sri chaitanya paper1

8/8/2019 Sri Chaitanya Paper1

1/33

G enera l Instructions : Max . Marks : 285

About your Institute :

Address :Sri Chaitanya IIT Academy.,A.P.ICON Central O ff ice - M ad hap ur - Hyd

Ph No : 04 0-6 45 66 17 7 Email : [email protected]

Super 10

IIT - JEE 2010

Mock Test 1Paper - 1

SOLUTIONS

Sri Chaitanya IIT Academy

Campus

Sr i Ch a i t a n y a I IT Ac a d e m y . , In d ia

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Sectoion - I contains 5 Q uestions each carr ies 3 Ma rks for W rong a ttempted Q uestion 1 mark w ill be ded ucted

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Sectoion - III contains 6 Q uestions each carr ies 4 Ma rks for W rong a ttempted Q uestion 1 mark w ill be ded uctedSectoion - IV contains 2 Q uestions each carr ies 8 Ma rks fo r W rong a ttempted Q uestion 0 mark w ill be ded ucted

Sectoion - V contains 5 Q uestions each carr ies 4 M arks for W rong a ttempted Q uestion -1 mar k will b e d educted

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2/33

Sri Chaitanya IIT AcademyCampusMock Test 1 / Pg 2

# P-1 CHEMISTRY - SOLUTIONS :

1. Ionic radii > 1.34A0

form ionic interstitial carbide

3. m-xylene is thermodynamically controlled

4. Ans : Thermodynamically controlled product

5. For isothermal process, pV = constant,

or p1V

1= p

2V

2

or p1V

= p

Tx 2V

or1 2T

p

p=

6. aND : d

6. As there compounds donot possess symmetry element

meso means C6H

5group present down

a)

C

C

CH3

CH3

H

NH2

Pt

C CHC6H5

C CHC6H5H

NH2

b)

C

C

C2H5

CH

CH3

O

O

Be

O

O

C

C

CH

CH3

C6H5

- tetrahedral

c) octahedral symmetrical bidentate ligand

d) sq - planar goemetry

7. B2H

6undergoes symmetrical cleavage

8. NaBH4/EtOH does not reduce the amide group




3/33

Sri Chaitanya IIT AcademyCampus

Mock Test 1 / Pg 3

9. Electron rich group migratefirst to electron defficient oxygen

10.

11. Below 15000C G of MgO is more negative than G of H2O3

12. As the temperature increases stability of oxides decreases

13. MgO requires more temp which can be clearly seen from elingam diagram at high temperature only C OG is

more ve than G of MgO

14. In present NaOEt base the carboanion form at more active position

15. Ans : B

16. LOA ------- base and carboanion forms at less substituted place

17. Ans : A - q,r,s B - p,q,r C - p,q,r D - q,r,s

Kohlrausch's law suggets independet migration of ions at infinite dilution of both the strong and weak electro-

lytes.

Delbye - Onsgar - Huckel helps for extrapolation of as C1/2 for all electrolytes

NaCl((aq) is strong electrolyte while NaCl(alc) is a weak electrolyte due to low permitivity of alcohol. Usually

CH3COOH is a weak electrolyte but in NH

3due to greater protophilic nature of NH

3, CH

3COOH behaves as

a strong electrolyte




4/33


18. A)

N

H

CH3-MgClCH3

Cl

CH4NCH3

Pd - C/A

KMnO4/H-

N

COOH

COOH

B)

Pd-C/

KMnO4/H+

KMnO4/H+

COOH

COOH

C)O

K

MnO

4CH3MgCl

COOH

O

CH3MgCl

D)

KMnO4

CH3MgCl

Mgl

HOOC - COOH+

HOOC - CH2 - COOH

19. Microsomic salt is NaNH4

HPO4.4H

2O

20. Only Sb+3 because

Bi+3 Sb+3 Sn+4 Cu+2

|H2S




5/33


Mock Test 1 / Pg 5

Bi2S

3, Sb

2S

3SnS

2CuS

soluble insoluble

Sb+3 & Sn+4 Bi2S3 &CuS

NH4Sx

Sb2

S3

+ (NH4)

2 ( )4 43 NH SbS

soluble complex

HClSnCl4 , SbCl3 (soluble)

H2O

SnCl4solution

SbOCl

( ) ( )2 4 4 32 2xSnS NH S NH SnS+

soluble complex

21.

4

2

//2 NaBH EtOHKCN EtOH KHBr H OPh CHO Ph CH C Ph Ph C C Ph

OH O O O

Ph CH CH Ph

OHOH

C

for C two activeisomers and one more compound total three isomers

22. Since there is only one hydrogen atom hence only one electron is being dexicited from 'n' to ground state to

obtain four lines in emission spectrum following transition is possible

( ) ( )11 2 3 4n excitedstate n n n n n ground state

23. Ans : 3




6/33


# P-1 PHYSICS - SOLUTIONS :

24. Solution : ANS : B

Since PVn = constant and alsop PV = RT taking 1 mole of the gas for simplicity,

dU = Cv

dT shere vC molar specific heat at constant volume.

Now the molar specific heat in a polytropic process PVn = constant is given by

( )

( )( )1 1 1 1vn RR R

Cn n

g

g g

- = - = - - - - ....................(1)

From this equation we see that 0 vC will be negative when n g< and n > 1 simultaneously (i.e.,) 1 n g< < .

Since g for all ideal gases is greater than 1, if 1n o r ng> < then Cv

will be positive.

25. ANS : C

m

q q

Initially, the tension is springs and strings in equilibrium is ( )1

2sin

mgT

q=

After spring and string is cut

eeeeeeeee

q

T

cosmg q sinmg qmg

q

In this case block can have all along the spring since t is extesible

costa g q\ = =

: sin2sin

n n

mga mg maq

q- =

( )22sin 1

2sinn

gm a q

q= - .




7/33


Mock Test 1 / Pg 7

T

cosmg qsinmg q

mg

q

let this case block cannot acceleration along string from sin sin2sin

mgmg Tormgq q

q> >

2 0sin 1 / 2 45orq q > >

costa g q=

0na =

on ( )22sin 12sin

n

ga q

q= -

2 22sint n

ga a a

q= + =

\ ratio is costa g q=

0na =

| cosa g q\ =

|1 25

2sin cos 24a

a q q= = .

26. ANS : D

For collision of m1 and m2 :

0 1 2mv mv mv= +

on ( )0 1 2 1v v v= +

and2 1

0

1

2 0

v v

v

-=

-

on ( )02 1 22

vv v- =




8/33


From (1) and (2),03

4

v= .

At the maximum extension of spring, angular speeds of m2 and m3 must be same.\ From conservation of angular momentum about centre 0 we have,

( )2203 .2 2

4

vm R m R w m R w= +

203 5

2

v RR w =

03

10

vw

R =

From conservation of energy,

( ) ( )2

2 22031 1 1 1 22 4 2 2 2

vm kx m Rw m Rw

= + +

03

4 5

v mx

k = .

27. 1 2= + B B B

0 0

0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1..... .....2 2 3 4 2 2 3 4

= + + + + +

I Ia a a a a a a a

0 1 1 12 1 .....

2 2 3 4

= + +

I

Ve

Ve+

04a 03a 02a 0a 0a 02a 03a 04a

( )Set 1 ( )Set 2

0

2

0

n

= lI

a.




9/33


Mock Test 1 / Pg 9

28.0

uurE between the capacitor plates

( )210 02 2

+=

0 02 2

Q x x

A A

+= +

E

Q x+ x

potential difference 00

2( 2 )

2 2

d Q xV E d Q x E

A C

+= = + = [ Here potential difference V = E finally]

2

2

EC Qx

= .

Answer2

= = Q

EC x

29. (C), (D)

The temperature gradient (which causes the heat to flow) is only along the length. The temperature at any point

of a given cross-section is the same and so 0.a bQ Q= =

In the steady state,1Q is constant 0 .

30. (A), (B), (D)

If P divides AB in ratio, 1 : 4, then the fundamental frequency corresponds to 5 loops, one loop in AP and 4 loops

in PB which corresponds to 5th harmonic of 1 kHz. Hence fundamental = 5 kHz.

If P be taken at mid-point, the third harmonic will have three loops in each half of the wire AB. Hence total no.

of nodes (including A and B) will be 5 + 2 = 7.

If P divides AB in the ratio 1 : 2, the fundamental will have three loops, corresponding to the frequency of 3 kHz.

For this string to vibrate with the fundamental of 1 kHz, the tension must b ( / 3).T

The wire AB will by symmetry, vibrate with the same fundamental frequency when P divides AB in the ratio

a : b or in the ratio b : a.




10/33


31. (A), (B), (C)

Cut-off wavelength 0 250nml =

Threshold frequency

8

0 9

0

3 10

250 10

cv Hz

l-

= =

151.2 10 Hz=

Work function of the metal0

1242

250

hc eV nmW

l= =

4.968 eV=

max

0

hc hcK

l l= +

max

12424.968 7.432 7.4

100

eV nmK eV eV eV

nm= - = ;

Potoelectric effect takes place only for light of wavelength less than 250 nm.

32.. (B), (C), (D)

Acceleration1 ( )dU x

am dx

-=

21 16(2 2) 16( 1) / 2

x x m s-

= - = - -

The particle executes S.H.M.

2 216 2 / 4 2

T sp p

w p w= = = =

At 1 , ( ) 0 ( . .) x m F x i e= = corresponds to equilibrium position

2 / 0.5 A m s A mw = =

The particle describes oscillatory motion from1 0.5x m= to 2 1.5x m= .

33 (A), (B)

2.00 0.01 2.00 0.5% I A A= =

100 0.2 100 0.2%R = W= W

Power dissipation 2 2(2.00) (100) 400P I R W = = =

Limiting error in the power dissipation

2P I R=




11/33


Mock Test 1 / Pg 11

2% (2 0.5)% (0.2)%

P I R

P I R

D D D= + = +

1.2%=

Power dissipation 400 1.2% 400 4.8W W= = .

34.. ANS : 11) B 12) C 13) C

Both reactions R and S must through the centre O. Hence centre of gravity of rod must lie vertically below O.

Drop perpendicular from O to AB.

^ ^

BO D A O D= = a

^ ^

D O G 90 D G O= - = q

Hencea AG AD GD

b GB BD GD

-= =

+

ODtanAOD ODtanGOD

ODtanBOD ODtanGOD

-=

+

a tan tan

b tan tan

a - q=

a + q

b a

tan tanb a

- \ q = a +

By lamis theorem

^ ^ ^R S W

sinBOG sinAOG sinAOB

= =

( ) ( )

R S W

sin sin sin2= =

a + q a-q a

Substituting appropriate values

R 8 5kgwt=




12/33


S 16 5kgwt=

1tan 1 / 2-

q =.

35. Ans : C

36. Ans : C

(37,38,39) When 'S' is open, all the charge will reside on the outer surface only. From

(Gaus's law 0=inq )When 'S' is closed, potential of earth = 0

Let 'q1' charge appear on 'A'.

10

3A

q qV K

R R

= + =

1

/ 3q q = . If a charge 'q' is given to A, a charge of '-q' appears on the inner surface by induction fromGauss law.

40. A :2( 1)

Rf

m=

-f decreases as m is increased. The present distance of object is more than the new focal length. Hence the

nature of image will be real and image size is also decreased.

B :2( 1)

Rfm

= -f increases as R is increased. Hence the present object distance is less than the new focal length. Hence the

image is virtual and the size of image is small than the previous.

C : Due to introduction of glass slab, effective object distance decreases as there will be normal shift. The new

object distance willbe less than the focal length. Hence the image is virtual and the size of image will be less than

the previous size.

D : There will be no refraction at the left spherical surface as there is no change in .m At second surface,

2 1 2 1 1 1.5 1 1.5

2

RV u R V R

m m m m- -- = - =

--

2

5

RV =- .

As V is negative, the image is virtual and the size of image will be less than the previous value.

41. ANS : A Q, S; B P, R; C S; D P.

F4

10

F = 0.5 x 40 = 20




13/33


Mock Test 1 / Pg 13

a = 2 ms -2

F = 24 x 2 = 28 N (ii)

case : 1F = 28 N a = 2 ms -2 f

1= 20 N

case : 2

F = 14 N a = 1 ms -2 f1

= 10 N

F

4

10

case : 1

F = 28 N a = 2 ms-2

f1 = 8 Ncase : 2

F = 14 N a = 1 ms -2 f1

= 4 N

42. Solution : ANS : B

F.B.D for A

Since A just starts kx mgm=kx

mgm

F.B.D for B

F kx mg mam- - = mgm

kx F

From work energy theorem

( )2 21 1

2 2Bkx mg x F x mvm- - + =

Since we require minimum force lets substitute VB

= 0.

21

2F x kx mg xm= + .

3

2F mgm= .




14/33


43.

4040

2( 1) 2(1.5 1)L

R f cm

m

- -= = = -

- -

202

M

R f cm= + = +

If f is the equivalent focal length of the equivalent mirror,

1 2 1

L M f f f - = -

1 2 110

40 20 f cm

f- = - = +

-

\ The given silvered lens behaves as a convex lens of focal length 10 cm. O acts as a virtual object.

1 1 1 1 1 1

500 10

51

V u f V + = + =

500 5 .V cm m = - = -44. 1 10Hz approximately Hz;

As the mass m moves by a length y, the pulley will shift by y. This should bring an extension of 2y on spring so

that a length y can be shared by the spring as well the string. The restoring force on spring

2F Ky= -

Tension on string

' 2T F Ky= =

The tension on the string attached to the mass is

' 4T T F Ky= + =

2 24eq

M mT

K Kp p= =

12

4000p=

1 14000 10

2 f Hz

T p= = ;

.




15/33


Mock Test 1 / Pg 15

45.

q-

q

dq

dq

Elemental dipole moment of -q and dq is dp.

.dp dqr =

/ 2

/ 2

.q

dp rd

p

p

qp

+

-

=

2qrP

p=

Substututing the values 4P units= .

46. (D)

From wedge constraint

( ) ( )A Ba a^ ^=

0 0

cos53 cos37 AX AY a a-

0cos53Ba=

5 /Ba m s= -

5Ba j= -r




16/33


# P-1 MATHEMATICS - SOLUTIONS :

47. Solution :

Let P be (a, ). As P lies on the directrix of the parabola y2 = 4ax, the chord of contact of P i.e AB (say) will

pass through the focus. Also from the property of tangent of parabola 90PSA = .

Let Q be the reflection of P in AB, then( )

32

h aa h a

+ = =

and 0 02

kk

+ = + =

Hence, locus of P will be x =3a.

48. Let21 9

: 04 4

L y m x + =

L intersects with C at P and Q (given)

Then, the equation of any circle passing through P and Q is 0C L+ =

Now let (1)

passes through A, then A will satisfy (1) ( ) ( ) ( )2 2 21 9

3 0 4 3 0 3 04 4

m + + + =

( )21 4

21 1 04 1

mm

+ + = = +

Hence, equation of the circle passing through A, P and Q is

40

1C L

m

= + .............(2)

We have to show that B lies on (2) and for that B should satisfy (2) for all values of m for which L intersects




17/33


Mock Test 1 / Pg 17

with C

( )22 4 21 90 3 4.0 3 0 0

1 4 4m

m

+ + = +

( ) ( )4 9

9 1 0 11 4

m mm

+ = +

0 = 0

Which is true

Hence, B always lies on the circle through A, P and Q

49. Solution : (a) For the graph, we will start with y = x2 and using transformation we will get the graph of |y| + (1

|x|)2 = 5. As the graph is symmertical about x and y-axis, we can find the area in the 1st quadrant and multi-

ply it with 4 to get the required area.




18/33


As the graph is symmetrical about x and y=axis, we can find the area in the 1 st quadrant and multiply it with

4 to get the required area.

Hence, the required area, ( )( )1 5 1 5

2

0 0

4 4 5 1 A ydx x dx

+ +

= =

( ) ( ) ( )

1 84 5 1 5 5 5 1 7 5 5 .

3 3sq units

= + + = +

50. Solution :

The given equation is 4 sinA sinB + 4 sin B sinc + 4sinCsinA =g

( ) ( ) ( ) ( ) ( ) ( )2cos 2cos 2cos 2cos 2cos 2cos 9 A B A B B C B C C A C A + + + + + =

or ( ) ( ) ( ) ( )

3

2 cos cos cos 9 2 cos cos cos 9 2. 62 A B B C C A A B C + + = + + =

( ) ( ) ( )cos cos cos 3 A B B C C A + +

But ( ) ( ) ( )cos 1,cos 1,cos 1 A B B C C A

Hence (1) is possible only if

cos (AB) = 1, cos (BC) = 1, cos (CA) = 1




19/33


Mock Test 1 / Pg 19

A = B = C

The triangle ABC is equilateral

51. Solution :

( )sin 12

n x as x

(sin)n is a positive fraction close to 1 but less than 1.

12 (sin x)n < 12 (but near to 1)

[12(sin x)n] = 0 12 1

22(sin x)n


20/33


( ) ( ) ( )4 1 1 Re

2

xy x y As

+ = =

= 4x (1x) y(1y)

Now as z lies in first quadrant satisfying 1z

0 1 0 1 x and y < < < 0

53. Solution :

The given equation may be written as( ) ( )

( ) ( )

sin cos sin cos

sin cos sin cos

+ +=

+ +

( )

( )

sin2 sin 2 2 sin cos

sin2 sin 2 2 sin cos

+

=

Using componendo dividendo, we get

( )

( )

( )

sinsin2 sin cos sin cos

sin 2 2 sin cos sin cos sin

++= =




21/33


Mock Test 1 / Pg 21

( ) ( )

( )

( )

sinsin2

2sin cos sin

+ =

( ) ( )sin2 2sin cos 0 + + =

sin2 sin2 sin2 0 + + =

54. Solution :

2 2 2

2cos2 1 tan 1 1 tan 1 tansin 1 tan sin sec .sin tan

x x x xdx dx dx dx x x x x x x

= = =+

Put 2 2 21 tan 2tan sec 2 x u x x dx udu = =

2

cos2.

sin tan .tan sec

x ud udx u

x x x x =

( ) ( )

2

2 21 1 1

u du

u u= +

( ) ( )

2

2 21 2

u du

u u=

2 2

1 2

1 2du

u u

=

2 22

2 1du du

u u=

1 2 1 12. log log

2 12 2 2

u uC

uu

+ += +

1 2 1 1log log

2 12 2

u uC

uu

+ += +

where 21 tanu x= and C = a constant




22/33


55. Solution :

( ).x a x b c d + =

Multiplying vectorially by c ,

( )x a c d c =

( )c x a d c =

( ) ( ). .c x a c a x d c =

Multiplying vectorially by a ,

( ) ( ) ( ).a c x a a d c =

( )( ).

a d cx a

a c

=

Again multiplying vectorially by a ,

( )( )

( ).

a d ca x a a

a c

=

( ) ( )( )

( ). .

.

a d ca a x a x a a

a c

=

( ) ( )( )( )

2 2

.

.

a a d ca x ax

a a c a

= +

56. Solution :

Let A (3, 1), B ( ), and C (h, k) be the vertices of the triangle .Let B ( ), lie on the angle bisector of B,

We have




23/33


Mock Test 1 / Pg 23

4 10 0 = + = .............(1)

Also mid-point of AB is3 1

,2 2

+

which lies on the median

( ) ( )3 3 5 1 59 0 + + =

or 3 5 55 0 + = ...........(2)

From (1) and (2), we find that 10, 5 = =

Now (h, k) lies on the median 6h + 10 k = 59 .............(3)

Also the line x 4y + 10 = 0 bisects angle 2B = (say)

( )

6 1 1 517 4 4 10tan 2 9 65

6 521 128 4 10

k

h h kk

h

= = = + =

+ +

...............(4)

From (3) and (4), we get h = 7/2, k = 8

Hence the equation of the three sides of the triangle are 6x 7y 25 = 0,

2x + 9y 65 = 0,

and 18x + 13y 41 = 0




24/33


Passage - I (Q.NO 57-59)

a) From the graph, it is clear that f (x) is continuous every where in the given interval

And clearly, f(x) is not differentiable at x = 1, 1

For x = 0

( ) [ ]0 0

' | 2 2 2x x

f x x = == + = and

( ) [ ]0 0

' | 2 2 2x x

f x x+ += == =

Hence f (x) is differentiable at x = 0

Clearly ( ) max| 1 1 f x at x= =

b) ( )( )

( ) ( )

( )( ) ( ) ( )

( ) ( )( ) ( )

( ) ( )

2

2

, 2 1

2 , 1 0

2 , 0 1

2 , 1 2

f x f x

f x f x f x f f x

f x f x f x

f x f x

+ < =

<

<

( )

( ) ( )

( ) ( )

, 2 1

2 , 1 0

2 , 0 2

f x x

f x f x x

f x f x x

+ < =

<

( )( )

( )( )( ) ( )

2 2

2 2

, 2 1

2 2 2 , 1 0

2 2 2 0 1

2 2 2 , 1 2

x x

x x x x x

x x x x x

x x x

+ + + < =

+ < + <

( ) ( )

( ) ( )

( )

2

2

, 2 1

2 2 2 , 1 0

2 2 2 , 0 1

2 , 1 2

x x

x x x x x

x x x x x

x x x

+ + + < =

+ < <

for continuity f (f(1)) = f(f (1+)) = 1




25/33


Mock Test 1 / Pg 25

f (f (0)) = f (f (0+)) = 0

f (f (1)) = f(f (1+)) = 1

Hence f (f (x)) is continuous every where in the given interval

Passage - II (Q.NO : 60 - 62)

Let P be (x, y) and QR be the tangent. Then equation of QR

( )dy

Y y X xdx

=

For R : Y = 0

: .0y

R xdy

dx

For Q : X = 0

: 0,dy

Q y xdx

dyx y

ydx AR xdy dy

dx dx

= =

anddy dy

BQ y x y xdx dx

= = . Given,3

2Area (OAPB) = 2Area ( )PAR + Area

( )PBQ

3 1 12 ..............(1)

2 2 2

y dy xy y x x

dy dx

dx

= +




26/33


2

2 23 2dy dy

xy y xdx dx

=

all three terms in (1) are positive as 0dy

dx< and x > 0, y > 0

2

2 23 2 0dy dy

x xy ydx dx

+ + =

2 0dy dy

x y x ydx dx

+ + =

Now if 0dy

x ydx

+ =

0dy dx

xy k y x

+ = =

9xy = ...........(2)

(as curve passes through (3, 3))

And if 2 0dy

x ydx

+ =

22 0dy dx

yx k y x

+ = =

( )2 27 ......... 3yx =

(b) Minimum distance of P on (2) is 3 2 corresponding to the point (3, 3) as (2) is a rectangular

hyperbola

For the minimum distance on (3), we can minimize

D = x2 + y2

227

yy= +




27/33


Mock Test 1 / Pg 27

For maxima and minima, 227

0 2dD

ydy y

= = +

3

3

2y =

and for

2

1/3 2

3, 0

2

d Dy

dy= > minima

Now, if 1/33

2y =

2 1/3 1 /327 2 9 23

x = =

Hence, the minimum distance of P

1/3

2 / 3 1/3

9 3 39 2

2 2d D units= = + =

63. Solution :

The letters are BEMNRU

Starting with B, 5! = 120 words

Starting with E, 5! = 120 wrods

Starting with M, 5! = 120 words

Starting with NB, 4! = 24 words

Starting NE, 4! = 24 words

Starting with NM, 4! = 24 words

Starting NR, 4! = 24 words

Starting N B, 3! = 6 words

Starting N E, 3! = 6 words

Number = 1 word

The desired rank is 120 3 24 4 2 6 1 469 + + + =

ii) R : Primes : 2, 3, 5

aaa 3

3

1

=




28/33


aaab 3

2 61

=

aabb 3

32

=

aabc 3

3 33

=

Total : 15

iii) Q : 100! = 297

. 524

..... = 1024

= 10012 ...... n = 12

iv) P : aaaa 9

a, b 0, a b :

aaab 9.8 = 72

aabb 6.9.8 = 432

0, 0a

:

a000, aa00, a00a, a0a0,

aaa0, aa0a, a0aa 7 9 63 =

Total : 9 + 72 + 432 + 63 = 576

Ans : A - s, B r, C q, D p

64. Solution :

A - r (adj A)1 =A

A

B - p A.adj A = A A I =

Taking determinants, |adj A| = |A|2

C - s : (adj adj A)1 =1A

A

D - q : adj A.adj adj A =2

adj A I A I =




29/33


Mock Test 1 / Pg 29

65. Ans : 2

Let A1

be the event that ball drawn from D is blue.

Then as min {m, n, p} 7 and only 7 balls are withdrawn, in every throw, the transferring of a ball fromA,B or C to D is equally likely.

Let Brbe the event that D contains r number of blue balls.

Then ( )1/7

r

rP A B =

and ( )

7

7 1 23 3

r r

r rP B C

=

( ) ( ) ( )7

1 1

0

. /r rr

P A P B P A R=

=

( )77

7

1

0

1 2

3 3 7

r r

r

r

rP A C

=

=

77 7

70

12 .

7.3

r

r

r

C r

=

=

................(1)

Now ( )7

7 7

0

1r

r

r

N x C x=

+ =

Differentiating with respect to x

( )7

6 7 1

0

7 1 rrr

x r C x =

+ = ...................(2)

Putting1

2x =

6 7 77 1 7 7 6

0 0

37 2 2 7.3

2

r r

r r

r r

r C r C +

= =

= =




30/33


From (1)

( ) ( )61 71

7.37.3

P A =

( )11

3P A =

Now let C be the event that D contains balls of all colours and equal number of red and green balls . Then

( )( ) ( ) ( )

2 22

1 7 7 7

7! 3 7! 57!. .7 72! 3! 1! 5!3! 1! 1

3 7 3 3P C A

= + +

7 7 7

7! 1 1 1 6! 14 140.

7.3 36 8 24 3 72 3

= + + = =

Hence the required probability ( )( )

( )

71

1

1

140

140 1403/ .

1 3 729

3

P C AP C A

P A6

= = = =

66. Use cosine rule in ABE

2 2

2

2 3cos

2 5

a b

a

=

2

2

1 31 cos

2 5

b

a

=

2

2

32 1 cos

5

b

a

=

Now let3

2 3 3

5

= =




31/33


Mock Test 1 / Pg 31

2sin2 sin3 2cos 4 cos 1 = =

2 4 16 1 5cos

8 4

+ = =

But3

5

is in IInd quadrant.

2

2

3 1 5 1 5 3 5cos 2 1

5 4 4 2

b

a

+ = = =

( )

2 2

2 2

3 5 2 9 5 6 5 43

2 3 5 2 3 5

a b

b a

+ + + ++ = + = =

+ +

67. Solution:

( ) ( ) ( )2 3

2 23 4 1 4 f x x x= + +

(given)

24 , Let t x= Clearly 0 2t

( ) ( ) ( )2 3

3 1F t t t = + +

for maxima and minima

( )' 0F t =

( ) ( )2

2 3 3 1 0t t + + =

23 8 3 0t t + =

( ) ( )3 1 3 0t t + =

1

3, 3t =




32/33


Also, ( ) 3'' | 10tF t = = maxima

and ( ) 1/ 3'' | 10tF t = = minima

as 3t hence, maximum value of F (t) will occur at the end points for which

F (0) = 10

F (2) = 28

Hence, maximum value of F (t) = 28 for t = 2

maximum value of f (x) = 28 for x = 0

68. Solution :

Let the given lines be OP and OQ and the point of intersection at P and Q be R : (h, k), then equation of PQ :

yk = 2(x + h)

For OP and OQ

y2 = 4x . 1

2 24

2

yk xy x

h

=

2 2

8 2 4 0 x hy xyk + = ..............(1)Given equation of OP and OQ is

2 25 3 0 x y xy+ + = ...............(2)

Now (1) and (2) represent the same lines

8 2 4

5 3

h k

= =

12

5h =

Hence, the required locus is 5x 12 = 0

69. Let a,b,c be the sides of the triangle so that .2

a cb

+= If a is the smallest side, then angle opposite this side

is smallest. If a is the smallest, then c is the largest side

angle opposite this side is greatest. Hence




33/33


Mock Test 1 / Pg 33

( )( ) ( )2 sin sin 2sin 2sina c b + = + = + = +

or 2sin cos 4sin cos2 2 2 2

+ + ==

( )cos 2cos ............ 12 2

+ =

Multiply both sides of (1) by 2cos ,2

so that

22cos 4cos cos2 2 2

=

( ) ( ) ( )1 cos 2 cos cos ............... 2 + = +

Now, multiply both side of (1) by 2cos2

+so that

22cos cos 4cos2 2 2

+ +=

( ) ( )cos cos 2(1 cos ........... 3 + = + +

From (2) and (3) we have

( ) ( ) ( )2 1 cos 2 1 cos 5 cos cos + + + + = +

or ( ) ( )4 1 cos cos 5 cos cos + = +

[ ]4 1 cos cos cos cos cos cos + = +

Hence ( )( )4 1 cos 1 cos cos cos = +


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