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JEE ADVANCED

EXAMINATION 2014

QUESTIONS WITH SOLUTIONS

PAPER - 1 [CODE - 9]

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)

PHYSICS

IIT-JEE 2014 Solutions by Motion Edu. Pvt. Ltd. Kota

1. A transparent thin film of uniform thickness andrefractive index n1=1.4 is coated on the convexspherical surface of radius R at one end of a longsolid glass cylinder of refractive index n2 = 1.5, asshown in the figure. Rays of light parallel to the axisof the cylinder traversing through the film from airto glass get focused at distance f1 from the film,while rays of light traversing from glass to air getfocused at distance f2 from the film. Then

(A) 1f 3R

(B) 1f 2.8R

(C) 2f 2R

(D) 2f 1.4R

Sol. A,CAir to Glass

1

1.4V -

1

= 1.4 1

R

.....(1)

1.5V

– 1

1.4V =

1.5 1.4R

.....(2)

From (1) and (2)

1f5.1

= R5.0

f1 = 3R

Glass to Air

1V4.1

– 5.1

= )R(5.14.1

= R1.0

....(3)

V1

– 1V4.1

= )R(4.11

= R4.0

.....(4)

2f1

= R5.0

f2 = 2R

2. A sudent is perfarming an expernment using aresonance column and a tuning fork of frequency244 s-1. He is told that the air in the tube has beenreplaced by another gas (assume that the columnremains filled with the gas). If the minimum hight atwhich resonance occurs is (0.350 0.005)m, thegas in the tube is

(useful information : 1/2167RT 640 J mole-1/2,1/2140RT = 590 J mole-1/2. The molar masses M is

garms are given in the options. Take the values of

10M

for each gas as given there.)

(A) 10 7Neon M=20, =20 10

(B) 10 3Nitrogen M=28, =28 5

(C) 10 9Oxygen M=32, =32 16

(D) 10 17Argon M=36, =36 32

Sol. Df = 244 Hz

4

= 0.35 = 1.4 m

V = 244 × 1.4 = 314.6 = M

)RT(

For Neon : V =

M

)RT(35

= 12RT

6

RT.M10

= 107

6

RT

For N2 : V = 20RT

For O2 : V = 32RT

57

= 7RT160

For argon : V = 36RT

35

= 108RT5

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3. A parallel plate capacitor has a dielectric slab ofdielectric constant K between it plates that covers1/3 of the area of its plates, as shown in the figure.The total capacitance of the capacitor is C whilethat of the portion with dielectric in between C1.When the capacitor is charged, the plate area cov-ered by the dielectric get charge Q1 and the rest ofthe area gets charge Qz. The electric field in thedielectric is E1 and that in the other portion is E2.Choose the correct option/options. ignoring edgeeffects.

(A) 1

2

E=1

E(B) 1

2

E 1=E K

(C) 1

2

Q 3=Q K (D)

1

C 2+K=C K

Sol. A,D

C0 0Ad

q = C0Vq C0

2

1

qq

= 2k

2

1

EE

= 1

C = 3k

C0 + 32

C0 = 3

)2K( C0

1CC =

K)2K(

4. One end of a taut string of length 3m along the xaxis is fixed at x = 0. The speed of the waves in thestring is 100 ms-1. The other end of the string isvibrating in the direction so that stationary wavesare set up in the string. the possible waveform ofthese stationary waves is (are)

(A) y (t) = A sin x 50 tcos6 3

(B) y (t) = A sinx 100 tcos3 3

(C) y (t) = A sin5 x 250 tcos6 3

(D) y (t) = A sin5 x2

cos 250 t

Sol. A,C,DV = 100 m/sL = 3mAt x = 0 nodeAt x = 3 antinodeSatisfying the condition in the equations onputting x = 3, sinkx should be equal to as itis an antinode.

sin 6x

6

350

sin 6x5

56

3250

sin 5 x2

25

250

5. At time t = 0, terminal A is the circuit shownin the figure is connected to B by a key and analternating current I(t) = I0cos(t) with I0=1A and=500 rad s-1 starts flowing in it with the initial di-

rection shown in the figure. At

7t . the key is

switched from B to D. Now onwards only A and Dare connected. A total charge Q flows from the bat-tery to charge the capacitor fully. If C=20F, R=10and the battery is ideal with emf of 50V, identify thecorrect statement (s).

(A) Magnitude of the maximum charge on the

capacitor before 7t6

is 1 × 10-3 C.

(B) The current in the left part of the circuit just

before

7t is clockwise.

(C) Immediately after A is connected to D, the cur-rent in R is 10A.(D) Q = 2×10-3C.


Sol. C,D

I = I0 cos t

V0 = 0Ic

= 610200.501

= 31010

1

= 2101 = 100

V = 100 cos (t – /2)V = 100 sin tq0 = CV = 20 × 10–6 × 100

= 2 × 10–3 sin × 76

= 2 × 10–3 sin 67

= 2 × 10–3 sin6

= –10–3 C

6. A light source, which emits two wavelengths1 = 400nm and 2 = 600 nm, is used in a Young’sdouble slit experiment. If recorded fringe widths for1 and 2 are 1 and 2 and the number of fringesfrom them within a distance y on one side of thecentral maximum are m1 and m2 respectively, then

(A) 2>1(B) m1>m2(C) From the central maximum, 3rd maximum

of 2 overlaps with 5th minimum of 1(D) The angular separation of fringes from 1

is greater than 2.Sol. A, B, C

= 400 nm 1 = (400)d

D

= 600 nm 2 = (600)d

D2>1

no. of fringes = y

Angular separation : BD d

Hence 1 2

Checking (C) 3×600× dD

= 9 × 400 × d2

D

1800 dD

= 1800 dD

The given option is true

7. In the figure, a ladder of mass m is shown leaning

against a wall. It is in static equilibrium making anangle with the horizontal floor. The coefficient offriction between the wall and the ladder is 1 and

that between the floor and the ladder is 2. Thenormal reaction of the wall on the ladder is N1 andthat of the floor is N2. If the ladder is about to slip,

then

(A) 1 = 0 2 0 and N2 tan mg2

(B) 1 0 2 0 and N1 tan mg2

(C) 1 0 2 0 and N2 1 2

mg1

(D) 1 0 2 0 and N1 tan mg2

Sol. C, D

N1 = 2N2N2 = mg + 1N1By torque balance:

l2

mg cos = (1N1 cos + N1 sin ) l

2mg

cos = 1N1 cos + N1 sin

1 = 0

2mg

cos = N1 sin

N1 tan = 2

mg

2 = 0 N1 = 0


8. Two ideal batteries of emf V1 and V2 and theseresistances R1, R2 and R3 are connected as shown inthe figure. The current in resistance R2 would bezero if

(A) V1 = V2 and R1 = R2 = R3(B) V1 = V2 and R1 = 2R2 = R3(C) V1 = 2V2 and 2R1 = 2R2 = R3(D) 2V1 = V2 and 2R1 = R2 = R3

Sol. A,B,D

VA = VB

V1 = iR1 ....(i)V2 = iR3 .....(ii)(i)/(ii)

2

1

VV

= 3

1

RR

9. Let E1(r), E2(r) and E3(r) be the respectiveelectric fields at a distance r from a point charge Q,an infinitely long wire with constant linear chargedensity , and an infinite plane with uniform surfacecharge density . If E1(r0) = E2(r0) = E3(r0) at a givendistance r0, Then

(A) 20Q 4 r

(B) 0r 2

(C) 1 0 2 0E r /2 2E r /2

(D) 2 0 3 0E r /2 4E r /2

Sol. C

01 2 2

00

r KQ 4KQE2 rr

2

0200

4K 8K2rrr

10. Heater of an electric kettle is made of a wire oflenght L and diameter d. It takes 4 minutes to raisethe temperature of 0.5 kg water by 40 K. This heateris replaced by a new heater having two wires of thesame material, each of length L and diameter 2d.The way these wires are connected is given in theoptions. Howe much time in munutes will it take toraise the temperature of the same amout of waterby 40 K?

(A) 4 if wires are in parallel(B) 2 if wires are in series(C) 1 if wires are in series(D) 0.5 if wires are in parallel

Sol. B,D

Since P = RV2

Series : Req = 2R

time half

Parallel : Req = 8R

time 81

times

11. A horizontal circular platform of radius 0.5 mand mass 0.45 kg is free to rotate about its axis.Two massless spring toy-guns, each carrying a steelball of mass 0.05 kg are attached to the platform ata distance 0.25b m from the centre on its eithersides along its diameter (see figure). Each gun si-multaneously fires the balls horizontally and perpen-dicular to the diameter in opposite directions. Afterleaving the platform, the balls have horizontal speedof 9 ms-1 with respect to the ground. The rotationalspeed of the platform in rad s-1 after the balls leavethe platform is

Sol. 0004Li = Lf0 = I – 2mvrI = 2mvr

= 2mvr

I =

5.05.045.09.009.02

× 2 = 4

12. A uniform circular disc of mass 1.5 kg and radius0.5 m is initially at rest on a horizontalfrictionless surface. Three forces of equalmagnitude F=0.5 N are applied simultaneously alongthe three sides of anequilateral triangle XYZ with itvertices on the perimeter ofthe disc (see figure). Onesecond after applying theforces, the angular speed of the disc is rad s-1 is


Sol. 0002

= I2

= 3F cos60 R

I

= 2F3

. )MR(2

2

= 2MRF3

= 2)5.0)(5.1()5.0(3

= 2 = t = 2 × 1 = 2

13. Consider an elliptically shaped rail PQ in the ver-tical plane with OP = 3 m and OQ = 4 m. A block ofmass 1 kg is pulled along the rail from P to Q with aforce of 18 N, which is always parallel to line PQ(see the figure given). Assuming no frictional losses,the kinetic energy of the block when it reaches Q is(n × 10) Joules. The value of n in (take accelerationdue to gravity = 10ms-2)

Sol. 0005

wmg + wF = KE

-1×4×10+18×5 = 21

×1×v2

-40+90 = 2

2V KE = 50

n = 0005

14. A galvanometer gives full scale deflection with0.006 A current. By connecting it to a 4990 resis-tance, it can be converted into a voltmeter of range

0 - 30 V. If connected to a 2n

246 resistance, it

becomes an ammeter of range 0 - 1.5 A. The valueof n isSol. 0005

ig = 0.006 Amp.

30 = 10006

[4990+ Rg]

4990 + Rg = 5000 Rg = 10

14946

49410060

10

.

.s

s = 24910

149460

n = 5

15. A thermodynamic system is taken from an initialstate i with internal energy Ui = 100 J to the finalstate f slong two different paths iaf and ibf, as sche-matically shown in the figure. The work done by thesystem along the paths af, ib and bf areWaf = 200 J, Wib = 50 J and Wbf = 100 J respectively.The heat supplied to the system along the path iaf,ib and bf are Qiaf, Qib and Qbf respectively. If theinternal energy of the system in the state b is Ub =200 J and Qiaf = 500 J, the ratio Qbf /Qid is


Sol. 0002

ib

bf

QQ

= ??

wiaf = 0 + 200 = 200Qiaf = 500 Thus Uiaf = 300 Uibf = 300But Uib = 100 Thus Ubf = 200 Qib = Wib + Uib = 50 + 100 = 150Qbf = Wbf + Ubf = 100 + (200) = 300

Ratio ib

bf

QQ

= 150300

= 2

16. To find the distance d over which a signal canbe seen clearly in foggy conditions, a railways engi-neer uses dimensional analysis and assumes thatthe distance depends on the mass density of thefog, intensity (power/area) S of the light from thesignal and its frequency f. The engineer finds that dis proportional to S1/n. The value of n isSol. 0003

L [ML–3]a [MT–3]b [T–1]c

L [Ma+b] [L–3a] [T–3b–c]a + b = 0 .....(1)–3a = 1a = –1/3 and b = 1/3 ......(2)n = 0003

17. Airplanes A and B are flying constant velocity inthe same vertical plane at angles 30° and 60° withrespect to the horizontal repectively as shown in

the figure. The speed of A is 1100 3ms . At time t= 0 s, an observer in A finds B at a distance of 500m. This observer sees B moving with a constantvelocity perpendicular to the line of motion of A. Ifat t = t0, A just escapes being hit by B, t0 in secondsis

Sol. 0005

AV

= 3100 cos 30 i + 3100 sin 30° j

= j250i100

BV

= x cos 60° i + x sin 60° j

= 2x

i + j23x

AB VV

= i1502x

+ j35023x

As A sees B at 90° to its line of motion hence

the angle between – x axis and BAV

= 60°

tan 60° = 2x150

35023x

150 – 2x

= 2x

– 50

x = 200

Hence BAV

= j350i50

|V| BA

= 22 )350()50(

= 50 × 2 = 100 m/s

Thus time to collide = 100500

= 5 sec

18. A rocket is moving in a gravity free space with aconstant acceleration of 2 ms-2 along + x direction(see figure). The length of a chamber inside therocket is 4 m. A ball is thrown from the left end ofthe chamber is +x direction with a speed of 0.3 ms-

1 relative to the rocket. At the same time, anotherball is thrown in -x direction with a speed of 0.2 ms-

1 from its right end relative to the rocket. The timein seconds when the two balls hit each other is


Sol. 0002 & 00084 = 0.2 × t × 1/2 × 2 × t2

t = 1.9~— 2 sec.

19. Two parallel wires in the plane of the paper aredistance X0 apart. A point charge is moving withspeed u between the wires in the same plane at adistance X1 from one of the wires. When the wirescarry current of magnitude l in the same direction,the radius of curvature of the path of the pointcharge is R1. In contrast, it the currents l in the twowires have directions opposite to each other, the

radius of curvature of the path is R2. If 0

1

X3

x , the

value of 1

2

RR is

Sol. 0003

Same direction:

B1 =

2

i0

101 xx

1x1

= )xx(x2)x2x(i

101

100

qVB = r

Mv2

r = qBMV

Opp. direction :

B2 =0i

2

1 0 1

1 1x x x

= 0i2

)xx(x

x

101

0

Thus 2

1

rr

= 1

2

BB

= )xx(xx

101

0

× 0 1 1

0 1

(x x )x(x 2x )

2

1

rr

= )x2x(x

10

0

= )x2x3(x3

11

1

= 3

20. During Searle’s experiment, zero of theVernier scale lies between 3.20×10-2 m and 3.25×10-

2 m of the main scale. The 20th division of the Ver-nier scale exactly coincides with one of the mainscale divisions. When an additional load of 2 kg isapplied to the wire, the zero of the Vernier scale stilllies between 3.20 ×10-2 m and 3.25×10-2 m of themain scale but now the 45th division of Vernier scalecoincides with one of the main scale divisions. Thelength of the thin metallic wire is 2 m and its cross-sectional area is 8 × 10-7 m2. The least count of theVernier scale is 1.0×10-5 m. The maximum percent-age error in the Young’smodulus of the wire isSol. 0004

Acc. to S7earle's experiment:

Amg

=

y

y =

Amg

ydy

=

d)(d

ydy

= 5

5

1025101

× 100

= 4 %= 0004


21. The correct combination of names of isomericalcohols with molecular formula C4H10O is/are:(A) tert-butanol and 2-methylpropan-2-ol(B) tert-butanol and 1, 1-dimethylethan-1-ol(C) n-butanol and butan-1-ol(D) isobutyl alcohol and 2-methylpropane-1-ol

Sol. ACDisomeric alcohol of C4H10O.

OH is ter. butanol or 2-methyl propan-

2-ol

OH is n-butanol or butan-1-ol

OH is iso butyl alcohol or 2-methyl

propan-1-ol.

22. The reactivity of compound Z with differenthalogens under appropriate conditions is givenbelow :

OH

C(CH )3 3Z

di halo substituted derivative when X = Br2 2

mono halo substituted derivative when X = l2 2

Tri halo substituted derivative when X = Cl2 2

X2

The observed pattern of electrophilicsubstitution can be explaind by :(A) the steric effect of the halogen(B) the steric effect of the tert-butyl group(C) the electronic effect of the phenolic group(D) the electronic effect of the tert-butylgroup

Sol. ABC

Products

I

OH

C(CH )3 3

Br

OH

C(CH )3 3

Br

and

Cl

OH

C(CH )3 3

Cl

Cl

are explained by steric effect of halogen &t-butyl group and electronic effect of phenolicgroup.

23. An ideal gas in a thermally insulated vesselat internal pressure = P1, volume = V1 andabsolute temperature = T1 expandsirreversibly against zero external pressure,as shown in the diagram. The final internalpressure, volume and absolute temperatureof the gas are P2, V2 and T2, respectively.For this expansion,

P =0ext

P =0ext

P , V , T1 1 1

P , V , T2 2 2

Irreversible

Thermal insulation(A) q = 0 (B) T2 = T1(C) P2V2 = P1V1 (D) P2V2

= P1V1

Sol. ABC(A) q = 0 adiabatic process(B) T2 = T1since Pext = 0, w = 0U = w + q = 0 + 0(C) P2V2 = P1V1since n and T are constant(D) PV = constantis applicable only for ideal gas in reversibleadiabatic process.

24. The correct statement(s) for the orthoboricacid is/are :(A) It behaves as a weak acid in water dueto self ionization.(B) Acidity of its aqueous solution increasesupon addition of ethylene glycol.(C) It has a three dimensional sructure dueto hydrogen bonding.(D) It is a weak electrolyte in water.

Sol. BD25. The pair(s) of reagents that yield

paramagnetic species is/are :(A) Na and excess of NH3(B) K and excess of O2(C) Cu and dilute HNO3(D) O2 and 2-ethylanthraquinol

Sol. ABCNa + NH3 Paramagnetic solution due topresence of solvated electronK + O2 KO2 (O2

– ion is paramagnetic, MOT)Cu + HNO3 NO + Cu(NO3)2 (both NO andCu(NO3)2 are paramagnetic)

O2 +

+ H2O2 (both are dimagnetic)

CHEMISTRY


26. In the reaction shown, below, the majorproduct(s) formed is/are :

(A)

O

NH2

O

CH3N

H

+ CH COOH3

(B) + CH3COOH

(C)

O

CH3N

H

O O

CH3NH

+ H O2

(D)

Sol. A

CH –C–O–C–CH3 3

••

SNAE

+ CH3COOH27. In a galvanic cell, the salt bridge

(A) does not participate chemically in thecell reaction.(B) stops the diffusion of ions from oneelectrode to another.(C) is necessary for the occurrence of thecell reaction.(D) ensures mixing of the two electrolyticsolutions.

Sol. AD

28. Upon heating with Cu2S, the reagent(s) thatgive copper metal is/are :(A) CuFeS2 (B) CuO(C) Cu2O (D) CuSO4

Sol. CCu2O + Cu2S Cu + SO2 (self reduction)

29. Hydrogen bonding plays a central role in thefollowing phenomena :(A) ice floats in water.(B) Higher Lewis basicity of primary aminesthan tertiary amines in aqueous solutions.(C) Formic acid is more acidic than aceticacid.(D) Dimerisation of acetic acid is benzene.

Sol. AD

30. For the reaction :I– + ClO3

– + H2SO4 Cl– + HSO4– + I2

The correct statement(s) in the balancedequation is/are :(A) Stoichiometric coefficient of HSO4

– is 6.(B) Iodide is oxidized(C) Sulphur is reduced(D) H2O is one of the products.

Sol. ABD +6 +6

6I– + ClO3– + 6H2 S O4 Cl– + 6H S O4

– + 3I2nf = 1 nf = 6 + 3H2O

31. A list of species having the formula XZ4 isgiven below.XeF4, SF4, SiF4, BF4

–, BrF4–, [Cu(NH3)4]

2+,[FeCl4]

2–, [CoCl4]2– and [PtCl4]

2–

Defining shape on the basis of the locationof X and Z atoms, the total number of specieshaving a square planar shape is.

Sol. 4XeF4, BrF4

–, [Cu(NH3)4]2+, [PtCl4]

2–

32. The total number(s) of stable conformerswith non-zero dipole moment for thefollowing compound is (are).

Cl

Br CH3

CH3

ClBr

Sol. 3

Cl

Br CH3

CH3

ClBr

µ 0,

Br CH3

Cl

Br

CH3 Clµ 0,

Br CH3

Cl

Cl Br

CH3

µ 0.


33. Consider the following list of reagents :Acidified K2Cr2O7, alkaline KMnO4, CuSO4,H2O2, Cl2, O3, FeCl3, HNO3 and Na2S2O3. Thetotal number of reagents that can oxidiseaqueous iodide to iodine is.

Sol. 6K2Cr2O7, CuSO4 , H2O2 , Cl2, O3, FeCl3These compound liberats iodine from iodideion.

34. Among PbS, CuS, HgS, MnS, Ag2S, NiS, CoS,Bi2S3 and SnS2, the total numer of BLACKcoloured sulphides is :

Sol. 7PbS, CuS, HgS, Ag2S, NiS, CoS, Bi2S3These compound are black in colour.

35. Consider all possible isomeric ketones,including stereoisomers of MW = 100. All theseisomers are independently reacted with NaBH4(NOTE : stereoisomers are also reactedseparately). The total number of ketones thatgive a racemic product(s) is/are.

Sol. 7Ketone with Mw = 100 is C6H12O

O

O O

O

O O

CH3

R

O

CH3

s

36. MX2 dissociated into M2+ and X– ions in anaqueous solution with a degree of dissociation(at of 0.5. The ratio of the observeddepression of freezing point of the aqueoussolution to the value of the depression offreezing point in the absence of ionicdissociation is .

Sol. 2MX2 M2+ + 2X–

1 – –1–0.5= 0.5 0.5 1

i

TT

thf

obsf

= 2

37. In an atom, the total number of electronshaving quantum numbers n = 4, |ml| = 1 and

ms = –12

is.

Sol. 6n = 4l = 0, 1, 2 , 3|ml| = 1 (only in p, d & f orbital)ms = –1/2

38. A compound H2X with molar weight of 80 g isdissolved in a solvent having density of 0.4 gm–1 . Assuming no change in volume upondissolution, the molality of a 3.2 molar solutionis.

Sol. 8Let’s take 1 litre of solution since its molarityis 3.2 it should contain 3.2 moles of solute.According to question on mixing 3.2 molessolute to one litre (1000 ml) solvent abovesolution can be prepared since no change involume upon dissolution.

m = 1000/Wn

here n = moles of soluteW = weight of solvent in gramsW = V × d= 1000 ml × 0.4 gm/ml= 400 gm

m = 3.2

400 /1000 = 8

39. If the value of Avogadro number is 6.023 ×1023 mol–1 and the value of Boltzmannconstant is 1.380 × 10–23 J K–1, then numberof significant digits in the calculated value ofthe universal gas constant is.

Sol. 4k = 1.380 × 10–23 J K–1 has 4 significant digitsand NA = 6.023 × 1023 mol–1 has 4 significantdigitshence R = k × NA has also four significantdigits

40. The total number of distinct naturallyoccurring amino acids obtained by completeacidic hydrolysis fo the peptide shown belowis :

O O O O

O

O

O

NN

N

NH

CH2CH2

H

NN N

N

H H

H HO O

Sol. 3Distinct naturally occuring amino acids mean-amine acids.

O

H N—C—C—OH2

H

H,

O

H N—C—C—OH2

H

CH2

,

O

H N—C—C—OH2

H


41. Let f : – ,2 2

R be given by

f(x) = (log (secx + tanx))3

Then(A) f(x) is an odd function(B) f(x) is an one-one function(C) f(x) is an onto fucntion(D) f(x) is an even function

Sol. A,B,C

f :

2,

2 R

f(x) = (log(sec x + tan x))3

f'(x) = 3(log (sec x + tan x))2 sec xf'(–x) = 3(log(sec x – tan x))2 sec x = f'(x) f'(x) is even f(x) is odd

42. Let a R and let f : R R be given byf(x) = x5 – 5x + aThen(A) f(x) has three real roots if a > 4(B) f(x) has only one real root if a > 4(C) f(x) has three real roots if a < – 4(D) f(x) has three real roots if –4 < a < 4

Sol. B,Df(x) : R R a Rf(x) = x5 – 5x + aConsider x5 – 5x + a = 0

or x5 – 5x = – alet f(x) = x5 – 5x g(x) = – a

= x(x4 – 5)f'(x) = 5x4 – 5 = 5(x2 + 1)(x2 – 1)F''(x) = 20x3

4

(5 , 0)1/4

(–5 , 0)1/4

–1 +1

–4

y

x

(1) –a < –4 one solution a > 4(2)–4 < –a < 4 3 solution 4 > a > –4(3) –a = 4, –4 2(4) –a > 4 one solution a < –4

43. Let f:(0, ) R be given by

f(x) 1x (t )t

1x

dte

t

.

Then(A) f(x) is monotonically increasing on [1, )(B) f(x) is monotonically decreasing on (0, 1)

(C) f(x) + f1x

= 0, for all x (0, )

(D) f(2x) is an odd function of x on RSol. A,C,D

f'(x) = x

e x1x

–

x1x

e

2x1x

= 0x

e.2x

ex

e x1x

x1x

x1x

(A) f(2x) = tdte

x

x

2

2

t1t

f(2x) = – f(2x) f is odd (D)

44. For every pair of continuous functions f, g :[0, 1] R such that max {f(x) : x [0, 1]} =max {g(x): x [0,1]}, the correctstatement(s) is (are):(A) (f(c))2 + 3f(c) = (g(c))2 + 3g(c) for somec [0,1](B) (f(c))2 + f(c) = (g(c))2 + 3g(c) for somec [0,1](C) (f(c))2 + 3f(c) = (g(c))2 + g(c) for somec [0,1](D) (f(c))2 = (g(c))2 for some c [0,1]

Sol. A,DLet Max. f(x) at x = c1& Max. g(x) at x = c2then f(c1) – g(c1) 0 &

f(c2) – g(c2) 0So f(c) – g(c) = 0 where c [c1, c2]By IVT

45. Let M and N be two 3 × 3 matrices such thatMN = NM. Further, if M N2 and M2 = N4,then(A) determinant of (M2 + MN2) is 0(B) there is a 3 × 3 non-zero matrix U suchthat (M2 + MN2)U is the zero matrix(C) determinant of (M2 + MN2) 1(D) for a 3 × 3 matrix U, if (M2 + MN2)Uequals the zero matrix then U is the zeromatrix

MATHEMATICS


Sol. A,BM2 – N4 = 0 (M + N2) (M – N2) = 0 ( MN = NM)as M – N2 0 |M + N2| = 0So A & B are correct options

46. A circle S passes through the point (0, 1)and is orthogonal to the circles (x – 1)2 + y2

= 16 and x2 + y2 = 1. Then(A) Radius of S is 8 (B) radius of S is 7(C) centre of S is (–7, 1)(D) centre of S is (–8, 1)

Sol. B,CLet S : x2 + y2 + 2gx + 2fy + c = 0pass (0, 1)1 + 2f + c = 0S & S1 orthogonal2(-1)(g) + 2(0) (f) = c – 15–2g = c – 15S & S2 or thogonal2(0) (g) + 2(0) (f) = c – 1

c – 1 = 0c = 1, g = 7 f = –1

S : x2 + y2 + 14x – 2y + 1 = 0r = 7C : (–7, 1)

47. Let M be a 2 × 2 symmetric matrix with inte-ger entries. Them M is invertible if(A) the first column of M is the transpose of

the second row of M(B) the second row of M is the transpose of

the first column of M(C) M is a diagonal matrix with nonzero en-

tries in the main diagonal(D) the product of entries in the main diago-

nal of M is not the square of an integerSol. C,D

(A) M a aa a

M is not invertible

(B) M = a aa a

M is not invertible

(C) a 00 b

M is invertible

(D) M = b aa c

|M| = bc – a2 clear M is

invertible.

48. Let x

, y

and z

be three vectors each ofmagnitude 2 and the angle between each

pair of them is 3

. If a is a nonzero vector

perpendicular to x

and y

× z

and b is a

nonzero vector perpendicular to y

and z

×

x

, then

(A) b (b · z)(z – x)

(B) a (a · y)(y – z)

(C) a · b –(a · y)(b · z)

(D) a (a · y)(z y)

Sol. A,B,CGiven 1y.x

1z.y

1x.z

Now a = (x

× (y

× z ))

a = ((x

. z )y

– (x

.y

) z )

a = (y

– z

)

Now a .y

= (y2 – y

. z )

a .y

= (2 – 1)

= a .y

a = (a

.y

) (y

– z ) (B)

Similarly b = (b

. z ) (z

–x ) (A)

a .b

= (a

.y

) (b. z ) {(y

– z

).(z –x

)}

= (a .y

) (b

. z ) {1 – 2 – 1 + 1}

a .b

= – (a

.y

) (b. z ) (C)

49. Let f : [a, b] [1, ) be a continuous func-tion and let g : R R be defined as

f(x) =

x

ab

a

0 if x a,

f(t)dt if a x b,

f(t)dt if x b

Then(A) g(x) is continuous but not differentiable at a(B) g(x) is differentiable on R(C) g(x) is continuous but not differentiable at b(D) g(x) is continuous and differentiable at ei-

ther a or b but not bothSol. A,C

f : [a, b] [1, )(A) g(x) is continuousat x = aClearly not derivableas f(a) 1 which is RHDwhile LHD is 0is (B) in wrongagain (C) is correct as at x = bcontinous But not derivable(D) is wrongSo, A & C


50. From a point P(), perpendiculars PQ andPR are drawn respectively on the lines y = x,z = 1 and y = – x, z = – 1. If P is such thatQPR is a right angle, then the possible value(s) of is(are)

(A) 2 (B) 1

(C) –1 (D) 2Sol. C

y = x, z = 1y = – x, z = – 1

x1

= y1

= z –1

0 =

Q(, , 1)

x–1

= y1

= z 1

0

=

R (–, , –1)

QP

· PR

= 1 ˆ( – )i ( – )j ( 1)k

ˆ( )i ( – )j ( 1)k = 0

( – )( + ) + ( – )( – ) + 2 – 1 = 02 + – – + 2 – – + + 2 – 1 =032 – 2 – 1 = 0 ...(i)

PQ

· (i j) = 0

– + – = 0 = ...(ii)

and PR

· (–i j) = 0

– ( + ) + – = 0 = 0 ...(iii)from (i) (ii) and (iii)(i) 32 – 22 – 1 = 0 2 = 1 = ± 1But at = 1P QSo answer is (C)

51. The value of 21 3 2 520

d4x (1 x ) dxdx

is

Sol. 0002

dx)x1(dxdx4

1

0

II

522

2

I

3

= 4x3 . dxd

(1 – x2)5 0

1

– dx)x1(dxd.x12 521

0

2

= – 1

0

422 dx)x1(5)x2(x12

= 0+60 0

1

422 dx)x1)(x(x2 (put 1 – x2= t)

= – 60 0

1

4 dt)t()t1( = 60

301

= 2

52. Let f : R R and g : R R be respectivelygiven by f(x) = [x] + 1 and g(x) = x2 + 1.Define h : R R by

h(x) = max {f(x),g(x)} if x 0,min {f(x),g(x)} if x 0

The number of points at which h(x) is notdifferentiable is

Sol. 0003||||||||||||||||||||||||||||||||||

|||||

|||g(x)

f(x)

(0,1)

f(x) = | x | + 1g(x) = x2 + 1

h(x) = max(f(x), f(g)) x 0min(f(x),g(x)) x 0

No of non differentiable points = 3.

53. Let n 2 be an integer. Take n distinct pointson a circle and join each pair of points by aline segment. Colour the line segment joiningevery pair of adjacent points by blue andthe rest by red. If the number of red andblue line segments are equal, then the valueof of n is

Sol. 0005n 2n-Blue lines

total lines = n2C

Red lines = n2C – n

n 12

3

4

5givenn

2C – n = n

n(n–1)2

= 2n

n2 – n = 4nn2 = 5nn = 5.


54. The slope of the tangent to the curve(y – x5)2 = x(1 + x2)2 at the point (1, 3) is

Sol. 0008(y – x5)2 = x(1 + x2)2 (1, 3)

2(y – x5).

4x5dxdy

=x.2(1 + x2).2x + (1 +

x2).1

2(y – x5)

4x5dxdy

= (1 + x2) (5x2 + 1)

2(3 – 1)

5dxdy

= (1 + 1) (5 + 1) = 12

dxdy

– 5 = 3

dxdy

= 8

Similar to teaching notes

55. Let a,b and c

be three non-coplanar unit

vectors such that the angle between every

pair of them is 3

. If

a b b c pa qb rc

, where p, q and r

are scalars, then the value of 2 2 2

2p 2q r

q

isSol. 0004

(a b)

+ (b c)

= pa

+ qb

+ rc

Taking dot product with a,b & c

0 = p2

+ q + r2

p + 2q + r = 0

0 + 12 = p +

q2

+ r2

2p + q + r = 2

12

+ 0 = p2

+ q2

+ r p + q + 2r = 2

p = 12

, q = –12

, r = 12

2 2 2

2p 2q r

q

=

1 112 2

12

= 4

56. Let a, b, c be positive integers such that ba

is an integer. If a, b, c are in geometric pro-gression and the arithmetic mean of a, b, c

is b + 2, then the value of 2a a 14

a 1

is

Sol. 0004a, ar, ar2

2a(1 r r )3

= ar + 2

a + ar + ar2 = 3ar + 6a + ar2 = 2ar + 6a(r – 1)2 = 6 a = 6, r – 1 = 1

2a a–14

a 1

= 36 6 –14

7

= 4.

57. Let f:[0, 4] [0, ] be defined by f(x) =cos–1 (cosx). The number of points x [0,4] satisfying the equation.

f(x) = 10 x

10

is

Sol. 0003f : [0, 4] [0, ]

f(x) = cos–1 (cos x) f(x) = 10

x10

3 10

4 x

10x10y

2

0

y=cos (cos x)–1

Number of solution = number of points ofintersection = 3

58. The largest value of the non-negative inte-ger a for which

1 x1 x

x 1

ax sin(x 1) a 1limx sin(x 1) 1 4

is

Sol. 0002

1 x

x 1

a(x 1) sin(x 1) 1Lim(x 1) sin(x 1) 4

1 x

x 1

sin(x 1)a 1x 1Limsin(x 1) 41

(x 1)

41

111a 2

(–a + 1)2 = 1–a + 1 = 1 or –a + 1 = –1a = 0 or a = 2Max value of a = 2


59. For a point P in the plane, let d1(P) andd2(P) be the distances of the point P fromthe lines x – y = 0 and x + y = 0 respectively.The area of the region R consisting of allpoints P lying in the first quadrant of theplane and satisfying 2 d1(P) + d2(P) 4 is

Sol. 0006x – y = 0 & x + y = 0

d1 = h–k

2 , d2 = h k

2

d2d1

x = y

2 d1 + d2 4

2 h – k

2 + h k

2

4

2 2 |h – k| + |h + k| 4 2

(i) only for 1st quad

h>0, k 0

(A) h k 2 2 h – k + h + k 4 2

2 h 2 2

(B) h < k 2 2 k – h + h + k 4 2

2 22 k 2 2

A (2 2) –( 2)

A = 8 – 2 = 6.

60. Let n1 < n2 < n3 < n4 < n5 be positive inte-gers such that n1 + n2 + n3 + n4 + n5 = 20.Then the number of such distinct arrange-ments (n1, n2, n3, n4, n5) is

Sol. 0007Only 7 possible combination and eachcombination can be arranged in only one way1, 2, 3, 4 101, 2, 3, 5 91, 2, 3, 6 81, 2, 4, 5 81, 2, 4, 6 71, 3, 4, 5 72, 3, 4, 5 6 Ans = 7

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