2k AD = ( α− 1) i + (1 − β) k ∴ A B AC A D = 0

[

α −1

p  q  Ans :   : p, q, ∈ R  q p − q 

Sol:

1 1 Clearly matrix A =   commute with 1 0 q  p matrix B =   q p − q i AB = BA

]

2

α −1 −1 α −1

0

2 2

=0

1− β

⇒ (α− 1) (β + 1) = 0 ⇒ ( 1 − α) (β −1) = 0 ⇒ (1 − α) (β +1) = 0

73. For a real number x, let [x] denote the greatest integer less than or equal to x-----Ans : one − one but NOT onto

70. Let f: (0, 1) → (0, 1) be a differentiable function such that f’ (x) ≠ 0----- 7 15  Ans :  ,  4   4  Sol:

Using L’ H rule, we get

Sol:

f(x) = 2x + [x] +sinxcosx 1 = 3x {x} + sin2x 2 ∴ f’(x) = 3 − 1 +cos2x > 0 ∴ f(x) is strictly increasing ⇒ One − one function

But due to the presence of [x] f(x) jumps at integral points. ⇒ f(x) is NOT onto 74. Let M be a 3 × 3 non singular matrix with det (M) = a -----Ans : α Sol:

adj (adjM) = |m| − M = Mα 1 1 ∴ M− adj (adjn) = M− mα = αΙ ∴k =α= 3 2

75. If y − x = 1 then the value of x

y

dy at x = 1 is-----dx

Ans : 2(1 − log2) Sol:

x

y

y =u x =v ⇒u+v=1 dy dv ⇒ = dx dx  dy  y x  x dy y  + log y  = x y  + log x  dx  x  y dx 

  y y   x   − y x log y   dy   x  ∴ =  dx  x  x  y   − x y log x     y  x=1 y=2 dy 2 − 2 log 2 ∴ = = 2(1 − log2) dx 1