JEE Advanced 2014 Solution Paper I

8/10/2019 JEE Advanced 2014 Solution Paper I

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IIT - JEE 2014 (Advanced)

CODE : 1


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(2) Vidyalankar : IIT JEE 2014 Advanced : Question Paper & Solution

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IIT JEE 2014 Advanced : Question Paper & Solution (Paper I) (3)

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PART I : PHYSICS

SECTION 1 : (One or More Than One Options Correct Type)

This section contains 10 multiple choice questions. Each question has four choices (A),

(B), (C) and (D) out of which ONE or MORE THAN ONE are correct.

1. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key andan alternating current I(t) = I0cos (t), with I0= 1A and = 500 rad s

1starts flowing in

it with the initial direction shown in the figure. At t =7

6

, the key is switched from B to

D. Now onwards only A and D are connected. A total charge Q flows from the battery tocharge the capacitor fully. If C = 20F, R = 10 and the battery is ideal with emf of 50V,identify the correct statement (s).

(A) Magnitude of the maximum charge on the capacitor before t =7

6

is 1 103C.

(B) The current in the left part of the circuit just before t =7

6

is clockwise.

(C) Immediately after A is connected to D, the current in R is 10A.(D) Q = 2 103C.

1. (C), (D)

If q represents the charge on capacitors upper plate:

I (t) = I0cos(t) =dq

dtq(t) = 0

I

sin(t)

Max charge =1

1A

500 rad s

= 2 103 C

Charge on upper plate at t =7

6

= 0I

sin7

6

= 0I

2

When capacitor is fully charged; charge on upper plate = 50 V 20F= 1 103C

Q = 1 102C 0I

2

= 2 103C

Voltage across capacitor when A and D are connected =3

6

1 10 C

20 10 F

= 50V

Total voltage across resistor = 100V

current =100v

10

= 10A

Current is anti clockwise.


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2. A light source, which emits two wavelengths 1= 400 nm and 2= 600 nm, is used in aYoung's double slit experiment. If recorded fringe widths for 1and 2are 1and 2andthe number of fringes for them within a distance y on one side of the central maximum arem1and m2, respectively, then(A)2> 1

(B) m1> m2(C) From the central maximum, 3rdmaximum of 2overlaps with 5thminimum of 1

(D) The angular separation of fringes for 1is greater than 2

2. (A), (B), (C)

= D

d

12 2 1n m

2

=

n2600 = 1400

m

2

3n2= m1For n2= 3m1 = 9which is 5thminima.

3. One end of a taut string of length 3m along the x axis is fixed at x = 0. The speed of thewaves in the string is 100 ms1. The other end of the string is vibrating in the y directionso that stationary waves are set up in the string. The possible waveform(s) of thesestationary waves is (are)

(A) y(t) =

x 50 t

A sin cos6 3

(B) y(t) =

x 100 t

A sin cos3 3

(C) y(t) =5 x 250 t

A sin cos6 3

(D) y(t) =

5 xA sin cos t

2

250

3. (A), (C), (D)

=k

1A B C D 100ms

= = = =

x = 3 is an antinode. This eliminates (B)

4. A parallel plate capacitor has a dielectric slab of dielectric constant Kbetween its plates that covers 1/3 of the area of its plates, as shown in

the figure. The total capacitance of the capacitor is C while that of theportion with dielectric in between is C1. When the capacitor is charged,the plate area covered by the dielectric gets charge Q1and the rest ofthe area gets charge Q2. The electric field in the dielectric is E1and thatin the other portion is E2. Choose the correct option/options, ignoringedge effects.

(A) 1

2

E

E= 1 (B) 1

2

E

E=

1

K

(C)1

2

Q

Q =

3

K (D) 1

C

C =

2

K

+


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4. (A), (D)

If c = oA

3d

then c1= kc c2= 2c

c1+ c2= C(k + 2) c = C c =c

k 2+

c1= kck 2+

c2= 2Ck z+

If charging voltage is V: charges will be in the ratio of capacities and as potentialdifference is same. Electric field should be equal.

5. Let E1(r), E2(r) and E3(r) be the respective electric fields at a distance r from a pointcharge Q, an infinitely long wire with constant linear charge density , and an infiniteplane with uniform surface charge density . If E1(r0) = E2(r0) = E3(r0) at a given distancer0, then

(A) Q = 204 r (B) r0=2

(C) E1(r0/2) = 2E2(r0/2) (D) E2(r0/2) = 4E3(r0/2)

5. (C)

202 0 00 0 00

20 02

0 00

1 q2 r q 2 r 2 r4 2 rr

1 qq 2 r r

4 2r

= = =

== =

1 0E (r / 2)

4= 0

E(r / 2)

2

6. A student is performing an experiment using a resonance column and a tuning fork offrequency 244 s1. He is told that the air in the tube has been replaced by another gas(assume that the column remains filled with the gas). If the minimum height at whichresonance occurs is (0.350 0.005) m, the gas in the tube is

(Useful information : 167RT = 640 j1/2mole1/2; 140RT = 590 J1/2mole1/2. THe

molar masses M in grams are given in the options. Take the values of10

Mfor each gas

as given there)

(A) Neon 10 7M 20,20 10

= =

(B) Nitrogen 10 3M 28,28 5

= =

(C) Oxygen10 9

M 32,32 16

= =

(D) Argon

10 17M 36,

36 32

= =

6. (D)

f = 244 Hz

4

= 0.356 0.005= 1.400 0.020= (341.6 4.88)

m

s

= f = =rRT

M

=100 rRT

100M

=

3

100RrT

100 M 10 kg / mol


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for neon :7

64010

= 448 ms1

for Nitrogen :3

5905

= 384 ms1

for Oxygen : 590

9

16 = 351.875 ms

1

for Argon : 640 17

32= 340 ms1

7. Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by anew heater having two wires of the same material, each of length L and diameter 2d. Theway these wires are connected is given in the options. How much time in minutes will ittake to raise the temperature of the same amount of water by 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

7. (B) (D)

MST =2

tR

M = 0.5 S T = 40K.

Q = 0.5 S 40 =2

4 minR

R =L

d

4

2 R1 =

2

L R=

4(Rd)

4

Req.seriesR R R+ =4 4 2

Q =2 2 4 = t

RR

2

t = 2 min

ReqParallel =

R R

R4 4 =R R 8+4 4

Q =2

4R

=2

tR

8

t = 0.5 min

8. In the figure, a ladder of mass m is shown leaning against a wall. It is instatic equilibrium making an angle with the horizontal floor. Thecoefficient of friction between the wall and the ladder is 1 and thatbetween the floor and the ladder is 2. The normal reaction of the wallon the ladder is N1and that of the floor is N2. If the ladder is about toslip, then

(A)1= 0, 20 and N2tan =mg

2

(B) 10, 2= 0 and N1tan =mg

2

(C) 10, 20 and N2 =1 2

mg

1+ (D)1= 0, 20 and N1tan =

mg

2


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8. (C) (D)

N2+ 1N1= MgN1 = 2N2N2+ 12N2= Mg

N2=1 2

Mg

1+

N1lsin = Mg cos 2

l

N1tan =Mg

2

9. 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 long solidglass cylinder of refractive index n2= 1.5, as shown inthe figure. Rays of light parallel to the axis of thecylinder traversing through the film from air to glassget focused at distance f1from the film, while rays oflight traversing from glass to air get focused atdistance f2from the film. Then(A) | f1| = 3R (B) | f1| = 2.8 R (C) | f2| = 2R (D) | f2| = 1.4R

9. (A) (C)

1.4 1 1.4 1=

R

1.4 4=

R

7R=

2

1.5 1.5 1.5 1.4=

7R +R

2

1.5 4 1=

R R

1.5 5=

R = 3R f1= 3R

1.4 1.5 1.4 1.5=

R

1.4 1=

R

= 14f1 1.4 1 1.4

= 14R R

1 1 +0.4=

10R +R

1

=

0.4 0.1 0.5+ =

R R R

R =0.5

= 2R f2= 2R

N2

N1 1N1

2N2


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10. Two ideal batteries of emf V1and V2and three resistances R1, R2and R3are connected asshown in the figure. The current in resistance R2would be zero if

(A) V1= V2and R1= R2= R3 (B) V1= V2and R1= 2R2= R3(C) V1= 2V2and 2R1= 2R2= R3 (D) 2V1= V2and 2R1= R2= R3

10. (A), (B), (D)

iR1+ i1R2 = V1i1R2(i i1) R3= V2i1(R2+ R3) iR3= V2i1(R2+ R3) iR3= V2R1i1R2+ iR1= V1R3i1(R1R2+ R1R3+R2R3) = V1R3V2R1

i1 =1 3 2 1

1 2 1 3 2 3

V R V R

R R R R R R

+ +

V1R3= V2R1(a) V1= V2 R1= R3

(b) V1= V2 R1 = R3(d) 2V1= V2 R3= 2R1

SECTION 2 : (One Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one

integer from 0 to 9 (both inclusive).

11.Airplanes A and B are flying with constant velocity in the same vertical plane at angles30and 60with respect to the horizontal respectively as shown in figure. The speed of A

is 100 3 ms1. At time t = 0 s, an observer in A finds B at distance of 500 m. This

observer sees B moving with a constant velocity perpendicular to the line of motion of A.If a t = t0, A just escapes being hit by B, t0in seconds is

i

i1

V1

i1

R1

i

V2

R3 i i1


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13. [2]

net = 31 = 3 F r sin 30 = 3 (0.5) (0.5)1

2=

3

8N-m

I =21.5(0.5) 3

2 16=

=I = 2 rad/s2

= t = 2 rad/s

14. Two parallel wires in the plane of the paper are distance X0 apart. A point charge ismoving with speed u between the wires in the same plane at a distance X1from one of thewires. When the wires carry current of magnitude I in the same direction, the radius ofcurvature of the path of the point charge is R1. In contrast, if the currents I 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

R

R

is

14. [3]

X1=0X

3

X2=02X

3

r =mu

qB

1 2

2 1

R B

R B=

B1=0

1 2

I 1 1

2 x x

= 0

0 0

I 3 3

2 x 2x

= 0

0

3 I

4 x

B2=0

1 2

I 1 1

2 x x

+

= 0

0

9 I

4 x

1 2

2 1

R B3

R B= =

15. To find the distance d over which a signal can be seen clearly in foggy conditions, arailways engineer uses dimensional analysis and assumes that the distance depends on themass density of the fog, intensity (power/area) S of the light from the signal and itsfrequency f. The engineer finds that d is proportional to S1/n. The value of n is

15. [3]d = vasbfc

d =a b c

2

kg w 1

m3 secm

d = ( )b c2 3a

32

ML T 1ML

secL

1 a b 3a 3b cL M L T+ =

3a = 1a = 13

0.5

30

X0

X1


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a + b = 0 b =1

3

3b c = 0 c = 1

d =1 1

13 3v s f

d s1/n

n = 3

16. A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a4990 resistance, it can be converted into a voltmeter of range 0 30 V. If connected to

a2n

249 resistance, it becomes an ammeter of range 0 1.5 A. The value of n is

16.[5]

0.006 (4990 + Rg) = 306 (4990 + Rg) = 30000Rg= 10

.006 10 = (1.5 006) 2n

249

2490 1.5001 249

2n .006= =

10 = 2nn = 5

17. Consider an elliptically shaped rail PQ in the vertical planewith OP = 3 m and OQ = 4 m. A block of mass 1 kg is pulledalong the rail from P to Q with a force of 18 N, which is alwaysparallel to line PQ (see the figure given). Assuming nofrictional losses, the kinetic energy of the block when it reachesQ is (n 10) Joules. The value of n is (take acceleration due togravity = 10 ms2)

17. [5]w = 18 5 = 90 joules, u = 1 10 4 = 40 jouleswF+ wg= K = 90 40 = 50n 10 = 50n = 5

18. A rocket is moving in a gravity free space with a constant acceleration of 2 ms2along + xdirection (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrownfrom the left end of the chamber in + x direction with a speed of 0.3 ms1relative to therocket. At the same time, another ball is thrown in x direction with a speed of 0.2 ms1from its right end relative to the rocket. The time in seconds when the two balls hit eachother is

G

Rg 4990Ig

Rg

R1.5 .006

.006


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18.[2]w.r.t. rocket,

tffo 1st=

( )2 0.30.3 sec.

2=

So it will keep on colliding in interval of 0.3 sec.

For other,4 = 0.2t t2t2+ 0.2 t 4 = 0

t = ( )0.2 0.04 16

4.01 0.1 1.9sec.2

+=

So they must collide between interval of 1.8 to 1.9 sec while 7 th trip of 1st ball. So itshould be close to 2 sec.

19.A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about itsaxis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are

attached to the platform at a distance 0.25 m from the centre on its either sides along itsdiameter (see figure). Each gun simultaneously fires the balls horizontally andperpendicular to the diameter in opposite directions. After leaving the platform, the ballshave horizontal speed of 9 ms1with respect to the ground. The rotational speed of theplatform in rad s1after the balls leave the platform is

19. [4]

2 .05 9 .25 =

2

.45 .52

2 .05 9 .25 =.45 .25

2

= 4 rad/sec

20.A thermodynamic system is taken from an initial state i with internal energy U i= 100 J tothe final state f along two different paths iaf and ibf, as schematically shown in the figure.The work done by the system along the paths af, ib and bf are W af= 200 J, Wib= 50 J andWbf= 100 J respectively. The heat supplied to the system along the path iaf, ib and bf are

Qiaf, Qiband Qbfrespectively. If the internal energy of the system in the state b is Ub= 200 Jand Qiaf=500 J, the ratio

bf

ib

Q

Qis

a

b

f

P


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20. [2]Waf = 200 jWbf = 100 jWib = 50 jWia = 0 jQiaf, Qib, QbfUb = 200 j Qiaf= 500 jQiaf = Uiaf+ Wiaf500 = Uiaf+ 200 j

Uiaf = 300 j Uibf = 300 j

Wibf = 150 j Qibf = 450 j = Qib+ Qbf

Ui = 100 j Ub= 200 jUjb = 100 j Wib= 50 j

Qib= 150 j

450 = Qib+ Qbf450 = 150 j + QbfQbf = 300 j

bf

ib

Q 3002.

Q 150= =

PART II : CHEMISTRY

SECTION 1 : (One or More Than One Options Correct Type)

This section contains 10 multiple choice questions. Each question has four choices (A),(B), (C) and (D) out of which ONE or MORE THAN ONE are correct.

21. The correct combination of names for isomeric alcohols with molecular formula C4H10Ois/are(A) tertbutanol and 2methylpropan2ol(B) tertbutanol and 1, 1dimethylethan1ol(C) nbutanol and butan1ol(D) isobutyl alcohol and 2methylpropan1ol

21. (A), (C), (D)

Factual.

22. The reactivity of compound Z with different halogens under appropriate conditions isgiven below :


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The observed pattern of electrophilic substitution can be explained by(A) the steric effect of the halogen(B) the steric effect of the tertbutyl group(C) the electronic effect of the phenolic group(D) the electronic effect of the tertbutyl group

22. (A), (B) , (C)

Factual.

23.In the reaction shown below, the major product(s) formed is/are

23. (A)

NH2

O

NH2

2

2 2

Ac O

CH Cl

NH2

O

NHCOCH3

(major)

24. An ideal gas in a thermally insulated vessel at internal pressure = P1 , volume = V1andabsolute temperature = T1expands irreversibly against zero external pressure, as shown inthe diagram. The final internal pressure, volume and absolute temperature of the gas areP2, V2and T2, respectively. For this expansion,

(A) (B)

(C) (D)


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29.For the reaction :I+ CIO3

+H2SO4CI+ HSO4

+ I2The correct statement(s) in the balanced equation is/are:(A) Stoichiometric coefficient of HSO4

is 6.(B) lodide is oxidized.

(C) Sulphur is reduced.(D) H2O is one of the products.29. (A), (B), (D)

3 2 4 2 2 4(R.A) (O.A)

6I ClO 6H SO Cl 3I 3H O SO + + + + + 6

30. 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) O2and 2-ethylanthraquinol

30. (A), (B), (C)

2 2K excess O KO (Superoxide)+ Paramagnetic

3 3 2 23Cu 8HNO 3Cu(NO ) 2NO 4H O+ + + Paramagnetic

SECTION 2 : (One Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one

integer from 0 to 9 (both inclusive).

31. 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 that give a racemic product(s) is/are

31. [5]

CnH2nO = 100 14n = 100 16 = 84 n = 6

I)| |

C C C C C C

O

II)| |

C C C C C C

O

III)|| |

C C C C C

CO

IV)||

|C

C C C C C

O

Et

OH

OH

2O

Et

O

O

H2O

2+ Dimagnetic


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NaBH4

CH3 - CH - C - C - C -COH dl+-

NaBH4 - CH - C - C - C

OH dl+-

C - C

NaBH4 - CH - C - C

OH dl+-

C - C

C

NaBH4 - C - C - C

OH dl+-

C - CH

C

NaBH4 - C - C

OH dl+-

C - CH

C

C

V)|

|| |

C

C C C C C

CO

I)

II)

III)

IV)

V)

32. A list of species having the formula XZ4is given below.

XeF4, SF4, SiF4, BF4

, [Cu(NH3)4]2+

, [FeCI4]2

, [CoCl4]2

and [PtCI4]2

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

32. [4]

XeF4, 4BrF , [Cu(NH3)4]

+2 , [PtCl4]2

33. Among PbS, CuS, HgS, MnS, Ag2S, NiS, CoS, Bi2S3 and SnS2, the total number ofBLACKcoloured sulphides is

33. [7]

PbS , CuS, HgS, Ag2S, NiS, CoS, Bi2S3

34. The total number(s) of stableconformers with non-zerodipole moment for the followingcompound is (are)


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42. For every pair of continuous functions f, g: [0, 1] such that

max {f(x) : x [0, 1]} = max {g(x) : x [0, 1]},the correct statement(s) is (are):(A) (f(c))2+ 3f(c) = (g(c))2+ 3g(c) for some c [0, 1](B) (f(c))2+ f(c) = (g(c))2+ 3g(c) for some c [0, 1]

(C) (f(c))2+ 3f(c) = (g(c))2+ g(c) for some c [0, 1](D) (f(c))2= (g(c))2for some c [0, 1]

42. (A), (D)Since f(x) and g(x) are continuous function and their maximum values are equal, theirgraphs will intersect at atleast one point in [0, 1]. f(c) = g(c) for some c [0, 1] Options (A) and (D) are correct.Options (B), (C) can be eliminated by taking f(x) = 1 and g(x) = 1

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

f(x) =

1(t )x

t1

x

e +

dt

t

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

(C) f(x) + f1

x

= 0, for all x (0, )

(D) f(2x) is an odd fuction of x on

43. (A), (C), (D)f : (0, ) R

f(x) =

1x tt

1/x

dte

t

+

f(x) =

1t

x t

1/x

edt

t

+

1 1x x

x x

2

e e 1

f '(x) 1x 1/ x x

+ +

= 1

xx 1 1f '(x) e

x x

+

= +

1x

x2ef '(x)

x

+

=

f(x) is monotonically increasing on [1, )

f(x) =

1x tt

1/x

e dt

+

. (i)


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11/x tt

x

1f e dt

x

+

=

1x tt

1/x

1f e dt

x

+

=

.. (ii)

equation (i) + (ii)

f(x) +1

fx

= 0

If we put x = 2xf(2x) + f(2x) = 0f(2x) = f(2x)It means f(2x) is odd function.

44. Let a and let f: be given by

f(x) = x55x + 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

44. (B), (D)f(x) = x55x + af(x) = 5x45 = 0x = 1f(1) = a + 4 and f(1) = a 4

f(x) has three real roots if 4 < a < 4and f(x) has one real root if a < 4 or a > 4.

45. Let f : [a, b] [1, ) be a continuous function and let g : be defined as

g(x) =x

a

b

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

(C) g(x) is continuous but not differentiable at b(D) g(x) is continuous and differentiable at either a or b but not both

45.(A), (C)

x a

Lt g(x) 0,

= a

x a a

Lt g(x) g(a) f (t)dt 0+

= = = .

Similarly,

b

x b x b aLt g(x) g(b) Lt g(x)dx f (t)dt + = = =

g(x) is continuous at both x = a and x = b

(a 4)

(a + 4)

1 1


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Also, g(a) = 0 and g(a+) = f(a)g(b) = f(b) and g(b+) = 0

But since f(a) and f(b) must be greater than equal to 1, g(x) is not differentiable at x = aand x = b

46. Let f : ,2 2

be given by

f(x) = (log (sec x + tan x))3.Then(A) f(x) is an odd function (B) f(x) is a one-one function(C) f(x) is an onto function (D) f(x) is an even function

46. (A), (B), (C)

( ) ( ){ }3

f x log sec x tan x= +

( ) ( ){ }

3f x log sec x tan x =

=

31

logsec x tan x

+

= ( ){ }3

log sec x tan x +

= f (x) odd function

( ) ( ){ } { }2 21f ' x 3 log sec x tan x . sec x tan x sec x

sec x tan x= + +

+

= ( ){ }

2

3 log sec x tan x sec x 0= + > as x ,2 2

one oneRange is RSo onto function

47.From a point P(, , ), perpendiculars PQ and PR are drawn respectively on the lines y =x, z = 1 and y = x, z = 1. If P is such that QPR is a right angle, then the possiblevalue(s) of is (are)

(A) 2 (B) 1

(C) 1 (D) 2

47. (C)

From (, , ), foot of the perpendicular on the linex 0

1

=

y 0

1

=

z 1

0

= r is (, , 1)

Similarly. Foot of the perpendicular on the line

x 0

1

=

y 0

1

=

z 1

0

+= r1is (0, 0, 1)

Since QP PR,() (0) + () (0) + (1) (+ 1) = 0 = 1. At = 1, the point P lies on the 1stline.

= 1.


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48.Let x

, y

and z

be three vectors each of magnitude 2 and the angle between each pair

of them is3

. If a

is a nonzero vector perpendicular to x

and y

z

and2

2 2 2

q

p 2q r+ +

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

)

48. (A), (B), (C)

Given | x | | y | | z | 2= = =

1x y | x | | y | cos 60 2 2 1

2 = = =

Similarly, y z 1 & z x 1 = =

Also, x (y z) (x z) y (x y) z =

a y z =

.. (i)Again, y (z x) (y x)z (y z)x =

b z x =

. (ii)

From (i)a y (y z) y =

a y y y z y =

a y 2 1 =

1a y =

Similarly1

b z =

1 1a y b z = =

1

(a y) (b z) =

.. (iii)

y z z xa b

=

y z z z y x z xa b

+ =

1a b =

. (iv)

Hence (A), (B), (C) are correct.

49.A circle S passes through the point (0, 1) and is orthogonal to the circles (x 1)2+ y2= 16and 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)

49. (B), (C)

Given circlesx2 + y22x 15 = 0 &x2+ y21 = 0


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52.Let n 2 be an integer. Take n distinct points on a circle and join each pair of points by aline segment. Colour the line segment joining every pair of adjacent points by blue andthe rest by red. If the number of red and blue line segments are equal, then the value of nis.

52.[5]

Number of red line segments = n(n 3)2

Number of blue line segments = n

n(n 3)

n n 52

= =

53.Let n1< n2< n3< n4< nsbe positive integers such that n1+ n2+ n3+ n4+ n5= 20. Thenthe number of such distinct arrangements (n1, n2, n3, n4, n5) is

53. [7]Under given restrictions, following arrangements are possible :

{1, 2, 3, 4, 10}{1, 2, 3, 5, 9}{1, 2, 3, 6, 8}{1, 2, 4, 5, 8}{1, 2, 4, 6, 7}{1, 3, 4, 5, 7}{2, 3, 4, 5, 6}

54. Let f : R R and g : R R be respectively given by f(x) = |x| + 1 and g(x) = x 2+ 1.Define h : R R by

h(x) =

max {f (x), g(x)} if x

min {f (x), g(x)} if x 0.

0,

>

The number of points at which h(x) is not differentiable is

54. [3]

h (x) is not differentiable at 3 points because sharp edge.

55.The value of1 2

3 22

0

d4x (1 x ) dx

dx

is

55. [2]

( )( )51 2 23

20

d 1 x dx4x

dx

= ( ) ( )

133 2 2

0

4x 10 1 x 9x 1 dx

put x2= t and 2xdx = dt

= ( )1

5 4 3 2

0

20 dt9t 28t 30t 12t t + + = 2

1

2

n

(1, 2)

(0, 1)

(1, 2)h (x)

g (x)

f (x)


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56.The slope of the tangent to the curve (y x5)2= x(1 + x2)2at the point (1, 3) is56.[8]

(yx5)2= x (1 + x2)2at x = 1(y 1)2= 4

(y 1) = 2y 1 = + 2 y = 3it means,

( )5 2y x x 1 x = +

( )2 5y x 1 x x= + +

( ) ( )2 4dy 1

1 x x 2x 5xdx 2 x

= + + +

at x = 1,dy

8dx

=

57.The largest value of the non-negative integer a for which1 x

1 x

x 1

ax sin (x 1) alim

x sin(x 1) 1

+ +

+ =

1

4is

57. [2]1 x

1 x

x 1

ax sin(x 1) a 1lim

x sin(x 1) 1 4

+ +=

+

Let, f(x) =sin(x 1) a(x 1)

(x 1) sin(x 1)

+

x 1

1 alimf(x)

2

= and g(x) =

1 x

1 x

x 1lim g(x) 2

=

2

1 a 1 1 a 1a 0,2

2 4 2 2

= = =

58.Let f: [0, 4] [0, ] be defined by f(x) = cos1(cos x). the number of points. x [0,4] satisfying the equation

f(x) =

10 x

10

is

58. [3]

f(x) = 1 x

10

There are three intersections.

0 2 3 4x

y

x


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JEE Advanced 2014 Solution Paper I

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Transcript of JEE Advanced 2014 Solution Paper I