1

PHYSICS

SECTION-I (MULTIPLE CHOICE QUESTIONS)

Section-I contains 20 Objective questions having 4 options each with only 1 correct option. For each question 4 Marks will be awarded for every correct answer and 1 Mark will be deducted for

every incorrect answer. 1. A stone is thrown vertically upwards from the top of a tower with a velocity u and it reaches

the ground with a velocity -3u. The height of the tower is

(A) u2/g (B) 3u2/g (C) 4u2/g (D) 6u2/g Solution (C):

2 2 2 2v u ( 3u) uh | s |

2.a 2( g)

24uh

g

2. The equation of projectile is y = 236x .x

4, where x and y are in meter. The horizontal range is

(A) 8 m (B) 10 m (C) 12 m (D) 15 m Solution (A):

23y 6x .x 0 x 0, 8m

4 Range 8m

3. A light string passing over a smooth light pulley connects two blocks of masses m1 and m2

(vertically). If the acceleration of the system is g/8, then the ratio of the masses is–

(A) 7

9 (B)

5

7 (C)

3

5 (D)

1

3

Solution (A):

1 1g

T m .g m .8

2 2g

m g T m .8

1

2

m 7

m 9

4. A ball is thrown with initial energy 100J at an angle to the horizontal. If its energy at the

topmost point of trajectory becomes 75J then angle of projection is- (A) 30º (B) 37º (C) 45o (D) 53o

Solution (A):

21mu 100J

2

2Topmost

1K m.(u.cos )

2

2100.cos 75 30

5. Bob of pendulum of length L is given horizontal velocity 7gL at lowest point. Mass of bob is

m. Tension in string when it makes angle 0o with vertically downward direction is- (A) 8mg (B) 13mg/2 (C) 5mg (D) 7mg/2

m1

m2

2

Solution (A):

07g v 20v

T mg m

6. Block A of mass m moving with velocity u has perfectly elastic head on collision with block B of mass 2m at rest. Velocity of A after collision is-

(A) u

3 (B)

u

3 (C)

2u

3 (D)

4u

3 Solution (A):

1 2m.V 2m.V m.u 1 2V 2V u

2 1 2 12u u

V V u V v3 3

7. The intensity ratio of two coherent sources of light is 4. They are interfering in some region and produce interference pattern. Then the ratio of intensity at maxima to intensity at minima is-

(A) 9

1 (B)

4

1 (C)

25

9 (D)

9

4 Solution (A):

2

max

min

I 4I I 9

I 14I I

8. A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is (A) MgL (B) MgL/3 (C) MgL/9 (D) MgL/18 Solution (D):

2 2

MgL MgL MgLW

182n 2(3) (n = 3 given)

9. Five blocks A, B, C, D and E are connected by five identical strings P,Q, R, S and T as shown frequency of P, Q, R, S and T in ratio 5 : 4 : 3 : 2 : 1. Ratio of mass of B and A is

(A) 1:3 (B) 3:5 (C) 5:7 (D) 7:9

A

B

C

D

E

P

Q

R

S

T

3

Solution (D):

n 1 n 1

n n

T

T

2n 1 n 1

n n

T

T

T m=(T/g)

P 5 25 9 Q 4 16 7 R 3 9 5 S 2 4 3 T 1 1 1 ME :MD :MC :MB :MA ::1:3:5:7:9

10. A man is sitting in a room at 2m from a wall W1, wants to see the full height of the wall W2 behind him 4 m high and 6 m away from the facing wall W1. What is the minimum vertical length of mirror on the facing wall required for the purpose?

(A) 4 m (B) 2 m (C) 3 m (D) 1 m Solution (D):

Conceptual

11. An astronomical telescope has an eyepiece of focal-length 5 cm. If the angular magnification in normal adjustment is 10, when final image is at least distance of distinct vision (25 cm) from eye piece, then angular magnification will be

(A) 10 (B) 12 (C) 50 (D) 60 Solution (B): In normal adjust

0 00

e

f f10 f 50

f 5

00

e 0 0 0

f 1 1 1 1 1f

u f D D u f

0 0

e

f f 51 10 1 12

f D 20

12. A plane mirror is moving with velocity ˆ ˆ ˆ4i 5j 8k. A point object in front of the mirror moves

with a velocity ˆ ˆ ˆ3i 4j 5k. Here k̂ is along the normal to the plane mirror and facing towards

the object. The velocity of the image is-

(A) ˆ ˆ ˆ3i 4j 5k (B) ˆ ˆ ˆ3i 4j 11k (C) ˆ ˆ ˆ3i 4j 11k (D) ˆ ˆ ˆ7i 9j 11k Solution (B): Conceptual

13. A point source of light S is placed at the bottom of a vessel containing a liquid of refractive index 5/3. A person is viewing the source from above the surface. There is an opaque disc D of radius 1 cm floating on the surface of the liquid. The centre of the disc lies vertically above the source S. The liquid from the vessel is gradually drained out through a tap. The maximum height of the liquid for which the source cannot be seen at all from above is

(A) 1.50 cm (B) 1.64 cm (C) 1.33 cm (D) 1.86 cm

4

Solution (C):

Suppose the maximum height of the liquid is h for which the source is not visible. Hence radius of the disc

2

h1 h 1.33cm

51

3

14. The diagram below shows the propagation of a wave. Which points are in same phase-

(A) F, G (B) C and E (C) B and G (D) B and F

Solution (D):

Points B and F are in same phase ass they are distance apart.

15. The potential energy between two atoms in a molecule is given by 12 6

a bU(x)

x x ; where a and

b are positive constants and x is the distance between the atoms. The atom is in stable equilibrium when

(A) 611a

x5b

(B) 6a

x2b

(C) x 0 (D) 62a

xb

Solution (D):

Condition for stable equilibrium dU

F 0dx

12 6

d a b0

dx x x

13 712ax 6bx 0

13 7

12a 6b

x x

62ax

b 6

2ax

b

16. Six identical balls are lined in a straight groove made on a horizontal frictionless surface as shown. Two similar balls each moving with a velocity v collide elastically with the row of 6 balls from left. What will happen

(A) One ball from the right rolls out with a speed 2v and the remaining balls will remain at

rest (B) Two balls from the right roll out with speed v each and the remaining balls will remain

stationary (C) All the six balls in the row will roll out with speed v/6 each and the two colliding balls will

come to rest (D) The colliding balls will come to rest and no ball rolls out from right Solution (B):

Conceptual

2

hr

1

5

17. A ball A collides elastically with another identical ball B at rest with velocity 10 m/s at an angle of 30° from the line joining their centres C1 and C2. Select the incorrect alternative. (Collision takes place in the gravity free space)

(A) Speed of ball A after collision is 5 m/s.

(B) Speed of ball B after collision is 5 3 m/s.

(C) Both the balls move at right angle after collision (D) Kinetic energy will not be conserved here because collision is not head on. Solution (D):

Velocity of ball A along and perpendicular to line of impact after collision (0,10sin30 )

Velocity of ball B along and perpendicular to line of impact before collision (0, 0) Velocity of ball B along and perpendicular to line of impact after collision (10cos30 ,0)

18. A bullet of mass m moving with velocity v strikes a suspended wooden block of mass M. If the block rises to a height h, the initial velocity of the block will be

(A) 2gh (B) M m

2ghm

(C)

m2gh

M m (D)

M m2gh

M

Solution (A):

Initial K.E. of block when bullet strikes to it 21(m M)V

2

Due to this K.E. block will rise to a height h. Its potential energy = (m M)gh.

By the law of conservation of energy

21(m M)V (m M)gh

2 V 2gh

19. A particle is moving on a circular path of radius r with uniform velocity v . The change in

velocity when the particle moves from P to Q is o( POQ 40 )

(A) o2vcos40 (B) o2v sin40 (C) o2v sin20 (D) o2vcos20

Solution (C):

Change in velocity = 2v sin( /2) 2vsin20

20. A particle moves with constant angular velocity in circular path of certain radius and is acted upon by a certain centripetal force F. If the angular velocity is doubled, keeping radius the same, the new force will be (A) 2F (B) 3F (C) 4F (D) F / 2 Solution (C):

2F m R 2F (m and R are constant)

If angular velocity is doubled force will becomes four times.

1C

2C

3010m / s

A

B

6

SECTION-II (FOR ANSWER WITH NUMERIC VALUE)

Section-II contains 5 Numerical questions with numerical value as answer. For each question 4 Marks will be awarded for every correct answer and 0 Mark will be deducted for every incorrect

answer. 21. A ball of mass 400gm is dropped from a height of 5m. A boy on the ground hits the ball

vertically upwards with a bat with an average force of 100 N so that it attains a vertical height

of 20 m. Find the time for which the ball remains in contact with the bat is 2[g 10m / s ]

Solution (0.1200):

Velocity by which the ball hits the bat

1 1v 2gh 2 10 5 .or 1v 10m/s 10m/s

velocity of rebound

2 2v 2gh 2 10 20 20m/s or 2v 20m/s

2 1m(v v )dv 0.4( 20 10)F m 100N

dt dt dt

by solving dt 0.12sec

22. A lens when placed on a plane mirror then object needle and its image coincide at 15cm. Find the focal length of the lens

Solution (15.0000): When the object is placed at focus the rays are parallel. The mirror placed normal sends them back. Hence image is formed at the object itself as illustrated in figure.

23. A convergent beam of light is incident on a convex mirror so as to converge to a distance 12 cm from the pole of the mirror. An inverted image of the same size is formed coincident with the virtual object. What is the focal length of the mirror Solution (6.0000):

Here object and image are at the same position so this position must be centre of curvature

R = 12 cm R

f2

7

T

mg

90 T

2mg

24. A mass 2m lies on a fixed, smooth cylinder. An ideal string attached to 2m passes over the cylinder to mass m. System is in equilibrium for if

1sin

n then n = ?

Solution (2.0000): T mg 0

T mg

T 2mgsin 0

mg 2mgsin 0

mg 2mgsin

1sin

2

n 2 25. With reference to the fig. shown, if the coefficient of friction at all the surfaces is 0.50, then

the force required to pull out the 6.0 kg block with an acceleration of 1.50 m/s2 will be-

F6kg

2kg

Solution (59.0000):

T f 2.0

F f f 6 1.5

max 1f N 10N

max 2f N 40N

F 10 40 6 1.5 F 59N

F

60N80N=N2

f

f

N1

8

CHEMISTRY SECTION-I

(MULTIPLE CHOICE QUESTIONS) Section-I contains 20 Objective questions having 4 options each with only 1 correct option. For each question 4 Marks will be awarded for every correct answer and 1 Mark will be deducted for every incorrect answer. 26. 7 gm of occupies 5.6 litre volume at STP. Assuming ideal gas nature, the value of x is

(A) 1 (B) 2 (C) 3 (D) None of these Solution (A):

i.e.,

27. An ammonia solution has a density of 0.910 g and is 25.0% by mass. What is the

molarity of the solution? (A) 12.1 M (B) 13.4 M (C) 14.5 M (D) 15.5 M

Solution (B):

of solution contains 325g NH

i.e.,

28. For which of the following transition energy change will be maximum

(A) 1 2n 2 to n 5 (B) 1 2n 3 to n 5

(C) 1 2n 2 to n 4 (D) 1 2n 3 to n 4

Solution (A):

From the given transitions; 2n 5 to 1n 2 has the highest energy

20 2 2

2 1

1 1E E Z

n n

29. The ionization energy of hydrogen atom (in the ground state) is x kJ. The energy required for

an electron to jump from 2nd

orbit to the 3rd

orbit will be (A) x/6 (B) 5x (C) 7.2 x (D) 5x/36 Solution (D):

I.E for H-atom 21 0E E E Z x kJ

Thus; 23 2 0

1 1E E E .Z

4 9

20E .Z 5 5x

kJ36 36

30. The quantum numbers of four electrons (e1 to e4) are given below:

n l m s e1 3 0 0 +1/2 e2 4 0 0 1/2 e3 3 2 2 -1/2 e4 3 1 -1 1/2

The correct order of decreasing energy of these electrons is:

(A) e4 e3 e2 e1 (B) e2 e3 e4 e1

(C) e3 e2 e4 e1 (D) e1 e4 e2 e3

xCO

7 5.6M 28g

M 22.4

12 x 16 28 x 1

3cm3NH

100g

25moles

1725/17

M 1000 13.4M100

0.910

9

P

V

1 2 3(T <T <T )

3T2T

1T

P

TK4 3 2 1(V <V <V <V )

4V

3V

2V

1VV

4P

3P

2P

1P

o o273 C T C

4 3 2 1(P P P P )

Solution (C):

Order of energy k l m n

(shell wise)1 2 3 4

s p d f (subshell wise)

31. At what temperature 2O molecules have same average momentum as helium molecule at

o27 C?

(A) 27 K (B) 37.5 K (C) 40 K (D) 54 K

Solution (B):

Av. Momentum gg

g

8RTM8RTM

M

2O2

He

M TAv. momentum of O 32 T1

Av. momentum of He M T 4 300

32T 4 300

1200T 37.5 K

32

32. In which of these graph the conditions indicated is incorrect? (A) (B) (C) (D)

Solution (B):

nRV T

P

; thus, greater the p lesser the slope.

33. When 0.1 M aqueous solutions of aluminium nitrate, magnesium nitrate, sodium nitrate and

urea, 2 2(NH ) CO, are arranged in order of increasing boiling point, which order is correct?

(A) 3 3 3 2 2 2 3Al(NO ) Mg(NO ) (NH ) CO NaNO

(B) 3 2 2 2 3 3 3Mg(NO ) (NH ) CO NaNO Al(NO )

(C) 2 2 3 3 2 3 3(NH ) CO NaNO Mg(NO ) Al(NO )

(D) 3 3 2 3 3 2 2NaNO Mg(NO ) Al(NO ) (NH ) CO

Solution (C):

34. Benzene and toluene forms an ideal solution. Vapour pressure of pure benzene is 100torr

while that of pure toluene is 50torr. If mole faction of benzene in liquid phase is 1

3. Then

calculate the mole fraction of benzene in vapour phase:

(A) 2

3 (B)

1

2 (C)

2

5 (D)

1

3

Solution (B):

1 2 200

P 100 503 3 3

o

Ben BenBen

x P 1y

P 2

10

35. A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statement is correct regarding the behaviour of the solution?

(A) The solution is non-ideal, showing +ve deviation from Raoult’s law (B) The solution is non-idea, showing –ve deviation from Raoult’s law (C) n-heptane shows +ve deviation while ethanol show –ve deviation from Raoult’s law (D) The solution formed is an ideal solution

Solution (A): Conceptual 36. The rate of disappearance of ammonia is 3.4 gm/litre-sec. when it dissociates to form

nitrogen and hydrogen. The rate of appearance of nitrogen will be: (A) 3.4 gm/litre sec (B) 1.7 gm/litre sec (C) 0.1 gm/litre sec (D) 2.8 gm/litre sec Solution (D):

233 2 2

d[N ]d[NH ]12NH N 3H

2 dt dt

2d[N ] 1 3.428

dt 2 17

37. For the reaction k

3A Products, the value of rate constant of reaction

3k 1 10 L/(mol min) the value of d[A]

dt in mol/L-sec when [A]=2 M is:

(A) 36.67 10 (B) 21.2 10 (C) 42 10 (D) 34 10

Solution (C):

2d [A]k[A]

3 dt

3 24d[A] 10 3 2

2 10dt 60

38. For a given reaction, energy of activation for forward reaction af(E ) is 80 kJ 1mol .

1H 40kJ mol for the reaction. A catalyst lowers afE to 20 kJ 1mol . The ratio of energy

of activation for reverse reaction before and after addition of catalyst is: (A) 1.0 (B) 0.5 (C) 1.2 (D) 2.0

Solution (D):

a a af b bE E H E 120kJ

After addition of catalyst, abE 120 60 60kJ

120 2Ratio

60 1

39. Two radioactive materials 1X and 2 X have decay constants 10 and respectively. If initially

they have the same number of nuclei, then the ratio of the number of nuclei of 1X to that of

2 X will be 1/e after a time

(A) 1/(10) (B) 1/(11) (C) 11/(10) (D) 1/(9) Solution (D):

tt 0N N e

1

10 tx 0N N e

2

tx 0N N e

1

2

x 9 t

x

N 1e

N e

t 1/9

11

40. Which of the following statements is incorrect

(A) On bombarding 147N Nuclei with -particle, the nuclei of the product formed after release

of proton would be 178O .

(B) Nuclide and it's decay product after -emission are called isobars

(C) Nuclide and it's decay product after -emission are called isodiaphers. (D) Half life of radium is 1580 years. Its average life will be 1097.22 years.

Solution (D):

14 4 17 17 2 8 1N O P

41. The correct order of second ionisation potential of C, N, O and F is: (A) C > N > O > F (B) O > N > F >C (C) O > F > N > C (D)F > O > N > C Solution (C): Conceptual

42. Successive ionisation energies of an element ‘X’ are given below (in K.Cal) : IP1 IP2 IP3 IP4

165 195 556 595

Electronic configuration of the element ‘X’ is -

(A) 1s2, 2s2 2p6, 3s2 3p2 (B) 1s2, 2s1

(C) 1s2, 2s2 2p2 (D) 1s2, 2s2 2p6, 3s2

Solution (D):

Difference is 2 3IF & IF is high i.e., element X has only 2 electron in outermost shell.

43. Among 2 2 2KO ,KAlO ,CaO and 2NO, unpaired electron is present in:

(A) 2 2NO and CaO (B) 2 2KO and KAlO (C) 2KO only (D) 2CaO only

Solution (C):

By molecular orbital theory 2O has one unpaired electron.

44. Given the correct order of initial T or F for following statements. Use T if statement is true and

F if it is false.

(I) 3 2 3 3(CH ) P(CF ) is non-polar and 3 3 3 2(CH ) P(CF ) is polar molecule

(II) 3 3ˆCH PCH bond angles are equal in 3 3 3 2(CH ) P(CF ) molecule

(III) 3PF will be more soluble in polar solvent than 4SiF

(A) TTF (B) FFT (C) FFF (D) FTT Solution (D):

3PF is pyramidal in shape & it is polar 4SiF is tetrahedral in shape & it is non-polar.

45. Consider the following reaction:

' '4 2 4 2MX X MX X

If atomic number of M is 52 and X and X’ are halogen and X’ is more electronegative than X. Then choose correct statement regarding given information:

P

3CF

3CF3H C

3H C

o120o120

o120

3CF

12

(A) Both X' atoms occupy axial position which are formed by overlapping of p and d-orbitals only

(B) All M X bond lengths are identical in both 4MX and '

4 2MX X compounds

(C) Central atom 'M' does not use any valence non-axial set of d-orbital in hybridization of final product

(D) Hybridization of central atom 'M' remains same in both reactant and final product Solution (C):

In 3 2sp d hybridization 2 2x yd

& 2z

d orbital are used which are oxide.

SECTION-II

(FOR ANSWER WITH NUMERIC VALUE) Section-II contains 5 Numerical questions with numerical value as answer. For each question 4 Marks will be awarded for every correct answer and 0 Mark will be deducted for every incorrect answer. 46. To determine the percentage composition of a mixture of anhydrous sodium carbonate and

sodium bicarbonate the following data is used: Weight of the mixture taken = 2g; Loss in

weight on heating = 0.11 gm. Calculate the % by mass of 3NaHCO in the mixture.

Solution (14.9): % by mass of NaHCO3 = 14.9 %

47. The following graph shows two isotherms for a fixed mass of an ideal gas(x). The r.m.s. speed

of molecules of gas(x) at temperature 1 2T and T are 1C and 2C respectively. The value of

2 1C /C is:

Solution (2):

2 2 211 2

1 1

3RT C T3RTC C

M M C T

According to graph for same volume 2

1

T 4

T 1

Hence 2

1

C2

C

48. Mole fraction of a non-electrolyte in aqueous solution is 0.07. If f 2K (H O) is o 11.86 mol kg,

depression in f.p., fT is:

Solution (7.78):

f1.86 0.07 1000

T 7.78K0.93 18

13

49. Reaction: A B follows zero order kinetics and initial concentration of A is 0.01 M.

If concentration of A is 0.008 M after 10 min, calculate half-life (in minute). Solution (25):

41/2

(0.01 0.008) ak 2 10 ; t 25 min.

10 2k

50. Calculate value of X Y Z

,10

here X is O N O bond angle in 3NO , Y is O N O bond angle

in 2NO and Z is F Xe F adjacent bond angle in 4XeF .

Solution (39):

X Y Z 120 180 9039

10 10

N

O O

O

X 120

O N O

Y 180

Xe

F

F F

F

Z 90

14

MATHEMATICS SECTION-I

(MULTIPLE CHOICE QUESTIONS) Section-I contains 20 Objective questions having 4 options each with only 1 correct option. For each question 4 Marks will be awarded for every correct answer and 1 Mark will be deducted for every incorrect answer. 51. Number of solutions of the equation tanx secx 2cosx lying in the interval [0, 2] is

(A) 0 (B) 1 (C) 2 (D) 3 Solution (C):

The given equation can be written as 1 sinx

2cos xcos x

1 + sin x = 2 cos2x = 2(1- sin2x) 2 sin2x + sin x – 1 = 0 (1 + sin x)(2 sin x – 1) = 0

sin x = 1 or 1/2

Now sin x =1, tan x and sec x not defined. Sin x = 1/2 x= /6 or 5/6.

The required number of solution is 2. 52. The line 2x – y + 1 = 0 is tangent to the circle at the point (2, 5) and the centre of the circle lies

on x – 2y = 4. The radius of circle is

(A) 3 5 (B) 5 3 (C) 2 5 (D) 5 2

Solution (A):

53. If both the roots of the equation 2 2x 2mx m 1 0 lies between –2 and 4, then m

(A)(–1, 3) (B) (1, 4) (C) (–2, 0) (D) none of these Solution (A):

2 42

f 2 f 4 0

m 1, 3

54. ABCD is a square whose sides is a. The equation of the circle circumscribing the square, taking AB and AD as axes of reference, is

(A) 2 2x y ax ay 0 (B) 2 2x y ax ay 0

(C) 2 2x y ax ay 0 (D) 2 2x y ax ay 0

Solution (C):

The circle passes through (0, 0), (a, 0), (0, a) and (a, a).

Thus, the required equation is 2 2x y ax ay 0.

55. The domain of 1sin 2x6

for all xR, is

(A) 1 1

,4 2

(B) 1 1

,4 4

(C) 1 1

,2 2

(D)1 1

,2 4

Solution (A):

Let 1f x sin 2x6

For f(x) to be defined,

1 1

1 2x 1 x2 2

(i) 1sin 2x 06

15

1sin 2x 2x sin6 6

1 1

2x x2 4

From (A) and (B), common values of x are given by 1 1

x4 2

56. If m is the slope of one of the lines represented by 2 2ax 2hxy by 0, then 2h bm ......

(A) 2a b (B) 2a b (C) 2h ab (D) 2h ab

Solution (D):

2 2

2h am m mm

b b

2a 2h m b a 2h

mmb b mb b

2m b a 2hm

2m b 2hm a 2 2m b 2hmb ab

2 2 2 2m b 2hmb h h ab

2 2h bm h ab

57. The maximum value of sin (cos x) is equal to

(A) sin 1 (B) 1 (C) sin 1

2

(D) sin 3

2

Solution (A):

cos x [1, 1] x R and sin x is increasing in ,

.

Maximum of sin (cos x) = sin 1

58. The domain of definition of the function 1

f(x)|x| x

is

(A) R (B) (0, ) (C) (–, 0) (D) none of these Solution (C):

If x 0, | x | – x > 0 x – x > 0, No solution

If x < 0, | x | – x > 0 –x – x > 0, 2x < 0 x (–, 0)

59. The period of sin 2 x 2sin 3 x 3sin5 x3 4

is

(A) 2 (B) 5 (C) 3 (D) none of these Solution (A):

Period of sin 2 x3

is 1

Period of sin 3 x4

is

2

3, Period of sin(5x) is

2

5

Required period is LCM of 1, 2

3and

2 LCM of 1, 2, 2 22

5 HCF of 1, 3, 5 1 .

60. The period of the function |sinx| |cos x|

f x|sinx cos x|

is

(A) 4

(B) 2 (C)

2

(D)

16

Solution (D):

Since period of |sin x| + |cos x| is 2

and period of |sin x – cos x| is so period of

|sinx| |cos x|

|sinx cos x|

is .

61. If the roots of the equation 2 2x a 8x 6a are real, then:

(A) a 2,8 (B) a 2,8 (C) a 2,8 (D) a 2,8

Solution (B):

2 2x a 8x 6a 2 2x 8x a 6a 0

For real roots D 0

2 28 4.1. a 6a 0 264 4 a 6a 0 24 16 a 6a 0

2a 6a 16 0 2a 6a 16 0

Now 2a 6a 16 0 a 2 a 8 0 a 2,8.

Hence given inequality is satisfied for a between the roots.

a 2,8 .

62. The domain of the function is

(A) (B)

(C) (D)

Solution (C):

We must have …(i)

& …(ii)

(i) gives

& (ii) gives

Domain is given by

63. If the equations 3 2ax 3bx 3cx d 0 and 2ax 2bx c 0 have a common root then

2

2 2

bc ad

ac b bd c

(A) 1

4 (B) 4 (C) 9 (D)

1

9

Solution (B): Let be the common root

3 2a 3b 3c d 0 …(i)

2a 2b c 0 …(ii)

21 2 b 2c d 0 …(iii)

Solving ii & iii : 2

22

bd c

ac b

&

2

bc ad

2 ac b

64. If 2a 4a 1 4, then the value of 3 2

22

a a a 1a 1

a 1

is equal to

(A) 2 (B) 1 (C) 4 (D) 3

4 2f x 10 x 21x

[5, ) 21, 21

5, 21 21,5 0

( , 5]

4 2x 21x 0

4 210 x 21x 0

x 0 or x 21 or x 21

5 x 5

5, 21 21,5 0

17

B

O A

C

Solution (C):

Given 2 2a 4a 1 4 a 1 4 1 a

2

2

2

a 1 1 a a 1y 4

a 1a 1

65. The foot of the perpendicular on the line 3x + y = drawn from the origin is C. If the line cuts the x-axis and y-axis at A and B respectively then BC : CA is

(A) 1 : 3 (B) 3 : 1 (C) 1 : 9 (D) 9 : 1 Solution (D):

tan (180° ) = slope of AB = 3

tan = 3

OC OC

tan , cotAC BC

BC tan

AC cot

= tan2 = 9.

66. The family of lines 3m x 2 m y m , where 0 passes through a fixed point

having co-ordinates

(A) (2, 1) (B) (0, 1) (C) (1, 1) (D) (2, 3) Solution (C):

Given equation can be written as x 2y 1 m 3x 2y 1 0 .

Dividing by l, it becomes m

x 2y 1 3x 2y 1 0

…(i)

The family of lines (i) passes through the point of intersection of x 2y 1 0 and

3x 2y 1 0.

x, y 1, 1

67. If two distinct chords, drawn from the point (p, q) on the circle (where

are bisected by the x-axis, then

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

Solution (D):

Let PQ be a chord of the given circle passing through P(p, q) and the coordinates of Q be (x, y). Since PQ is bisected by the x-axis, the mid-point of PQ lies on the x-axis which gives y = -q

Now Q lies on the circle 2 2x y px qy 0

So

Which gives two values of x and hence the coordinates of two points Q and R (say), so that the chords PQ and PR are bisected by x-axis. If the chords PQ and PR are distinct, the roots of (i) are real distinct.

the discriminant

2 2x y px qy

( 0)pq

2 2p q 2 28p q 2 28p q 2 28p q

2 2 2 0x q px q 2 22 0x px q

2 2 2 28 0 8p q p q

18

68. Let PS be the median of the triangle with vertices P(2, 2), Q(6, 1) and R(7, 3). The equation of

the line passing through (1, 1) and parallel to PS is (A) 2x 9y 7 0 (B) 2x 9y 11 0

(C) 2x 9y 11 0 (D) 2x 9y 7 0

Solution (D):

Points 6 7 1 3

S ,2 2

13

S ,12

PS

2 1 2m

13 92

2

Hence equation of the required line:

2

y 1 x 19

2x 9y 7 0

69. The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices (0, 0), (0, 21) and (21, 0), is:

(A) 133 (B) 190 (C) 233 (D) 105 Solution (B):

Here the given vertices of are 0,0 , 0,21 and 21,0 . For integral points inside the AOB

x 0 , y 0 & x y 21 , where x and y are integers.

Hence for x 1 , y 1,2,3,.........,19

x 2 , y 1,2,3,.........,18

x 3 , y 1,2,3,.........,17

............................................ ............................................

x 20 , y 1

Hence number of points exactly in the interior of the triangle is

19 2019 18 17 ... 3 2 1 190

2n

.

70. The range of values of for which all the roots of the equation

are imaginary is

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

Solution (C):

The given equation can be written as

' 'a

2

2 2 41 1 1 1a x x a x x

, 2 2, 2,2 2,

2 21 1 0x x x ax

19

SECTION-II (FOR ANSWER WITH NUMERIC VALUE)

Section-II contains 5 Numerical questions with numerical value as answer. For each question 4 Marks will be awarded for every correct answer and 0 Mark will be deducted for every incorrect

answer.

71. If , be the roots of 2x px q 0 and , are the roots of 2x px r 0 , q r 0 , then

( )( )

( )( )

is equal to

Solution (1):

Here, p

2 2( )( ) ( ) ( ) r r q r

Similarly ( )( ) q r

So, ratio is 1.

72. If the roots of the equation 2x bx c 0 are two consecutive integers, then 2b 4c equals to

3

4

then is equal to

Solution (1.33):

Let roots be and

21 4 1

2b 4c 1

73. Let f x be a function such that f x 1 f x 1 3.f x x R . If f 5 10 . Then the value

of 99

r 1

f 5 12r

is 44K then K is equal to

Solution (22.5):

f x 1 f x 1 3.f x ...(i)

Putting x x 1

f x f x 2 3.f x 1 ...(ii)

Again putting x x 1

f x 1 f x 3 3.f x 2 ...(iii)

+ (iii) f x 1 f x 3 f x 1 ...(iv)

Continuing like this we get:

f x is a periodic function with period ‘12’

Then 99

r 1

g x f 5 12r 99 f 5

= 990.

74. Largest integral value of m for which the quadratic expression 2y x (2m 6)x 4m 12 is

always positive, x R , is Solution (0):

D 0 3 m 1 m 0

75. Let ‘k’ be sum of all ‘x’ in the interval such that 23cot x 8cot x 3 0. Then the value

of k

2is

Solution (2.5): 2cot x u: 3u 8u 3 0

0,2

20

Both roots are real and product of roots = 1.

But cot x bijection in 0, . Let 1 2x ,x are roots such that 1 20 x x .

But 1 2cot x ,cot x are both negative.

1 2x ,x2

But 1 2x x 2

1 2cot x cot x 1

1 13

cot x cot x 12

1 23

x x2

similarly 3 4

7x x

2

k 5