SECTION - IStraight Objective Type

This section contains 5 multiple choice questions. Each question has choices (A), (B), (C) and (D), outof which ONLY ONE is correct.

1. A solid insulating sphere of radius a carries a net positive charge 3Q, uniformly distributed throughout itsvolume. Concentric with this sphere is a conducting spherical shell with inner radius b and outer radius c,and having a net charge �Q, as shown in Figure. Electric field varies with distance r from the centre as :

(04

1K

)

-Q

3Q

a

b

c

(A*) (B)

(C) (D)

Ans. (a)Sol. For r < a E r b < r < c E = 0

a < r < b E 2r

1c < r E 2r

1.

2. Flux () as a function of distance (r) from centre of, uniformly charged solid sphere of charge Q and radius R,is best represented by

(A) (B)

(C*) (D)

Ans. (c)

Sol. for inside r < R

33

0

0

Qin Q.r

R

for r > R 0 0

Qin Q

3. Figure shows a uniform and positively charged hemispherical shell. A positive point charge q0 is shifted

slowly from A to C via B , then select correct alternative :

(A*) work done by external agent from A to B is more as compared to work from B to C(B) work done by external agent from A to B is less as compared to work from B to C(C) work done by external agent from A to B is equal to work from B to C(D) Nothing can be concluded with the given information.

Sol. Direction of electric field is perpendicular to surface of circular base, so work done by external agent from Bto C is zero.

4. The potential difference across 8 resistance is 48V in the circuit shown in figure. The value of appliedpotential difference across x and y point will be :

(A*) 160 V (B) 128 V (C) 80 V (D) 62 VSol. R

eq = 10 + 6 + 3 + 1 = 20

Current through 8

=8

48 = 6A

Current through 24

=2448

= 2A

Total current = 6 + 2 = 8APotential difference across x and y = 8 × 20 = 160 V.

5. Three similar cells, each of emf 2V and internal resistance r send the same current through an externalresistance of 2, when connected in series or in parallel. Then the magnitude of current flowing through theexternal resistance is :(A*) 0.75 A (B) 1 A (C) 1.5 A (D) zero

Sol. In series = 2r323

In parallel = 23/r

2

but2r3

6

= 6r

6

r = 2

then = 223

23

= 0.75 Amp.

SECTION - IIMultiple Correct Answers Type

This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A),(B), (C) and (D), out of which ONE OR MORE is/are correct.

6. Two point charges are placed at a and b at a certain distance from each other. Assuming the field strength is

positive in the direction coinciding with the positive direction of the x axis 1 2

(A) Charge at a is positive and at b negative

(B*) Magnitude of charge at a is greater than that of charge at b

(C*) Both charges at a and b are negative

(D) Magnitude of charge at b is greater than that of charge at a

Ans. (bc)

Sol. 21

KQ

= 22

bKQ

1 >

2So, Q

a > Q

b .

7. As the constant current enters into a resistor, power dissipiated in resistor is I2R, then for the given figure:

I R

(A*) resistor remains neutral, as equal charge flow in and out.(B) charge flow, in and out are not same(C*) rate of decrease in potential energy of charge entering at point and exit from other

appears as I2R.(D) potential energy always increases, while charge flow in or out of wire.

Ans. (ac)Sol. l

in = i

out .

8. In the given circuit the point A is 9V higher than point B.

(A) R = 1(B*) R = 7(C*) Potential difference between B and D is 30 V.(D) Potential difference between B and C is 15 V.

Ans. (bc)

Sol.24 15 6

IR 1 2 1

A BV V 6 Ir

339 6

R 4

R 7

BCV 15 3 2 9V

BDV 30V

9. A solid cuboid is made of homogeneous material with shortest edge length is 2

rd3

of the longest one, If Rmax

,

Rmin

are the maximum and minimum resistances b/w two parallel faces and Imax

and Imin

are maximum &minimum current, when potential difference V is applied b/w parallel faces each. Then :

(A*)max

min

R 9R 4

(B)min

max

R 2R 3

(C)max

min

I 2I 3

(D*)min

max

I 4I 9

Ans. (ad)

Sol. Let 2

,b,3

are sides

Then max.

R2

b.3

max

min

R 9R 4

min

2.

3Rb

max min

min max

I R 4I R 9

10. A spherical planet has uniform density ��. The minimum time period Tmin

for a satellite in orbit around it,

(A) depends on mass of planet (B*) Independent of radius of planet

(C*) depends on density of planet (D) depends on radius of planet

Ans. (bc)

Sol. For Minimum time period radius of orbit equals �R� of planet

22

2min

GMm mV m 2 RR R TR

2 2

2

R.4 RGM

T

2min

33

T

R4R

34

..G

2 3T

G

min

3T

G

SECTION - IIIInteger Answer Type

This section contains 10 questions. The answer to each of the questionsis a single digit integer, ranging from 0 to 9. The appropriate bubble belowthe respective question number in the ORS have to be darkened.

11. The electric field in a region is radially outwards with magnitude E = r/. In a sphere of radius R centered

at the origin, calculate the value of charge in coulombs if

5

V/m2 and m103

R3/1

.

Ans. 6

Sol. q = ds.E0

= 2

00 r4

r

= 4r3 = 6.

12. A charged particle of charge Q and mass m is projected towards centre of fixed uniformly charged annulardisk as shown in figure. Disk is fixed. The minimum speed of particle so that it can reach the centre of disc,

from infinity, is 0

QRx

m

then x is : ( = surface charge density, inner radius is 2R, outer radius is 6R)

u

Ans. x = 2

Sol.2

0 0

1 6R. .2Rmu 0 0 .Q

2 2 2

2

0

1 .4Rmu .Q

2 2

0 0

4 QR QRu 2

m m

x = 2 Ans.

13. The given circuit consists of four resistors R1,R

2,R

3,R

4 and an ideal battery of 100 V. the direction of current

in circuit is shown and is 4A in magnitude. The potentials of points x,y and z are 48 V, 12 V and �20 V

respectively. Find value of 42

31

RR

RR15.

Ans. 8

Sol. R1 =

i4880

= i

32

R2 =

i1248

= i

36

R3 =

i012

= i

12

R4 =

i)20(0

= i

20

42

31

RR

RR15 =

2036123215

= 8.

14. If earth has uniform density, and radius �R�. The value of acceleration due to gravity at distance d above

the surface is same as acceleration due to gravity at distance d below the surface. If d = R2

1x

,

then find x.Ans. 5

Sol.

2

h 2

Rg g.

R d

h = d

d 0d

g g 1R

2

2

R d1

RR d

22R .R R d R d

3 2 2R R d 2dR R d

3 3 2 2 2 3 2R R Rd 2dR dR d 2d R

2 2 3d R dR d

2 2d d.R R 0

5 1d R

2

15. A thin uniform wire AB of length 50 cm and resistance 1 is connected to the terminals of a battery of emf

1 = 2.2 V and internal resistance 0.1. If the terminals of another cell (assume ideal) are connected to two

points 25 cm apart on the wire AB without altering the current in the wire AB, then emf 2 of cell in volts is :

Ans. x = 1

Sol. current 1 2.2i 2A

r R 0.1 1

resistance of 25 cm wire (25 cm) 25 1

150 2

so emf 21

2A 1V2

16. Two conducting fixed spheres of radii R and 2R are separated by distance r (r >>R) initially, and having equalcharges Qeach. Now they are connected to each other by conducting wire and then disconnected. If �F� was

the magnitude of electrostatic force between them before connecting and xF/9 is the magnitude of electro-static force between them after disconnecting then find x.

Ans. 8

Sol. Q1, =

3Q2

QR3R2

Q2, =

3Q4

QR3R4

2

2

kQF

r ---- (1)

F' = 2r3Q4

.3Q2

.k = 2

2

r9

kQ8

17. A dipole 3

)j�2i�(P

10�9 C m is placed at origin. Calculate potential in volt at a point having coordinate

(1m , 2 m)

Ans. 3

Sol.)33( )3(

10)21(109 9�9 = 3 V..

18. The system of four point charges are arranged to be in equilibrium (see figure) under only mutualelectrostatic forces. Find magnitude of Q(in C). (Neglect gravity).

Ans. 1

Sol.

F2 � 2F

1 cos 30º = 0

2R

)Q()q(K = 2

2

)R3(

)Kq(2 cos 30º

Q = 3

q =

3

3C = 1 C.

19. Find the equivalent capacitance (in F) of the circuit between the points �A� and �B�. All capacitors are in

microfarad.

Ans. 2

Sol.

by wheat stone balanced bridge method

32

+ 34

= 2 F

Ans. 2 F20. 10 identical capacitors are connected as shown. The capacitance of each capacitor is 30 F. If the equivalent

capacitance between A and B is 10y F, then find y :

A B

Ans. 6Sol. (Easy) The equivalent circuit is

CAB

= 60 F

A B

Page # 2

Paper - 2

SECTION - IStraight Objective Type

This section contains 5 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of whichONLY ONE is correct.

21. The density of an unknown gas X is 3

80 g/L at 600 K and 100 atm. The rate of diffusion of gas X is 0.25 times rate

of diffusion of Helium under identical conditions. Which of the following conclusions is incorrect for gas X ?(A) Molar mass of gas X is 64.(B*) The gas X behaves ideally under the given conditions.(C) The gas X is showing positive deviation from ideal gas behaviour.(D) The gas X diffuses faster than SO

3 under identical conditions.

Sol.He

x

rr

= 41

= xM

4

161

= xM

4 Mx = 64

Z = dRTPM

600

121

380

64100

= 4.8 > 1

(Real gas and positive deviation)

22. (I) (II)

The graphs I and II are plotted for same real gas for 1 mole at constant T. The dashed line represent ideal gas in(II) graph. Which of the following is correctly matched ?(A) X C (B) Y b (C) Z a (D*) None of these

23. In which of the following reaction H < 0 (exothermic)(A) Ne(g) + e Ne� (g) (B) Na(g) Na+ (g) + e�

(C) O�(g) + e� O� � (g) (D*) Mg++(g) + e Mg+ (g)

Organic SCQ (2)

24. X OH,Zn/O 23 H�C�CH�C�CH�C�H

||||||OOO

22 + HCHO

The structure of X will be :

(A) (B*) (C) (D)

25. An alkene having molecular formula C6H12 gives a hydrocarbon (x) C6H14 on catalytic hydrogenation and compound(x) gives two monochloro structural isomeric product (y) and (z). The structure of alkene is :

Page # 3

(A) CH3�CH2�CH2�CH2�CH=CH2 (B)

3

23

3

CH|

CHCH�C�CH|CH

(C) CH3� C|CH3

CH�CH2�CH3 (D*)

SECTION - IIMultiple Correct Answer Type

This section contains 5 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of whichONE OR MORE is/are correct.

MCQ (3)26. The equilibrium constant 'K

p' of the following reaction

N2 + O

2(g) 2NO(g)

is 3 × 10�5 at TK. When the temperature is changed to 4 T K, the equilibrium constant 'Kp' doubles. Which of the

following statement is/are correct ?(A*) The reaction is endothermic(B) The reaction shifts in forward direction on addition of He(g)(C*) If the volume of container is constant, then the partial pressure of NO at 4T K is higher as compared to atTK, under equilibrium condition.

(D*) 2 log )Plog()Plog(P22 ONNo at equilibrium

27. Which of the following statement is/are correct ?(A*) The highest value of principle quantum number 'n' is 4 for ground state configuration of Cr.(B*) 5s, 4p and 3d have same value of (n + l) but different energy.(C*) Number of radial nodes in 4p and number of angular nodes in 3d are equal.(D*) In

29Cu+, 14 electrons have spin + 1/2

28. Which of the following statements is/are correct for a real gas at a temperature below its critical temperature ?(A*) The gas can be liquified on applying pressure at constant temperature(B) The compression curve of the gas (P vs V) has negative slope only

(C*) In low pressure region, the gas (vapor) can be described by the vander Waal's equation Z = 1� mRTV

a.

(D) The P vs V graph resembles a hyperbola. (like that of an ideal gas)

Organic MCQ (2)29. In which of the following pairs of first species is more stable than second.

(A*) , (B*) ,

(C) , (D) ,

Page # 4

Sol. �ve charge is more stable on more electronegative atom and +ve charge should on less electronegative atom.

Species with complete octet around each atom in reasonating structure is more stable. Linear conjugation ismore stable than cross conjugation.

30. Which of the following statements are correct ?(A*) Inductive effect causes permanent polarity in the molecule.(B*) Free radicals have odd electron in a p-orbital.(C) All C � C bond lengths are same in Buta-1,3-diene.

(D*) Hyperconjugation of an alkyl group is an electron releasing effect.Sol. All C � C bond lengths are not same in Buta-1,3-diene.

CH =CH�CH=CH2 2 CH �CH=CH�CH2 2 CH �CH�CH�CH2 2

( )I ( )II

+ �

(I) and (II) are resonating structure and (I) is more stable than (II). So, (I) is more contributing.

SECTION - IIIInteger Answer Type

This section contains 10 questions. The answer to each of the questions is asingle digit integer, ranging from 0 to 9. The appropriate bubble below therespective question number in the ORS have to be darkened.

SS (Single Integer) (7)31. In the reaction :

K4[Fe(CN)

6] + KMnO

4 + H

2SO

4 K

2SO

4 + MnSO

4 + CO

2 + NO + Fe

2 (SO

4)

3, the number of elements

undergoing change in oxidation number is / are :Ans. 4

Sol. 32FeFe

32

)CN( 42CO

+

2NO

74KMnO

6

4MnSO

32. The molarity of aqueous CaCl2 solution is 0.35 mol/L. What volume should be taken (in mL) to dispense 0.14

moles of chloride ions ? Give answer by dividing by 100.Ans. 2

Sol. 0.35 × 2 = V14.0

V = 7.0

14.0 =

51

L

= 200 ml.

33. By how many of the following ways, the concentration of CO2 can be increased at equilibrium ?

CO(g) + H2O(g) CO

2(g) + H

2(g) H = + ve

(a) By increasing temperature (V = constant)(b) By decreasing volume (T = constant)(c) By increasing volume (T = constant)(d) By adding more CO(g) (V, T = constant)(e) By removing H

2(g) (V, T = constant)

Page # 5

(f) By adding inert gas (P, T = constant)(g) By adding inert gas (V, T = constant)

Ans. 4Sol. (a, b, d, e)

34. The t1/2

of a zero order reaction is 4s when initial concentration of A is 8 mol/L. The concentration of B after 2 secfrom the start of reaction is : (in mol / L)

Given reaction : 2A B.Ans. 1Sol. C

t = C

O � kt

= 8 � 42

8

× 2

= 8 � 2

= 6 mol / L.Concentration of A reacted 2 mol / L.Concentration of B reacted 1 mol / L.

35. A 10 M solution of H2C

2O

4 was prepared, having 45% w/w of H

2C

2O

4. Find specific gravity of H

2C

2O

4.

Ans. 2

Sol.90

10d45 = 10

d = 2.

36. Total minimum moles of Fe and O2 required to produce equal integral moles of FeO and Fe

2O

3 are :

Ans. 5

Sol. Fe + O2

FeO + Fe2O

3

x y a aPOAC on Fe POAC on Ox = a + 2a 2y = a + 3ax = 3a y = 2ax + y = 5a ; for minimum, a = 1 ; x + y = 5.

37. The total spins of 25

Mnx+, 23

Vy+ and 24

Crz+ are ± 5/2, ±1 and ±23

respectively. The value of x + y + z is :

Ans. 8Sol.

25Mn2+,3d5 4s0

;

23V3+,3d2 4s0

;

24Cr3+,3d3 4s0.

Organic SS (Single Integer) (3)38. How many groups (each attached with benzene ring) show + M effect?

Ans. 5

Page # 6

Sol. have + M group.

39. Number of electrons in resonance in the following structure is.

Ans. 8

40. Number of structural monochloro products of the following hydrocarbon.

Ans. 5

SECTION - IStraight Objective Type

This section contains 5 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of whichONLY ONE is correct.

41. Domain of the function f(x) =

2x3xsinlog 21 is

(A) (�, �2) (�1, ) (B)

253

,2

53

(C*)

2,

253

253

,1 (D) none of these

Sol. f(x) = log

2x3xsin 21�

0 < x2 + 3x + 2 1 (x + 1) (x + 2) > 0 and x2 + 3x + 1 0 x (� , � 2) (� 1, ) and x3 + 3x + 1 0

x (� , � 2) (� 1, ) and

253�

,2

5�3�

x

2,�

25�3�

253�

,1�

42. sin�1sin3 + cos�1 cos5 + cot�1 cot 43

=

(A) 3 + 421

(B) + 43

(C*) 3 � 4

29(D) none of these

Sol. ( � 3) + (2 � 5) + 43

= 3 � 4

29

43. The sum of the series tan�1

42 111

2 + tan�1

42 221

4 + tan�1

42 331

6 + ....infinite terms, is

(A) 2 (B) (C*) 2

(D) 4

Sol. General term is

tan�1 42 rr1

r2

= tan�1

)1r(r1

r222

= tan�1 )rr)(rr(1

)rr()rr(22

22

= tan�1 (r2 + r) � tan�1 (r2 � r)

tan�1 (12 + 1) � tan�1 (12 � 1) + tan�1 6 � tan�1 2 + tan�1 12 � tan�1 6 + ......

= tan�1 = 2

44. The complete solution set of the in-equation x + 18 < x2 , is

(A*) (�, �14) (B) (�, �18) (C) (�18, �14) (D) (�, 2)

Sol. x2 > x + 18 ..... (i)

2 � x 0 x 2 ..... (ii)Case I x + 18 0 x �18 ..... (iii)equation (i) always holds

x (�, �18]

Case II x + 18 0 x �18 ..... (iv)2 � x > x2 + 36x + 324x2 + 37x + 322 < 0(x + 14) (x + 23) < 0x (�23, �14) ..... (v)

from equation (ii), (iv) & (v), we getsolution of equation (i) x (�, �14)

45. System of equations x + 3y + 2z = 6,x + y + 2z = 7,x + 3y + 2z = , has infinitely many solutions, if

(A) = 2, 6 (B*) = 4, = 6 (C) = 5, = 7 (D) = 3, = 5

Sol. x + 3y + 2z = 6 ............(i)x + y + 2z = 7 ............(ii)x + 3y + 2z = ............(iii)If = 4, = 6

x + 3y = 6 � 2z

x + 4y = 7 � 2z

y = 1 and x = 3 � 2z

substituting in equation (iii)3 � 2z + 3 + 2z = 6 is satisfied

infinite solutions

SECTION - IIMultiple Correct Answer Type

This section contains 5 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of whichONE OR MORE is/are correct.

46. If A and B are different matrices satisfying A3 = B3 and A2B = B2A, then(A*) det(A2 + B2) is necessarily equal to zero.(B*) det(A � B) may be zero.

(C*) det(A2 + B2) as well as det(A � B) may be zero.

(D) exactly one of det(A2 + B2), det(A � B) is zero.

Sol. (ABC)Given that A3 = B3 ...(i)and A2B = B2A ...(ii)On subtracting Eqs. (i) and (ii), we get

A3 � A2B = B3 � B2A A2(A � B) = � B2(A � B)

(A2 + B2)(A � B) = 0

det(A2 + B2) = 0 or det (A � B)= 0. A�B 0

47. Let a and c be odd prime numbers and b be an integer. If the quadratic equation ax2 + bx + c = 0 has rationalroots, then roots of equation are

(A) 1 (B*) �1 (C*) ac

� (D) ac

Sol. Since ax2 + bx + c = 0 has rational rootsb2 � 4ac is square of an integer

i.e. b2 � 4ac = k2 for some integer k b2 � k2 = 4ac (b � k) (b + k) = 4ac ... (1)Since b and k are integer, therefore, (b � k) and (b + k) are either both even.

since 4ac is even. b � k = 2a and b + k = 2c or{ b � k = 2c and b + k = 2a}

on adding, 2b = 2a + 2ci.e. a � b + c = 0

x = � 1 is a root of ax2 + bx + c = 0

Product of roots = ac

� 1 × other root = ac

other root = � ac

Roots are � 1 & ac

�

48. If cos x + cos y = a, cos 2x + cos 2y = b, cos 3x + cos 3y = c, then

(A*) cos2x + cos2y = 1 + 2b

(B*) cos x cos y = 2

a2

�

42b

(C*) 2a3 + c = 3a (1 + b) (D) a + b + c = 3abcSol. (cos x + cos y)2 = a2

cos2x + cos2y + 2 cos x cos y = a2 .....(i)cos 2x + cos 2y = b2 cos2x � 1 + 2 cos2y � 1 = b

2[cos2x + cos2y] = b + 2 ......(ii)

cos2x + cos2y = 2b

+ 1

From (i) and (ii),

2 cos x cos y = a2 �

22b

cos x cos y = 2

a2

�

42b

cos 3x + cos 3y = c4 cos3x � 3 cos x + 4 cos3y � 3 cos y = c

4[cos3x + cos3y] � 3 [cos x + cos y] = c

4[(cos x + cos y) (cos2x + cos2y � cos x cos y)] � 3 (cos x + cos y) = c

4

22b

�a21

�2

2ba 2

� 3a = c

2ab + 4a � 2a3 + ab + 2a = 3a + c 2a3 + c = 3a(1 + b).

49. Which of the following is a rational number ?

(A*) sin

31

tan3tan 11 (B*) cos

43

sin2

1

(C*) log2

863

sin41

sin 1 (D) tan

35

cos21 1

Sol. (A) sin

31

tan3tan 11 = sin 2

= 1

(B) cos

43

sin2

1 = cos

43

cos 1 = 43

(C) sin

863

sin41 1

Let sin�1 863

= . Then 0 < < 2

and sin = 863

cos = 81

we have cos 2

= 43

2cos1

sin 4

= 2

2cos�1

= 22

1

Now log2 sin

863

sin41 1 = log

2

22

1 = �

23

(D) Let cos�1 35

= Then 0 < < 2

and cos = 35

tan 2

= 2

5�3 which is irrational

50. The domain of f(x) = ]2[log

1

x contains the set (where [] is greatest integer function)

(A) (1, 3) (B*) (0,1) (C*) (1, 2] (D) [2, )

Sol. f(x) = ]2[log

1

x

for 2logx to be defined, x > 0, x 1

for f(x) to be defined [ 2logx ] 0

2logx [0, 1)

logx2 < 0 or logx2 1Hence domain is (0, 1) (1, 2]

SECTION - IIIInteger Answer Type

This section contains 10 questions. The answer to each of the questions is asingle digit integer, ranging from 0 to 9. The appropriate bubble below therespective question number in the ORS have to be darkened.

51. If a function f(x) = ax3 + bx2 + cx + 1, where a, b and c are integers and a > 0 is such that

f sin18

= 0, then find the value of f(1).

Ans. 3

Sol. sin18

= sin 10°, sin 30° = 12

Also , sin 30° = 3sin 10° � 4 sin3 10°

12

= 3sin10° � 4sin310°

8 sin3 10° + 0 sin210° � 6sin 10° + 1 = 0 ...(i)

Given f(sin 10°) = 0

a sin3 10° + bsin2 10° + c sin 10° + 1 = 0...(ii)

On comparing Eqs. (i) and (ii), we geta = 8, b = 0, c = � 6

Hence f(1) = a + b + c + 1f(1) = 3.

52. If A is 6 × 6 matrix and )A|A(|adj|A| = |A|n , then (n�1)/8 is

Ans. 5Sol. A is square matrix of order 6

|adj A| = |A|5

|adj (|A| A)| = ||A| A|5

|adj (|A| A)| = (|A|6 |A|)5 { |A| = 6 |A|} |adj |A| A| = |A|35

Now, ||A| adj |A| A| = |A|6 |adj |A| A|= |A|6 |A|35 = |A|41

n = 41

53. If 56 5log.alog.xlog a10aa �

10x

log10

3 = 2logxlog 41009 , then find the value of x/20.

Ans. 5

Sol. )1�x(logalog

5log.alog

1010

1010

3�)x(.56

= 21

xlog21

109

56

5log10x � 3

3 xlog10

= )3( xlog10 . 3

56

5log10x = 3

10 . xlog103

xlog105 = 9

25 xlog103

xlog10

35

=

2

35

log10x = 2

x = 100 20x

= 5

54. If the both roots of the equation x2 � 6kx + 9k2 � 2k + 2 = 0 exceed 3 for k

,

9 , then find the smallest value

of � 5.

Ans. 6Sol. Let f(x) = x2 � 6kx + 9k2 � 2k + 2 as both roots of f(x) = 0 are greater than 3, we can take D 0, if (3) > 0

and � a2

b > 3.

(i) Consider D 0 : (�6k)2 � 4 (9k2 � 2k + 2) 0 8k � 8 0 k 1 i.e. k [1, )......(i)

(ii) f (3) > 0 (9 � 18k + 9k2 � 2k + 2) > 0

9k2 � 20k + 11 > 0

(9k � 11) (k � 1) > 0

9

11�k (k � 1) > 0

k (�, 1)

,

911

, ......(ii)

(iii)2k6

> 3

k > 1 k (1, ) ........(iii)

From (i), (ii) and (iii), we get k

,

911

55. The number of solutions of y = ex and y = | n |x| | is / areAns. 3

Sol.

Number of solutions are 3

56. Find the number of solutions of the equation 2

3xxcos1xxsin2 2121

.

Ans. 1

Sol.2

3xxcos1xxsin2 2121

is possible only when

x2 + x + 1 = 1 and x2 � x = 0

x = 0 is the only one solution

57. Given f(x) = 4 2x � 7x 9

x � (3 / x) 1

. Its zeros are of the form

a bc

, where a, b and c are smallest positive

integers. Then the value of 2

cba , is

Ans. 8

Sol. f(x) =

1x3

�x

1x3

�x1�x3

�xx

1x3

�x

7�x

9xx 2

222

=

1�

x3

�xx2

f(x) = 0, givesS x2 � x � 3 = 0

x = 2

1312

1211

a = 1, b = 13, c = 2 a + b + c = 16.

58. Range of the function f(x) = cos�1 (� {x}), where {.} is fractional part function, is

, , N, then find the value

of Ans. 2Sol. 0 {x} < 1 i.e. � 1 < � {x} 0

2

cos�1(� {x}) <

the range is

,

2

59. If ,, are roots of x3 + x + 3 = 0, then the value of

1�1 is equal to (where means continued product)

Ans. 5

Sol. Put y = x1x�1

i.e. x =

1y1�y

�

3

1y1�y

�

1y1�y

� + 3 = 0

i.e. � (y � 1)3 � (y � 1) (y + 1)2 + 3(y + 1)3 = 0

whose roots are

1�1

,

1�1

,

1�1

1�1

= 5

60. If [x] � {x} = 3x, where [ ], { } represents greatest integer function and fractional part function respectively,

then value of |6|, where is sum of values of x satisfying it, isAns. 3Sol. Case- when x equation becomes

x � 0 = 3x

x = 0Case-when x x = m + f equation becomesm � f = 3m + 3f

4f = �2m f = 4m2�

= � 2m

0 < f < 1 0 < �2m

< 1 �2 m < 0

m = � 1 and f = 21

x = � 21

now sum of values of x is equal to �21

|6| =

21

�6 = 3