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JEE Main Online test_09-04-19_MorningChemistry

Single Choice Questions :1. The major product of the following reaction is :

CH3CH = CHCO2CH3 4LiAlH

(1) CH3CH2CH2CH2OH (2) CH3CH2CH2CHO (3) CH3CH = CHCH2OH (4) CH3CH2CH2CO2CH3Ans. (3)Sol. LiAlH4 reduces only ester and not the alkene.

2. The increasing order of reactivity of the following compounds towards aromatic electrophilic substitutionreaction is :

(1) B < C < A < D (2) D <B < A < C (3) A < B < C < D (4) D < A < C < BAns. (4)Sol. Rate of ESR, depends upon electron density of benzene ring.

3. The major product of the following reaction is :

(1)

3

(2)

(3) (4)

Ans. (4)Sol. Elimination second order (E2) followed by polymerisation reaction.


4. Excessive release of CO2 into the atomosphere results in :(1) global warming (2) formation of smog(1) polar vortex (2) depletion of ozone

Ans. (1)Sol. CO2 is a green house gas which increases the atmospheric temperature causing global warming.

5. Magnisium powder burns in air to give :(1) MgO only (2) MgO and Mg3N2(3) Mg(NO3)2 and Mg3N2 (4) MgO and Mg(NO3)2

Ans. (2)Sol. Mg reacts with O2 & N2 in air to form MgO & Mg3N2.

6. Aniline dissolved in dilute HCl is reacted with sodium nitrite at 0ºC. This solution was added dropwise to asolution containing equimolar mixture of aniline and phenol in dil. HCl. The structure of the major productis :

(1) (2)

(3) (4)

Ans. (2)

Sol.

7. For any given series of spectral lines of atomic hydrogen, let max minv be the difference in maximum

and minimum frequencies in cm–1. The ratio Lyman Balmer/ is :

(1) 9 : 4 (2) 5 : 4 (3) 4 : 1 (4) 27 : 5Ans. (1)

Sol.max min

max min

Lyman

Balmer

C E

h

= max min

max min

Lyman 1 1 2

2 2 3Balmer

E E E EE E E E

=

1 1 1 113.6 13.61 1 41 1 1 113.6 13.64 4 9

=

11 0 1 1/ 4 941 1 1 1/ 9 404 4 9


8. The organic compound that gives following qualitative analysis is : Test Inference(a) Dil. HCl Insoluble(b) NaOH solution soluble(c) Br2/water Decolourization

(1) (2) (3) (4)

Ans. (4)

Sol.

9. The number of water molecule(s) not coordinated to copper ion directly in CuSO4.5H2O, is :(1) 1 (2) 2 (3) 4 (4) 3

Ans. (1)

Sol.

10. The element having greatest difference between its first and second ionization energies, is :(1) Ca (2) K (3) Ba (4) Sc

Ans. (2)Sol. Alkali metals have maximum difference in IE2 & IE1.


CH3C CH (i)DCl(1 equiva.)(ii)D

I

(1) CH3CD(Cl)CHD(I) (2) CH3CD2CH(Cl)(I)(3) CH3C(I)(Cl)CHD2 (4) CH3CD(I)CHD(Cl)

Ans. (3)Sol. According to Morkovnikov addition.


12. The aerosol is a kind of colloid in which :(1) solid is dispersed in gas (2) gas is dispersed in solid(3) gas is dispersed in liquid (4) liquid is dispersed in water

Ans. (1)

13. The correct IUPAC name of the following compound is :

(1) 2-methyl-5-nitro-1-chlorobenzene (2) 3-chloro-4-methyl-1-nitrobenzne(3) 2-chloro-1-methyl-4-nitrobenzene (4) 5-chloro-4-methyl-1-nitrobenzene

Ans. (3)Sol. Correct IUPAC name is : 2-chloro-1-methyl-4-nitrobenzene

14. The standard Gibbs energy for the given cell reaction in kJ mol–1 at 298 K is :Zn(s) + Cu2+ (aq) Zn2+(aq) + Cu(s), E° = 2 V at 298 K(Faraday's constant, F = 96000 C mol–1)(1) –192 (2) –384 (3) 384 (4) 192

Ans. (2)Sol. G° = –nFE°

G° = –(2) (96000)(2) = –384 kJ

15. Match the catalysts (Column-I) with products (Column-II).Column- I Column-II(Catalyst) (Product)

(A) V2O5 (i) Polyethylene(B) TiCl4/Al(Me)3 (ii) ethanal(C) PdCl2 (iii) H2SO4

(D) Iron Oxide (iv) NH3

(1) (A)-(iii); (B)-(i); (C)-(ii); (D)-(iv) (2) (A)-(ii); (B)-(iii); (C)-(i); (D)-(iv)(3) (A)-(iv); (B)-(iii); (C)-(ii); (D)-(i) (4) (A)-(iii); (B)-(iv); (C)-(i); (D)-(ii)

Ans. (1)Sol. V2O5 H2SO4

TiCl4(Al(CH3)3 PolyethylenePdCl2 EthanalIron oxide NH3


16. For a reaction, N2(g) + 3H2(g) 2NH3(g) ; identify dihydrogen (H2) as a limiting reagent in the followingreaction mixtures.(1) 56 g of N2 + 10 g of H2 (2) 35 g of N2 + 8 g of H2(3) 14 g of N2 + 4 g of H2 (4) 28 g N2 + 6 g of H2

Ans. (1)

Sol. Mole of N2 Mole of H22Moleof N

S.C.2Moleof H

S.C.

(1)56 228

102

= 52 21

53 = 1.66 H2 is LR

(2)3528 = 1.25

8 42

1.25 1.251

4 1.333 N2 is LR

(3)14 0.528

4 22

0.5 0.51

2 0.673 H2 is LR

(4)28 128

6 32

1 11

3 13 both are LR

17. Which of the following statements is not true about sucrose?(1) It is also named as invert sugar(2) The glycosidic linkage is present between C1 of -glucose and C1 of -fructose(3) It is a non reducing sugar(4) On hydrolysis, it produces glucose and fructose

Ans. (2)Sol. C1 – C2 glycosidic linkage is present in sucrose.

18. Consider the van der Waals constants, a and b, for the following gases.Gas Ar Ne Kr Xea/ (atm dm6 mol–2) 1.3 0.2 5.1 4.1b/ (10–2 dm3 mol–1) 3.2 1.7 1.0 5.0Which gas is expected to have the highest critical temperature?(1) Kr (2) Ne (3) Ar (4) Xe

Ans. (1)


(1) (2) (3) (4)

Ans. (3)Sol. Oxidation of alkyl chain.


20. The given plots of represent the variation of the concentration of a reactant R with time for two differentreaction (i) and (ii). The respective orders of the reactions are :

(1) 0, 2 (2) 1, 0 (3) 1, 1 (4) 0, 1Ans. (2)Sol. Kt = lnA0 – ln[A]t First order (graph-(i))

[A]t = A0 – Kt Zero order (graph-(ii))both graphs have negative slope.

21. The correct order of the oxidation states of nitrogen in NO, N2O, NO2 and N2O3 is :(1) NO2 < NO < N2O3 < N2O (2) NO2 < N2O3 < NO < N2O(3) N2O < NO < N2O3 < NO2 (4) N2O < N2O3 < NO < NO2

Ans. (3)

Sol.

22. The ore that contains the metal in the form of fluoride is :(1) cryolite (2) sphalerite (3) malachite (4) magnetite

Ans. (1)Sol. Cryolite Na3[AlF6] Magnetite Fe3O4

Malachite CuCO3.Cu(OH)2 Sphalerite ZnS

23. The degenerate orbitals of [Cr(H2O)6]3+ are :

(1) 2zd and dxz (2) dyz and 2z

d (3) 22x yd

and dxy (4) dxz and dyz

Ans. (4)

Sol.

3+

xy yz zx

3

24. C60, an allotrope of carbon contains :(1) 12 hexagons and 20 pentagons (2) 18 hexagons and 14 pentagons(3) 16 hexagons and 16 pentagons (4) 20 hexagons and 12 pentagons

Ans. (4)Sol. C60 contains twenty (20) six-membered rings & 12 five membered ring.


25. Among the following, the test of parameters that represents path functions, is :(A) q + w (B) q (C) w (D) H – TS(1) (A), (B) and (C) (2) (A) and (D) (3) (B), (C) and (D) (4) (B) and (C)

Ans. (4)Sol. (a) q + W = U (Change in internal energy is a state function)

(b) W = Work is a path function(c) q = Heat change is a path function(d) H – TS = G (Gibbs free energy) is a state function

26. The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution of0.01M BaCl2 in water. Assuming complete dissociation of the given ionic compounds in water, the concentrationof XY (in mol L–1) in solution is :(1) 16 × 10–4 (2) 4 × 10–2 (3) 6 × 10–2 (4) 4 × 10–4

Ans. (3)Sol. BaCl2 xy = iCRT

= 3(0.01)RT ... (i) 4 = 2.C(RT) ... (ii) (i) (ii)

1 3 0.01 R T4 2 C R T

C = 0.06 M


(1) (1) (3) (4)

Ans. (2)Sol. Halogenation followed by elimination reaction to give canjugated product.

28. Among the following, the molecule expected to be stablized by anion formation is :C2, O2, NO, F2

(1) F2 (2) NO (3) C2 (4) O2Ans. (3)Sol. In C2 electron is added in bonding molecular orbital. So C2 will stabilize after adding an electron.

In F2, O2 and NO electron is added in antibonding molecular orbital.


29. The one that will show optical activity is : (en = ethane-1,2-diamine)

(1) (2) (3) (4)

Ans. (4)

Sol. does not have plane of symmetry. Hence, is optically active.

30. Liquid 'M' and liquid 'N' form an ideal solution. The vapour pressures of pure liquids 'M' and 'N' are 450 and 700mm Hg, respectively, at the same temperature. Then correct statement is :(xM = Mole fraction of 'M' in solution; xN = Mole fraction of 'N' in solution; yM = Mole fraction of 'M' in vapourphase; yN = Mole fraction of 'N' in vapour phase)

(1) M M

N N

x yx y

(2) M M

N N

x yx y

(3) (xM – yM) < (xN – yN) (4) M M

N N

x yx y

Ans. (1)

Sol. 0m m m Tp X Y p ..... (i)

0n n n Tp X Y p .....(ii)

0m m m T0n n n T

p X Y pp X Y p

m m

n n

X Y450700 X Y

m m

n n

X Y700X 450 Y

m m

n n

X YX Y


JEE Main Online test_09-04-19_MorningPhysics

Single Choice Questions :1. Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength

of the 2nd Balmer line (n = 4 to n = 2) will be(1) 889.2 nm (2) 388.9 nm (3) 642.7 nm (4) 488.9 nm

Ans. (4)

Sol.2

21 1 1 5RzRz –660 4 9 36

221 1 1 3RzRz –

4 16 16

316

365

660

= 488 nm

2. A concave mirror for face viewing has focal length of 0.4 m. Th distance at which you hold the mirror from yourface in order to see your image upright with a magnification of 5 is(1) 1.60 m (2) 0.24 m (3) 0.16 m (4) 0.32 m

Ans. (4)

Sol. mirror formula 1 1 1v u f

given that, v 5u

1 1 15u u 0.4

u = 0.32 m

3. An NPN transistor is used in common emitter configuration as an aplifier with 1 k load resistance. Signalvoltage of 10 mV is applied across the base-emitter. The produces a 3 mA change in the collector current and15 A change in the base current of the amplifier. The input resistance and voltage gain are(1) 0.33 k, 300 (2) 0.67 k, 200 (3) 0.67 k, 300 (4) 0.33 k, 1.5

Ans. (3)Sol. Vinput = rinput × iinput

10 × 10–3 = rinput × 15 × 10–6 rinput = 0.67 k.

Voltage gain = 3001010

10310003–

3–


4. A rigid square loop of side 'a' and carrying current I2 is lying on a horizontal surface near a long current I1carrying wire in the same plane as shown in figure. The net force on the loop due to the wire will be

(1) Repulsive and equal to

2210 (2) Attractive and equal to

3

210

(3) Repulsive and equal to

4210 (4) Zero

Ans. (3)

Sol. 0 1AB 2

µ iF i a

2a

(away from wire)

FBC = FAD = 0

0 1CD 2

µ iF i a

4a

(towards wire)

0 1 2net

µ i iF

4

(away from wire)

5. A solid sphere of mass 'M' and radius 'a' is surrounded by a uniform concentric spherical shell of thickness2a and mass 2M. The gravitational field at distance '3a' from the centre will be

(1) 2a9GM2

(2) 2a9GM

(3) 2a3GM

(4) 2a3GM2

Ans. (3)

Sol.2 2

due to inner sphere due to outer shell

GM 2GME(3a) (3a)

2

GM3a


6. A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance betweenE and C is (E is mid-point of arm CD)

(1) R647

(2) R (3) R43

(4) R161

Ans. (1)Sol. let side length is a.

Resistance per unit length R4a

BER a R

R4a 2 8

EADCBR 7a 7RR4a 2 8

eq

R 7R7R8 8R

R 7R 648 8

7. Determine the charge on the capacitor in the following cirucuit

(1) 60 C (2) 200 C (3) 10 C (4) 2 CAns. (2)Sol. On simplifying given circuit in steady state

i = 8 amperes.By using current division formula, current through 2 branch is 2 Amperes.Therefore, potential differene across 10 resistance is 2 × 10 = 20 volts. Charge on capacitor,

Q = C × V = 10 × 10–6 × 20 = 2 × 10–4 Coulomb.

8. If 'M' is the mass of water that rises in a capillary tube of radius 'r', then mass of water which will rise in acapillary tube of radius '2r' is

(1) 4 M (2) 2 M (3) 2M

(4) M

Ans. (2)Sol. Force inside capillary tube

2 rT = Mg M r


9. A rectangular coil (Dimension 5 cm × 2.5 cm) with 100 turns, carrying a current of 3A in the clock-wisedirection, is kept centered at the origin and in the X-Z plane. A magnetic field of 1 T is appied along X-axis. Ifthe coil is tilted through 45º about Z-axis, then the torque on the coil is(1) 0.55 Nm (2) 0.42 Nm (3) 0.27 Nm (4) 0.38 Nm

Ans. (3)

Sol. = BINA sin 45º = 1 × 3 × 100 × 12.5 × 10–4 × 21

= 0.27 N-m

10. The electric field of light wave is given as CNxt1062–

105x2cos10E 14

7–3–

. This light falls on a

metal plate of work function 2 eV. The stopping potential of the photo-electrons is

Given, E (in eV) = )Åin(12375

(1) 0.72 V (2) 2.48 V (3) 0.48 V (4) 2.0 VAns. (3)Sol. From equation = 5 × 10–7 m = 5000 Å

maxhC

kE

max12375

2 0.48 kE5000

11. In the density measurement of a cube, the mass and edge length are measured as (10.00 ± 0.10) kg and(0.10 ± 0.01) m, respectively. The error in the measurement of density is(1) 0.31 kg/m3 (2) 0.01 kg/m3 (3) 0.10 kg/m3 (4) 0.07 kg/m3

Ans. (1)

Sol. Density of cube 3M

Maximum permissible error M 3

M

1.001.03

101.0

= 10031

103

1001

31.0


12. The pressure wave, P = 0.01 sin [1000t – 3x] Nm–2, corresponds to the sound produced by a vibrating bladeon a day when atmospheric temperature is 0ºC. On some other day when temperature is T, the speed ofsound produced by the same blade and at the same frequency is found to be 336 ms–1. Approximate valueof T is(1) 12ºC (2) 4ºC (3) 11ºC (4) 15ºC

Ans. (2)

Sol. wave no. kv

vk

= 1000, k = 3

velocity of sound RTvM

11

2 2

Tvv T

2 22

2 12 21

v 336T T 273v 1000

3

T2 = 277.38 k = 4.38ºC

13. A ball is thrown vertically up (taken as +z-axis) from the ground. The correct momentum-height (p-h) diagramis

(1) (2) (3) (4)

Ans. (4)Sol. Since

V2 = U2 + 2as

gh2–UmP 2

2

2

(Parabola)

14. A moving coil galvanometer has resistance 50 and it indicates full deflection at 4 mA current. A voltmeteris made using this galvanometer and a 5 k resistance. The maximum voltage, that can be measured usingthis voltmeter, will be close to(1) 20 V (2) 10 V (3) 40 V (4) 15 V

Ans. (1)Sol. V = ig (G + S) = 4 × 10–3 × 5050 = 20.2 volts


15. A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary waveis Y = 0.3 sin (0.157x) cos (200t). The length of the string is (All quantities are in SI units)(1) 80 m (2) 20 m (3) 60 m (4) 40 m

Ans. (1)

Sol. m40157.02

24

= 80 m

16. An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in

its gaseous phase is , m is its mass and kB is Boltzmann constant, then its temperature will be

(1) B

2

k3m

(2) B

2

k7m

(3) B

2

k5m

(4) B

2

k6m

Ans. Ans. given by NTA is Option-4

Sol.mKT3

rms

KT3mT

2

17. The magnetic field of a plane electromagnetic wave is given by )tkzcos(jB)]t–kz[cos(iBB 10 ,

where B0 = 3 × 10–5 T and B1 = 2 × 10–6 T. The rms value of the force experienced by a stationary chargeQ = 10–4C at z = 0 is closest to(1) 0.9 N (2) 3 × 10–2 N (3) 0.6 N (4) 0.1 N

Ans. (3)

Sol. )i(–E)j(–EE 10

26–25–8 )102()103(103|E|

N9.0|E|q|F| 0

N6.02

F|F| 0

rms

18. The stream of a river is flowing with a speed of 2 km/h. A swimmer can swim at a speed of 4 km/h. Whatshould be the direction of the swimmer with respect to the flow of the river to cross the river straight?(1) 150º (2) 90º (3) 120º (4) 60º

Ans. (3)Sol. 4 cos = 2

= 60º = 120º


19. Following figure shows two processes A and B for a gas. If QA and QB are the amount of heat absorbed bythe system in two cases, and UA and UB are changes in internal energies, respectively, then

(1) QA > QB, UA = UB (2) QA > QB, UA > UB(3) QA < QB, UA < UB (4) QA = QB, UA = UB

Ans. (1)Sol. Since U is a state function

U1 = U2For process (1) Q1 = U1 + W1For process (2) Q2 = U1 + W2Since U1 = U2 and W1 > W2 (area under pv graph)

Q1 > Q2

20. A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its th

n1

part is

hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work doneshould be

(1) 2nMgL2

(2) nMgL (3) 2nMgL

(4) 2n2MgL

Ans. (4)

Sol.

mass of the chain of hanged part mn

Initial potential energy of the chain mgn 2n

i 2

mgU2n

when whole chain is on the table. Final potential energy of the chainUf = 0

nxt f i 2 2

mg mgw U U I 02n 2n


21. A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in theoriginal direction but with one fourth of its original speed. What is the mass of the second body?(1) 1.8 kg (2) 1.0 kg (3) 1.2 kg (4) 1.5 kg

Ans. (3)

Sol.

from momentum conservation

vmv m MV '4

e = 1, v

v v '4

3M m 1.2kg5

22. A system of three charges are placed as shown in the figure :

If D >> d, the potential energy of the system is best given by

(1)

2

2

0 DqQd2

dq–

41

(2)

2

2

0 DqQd–

dq–

41

(3)

2

2

0 DqQd

dq

41

(4)

2

2

0 D2qQd–

dq–

41

Ans. (2)

Sol.2

2

kq kQPE (qd)sin

d D

2

2

kq kQqdPE

d D


23. A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-

viscous liquid. The density of the liquid is 161

th of the material of the bob. If the bob is inside liquid all the

time, its period of oscillation in this liquid is

(1) 141T2 (2)

141T4 (3) 10

1T2 (4) 151T4

Ans. (4)

Sol. initially, time period, iLT 2g

when the bob is dipped in liquid of density 16

,

v = volume of bob

netvF v g g16

eff15

(v )g gv16

eff15g g16

eff

LT' 2

g

LT ' 215g16

4TT'

15


24. The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane :

(i) a ring of radius R, (ii) a solid cylinder of radius 2R

and (iii) a solid sphere of radius 4R

. If, in each case, the

speed of the center of mass at the bottom of the incline is same, the ratio of the maximum heights they climb(1) 14 : 15 : 20 (2) 4 : 3 : 2 (3) 10 : 15 : 7 (4) 2 : 3 : 4

Ans. (3) [Correct Ans. is Option-1]Sol. For ring

112

202

1201 ghm

Rv

Rm21vm

21

gh1 = v20

For cylinder

222

20

222

02 ghmRv4

8Rm

21vm

21

; 202 v

43gh

For sphere

33

20

232

03 ghmRv16

16Rm

52

21vm

21

; 203 v

107gh

h1 : h2 : h3 = 20 : 15 : 14

25. The total number of turns and cross-section area in a solenoid is fixed. However, its length L is varied byadjusting the separation between windings. The inductance of solenoid will be proportional to

(1) 2L1

(2) L (3) L1

(4) L2

Ans. (3)

Sol. LRNN 20

; 1L

26. The figure shows a Young's double slit experimental setup. It is observed that when a thin transparent sheetof thickness t and refractive index is put in front of one of the slits, the central maximum gets shifted by adistance equal to n fringe widths. If the wavelength of light used is , t will be

(1) )1–(aD2

(2) )1–(anD2

(3) )1–(a

nD

(4) )1–(aD

Ans. (3) Bonus

Sol. dD)1–(t

dDn

1–nt


27. A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energyas a function of , where is the angle by which it has rotated, is given as k2. If its moment of inertia is I thenthe angular acceleration of the disc is

(1) k2

(2) 2

k(3)

k

(4) 4

k

Ans. (1)

Sol.

k2k

21 22

k2

dd

28. A capacitor with capacitance 5 F is charged to 5 C. If the plates are pulled apart to reduce the capacitanceto 2 F, how much work is done?(1) 2.55 × 10–6 J (2) 2.16 × 10–6 J (3) 6.25 × 10–6 J (4) 3.75 × 10–6 J

Ans. (4)

Sol. Energy of a capacitor (U) = C2

Q2

Work done = U = 2 2

2 1

Q Q–2C 2C =

2

2 1

Q 1 1–2 C C = 3.75 × 10–6 J

29. A signal Acost is transmitted using 0sin0t as carrier wave. The correct amplitude modulated (AM) signalis(1) (0 + A)cost sin0t (2) 0sin[0(1 + 0.01 A sint)t]

(3) 0sin0t + 2A

sin (0 – )t + 2A

sin(0 + )t (4) 0sin0t + Acost

Ans. (3)

Sol. ym = (m + Acost)sin0t = 0sin0t + 2A

sin(0 – )t + 2A

sin(0 + )t


30. For a given gas at 1 atm pressure, rms speed of the molecules is 200 m/s at 127ºC. At 2 atm pressure andat 227ºC, the rms speed of the molecules will be

(1) 5100 m/s (2) 580 m/s (3) 100 m/s (4) 80 m/s

Ans. (1)Sol. Given

rms3RTvM

Tv rms . Let at T = 227ºC, Vrms = V2.

At t = 127ºC = 400 K, Vrms = 200 m/s

3R 400 200M

.....(i)

At t = 227ºC = 500 K, 23R 500 VM

.....(ii)

Dividing eqn. (i) by (ii),

L

3R 400 200M3R 500 VM

LV200

52

VL = 5100 m/s


JEE Main Online test_09-04-19_MorningMathematics

Single Choice Questions :1. A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the

committee is formed with at least 6 males and n is the number of ways the committee is formed with at least3 females, then :(1) m + n = 68 (2) m = n = 78 (3) m = n = 68 (4) n = m = –8

Ans. (2)Sol. m = 8C6

5C5 + 8C75C4 + 8C8.

5C3

= 78n = 5C3.

8C8 + 5C48C7 + 5C5

8C6

= 10 + 5(8) + 1(28) = 78

2. Let and be the roots of the equation x2 + x + 1 = 0. Then for y 0 in R, y 1

y 11 y

is equal to

(1) y(y2 – 1) (2) y3 (3) y3 – 1 (4) y(y2 – 3)Ans. (2)Sol. + = – 1 = 1

1 1 2 3

y 1 y y yy 1 R R R R y 1

1 y 1 y

= y ((y + ) (y + ) –1 – (+ ) + + – (y + ))= y (y2 + (+ )y + –1 + + – (+ )y – 2 – 2)= y (y2) = y3

3. Let 10

10

k 1

f(a k) 16(2 – 1)

, where the function f satisfies f(x + y) = f(x) f(y) for all natural numbers x, y and

f(1) = 2. Then the natural number 'a' is :(1) 3 (2) 16 (3) 2 (4) 4

Ans. (1)Sol. f(x + y) = f(x) f(y)

f(1) = 2hence f(x) = 2x

Consider 10

10

K 1

f(a K) 16(2 1)

f(a + 1) + f(a + 2) + ........ + f(a + 10) = 16 (210 –1)2a+1 + 2a + 2 + ........... + 2 a + 10 = 16(210 – 1)2a (2 + 22 + ....... + 210) = = 16(210 – 1)

2a .10(2 1)22 1

= 16 (210 – 1)

2a + 1 = 24

a = 3


4. If the function f defined on ,6 3

by f(x) =

2 cos x – 1, xcot x – 1 4

k, x4

is continuous, then k is equal to :

(1) 2 (2) 12

(3) 12

(4) 1

Ans. (2)

Sol.x

4

2 cos x 1k lim

cot x 1

2x

4

2 sinxlimcosec x

3

x4

lim 2(sinx)

31 1222

5. Let f(x) = 15 – |x – 10| ; x R. Then the set of all values of x, at which the function, g(x) = f(f(x)) is notdifferentiable, is :(1) {5, 10, 15} (2) {5, 10, 15, 20} (3) {10, 15} (4) {10}

Ans. (1)Sol. f(x) = 15 – |x – 10|

f(x) = 15 – |15 – |x – 10| – 10|= 15 – |5 – |x – 10||

=

x 10 : x 520 x :5 x 10x : 10 x 1530 x : x 15

Non. Diff. at x = 5, 10, 15


6. The value of /2 3

0

sin x dxsinx cosx

is :

(1) – 12

(2)

– 28

(3)

– 14

(4)

– 24

Ans. (3)

Sol. I = /2 3

0

sin x dxsinx cos x

...(1)

I =

3/2

0

sin x2 dx

sin x cos x2 2

...(2)

Adding equation (1) and (2)

2I = /2 3 3

0

sin x cos x dxsinx cosx

2I = /2

2 2

0

(sin x cos x sin xcos x) dx

= /2 /2

0 0

11. dx sin2x dx2

I = /2

0

1 cos2x4 4 2

I = 1 1( 1 1)

2 4 4 4

7. Let S be the set of all values of x for which the tangent to the curve y = f(x) = x3 – x2 – 2x at (x, y) is parallelto the line segment joining the points (1, f(1)) and (–1, f(–1)), then S is equal to :

(1) 1– , – 13

(2) 1 , 13

(3) 1– , 13

(4) 1 ,–13

Sol. f(1) = 1 – 1 – 2 = –2f(–1) = –1 –1 + 2 = 0

f(1) f( 1) 2 0m 11 1 2

dydx = 3x2 – 2x – 2 ; 3x2 – 2x – 2 = –1

3x2 – 2x – 1 = 0 ; (x – 1)(3x + 1) = 0

x = 1, 13


8. If the tangent to the curve, y = x3 + ax – b at the point (1, –5) is perpendicular to the line, –x + y + 4 = 0, thenwhich one of the following points lies on the curve?(1) (–2, 1) (2) (–2, 2) (3) (2, –2) (4) (2, –1)

Ans. (3)Sol. f(1) = –5

1 + a – b = –5b – a = 6 ......(1)Slope of tangent at P

= f '(1)= 3 + a = –1

a – 4 .....(2)

b 2 y = x3 – 4x – 2

9. If the fourth term in the Binomial expansion of 8

6log x2 x

x

(x > 0) is 20 × 87, then a value of x is :

(1) 8 (2) 83 (3) 8–2 (4) 82

Ans. (4)

Sol. T4 = 8

3 3log x6 73

2C . x 20(8 )

x

8

3 3log x 78 8

2log x log (8 )x

28 8

23log 3(log x) 7x

Let log8x = t 31 t3

+ 3t2 = 7

3t2 – 3t – 6 = 0t2 – t – 2 = 0t = 2, –1log8x = 2, –1

x = 64, 18


10. Slope of a line passing through P(2, 3) and intersecting the line, x + y = 7 at a distance of 4 units from P, is:

(1) 1– 51 5

(2) 1– 71 7

(3) 7 – 17 1

(4) 5 – 15 1

Ans. (2)

Sol. sin= 2 1

4 2 2

tan= 17

tan= t 11 t

1 t 11 t7

t – 1 = 7t 7 or 1 – t = 7t 7

t = 7 1

7 1

or

1 7 t1 7

11. Let S = { [–2, 2] : 2cos2 + 3sin = 0}. Then the sum of the elements of S is :

(1) 13

6

(2) 2 (3) (4) 53

Ans. (2)Sol. 2cos2+ 3 sin= 0

2 – 2sin2+ 3sin= 0 2sin2– 3sin– 2 = 0(2sin+1) (sin–2) = 0

sin = – 12

or sin= 2

Hence sin= – 12

, we get = n+ (–1)n 6

Hence solution in the interval 2 , 2 are 5 7 11, , ,6 6 6 6

sum of all possible value of is 2

12. If 1 1 1 2 1 3 1 n – 1 1 78...........

0 1 0 1 0 1 0 1 0 1

, then the inverse of 1 n0 1

is :

(1) 1 0

12 1

(2) 1 0

13 1

(3) 1 –130 1

(4) 1 –120 1

Ans. (3)

Sol.1 1 1 2 1 3 1 4 1 n 1

.......0 1 0 1 0 1 0 1 0 1

= 1 3 1 3 1 4 1 n 1

....0 1 0 1 0 1 0 1


= 1 6 1 4 1 n 1

....0 1 0 1 0 1

= 1 x0 1

where x is (n –1)th term of sequence1, 3, 6, 10, ......

Whose general term is n(n 1) (n 1)n,so x

2 2

(n 1)n 1 7812

0 10 1

So,n(n 1)

2

= 78 n2 – n – 156 = 0

n = 13, – 12

So A = 1 130 1

A–1 = 1 130 1

13. If the function f : R – {1, –1} A defined by f(x) = 2

2x

1– x, is surjective, then A is equal to :

(1) R – {–1} (2) R – [–1, 0) (3) R – (–1, 0) (4) [0, )Ans. (2)

Sol. y = 2

2

x1 x

y – yx2 = x2

(y + 1)x2 = y

x2 = y 1

y 1

y R [ 1,0)

14. For any two statements p and q, then negation of the expression p (~ p q) is :(1) p q (2) ~ p ~ q (3) ~ p ~ q (4) p q

Ans. (2)

Sol. ~ p ~ p q ~ (p q) ~ p ~ q


15. If the line y = mx + 7 3 is normal to the hyperbola 2 2x y–

24 18 = –1, then a value of m is :

(1) 25 (2)

152

(3) 35 (4)

52

Ans. (1)

Sol.2 2x y 1

24 18

Equation of normal is slope form

y = mx – 2 2

2 2 2

(a b )m

a b m

2 2

2 2 2

(a b )m 7 3a b m

2

(24 18)m7 3

24 18m

42m = 27 3 24 18m

(6m)2 = 3(24 – 18m2)36m2 = 72 – 54 m2

90 m2 = 72

m2 = 72 490 5

±2m5

16. If one end of a focal chord of the parabola, y2 = 16x is at (1, 4), then the length of this focal chord is :(1) 20 (2) 25 (3) 24 (4) 22

Ans. (2)Sol. Length of focal chord is = 4a cosec2

: inclination angle of chord.y2 = 16x a = 4Focal chord passes through (4, 0) and (1, 4)

tan= 43

cosec= 54

length of focal chord = 1625 2516


17. Let the sum of the first n terms of a non-constant A.P., a1, a2, a3, ........ be 50n + n(n – 7)

2A, where A is a

constant. If d is the common difference of this A.P., then the ordered pair (d, a50) is equal to :(1) (50, 50 + 45A) (2) (A, 50 + 46A) (3) (A, 50 + 45A) (4) (50, 50 + 46A)

Ans. (2)Sol. Let a be first term

d be common difference

2n 2a (n 1)d 100n (n 7n)A2 2

(20 – d)n + n2d = n2A + (100 – 7A)n2a – d = 100 – 7A

d A

2a = 100 – 6A

a 50 3A

a50 = a + 49d = 50 – 3A + 49A = 50 + 46A

18. The area (in sq. units) of the region A = [(x, y) : x2 y x + 2} is :

(1) 316 (2)

103 (3)

92

(4) 136

Ans. (3)

Sol.2

2

1

A (x 2) x dx

22 3

1

x x2x2 3

= 8 1 12 4 23 2 3

= 92


19. If the line, x – 1 y 1 z – 2

2 3 4

meets the plane, x + 2y + 3z = 15 at a point P, then the distance of P from

the origin is :

(1) 7/2 (2) 9/2 (3) 5 /2 (4) 2 5Ans. (2)

Sol.x 1 y 1 z 2

2 3 4

(x, y, z) (2 + 1, 3–1, 4+ 2)Satisfying the plane.2+1 + 2 (3–1) + 3 (4+ 2) = 1520 = 10

= 12

Point is 12, ,42

Hence distance of point from origin is 92

20. If a tangent to the circle x2 + y2 = 1 intersects the coordinate axes at distinct points P and Q, then the locusof the mid-point of PQ is :(1) x2 + y2 – 4x2y2 = 0 (2) x2 + y2 – 2x2y2 = 0 (3) x2 + y2 – 2xy = 0 (4) x2 + y2 – 16x2y2 = 0

Ans. (1)

Sol. Equation of line PQ = x y2h 2k

= 1

It is tangent to circle P = r

2 2

1 11 1

2h 2k

2 2

1 1 4h k


21. All the points in the set S = i : R (i –1)

– i

lie on a :

(1) Circle whose radius is 2 (2) Straight lien whose slope is 1.

(3) Circle whose radius is 1. (4) Straight line whose slope is –1.Ans. (3)

Sol.izi

2( i)( i)z

1

2

21 2 i

1

2

21x1

2

2y1

2 222 2

2 21 2x y1 1

x2 + y2 = 1circle, radius = 1

22. The integral 2/3 4/3sec x cosec x dx is equal to :

(Here C is a constant of integration)

(1) –3 tan–1/3x + C (2) –3cot–1/3x + C (3) 3tan–1/3x + C (4) 3–4

tan–4/3x + C

Ans. (1)

Sol. 4/3 2/3(sinx) (cos x) dx Put tanx = t

= 4/3 2(cot x) sec xdx sec2x dx = dt

= 1/3

4/3 tt dt c1/ 3

dx = cos2x dt

= – 3 (tanx)–1/3 + c


23. The solution of the differential equation xdydx + 2y = x2(x 0) with y(1) = 1, is :

(1) y = 3

2x 15 5x

(2) y = 2

2x 34 4x

(3) y = 3

24 1x5 5x

(4) y = 34

x2 + 21

4x

Ans. (2)

Sol. x 2dy 2y xdx

Multiply by x

x2 dydx

+ 2xy = x3

2 3d x y x

dy

x2y = 4x c

4

as y (1) = 1

1 = 1 c4

c = 34

x2y = 4x 3

4 4

y = 2

2

x 34 4x

24. Let ˆ ˆ3i j and ˆ ˆ ˆ2i – j 3k

. If 1 2–

, where 1

is parallel to

and 2

is perpendicular to

,

then 1 2

is equal to :

(1) 1 ˆ ˆ ˆ(3i – 9 j 5k)2

(2) ˆ ˆ ˆ–3i 9 j 5k (3) ˆ ˆ ˆ3i – 9 j – 5k (4) 1 ˆ ˆ ˆ(–3i 9 j 5k)2

Ans. (4)

Sol. 1 2–

1 1ˆ ˆ | |

1 ˆ ˆ( )

2 ˆ ˆ( ) –

1 2 ˆ ˆ( )

5 ˆ ˆ ˆ ˆ ˆ(3i j) (2i j 3k)10

1 ˆ ˆ ˆ( 3i 9 j 5k)2


25. If the standard deviation of the numbers –1, 0, 1, k is 5 where k > 0, then k is equal to :

(1) 543

(2) 6 (3) 2 6 (4) 1023

Ans. (3)Sol. – 1 , 0, 1, K

mean = K4

S.D

2 2 2 2K K K 3K1 14 4 4 4 5

4

2212K 2 20 K 24

16

26. The value of cos210º – cos10º cos50º + cos250º is :

(1) 32

(1 + cos 20º) (2) 34

+ cos 20º (3) 3/2 (4) 3/4

Ans. (4)

Sol. cos210º – 22 cos10ºcos50º 1 sin 50º2

cos210º – sin250º + 1 – 1 cos60º cos40º2

cos 60º cos 40º + 1 – 12

1 cos40º2

1 – 1 34 4

27. If f(x) is a non-zero polynomial of degree four, having local extreme points at x = –1, 0, 1 ; then the setS = {x R : f(x) = f(0)}contains exactly :(1) Two irrational and two irrational numbers. (2) Four irrational numbers.(3) Two irrational and one rational number. (4) Four rational numbers.

Ans. (3)Sol. f '(x) = k(x + 1)x (x –1)

= k(x3 – x)

f(x) = 4 2x kxk c

4 2

f(x) = f(0)

4 2x kxk c c4 2

4 2x xk 04 2

x2(x2 – 2) = 0

x = 0, 2, 2


28. Four persons can hit a target correctly with probabilities 1 1 1 1, , and2 3 4 8 respectively. If all hit at the target

independently, then the probability that the target would be hit, is :

(1) 2532 (2)

732 (3)

1192 (4)

25192

Ans. (1)Sol. Let persons be A, B, C, D

P(Hit) = 1 – P(none of them hits)

1 P(A B C D)

1 P(A) P(B) P(C) P(D)

2 3 712 3 4 8

2532

29. A plane passing through the points (0, –1, 0) and (0, 0, 1) and making an angle 4

with the plane y – z + 5

= 0, also passes through the point :

(1) ( 2 , –1, 4) (2) (– 2 , 1, –4) (3) ( 2 , 1, 4) (4) (– 2 , –1, –4)Ans. (3)Sol. Let ax + by + cz = 1 be the equation of plane

0 – b + 0 = 1 b = –10 + 0 + c = 1 c = 1

cos= a b

| a | | b |

2

1 | 0 1 1|2 a 1 1 0 1 1

a2 + 2 = 4 a = ± 2

± 2 x – y + z = 1

Now for negative sign – 2 2 – 1 + 4 = 1

30. Let p, q R. If 2 – 3 is a root of the quadratic equation, x2 + px + q = 0, then :

(1) p2 – 4q + 12 = 0 (2) q2 + 4p + 14 = 0 (3) q2 – 4p – 16 = 0 (4) p2 – 4q – 12 = 0Ans. BonusSol. Remarks : Bonus (p and q should belongs to set of rational number)

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