86
EDUCATIVE COMMENTARY ON JEE 2015 ADVANCED MATHEMATICS PAPERS (Last Revised on May 25, 2019) Contents Paper 1 3 Paper 2 40 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 the IITs and for more than half a century they did this through the JEE which acquired a world wide reputation as one of the most challenging tests for entry to an engineering programme. Having cleared the JEE was often a passport for many lucrative positions in all walks of life (many of them having little to do with engineering). It is no exaggeration to say that the coveted position of the IIT’s was due largely to the JEE system which was renowned not only for its academic standards, but also its meticulous punctuality and its unimpeachable 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, but all NIT’s and other engineering colleges who would want to come under its umbrella. 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 IITs and eventually of their reputation. Finally a compromise was reached that the 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), which would be conducted solely by the IITs, exactly as they conducted their JEE in the past. So, in effect, the JEE (Advanced) from 2013 took the role of the JEE in the past except that the candidates appearing for it are selected by a procedure over which the IITs have no control. So, this arrangement is not quite the

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Page 1: EDUCATIVE COMMENTARY ON JEE 2015 ADVANCED … · EDUCATIVE COMMENTARY ON JEE 2015 ADVANCED MATHEMATICS PAPERS (Last Revised on May 25, 2019) Contents Paper 1 3 Paper 2 40 Concluding

EDUCATIVE COMMENTARY ON

JEE 2015 ADVANCED MATHEMATICS PAPERS

(Last Revised on May 25, 2019)

Contents

Paper 1 3

Paper 2 40

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 positionof the IIT’s 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 NIT’s 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 havea 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 procedureover which the IITs have no control. So, this arrangement is not quite the

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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 havebeen sufficient time to make the switch-over). But this has not happenedeven in 2015 and after a change of regime at the Union Government. TheJEE (Advanced) 2015 is completely multiple choice type and its patterndiffers little from that of the JEE (Advanced) 2014. Questions with singledigit answers come before rather than after those with one or more correctanswers. Also negative marking has been revived for the latter.

Academically (and socially), the JEE (Advanced) has the same status asthe JEE in the past. So, from 2013, the Educative Commentary to JEEMathematics Papers is confined only to JEE (Advanced). This year, thenumbering of the questions will be that in Code 8 of the question papers.As in the past, unless otherwise stated, all the references made are to theauthor’s book Educative JEE (Mathematics) published by Universities Press,Hyderabad. The third edition was made available online on the author’s bloghttp://www.mathjeecommentary.blogspot.in but has now been withdrawn asthe book is easily available in 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 the time relevance was reduced.

As an experiment, a draft version of this year’s commentary on both thepapers was uploaded. Those readers who noticed any errors in it were invitedto send an email to the author at [email protected] or send an SMSto the author at 9819961036. Alternate solutions and any other commentswere also solicited. This really paid off. Several readers, notably SiddheshNaik and Deepanshu Rajvanshi pointed out several corrections and suggestedsome additions. The elegant solution to Q.55 in Paper 1, where the givendeterminant is expanded by writing it as a product of two determinants wasgiven by Siddhesh Naik. I am grateful to all these and other readers.

2

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PAPER 1

Contents

Section - 1 (One Integer Value Correct Type) 3

Section - 2 (One or More than One Correct Choice Type) 13

Section - 3 (Matching the Pairs Type) 32

SECTION 1

One Integer Value Correct Type

This section contains eight questions. The answer to each question is aSINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.

Marking scheme : +4 If the bubble corresponding to the answer is dark-ened, 0 In all other cases

Q.41 Let the curve C be the mirror image of the parabola y2 = 4x withrespect to the line x+y+4 = 0. If A and B are the points of intersectionof C with the line y = −5, then the distance between A and B is

Answer and Comments: 4. Call the line y = −5 as L.

O

L

x

y

A

B

L

x + y + 4 = 0

C *

*

*

*

C

A B

A straightforward approach would be to first identify C which is given

3

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to be the reflection of the parabola y2 = 4x in the line x + y + 4 = 0.Call this parabola C∗. Then C and C∗ are reflections of each other andhence are congruent. So C is also a parabola. To find its equation, wewould have to start with a typical point (x0, y0) on C, find its reflection,say (x∗

0, y∗0), in the line x+ y + 4 = 0 and then put y0∗2 = 4x0∗2. The

formula for the reflection of a point (x0, y0) into a line ax+ by + c = 0is possible but rather complicated.

A better approach is to realise that under reflections all distancesare preserved. So, instead of taking the points of intersection of theparabola C and the line L, we may as well take the points of intersectionof the parabola C∗ and the reflection, say L∗, of the line L into the linex+ y + 4 = 0. L∗ can be found almost by inspection. The line L cutsthe line x+ y + 4 = 0 at the point P = (1,−5) at an angle 45 degrees.So, L∗ must be the line through P making an angle of 45 degrees withthe line x+ y + 4 = 0. Clearly this line is x = 1. This line L∗ happensto lie along the latus rectum of the parabola C∗ : y2 = 4x. Hence itsintercept with the parabola has length 4. But even if we miss this,the points of intersection, say A∗ and B∗ of L∗ with C∗ can be foundby merely solving the two equations y2 = 4x and x = 1. They are(1,±2) and so the distance between them is 4. This is also the distancebetween A and B since A∗, B∗ are the reflections of A,B.

This is an excellent problem which tests the ability of realisinghow not to do a problem in the most straightforward way and, instead,look for alternate ways. Once the idea of transforming the problem tofinding the intersections of C∗ and L∗ strikes, the calculation requiredis minimal.

Q.42 The minimum number of times a fair coin needs to be tossed, so thatthe probability of getting at least two heads is at least 0.96, is

Answer and Comments: 8. Yet another problem where it is mucheasier to find the answer by transforming the problem. In the presentcase, the transformation is to consider the complementary probability,say q of the given event. The complementary event here is that atmost one head appears. This falls into two mutually exclusive cases, noheads and exactly one head. If there are n tosses, then their respective

probabilities are (1

2)n and n × 1

2× (

1

2)n−1 = n(

1

2)n. Together, q =

4

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(n + 1)(1

2)n. The problem asks for the least integral value of n for

which q < 0.04. This reduces to

25(n+ 1) < 2n (1)

As n grows, 2n grows much more rapidly than n+1. But the least suchn has to be found by trial and error. The L.H.S. is at least 50 and son has to be at least 6. By trial we get that 8 is the least integer forwhich (1) holds.

This is also a good problem. But the technique of complemen-tary probability is fairly common as compared with using properties ofreflections in the last problem.

Q.43 Let n be the number of ways in which 5 boys and 5 girls can stand in aqueue in such a way that all the girls stand consecutively in the queue.Let m be the number of ways in which 5 boys and 5 girls can stand ina queue in such a way that exactly four girls stand consecutively in the

queue. Then the value ofm

nis

Answer and Comments: 5. To find n think of the 5 girls linedtogether as a single object. Then the number of ways to arrange thisobject along with the 5 other objects (the boys) is 6!. But the 5 girlscan form a single object in 5! ways. So

n = 6!5! (1)

To find m, we consider a single object consisting of four girls in arow. This object can be formed in 5 × 4! = 5! ways. Now we have7 objects, this object with four girls, the remaining girl and the 5boys. They can be arranged in 7! ways. Hence the total number ofarrangements in which at least 4 girls are together is 7!. But we haveto exclude those in which all 5 girls are together. That is alreadycounted as n. Moreover, each such arrangement has to be excludedtwice because the excluded girl can be the one at the head or at thetail. [(G1G2G3G4G5) gets excluded twice, once as G1(G2G3G4G5), andthen again as (G1G2G3G4)G5.] Hence

m = 7!× 5!− 2n = 7!× 5!− 2× 6!× 5! (2)

5

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Hence

m

n=

7!− 2× 6!

6!= 7− 2 = 5 (3)

A simple but interesting problem. The ideas of transforming theproblem (by means of forming a single object of the girls) and of com-plementary counting are present here too. But what makes the problemtricky is that each unwanted arrangement appears twice. Those who

miss this will get7!− 6!

6!= 6 as the answer. And since this is also an

integer between 0 and 9, there is no built in alert.

There is an alternate and a less tricky way to calculate m. Asbefore, form two groups of girls, one with 4 girls and the other withjust one girl. Now arrange the 5 boys on a row in any of the 5! ways.In each such arrangement, there are six possible places where the twogroups of girls can go, one before the first boy, one after the last boyand four in between two consecutive boys. We have to insert the twogroups of girls into these 6 places so that they do not go into the sameplace. This can be done in 6× 5 = 30 ways. This would give

m = 5!× 5!× 30 (4)

which is the same as (2).

Q.44 If the normals of the parabola y2 = 4x drawn at the end points of itslatus rectum are tangents to the circle (x − 3)2 + (y + 2)2 = r2, thenthe value of r2 is

Answer and Comments: 2. Clearly r is the distance of the point(3,−2) (the centre of the circle) from either of the two normals. Theproblem is an unnecessarily clumsy way of asking the distance of thispoint from these normals. Actually, either one of the two ends of thelatus rectum will suffice since we are given that both the normals atboth the ends are equidistant from (3,−2). We choose the end A =

(1, 2) for this purpose. The slope of the tangent at this point isdy

dx

where y2 = 4x. This givesdy

dx=

2

y= 1 at A. Hence the slope of the

6

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normal is −1. The equation of the normal at A is y − 1 = −(x − 1)i.e.

x+ y − 3 = 0 (1)

The distance r of the point (3,−2) is|3− 2− 3|√

1 + 1=

√2. Hence r2 = 2.

An extremely straightforward problem. A disappointment on thebackdrop of the last three problems. Such problems hardly belong toan advanced test. Some students may be unnecessarily tempted to useformulas for equations of normals in terms of their slopes. The directapproach above is far better.

Q.45 Let f : IR −→ IR be a function defined by f(x) =

{

[x], x ≤ 20, x > 2

where [x] is the greatest integer less than or equal to x. If I =∫ 2

−1

xf(x2)

2 + f(x+ 1)dx,

then the value of (4I − 1) is

Answer and Comments: 0. The function f changes its formula atx = 2. Since the integrand involves f(x2) and f(x + 1), we have tokeep track of when x2 exceeds 2 and also when x+ 1 exceeds 2 in theinterval of integration, viz. [−1, 2]. The former happens at x =

√2

and the latter at x = 1. Moreover [x] has discontinuities at 0 and −1.So we have to split the interval of integration into four subintervals,[−1, 0], [0, 1], [1,

√2] and [

√2, 2] and integrate over each one of them

and add the four integrals. We do so one-by-one. Call the integrandas g(x) in all cases and denote the four integrals by I1, I2, I3 and I4respectively.

On both the intervals [−1, 0] and [0, 1] 0 < x2 < 1 except possiblyat the end-points and so f(x2) = [x2] = 0. Hence the first two integralsare 0. In the third and the fourth intervals, x + 1 exceeds 2 except atthe point 1 and hence f(x+1) = 0. So in both the cases the integrand

g(x) simplifies to1

2xf(x2). On [1,

√2] this becomes

x

2while on [

√2, 2],

it vanishes. So the whole integral I is merely∫

√2

1

x

2dx =

x2

4

√2

1=

1

4.

Therefore 4I − 1 = 0.

7

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A tedious problem which tests little more than some carefulness onthe part of the candidate. The integration part itself is trivial.

Q.46 A cylindrical container is to be made from certain solid material withthe following constraints : It has a fixed inner volume of V mm3, has a 2mm thick solid wall and is open at the top. The bottom of the containeris a solid circular disc of thickness 2 mm and is of radius equal to theouter radius of the container. If the volume of the material used tomake the container is minimum when the inner radius of the container

is 10 mm, then the value ofV

250πis

Answer and Comments: 4. A typical problem about the minimisa-tion of a function of one variable. The choice of this variable is ours.We could take it to be the either the inner or the outer radius of thebase or even its inner or outer height. But since the last part of thedata is in terms of the inner radius of the base, that is a more naturalchoice. So let r be the inner radius of the base and h the inner heightof the cylinder. Then its (inner) volume V is

V = πr2h (1)

As V is fixed, this equation allows us to express the inner height h interms of r. The volume, say W , of the material used is the differenceof the volumes of two coaxial cylinders, the inner one of radius r andheight h and the outer one of radius r + 2 and height h + 2 (and noth+ 4 as the container has no top). Therefore

W = π(r + 2)2(h+ 2)− πr2h

= π[(r2 + 4r + 4)(h+ 2)− r2h]

= π[h(4r + 4) + 2(r + 2)2]

=V

r2(4r + 4) + 2π(r + 2)2 (2)

by (1).

As V is a constant, this expresses W as a function of the single variable

r. We are given that it is minimum when r = 10. SodW

drmust vanish

at r = 10. By a direct calculation,

dW

dr= −4V

r2− 8V

r3+ 4π(r + 2) (3)

8

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As this vanishes for r = 10, we get

4V

100+

8V

1000= 48π (4)

which simplifies to48V

1000= 48π and gives

V

250π= 4.

Although this is a problem about minimising a function of onevariable, it is of a different spirit. Here we are not asked to minimiseW . Rather, we are given where it is minimum. That simplifies thework. If we were asked to find where W is minimum, we would haveto solve a cubic equation and then hunt for the minimum among thecritical points. That makes the problem reasonable. Also the fact thatthe answer is an integer between 0 and 9 serves as an alert in case thereare any computational mistakes. This is a good feature of the problem.(A problem of a similar spirit was asked in Jee 2013 Advanced Paper1. An open box was to be formed by folding a rectangle after removingsquares of the same size from its four corners and we were given the sideof the square for which this volume is maximum. There is, of course,nothing wrong in asking problems similar to those in the past years.But one wishes that the repetition would not have occurred so soon.Many but not all students must have studied the 2013 papers and thosewho did would have an easier time in understanding the problem thanthe others. It is all right to repeat an idea that was used last year, orone that was used a decade ago. But a gap of two years can be unfair.)

Q.47 Let F (x) =∫ x2+π/6

x2 cos2 t dt for all x ∈ IR and f : [0, 1/2] −→ [0,∞)

be a continuous function. For a ∈ [0, 1/2], if F ′(a) + 2 is the area ofthe region bounded by x = 0, y = 0, y = f(x) and x = a, then f(0) is

Answer and Comments: 3. The problem asks the value of thefunction f(x) at x = 0. But this function is not given explicitly. In-stead, we are given that f is defined on [0, 1/2] and takes only non-negative values. We are also told something about the area boundedby x = 0, y = 0, x = a and y = f(x). This is precisely the area underthe graph of y = f(x). So, the second piece of information means that

∫ a

0f(x) dx = F ′(a) + 2 (1)

9

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where the R.H.S. needs to be calculated from the first part of theproblem. (We have to assume that (1) holds for all a ∈ [0, 1/2]. Thisis not clearly stated in the problem. But the problem cannot be solvedwithout such an assumption.)

Our interest is in f(x). We can get it from (1) by differentiating boththe sides (w.r.t. a) using the second form of the FTC (FundamentalTheorem of Calculus). We then get,

f(a) = F ′′(a) (2)

for all a ∈ [0, 1/2]. So our problem now reduces to find F ′′(0).

To find F ′′(x), we need to differentiate F (x) twice. The functionF (x) is defined by an integral in which both the upper and the lowerlimits are functions of x. Therefore, by the generalised form of thesecond fundamental theorem of calculus,

F ′(x) = 2 cos2(x2 + π/6)d

dx(x2 + π/6)− 2 cos2 x

d

dx(x)

= 4x cos2(x2 + π/6)− 2 cos2 x (3)

We can get F ′′(x) by differentiating the R.H.S. Instead of doing thismechanically, let us observe that our interest is only in F ′′(0). Sothere is no need to consider those terms in the derivative which aresure to vanish at 0. The derivative of cos2 x is one such term since itwill involve a factor sin x. As for the derivative of the first term, viz.4x cos2(x2 + π/6), when we apply the product rule, the factor 4x willvanish at x = 0. So there is no need to take the derivative of the secondfactor cos2(x2 + π/6). The only term in the derivative that remains tobe considered is 4 cos2(x2 + π/6), evaluated at x = 0. This comes out

to be 3 since cos(π/6) =

√3

2.

Problems based on the second form of the FTC are fairly commonin JEE. The present problem is more a test of a candidate’s abilityto analyse a problem correctly and focus on the essence so as to weedout unnecessary work. In this sense it is a very good problem. Manycandidates will be tempted to simplify the integrand of F (x) to 1 +cos 2t. But such a simplification has no role in the solution. In thisrespect, the problem is a bit tricky too.

10

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Note that the second term (viz. 2) on the R.H.S. of (1) plays norole in the solution since we are only dealing with the derivative of theR.H.S. It would thus appear that in the data of the problem, the area ofthe region could as well have been given as F ′(a) + k for any constantk. But then there would have been an inconsistency. From (3), weknow that F ′(0) = −2. On the other hand, by putting a = 0 in (1), wewould have gotten F ′(0) = −k. So, the data would be consistent onlyif k = 2. A somewhat similar inconsistency in the data had occurredin a problem in JEE 2011 Paper 1 and was commented upon. (SeeQ.22 of that year’s commentary.) It was given in the statement of the

problem that 6∫ x

1f(t)dt = 3xf(x)− x3 for all x ≥ 1. Also f(1) was

given to be 2. Together we get a contradiction that 0 = 5 (by puttingx = 1). The lapse could have been corrected by changing the data to

6∫ x

1f(t)dt = 3xf(x)− x3 − 5. This would not affect the rest of the

problem or its solution. Indeed that is probably what led to the lapse.

This year the paper-setters have been careful. The problem couldhave been made a little more interesting by giving in the data that thearea of the region was F ′(a) + k for some constant k and then askingthe candidates to find the value of f(0) + k. That would have forcedthe candidates to determine k as 2 from (1) and (3). The answer tothe problem would have been 5 instead of 3, still a single digit number.

Q.48 The number of distinct solutions of the equation

5

4cos2 2x+ cos4 x+ sin4 x+ cos6 x+ sin6 x = 2

in the interval [0, 2π] is

Answer and Comments: 8. Superficially, this is a problem of solvinga trigonometric equation. But there is no way to do this unless we firstsimplify the expression, say E on the L.H.S. The key idea is to notethat all terms are expressible in terms of sin2 x and cos2 x. So the goodold identity sin2 x+cos2 x = 1 may be useful. Indeed, if we square this,we get

sin4 x+ cos4 x = 1− 2 sin2 x cos2 x (1)

11

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while if we take the cubes of both the sides, we get

sin6 x+ cos6 x = 1− 3 sin4 x cos2 x− 3 sin2 x cos4 x

= 1− 3 sin2 x cos2 x(sin2 x+ cos2 x)

= 1− 3 sin2 x cos2 x (2)

With these substitutions, the expression E on the L.H.S. of the givenequation becomes

E =5

4cos2 2x+ 2− 5 sin2 x cos2 x (3)

and so the equation simplifies to

cos2 2x− 4 sin2 x cos2 x = 0 (4)

or equivalently,

cos2 2x− sin2 2x = 0 (5)

and still further to

cos 4x = 0 (6)

which has eight solutions in the interval [0, 2π]. Specifically, the solu-tions are of the form where 4x = π/2 and where 4x = 3π/2. But theproblem only asks for the number of solutions.

This is a fairly easy problem once the idea of taking the powers ofthe basic identity for sin2 x + cos2 x strikes. See Comment No. 14 ofChapter 7 for a problem where a similar trick is used.

12

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SECTION 2

This section contains TEN 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, 0 if none of the bubbles is darkened and −2 inall other cases.

Q.49 Let y(x) be a solution of the differential equation (1 + ex)y′ + yex = 1.If y(0) = 2, then which of the following statements is (are) true?

(A) y(−4) = 0

(B) y(−2) = 0

(C) y(x) has a critical point in the interval (−1, 0)

(D) y(x) has no critical point in the interval (−1, 0).

Answer and Comments: (A), (C). This is an extremely standardproblem of solving a first order linear differential equation. Normally,one would begin by recasting the equation in the standard form y′ +p(x)x = q(x) and find an integrating factor. In the present problem,that is hardly necessary. The equation is exact as it stands, becausethe L.H.S. is simply the derivative of y(1 + ex). So, integrating boththe sides, the general solution is

y(1 + ex) = c+ x (1)

where c is an arbitrary constant. The initial condition y(0) = 2 gives4 = c. So the function y is given by

y(x) =4 + x

1 + ex(2)

Clearly y(−4) = 0 while y(−2) 6= 0. For critical points, we need the

derivativedy

dx.

dy

dx=

(1 + ex)− (4 + x)ex

(ex + 1)2(3)

13

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At a critical point the numerator must vanish. This gives the equation

3ex + xex − 1 = 0 (4)

We cannot solve this equation explicitly. Nor is it needed. All we areasked is if it has a solution in the interval (−1, 0). For this we apply theIntermediate Value Property. Call the L.H.S. as g(x). It is continuous

everywhere. Also g(−1) =2

e− 1 < 0 since e > 2. On the other hand

g(0) = 2 > 0. So, by the Intermediate Value Property, g(x) has atleast one root in (−1, 0). Therefore y(x) has at least one critical pointin (−1, 0).

The problem is a combination of two unrelated parts. First, solvingan initial value problem and secondly testing if a given function has aroot in an interval. Both are very standard. There is little point inasking such questions in an advanced test.

Q.50 Consider the family of all circles whose centers lie on the straight liney = x. If this family of circles is represented by the differential equation

Py′′+Qy′+1 = 0, where P,Q are functions of x, y and y′ (here y′ =dy

dx

and y′′ =d2y

dx2), then which of the following statements is (are) true?

(A) P = y + x (B) P = y − x(C) P +Q = 1− x+ y + y′ + (y′)2 (D) P −Q = x+ y − y′ − (y′)2

Answer and Comments: (B), (C). This problem is about findingthe differential equation of a family of curves. In the present case, atypical member of the given family is a circle of the form

(x− h)2 + (y − h)2 = r2 (1)

where h and r are arbitrary constants. So this is a two parameterfamily and the differential equation representing it will be of order 2.To get it we differentiate (1) to get

2(x− h) + 2(y − h)y′ = 0 (2)

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Solving this for h we get

h =x+ yy′

y′ + 1(3)

One more differentiation yields

(y′ + 1)(1 + yy′′ + y′2)− (x+ yy′)y′′

(y′ + 1)2= 0 (4)

Hence the differential equation representing the given family of circleis

(y′ + 1)(1 + yy′′ + y′2)− (x+ yy′)y′′ = 0 (5)

When expanded, this may contain terms involving the product y′y′′. Inthe statement of the problem, the coefficients P and Q are not allowedto contain y′′. So we have to recast this equation collecting all theterms in y′′ together. That gives

(y − x)y′′ + (1 + y′ + y′2)y′ + 1 = 0 (6)

(Luckily, the terms involving y′y′′ have cancelled.) Comparing this withthe form given in the statement of the problem,

P = y − x (7)

and Q = 1 + y′ + y′2 (8)

Hence (B) and (C) are correct.

A very mechanical problem. The solution essentially ends at (5).The remainder is a useless addendum. It might have served some pur-pose if there were any terms involving y′y′′ because then the candidatewould have to think whether to include them as multiples of y′ or ofy′′. But since these terms get cancelled, this fine thinking is not testedanyway.

Q.51 Let g : IR −→ IR be a differentiable function with g(0) = 0, g′(0) = 0

and g′(1) 6= 0. Let f(x) =

x

|x|g(x), x 6= 0

0, x = 0and h(x) = e|x| for all

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x ∈ IR . Let (f ◦ h)(x) denote f(h(x)) and (h ◦ f)(x) denote h(f(x)).Then which of the following is (are) true?

(A) f is differentiable at x = 0 (B) h is differentiable at x = 0(C) f ◦ h is differentiable at x = 0 (D) h ◦ f is differentiable at x = 0

Answer and Comments: (A), (D). This is a question about thedifferentiability of two functions and also of their composites. Thecomposite of two differentiable functions (when defined) is always dif-ferentiable. But sometimes the composite may be differentiable evenwhen one of the functions fails to be so. An extreme counterexampleis when one of the functions is constant. Then the composite is alsoa constant and differentiable regardless of the other function. So suchcases have to be handled carefully.

Let us begin with the differentiability of the function h(x) = e|x|

at x = 0. This is the composite of the absolute value function and theexponential function. The former is not differentiable at 0. But that isno reason to hastily declare that the composite e|x| is not differentiableat 0 because as we just saw, the composite of a differentiable functionwith a non-differentiable one can be differentiable sometimes. But wecan put the non-differentiability of |x| to use as follows. We want to

consider whether limx→0

e|x| − e0

xexists. For x 6= 0 we rewrite this ratio

as

e|x| − e0

x=

e|x| − 1

|x| × |x|x

(1)

As x → 0, the first factor tends to 1, this being the right handedderivative of ex at x = 0. But the second factor is 1 for x > 0 and −1for x < 0. So the product tends to 1 as x → 0+ and to −1 as x → 0−.Therefore the product of the two ratios tends to 1 as x → 0+ and to −1as x → 0−. Hence h is not differentiable at 0. (There is a slicker wayto see this by observing that |x| = ln(h(x)). The logarithm function isdifferentiable everywhere in its domain. So, if h were differentiable at0, then this composite |x| would be differentiable at 0, a contradiction.)

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Next, we turn to the differentiability of f at 0. Note that

f(x) =

g(x), x > 00, x = 0

−g(x), x < 0(2)

Therefore, f ′+(0), i.e. the right handed derivative of f at 0, will equal

g′+(0) which is 0 since it is given that g′(0) = 0. Similarly, f ′−(0) will

equal −g′−(0) which is also 0. Hence f is differentiable at 0. (It wouldnot be so if g′(0) were non-zero.)

We now tackle the differentiability of the composite functions f ◦ hand h ◦ f at 0. Note that h(x) is always positive and so by (2)

(f ◦ h)(x) = f(h(x)) = g(h(x)) (3)

for all x. Further h(0) = 1. So, by a reasoning similar to that in (A),(f ◦ h)′+(0) would equal g′+(h(0)), i.e. g′+(1) = g′(1) as g is given tobe differentiable at 1, while (f ◦ h)′−(0) would equal −g′−(1) = −g′(1).Since g′(1) 6= 0, these two numbers are unequal and so f ◦ h is notdifferentiable at 0.

Finally, we consider the differentiability of h◦ f at 0. By definition,

(h ◦ f)(x) = e|f(x)| (4)

for all x. Also, by (2), |f(x)| = |g(x)| for all x. Hence e|f(x)| is thesame as e|g(x)|. The exponential function is differentiable everywhere.So if we can show that |g(x)| is differentiable at 0, then it would followthat e|g(x)|, and hence h ◦ f is differentiable. We are given that g isdifferentiable at 0. This does not by itself imply that |g| is differentiableat x, as one sees from the fact that x is differentiable at 0 but |x| isnot. But now we are also given that g(0) = 0 and g′(0) = 0. So, forx 6= 0 we can write

|g(x)| − |g(0)|x

= ±∣

g(x)

x

(5)

with the + sign holding for x > 0 and the minus sign holding for

x < 0. But as we are given thatg(x)

x→ 0 as x → 0, it follows

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that

g(x)

x

→ 0 as x → 0, no matter from which side. This proves

that |g(x)| is differentiable at 0, and, as noted earlier that h ◦ f isdifferentiable at 0.

This is a very good conceptual question which demands very finethinking and little computations from the candidate. Unfortunatelysuch questions are unsuitable as multiple choice questions because thereis no way to test if the candidate has understood the subtlety of theproblem, even if he has ticked the correct answer. For example, acandidate who rules out (B) on the superficial ground that |x| is notdifferentiable at 0 cannot be distinguished from one who really provesit as above.

Negative credit is very unfair for such questions. There is noreason why a candidate who has done the fine thinking for, say (B),should be penalised for failure to do a similar thinking for (D). Mostcandidates who are adept at the JEE strategy will stay away from suchquestions and in the time saved, safely bag 12 points by doing someabsolutely routine questions. So, such questions serve little purpose inthe selection when they are clubbed together with a large number ofmediocre questions.

Q.52 Let f(x) = sin(

π

6sin

(

π

2sin x

))

and g(x) =π

2sin x for all x ∈ IR. Let

(f ◦ g)(x) denote f(g(x)) and (g ◦ f)(x) denote g(f(x)). Then whichof the following is (are) true?

(A) Range of f is[

−1

2,1

2

]

(B) Range of f ◦ g is[

−1

2,1

2

]

(C) limx→0

f(x)

g(x)=

π

6(D) There is an x ∈ IR such that (g ◦ f)(x) = 1

Answer and Comments: (A), (B), (C). Yet another question in-volving the composites of two functions. In both the questions thepaper-setters have been careful enough to indicate which of the twopossible interpretations of a composite they have in mind. Althoughthe interpretation given here is more standard, candidates who followthe other interpretation should not suffer solely for that reason.

Now, coming to the question itself, all parts except (C) are based onthe images of various intervals under the sine function. and the fact that

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the range of the composite is the image of the range of the first function.

Since sin x maps IR onto [−1, 1],π

2sin x maps it onto [−π

2,π

2]. The sine

function maps this interval onto [−1, 1] and soπ

6sin(

π

2sin x)) has range

[−π

6,π

6]. Under the sine function this interval goes to [−1/2, 1/2] which

proves (A). In (B), we first have to find the range of g which is [−π

2,π

2].

Under the sine function, this interval has the same image viz. [−1, 1]as the image of the sine function over the entire IR. So f ◦ g has thesame range as f , which is [−1/2, 1/2]. In (D), the range of g ◦ f is theimage under g of the range of the f which we already know to be theinterval [−1/2, 1/2]. So (D) will be true if and only if there is some

x ∈ [−1/2, 1/2] for which g(x) = 1, i.e. sin x =2

π. As the sine function

is strictly increasing on the interval [−1/2, 1/2], the answer dependsupon which of the two numbers sin(1/2) and 2/π is bigger. To do thiswithout calculators, we use the inequality sin x < x for all x > 0. Inparticular, sin 1/2 < 1/2. But 1/2 < 2/π since π < 4. So (D) is false.

Part (C) is of a totally different spirit than the others. The limitin question can be calculated rather mechanically using the L’Hopital’s

rule. But a better way out is to put u = g(x) =π

2sin x and v = π

6sin u.

Then u → 0 and v → 0 as x → 0. Therefore,

f(x)

g(x)=

sin(π6sin u)

u

=sin(π

6sin u)

π6sin u

×π6sin u

u

=sin v

v× π

6× sin u

u(1)

Both the first and the last factors tend to 1 and so the limit isπ

6.

A simple but highly repetitious problem based on the range of thesine function. Part (D) requires the approximate value of π. Part (C)is a useless addendum.

Q.53 Let △PQR be a triangle. Let ~a =−→QR,~b =

−→RP and ~c =

−→PQ. If

|~a| = 12, |~b| = 4√3 and ~b · ~c = 24, then which of the following is (are)

true?

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(A)|~c|22

− |~a| = 12 (B)|~c|22

+ |~a| = 30

(C) |~a×~b+ ~c× ~a| = 48√3 (D) ~a ·~b = −72

Answer and Comments: (A), (C), (D). A routine problem aboutcomputations involving vectors, their lengths and various products.The lengths of the two sides of the triangle are given. If we were giventhe dot product of the vectors representing them, viz. ~a and ~b, wewould also know the angle between them and then we would know thetriangle completely. But we are not given that. We are given |~a|, |~b|,but not ~a ·~b. Instead, we are given ~b · ~c. We are also given |~b|. So, if

we could get |~b+ ~c|, we would know |~c| and that would also determinethe triangle completely.

The basic idea is thatsince the vectors ~a,~b and ~care the sides of the triangle(directed appropriately), theirvector sum is 0. That is,

~a +~b+ ~c = ~0 (1)

P

R

bQ

θ/2θ/2

a O 663030

c

Therefore

~b+ ~c = −~a (2)

Taking lengths of both the sides

|~b|2 + |~c|2 + 2(~b · ~c) = |~a|2 = 144 (3)

This gives

|~c| =√144− 48− 48 =

√48 = 4

√3 (4)

Having known |~a| = 12 and |~c| = 4√3 we immediately dispose of (A)

and (B).

For the other two statements, we need to know more about thetriangle. Since |~b| = |~c| the triangle PQR is isosceles with PQ = PR.

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Let the angle between these sides be θ. Then θ is the angle betweenthe vectors −~b and ~c. So,

cos θ =−~b · ~c|~b| |~c|

=−24

48= −1

2(5)

which gives θ = 120◦. As the triangle is isosceles, the remaining twoangles are 30◦ each. Since the angle between the vectors ~b and −~a is30◦ the angle between the vectors ~a and ~b is 150◦. So we get

~a ·~b = |~a| |~b| cos 150◦ = −12× 4√3×

√3

2= −72 (6)

which shows that (D) is true. Finlly, for (C), we note that by (1)~b = −~a − ~c. Since ~a × ~a = ~0, and the angle between ~a and ~c is also150◦, we have

|~a×~b+ ~c× ~a| = ~a× (−~a− ~c) + ~c× ~a

= |2~c× ~a|= 2|~c| × |~a| × sin 150◦

= 4√3× 12 = 48

√3 (7)

Hence (C) is true too.

Our solution is purely geometric. The second half of it could havebeen shortened a little by resolving the vectors along a suitable pairof mutually orthogonal unit vectors. We take the midpoint of the sideQR as the origin O and two unit vectors ~i and ~j along OR and OP

respectively. Since 6 OPQ = 6 OPR = 60◦ we have OP =1

2PQ =

2√3. This allows us to express all the three vectors ~a,~b and ~c as linear

combinations of the mutually orthogonal vectors ~i and ~j as

~a = 12~i (8)

~b = −6~i+ 2√3 ~j (9)

and ~c = −6~i− 2√3 ~j (10)

By a direct calculation,

|~a×~b+ ~c× ~a| = |~a× (~b− ~c)|= |12~i× 4

√3 ~j| = 48

√3 (11)

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which shows that (C) is correct. Also

~a ·~b = 12~i · (−6~i+ 2√3 ~j) = −72 + 0 = −72 (12)

so that (D) is also true.

In fact, the entire solution could have been algebraic. We take ~i tobe a unit vector along ~a and ~j to be a unit vector perpendicular to ~i.Then, ~a = 12~i and ~b = α~i + β~j where α, β are some scalars. Then by(1), ~c = (−12− α)~i− β~j. Since ~b = 4

√3, we have

α2 + β2 = (4√3)2 = 48 (13)

Similarly, ~b · ~c = 24 gives

−α(α + 12)− β2 = 24 (14)

Solving this system simultaneously, we get

α = −6, β = 2√3 (15)

This way we get (9) and (10) more efficiently. Having known all threevectors in terms of ~i and ~j, all the four statements can be tested oneby one. Thus we see that in the present problem the purely algebraicsolution is fastest. But the gain is not so significant as the problemitself is simple.

The problem is simple, once the essential idea, viz. Equation (1)strikes. Unfortunately, there is too much numerical work. Even a singlemistake is costly. So this problem is more a test of speed and numericalaccuracy than reasoning.

Q.54 Let X and Y be two arbitrary, 3×3, non-zero skew-symmetric matricesand Z be an arbitrary 3 × 3 non-zero symmetric matrix. Then whichof the following matrices is (are) skew-symmetric?

(A) Y 3Z4 − Z4Y 3 (B) X44 + Y 44

(C) X4Z3 − Z3X4 (D) X23 + Y 23

Answer and Comments: (C), (D). Parts (B) and (D) are basedon some simple properties of skew-symmetric and symmetric matrices.

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Specifically, we need : (i) the sum of two symmetric (skew-symmetric)matrices is symmetric (respectively, skew-symmetric) and (ii) all pow-ers of symmetric matrices are symmetric (iii) all even powers of skew-symmetric matrices are symmetric while all odd powers of them areskew-symmetric. With these properties, we have X44, Y 44 and hencetheir sum symmetric and similarly X23 + Y 23 skew-symmetric. Notethe analogy of these properties with the properties about the signs ofreal numbers. For example, the sum of two positive (negative) realnumbers is also positive (negative). Also the even powers of a negativenumber are positive while the odd powers are negative.

However, in (A) and (C), we are dealing with the products ofsymmetric and a skew-symmetric matrices. Here the analogy with thereal numbers breaks down. For example, the product of two negativenumbers is positive. But little can be said about the product of twoskew-symmetric matrices unless they commute with each other. So,(A) and (C) have be handled by directly taking transposes and usingthe elementary properties of transposes, viz. the anticommutativityand the self-reciprocity. (In simpler terms, this means (AB)T = BTAT

and (AT )T = A, for all A,B.)

With these rules in mind, we have

(Y 3Z4 − Z4Y 3)T = (Y 3Z4)T − (Z4Y 3)T

= (Z4)T (Y 3)T − (Y 3)T (Z4)T

= −Z4Y 3 + Y 3Z4 (1)

because Y 3 is skew-symmetric and Z4 is symmetric. So, Y 3Z4 −Z4Y 3

is symmetric. If it were to be skew-symmetric too, it would have tovanish, which means that Y 3 and Z4 must commute with each other.As this is not given, we discard (A).

By an analogous computation,

(X4Z3 − Z3X4)T = (Z3)T (X4)T − (X4)T (Z3)T = Z3X4 −X4Z3

since Z3 and X4 are both symmetric. This shows that X4Z3 − Z3X4

is skew-symmetric.

Because of their limited scope, questions about matrices tend to berepetitious. The present one is a little unusual. The calculations are

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simple once the essential idea is understood. However, in (A), there is apossibility that the matrix may be skew-symmetric as well, as pointedout above. It is given in the question that the matrices Y, Z are non-zero. But that does not preclude the possibility that Z4 commuteswith Y 3. For example, Z could be the identity matrix. It would havebeen better if the question had asked to identify those matrices thatare necessarily skew-symmetric.

Q.55 Which of the following values of α satisfy the equation∣

(1 + α)2 (1 + 2α)2 (1 + 3α)2

(2 + α)2 (2 + 2α)2 (2 + 3α)2

(3 + α)2 (3 + 2α)2 (3 + 3α)2

= −648α ?

(A) −4 (B) 9 (C) −9 (D) 4

Answers and Comments: (B), (C). The given determinant, say Dis a polynomial in α. It would be horrendous to compute D by directexpansion. But if we subtract R2 from R3 and then R1 from R2 we get

D =

1 + 2α + α2 1 + 4α + 4α2 1 + 6α + 9α2

3 + 2α 3 + 4α 3 + 6α5 + 2α 5 + 4α 5 + 6α

(1)

Next, we subtract the middle row from the other two to get

D =

α2 − 2 4α2 − 2 9α2 − 23 + 2α 3 + 4α 3 + 6α

2 2 2

(2)

We now add the last row to the first to get

D =

α2 4α2 9α2

3 + 2α 3 + 4α 3 + 6α2 2 2

(3)

To simplify D further, we subtract the first column from the other twoto get

D =

α2 3α2 8α2

3 + 2α 2α 4α2 0 0

(4)

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Now D is simple enough to be expanded by its last row. We get

D = 2(12α3 − 16α3) = −8α3 (5)

So the given equation, viz. D = −648α reduces to α3 = 81α whoseroots are 0, 9 and −9.

In the past there used to be many interesting problems based onevaluation of determinants by various manipulations. But they used tobe full length questions, allowing five to ten minutes. In the present set-up, there are hardly 3 minutes. A saving feature is that the candidatesdo not have to explain their work. In the present problem, it is unlikelythat a candidate would get the correct answer by mere substitution. Sothis is a good, classic type problem on determinants.

There is an alternate, albeit a trickier solution to the problem.We first expand all the squares and then write the determinant as theproduct of two determinants, viz.

D =

1 + 2α+ α2 1 + 4α+ 4α2 1 + 6α+ 9α2

4 + 4α+ α2 4 + 8α+ 4α2 4 + 12α + 9α2

9 + 6α+ α2 9 + 12α + 4α2 9 + 18α + 9α2

=

1 1 14 2 19 3 1

1 1 12α 4α 6αα2 4α2 9α2

= 2α3

1 1 14 2 19 3 1

1 1 11 2 31 4 9

(6)

Both the determinants can be evaluated directly or by subtracting thelast column from the others for the first and subtracting the first rowfrom the remaining ones for the second. Their values are 2 and −2.(It is not an accident that their values are the negatives of each other,because if we interchange the first and the last column of the first de-terminant and then take its transpose, we get the second determinant.)So, we finally get D = −8α3 which is the same as (5). The rest of thesolution is the same.

More generally one can consider a determinant D of the form

D = D(x, y, z, a, b, c) =

(a + x)2 (a+ y)2 (a + z)2

(b+ x)2 (b+ y)2 (b+ z)2

(c+ x)2 (c+ y)2 (c+ z)2

(7)

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where x, y, z, a, b, c are variables. When fully expanded this is a homo-geneous polynomial in these six variable in which each term has totaldegree 6. By expanding the squares and taking steps similar to thoseabove, one can show that

D(x, y, z, a, b, c) = 2

a2 a 1b2 b 1c2 c 1

1 1 1x y zx2 y2 z2

(8)

The second determinant is a Vandermonde determinant (see Exercise(3.26)) and hence has value (y − x)(z − y)(z − x). If we interchangethe first and the last columns of the first determinant and take itstranspose, that is also a Vandermonde determinant. As a result, weget

D = −2(a− b)(b− c)(c− a)(x− y)(y − z)(z − x) (9)

There is also an easier way to see this if we observe from (7) that Dvanishes if, say a = b. Hence (a − b) is a factor of D. And so are(b − c), (c − a), (x − y), (y − z) and (z − x). So the product of thesesix factors also divides D(x, y, z, a, b, c). But this product is already apolynomial of total degree 6 in x, y, z, z, b, c. Hence we must have

D(x, y, z, a, b, c) = k(a− b)(b− c)(c− a)(x− y)(y − z)(z − x) (10)

for some constant k. The value of k can be determined by giving somespecial, simple values to a, b, c, x, y, z, e.g. a = x = 0, b = y = 1 andc = z = −1. (A similar technique is also possible for evaluating theVandermonde determinant.)

There is a certain formal resemblance between the determinant in(7) and the determinant

D(P,Q,R,A,B, C) =

cos(A− P ) cos(A−Q) cos(A−R)cos(B − P ) cos(B −Q) cos(B −R)cos(C − P ) cos(C −Q) cos(C −R)

(11)

A 1994 JEE problem (see Comment No. 22 of Chapter 2) asked toshow that this determinant vanishes for all A,B,C, P,Q,R. Again, adirect expansion is simply ruled out. But a solution is possible using

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elementary row operations and trigonometric identities. The best solu-tion, however, is to expand all the entries of the determinant and thenshow that it equals the product

cosA sinA 0cosB sinB 0cosC sinC 0

cosP cosQ cosRsinP sinQ sinR0 0 0

(12)

Q.56 In IR3, consider the planes P1 : y = 0 and P2 : x + z = 1. Let P3 be aplane, different from P1 and P2 which passes through the intersectionof P1 and P2. If the distance of the point (0, 1, 0) from P3 is 1 and thedistance of a point (α, β, γ) from P3 is 2, then which of the followingrelations is (are) true ?

(A) 2α + β + 2γ + 2 = 0 (B) 2α− β + 2γ + 4 = 0(C) 2α + β − 2γ − 10 = 0 (D) 2α− β + 2γ − 8 = 0

Answer and Comments: (B), (D). This is a problem about aparametrised family of planes. Let L be the line of intersection ofthe planes P1 and P2. Write the equations of P1 and P2 in the formE1 = 0 and E2 = 0 where E1, E2 are linear expressions in x, y, z.Then the equation of every plane passing through L is of the formλE1 + µE2 = 0 for some values of the parameters λ and µ. We candispense with one of the parameters, say µ and consider an equation ofthe form λE1 +E2 = 0. This will represent all possible planes throughL for various values of λ, except the plane P1. (For the plane P1,we need λ = 1 and µ = 0.) Similarly, the equation E1 + µE2 = 0will represent all planes through L except P2. In the present case weare given that P3 is different from both P1 and P2. So we are free totake either approach. (See Comment No. 13 of Chapter 9) for moreexamples of this technique.)

We take E1 as y, E2 as x+ z − 1. Then the equation of P3 is of theform

x+ z − 1 + λy = 0 (1)

for some value of λ. To determine it, we use the condition that thedistance of the point (0, 1, 0) is 1. This gives

0 + λ+ 0− 1√1 + λ2 + 1

= 1 (2)

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which simlifies to

(λ− 1)2 = λ2 + 2 (3)

and determines λ as −1

2. So the equation of the plane P3 is x + z −

1− 1

2y = 0 or, equivalently,

2x− y + 2z − 2 = 0 (4)

We are further given that the distance of the point (α, β, γ) from P3 is1. This means

2α+ β + 2γ − 2√4 + 1 + 4

= 2 (5)

This means

2α− β + 2γ − 2 = ±6 (6)

The two signs correspond to (D) and (B) respectively.

A routine problem once the idea of a parametrised family strikes.A discerning student will observe that normally, (2) would reduce to aquadratic in λ, which is consistent with the fact in the family of planescontaining the line L, there are two planes whose distance from thepoint (0, 1, 0) is 1. In the present case, one of these two planes is P1

itself and is discarded by the data. Since we took the equation of P3

in the form λE1 + E2 = 0, which represents all planes containing Lexcept P1, (2) degenerated into a linear equation in λ. Had we takenthe equation of P3 in the form

y + µ(x+ z − 1) = 0 (7)

where µ is a parameter, then instead of (2) we would have gotten

1− µ√1 + 2µ2

= 1 (8)

which would reduce to the quadratic µ2 + 2µ = 0, having 0 and −2 asits roots. The root µ = 0 gives the plane P1 and has to be discarded.

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The second root, viz. −2 will give the equation of the plane P3 as2x − y + 2z − 2 = 0 which is the same as (4). So the answer doesnot change if we take the other parametrisation of the family, but thework involved does slightly. So, we were rather lucky to start with theequation λE1+E2 = 0 rather than with E1+µE2 = 0. A sharp studentwill, however, not leave this choice to luck. He will observe that P1 isalready at a distance 1 from (0, 1, 0) and since it is to be excluded, itis safe (and numerically easier) to take the equation of P3 as (1) ratherthan (7). But in the present problem the advantage gained is minor.Most candidates would anyway prefer to start with (1) rather than (7)because it is simpler. If the advantage were substantial and the easieroption not so tempting, then this would have been a good problemwhich rewards the sharp candidates.

Q.57 In IR3, let L be a straight line passing through the origin. Supposethat all the points on L are at a constant distance from the two planesP1 : x + 2y − z + 1 = 0 and P2 : 2x − y + z − 1 = 0. Let M be thelocus of the feet of the perpendicular drawn from the points on L tothe plane P1. Which of the following points lie(s) on M ?

(A) (0,−5/6,−2/3) (B) (−1/6,−1/3, 1/6)(C) (−5/6, 0, 1/6) (D) (−1/3, 0, 2/3)

Answer and Comments: (A), (B). The first part of the data simplymeans that the line L is parallel to both P1 and P2 and hence to theline of their intersection. From the two equations of the planes, viz.

x+ 2y − z = −1 (1)

and 2x− y + z = 1 (2)

we see that the direction numbers of their line of intersection are1,−3,−5. (This is a standard result. For those who don’t know it,these are the components of the vector u×v where u = i+2j−k andv = 2i− j + k are normals to the planes P1, P2 respectively.) Since Lpasses through the origin, its parametric equations are

x = t, y = −3t, z = −5t (3)

where t is a parameter.

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As L is parallel to the plane P1, the feet of the perpendiculars fromthe points on L to the plane P1 will form a line M parallel to L. Todetermine M , it suffices to know any one point P0 = (x0, y0, z0) on it.We take it to be the foot of the perpendicular from the point (0, 0, 0)(which is given to lie on L) to P1. To determine this point we needthree equations in x0, y0, z0. one of them comes of course, from theequation of the plane P1. That is

x0 + 2y0 − z0 = −1 (4)

The other two equations come from the fact that the vector−→OP0=

x0i+ y0j+ z0k is perpendicular to P1 and hence parallel to the normalvector u = i+ 2j− k. This gives

x0 = r, y0 = 2r, z0 = −r (5)

for some real number r. Substituting this into (4), we get r = −1/6.Hence P0 = (−1/6,−1/3, 1/6). Therefore the locus M is the line

x+ 1/6

1=

y + 1/3

−3=

z − 1/6

−5= λ (6)

where λ is any real number. λ = 0 gives (−1/6,−1/3, 1/6) as a point onM . For the point (0,−5/6,−2/3) to lie on M , we must have λ = 1/6from x + 1/6 = λ. This value also satisfies the other two equationsin (6). But for the other two given points, we have y = 0, whenceλ = −1/9. But that would make x = −1/9 − 1/6 = −5/18. So thepoints in (C) and (D) do not lie on M .

A fairly simple problem once the idea strikes that L is parallel tothe line of intersection of the two given planes. In the conventionalexamination the solution would end with (6), i.e. finding the equationof the locus M . Asking which of the given four points satisfy (6) issheer arithmetic and prone to numerical errors. Also it is anybody’sguess what is the reason for making the candidates do this work fourtimes. The only answer is that the constraints on the paper-settersstipulate that every MCQ must have four choices. Thank God it wasnot 10.

Q.58 Let P and Q be distinct points on the parabola y2 = 2x such thata circle with PQ as a diameter passes through the vertex O of the

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parabola. If P lies in the first quadrant and the area of the triangle△OPQ is 3

√2, then which of the following are the coordinates of P ?

(A) (4, 2√2) (B) (9, 3

√2) (C) (1/4, 1/

√2) (D) (1,

√2)

Answer and Comments: (A), (D). There is a minor anomaly in thewording of the problem. Given any two distinct points P and Q, thereis only one circle having PQ as a diameter. So instead of saying ‘acircle with PQ as a diameter’, the wording should have been ‘the circlewith PQ as a diameter. This might be just a lapse on the part of thepaper-setters. But it might confuse a discerning student, maybe onlyfor a few seconds. But in a severely competitive test even a few secondsare precious.

Now, coming to the problem itself, take the points P and Q in theparametric form as

P = (t212, t1) (1)

and Q = (t222, t2) (2)

Then the slopes of OP and OQ are2

t1and

2

t2respectively. As O lies

on the circle with PQ as a diameter, we have OP ⊥ OQ which means4

t1t2= −1 and hence

t1t2 = −4 (3)

We are further given that the area of the triangle △OPQ is 3√2. Since

OPQ is right angled at O, we get

3√2 =

1

2× OP ×OQ

=1

2× |t1|

2

t21 + 4× |t2|2

t22 + 4

=1

2

(t21 + 4)(t22 + 4) (4)

using (3). Squaring both the sides

(t21 + 4)(t22 + 4) = 72 (5)

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Expanding and using (3) again, this gives

t21 + t22 = 10 (6)

Hence (t1 − t2)2 = 10− 2t1t2 = 18. This gives

t1 − t2 = ±3√2 (7)

We are given that P is in the first quadrant. This makes t1 positiveand hence t2 negative by (3) again. So the positive sign must hold.Now that we know both t1 − t2 and t1t2, we can find t1 by solving theequation

t1 +4

t1= 3

√2 (8)

which becomes a quadratic

t21 − 3√2t1 + 4 = 0 (9)

The roots are3√2±

√2

2i.e. 2

√2 and

√2. The corresponding point

P = (t212, t1) then is (4, 2

√2) and (1,

√2).

A routine problem, once the idea of taking the points in theirparametric forms strikes.

SECTION 3

This section containsTWO questions. Each question contains two columnsColumn I and Column II. Column I has four entries (A), (B), (C) and(D). Column II has five entries (P), (Q), (R), (S) and (T). Match the en-tries in Column I with the entries in Column II. One or more entries inColumn I may match with one or more entries in Column II.

Marking scheme: For each entry in Column I, 2 points if fully correct, 0points if not attempted and −1 points in all other cases.

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Q.59 Column I

(A) In IR2, if the magnitude of the projection vectorof the vector αi + βj on

√3 i + j is

√3 and if

α = 2 +√3β, then possible value(s) of |α| is

(are)

(B) Let a and b be real numbers such that the func-

tion f(x) =

{

−3ax2 − 2, x < 1bx+ a2, x ≥ 1

is differen-

tiable for all x ∈ IR, then the possible value(s)of a are

(C) Let ω 6= 1 be a complex cube root of unity. If(3− 3ω+2ω2)4n+3+(2+3ω− 3ω2)4n+3+(−3+2ω + 3ω2)4n+3 = 0, the possible value(s) of n is(are)

(D) Let the harmonic mean of two positive real num-bers a and b be 4. If q is a positive real numbersuch that a, 5, q, b is an arithmetic progression,then the value(s) of |q − a| is (are)

Column II

(P) 1

(Q) 2

(R) 3

(S) 4

(T) 5

Answers and Comments: (A;P,Q), (B;P,Q), (C; P,Q,S,T), (D; Q,T).The four parts in Column I are quite independent of each other. Butso is the marking for them. This is a merciful departure from the pastwhere credit would be given only if all parts of Column I were answeredcorrectly. But then one wonders what was the point of grouping thesequestions in such an elaborate manner especially when all of them havenumerical answers. Why not ask four separate questions, each havingfive possible answers? This is again a silly constraint on the paper-setters regarding the format of the question paper.

Now, coming to the questions in Column I, in (A), a unit vector in

the direction of the vector√3 i+ j is

√3

2i+

1

2j. Hence the projection

of αi + βj has magnitude |√3

2α +

1

2β|. Equating this with

√3 and

squaring gives

(√3 α+ β)2 = 12 (1)

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We are further given that β =α− 2√

3. Substituting this into (1) we get

(4α− 2)2 = 36 (2)

which gives 4α − 2 = ±6. So the possible values of α are α = 2 andα = −1. Hence |α| = 1 or 2.

In (B), there are two unknowns, a and b. To determine themwe need two equations in them. One is provided by the fact thatevery differentiable function is continuous. Continuity of f(x) at x = 1implies

−3a− 2 = b+ a2 (3)

As for differentiability of f(x), the right handed and the left handedderivatives at x = 1 are −6a and b respectively. So we get

b = −6a (4)

Eliminating b from these two equations gives a quadratic in a, viz.

a2 − 3a + 2 = 0 (5)

whose possible solutions are 1 and 2.

In (C), we are dealing with a sum of the (4n+3)-th powers of threequadratic expressions in ω, specifically,

E1 = 3− 3ω + 2ω2 (6)

E2 = 2 + 3ω − 3ω2 (7)

and E3 = −3 + 2ω + 3ω2 (8)

We observe that the same coefficients, viz. 3,−3 and 2 appear in thesepolynomials. This suggests that the polynomials must be related toeach other in some simple way. (If no such relationship exists, then theproblem would be extremely hard.) Once this idea strikes, finding theactual relationship is easy. Since ω3 = 1, ω4 = ω, ω5 = ω2 we see that

E1 = ωE3 and E2 = ω2E3 (9)

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So the given equation simplifies to

(ω4n+3 + (ω2)4n+3 + 1)E4n+33 = 0 (10)

We also have ω2+ω+1 = 0. (This is a well known result and follows byfactorising ω3−1 and noting that ω 6= 1.) So E3 = −3+2ω−3−3ω =−6 − ω which is non-zero. Hence E4n+3

3 is also non-zero. So, only thefirst factor of the L.H.S. of (10) vanishes. Further, since ω3 = 1, ω3n

and (ω2)3n also equal 1 each. So the equation simplifies to

ωn + ω2n + 1 = 0 (11)

Powers of ω repeat in a cycle of 3. When n is a multiple of 3, theequation becomes 1+1+1 = 0 which is impossible. However, for othervalues of n, it reduces to ω + ω2 + 1 = 0 (for n = 1, 4, 7, . . .) or toω2 + ω + 1 = 0 (for n = 2, 5, 8, . . .) both of which are true. Hence nmust not be divisible by 3. In Column II, the possible values of n arethose other than 3.

Finally, in (D), the first part gives the equation

ab = 2(a+ b) (12)

We are also given that a, 5, q, b are in an A.P. This gives two moreequations, viz.

q + a = 10 (13)

and b+ 5 = 2q (14)

Eliminating q,

b = 2(10− a)− 5 = 15− 2a (15)

Putting this into (12) we get

a(15− 2a) = 2a+ 30− 4a (16)

which simplifies to a quadratic in a, viz.

2a2 − 17a+ 30 = 0 (17)

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whose roots are17± 7

4, i.e. 6 and 5/2. Corresponding values of q are 4

and 15/2. So |q− a| = 2 in the first case and |q− a| = 5 in the second.

(C) is the only interesting problem in the bunch. The others reduceto solving systems of two equations in two unknowns. There is nothingvery great either in formulating these equations or in solving them.These parts hardly belong to an advanced test.

Q.60 Column I

(A) In a triangle △XY Z, let a, b and c be thelengths of the sides opposite to the angles X, Yand Z respectively. If 2(a2 − b2) = c2 and

λ =sin(X − Y )

sinZ, then possible values of n for

which cos(nπλ) = 0 is (are)

(B) In a triangle △XY Z, let a, b and c be thelengths of the sides opposite to the angles X, Yand Z, respectively. If 1 + cos 2X − cos 2Y =

2 sinX sinY , then possible value(s) ofa

bis (are)

(C) In IR2, let√3 i+ j, i+

√3 j and βi+(1−β)j be

the position vectors of X, Y and Z with respectto the origin O, respectively. If the distance of Zfrom the bisector of the acute angle of OX with

OY is3√2, then possible value(s) of |β| is (are)

(D) Suppose that F (α) denotes the area of the regionbounded by x = 0, x = 2, y2 = 4x and y =|αx−1|+ |αx−2|+αx, where α ∈ {0, 1}. Thenthe value(s) of F (α) +

8

3

√2 when α = 0 and

α = 1 is (are)

Column II

(P) 1

(Q) 2

(R) 3

(S) 5

(T) 6

Answers and Comments: (A; P,R,S), (B; P), (C; P,Q), (D; S,T).

In this question, items (A) and (B) have a common setting. Sincea, b, c are proportional to the sines of their opposite angles, the equation

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2(a2 − b2) = c2 in (A) translates into

2(sin2X − sin2 Y ) = sin2 Z (1)

By a well-known identity the L.H.S. factorises and we get

2 sin(X − Y ) sin(X + Y ) = sin2 Z (2)

which further simplifies to

sin(X − Y )

sinZ=

1

2(3)

since in any triangle △XY Z, sin(X + Y ) = sin(π −Z) = sinZ. So we

get λ =1

2and the equation to be solved reduces to

cos(nπ/2) = 0 (4)

which is possible only when n is an odd integer. The odd integers inColumn II are 1, 3 and 5.

In (B), the condition given is in terms of the angles and we have

to find the ratioa

bwhich equals

sinX

sin Y. Let us first recast the given

condition in terms of sinX and sinY .

1 + cos 2X − 2 cos 2Y = 2 cos2X − 2(1− 2 sin2 Y )

= (2− 2 sin2X)− 2 + 4 sin2 Y

= 4 sin2 Y − 2 sin2X (5)

Hence the condition given translates as

2 sin2 Y − sin2X = sinX sinY (6)

Dividing by sin2 Y and callingsinX

sinYas λ (which is the ratio we are

interested in), we get a quadratic in λ, viz.

λ2 + λ− 2 = 0 (7)

whose roots are λ = 1 and −2. As λ, being the ratio of two sides cannotbe negative, we get λ = 1.

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In (C), the vectors are involved only superficially. The first partof the data is just another way of saying that the points X, Y, Z are(√3, 1), (1,

√3) and (β, 1 − β) respectively. Even without drawing a

diagram it is obvious that the pointsX and Y are symmetrically locatedw.r.t. the line y = x. Also OX and OY are inclined at angles 30◦ and60◦ respectively. So, the line y = x is the acute angle bisector 6 XOY .

We are given that the distance of Z from this line y − x = 0 is3√2.

This implies

|(1− β)− β|√1 + 1

=3√2

(8)

This means 2β − 1 = ±3 and hence β equals 2 or −1. So |β| equals 2or 1.

Finally, (D) consists of two separate problems, one for α = 0 andthe other for α = 1. The first one is easier because in this case y = 3and so, F (0) is the area bounded by x = 0, x = 2, y2 = 4x and y = 3.It is the shaded area OABCO in the figure below.

O

A

BC y = 3

(2, 0)

(0, 3)

x

y

By a direct calculation,

F (0) =∫ 2

03− 2

√x dx = 3x− 4

3x3/2

2

0= 6− 8

√2

3(9)

So, F (0) +8√2

3= 6 which tallies with (T) in Column II.

On the other hand, when α = 1,

y = |x− 1|+ |x− 2|+ x (10)

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For x ∈ [0, 2], |x− 2| = 2− x and so (10) simplifies a little to

y = |x− 1|+ 2 (11)

whose graph for 0 ≤ x ≤ 2 is the union of two straight line segments,one from C = (0, 3) to A = (1, 2) and the other from A = (1, 2) toB = (2, 3). So this time F (1) is the sum of the two shaded areasOACO and ADBA shown in the figure below.

O

BC y = 3

(2, 0)

(0, 3)

x

y

A

D

Again, by a direct calculation,

F (1) =∫ 1

03− x− 2

√x dx+

∫ 2

1x+ 1− 2

√x dx (12)

By a routine integration which we skip, F (1) comes out to be 5− 8√2

3.

So F (1) +8√2

3= 5.

Parts (A) to (C) are straightforward, but too elementary to beasked in an advanced test. The wording of (D) is clumsy and manycandidates might not understand the problem and might have skippedit. They are the clever ones, because those who do struggle success-fully to realise that the problem involves the calculation of two (inreality three) unrelated areas will pay a heavy price in terms of timeand the strong possibility of some numerical error. If the idea wasmerely to give a problem about identifying and finding the area of aplane region, the first half of the problem (where α = 0) would have

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served the purpose quite well. Adding one more part which hardlytests anything new (except the ability to draw the graph of the func-tion y = |x − 1| + |x − 2| + x) is nothing short of torture. It is suchsadistic problems which make success at the JEE a matter of adopt-ing a clever strategy, whose prime rule is to simply stay away from aproblem which is clumsily worded and utilise the time saved on routineproblems requiring mediocre intelligence.

PAPER 2

Contents

Section - 1 (One Integer Value Correct Type) 40

Section - 2 (One or More than One Correct Choice Type) 54

Section - 3 (Paragraph Type) 71

SECTION 1

One Integer Value Correct Type

This section contains eight questions. The answer to each question is aSINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.

Marking scheme : +4 If the bubble corresponding to the answer is dark-ened, 0 In all other cases

Q.41 Suppose that all the terms of an arithmetic progression (A.P.) are nat-ural numbers. If the ratio of the sum of the first seven terms to thesum of the first eleven terms is 6 : 11 and the seventh term lies between130 and 140, then the common difference of this A.P. is

Answer and Comments: 9. An A.P. is determined by two numbers,viz. its initial term, say a, and its common difference, often denotedby d. To determine them, we need two equations in a and d. Theproblem gives only one. In such cases some additional restrictions on

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the variables such as that they are integers or that they lie in somespecific intervals is needed to determine the unknowns. The presentproblem is of this type. (See Exercise (4.24) for an example wherethere are fewer equations than unknowns and still a unique solution.)

With the notations just introduced, the sum of the first n terms(often denoted by Sn) is

Sn = na + d(1 + 2 + . . .+ (n− 1)) = na +n(n− 1)

2d (1)

We are given that S7

S11= 6

11. Because of (1) this means

7a+ 21d

11a+ 55d=

6

11(2)

which yields 7a+ 21d = 6a+ 30d and hence

a = 9d (3)

This single equation cannot determine d (or a) uniquely. We now usethe second condition. The seventh term is a + 6d. So we are given

130 < a + 6d < 140 (4)

which by (2) becomes

130 < 15d < 140 (5)

But d is given to be integer. The only multiple of 15 between 130 and140 is 135. Setting 135 = 15d gives d = 9.

A simple, but thought provoking problem. The computationsneeded are minimal and can be done quickly. Such questions are idealfor multiple choice tests because the answer is not likely to come byguessing.

Q.42 The coefficient of x9 in the expansion of (1+ x)(1+ x2)(1+ x3) . . . (1+x100) is

Answer and Comments: 8. The problem is superficially algebraic.But in reality it is combinatorial. When the product, say P is fully

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expanded it will be a sum of 2100 terms, each being a product of 100terms of the form u1u2 . . . u100 where each ui has two possibilities , either1 or xi, for i = 1, 2, . . . , 100. We can combine these two possibilitiesby saying that ui = xni where the exponent ni is either 0 or i. So,u1u2 . . . u100 will equal un1+n2+...+n100 . Clearly only the positive termsin the exponent matter. So each occurrence of x9 will result from asequence of distinct positive integers in ascending order adding to 9.

Once this is understood, the problem reduces to finding the num-ber of ways to express 9 as a sum of distinct integers (arranged in anascending order), ranging from 1 to 100. As 9 is a small number, thesepossibilities can be counted by classifying according to the number ofterms in the summation. There is only one way to express 9 as a sumof a single integer, viz. 9 = 9. To find the number of ways to write 9as a sum of two distinct integers in ascending order, we go systemati-cally as 1 + 8, 2 + 7, 3 + 6 and 4 + 5. If there are three summands, say9 = a + b + c, with 0 < a < b < c, note that a + b is at least 3. Soc can be only 6, 5 or 4 since with c = 3, a < b < c implies a + b canbe at most 3. For c = 4, we have a + b = 5. But since a < b < c thiscan happen only a = 2, b = 3. For c = 5, a + b will equal 4 only fora = 1, b = 3 and finally for c = 6, the only possibility is a = 1, b = 2.

There is no need to go further because the sum of any four distinctpositive integers will be at least 10. So 9 can be expressed as a sum ofdistinct positive integers in an ascending order in 1 + 4 + 3 = 8 ways.

The problem could have been formulated as a combinatorialproblem. Suppose there are 100 boxes, numbered 1 to 100. Then theproblem asks for the number of ways to put nine identical balls intothese boxes, so that each box is either empty or contains as many ballsas its number.

In the present case, we have translated the algebraic probleminto a combinatorial one. It could have as well been done the otherway. The work done is essentially the same with either approach. Butthere are situations where converting a combinatorial problem into analgebraic one pays off. Consider, for example, the problem of countingselections with repetitions. So, let an,k be the number of ways to choosek objects from n types of objects with repetitions allowed freely. Thisis equivalent to placing k identical balls into n distinct boxes, there

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being no restriction as to how many balls can go into any box. There

is a tricky way to show that an,k equals

(

k + n− 1

k

)

by thinking of

each such placement as an arrangement of k balls and n− 1 inter-boxpartitions. (See Comment No. 5 of Chapter 1.)

The algebraic version of this problem can be constructed as follows.Each of the n boxes can hold at least 0 and at most k balls. So an,k isthe coefficient of xk in the expansion of (1 + x+ x2 + . . .+ xk)(1 + x+. . .+xk) . . . (1+x+ . . .+xk), there being n factors in all. Equivalently,an,k is the coefficient of xk in (1 + x+ x2 + . . .+ xk)n. This conversiondoes not help much by itself. But, instead of taking the polynomial1 + x+ x2 + . . .+ xk, let us take the entire infinite series 1 + x+ x2 +. . .+xk+xk+1+xk+2+ . . .. This may appear useless because the extraterms we are adding cannot contribute to any selection. (There is noway to write k as a sum of non-negative integers if one or more termsis greater than k.) But now the advantage is that the infinite series

1+x+x2+ . . .+xk+ . . . can be identified as the geometric series1

1− x.

Doing this for each of the factors, we see that an,k is the coefficient of xk

in(

1

1− x

)n

. This too, is not of much use by itself. But if rewrite this

expression as (1−x)−n and expand it using the binomial theorem wherethe exponent is the negative integer −n, we see that the coefficient of

xk is (−1)k(

−nk

)

= (−1)k(−n)(−n− 1) . . . (−n− k + 1)

k!which comes

out to ben(n + 1) . . . (n+ k − 1)

k!which is nothing but the binomial

coefficient

(

n+ k − 1

k

)

.

What makes this solution possible is the rich machinery of algebra,including power series. The combinatorial solution was elementary butrather tricky. The situation is analogous to the relationship betweenpure geometry and coordinate geometry. Pure geometry solutions areelegant but sometimes tricky. When coordinates are introduced thepowerful machinery of algebra makes the problem more amenable, ifsomewhat dull. (See, for example, the second proof of the concurrencyof the three altitudes of a triangle, given in Comment No. 3 of Chapter8.)

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Even without the powerful machinery of power series, the algebraicrecasting of a combinatorial problem in terms of a suitable polynomialis sometimes useful when the answer is less demanding than finding thecoefficient of a specific power of x. Consider, for example, the problemof placing n identical balls into, say, r distinct boxes with the restrictionthat each box can contain at most k balls. If we denote this number byan, then as we just saw, an is the coefficient of xn in the polynomial,say f(x) = (1 + x + x2 + . . . + xk)r and there is no easy of findingit except for some select values of n. For example, we can say thatark = 1 and an = 0 for n > rk because f(x) is a polynomial of degreerk and leading coefficient 1. But this is something obvious by commonsense anyway. Similarly, we can tell a0 = 1 and a1 = r equally easilywith or without the help of the polynomial f(x).

But, suppose that our problem is not to calculate an for a particularvalue of n, but to find the entire sum, say S1 = a0+a1+a2+. . .+ark. We

could have written S1 ostensibly as an infinite sum S1 =∞∑

n=0

an because

all except finitely many terms of this series vanish. Since f(x) = a0 +a1x + a2x

2 + . . . arkxrk, we have S1 = f(1). From the factorisation of

f(x) as (1 + x + x2 + . . . + xk)r we immediately get S1 = (k + 1)r.Combinatorially, S1 is the number of all possible ways of putting anynumbers of identical balls into r distinct boxes so that each box containsat most k balls. Here, too, a direct combinatorial argument is easybecause, for each of the r boxes, there are k+1 possibilities dependingupon how many balls go into it. So again, this example does not quitebring home the power of algebraic codification.

To do so, consider S2 = a0+a2++a4+ . . .+a2m+ . . ., i.e. the sumof the coefficients of all even degree terms in f(x). Combinatorially, S2

is the number of all possible ways to put any numbers of balls into theboxes as before with the additional restriction that the total numberof balls is even. (It is not required that only even numbers of ballsgo into the individual boxes. The restriction of evenness is only onthe total number of balls.) This time, the combinatorial count is notas immediate as for S1. But the algebraic one is simple. If we addf(1) and f(−1), the even powers add up while the odd ones cancel

out. In other words, S2 = a0 + a2 + a4 + . . . =f(1) + f(−1)

2. This is

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true for any polynomial. In our example, f(1) = S1 = (k + 1)r whilef(−1) = (1− 1 + 1− . . .+ (−1)k+1)r which equals 0 or 1 according as

k is odd or even. Hence S2 =(k + 1)r

2or

(k + 1)r + 1

2depending upon

whether k is odd or even.

The essence of the calculations of S1 and S2 was, respectively, thatfor every non-negative integer n, 1n = 1 for all n while 1n+(−1)n equals2 or 0 depending upon whether n is a multiple of 2 or not. Note that1 and −1 are the square roots of 1. In the solution to Part (C) of Q.59of Paper 1, we proved that 1n + ωn + (ω2)n equals 3 or 0 dependingupon whether n is a multiple of 3 or not. So, for any polynomialf(x) = a0+a1x+a2x

2+. . ., the sum, say S3 = a0+a3+a6+. . .+a3m+. . .

would equalf(1) + f(ω) + f(ω2)

3.

The picture is now quite clear. Even if we may not be able to identifythe coefficients an individually for all n, we can calculate the sums of thecoefficients whose suffixes are in an A.P. by adding the values of f(x) atthe d complex d-th roots of unity where d is the common difference ofthis progression. Sometimes such sums have some appeal and providean unexpected solution to a problem where a direct combinatorial countmay be laborious. Consider, for example, the problem of finding thenumber of 6-digit numbers whose digits add to a number of the form4p + 1. This is equivalent to counting the number of ways to putidentical balls into 6 boxes the first of which can hold 1 to 9 balls andthe remaining 0 to 9 balls each, so that the total number of balls is ofthe form 4p+1. Then the number we want is the sum a1+a5+a9+ . . .for the polynomial f(x) = (x+ x2 + . . . + x9)(1 + x+ x2 + . . .+ x9)5.Equivalently, this is the sum b0+b4+b8+ . . . for the polynomial g(x) =b0+b1x+b2x

2+. . . = (1+x+x2+. . .+x8)(1+x+x2+. . .+x9)5. By our

work, the count isg(1) + g(i) + g(−1) + g(−i)

4. By direct substitutions

of the powers of these complex numbers into the factors of g(x), we haveg(1) = 900000, g(−1) = 0, g(i) = (1+i)5 = −4−4i and g(−i) = −4+4i.

So the desired number is9000000− 4− 4

4= 249998.

Q.43 Suppose that the foci of the ellipsex2

9+

y2

5= 1 are (f1, 0) and (f2, 0)

where f1 > 0 and f2 < 0. Let P1 and P2 be two parabolas with a com-

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mon vertex at (0, 0) and with foci at (f1, 0) and (2f2, 0), respectively.Let T1 be a tangent to P1 which passes through (2f2, 0) and T2 be atangent to P2 which passes through (f1, 0). If m1 is the slope of T1 and

m2 is the slope of T2, then the value of

(

1

m21

+m22

)

is

Answer and Comments: 4. The problem is about the tangents tothe two parabolas. The role of the ellipse is only to specify the foci of

the two parabolas. The eccentricity e of the given ellipse is

1− 5

9=

2

3and so the foci are at (±3e, 0) i.e. f1 = 2 and f2 = −2.

Since both the parabolas P1 and P2 have their vertices at (0, 0), andtheir foci are at (2, 0) and (−4, 0) respectively, their equations are

y2 = 8x (1)

y2 = −16x (2)

respectively. We are given that the line y = m1(x + 4) touches theparabola y2 = 8x. So it intersects the parabola in two coincidentpoints. Therefore, the quadratic m1(x+ 4)2 − 8x = 0 has discriminant0. On simplification, this gives (8m2

1 − 8)2 = 64m41 and hence

m21 =

1

2(3)

(Incidentally, this shows that there are two lines through (−4, 0) thattouch the parabola P1. The problem does not specify which of the twois to be chosen. But the answer does not depend on the choice since itinvolves only m2

1.)

By an entirely analogous computation, the slope m2 of T2 satisfies

m22 = 2 (4)

Hence the expression1

m21

+m22 equals 4.

Those who know the equation of a tangent to a parabola in terms ofits slope can shorten the work by observing that a tangent to y2 = 8xhaving slope m1 has its equation of the form

y = m1x+2

m1(5)

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and get (3) from the fact that this tangent passes through (−4, 0). Sim-ilarly an alternate derivation of (4) is possible. We have intentionallygiven a derivation from first principles, based on the concept of a tan-gent as a line which intersects a curve in two coincident points, to showthat even if you do not remember a whole lot of formulas, things canoften be salvaged if you go by the basic principles.

A very straightforward problem about identifying the tangents toa parabola passing through a given point. Perhaps the paper-settersrealised that the problem is too straightforward and hence gave it atwist by first making the candidates identify the points through whichthe tangents are to pass. In the old days, instead of making the problemnumerical, it would have been asked to show that if two parabolas havetheir vertices at the centre of an ellipse and their foci at the foci of thatellipse then the tangents to either of them passing through the focusof the other are mutually orthogonal. Such geometric results expressedsolely in words, have their own beauty.

Q.44 Let m and n be two positive integers greater than 1. If

limα→0

(

ecos(αn) − e

αm

)

= −e

2then the value of

m

nis

Answer and Comments: 2. As α → 0 both the numerator andthe denominator tend to 0. So the limit, say L, in the question if

of the indeterminate form0

0. It is, therefore, tempting to apply the

L’Hopital’s rule. But if m > 1 this rule will have to be applied againand again. Let us temporarily assume that m = 1 and allow n to takenon-integral values as well. Define the function f(α) = ecos(α

n) − e.Then L’Hopital’s rule implies that

L = limα→0

f ′(α)

1(1)

provided this limit exists.

Now, by a direct computation, for α 6= 0 we have,

f ′(α) = − sin(αn)ecos(αn)nαn−1 (2)

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It is tempting to think that this is 0 at α = 0 because of the first factor,viz., sin(αn). This is valid for n > 0. But we have to be careful aboutthe last factor, viz. αn−1. If n < 1, then this factor tends to ∞ asα → 0+. The middle factor ecos(α

n) creates no problem as it tends to 1when α → 0+ as long as n > 0. Nor does n which is a fixed number.

To see the result of the battle between the first and the last factorwhen 0 < n < 1, let us rewrite (2) as

f ′(α) = −sin(αn)

αnnecos(α

n)α2n−1 (3)

Now the first factor tends to 1 as α → 0. So the fate of the limitdepends on the last term, viz. α2n−1. If n > 1/2, then it will makef ′(α) tend to 0. But if 0 < n < 1/2, then it will tend to ∞ as α → 0+.

We conclude that if m = 1, then it is only for n = 1/2 that the

limit in question will exist. When it does, its value will be −e

2. In the

problem, of course, n is given to be an integer. But the argument above

suggests that if at all the ratiom

nhas some fixed value, it is probably

2.

An intelligent and a smart gambler will leave the solution at thispoint. But a scrupulous person can make the argument valid as follows.

Let k =m

n. Call αm as β. Then n = m/k and therefore, αn = β1/k.

Moreover β → 0 as α → 0. So, by a change of variable,

L = limβ→0

ecos(β1/k) − e

β(4)

We are now back to the old game with α replaced by β and n replaced by1/k. By the argument above, L will equal a finite non-zero value (whichwill be necessarily −e/2 ) only when 1/k = 1/2. This is equivalent tosaying that k = 2.

A mature person will approach the problem by considering thecomparable orders of magnitude. For example, when u → 0, sin u is of

the same order as u becausesin u

utends to a finite non-zero limit as

u → 0. Similarly, cosu−1 is of the order of u2 becausecosu− 1

u2tends

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to the non-zero limit −1

2. Similarly, eu − 1 is of the order of u. Now

factor out e from the numerator of the given expression and consider

L′ = limα→0

ecos(αn)−1 − 1

αm(5)

Then clearly, L = eL′ provided L′ exists. By what we just said, thenumerator is of the order of cos(αn)− 1, which, in turn, is of the orderof α2n. The denominator, on the other hand, is of the order of αm. Sothe ratio will tend to a finite non-zero limit only when the numeratorand the denominator have the same orders and that happens preciselywhen m = 2n.

If need arises, this reasoning based on comparable orders of mag-nitude can be made precise by dividing and multiplying the expressionon the R.H.S. of (5) by cos(αn)− 1 and then again by α2n. But that isessentially a clerical work.

This is a very good, thought oriented problem. But since it isa multiple choice question where no reason has to be given, a smartcandidate may get the correct answer by the sneaky path, i.e. byassuming m = 1.

Q.45 If α =∫ 1

0e9x+3 tan−1 x

(

12 + 9x2

x2 + 1

)

dx where tan−1 x takes only princi-

pal values, then the value of loge(1 + α)− 3π

4is

Answer and Comments: 9. This is plainly a question about evalu-ating a definite integral. Since the answer is to be an integer between0 and 9, the last part operates as a hint to the value, viz. 1 + α is ofthe form ek+3π/4 for some integer k. Although this does not help inevaluating the integral, it serves to alert against numerical errors. Inthe conventional form the question would have merely asked the valueof the integral.

The integral itself is easy if we use the substitution

u = 9x+ 3 tan−1 x (1)

which yields

du

dx= 9 +

3

1 + x2=

12 + 9x2

x2 + 1(2)

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Hence we have

α =∫ 9+3π/4

0eu du

= e9+3π/4 − 1 (3)

So ln(1 + α) = 9− 3π/4.

Too simple a problem once the correct substitution strikes. Andthere is little choice about the correct substitution. The paper-setters

at least could have given the integrand as (e9)x(

12 + 9x2

x2 + 1

)

(etan−1 x)3,

when in many other questions (for example, Q. 43 above) they havegiven the data in an unnecessarily twisted form.

Q.46 Let f : IR −→ IR be a continuous, odd function which vanishes ex-

actly at one point and f(1) =1

2. Suppose F (x) =

∫ x

−1f(t) dt for

all x ∈ [−1, 2] and G(x) =∫ x

−1t|f(f(t))| dt for all x ∈ [−1, 2]. If

limx→1

F (x)

G(x)=

1

14, then the value of f(1/2) is

Answer and Analysis: 7. The problem is a combination of L’Hopital’srule and the second Fundamental Theorem of Calculus (FTC). To ap-

ply the former to the limit limx→1

F (x)

G(x)we must first ensure that it is

of the0

0form. As both F and G are continuous (being functions de-

fined by integrals), this amounts to verifying that F (1) and G(1) bothvanish. This is not given. In fact, all we know is

F (1) =∫ 1

−1f(t) dt (1)

and G(1) = =∫ 1

−1t|f(f(t))| dt (2)

Since f is given to be an odd function, the integral on the R.H.S. of(1) vanishes. For the same to happen for (2), we must ensure thatx|f(f(x))| is an odd function of x. For this we note that f(f(x)) is an

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odd function of x, being the composite of two odd functions. Henceits absolute value |f(f(x))| is an even function. Finally, x is an oddfunction of x and so the product x|f(f(x))| is an odd function of x,being the product of an even function and an odd function. So, theintegral on the R.H.S. of (2) vanishes as well.

We are now justified in applying the L’Hopital’s rule. By the secondFTC, F ′(x) = f(x) and G′(x) = x|f(f(x))| for all x ∈ [−1/2, 2]. Hence

limx→0

F (x)

G(x)= lim

x→1

F ′(x)

G′(x)=

F ′(1)

G′(1)=

f(1)

1× |f(f(1))|

=1/2

|f(f(1))|

=1/2

|f(1/2)| (3)

We are given that this limit is1

14. So, we now get

|f(1/2)| = 14

2= 7 (4)

We are still not quite through, because the problem asks for f(1/2)and we only know that f(1/2) = ±7. A sneaky reasoning would bethat since the correct answer has to be an integer from 0 to 9, −7 isexcluded. But, for an honest answer, we must show that f(1/2) > 0.For this we apply the Intermediate Value Property (IVP). f(x) is givento be continuous everywhere. Also f(1) = 1/2 > 0. So, if f(1/2) werenegative, then by the IVP, f would vanish at least once in the interval(1/2, 1). But, being an odd function, f already vanishes at 0. As weare given that f vanishes at exactly one point, we get a contradiction.

So, at long last we have fully justified our answer, viz. f(1/2) = 7.To arrive at it legitimately, we had to give a lot of reasoning. First toensure the applicability of the L’Hopital’s rule. Many students applythis rule without ensuring this. Secondly, many students might notbother to justify that f(1/2) > 0. And there is no way to tell ifthey have really done the reasoning. So, this question rewards thesloppy students and punishes the scrupulous ones. In the conventionalexaminations, the two could be differentiated because reasonings wouldhave to be given.

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The essential idea of the question is the combination of L’Hopital’srule and the FTC. Properties of odd functions have also been testedalong the way. But being an MCQ, all this fine creativity on the partof the paper-setters is wasted. (In fact, the possibility that a studentmight pick 7 as the correct answer simply because of the number 14appearing in the problem cannot be entirely discounted. A safe gamble,as there is no negative credit for a wrong answer.)

Q.47 Suppose that ~p, ~q and ~r are three non-coplanar vectors in IR3 . Let thecomponents of a vector ~s along ~p, ~q and ~r be 4, 3 and 5, respectively.If the components of this vector ~s along (−~p + ~q + ~r), (~p− ~q + ~r) and(−~p− ~q + ~r) are x, y and z respectively, then 2x+ y + z is

Answer and Comments: 9.. An extremely straightforward problemabout resolving a vector along three linearly independent vectors. Weare given

~s = 4~p+ 3~q + 5~r (1)

But we are also given that

~s = x(−~p + ~q + ~r) + y(~p− ~q + ~r) + z(−~p − ~q + ~r) (2)

= (−x+ y − z)~p + (x− y − z)~q + (x+ y + z)~r (3)

As ~p, ~q, ~r are non-coplanar and hence linearly independent, the coeffi-cients in the two resolutions must match. This gives a system of threeequations in the three unknowns, viz.

x+ y − z = 4, x− y − z = 3, and x+ y + z = 5 (4)

Solving this system is child’s play and gives x = 4, y = 9/2, z = −7/2.A straight substitution gives 2x+ y + z = 8 + 1 = 9.

The word ‘component’ in the statement of the problem is likelyto confuse some candidates. Sometimes, by a component of a vector ~salong a vector ~p one understands, ‘component along the direction of ~p’,i.e. the component along a unit vector in the direction of ~p. This usageis common in expressions like ‘the tangential and normal componentsof acceleration’ used in physics. With this meaning, instead of (1) wewould have

~s =4~p

||~p|| +3~q

||~q|| +5~r

||~r|| (5)

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Similarly, Equation (2) will have to be replaced by a more complicatedequation. This would make the problem not only more complicated,but impossible to solve since the lengths of the vectors ~p, ~q and ~r andof their various linear combinations are not given.

But the paper-setters should not be blamed for this. The interpreta-tion of ‘component’ as they have intended is fairly standard. With thealternate interpretation, the components are generally taken to be vec-tors along the respective directions. Nevertheless, the JEE authoritieshave decided to award four points to every candidate for this question.

For those who interpret the word ‘component’ as we have, thequestion is a cakewalk. It is difficult to believe that such a simplequestion is asked in an advanced test. It may be argued that this isan atonement for some other difficult questions. But when these twotypes of questions are combined, the selection takes place mostly onthe basis of such mediocre questions. That defeats the very purpose ofselection. And, for this, the paper-setters do deserve some criticism.

Q.48 For any integer k, let αk = cos

(

7

)

+ i sin

(

7

)

where i =√−1.

Then the value of the expression

12∑

k=1|αk+1 − αk|

3∑

k=1|α4k−1 − α4k−2|

is

Answer and Comments: 4. In the Argand diagram, the complexnumbers lie at the vertices of a regular polygon with 14 sides inscribedin the unit circle {z : |z| = 1}. (This happens because all the givencomplex numbers are 14-th roots of unity.) For every k the verticesαk+1 and αk are consecutive. So, |αk+1 − αk| equals the length, sayd of each side of the heptagon. So the numerator of the expression is12d. In the denominator also we are summing the distances betweenconsecutive vertices, except that in the denominator there are only 3

terms. Hence the denominator is 3d. So the given ratio equals12d

3d= 4.

A good problem based on the geometric representation of complexnumbers. Note that the number 7 has little role in the problem. Itcould have been any positive integer r. Then the vertices would repeat

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in a cycle of length 2r and even if the summation extended to morethan 2r terms, the calculations would be the same.

The problem is reminiscent of a 1994 JEE problem which was a sortof a converse to the present problem. There we were given a regularn-gon with vertices A1, A− 2, . . . , An and the relationship between thelengths of its sides and some of its diagonals. Specifically, it was giventhat

1

A1A2

=1

A1A3

+1

A1A4

(1)

and the problem asked to determine n. This can be reduced to solvinga trigonometric equation. (Ironically, the answer to that problem wasalso 7. See Comment No. 1 of Chapter 10.) However, the regularityof the n-gon permits us to conclude from (1) a similar relationship forany four consecutive vertices of the polygon, that is

1

AiAi+1=

1

AiAi+2+

1

AiAi+3(2)

for every i, with the understanding that An+1 = A1 etc. This fact, alongwith an application of what is called Ptolemy’s theorem about cyclicquadrilaterals, gave an unexpected solution to the problem. (See againComment No. 2 of Chapter 10.) A solution using complex numbers isalso possible by taking the vertices to lie at the complex 2n-th roots ofunity. (See Comment No. 3 of the same chapter.) It is possible thatthis solution has spurred the present problem. Of course, the presentproblem is much simpler than the 1994 problem. How one wishes torevive the golden days when such nice problems could be asked as fulllength questions! All we have today is their weak imitations.

SECTION 2

This section contains EIGHT questions. Each questions has FOUR op-tions (A), (B), (C) and (D). ONE OR MORE THAN ONE of these fouroption(s) is(are) correct.

Marking scheme : +4 If fully correct, 0 if not attempted and −2 In allother cases.

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Q.49 Let f, g : [−1, 2] −→ IR be continuous functions which are twice differ-entiable in the interval (−1, 2). Let the values of f and g at the points−1, 0 and 2 be given by the following table:

x = −1 x = 0 x = 2f(x) 3 6 0g(x) 0 1 -1

In each of the intervals (−1, 0) and (0, 2) the function (f − 3g)′′ nevervanishes. Then the correct statement(s) is (are)

(A) f ′(x)− 3g′(x) = 0 has exactly three solutions in (−1, 0) ∪ (0, 2)

(B) f ′(x)− 3g′(x) = 0 has exactly one solution in (−1, 0)

(C) f ′(x)− 3g′(x) = 0 has exactly one solution in (0, 2)

(D) f ′(x)−3g′(x) = 0 has exactly two solutions in (−1, 0) and exactlytwo solutions in (0, 2)

Answer and Comments: (B), (C). All the four statements are aboutthe zeros of h′(x) where h(x) is defined by

h(x) = f(x)− 3g(x) (1)

From the table, we get

h(−1) = 3, h(0) = 3 and h(2) = 3 (2)

Rolle’s theorem implies that h′(x) vanishes at least once in each of theintervals (0, 1) and (1, 2). If it had more zeros in either of these twointervals, then by Rolle’s theorem again (applied to h′(x) this time),h′′(x) would vanish at least once. But we are given that this is not so.Hence (B) and (C) are correct. That automatically precludes (A) and(D).

A simple problem based on Rolle’s theorem. One fails to see whatwas the idea of introducing the two functions f(x) and g(x) when theentire problem is about the function h(x) = f(x) − 3g(x). But then,

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instead of the table, we would have to be given (2) and the paper-settersprobably thought that that would make the solution too obvious. So,they worded the problem in terms of two functions. But nothing morethan elementary arithmetic is needed to convert the table to (2). Thequestion would have been interesting if some other pieces of informationwere given from which (2) would follow, but not in such a direct manner.(Contrast this with Q.46 above, where for an honest solution so manyproperties of even and odd functions, and also the FTC had to be used.)Since both the questions have equal marks, between two competingcandidates with severely limited time, the mediocre one who opts forthe present question will easily beat his possibly more intelligent andscrupulous rival who might not have time left for this problem.

Q.50 Let f(x) = 7 tan8 x+7 tan6 x− 3 tan4 x− 3 tan2 x for x ∈ (−π/2, π/2).Then the correct expression(s) is (are)

(A)∫ π/4

0xf(x) dx =

1

12(B)

∫ π/4

0f(x) dx = 0

(C)∫ π/4

0xf(x) dx =

1

6(D)

∫ π/4

0f(x) dx = 1

Answer and Comments: (A) and (B). The paper-setters have care-fully avoided the end points ±π/2 from the domain because the tan-gent function is undefined at these points. Anyway that does not mat-ter because in the problem the interval of integration is [0, π/4] in allstatements. And there is no difficulty at the end points here. (If theintegrand tends to ±∞ at either end points, the integral becomes whatis called an improper integral. Occasionally, improper integrals areasked in the JEE. See Comment No. 16 of Chapter 18 and Exercise(18.16). But they are not a part of the syllabus.)

Now, coming to the problem itself, the identity

tan2 x+ 1 = sec2 x (1)

simplifies the function to

f(x) = 7 tan6 x sec2 x− 3 tan2 x sec2 x (2)

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The substitution

u = tanx (3)

transforms the integrals in (B) and (D) to

∫ π/4

0f(x)dx =

∫ 1

07u6 − 3u2 du

= u7 − u3∣

1

0= 1− 1 = 0 (4)

The calculation of the integral in (A) and (C) is less direct. The samesubstitution gives

∫ π

0xf(x)dx =

∫ 1

0tan−1 u(7u6 − 3u2) du (5)

As we already know an antiderivative of the second factor of the inte-grand, we resort to integration by parts and get

∫ π/4

0xf(x) dx = tan−1 u(u7 − u3)

1

0−∫ 1

0

1

u2 + 1(u7 − u3) du (6)

= 0−∫ 1

0u3(u2 − 1) du

=∫ 1

0u3 − u5 du

=1

4− 1

6=

1

12(7)

The problem is a godsend for the huge mediocrity that banks onproblems asking a direct evaluation of some integral (instead of askingsome areas which first have to be written as integrals). The identity(1), the substitution (3) and the factorisation of the last integrand in(6) are so obvious, that a candidate who has cleared JEE Main can besafely presumed to know them. So this problems tests no new ability.It hardly belongs to an advanced test.

Q.51 Let f ′(x) =192x3

2 + sin4 πxfor all x ∈ IR with f(

1

2) = 0.

If m ≤∫ 1

1/2f(x) dx ≤ M , then the possible values of m and M are

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(A) m = 13, M = 24 (B) m =1

4, M =

1

2(C) m = −11, M = 0 (D) m = 1, M = 12

Answer and Comments: (D). We are not given f(x) directly. In-stead we are given f ′(x) and f(1/2). In theory this determines f(x)uniquely. In fact

f(x) =∫ x

1/2f ′(t) dt =

∫ x

1/2

192t3

2 + sin4 πtdt (1)

But this is of little help because we are not in a position to evaluatethe last integral in a closed form. Fortunately, the problem does not

ask for the exact value of the integral∫ 1

1/2f(x) dx, but only for some

estimates, i.e. some lower and upper bound on it. And these can beobtained from some estimates on the integrand f(x), which, in turn,can be obtained from properties of f ′(x) for x ∈ [1/2, 1]. Note thatf ′(x) is a ratio of two positive quantities. The numerator increasesstrictly on the interval [1/2, 1] (in fact it increases on the entire realline). But the denominator decreases on [1/2, 1] because of the termsin4 πx since the sine function decreases on the interval [π/2, π]. So the

maximum of f ′(x) on [1/2, 1] is f ′(1) =192

2 + 0= 96 while its minimum

on [1/2, 1] is f ′(1/2) =24

2 + 1= 8.

Thus we have proved that

8 ≤ f ′(x) ≤ 96 (2)

for all x ∈ [1/2, 1]. From this, we now derive estimates for f(x) us-ing the Fundamental Theorem of Calculus (FTC), and the fact thatf(1/2) = 0. We replace x by t in (2) and integrate over [1/2, x] to get

∫ x

1/28 dt ≤

∫ x

1/2f ′(t) dt ≤

∫ x

1/296 dt (3)

BY FTC, the middle integral is simply f(x)−f(1/2) = f(x), while thefirst and the last integrals are 8(x− 1/2) and 96(x− 1/2) respectively.So we have

8x− 4 ≤ f(x) ≤ 96x− 48 (4)

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for all x ∈ [1/2, 1]. Integrating again,∫ 1

1/28x− 4 dx ≤

∫ 1

1/2f(x) dx ≤

∫ 1

1/296x− 48 dx (5)

By an easy computation, the first integral equals (4x2 − 4x)∣

1

1/2= 1

while the last integral equals (48x2 − 48x)∣

1

1/2= 24 − 12 = 12. These

are exactly the lower and upper bounds on the given integral as givenin (D).

This is a good problem which tests the ability to estimate when theexact evaluation is not possible. Apart from some elementary knowl-edge of trigonometric functions, the entire solution is based on the factthat inequalities are preserved by integrals, that is to say, if f(x) ≤ g(x)

for all x ∈ [a, b], then∫ b

af(x) dx ≤

∫ b

ag(x) dx. It is precisely problems

of this kind and not those of the type of the last problem that aresuitable for an advanced test.

Q.52 Let S be the set of all non-zero real numbers α such that the quadraticequation αx2−x+α = 0 has two distinct real roots x1 and x2 satisfyingthe inequality |x1 − x2| < 1. Which of the following intervals is (are) asubset(s) of S?

(A) (−1/2,−1/√5) (B) (−1/

√5, 0) (C) (0, 1/

√5) (D) (1/

√5, 1/2)

Answer and Comments: (A), (D). For the equation to be quadratic,its leading coefficient α must be non-zero. When this is the case the

two roots are1±

√1− 4α2

2α. For the roots to be real and distinct, we

must have

4α2 < 1 (1)

Also the difference of the roots x1 and x2 is

|x1 − x2| =√1− 4α2

|α| (2)

So the second condition about the roots reduces to

1− 4α2 < α2 (3)

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(1) is satisfied if and only if α ∈ (−1/2, 1/2) while (3) holds if and onlyif α ∈ (−∞,−1/

√5) ∪ α ∈ (1/

√5,∞). The intersection of these two

sets is (−1/2,−1√5) ∪ (1

√5, 1/2).

A very simple problem about the roots of a quadratic.

Q.53 If α = 3 sin−1( 611) and β = 3 cos−1(4

9), where the inverse trigonometric

functions take only the principal values, then the correct option(s) is(are)

(A) cos β > 0 (B) sin β < 0 (C) cos(α + β) > 0 (D) cosα < 0

Answer and Comments: (B), (C), (D). Denote sin−1(6

11) and cos−1(

4

9)

by θ and φ respectively. Then α = 3θ and β = 3φ. We are given

sin θ =6

11(1)

and cosφ =4

9(2)

A straightforward approach is to first calculate cos β, sin β etc. from(1) and (2) using the formulas for the sines and cosines of triples ofangles. But the calculation of sin β = sin 3φ would require the value

of sin φ. It can be calculated from cos φ =4

9. But that will involve a

radical and hence the calculation will be complicated. The calculationsfor (C) will be even more complicated.

So we look for an easier alternative. If we look carefully, we areinterested only in the signs of the various expressions and not so muchin their exact values. And this can be done by locating the quadrantsin which the angles lie. We already know that both θ and φ lie in thefirst quadrant, i.e.

0 < θ <π

2and 0 < φ <

π

2(3)

But this information is inadequate to locate the quadrants in which3θ, 3φ and 3(θ + φ) lie. So we need sharper inequalities for θ and φthan (3). How sharp should they be? If θ lies in an interval [a, b],then 3θ lies in [3a, 3b] whose length is thrice that of the first interval.Since the angular measure of each quadrant is π/2, to determine the

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quadrants in which α, β lie, we should trap θ and φ in intervals of lengthπ/6 or less.

Let us first tackle φ. We are given that cosφ =4

9which is only

slightly less than1

2= cos(

π

3). As the cosine function is decreasing in

the first quadrant, we have

φ >π

3(4)

Thus we have found a sharper lower bound on φ than in (3). To get

an upper bound, we need an angle whose cosine is slightly less than4

9.

But here we need not be very choosy. Since we want to trap φ in aninterval of length π/6, in view of (4), the upper bound π/2 which wealready know from (3) will do. So we now have

π

3< φ <

π

2(5)

The determination of a narrower interval containing θ is similar.

We are given that sin θ =6

11which is slightly bigger than

1

2. Since

sin(π/6) =1

2, we get π/6 as a lower bound on θ. As for an upper

bound, we try π/4 knowing that sin(π/4) =1√2. And, indeed, it works

because6

11does happen to be less than

1√2as we see by squaring both

the numbers. So we get

π

6< θ <

π

4(6)

We now have all the ammunition to fire the four shots. Since β = 3φ,because of (5) β ∈ (π, 3π/2). So β lies in the third quadrant whereboth the sines and cosines are negative. This shows that (A) is falsewhile (B) is true. In a similar vein, because of (6), α lies in (π/2, 3π/4)which is a part of the second quadrant. So (D) is true. For (C), weneed both (5) and (6). Multiplying them by 3 and adding, we get

2< α + β <

4(7)

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The interval (3π/2, 9π/4) cuts across two quadrants, specifically, thefirst and the fourth quadrant. But since the cosine function is positivein both these quadrants, we see that (C) is true.

Like Q, 41 in Paper I, this question is also an excellent example oftesting the ability to quickly discard the most obvious approach. Thepaper-setters have given just the right degree of hint by specifying thatall the inverse trigonometric functions are to be taken to have theirprincipal values.

Q.54 Let E1 and E2 be two ellipses whose centers are at the origin. The majoraxes of E1 and E2 lie along the x-axis and the y-axis, respectively. LetS be the circle x2 + (y − 1)2 = 2. The straight line x+ y = 3 touchesthe curves S,E1 and E2 at P,Q and R, respectively. Suppose that

PQ = PR =2√2

3. If e1 and e2 are the eccentricities of E1 and E2,

respectively, then the correct expression(s) is (are)

(A) e21 + e22 =43

40(B) e1e2 =

√7

2√10

(C) |e21 − e22| =5

8(D) e1e2 =

√3

4

Answer and Comments: (A), (B). A highly computational problemin coordinate geometry. The equations of the ellipses E1 and E2 arenot given. We are only given that their major axes lie along the x- andthe y-axes respectively. Hence the equation of E1 is of the form

x2

a2+

y2

b2= 1 (1)

where a > b > 0. Similarly, the equation of E2 is of the form

x2

c2+

y2

d2= 1 (2)

where 0 < c < d.

Let us first determine the points of contacts of the line (say L), x+y = 3with S,E1 and E2. Putting

y = 3− x (3)

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into the equation of S gives x2 + (2 − x)2 = 2 or 2x2 − 4x + 2 = 0,which has x = 1 as its only root, confirming that L is a tangent to S.From x = 1, y comes out to be 2 by (3). The point of contact, P is

P = (1, 2) (4)

The points Q and R will have to be expressed in terms of the unknownsa, b, c, d.

Putting y = 3− x into (1) gives a quadratic

(a2 + b2)x2 − 6a2x+ a2(9− b2) = 0 (5)

For tangency, this quadratic has discriminant 0 which simplifies to

a2 + b2 = 9 (6)

Also the quadratic has only one root, viz.3a2

a2 + b2=

a2

3. Hence the

point of contact comes out to be

Q = (a2

3,9− a2

3) (7)

As we are given PQ =2√2

3, we get

(a2

3− 1)2 + (

9− a2

3− 2)2 =

8

9(8)

which simplifies to a2 − 3 = ±2. So a2 = 5 or 1 and correspondingly,from (6), b2 = 4 or 8. But we are given that a > b. So we havea2 = 5, b2 = 4. Since the eccentricity e1 of the first ellipse is given by

b2 = a2(1− e21) (9)

we get

e1 =

1− 4

5=

1√5

(10)

The calculation of e2 is entirely similar. We merely replace a, b by c, drespectively. Since PQ = PR, (8) also holds with a replaced by c. So

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once again we get c2 = 5 or 1 and correspondingly, d2 = 4 or 8. But thistime we are assuming d > c. So we have to take the second possibility,viz. c = 1 and d = 8. Then e2 is given by

e2 =

1− c2

d2=

1− 1

8=

√7

2√2

(11)

Now that we know both e1 and e2, by a direct calculation we get

e1e2 =

√7

2√10

, e21 + e22 =1

5+

7

8=

43

40, |e21 − e22| = |1

5− 7

8| = 27

40

The point P could as well have been given directtly. After determin-ing it, Q and R can be determined by solving x+ y = 3 simultaneouslywirh (x − 1)2 + (y − 2)2 = 8

9. They come out to be (5/3, 4/3) and

(1/3, 8/3) although we do not know which one of them is Q. There isa short cut (pointed out by Kaustuv Lahiri) to determine the eccen-tricty because it depends only on the ratio of the axes. Take E1. Theequation of the tangent to it at a point (x0, y0) is

xx0

a2+

yy0b2

= 1 (12)

Comparing this with x + y = 3 gives y0x0

= b2

a2. The first ratio is 4

5

or 8 depending on whether (x0, y0) is Q or R. Since b < a, the firstpossibility holds. From b2

a2= 4

5, e1, the eccentricity of E1 comes out to

be as in (10). By a similar reasoning, for E2,d2

c2= 8 and e2 comes out

as in (11).

The symmetry of the data allows us to apply most of the workfor the first ellipse to the second ellipse. The only difference comes inthe last step because of the interchange of the directions of the majorand the minor axes. In this problem, there does not seem to be anyway to directly compute, say e1e2, without first identifying e1 and e2individually (as there would have been if we could write a quadraticwhose roots are e1 and e2).

Q.55 Consider the hyperbola H : x2 − y2 = 1 and a circle S with centreN(x2, 0). Suppose H and S touch each other at a point P (x1, y1) withx1 > 0, y1 > 0. The common tangent to H and S at P intersects the

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x-axis at a point M . If (l, m) is the centroid of the triangle △PMN ,then the correct expression(s) is (are)

(A)dl

dx1

= 1− 1

3x21

for x1 > 1 (B)dm

dx1

=x1

3√

x21 − 1

for x1 > 1

(C)dl

dx1

= 1 +1

3x21

for x1 > 1 (D)dm

dy1=

1

3for y1 > 0

Answer and Comments: (A), (B), (D). The notations in the problemare slightly confusing. Usually, P (x1, y1) would denote a fixed point inthe plane. In the present problem, it is a variable point in the firstquadrant which moves on the hyperbola H . But instead of beginningthe problem by saying so, the point (x1, y1) is specified as the pointof contact of this hyperbola with a circle S whose centre is given asN = (x2, 0). So, it is to be assumed that x2 is also not fixed. So, thepicture we gather is that x1 and y1 are functions of the variable x2.But the problem asks for derivatives w.r.t. the variables x1 and y1.

O M N

P S

H

x

y

Note that instead of defining P in terms of N , we can as wellexpress N in terms of P . In fact, the circle S has little role to playexcept to tell us that the line PN is normal to S and hence also to Hat P , because S and H have a common tangent at P . So, we couldas well define N as the point where the normal to H at P meets thex-axis. The point M , on the other hand, is specified directly in termsof (x1, y1), as the point where the tangent to H at P cuts the x-axis.

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So the essence of the data of the problem is this. Suppose P (x1, y1)is a variable point in the first quadrant portion of the hyperbola H :x2 − y2 = 1, M and N are, respectively, the points where the tangentand the normal to H at P cut the x-axis. And finally, (l, m) is thecentroid of the triangle △PMN . The coordinates of M and N andhence those of the centroid will be functions of the variables x1 and y1.But these two variables are not independent of each other. Either ofthem can be expressed in terms of the other by the equations

y1 =√

x21 − 1 (1)

and x1 =√

y21 + 1 (2)

As a result, we can express l both as a function of x1 and also of y1.The same goes for m. The problem then asks about the derivatives ofthese functions.

Let us now turn to the problem itself. The equation of the tangentto H at (x1, y1) is

xx1 − yy1 = 1 (3)

Its point of intersection with the x-axis is (1

x1

, 0). So, we have

M = (1

x1

, 0) (4)

As the slope of the tangent isx1

y1, that of the normal at P is −y1

x1.

Hence the equation of the normal to H at P is

y − y1 +y1x1

(x− x1) = 0 (5)

which cuts the x-axis at the point (2x1, 0). Hence

N = (2x1, 0) (6)

As (l, m) is the centroid of △PMN , we have

l =1

3(3x1 +

1

x1) = x1 +

1

3x1(7)

and m =y13

(8)

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From (7) we getdl

dx1= 1 − 1

3x21

which shows that (A) is true and (C)

is false. Also from (8), we see that (D) is true. For (B), we first needto express m as a function of x1. For this we combine (8) with (1) toget

m =

x21 − 1

3(9)

Differentiating,

dm

dx1

=x1

3√

x21 − 1

(10)

which proves that (B) is true.

A simple problem except possibly for its clumsy framing. Onewonders if it was intentional. If so, the idea was probably to test acandidate’s ability to grab the essence of a problem from a nebulousdata. But it does look silly. It is like a person telling you that he methis father’s wife’s daughter, instead of saying that he met his sister(possibly a step-sister)!

Q.56 The option(s) with the values of α and L that satisfy the followingequation is (are)

∫ 4π

0et(sin6 αt+ cos4 αt) dt

∫ π

0et(sin6 αt+ cos4 αt) dt

= L

(A) α = 2, L =e4π − 1

eπ − 1(B) α = 2, L =

e4π + 1

eπ + 1

(C) α = 4, L =e4π − 1

eπ − 1(D) α = 4, L =

e4π + 1

eπ + 1

Answer and Comments: (A), (C). The given expression is a ratio oftwo integrals, say I and J . Both the integrals have the same integrand,say

f(t) = et(sin6 αt+ cos4 αt) (1)

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The numerator I can be written as

I = I1 + I2 + I3 + I4 (2)

where

I1 =∫ π

0f(t)dt, I2 =

∫ 2π

πf(t)dt I3 =

∫ 3π

2πf(t)dt and I4 =

∫ 4π

3πf(t)dt (3)

Clearly, I1 = J . We need to relate I2, I3 and I4 to I1. For this we needthe periodicity of the trigonometric functions. The functions sin2 αtand cos2 αt and hence all their powers are periodic with a period of πfor both α = 2 and α = 4 (in fact, for all integral values of α) because

sin(α(t+ π)) = (−1)α sinαt (4)

and cos(α(t+ π)) = (−1)α cosαt (5)

for all t. However, the first factor of the integrand, viz. et is notperiodic. Still we have the identity

et+π = eπet (6)

for all t. Using (4), (5) and (6), we get

f(t+ π) = eπf(t) (7)

for all t. The simple substitution u = t− π now gives

I2 =∫ 2π

πf(t) dt

=∫ π

0f(u+ π) du

=∫ π

0eπf(u) du

= eπI1 (8)

The same reasoning gives I3 = eπI2 and I4 = eπI3. So the terms of theR.H.S. of (2) are in a geometric progression with common ratio eπ andthe first term I1. Hence

I = I1e4π − 1

eπ − 1(9)

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This holds regardless of what the integer α is. So both (A) and (C) arecorrect.

Problems involving integrals of periodic functions have been askedin the JEE (e.g. see Exercise (18.8)). But the present one is a novelone in that it combines periodic functions and a simple property of theexponential functions. Once the idea strikes, the computations involvedare minimal. So this is a very good problem.

Although it is well beyond the JEE level, we remark that similarcalculations arise in finding what is called the Laplace transform ofa periodic function. Suppose f(t) is a continuous function defined forall t ≥ 0. Then assuming that the growth of |f(t)| as t → ∞ is not toowild, it can be shown that the improper integral

F (s) =∫ ∞

0e−stf(t)dt (10)

is convergent for all sufficiently large values of the real parameter s.We now have a new function, F (s), of a new variable s. It depends onthe function f(t) and is often denoted by L(f). As easy examples, weinvite you to show, using integration by parts, that

L(tn) =n!

sn+1, n = 0, 1, 2, . . . (11)

L(eαt) =1

s− α(12)

L(cosωt) =s

s2 + ω2(ω ∈ IR) (13)

and L(sinωt) =ω

s2 + ω2(ω ∈ IR) (14)

What makes the Laplace transforms very useful is the result, easilyproved by integration by parts, that for a continuously differentiablefunction f(t),

L(f ′) = sL(f)− f(0) (15)

As a consequence, a differential equation for a function f(t) is trans-formed into an algebraic equation for its transform F (s). Algebraicequations are easier to solve. So just as logarithms are useful in arith-metic because they transform the multiplication of two positive real

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numbers (which is time consuming) into the addition of their loga-rithms (which is much easier), Laplace transforms are very useful insolving differential equations.

Now suppose f(t) is a periodic function with a period T > 0. Thatmeans

f(t+ T ) = f(t) (16)

for all t ∈ IR. Then by taking steps similar to those in the solution of

the present problem, for every s, the integral∫ ∞

0e−stf(t) dt can be split

as an infinite series of integrals over intervals of the form [(n−1)T, nT ]for n = 1, 2, 3, . . .. As a result, we get

F (s) = I1 + I2 + . . .+ In + . . . (17)

where

In =∫ nT

(n−1)Te−stf(t)dt (18)

(17) and (18) are valid for any f(t). But if f(t) is periodic with a periodT , then by taking steps similar to the derivation of (8) above, we get,that for every fixed s,

In+1 =∫ (n+1)T

nTe−stf(t) dt

=∫ nT

(n−1)Te−s(u+T )f(u+ T ) du

=∫ nT

(n−1)Te−sT e−suf(u) du

= e−sT In (19)

In other words {In}n≥1 is a geometric progression with common ratioe−sT . So the the R.H.S. of (17) is an infinite geometric series. As aresult, we get

F (s) =I1

1− e−sT(20)

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Note that even though F (s) is defined through an improper integral,the integral I1 is not improper. So periodicity of f(t) allows us toexpress its Laplace transform L(f) in terms of a proper integral overan interval whose length equals the period of f(t).

The JEE paper-setters are undoubtedly familiar with Laplacetransforms and chances are that the present problem was spurred as aminiature version of (20).

SECTION 3

This section contains TWO paragraph. Based on each paragraph, therewill be TWO questions Each question has FOUR options (A), (B), (C) and(D). ONE OR MORE THAN ONE of these four option(s) is(are) correct Foreach question, darken the bubble(s) corresponding to all the correct option(s)in the ORS.Marking scheme :+4 If only the bubble(s) corresponding to all the correct option(s) is(are)darkened0 If none of the bubbles is darkened-2 In all other cases.

PARAGRAPH 1

Let F : IR −→ IR be a thrice differentiable function.Suppose that F (1) = 0, F (3) = −4 and F ′(x) < 0for all x ∈ (1/2, 3). Let f(x) = xF (x) for all x ∈ IR.

Q.57 The correct statement(s) is (are)

(A) f ′(1) < 0 (B) f ′(2) < 0(C) f ′(x) 6= 0 for any x ∈ (1, 3) (D) f ′(x) = 0 for some x ∈ (1, 3)

Answer and Comments: (A), (B), (C). The problem asks aboutf(x) and its derivative f ′(x). We are given

f(x) = xF (x) (1)

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As F (x) is differentiable, so is f(x) and

f ′(x) = F (x) + xF ′(x) (2)

Hence

f ′(1) = F (1) + F ′(1) = 0 + F ′(1) = F ′(1) < 0 (3)

since F ′(x) < 0 for all x ∈ (1/2, 3). So, (A) is true.

For (B),

f ′(2) = F (2) + 2F ′(2) (4)

We are not given the values of either F (2) or F ′(2). But we are giventhat F ′(x) < 0 for all x ∈ (1/2, 3). So F (x) is strictly decreasingon [1/2, 3]. (This assertion requires the use of Lagrange’s Mean ValueTheorem. But it is possible that this subtlety is missed by many can-didates, and, being an MCQ, there is no way to tell.) Since F (1) = 0and 2 > 1, we get F (2) < F (1) = 0. Also since 2 ∈ (1/2, 3), F ′(2) < 0straight from the hypothesis. So by (3), (B) is also true.

The options (C) and (D) are the logical negations of each other.So exactly one of them is true. The reasoning given in (B) for x = 2applies for any x ∈ (1, 3) and so from (2), f ′(x) < 0 for all x ∈ (1, 3),as both the terms on the R.H.S. are negative. So (C) is true, andautomatically, (D) is false.

Q.58 If∫ 3

1x2F ′(x) dx = −12 and

∫ 3

1x3F ′′′(x) dx = 40, then the correct ex-

pression(s) is (are)

(A) 9f ′(3) + f ′(1)− 32 = 0 (B)∫ 3

1f(x) dx = 12

(C) 9f ′(3) + f ′(1) + 32 = 0 (D)∫ 3

1f(x) dx = −12

Answer and Comments: (C), (D). All options involve exact val-ues of some integrals and not just their sign determinations. So, theinequalities given in the paragraph are not likely to be of much help.From the nature of the integrands, it is obvious that integration by

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parts is called for. But when we form∫

f(x)dx =∫

xF (x) dx nothing

is given to us about an antiderivative of F (x). So we have to start froman antiderivative of the first factor, viz. x of the integrand. We thenget

∫ 3

1f(x) dx =

∫ 3

1xF (x) dx

= (x2

2F (x))

3

1−∫ 3

1

x2

2F ′(x) dx (5)

The value of the integral on the R.H.S. is given to us as −6. By a direct

substitution, the first term is9

2F (3)− 1

2F (1). The values of F (3) and

F (1) are given as −4 and 0 respectively. Substituting these, the R.H.S.of (5) comes out to be −18 − 0 + 6 = −12. Hence (D) is correct and(B) false.

To determine which (if any) of the remaining options is true, we usethe second piece of information, viz.

∫ 3

1x3F ′′(x) dx = 40 (6)

The integral on the L.H.S. can be evaluated by parts as

∫ 3

1x3F ′′(x) dx = (3x2F ′(x))

3

1−∫ 3

13x2F ′(x) dx (7)

Again, we are given the value of the integral on the R.H.S. as −36. So,by a direct substitution we get

27F ′(3)− F ′(1) = 40− 36 = 4 (8)

Since the options (A) and (C) are in terms of f ′ rather than F ′, wemust convert F ′(3) and F ′(1) suitably. Applying (2) and the fact thatF (3) = −4, we have

F ′(3) =f ′(3)− F (3)

3=

f ′(3) + 4

3(9)

and similarly, since F (1) = 0,

F ′(1) =f ′(1)− F (1)

1= f ′(1) (10)

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Putting these values into (8), we get

9f ′(3) + 36− f ′(1) = 4 (11)

which means (C) is true and (A) false.

A simple, but extremely weird pair of questions. Conceptually, thetwo questions have little in common. But because of the constraint inpaper-setting, they have been artificially clubbed together into a sin-gle paragraph. The first question is based on applying the Lagrange’sMean Value Theorem to get some inequalities. The second one is basedon integration by parts. The work involved is repetitious and the arith-metic required is arbitrary and prone to numerical errors.

PARAGRAPH 2

Let n1 and n2 be the number of red and black balls,respectively, in box I. Let n3 and n4 be the number

of red and black balls, respectively, in box II.

Q.59 One of the two boxes, box I and box II was selected at random and aball was drawn at random from out of this box. The ball was found tobe red. If the probability that this red ball was drawn from box II is1

3, then the correct option(s) with the possible values of n1, n2, n3 and

n4 is (are)

(A) n1 = 3, n2 = 3, n3 = 5, n4 = 15

(B) n1 = 3, n2 = 6, n3 = 10, n4 = 50

(C) n1 = 8, n2 = 6, n3 = 5, n4 = 20

(D) n1 = 6, n2 = 12, n3 = 5, n4 = 20

Answer and Comments: (A), (B). This is a problem of conditionalprobability. We are first selecting a box and then drawing a ball from

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the selected box. We want to find the probability that the ball drawnis from the second box, given that its colour is red.

There is a chain of two processes here, first selection of a box andthen the draw of a ball. So for a visual representation of the thoughtprocess, tree diagrams are more convenient than Venn diagrams aselaborated in Comment No. 11 of Chapter 22. The appropriate treediagram for the present problem is shown below.

A

1/ 2 1/ 2

I II

r b r b

p p p p1 2 3 4

Here A is the starting node. From there, there are two branches lead-ing to the two nodes I and II corresponding to the two boxes. Eachbranch has probability 1/2. From each of these two nodes, there aretwo branches leading to the nodes marked r and b depending upon thecolour of the ball drawn. Their probabilities are marked as p1, p2, p3, p4.They can be easily calculated from the data of the problem as

p1 =n1

n1 + n2, p2 =

n2

n1 + n2, p3 =

n3

n3 + n4and p4 =

n4

n3 + n4(1)

Now let R be the event that the ball drawn is red. Then from thediagram above,

P (R) =1

2p1 +

1

2p3

=1

2

n1

n1 + n2+

1

2

n3

n3 + n4(2)

Let E2 be the event that the second box was chosen. Of course P (E2) =1

2. But the desired probability is not this, but rather the conditional

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probability P (E2|R), i.e. the probability that the ball drawn is fromthe second box given that it is red. By Bayes theorem (see CommentNo. 5 of Chapter 22), or more easily, by common sense,

P (E2|R) =P (E2 ∩ R)

P (R)(3)

The denominator was already calculated in (2). It is clear that thenumerator is simply the second term of the R.H.S. of (2). So, thedesired probability, say p, is

p =

n3

n3 + n4n1

n1 + n2

+n3

n3 + n4

(4)

The conceptual part of the problem ends here. In a numericalproblem, we would be given the values of n1, n2, n3 and n4 and asked

to calculate p from (4). But here we are given that p =1

3and are

asked to see for which of the given sets of values of n1, n2, n3 and n4,

the answer comes out to be1

3. So, in essence, we are asked to repeat

the arithmetical work four times. A slight simplification is possible.

Note that p can be written asp3

p1 + p3. So this will equal

1

3if and only

if p1 = 2p3. In (A), p1 = 1/2, p3 = 1/4 and so (A) is true. In (B),p1 = 1/3, p2 = 1/6 and so (B) is also true. In (C), p1 = 4/7, p3 = 1/5and so (C) is false. Finally, in (D), p1 = 1/3, p3 = 1/5 and so (D) isfalse.

A very standard problem of computing conditional probability.But one wonders what is gained by forcing the candidate to do thelast bit of arithmetic four times. Why not give only one set of valuesof n1, n2, n3 and n4 and give four numerical alternatives for the valueof p? This used to be the standard practice in the past. But therehave been cases where, because of a carelessness on the part of thepaper-setters, the fake answers could be eliminated easily without doingthe problem honestly, for example, when the fake answer was given asa rational number whose denominator did not divide the size of thesample space. It is probably to avoid this kind of a criticism that thistime the paper-setters have chosen to torture the candidates.

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Q.60 A ball is drawn at random from box I and transferred to box II. If the

probability of drawing a red ball from box I, after this transfer, is1

3,

then the correct option(s) with the possible values of n1 and n2 is(are)

(A) n1 = 4 and n2 = 6 (B) n1 = 2 and n2 = 3(C) n1 = 10 and n2 = 20 (D) n1 = 3 and n2 = 6

Answer and Comments:. (C), (D). It is given that a ball, randomlydrawn from box I, is transferred to box II. But the second part of theproblem deals only with what is left in box I. So what happens tothis ball after its removal from box I is irrelevant. The paper-setterscould simply have said that one ball at random was removed from boxI, or kept the data as it is except that the second ball is drawn frombox II. That would make the transfer relevant and the problem moreinteresting.

Let p1 and p2 be the probabilities that the ball removed was redand black respectively. Clearly,

p1 =n1

n1 + n2and p2 =

n2

n1 + n2(5)

Now, if the ball removed is red, then box I is left with n1 − 1 red andn2 black balls. The probability of drawing a red ball from this box isp3, given by

p3 =n1 − 1

n1 + n2 − 1(6)

Similarly, if the ball removed is black, then box I is left with n1 redand n2−1 black balls and the probability, say p4, of drawing a red ballfrom it now is

p4 =n1

n1 + n2 − 1(7)

With the help of a tree diagram, if necessary, the probability, say p,that the second ball drawn from box I is red, is

p = p1p3 + p2p4

=n1(n1 − 1)

(n1 + n2)(n1 + n2 − 1)+

n2n1

(n1 + n2)(n1 + n2 − 1)

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=n1(n1 + n2 − 1)

(n1 + n2)(n1 + n2 − 1)

=n1

n1 + n2

(8)

We are given that p =1

3. This will happen if and only if n2 = 2n1.

Clearly this is the case in options (C) and (D) but not in the others.

This problem is much simpler than the last one. In fact, a perceptivestudent will hardly fail to notice that p is the same as p1, i.e. the sameas the probability of drawing a red ball from the box I at the start. It isunlikely that he will have the time to ponder if this is just a coincidenceor can be somehow justified. But if he does, he can paraphrase theproblem to say that two balls were drawn from box I, one after theother, and the problem asks for the probability that the second ball isred. How do we justify that this is the same as the probability that thefirst ball is red?

Intuitively, the answer is that we can interchange the two ballsdrawn. For a formal proof, let n = n1 + n2. Let us suppose that thered balls as well as the black balls are distinct from each other. (Thisdoes not affect the probabilities of the events involved.) Let X be theset of all these n balls. Our sample space, say S is the set of all orderedpairs (x, y) ∈ X×X such that x 6= y. Clearly S has n(n−1) elements.Now let F1 and F2, respectively, be the favourable subsets for the eventthat the first ball is red and the event that the second ball is red. Insymbols,

F1 = {(x, y) ∈ S : x is red} (9)

and F2 = {(x, y) ∈ S : y is red} (10)

As the sample space is the same, proving that the two events have thesame probability is equivalent to showing that the sets F1 and F2 havethe same number of elements. This can be done by defining a bijectionf : F1 −→ F2 which interchanges the two balls drawn. Formally,f(x, y) = (y, x). Then f is its own inverse and hence a bijection.

Note that this argument will also apply if we draw 3 balls, oneafter the other without replacement. The probability that the first onebe red equals the probability that the second one be red and also the

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probability that the third one be red. Indeed this will hold even if thebox contains balls of more than two colours and we draw any numberof balls from it as long as this number does not exceed the total numberof balls in the box. (If it does, then the sample space is vacuous andthe probability is undefined.)

As in every year, the MCQ format makes it impossible to reward acandidate who has the ability to do this extra bit of thinking. In fact,he is advised against it because it is a sheer waste of his precious time.

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CONCLUDING REMARKS

The paper-setters have been careful to avoid mathematical mistakes andambiguities. For example, in both the questions involving the compositeof two functions, the order in which the two functions act has been clearlyspecified. In Q. 47 of Paper 1, the symbol a is used to denote a variable.This should have been made clear, because normally the symbols a, b, c etc.are used for constants and x, y, z, t etc. for variables. Similarly, in Q. 55 ofPaper 2, it should have been made clear that x2 and x1 are variables. InQ.54 of Paper 1, instead of saying ‘true’ it would have been better to say‘necessarily true’ to avoid the degenerate possibility of a zero matrix whichis symmetric as well as skew-symmetric. But these omissions are unlikely tohave caused any serious confusion.

Except for a slight lapse of the article in Q. 58 of Paper 2, (saying, ‘a circlewith a given diameter’ instead of saying ‘the circle’) there are no anomaliesin the wording of the problems. As pointed out in Q.60 of Paper 2, the in-formation that the ball removed from box I was put into box II is irrelevant.But again, these are things to be forgotten after a slight chuckle. On thecontrary, as commented in Q. 47 of Paper 1, the paper-setters have com-mendably avoided an inconsistency in the data which could have resulted ifthere was a repetition of carelessness in the drafting of a similar question inJEE 2011.

A controversy did arise about the word ‘component’ used in Q.47 of Paper2. In a conventional examination, from the answerbook of a candidate, itwould be possible to see which interpretation he had taken. And then adecision could be taken as to how much credit be given. The MCQ formatmakes this impossible. The JEE organisers have decided to award full marksto every candidate. As a result, those who interpreted and solved the problemcorrectly are at a loss. Fortunately, it is not a serious loss because theproblem was very trivial anyway. The JEE Advanced organisers deserve tobe commended for showing this openness of mind. (In an earlier version ofthis commentary, the JEE Advanced organisers were wrongly criticised. Themistake is regretted.)

Given the constraints imposed on them, the paper-setters have come upwith some good problems. The two that stand out are Q.41 in Paper 1(about the reflection of a parabola) and Q.53 of Paper 2 (about the signdetermination of the trigonometric functions of angles that are triples of somegiven angles). Q. 51 of Paper 1 (about the differentiability of the composite of

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two functions), Q.44 (about orders of magnitude) and Q.46 (about functionsdefined by integrating odd functions) of Paper 2 are also very good questionsof the conceptual type. But their ability to separate the men from the boysis marred by the multiple choice format.

Some other good questions in Paper 1 are Q.43 (about boys and girlsstanding in a row), Q.47 (about a function defined by integrating cos2 t), Q.54(about symmetric and skew-symmetric matrices), Q.55 (about expanding adeterminant of order 3) and Q.59(C) (about the complex cube roots of unity),while those in Paper 2 include Q.41 (about the sums of the terms in anarithmetic progression), Q.48 (about representing complex roots of unity asvertices of a regular polygon), Q.51 (about finding lower and upper bounds onan integral), Q.54 (about the eccentricities of two ellipses) and Q.56 (aboutintegrating products of periodic functions with the exponential function).

Questions which are too straightforward or familiar and hence have noplace in an advanced test include Q.42 (about complementary probabilitywith a coin toss), Q.46 (about minimising the material needed for a cylin-drical container), Q.49 and Q.50 (both on differential equations) and Q.56(about parametrised family of planes) in Paper 1 and Q.47 (about resolv-ing a vector), Q.49 (about applications of Intermediate Value Property andRolle’s theorem), Q.50 (a straight, simple integral), Q.52 (about the roots ofa quadratic) and both the probability problems (Q.59 and 60) in Paper 2.There are also numerous questions which reduce to writing down and solvinga system of equations, often in two unknowns. As there is nothing very greatin writing these equations or in solving them, such questions only reward themediocrity.

Many of these questions could have been dropped and room made toaccommodate certain areas that are singularly absent, such as number theory,surds, logarithms and inequalities. This is probably the first year whenthere is no question based on the A.M.-G.M. inequality. It is true thatthe multiple choice format precludes all proofs and that includes proofs byinduction, combinatorial identities, triangular optimisation and many resultsin theoretical calculus. Still, in the past the paper-setters have managed togive at least a token place to the areas just mentioned.

There is also a tendency apparent in many questions to force the can-didates to do extra, repetitious drudgery after the conceptual part of theproblem is over. For example, Q.54 in Paper 2 logically ends with the deter-mination of the eccentricities of the two ellipses. Subsequent work is sheerlyclerical. Similarly, in Q.59 and 60 about probabilities, instead of giving some

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particular values of the parameters n1, n2, n3 and n4 and asking for a numer-ical answer, which is the standard practice, the paper-setters have given foursets of such values and asked the candidates to identify those sets of valuesin which the answer will come out to be a given number. One really failsto see the purpose of such a twist when what it tests is not even remotelyrelated to any probability, It is sheer arithmetical drudgery. In Q.53 of Paper1, once the vectors ~a,~b and ~c are identified correctly, little purpose is servedby asking the candidates to compute the various expressions involving theirdot or the cross products.

Some questions demand work that is far in excess of what is justified bytheir credit. A notable example is Q.60(D) where a candidate has to sketchthree regions and find their areas all for 2 marks.

There is no significant difference either in the standard or the topics cov-ered in the two papers. One really wonders the rationale behind having twopapers each having all three subjects. Why not have three separate papers,one for each subject and give the paper-setters of each subject the freedomto choose the format of that paper as demanded by the peculiarities of thesubject, rather than impose the same format for all the three subjects? InChemistry, many of the questions are memory oriented. You simply have toknow the answer beforehand. There is no way you can get it by deductivereasoning. For example, questions that ask for the compounds that resultfrom some chemical reactions.

As a sample, as many as eight questions out of the 20 questions in Chem-istry in Paper 2 of this year’s advanced JEE directly ask to identify theproducts of some reactions. Q.39 in Paper 1 is for eight marks and asks toidentify the radicals (from a given list) occurring in each of the five givenores, viz. siderite, malachite, bauxite, calamine and argentite. This is sheermemory. If you know the chemical composition of these ores you get 8 markswithin a few seconds. If you don’t, just forget the question. No amountof deductive reasoning can help you. Its counterpart in mathematics, viz.,Q.59 in Paper 1, requires you to solve four completely unrelated problems,each requiring some thought and moderate computation. There is simply nocomparison between the two.

One can argue that even in chemistry, there are questions where somethought and numerical work is needed. This is true. But the thrust in thesequestions is on knowing some reaction. For example, in one of the questionsin Paper 2, a closed vessel with rigid walls containing 1 mol radioactiveuranium and one mol of air is given. The question then asks what will be

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the ratio of the final pressure to that of the initial one after the uraniumdecays completely. The answer comes from the fact that in the radioactivedecay, every atom of uranium produces one atom of lead and eight atoms ofhelium. This is pure memory. The ‘thought’ needed afterwards is that nowin the vessel we have 8 mols of gas in addition to the original 1 mol of air.As the volume and the temperature are the same, the new pressure will be9 times the original. This bit of mathematics is childishly simple. The samething is true about a question in which diborane reacts with methanol.

The mathematical equivalent of such a single idea question, would be, forexample, to give a right angled triangle with hypotenuse 5 and one side 4 andask to find the third side, or to ask in how many ways 5 boys and 5 girls canstand in a row. In both the cases, the answers (viz.

√52 − 42 =

√25− 16 =√

9 = 3 and 10! respectively) come with the application of single formulas.But such questions are not asked in mathematics even in the eliminationround. The second question appears as a part of Q.43 in Paper 1 this year.But it is only the beginning and not the end. In fact, the real problem then isto count in how many of these 10! arrangements, all the 5 girls stand togetherand also in how many of them exactly four girls stand together. This cannotbe done just by a hand counting. It requires a thought. Once the correctthought is grasped, the arithmetic needed is elementary. But the thoughthas to come from the candidate. No manual can help.

This is certainly not to deride Chemistry as a discipline of science. Butthe point is to decide which qualities are more important when they are beingtested by an examination like the JEE whose avowed purpose is to select themost gifted candidates. It would be great if the IIT’s get a sufficient numberof candidates who are good in all respects. That is, they know the chemicalcomposition of bauxite as well as have the ability to solve the problem aboveof 5 boys and 5 girls. Unfortunately, this does not happen and so the IIT’shave to make a choice. So, suppose the IIT’s have to choose between twostudents, say A, who can do the problem of 5 boys and 5 girls but does notknow, or has forgotten what bauxite is, and another candidate B who knowswhat bauxite is but cannot do the thinking needed for the problem of 5 boysand 5 girls. Whom should they choose? Clearly, the choice should be A,because he can be taught what bauxite is, or, when need arises, can easilylook it up. But, although B can also be taught how to do this particularproblem, the ability to do such thinking originally cannot be inculcated intohim as it is largely an innate quality.

In Physics the picture is better than in Chemistry. There are virtually no

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questions which are purely memory oriented. Most questions require somework to arrive at the answer. But usually, this work consists of applying one,or in some cases, two standard formulas. Once you know the right formulasto apply, the rest of the work is purely mathematical. For example, in oneof the Physics questions in Paper 2 this year, two solid spheres of equalradii R but with different mass density functions are given and we have tofind the ratio of their moments of inertia about axes passing through theirrespective centres. This is a standard application of integrals. At one time itwas taught in mathematics under mechanics. There is one question where aparallel plate capacitor is given, the plates being separated by two dielectricslabs of equal thickness but different relative permittivities. We are askedto find its capacitance given its capacitance in air. A diagram is also given.That makes it abundantly clear that we have to apply the formula for thecapacitance of two capacitors in series. Once this is realised, the mathematicsinvolved is straightforward.

In fact, in some questions in the Physics paper, the relevance with physicsis superficial. Take, for example, this question from Paper 2 where we aregiven that the energy of a system as a function of time t is E(t) = A2e−αt,α being a given constant. We are further given the percentage errors in themeasurements of A and of t and are asked to calculate the percentage errorin E(t) at t = 5. Now, this is a purely mathematical question, involvingfirst order approximation using derivatives. (Ironically, that is usually donein the first mathematics course in the IIT’s after the successful candidatesenter them!) There is hardly any physics in this problem except the name.The problem could as well have been posed as a problem in economics byreplacing the phrase ‘the energy of a system at time t’ by a phrase like ‘thedemand for some commodity at time t’.

The point to note is that once a question in chemistry is understood, ittakes very little time to answer it, provided you know the answer. In physics,it takes some work to arrive at the answer. But that work is usually somestandard mathematical computation. Moreover, the physics paper-settersusually make it easier for the candidates to understand the question, bysupplying appropriate diagrams. This year, for example, the Physics sectionof Paper 2 has as many as 10 diagrams for the 20 questions.

Contrast this with mathematics. There is not a single diagram in themathematics sections of either of the two papers! And this year is not anexception. In fact, some questions seem to be intentionally obscure. Forexample, Q.55 of Paper 2 about the tangent and the normal to a hyperbola.

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This problem as well as many other problems where the data is geometricwould be easier to understand if appropriate diagrams were given. But thenthe trouble is that they will also be easier to solve. For example, if a diagramwere given for Q.41 of Paper 1 (about the reflection of a parabola) that wouldalso give an unwarranted hint to the solution. And that will kill the beautyof the problem. One of the qualities considered desirable in mathematics isthe ability to read between the lines.

Summing up, Chemistry questions test the knowledge of facts, Physicsquestions test the ability to apply some standard formulas. But Mathematicsis inherently different. A mathematical equivalent of the question about thecapacitor considered above, would be, for example, a question asking forthe area of a triangle, given its two sides and one angle. None of the goodquestions listed above are of this type. There is an element of art and sportin mathematics which is generally absent in physics and chemistry which aresterile sciences. It becomes very difficult for the paper-setters in mathematicsto cater to this element if they are forced to frame the questions in thesame format as physics and chemistry. Designing and solving a truly goodmathematics problem is an artistic experience. No wonder some of the goodproblems in the old JEE’s are remembered even after decades. (Two suchproblems from the 1994 JEE have been referred to in this commentary, onein Q.55 of Paper 1 and the other in Q.48 of Paper 2.) Can that be said aboutany of the mathematics problems of the recent JEE’s?

True improvement would come only if the unique position of mathematicsis recognised by the policy makers. When the candidates are already screenedby JEE Main and the number of those eligible for the Advanced test is re-duced to a fairly manageable size, there is no reason to duplicate the selectionthrough another test of the same level and the same format. The AdvancedJEE ought to be qualitatively different from the Main one. At present it isnot. Because of the constraints on the paper-setters in mathematics, theyare forced to ask questions which fit only in the screening round but hardlyqualify for the final selection.

Ideally, at the Advanced JEE, there should be two papers of three hourseach, one solely for mathematics and the other for a combination of physicsand chemistry in the proportion, say 2 : 1. Since the Main examinationalready tests the degree of exposure to the three subjects, there should be nocompulsion to cover the entire syllabus in the Advanced papers. Moreover,mathematics paper-setters should have the freedom, as they did many yearsago, to design questions requiring different amounts of work and allot the

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credit proportionately. The evaluation should be manual, where the qualityof thinking of a candidate can be assessed, as he has to give reasoning for hiswork.

One can only pray that this will happen in near future.

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