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EDUCATIVE COMMENTARY ON

JEE 2016 ADVANCED MATHEMATICS PAPERS

(Revised Draft uploaded June 8, 2016)

Contents

Paper 1 3

Paper 2 38

Concluding Remarks 80

The year 2013 represented a drastic departure in the history of the JEE.Till 2012, the selection of the entrants to the IITs was entirely left to theIITs and for more than half a century they did this through the JEE whichacquired a world wide reputation as one of the most challenging tests for entryto an engineering programme. Having cleared the JEE was often a passportfor many lucrative positions in all walks of life (many of them having little todo with engineering). It is no exaggeration to say that the coveted positions

of the IITs was due largely to the JEE system which was renowned notonly for its academic standards, but also its meticulous punctuality and itsunimpeachable integrity.

The picture began to change since 2013. The Ministry of Human Re-sources decided to have a common examination for not only the IITs, butall NITs and other engineering colleges who would want to come under itsumbrella. This common test would be conducted by the CBSE. Serious con-cerns were raised that this would result in a loss of autonomy of the IITsand eventually of their reputation. Finally a compromise was reached thatthe common entrance test conducted by the CBSE would be called the JEE(Main) and a certain number of top rankers in this examination would have

a chance to appear for another test, to be called JEE (Advanced), whichwould be conducted solely by the IITs, exactly as they conducted their JEEin the past.

So, in effect, the JEE (Advanced) from 2013 took the role of the JEE in thepast except that the candidates appearing for it are selected by a procedure

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over which the IITs have no control. So, this arrangement is not quite the

same as the JEE in two tiers which prevailed for a few years. It was hopedthat now that the number of candidates appearing for the JEE (Advanced)is manageable enough to permit evaluation by humans, the classic practiceof requiring the candidates to give justifications for their answers would berevived at least from 2014, if not from 2013 (when there might not have beensufficient time to make the switch-over). But this did not happen in 2015even after a change of regime at the Union Government and has not happenedfor 2016 either. A slight improvement is the provision of some partial creditwhere a candidate marks some but not all correct answers provided thatno incorrect is marked. Another welcome feature is that the total number

of questions has been reduced and, more importantly, matching pairs hasbeen dropped. This often resulted in increasing the number of questionstremendously.

As in the past, unless otherwise stated, all the references made are to theauthors bookEducative JEE (Mathematics)published by Universities Press,Hyderabad. The third edition of this book is now available in the market.

Because of the multiple choice format and many other constraints in paper-setting, interesting questions in mathematics are getting rarer. The contin-uation of these annual commentaries has been possible largely because ofthe keen interest shown by the readers. These commentaries are preparedsingle-handedly and hence are prone to mistakes of spelling, grammar and oc-casionally, wrong symbols (but, hopefully, not mistakes of reasoning!). Manyalert readers in the past had pointed out some such errors. They were cor-rected and the corrected versions were uploaded from time to time. But bythat time their immediate relevance was reduced.

As an experiment, last year, a draft version of the commentary on both thepapers was uploaded. Those readers who noticed any errors in it were invitedto send an email to the author at kdjoshi314@gmail.com or send an SMSto the author at 9819961036. Alternate solutions and any other commentswere also solicited. This really paid off. So this year too, the experimentis repeated. The readers may send their comments to the author by e-mail

(kdjoshi314@gmail.com) or by short mobile messages (9819961036).As in the past, although all candidates have to answer the same questions,

multiple codes are generated by permuting them. So the answer keys haveto be prepared keeping in mind these codes. Since the present commentarygives not only the answers but detailed solutions, the code hardly matters.But for easy reference, Paper 1 is in Code 1 while Paper 2 is in Code 9.

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PAPER 1

Contents

Section - 1 (Only One Answer Correct Type) 3

Section - 2 (One or More than One Correct Choice Type) 11

Section - 3 (Single Digit Integer Answer Type) 30

SECTION - 1 (Maximum Marks : 15)

This section contains FIVE questions each of which has FOUR optionsout of which ONLY ONE is correct.

Marking scheme :+3 If only the bubble corresponding to the correct answer is darkened0 If no bubble is darkened1 In all other cases.

Q.37 Let6

< 1 and 2> 2, then 1+ 2 equals(A) 2(sec tan ) (B) 2 sec (C) 2tan (D) 0

Answer and Comments: (C). By the quadratic formula, the rootsof the first equation are sec sec2 1 i.e. sec +tan and sec tan . Since tan is negative, the first root is smaller than the second.So we get

1= sec tan and 1= sec + tan (1)By a very similar calculation, the roots of the second quadratic are tan sec . This time, however, sec is positive and so

2=tan + sec and 2 =tan sec (2)From (1) and (2), 1+ 2 =2tan .

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A very straightforward problem, requiring only the quadratic for-

mula and the trigonometric identity sec2 = tan2 +1. The only catchis to determine which of the roots is larger in each case and this requiresthat since lies in the fourth quadrant, sec is positive and tan neg-ative. Using periodicity of trigonometric functions, the problem couldhave been made to demand a little more work by giving the coefficientsofxin the quadratics as, say2 sec(3) and 2 tan(2) and giving tolie in some other interval. But that would not change the spirit of theproblem. The four given options correspond to all possible choices ofthe larger roots. So, a candidate who makes a wrong choice will not bealerted.

Q.38 A debate club consists of 6 girls and 4 boys. A team of 4 members isto be selected from this club including the selection of a captain (fromamong these 4 members) for the team. If the team has to include atmost one boy, then the number of ways of selecting the team is

(A) 380 (B) 320 (C) 260 (D) 95

Answer and Comments: (A). Another very straightforward prob-lem. Let us first form only the teams (without worrying about thecaptain). There are two types of teams, those with no boys and those

with only one boy. Clearly, there are

62

= 15 teams of the first type.

For the second type, the boy can be chosen in 4 ways and for each such

choice, the three girls in the team can be picked in

6

3

= 20 ways. So

there are 80 teams of this type. Together there are 95 teams. For eachsuch team, the captain can be chosen in 4 ways. So, in all there are95 4 = 380 teams, where two teams with the same members but withdifferent captains are to be distinguished. This is perhaps the onlytesting point in an otherwise dull problem. But in verbal problems,

there is sometimes a risk that the correct interpretation may not striketo a candidate because of language deficiency.

There is an alternate way to do the problem. First fix the captain.This can be done in 10 ways. Now ask the captain to choose theremaining team members. If the captain is a boy, all three remainingmembers are girls and so the number of teams with his captaincy is

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63= 20. As there are 4 boys, there are 80 teams with a male captain.

If the captain is a girl, she can choose either three girls in

5

3

= 10

ways or two girls and one boy in

5

2

4 = 40 ways. So, for every girl,

there are 50 possible teams where she is the captain. As there are 6girls in all, the number of teams with a female captain is 300. So thetotal count is 80 + 300 = 380, the same answer as before.

The second solution is a little more complicated. Although for asimple problem like this such a comparison is not of much value, the

approach taken in the second solution deserves to be commented. Inessence, the problem involves selection at two stages, once the selectionof 4 members from 10 (subject to the given restrictions) and then asecond selection, viz., that of the captain from those selected at thefirst round. The second solution reverses this order. More generally,suppose we have n objects from which k objects are to be chosen inthe first round and then from these k chosen objects, r objects are tobe chosen at the second round, where we assume that r n. Thenby the first method the number of selections is

n

k

k

r

while by

the second method, the answer is

nr

n rk r

. Equating the two

counts we get a combinatorial proof of the binomial identity

n

k

k

r

=

n

r

n rk r

(1)

(An algebraic proof is also possible by simply expanding both the sides.But the combinatorial proof tells you the logic.)

An interesting special case of (1) arises when r = 1. (This is

also the case in our present problem, except that now there are somerestrictions on the selections.) In this case, we have

k

n

k

= n

n 1k 1

(2)

from which it follows that if n is relatively prime to k, then it must

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divide nk. (See also the last solution to the JEE 1998 problem in

Comment No. 5 of Chapter 4.)

Q.39 Let S={x(, ) : x= 0, 2}. The sum of all distinct solutions

of the equation

3sec x + cosecx + 2(tan x cot x) = 0 in the set S is

(A) 79

(B) 29

(C) 0 (D) 5

9

Answer and Comments: (C). As we are given that x= 0, /2,all the terms in the equation are defined for every x in the set S.

Multiplying the equation throughout by sin x cos x (which is non-zeroon S), we get an equivalent equation

3sin x+ cos x + 2(sin2 x cos2 x) = 0 (1)

Since both sin x and cos x are present, this cannot be regarded as aquadratic equation either in sin xor in cos x. So, we have to solve it asa trigonometric equation. We first rewrite it as

3sin x + cos x= 2 cos 2x (2)

If we divide both the sides by 2, and notice that3

2 and

1

2are familiar

expressions in trigonometry, we get a clue to the solution. Indeed thesenumbers are sin(/3) and sin(/6) respectively. So (2) reduces to

cos(x 3

) = cos 2x (3)

whose general solution is

2x= 2n (x 3

) (4)

where n is an integer. The two possibilities that arise are

x = 2n 3

(5)

and x = 2n

3 +

9 (6)

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For x to lie in (

, ), n must equal 0 in (5) and either 0 or

1 in

(6). So the given equation has four solutions in S, viz. 3

,9

,79

and

59

. These add up to 0.

From the answer, it is tempting to think that there is a cleverer wayto get it if we can show that the solutions are symmetrically locatedabout 0, without actually identifying them. This, for example, wouldhave been the case if the L.H.S. of the given equation were an oddfunction ofx. As this is not the case, there is probably no easier way.So this is not one of those interesting types of problems where you canget the sum of the roots of an equation without actually identifying theroots.

Q.40 A computer producing factory has only two plants T1 and T2. PlantT1 produces 20% and plant T2 produces 80% of the total computersproduced. 7% of the computers produced in the factory turn out to bedefective. It is known that

P(computer turns out to be defective given that it is produced in plantT1) = 10 P(computer turns out to be defective given that it is producedin Plant T2).

where P(E) denotes the probability of an event E. A computer pro-duced in the factory is randomly selected and it does not turn out tobe defective. Then the probability that it is produced in Plant T2 is

(A) 36

73 (B)

47

79 (C)

78

93 (D)

75

83

Answer and Comments: (C). The very words given that suggestthat this is a problem on conditional probability. The paper-settersdeserve to be congratulated for asking a probability problem which doesnot involve drawing balls from urns or throwing dice. Instead, they haveposed the problem in a setting that would easily appeal to the younggeneration. Superficial as these garbs are, a faulty computer is surelyfar less deadly than a cancerous tumour or a poisonous mushroom!

One wishes, however, that the paper-setters were also careful ingiving the fake answers. A layman often thinks of probabilities in termsof percentages and this can also be justified mathematically. So, let us

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say, that 100 computers are produced by the factory, with 20 coming

from plant T1 and 80 from T2. We are given that 7 out of these 100computers are faulty. The computer chosen at random is among theremaining 93 ones. Out of these some, say x are fromT1andy fromT2.

The desired answer then is the ratio y

93. As only one of the answers

has a denominator which divides 93, it must be the right one. So evenwithout calculatingy, a clever candidate can identify the answer. Topreclude such a sneaky cakewalk, at least one of the fake answers shouldhave had a denominator like 31.

For an honest solution, one has to calculate y. Let p be thepercentage of faulty computers in T2. Then the percentage of faultycomputers inT1is given to be 10p. So the numbers of faulty computersinT1and T2are, respectively, 200pand 80prespectively. The data givesthe equation

280p= 7 (1)

So we get p = 1

40. We have y = 8080p = 78. Hence the correct

answer is 78

93.

In a systematic (albeit somewhat pedantic) approach, the solutioncan be paraphrased using Bayes theorem on conditional probability.Let T1 and T2 also denote the events that the computer comes fromplants T1 and T2 respectively. These are complementary events withP(T1) = 1/5 and P(T2) = 4/5. Let F be the event that a computerfrom the factory is faulty. We are given that P(F) = 0.07. Let F

be the complementary event of F. Then P(F) = 0.93, The desiredprobability is P(T2/F

). By Bayes theorem, we have

P(T2/F) =

P(T2 F)P(F)

=P(T2 F)

0.93 (2)

Thus the problem is now reduced to calculating P(T2 F). Since T2is the disjunction of the mutually exclusive events T2 F and T2 F,and we are given P(T2), we get

P(T2 F) =45 P(T2 F) (3)

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This reduces the problem to calculating P(T2

F). For this, we use

the data that

P(F/T1) = 10P(F/T2) (4)

By Bayes theorem again and the given values ofP(T1) andP(T2), thisbecomes

P(F T1)1/5

=10P(F T2)

4/5 (5)

i.e.

P(F T1) =5

2 P(F T2) (6)But F is the disjunction of the mutually exclusive events F T1 andF T2. As we are given that P(F) = 7/100, from (6) we get

7

100=P(F T1) + P(F T2) =7

2P(F T2) (7)

Thus we have finally obtained the value ofP(F T2) as 150

. Putting

this into (3) we get P(T2 F) = 45 1

50 =

39

50. Putting this into (2),

we get P(T2/F) =7893 , the same answer as before.

While the second solution is the polished one, it is the first solutionwhich enables even a layman to solve the problem and often morequickly. He can aid his thought process by thinking of two disjointpiles of 20 and 80 computers, with the faulty computers marked withsome red tag. Once he correctly translates the data that the percentageof such computers in the first pile is 10 times that in the second intoan equation (viz. (1)), not much is left in the problem.

Q.41 The least value of

IR for which 4x2 +1

x1, for all x >0 is

(A) 1

64 (B)

1

32 (C)

1

27 (D)

1

25

Answer and Comments: (C). This problem is in sharp contrast withthe last one. The last problem had a very lengthy statement but was

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very easy to understand and even to solve for a layman. The present

problem has a one line statement. But it takes some patient thinkingto understand what is asked.

As a starter, the requirement that 4x2 +1

x 1 for all x > 0 is

equivalent to saying that the minimum value of the function 4x2 +1

xover the interval (0, ) be at least 1. This is hardly profound. But itgives the comforting clue that this is essentially a problem of findingthe minimum of a function over an interval. The catch is that this is adifferent function for every real and so its minimum may change as changes. The problem asks us to find the least value of for which

this minimum is at least 1. Clearly, has to be positive. For < 0,

4x2 +1

x is negative for large x. For = 0, the function is simply

1

xwhich is less than 1 for x >1. So we assume >0.

Once this is understood, the problem is routine. Letf(x) be the

function f(x) = 4x2 +1

x. Differentiation gives

f(x) = 8x 1x2

(1)

which vanishes only when 8x3 = 1, i.e. when x = 12

1/3. At this

point the derivative changes sign from negative to positive and so the

functionf(x) has its minimum atx=1

21/3. We denote the minimum

value byMto stress that it depends on . As cube roots are involved,in the calculation ofM, it helps to rewrite f(x) as

f(x) =4x3 + 1

x (2)

By a direct calculation,

M = f( 1

21/3)

=48

+ 112

1/3

= 31/3 (3)

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The requirement that M

1 is satisfied if and only if3

1

27

. So

1

27is the least value of for which the given condition holds.

When the product of two positive functions is a constant, the A.M.-G.M. inequality gives a purely algebraic solution to the problem ofminimising a sum of two positive functions. But as pointed out at theend of Comment No. 6 of Chapter 6, to apply this inequality sometimesone has to recast the sum as the sum of three functions instead of twoso that their product is a constant. (See also the end of CommentNo. 11 of Chapter 13 for minimisation of 2 cos x+ sec2 x.) So, in ourproblem, after it is established that > 0, one can rewrite f(x) as

4x2 + 12x

+ 12x

and then the A.M.-G.M. inequality gives that f(x)31/3 with equality holding when 4x2 =

1

2x i.e. whenx =

1

21/3. This

leads to a quicker, albeit trickier, calculation ofM. (The problem isshockingly similar to the one mentioned at the end of Comment No. 6,Chapter 6.)

In essence there are two minimisation problems here. The firstone is a bit tedious and prone to numerical errors. The second one iseasy. Fortunately, if a good student (who has understood the problem

correctly) makes a mistake in the first part, it is unlikely that his answerwill tally with any of the given ones and he will be alerted. Unlike thelast problem, there is no sneaky way to guess the answer. Becausecube roots are involved, a clever student might make a wild guess thatthe correct answer is a perfect cube. But the paper-setters have takenthe precaution to include a fake answer which is also a perfect cube.So this is a rare problem where the multiple choice format helps thesincere students without rewarding the insincere ones.

SECTION - 2: (Maximum Marks : 32)

This section contains EIGHT questions.Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE

THAN ONE of these four option (s) is (are) correct.Marking scheme :

+4 If only the bubble(s) corresponding to all the correct option(s) is (are)darkened

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+1 for darkening a bubble correspondingto each correct answer provided

NO incorrect option is darkened0 if none of the bubbles is darkened2 in all other cases.

Q.42 Consider a pyramid OPQRS located in the first octant (x 0, y0, z 0) with O as the origin, and OP and OR along the x-axis andthey-axis respectively. The baseOPQR of the pyramid is a square withOP= 3. The point Sis directly above the midpoint Tof diagonalOQsuch that T S= 3. Then

(A) the acute angle betweenOQ and OS is

3(B) the equation of the plane containing the triangleOQSis xy = 0(C) the length of the perpendicular from P to the plane containing

the triangle OQS is 3

2

(D) the perpendicular distance fromO to the straight line containing

RS is

15

2

Answer and Comments: (B, C, D). The first step is to visualise thepyramid correctly. It is a solid figure and therefore only a rough sketchis possible on a plane paper as shown below.

O

x

y

z

P

Q

R

S

3 3

3

33

T

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But because of coordinates, if we can correctly identify the coordinates

of all the vertices, all questions can be answered using appropriateformulas from coordinate geometry. We are already given that O =(0, 0, 0), P = (3, 0, 0). As the base OPQR is a square in the firstoctant and R lies on the y -axis, the base must lie in the xy-plane withR= (0, 3, 0) and Q= (3, 3, 0). The midpoint T ofOQ is (3/2, 3/2, 0).All these points are in the xy-plane. The vertexS lies directly aboveTand at a distance 3 from it. So S= (3/2, 3/2, 3).

With this spadework, we answer the questions one-by-one. For

(A), we can consider the vectors

OQ= 3i + 3j and

OS=3

2i+

3

2j+ 3k

and find the angle between them by taking their dot product. Butit is much easier to do this by simple trigonometry. The angle, say between OQ and OS is the same as the angle SOT in the right

angled triangle SOT. Hence tan = ST

OT =

ST12

OQ =

3

3

2/2=

2.

Since tan(/3) =

3= tan we see that (A) is false.For (B) it is hardly necessary to find the equation of the plane

containing the triangle OQS. All we need to verify is whether thecoordinates of the three points O, Q and Ssatisfy the given equation.They certainly do and so (B) is correct. For (C), we see from the

diagram that the perpendicular from P to the plane containing thetriangle OQS isP T. Also the triangle OT Pis a right angled isosceles

triangle. So,P T =OTwhich we already calculated as 3

2. Hence (C)

is also correct.

For (D), we first find the area, say , of the triangle ORS. Thiscan be done by taking OR as a base of length 3 and calculating thecorresponding altitude. But vectors are more convenient, for we get

= 1

2| OR OS|

= 1

2

i j k0 3 032

32

3

=

1

2|9i 9

2k|= 9

4

5 (4)

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Next, we find RS =9

4+

9

4+ 9 =27

2 = 33

2. By dividing 2 byRSwe get that the perpendicular distance ofOfrom the line containing

RS is 9

5

2

2.3

3=

15

2 . Hence (D) is true. An alternate approach is

to get the equations of the line RS. Since R = (0, 3, 0) and S =(3/2, 3/2, 3), the parametric equations of the line RSare

x 03/2

=y 33/2=

z 03

=t (1)

Therefore a typical point Xon the line RSis of the form

X= ( 32

t, 3 32

t, 3t) (2)

for some t. We want t for whichOXRS. Taking the dot product ofthe vectors

RSand

OX, this requirement comes out to be

3

2(

3

2t) + (3

2)(3 3

2t) + 3(3t) = 0 (3)

which simplifies to 27

2t =

9

2 giving t =

1

3. Therefore, from (2)X =

(1

2,5

2, 1). Hence the perpendicular distance OX of O from the line

RS is

1

4+

25

4 + 1 =

30

4 =

15

2. Hence (D) is correct. Those

who remember the formula for the perpendicular distance of a pointfrom a parametrically given line can save some time here. Yet anotherapproach is to minimise the distance OX as a function of t. This

reduces to minimising the quadratic function 27

2 t2 9t+ 9. Taking

derivative, the minimum occurs at t =1

3. The rest of the work remains

the same.

The problem is not as lengthy as it appears at first sight. Once adiagram is drawn, the first three statements can be disposed off almostby inspection. It is only Part (D) that demands some work. The paper-setters this year have been merciful enough to give some partial creditfor a candidate who solves the first three parts correctly but leaves thelast one. This is a welcome change from the past.

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Q.43 Let f : (0,

)

IR be a differentiable function such that f(x) =

2 f(x)x

for allx(0, ) and f(1)= 1. Then

(A) limx0+

f(1

x) = 1

(B) limx0+

xf(1

x) = 1

(C) limx0+

x2f(x) = 1

(D)|f(x)| 2 for allx(0, 2)

Answer and Comments: (A). Let y = f(x). Then we are given thaty satisfies the differential equation

dy

dx+

y

x= 2 (1)

Multiplying both the sides by x, this can be rewritten as

d

dx(xy) = 2x (2)

whose general solution is

xy= x2 + c (3)

wherec is an arbitrary constant. (This could also have been obtained bynoting that (1) is a linear differential equation with integrating factor

e

dx

x = eln x = x. But a direct approach, when feasible, is alwaysbetter.)

No intial condition is given and so the value ofc cannot be determined.From f(1)= 1 we can only infer that c= 0. As the function is givento be defined only for x >0, we can divide by xand get

y = f(x) =x+c

x (4)

The question asks which of the given four statements is(are) true forthis function. We tackle them one-by-one.

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By direct differentiation of (4),

f(x) = 1 cx2

(5)

for all x > 0. So f(1

x) = 1cx2 which tends to 1 as x tends to 0

from the right. So (A) is true. For (B), xf(1

x) =x(

1

x+cx) = 1 +cx2

which tends to 1 as x 0+. So (B) is false. In (C), x2f(x) =x2(1 c

x2) = x2 c which tends toc as x 0+. As c= 0, (C) is

false. Finally for (D), from (4), the function f(x) is clearly unbounded

in a neighbourhood of 0. So (D) is false too.The question is not difficult, but puzzling because its purpose is not

clear. It is a combination of two problems, solving a differential equa-tion and then answering some arbitrary questions about the solution.The first part is too familiar and the second is completely unrelated.The problem would have been interesting if the properties of the func-tion could be derived directly from the differential equation withoutsolving it. Combining two mediocre problems only results in a clumsyand not in an interesting problem.

Q.44 Let P = 3 1 22 0

3 5 0, where IR. Suppose Q = [qij] a matrix

such that P Q= kI where kIR, k= 0 andIis the identity matrix oforder 3. Ifq23 =k

8 and det(Q) =

k2

2, then

(A) = 0, k= 8

(B) 4 k+ 8 = 0(C) det(Padj(Q)) = 29

(D) det(Qadj(P)) = 213

Answer and Comments: (B, C). The matrix P has one unknownentry, viz. . The given relationship between P and Q, viz.,

P Q= kI (1)

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introduces one more unknown, viz. k. Taking determinants of both

the sides,

det(P)det(Q) =k3 (2)

(since I has order 3). But we are also given that det(Q) = k2

2 and

k= 0. Putting this into (1), we getdet(P) = 2k (3)

By an easy direct calculation, det(P) = 15 3 + 20 = 12 + 20. Sowe now have one equation in the two unknowns and k, viz.

6+ 10 = k (4)

We need one more equation in and k to determine their values. Thisis given indirectly by specifying that the entry q23 of the matrix Q is

k8

. To make out anything from this we need to know the entries ofQ

from those ofP. If we take the adjoint ofP, then we have

Padj(P) = det(P)I= (12 + 20)I (5)

Ask= 0, P is non-singular. Hence multiplying (1) and (5) byP1 weget

Q= kP1 (6)

and

adj(P) = (12 + 20)P1 (7)

respectively. Combining these two equations we can express Q in termsof adj(P) as

Q= k

12+ 20adj(P) (8)

It is easy to calculate the entries of adj(P). We are interested onlyin the entry in the second row and the third column of it. It is thecofactor of the entry in the third row and the second column ofP, i.e.

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the cofactor of the entry

5 in the third row. To obtain it, we delete

the third row and the second column ofP, take the determinant of the

the resulting 2 2 submatrix

3 22

, and multiply by (1)3+2, i.e.

by1. This comes out to be34. So the dataq23 =k8

translates

into

k8

= k

12+ 20(3 4) (9)

Ask= 0, we get2(3+ 4) = 3 + 5 (10)

This determines uniquely as1 and shows that option (A) is incor-rect. Putting =1 into (4) gives k = 4. With these values we seethat (B) is true.

The remaining two options deal with the determinants of the adjointsofP and Q. From (3) we know that det(P) = 8. Since Padj(P) = 8I,

we get det(adj(P)) = 82 = 26. SinceQ has determinantk2

2= 8, we get

that the determinant of the adjoint ofQ is also 26. So, both Padj(Q)and Qadj(P) have the same determinants, viz. 8 26 = 29, showingthat (C) is true and (D) false.

Problems on adjoints of matrices are rarely asked. Computation ofthe entire adjoint of ann nmatrix is a laborious job even forn as lowas 3. The present problem requires the calculation of only one entry inthe adjoint and so the work is reasonable. However, some candidatesare likely to confuse adjoints with Hermitian adjoints of matrices, whichare totally unrelated to each other.

Knowledge of the adjoint of a matrix is essential in answering (C)and (D) because the very statement of these options involves adjoints.However, in answering (B), knowledge of the adjoint was only instru-

mental and not essential. A direct calculation of is possible by abrute force method using the data that q23 =k

8. Since q23 appears

in the third colum ofQ, we calculate the third column of the product

P Q and equate it with the column vector

00k

. This gives a system

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of three equations in q13, q23, q33 (out of which q23 is already known as

k8

), viz.

3q13+k

8+ 2q33 = 0 (11)

2q13+ q33 = 0 (12)

and 3q13+5k

8 = k (13)

From (13), q13 = k

8. Putting this into (12) gives q33 = k

4. Putting

these into (11) we get

3k

8 +

k

8 k

2= 0 (14)

On simplification and cancellation ofk (which is given to be non-zero,this gives =1. Combining this with (4) (which too, did not needadjoints) we get k = 4 and so we are in a position to say that (B) istrue. Of course, this derivation is not as elegant as the earlier one basedon adjoints. But it is elementary and allows those who do not knowadjoints to pocket some partial credit (which has been introduced forthe first time) if they do not answer the remaining parts.

Q.45 In a triangleX Y Z, letx, y,zbe the lengths of the sides opposite to the

angles X , Y , Zrespectively and let 2s= x+y+z. Ifs x

4 =

s y3

=s z

2

and the area of the incircle of the triangle XY Z is 8

3, then

(A) area of the triangleXY Z is 6

6

(B) the radius of the circumcircle of the triangleX Y Z is 35

6

6.

(C) sinX

2 sin

Y

2 sin

Z

2 =

4

35

(D) sin2

X+ Y

2

=

3

5

Answer and Comments: (A, C, D). A triangle XY Z is completelydetermined by its three sides x,y,z. So, if we know all three sides

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we can calculate any attribute, be it the angles, the circumradius, the

inradius or the area of the triangle. In the present problem, the sidesare not specified. The given system of equations, viz.

s x4

=s y

3 =

s z2

(1)

can only determine the ratios of the sides to each other. That is, the tri-angle is uniquely determined upto similarity. And this is good enoughfor answering (C) and (D). However, for (A) and (B) we need the ad-ditional piece of data that that the area of the incircle of the triangle is8

3 which is a twisted way of saying that its inradius, generally denoted

byr, equals

83

. (Incidentally, area of a circle is the familiar but old

language. Nowadays, it is more customary to call it the area boundedby a circle.)

Let us see first what we get by setting each of the ratio in (1) equal tosomek. Adding the three resulting equations we get 3s(x+y+z) = 9ki.e. s= 9k. From this we get

x= 5k, y= 6k, z= 7k (2)

This is sufficient to tackle (C) and (D) using some well-known formulasfor the sines and cosines of half the angles of a triangle in terms of itssides. Specifically, for (C) we use

sinX

2 =

(s y)(s z)

yz (3)

and two similar formulas for sinY

2 and sin

Z

2. If we multiply all these

three formulas, every factor in the numerator as well as in the denom-inator appears twice and so the radicals drop out. Also the k factorcancels out. Thus

sinX

2 sin

Y

2 sin

Z

2 =

(s x)(s y)(s z)xyz

= 4 3 2

5 6 7=

4

35 (4)

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Hence (C) is true. (D) is even easier because sinX+ Y

2

= cosZ

2

which,

by yet another well-known formula, equals

s(s z)

xy =

9 25 6=

3

5.

Squaring, we see that (D) is also true.

As remarked earlier, for (A) and (B), we need the additionalinformation that

r=

8

3 (5)

To put this piece of information to use, we need to express r in terms

of the sides x, y,z. There are various formulas for this. For example

r = (s x)tanX2

= (s x)(s y)(s z)

s(s x) (6)

=

(s x)(s y)(s z)

s (7)

With our present data, after squaring both the sides we have

83

=4k 3k 2k9k

=83

k2 (8)

which meansk= 1. Hence we now have x= 5, y= 6, z= 7 and s= 9.Herons formula for the area of a triangle, viz.

=

s(s a)(s b)(s c) (9)gives =

9 4 3 2 = 66. Thus (A) is correct.

Finally, for (B), there are many formulas for the circumradius Rof a triangle. But as we already know the area , the best formula to

use isR=

xyz

4 (10)

As all the values on the R.H.S. are known, we get

R = 5 6 7

4 66 = 35

4

6(11)

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So (B) is false.

The problem is a test of the knowledge of the umpteen numberof trigonometric formulas for a triangle. But more importantly, it is atest of the ability to select the right formula when several formulas areavailable.

Q.46 A solution curve of the differential equation

(x2 + xy+ 4x + 2y+ 4)dy

dx y2 = 0. x >0

passes through the point (1, 3). Then the solution curve

(A) intersectsy = x + 2 exactly at one point

(B) intersectsy = x + 2 exactly at two points

(C) intersectsy = (x + 2)2

(D) does NOT intersecty = (x+ 3)2

Answer and Comments: (A, D). The first three options all deal withthe curves involving x + 2. This is a built-in hint that the substitutionu= x+2 may simplify the problem. This is borne out by the coefficient

of dydx in the given differential equation because it factors out as

(x+ 2)(x+ 2 +y). Since du

dx = 1,

dy

dx is the same as

dy

du. So with this

substitution. the differential equation becomes

dy

du=

y2

u2 + uy (12)

The R.H.S. is the ratio of two homogeneous polynomials of equal de-grees (viz. 2) and so we use the substitution y=vu.

dydu

=u dvdu

+ v (13)

and (12) becomes

udv

du+ v=

v2u2

u2 + u2v=

v2

1 + v (14)

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which simplifies to

udv

du=

v2

1 + v v= v

1 + v (15)

Separating the variables,

(1 +1

v)dv=du

u (16)

Integrating we get the general solution as

v+ ln v+ ln u= c (17)

where c is some constant. In terms ofy this becomes

ln y+y

u=c (18)

It is customary at this stage to go back to the original variables andrewrite the solution as

ln y+ y

x+ 2=c (19)

However, since the options involve x+ 2, it is better to keep (18) as

it is and work with y and u instead of with y and x. After all thisonly amounts to shifting the origin to the point (0, 2) and so does notaffect the intrinsic geometric properties of any curve. In terms of thenew coordinates, the point (1, 3) becomes (3, 3) and we are given thatit lies on the curve (18). This gives ln 3 + 1 =c and so the equation ofthe solution curve (in the new coordinates uand y) is

ln y+y

u= 1 + ln 3 (20)

We cannot solve this explicitly for y. So we have to resort to ad-hoc methods to test the various options. In (A) and (B), we have to

take the points of intersection of the curve (20) with the straight liney =u. Solving simultaneously, we have ln y = ln 3, i.e. y = 3. So (20)intersects the line y = u at the unique point (3, 3). Therefore (A) istrue and (B) is false. (The existence of the point of intersection wasalready given in terms of the initial condition. But now we have provedits uniqueness.)

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In (C), we have to take the intersection of (20) with the parabola

y= u2. At every such point of intersection we shall have

ln u2 + u= 1 + ln 3 (21)

The requirement that x >0 translates into u >2. Consequently,u2 >4and so the L.H.S. of (21) is bigger than ln 4 + 2 and hence bigger thanthan the R.H.S. So equality can never hold in (21) for u >2. Thereforethe solution curve and the part of the parabola y = u2, u > 2 cannotintersect. Thus (C) is false.

For (D), the method is similar but a little more complicated. Thepoints of intersection, if any, of (20) with y = (u + 1)2 must satisfy

2ln(u+ 1) +(u+ 1)2

u = 1 + ln 3 (22)

which simplifies to

2ln(u+ 1) + u+1

u=1 + ln 3 (23)

Foru >2, the first term is bigger than 2 ln 3 and hence than ln 3 (as ln 3is positive). The two remaining terms of the L.H.S. are both positivefor u >2 and hence certainly bigger than

1. So equality cannot hold

in (23) for u >2. Hence the solution curve and the curve y = (u + 1)2

do not intersect. So (D) is true.

This problem is of the same spirit as Q.43 above. In both, the firstpart was to solve a given differential equation and the second part wasto test the properties of the solution. But both the parts of the presentproblem are considerably more complicated than their counterparts inQ.43. It is ironic that both the questions are given the same credit.

Q.47 Let f : IR IR, g : IR IR and h : IR IR be differentiablefunctions such thatf(x) =x3 + 3x + 2,g(f(x)) =x and h(g(g(x))) =x

for all xIR. Then(A) g(2) =

1

15 (B) h(1) = 666 (C) h(0) = 16 (D) h(g(3)) = 36

Answer and Comments: (B, C). The relationship g (f(x)) =x doesnot automatically mean that the function g is the inverse function of

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f. For this to happen, one must first show that the functionf(x) =

x3 + 3x + 2 is one to one and onto. The first part follows by LagrangesMean Value Theorem, because f(x) = 3x2 + 3 > 0 for all x IRwhich means that f(x) is strictly increasing and hence one to one onIR. That it is onto follows from the fact that f(x) as x .So, the function findeed has an inverse function f1 and the equalityg(f(x)) =x is sufficient to conclude that g = f1.

There is no easy explicit formula for g(x) because such a formulawould require solving the cubic equation x3 + 3x+ 2 = y for everyyIRand there is no easy way to do this. However, by inspection wesee that f(0) = 2 and so g(2) = 0. We are asked g(2). From the chain

rule

g(f(0))f(0) = 1 (1)

Hence g(2) =g (f(0)) = 1

f(0)=

1

3. So we see that (A) is false.

The remaining three statements are about the functionh. Sinceg =f1 it is a bijection (i.e. one-to-one and onto). Therefore the compositefunction g g : IRIR is also a bijection. Since h(g(g(x))) =x forallxIR, we get that h is the inverse function ofg g. Sinceg1 =f,by the formula for the inverse of a composite function we get

h= (g g)1 =g1 g1 =f f (2)Hence by the chain rule

h(1) =f(f(1))f(1) (3)

Since we have f(1) = 6 and f(6) = 3 36 + 3 = 111 from (3) we seethat h(1) = 666. So (B) is true.

The truth of (C) is even easier to test. Since h(0) = f(f(0)) =f(2) = 8 + 6 + 2 = 16, (C) is true.

Finally, for (D), there is no formula for g(3). However, we can use(2) to get

h g = (f f) g (4)= f (f g) (5)= f (6)

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where we have used the associativity of the composition of functions

in going from (4) to (5) and the fact that f and g are inverses of eachother in going from (5) to (6). Therefore h(g(3)) =f(3) = 27 +9 + 2 =38= 36. So (D) is false.

A very interesting and well-designed problem requiring only simpleproperties of composition of functions, the inverse functions and thechain rule. Unfortunately, the Multiple Choice format masks the dif-ference between those who rigorously prove that fhas an inverse andthose who simply assume it.

The paper-setters have also done a commendable educative jobthrough this problem. At the undergraduate level (and definitely moreso at the JEE level) a function is invariably mistaken with the formulafor it. This happens primarily because nearly all the problems dealwith real valued functions of a real variable defined by some formulas.Non-numerical functions such as the father function rarely figure in anyproblems. Actually, many concepts about functions can be illustratedthrough such real life examples. For example monogamy is equivalentto saying that the husband function (defined on the set of marriedwomen) is one-to-one. But such questions are rarely asked in JEE.

Even though all the three functions f, g, h in the present problemare fromIRtoIR, their nature is such that there is no explicit formulafor g(x). So the derivations of (2) and (6) would have been very com-plicated if the functions were tied down to their formulas rather thanbeing treated as independent entities.

Q.48 The circle C1 : x2 +y2 = 3, with centre at O intersects the parabola

x2 = 2y at the point P in the first quadrant. Let the tangent to thecircle C1 at Ptouch two other circles C2 and C3 at the points R2 andR3 respectively. Suppose C2 and C3 have equal radii 2

3 and centres

Q2 and Q3 respectively. IfQ2 and Q3 lie on the y-axis, then

(A) Q2Q3= 12(B) R2R3= 4

6

(C) area of the triangleOR2R3 is 6

2

(D) area of the triangleP Q2Q3 is 4

2

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Answer and Comments: (A, B, C).

.

.

C

C

C

x

y

O

Q

Q

3

R

R

2

2

3

2

3

1 P

M

Before solving the problem, its long-winded statement deservessome rebuttal. It reminds one of the unending narratives of illiterate

villagers. What is the great idea in specifying the point Pas the pointof intersection of the circlex2+y2 = 3 and the parabolax2 = 2ylying inthe first quadrant, instead of giving it directly as (

2, 1)? A candidate

who has successfully cleared JEE Main can be presumed to know howto find this point. (If not, that would raise serious doubts about thevalidity of JEE Main!). Nothing is gained by putting such mediocreroadblocks (which are irrelevant to the main core of the problem) rightat the start, except to increase the chances of a wrong start because of anumerical slip and, at any rate, the loss of some precious time. Soundslike a person saying that he met his mothers husbands daughter when

what he means is that he met his sister.Also, it would be best to describe all three circles together in

a single statement as C1, C2, C3 are circles with their centres at thepointsO, Q2, Q3 respectively on the y-axis, withO = (0, 0) and havingradii

3, 2

3 and 2

3 respectively. They have a common tangent

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which touches them at P, R2 and R3 respectively with P = (

2, 1).

This covers all the data but in a more succinct and systematic manner.Now coming to the solution itself, the pointPcan be easily identified

as (

2, 1) by solving the equations x2 + y2 = 3 and x2 = 2y simultane-ously and choosing the solution for which both x, y are positive. Theequation of the tangent to C1 at this point (

2, 1) is

2x+ y= 3 (1)

If a circle with centre at a point (0, b) (say) on they-axis and radius 2

3is to touch this line, then equating the radius with the perpendiculardistance from this point, we getb 33

= 2

3 (2)

which simplifies to|b3|= 6 and has two solutions, b = 9 and b =3.We take Q2 as (0, 9) and Q3 as (0, 3). Then the distance Q2Q3 is 12and so (A) is correct.

Similarly we can determine the distance R2R3 by identifying thecoordinates of the points R2 and R3. But there is a better way. Let Mbe the midpoint ofQ2Q3. ThenM is also the midpoint ofR2R3 andso

R2R3= 2MR2= 2

MQ22 R2Q22 = 236 12 = 46 (3)Hence (B) is true. Now that we know R1R2, to get the area of thetriangle OR2R3 we only need the perpendicular distance of O fromR2R2. This distance is the radius ofC1 because R2R3 touches C1 andC1 is centred at O . Therefore

area ofOR2R3=12 4

6

3 = 6

2 (4)

which shows that (C) is also correct.

Finally, for the triangle P Q2Q3 too, we already know Q2Q3 = 12.We need the perpendicular distance ofP from Q2Q3, But as the lineQ2Q3 is the y-axis, this distance is simply the x-coordinate ofP, viz.

2. Hence

area ofP Q2Q3=12 12

2 = 6

2 (5)

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which shows that (D) is not correct.

Like Q.42, although the problem can be done using coordinates,pure geometry methods work better. Of course, the success of thesemethods hinges on drawing a good diagram. In the golden days of JEE,sometimes the question would specifically ask the candidates to drawa diagram. And even when it did not, the marking scheme would oftenallow some partial credit if a diagram was drawn correctly but somemistake made later. Like the ability to express ones thoughts verbally,the equally important ability to correctly draw a diagram is a casualtyof the Multiple Choice format.

Q.49 Let RSbe the diameter of the circle x2

+ y2

= 1, where Sis the point(1, 0). Let Pbe a variable point (other than R and Son the circle)and let the tangents to the circle at Sand Pmeet at the point Q. Thenormal to the circle atPintersects a line drawn through Q parallel toRSat a point E. Then the locus ofEpasses through the point(s)

(A)

1

3,

13

(B)

1

4,1

2

(C)

1

3, 1

3

(D)

1

4, 1

2

Answer and Comments: (A, C). Once again, a good diagram makes

O x

y

SR

P

Q

E

you not only understand the

problem correctly but may alsoinspire solutions. Take P as(cos , sin ). Then the equa-tions of the tangent and thenormal to the circle at P arex cos +y sin = 1 and y =tan x respectively while thetangent at Sis simply the linex= 1.

Solving x cos + y sin = 1 and x= 1 simultaneously, we get

Q= (1,1 cos

sin ) = (1, cosec cot ) (1)

Hence the equation of the line through Q parallel to RS is

y= cosec cot (2)

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Solving this simultaneously with y= tan x gives the point Eas

E= (cosec cot cot2 , cosec cot ) (3)To get the locus ofEwe need to eliminate between the equations

x = cosec cot cot2 (4)and y = cosec cot (5)

To do this, we rewrite (5) as cosec = y+ cot . Squaring both thesides we get

1 =y2 + 2 cot y =y2 + 2 cot cosec 2cot2 = y2 + 2x (6)So the equation of the locus of the point E is y2 + 2x= 1. By direct

substitution this is satisfied ifx=1

3 and y = 1

3. So, both (A) and

(C) are on the locus. However, when x=1

4, we get y = 1

2. Hence

the points in (B) and (D) are not on it.

SECTION - 3 : (Maximum Marks : 15)

This section contains FIVE questions, each having a single correct answerwhich is an integer ranging from 0 to 9 both inclusive.

Marking scheme:+3 If only the correct bubble is darkened0 In all other cases.

Q.50 The total number of distinct xIR for whichx x2 1 + x3

2x 4x2 1 + 8x3

3x 9x2 1 + 27x3

= 10 is

Answer and Comments: 2. Clearly the question is about the num-ber of real solutions of a polynomial equation p(x) = 10 where thepolynomial p(x) is specified as a determinant, viz.

p(x) =

x x2 1 + x3

2x 4x2 1 + 8x3

3x 9x2 1 + 27x3

(1)

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So the first task is to evaluate this determinant. Taking out x and x2

as common factors from the first and the second columns respectivelywe have

p(x) =x3q(x) (2)

where

q(x) =

1 1 1 + x3

2 4 1 + 8x3

3 9 1 + 27x3

(3)Subtracting suitable multiples of the first row from the remaining rows,

q(x) =

1 1 1 + x3

0 2 6x3 10 6 24x3 2

= 48x3 4 36x3 + 6= 12x3 + 2 (4)

Hence the equation to be solved is x3(12x3 + 2) = 10 which reduces to

6x6 + x3 5 = 0 (5)

Factoring the L.H.S. as (6x3

5)(x3

+ 1) we see that the only solutionsare those where x3 =

5

6 or x3 =1. Each of these equations has only

one real solution, Hence the given equation has two real solutions.

A clever student can save some time by dispensing with the actualroots of (5). Treating it as a quadratic in x3, as the discriminant ispositive, there are two distinct real roots. Each of these has only onereal cube root and hence (5) has exactly 2 solutions.

The starting step in the expansion of the determinant (1) is toosimple to be missed by anybody. However, although the evaluation

of the determinant in (3) as given here is simple enough, an alternatesolution is to split the third column and write the determinant as asum of two determinants, viz.

q(x) =

1 1 12 4 13 9 1

+

1 1 x3

2 4 8x3

3 9 27x3

(6)

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If we cyclically permute the columns of the first determinant, it becomes

a Vandermonde determinant (see Exercise (3.26)). If we take out thefactor x3 from the third column of the second determinant and thefactors 2 and 3 from its second and third rows respectively, it alsobecomes a Vandermonde determinant. Thus,

q(x) =

1 1 11 2 41 3 9

+ 6x3

1 1 11 2 41 3 9

(7)These two Vandermonde determinants are the same and their commonvalue is (21)(31)(32) = 2. This gives an alternate derivationof (4). But the direct derivation is faster.

A fairly simple problem on determinants.

Q.51 Let m be the smallest positive integer such that the coefficient ofx2 inthe expansion of (1 +x)2 + (1 +x)3 +. . .+ (1 +x)49 + (1 +mx)50 is(3n + 1) 51C3 for some positive integer n. Then the value ofn is

Answer and Comments: 5. Combinatorial identities have faded outfrom the JEE papers after the multiple choice format came in. But inthe present problem the paper-setters have managed to squeeze in one

such identity. The identity itself is not new. Indeed it is exactly thesame identity that appears as the first hint to the Main Problem ofChapter 5, viz. that for any positive integers mand n with mn,

n

m

+

n 1

m

+

n 2

m

+ . . . +

m + 1

m

+

m

m

=

n+ 1

m + 1

(1)

the present problem, we let S be the coefficient of x2 in the givenpolynomial. Then by the binomial theorem,

S=

2

2

+

3

2

+ . . . +

49

2

+

50

2

m2 (2)

The last term in the sum is irregular. But the remaining terms can besummed by writing them in the reverse order and applying (1) withn= 49 and m= 2. Hence

S=

50

3

+

50

2

m2 (3)

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Thus the data reduces to the equation

50

3

+

50

2

m2 = (3n + 1)

51

3

(4)

We now expand the binomial coefficients and cancel the factor 50 49which appears in all numerators to get

8 +m2

2 = (3n + 1)

17

2 (5)

which reduces to

m2 = 51n + 1 (6)

The problem is now reduced to a problem in number theory. We aregiven thatm is the smallest positive integer for which there exists somepositive integer n such that the equation above holds true. Simplystated, we want the smallest positive integer n for which 51n+ 1 is aperfected square. This can be done by trial. For n= 1, 2, 3 or 4 we donot get a perfect square. But for n= 5, 51n + 1 = 256 = (16)2. So thecorrect answer is 5.

A slightly different approach (although not entirely free of trial anderror) is to rewrite (6) as

(m+ 1)(m 1) = 51n (7)

The problem now is to find the least positive integer n for which 51ncan be factored as the product of two numbers that differ from eachother by 2. If 51 had too many factors then the search would take alot of time. But fortunately 51 has only two prime factors, viz. 17 and3. Rewriting 51nas 17 3n we see that n= 5 fits the table.

This is a good problem and an easy one once you hit upon the right

combinatorial identity to apply. In this respect the problem resemblesa trigonometric problem which unfolds itself once you pull the righttrigonometric identity. The difference is that the world of trigonometricidentities useful at the JEE level is limited and well trodden. That canhardly be said of combinatorial identities. They are used less frequentlyand so while it is reasonable to give the candidate a binomial identity

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and ask him to prove it, it is a little unreasonable to expect him to

pull it from his memory kit. Doing so brings mathematics closer tochemistry where you simply have to remember what cassiterite is anore of. No amount of logical thinking will help you.

Q.52 The total number of distinct x[0, 1] for which x0

t2

1 + t4dt= 2x 1

is

Answer and Comments: 1. Consider the difference function

f(x) = x

0

t2

1 + t4

dx

2x + 1 (8)

defined for x[0, 1]. Our job is to find out how many zeros f(x) hasin the interval (0, 1]. As in Q.50, our interest is only in the numberof zeros and not in identifying them. So there may be an easier waywithout evaluating the integral explicitly. And indeed there is. Clearly

f(0) = 1 > 0 while f(1) = 10

t2

1 + t4 1< 0 because the interval of

integration has length 1 and the integrand is less than 1 at all its points.As functions defined by integrals are continuous, f(x) is continuous.Therefore by the Intermediate Value Property, f(x) vanishes at least

once in the interval (0, 1).If we could prove that f(x) is strictly decreasing in the interval

[0, 1] that would mean that there are no more zeros. The most standardmethod to test this is to check the sign off(x). By the second formof Fundamental Theorem of Calculus, we have

f(x) = x2

1 + x4 2 (9)

As already noted, the first term is always less than 1 for all x[0, 1].So f(x) < 0 throughout (0, 1) and hence by Lagranges Mean Value

Theorem, f(x) is strictly decreasing in (0, 1). Hence there cannot bemore than one zeros off(x) in [0, 1]. As we already know that there isat least one zero, the correct count is 1.

Problems where the number of zeros of a strictly monotonic functionhave to be determined using Intermediate Value Property and the La-grange Mean Value theorem are fairly common in JEE. The additional

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feature in the present problem is that the derivative is obtained using

the second form of the fundamental theorem of calculus. That too isquite common.

Q.53 Let , IR be such that limx0

x2 sin x

x sin x = 1. Then 6( + ) equals

Answer and Comments: 7. The limit in question is of the 0

0 form.

So application of lHopitals rule is indicated. It gives

1 = limx0

2x sin x + x2 cos x

cos x (1)

Here the denominator tends to 1 while the numerator tends to 0as x0. If 1= 0, then the limit would be 0, a contradiction. Sowe get = 1.

To determine , we could apply lHopitals rule again. But thereis a better way out. The denominator 1 cos x equals 2 sin2(x/2) Butnear 0, sin xis of the same order as x near 0 in the sense that the ratiosin x

x tends to 1 as x 0. So when = 1, the denominator of the

function on the R.H.S. of (1) is of the same order as 2(x/2)

2

=

x2

2. Asfor the numerator, sin x is of the order of x while cos x tends to1. So the numerator is of the order of 2x2 +x2 = 3x2. Cancelling

the factor x2 the limit on the R.H.S. of (1) is 3

1/2 = 6. So we get

6( + ) = 6 + 1 = 7.

The argument can be presented rigorously by rewriting the ratioon the R.H.S. of (1) (with = 1). We divide both the numerator andthe denominator byx2. Then the ratio equals

2sinx

x

+ cos x

(1/2)( sin(x/2)x/2

)2(2)

Then as x 0, each ratio in this expression tends to 1 and so doescos x. Therefore this ratio tends to 6.

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For those familiar with the infinite series expansion of sin x an

argument can be given by rewriting the ratio as

x2

x 3x3

3! +

5x5

5! . . .

( 1)x + x3

3! x

5

5!+ . . .

For the ratio to tend to a finite non-zero limit, the dominating powersin the denominator and the numerator must be the same. This forces = 1, and, afterwards, the limit equals the ratio of the dominating

terms, which is 6.A fairly simple problem. Although a rigorous justification is time

consuming, in a multiple choice format it is only the numerical workthat matters and the computations involved are light.

Q.54 Let z =1 + 3i

2 where i =

1 and r, s {1, 2, 3}. Let P = (z)r z2s

z2s zr

and Ibe the identity matrix of order 2. Then the total

number of ordered pairs (r, s) for which P2 =I is

Answer and Comments: 1. It is more customary to denote zby .It is called the complex cube root of unity and in the polar form it equalse2i/3. Because of the factorisation ofz3 1, it satisfiesz2 + z+ 1 = 0.Also powers ofzrecur with a cycle of 3. That is, z1 =z2 =z5 =z8 =. . . etc. Most of the problems about complex cube roots of unity canbe handled with these two basic properties. (See Comment No. 12 inChapter 7 for a few such problems. Also see Part (C) of Q.59 in thefirst paper of JEE Advanced 2015.)

In the present case, a direct computation gives

P2 =

(z)r z2sz2s zr

(z)r z2s

z2s zr

=

z2r + z4s (1)rzr+2s + zr+2s

(1)rzr+2s + zr+2s z4s + z2r

(1)

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Equating this matrix with

I results into a system of two equations,

viz.

z2r + z4s = 1 (2)and (1)rzr+2s + zr+2s = 0 (3)

Since zand its powers are all non-zero, the second equation can holdonly when (1)r =1, i.e. when r is odd. In the given set{1, 2, 3},r has to be 1 or 3. We determine the possible values of s in eachcase separately from the first equation.. When r = 1, we must have1 +z2 +z4s = 0. This meansz4s =z. So, 4s 1 is divisible by 3. Inthe set{1, 2, 3} this is possible only for s = 1. So (1, 1) is a possiblepair.

In the other possibility, viz. r= 3 we have z2r =z6 = 1 and so byequation (2) above z4s =2. But this is not possible for anys because|z|= 1 and so all powers ofzare also of absolute value 1.

So there is only one pair. The problem is much simpler and henceless interesting than last years problem referred above. Apparentlythe question was motivated by a perceived compulsion to ask somequestion about complex numbers.

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PAPER 2

Contents

Section - 1 (Only One Correct Answer Type) 38

Section - 2 (One or More than One Correct Answers Type) 55

Section - 3 (Paragraph Type) 74

SECTION - 1 : (Maximum Marks : 18)

This section contains SIX questions, each having FOUR answers only oneof which is correct.

Marking scheme :+3 If only the bubble corresponding to the correct answer is darkened.0 If no bubble is darkened1 in all other cases.

Q.37 Let P =

1 0 04 1 0

16 4 1

and I be the identity matrix of order 3. If

Q= [qij] is a matrix such that P50 Q= I, then q31+ q32

q21equals

(A) 52 (B) 103 (C) 201 (D) 205

Answer and Comments: (B). Whatever be the other criticism lev-elled upon the JEE paper-setters, even their worst critic would notaccuse them of being so inhuman as to require the candidates to calcu-late the fiftieth power of a 3 3 matrix Pby actual multiplication 49times. Multiplication of even one pair of matrices is such a horrendous

job. So brute force is not the right method. There are two standardways to proceed to find such a high power. One is to calculate somelower powers by actual multiplication, notice some pattern in their en-tries, make a guess about the n-th power and finally prove the guessby induction onn (or by some other method). Since JEE doesnt careabout proofs any more, the last step is irrelevant. So he who makes a

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correct guess gets the reward.

If we follow this strategy then by a direct calculation, the first threepowers come out to be

P =

1 0 04 1 0

16 4 1

P2 =

1 0 08 1 0

16 + 32 8 1

and P3 =

1 0 012 1 0

16 + 32 + 48 12 1

(1)

The pattern is sufficiently clear. Only three entries change. Two ofthese are equal and are in an A.P. The third one, sitting in the bottomleft corner is 16 times the sum of the firstn integers, for which we havea formula. So we guess that for every positive integer n,

Pn =

1 0 0

4n 1 0

8n(n+ 1) 4n 1

(2)

An honest student would try to prove this by induction as follows. Thetruth for n = 1 is evident from the very definition ofP. As for theinductive step,

Pn+1 = PnP

=

1 0 04n 1 0

8n(n + 1) 4n 1

1 0 04 1 0

16 4 1

= 1 0 04n + 4 1 0

8n(n + 1) + 16n + 16 4n + 4 1

=

1 0 04(n + 1) 1 0

8(n + 1)(n + 2) 4(n + 1) 1

(3)

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This completes the proof of the inductive step. Hence we have now

established that (2) holds for every positive integer n. Putting n= 50we get

P50 =

1 0 0200 1 0

400 51 200 1

(4)

To finish the solution we are given that

Q= [qij] =P50 + I (5)

Mercifully, unlike matrix multiplication, matrix addition is done en-trywise. To make the things even simpler, the entries of Q we areinterested in are all outside the diagonal and all non-diagonal entriesof the identity matrix are 0. So we simply read out the correspondingentries ofP50 and get

q31 = 400 51q32 = 200

and q21 = 200 (6)

So, by sheer arithmetic, q31+ q32

q21=

400 51 + 200200

= 102 + 1 = 103.

Another method of finding high powers of a matrix without bruteforce is to relate it to some other matrices whose high powers are knownfrom definition or because of their relationship with some other entitiessuch as complex numbers (see Comment No. 20 of Chapter 2). Twoclasses of such matrices are (i) nilpotent matrices and (ii) idempotentmatrices. By definition, an n n matrix A is nilpotent ifAr is thezero matrixO for some positive integer r while it is called idempotentifA2 =A. Note that ifA is nilpotent then all powers ofAafter somestage are O while ifA is idempotent then all powers are equal to A

itself.In the present problem, we write P asI+ Awhere I is the identity

matrix of order 3 andA is the matrix

0 0 04 0 0

16 4 0

. By a direct calcu-

lation, we see that A3 =O, the zero matrix. SoA is nilpotent. (This

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is true of every n

n matrix which has zero entries on and above its

main diagonal. But the proof is a bit tedious. An elegant proof can begiven using what are called linear transformations. But that is wellbeyond our scope.)

Since P =I+ A, P50 = (I+ A)50. We have to be wary in hastilyexpanding this by the binomial theorem. The binomial theorem is validfor the power of a sum of two real (or complex) numbers because themultiplication for real or complex numbers is commutative. In fact, themost direct proof is by writing (a +b)n as a sum of 2n terms of the formx1x2 . . . xn where each xi equals aor band then grouping together thelike terms using the fact because of commutativity of multiplication,

every term in which exactly r factors are a and the remaining nrare b equals arbnr, there being in all

n

r

terms of this type for every

r= 0, 1, 2, . . . , n. But matrix multiplication is not commutative. So, ifAand B are two square matrices of the same order then we can expand(A+B)2 only as (A+B)(A+B) = A2 +AB + BA+B2. However,if A and B commute with each other, then we can write (A+ B)2

as A2 + 2AB +B2. More generally, the binomial theorem holds for(A+B)n for any positive integer n if the matrices A and B commutewith each other.

Coming back to P50 = (I+ A)50, as the identity matrix commuteswith every matrix we can apply the binomial theorem to conclude

P50 =I50 +

50

1

I49A +

50

2

I48A2 + . . . + InrAr + . . . + A50 (7)

All powers ofIequalI. As for powers ofA, all powers afterA2 vanish.So only the first three three terms in the expansion above matter andgive

P50 =I+ 50A + 1225A2 (8)

Hence

Q= P50 + I= 2I+ 50A + 1225A2 (9)

By a direct calculation,

A2 =

0 0 04 0 0

16 4 0

0 0 04 0 0

16 4 0

=

0 0 00 0 0

16 0 0

(10)

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We substitute these in (9). As before, we get q31, q32 and q21 by adding

the corresponding entries on the R.H.S. This gives q31 = 0 + 50.16 +1225.16 = 1275.16 = 40051, q32 = 0 + 50.4 + 0 = 200 and q21 =0 + 50.4 = 200. These are the same values that we got in (6). So theanswer is the same.

A good problem about matrices. But a bit above the JEE level.Nilpotent matrices are studied only in degree courses. So most candi-dates will be forced to try only the first approach and sadly one again,the scrupulous candidates who not only guess but prove (2) by induc-tion on n will lose in terms of time. The question therefore deserves abetter place in the conventional examination as a question on a proof

by induction.In passing we mention two other instances where the high powers

of a matrix can be calculated without brute force. First, analogouslyto (7), ifP =I+A where A is an idempotent matrix, then from thebinomial theorem and the fact that Ar =A for all r1, we get

Pn = (I+ A)n =I+ (2n 1)A (11)

As the other example, ifP is a 2 2 matrix of the form

x yy x

,

then P corresponds to the complex number z=x+iy. By writing zas rei, we get (using DeMoivres rule) that

Pn =rn

cos n sin nsin n cos n

(12)

Dont be surprised if problems based on (11) or (12) are asked in thecoming JEEs ! But now that the secret is out, maybe they wont beasked!

Q.38 Area of the region{(x, y) IR2 : y |x + 3|, 5y x+ 9 15} isequal to

(A) 1

6 (B)

4

3 (C)

3

2 (D)

5

3

Answer and Comments: (C). A typical problem of finding the areabetween two curves of the form y =f(x) and y =g(x) by integrating

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the difference function

|f(x)

g(x)

|. When the vertical boundaries of

the region are specified in the form x = a andx = b for some a, bIR,with a < b, the integration is over the interval [a, b]. But when this isnot the case, the first task is to find the points of intersection of the twogiven curves because the x-coordinates of these points of intersectiongive us the interval over which to integrate.

O x

A

B

CE

D F

y

L

G

H

The present problem falls in the latter category. The upper boundaryof the region is the straight line (shown by L in the figure above)

5y= x+ 9 (1)

while the lower boundary is given as the curve

y=

|x+ 3| (2)

Because of the absolute values sign, we have to consider two cases: (i)where x+ 3 > 0, i.e. x 3 and (ii) where x+ 3 < 0, i.e. x 3.For (i), i.e. for x 3, (2) can be rewritten as

y2 =x+ 3 (3)

which is a parabola with vertex atE= (3, 0). We are interested onlyin the upper half of this parabola. It cuts the line (1) at points wherey2 5y+ 6 = 0, i.e. wherey = 2 or y = 3. The corresponding valuesofx are x= 1 and x= 9. So the points of intersection of (1) and the

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upper half of (3) are (2, 1) and (9, 3). These are shown as B and H

in the figure above. Note that in between these two points, the linelies below the parabola and so the portion between them is not in ourregion. For x >9, the line again lies above the parabola. This is truefor all x > 9 and would result in an unbounded region. But we arespecifically given that x+ 6 < 15 and that precludes any part of thisunbounded region.

Summing up, the subregion for which x3 shown as the shadedpart bounded by E B F Ehas area A1 where

A1 = 1

3

x + 9

5

x+ 3 dx

=

(x+ 9)2

10 2

3(x + 3)3/2

13

= 100

10 36

1016

3 =

16

15 (4)

Let us now turn to the other subregion (bounded by E F D E).Here the calculations are very similar. The upper boundary is the samelineL. But the lower boundary is the upper half of the parabola

y

2

=x 3 (5)which is the mirror image of the parabola (3). It has the same vertex,viz. E = (3, 0). To find out where it cuts the line (1), we have tosolve the equation y2 + 5y = 6 which has two roots, 1 and6. Weignore the second value as it will give a point of intersection which lieson the lower half of the parabola. The first value gives (4, 1) as the(unique) point of intersection of the line (1) with the upper half of theparabola (5). It is shown as D in the figure above. So, the area of thesecond subregion, say A2 is

A2 = 34

x+ 9

5 x 3 dx

=

(x + 9)2

10 +

2

3(x 3)3/2

34

= 36

1025

102

3=

13

30 (6)

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Adding (4) and (6) we get the total area of the region as 45

30

=3

2

.

The integration part of the problem is too straightforward. The realtask is to identify the region correctly. So essentially, this is a problemin coordinate geometry. The paper-setters have been careful enoughto stipulate the condition x+ 9 15 which may appear irrelevant atfirst sight. (of course, it would have looked more natural to give thiscondition as x 6.) Also even though square roots are involved inthe definition of the function, the numerical data has been carefullychosen to avoid any surds in the answer. This reduces the chances ofnumerical error.

Q.39 The value of13k=1

1sin

4

+ (k1)6

sin

4

+ k6

is equal to

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

Answer and Comments: (C). Just as finding the fiftieth power of amatrix is a torture if done directly, so is adding 13 terms, each of whichis the reciprocal of the product of two factors. The angles appearing inthe problem are in an A.P. and there are formulas for the sum of theirsines (see Comment No. 6 of Chapter 7). But these are inapplicable

here. So here we apply a general rule of thumb. When there is noobvious way to find the sum of a series of a large number of terms,recast the series as a telescopic series. Problems based on this trickare popular with paper-setters of competitive examinations and haveappeared several times in the JEE too. See for example, the JEE 1999problem in Comment No. 11 of Chapter 5. Also see the comments onQ.28 of Paper I of JEE 2010 and Q.47 of Paper I of JEE 2013.

So our job is to express the k-th term of the series in the formAk Bk in such a way that Ak+1 will equal Bk for k= 1, 2, . . . , 12. Inthe present problem the k-th term is the reciprocal of the product of

two sines, i.e. of the form sin sin . It is tempting to rewrite this as

half the difference of two cosines, viz. as1

2(cos()cos(+)). And

this would have worked had the term sin sin been in the numerator(with the denominator a constant). But unfortunately, it is in thedenominator.

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So, we look for identities that have the product of two sines in the

denominator. One such identity is

cot + cot =cos

sin +

cos

sin =

sin(+ )

sin sin (1)

But as our interest is to express the general term of the series as adifference and not as a sum of two terms, we replace by in (1)and get

cot cot =cos sin

cos sin

=sin( )

sin sin (2)

So, we apply this identity with = (k 1) 6

+

4 and = k

6 +

4.

Then is simply 6

which is independent ofk. Dividing both the

sides of (2) by sin( ) which equals 12

, we get

1

sin4

+ (k1)6

sin

4

+ k6

= 2cot(k 1) 6

2cot

k

6

(3)

The job is almost done. Now it is only a matter of adding these terms

fork = 1 tok = 13 and noting that the R.H.S. forms a telescopic series.Thus if we denote the given sum by S, we have

S =13k=1

1

sin4

+ (k1)6

sin

4

+ k6

=13k=1

2 cot((k 1) 6

+

4) 2cot(k

6+

4)

= 2 c ot

4 2cot

13

6 +

4

= 2 c ot4 2cot2912 (4)

To finish the solution we only need to calculate the values of these twocotangents. The first one is 1. For the second one, we need to work a

little. First, since cot is periodic with period , cot29

12 is the same as

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cot5

12

. In terms of degrees, this equals cot 75 or equivalently, tan 15.

This value is not needed as frequently as tan of some other angles suchas 30 or 45. To calculate it, we let u = tan15. Then the identity fortan2 gives

tan30 = 1

3=

2u

1 u2 (5)

This gives a quadratic for u, viz.

u2 + 2

3u 1 = 0 (6)whose (positive) root is 2

3. Putting these values into (4) we finally

get S= 2 2(2 3) = 2(3 1).A tricky but otherwise an easy problem once you get the trick. One

wishes that the numerical data was so arranged as not to require thevalue of tan 15. This value is irrelevant to the main theme. Having tocalculate it gives an advantage to those who remember such rarely usedvalues listed in some tables over those who prefer to remember only afew standard values and derive the rest as and when needed. So, likeQ.51 in Paper I, the last part of the question brings it closer in spiritto a question in chemistry.

Q.40 Let bi>0 for i = 1, 2, . . . , 101. Suppose log b1, log b2, . . . , log b101 are inArithmetic Progression (A.P.) with common difference loge 2. Supposea1, a2, . . . , a101 are in A.P. such that a1 = b1 and a51 = b51. If t =b1+ b2+ . . . + b51 and s= a1+ a2+ . . . + a51, then

(A) s > t and a101> b101 (B) s > t and a101< b101(C) s < t and a101 > b101 (D) s < t and a101< b101

Answer and Comments: (B). The first statement is an indirectway of specifying that the numbers b1, b2, . . . , b101 are in a geometric

progression (G.P.) with common ratio 2. Any candidate who knowsthe most basic property of logarithms, viz.

ln y ln x= lnyx

(1)

will know this. And once again, candidates who have cleared the JEEMain can be presumed to belong to this category. So, one fails to

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know what is gained by giving this twist to the data. Or is it that the

paper-setters are allergic to geometric progressions?Anyway, coming to the problem, the question is about the relative

rates of growths of geometric progressions and arithmetic progressions.An A.P. is like a car which runs at a uniform speed, so that it cov-ers equal distances in time intervals of equal durations. A geometricprogression (with common ratio greater than 1), on the other hand, islike a car which keeps accelerating. Suppose we track two such cars,A and B respectively by recording their positions at regular time in-tervals. Suppose that at the start both the cars are together. (Thatdoes not mean that they are at rest. They are both moving but pass

the same roadmark simultaneously.) Suppose we notice that half anhour later they are together again. What do we make out? Evidently,their average speeds over this half hour are the same. But in the caseofA, this is also its uniform speed every moment. For B , on the otherhand, intially the speed is less than this average speed and so the caris lagging behind A. But later its speed picks up and the distance be-tween them begins to decrease and becomes 0 half an hour after thestart. Obviously, at this point the speed ofB is much higher than thatof the (constant) speed ofA. So, after this half an hour, B will beat Athroughout and the gap between them will go on increasing forever.

All this is clear by sheer common sense. And a candidate whosecommon sense is sound can easily tell that (B) is the correct option. Infact, he can predict that till n = 50, an> bn (the carB lagging behindA). So not only s > t but even term by term, the summands ofs arebigger than the corresponding summands oft, except the first and thelast summands which are equal for both. In other words, sbeatstnotonly in the aggregate but in each term. And for alln > 51, an < bn(car B is ahead of car A).

The present problem asks you to put this common sense into amathematical form. an and bn are like the positions of the cars at the

end of the n-th time interval (or to be fussy, at the end of the (n 1)-th time time interval from the start, because the progressions beginwitha1andb1 rather than with a0 andb0 which would have been morenatural). Letd be the common difference of the A.P. a1, a2, . . . , a100.(In terms of the analogy above d is like the uniform speed of the car

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A.) Then

an = a1+ (n 1)d (2)while bn = b12

n1 =a12n1 (3)

for all n = 1, 2, . . . , 101. Equating a51 with b51 gives the value ofd as

d= a1250 1

50 (4)

With this, (2) becomes

an =a1+ a1(n 1)250

1

50 (5)

We now attempt to prove that an > bn for n 51. Since a1 > 0, it cancels out and these comparisons reduce toshowing that

250 150

> 2n1 1

n 1 for 2n50 (6)

and 250 1

50 51 (7)

Note that equality would hold in both (6) and (7) for n= 51. This isa consequence of the data that a1 =b1 and a51 =b51. If we can provethese inequalities, that will surely imply (B) becauses = a1 + a2 + . . . +a50+ a51 while t= b1+ b2+ . . . + b51.

Unfortunately, a direct proof of these inequalities is not easy. Thepaper-setters have apparently realised this and made the problem lessambitious by requiring to prove merely that s > tand a101 < b101. Thisis a lot easier because of the formulas for the sums of the terms in anA.P. or G.P. Indeed, from the formula for the sum of the terms in anA.P., we get

s = a1+ a2+ . . . + a51= 51a1+50.51

2 d

= 51a1+250 1

50 25 51a1

= 51a1250 + 1

2 (8)

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As fort, it is the sum of the first 51 terms of a G.P. with common ratio

2 and the first term a1. So

t= a1(251 1) (9)

Clearly,t < a1251 while from (8),s > a12

4951> a1251 because surely51> 4. Put together, s > t.

The comparison ofa101 with b101 is much easier. Indeed, from (5),a101 = a1+ 2a1(2

50 1) = a1(251 1) < a1251 which is a lot smallerthan a12

100 which is precisely b101.

Although that completes the solution, one nagging question remains.

Why is it so difficult to prove the inequalities (6) and (7) when theircounterparts in the analogy given above are so obvious by commonsense, viz. that during the first half an hour, the accelerating car Balways lags behind the car Awhile thereafter B leads A?

The explanation is that there is a suble flaw in the analogy. Themotion of a car is continuous. Its position is a function of time t whichis a continuous variable. On the other hand, progressions, and moregenerally all sequences are functions of a discrete variable n. As aresult, the speed of a moving car has no analogue for progressions,since speed is defined in terms of derivatives and derivatives (and more

generally limits) are meaningless for functions of discrete variables. Ifmangos are falling off a tree, the instantaneous rate of their fall hasno meaning. But if these mangos are squeezed and made into a pulpand the pulp is poured from a jar then the instantaneous rate of flowof the pulp is a well-defined entity. This happens because the numberof mangos fallen is a discrete variable while the volume of the mangopulp is a continuous variable.

However, sometimes we can extend a discrete variable by allow-ing it to assume all real values (or all non-negative real values) whenthe nature of a function is such that it makes sense for a continuous

variable as well. Doing so allows us to apply the powerful machineryof continuous mathematics such as theorems like the Lagranges MeanValue Theorem. And then it may be possible to prove some resultsabout sequences which would not be easy to prove directly.

We illustrate this technique by giving a proof of the inequality(6). The proof of (7) is similar and left to the reader. Let us first get

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rid ofa1 which has no role. So we take a1 = 1. Also let us take the

progressions withn starting from 0 instead of 1 (which is more naturalfor many purposes.) In effect this means that we call n 1 as our newn. So instead of (3) we now have bn= 2

n. Now, we replace the discretevariable n (taking values 0, 1, 2, . . .) by a continuous variable xtakingall non-negative real values. The function 2x makes sense for all suchvalues. We call it f(x). Thus,

f(x) = 2x (10)

The inequality (6) now translates as

f(50) f(0)50 0 >

f(x) f(0)x 0 for 0< x 0 for all x 0. Nowconsider the points P1= (0, f(0)),P2= (50, f(50)) and P3 = (x, f(x))on the graph ofy = f(x). Then because of the concave upwards natureof the function, the slope of the chord P1P3 is less than the slope of thechordP1P2 (see Comment No. 18 of Chapter 13). But thats precisely(11).

This is an illustration of how results from continuous mathematicscan be used to prove results in discrete mathematics. Let us now goback and see if we can prove inequalities (6) and (7) without recourseto methods from continuous mathematics. Note that (6) and (7) willboth follow if we prove, more generally, that

2m1 1m

>2n1 1

n (12)

wheneverm, n are integers with m > n2. In discrete mathematics,we do not have derivatives. But we have one important concept which

has no analogue in continuous mathematics. For a real number x, thereis no such thing as the next bigger real number. But, for an integer x,the next bigger integer isx+1. It is often called the successorofx. Asa result, we can go from one integer to another with finitely many unitjumps. The successor function, restricted to the set of natural numbersis the very basis of mathematical induction. It also helps in proving

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inequalities. Instead of comparing what happens at two arbitrarily

separated positive integers, it suffices to compare what happens at twoconsecutive integers. Suppose, for example, that we are able to provethat

2k 1k

>2k1 1

k 1 (13)for every positive integer k 2. Then by applying this with k =n, n+ 1, . . . , m 1, we get (12) by repeated use of what is called thetransitivity property (which simply means that whenever a < b andb < c we have a < c). See Comment No. 9 of Chapter 6 for a similar

application of transitivity, to prove that (1 + 1

m)m >(1 +

1

n)n when-

ever m > n.

To prove (13), we write 2k as 2.2k1 and multiply both the sides byk(k 1) which is positive. Then after cancellation of common terms,(13) reduces to showing that k2k1 > 2k 1 which is true since theL.H.S. is at least 2.2k1.

This is an excellent problem which rewards candidates who havegood intuition and sound common sense. In fact such candidates willget the answer without doing any numerical work. For a rigorous so-lution, numerical work is indeed needed. The paper-setters have taken

care to see to it that this work is manageable by making the problemless ambitious than