EDUCATIVE COMMENTARY ON

JEE 2012 MATHEMATICS PAPERS

Contents

Paper 1 2

Paper 2 36

Concluding Remarks 60

The pattern of JEE 2012 is the same as that of the previous year. Thenumber of questions in Paper 1 has been reduced to a more reasonable 20 for eachsubject in each of the two papers. Further, negative marking is restricted only tothose questions where only one of the alternatives is correct. Negative markingfor questions where more than one alternatives are correct often resulted ina draconian penalty when a candidate could identify some but not all correctanswers. An even more welcome change is that the ‘match the pairs’ typesof questions have been dropped. In the past such questions often artificiallyclubbed together several unrelated problems the only commonality being thattheir numerical answers were the same.

As in the commentaries in the last few years, the questions in every sec-tion in each paper are listed here in a random order except for those based onparagraphs. Because of staggering of the questions so as to prepare differentversions of the same question paper (to curb copying), no particular order ofthe questions is sacrosanct. As in the past, unless otherwise stated, all the ref-erences made are to the author’s book Educative JEE (Mathematics) publishedby Universities Press, Hyderabad.

At the time of writing this commentary, the fate of the JEE is uncertain.Despite stiff opposition from certain corners, there is a move to abolish it to-tally and base the admissions to all engineering institutes on a single commonexamination. The real JEE was already put on the death bed when it was madetotally objective. If the proposal comes into effect, it will be totally dead andso this year may also be the year of saying good bye!

2

PAPER 1

Contents

Section I (Single Correct Choice Type) 2

Section II (Multiple Correct Choice Type) 18

Section III (Integer Type)

SECTION I

Single Correct Choice Type

This section contains ten multiple choice questions. Each question has 4choices out of which ONLY ONE is correct. A correct answer gets 3 pointsand an incorrect one −1 point.

Q.1 If limx→∞

(

x2 + x+ 1

x+ 1− ax− b

)

= 4, then

(A) a = 1, b = 4 (B) a = 1, b = −4 (C) a = 2, b = −3 (D) a = 2, b = 3

Answer and Comments: (B). The given function, say f(x) equals

f(x) =(1− a)x2 + (1 − a− b)x+ (1− b)

x+ 1(1)

which is a rational function of x, i.e. a ratio of two polynomials in x.The behaviour of such a function as x → ∞ depends on the degrees ofthe numerator and the denominator, say p and q respectively. If p > qthe limit does not exist; if p < q the limit is 0 and when p = q the limitexists and equals the ratio of the leading coefficients of the numerator andthe denominator. In the present problem, the third possibility is givento hold. So the numerator must be of degree 1 (which is the degree ofthe denominator) and moreover the coefficient of x in the numerator mustequal 4 times that of x in the denominator. This gives a simple system oftwo equations in two unknowns, viz.,

1− a = 0 and 1− a− b = 4 (2)

solving which we get a = 1 and b = 1− a− 4 = −4.This is a good problem, requiring very little computation once you get

the key idea. In working out the problem mentally, it is hardly necessaryto write down the steps (1) and (2) elaborately. Taking the expression

as it stands, we can split the numerator and write the ratio asx2

x+ 1+ 1

3

(a commonly adopted trick while finding antiderivatives of rational func-tions). The second term is a constant and does not affect the finiteness of

the limit. The first factor, viz.x2

x+ 1is of the order of x as x → ∞ and

the limit of the whole expression cannot be finite unless this troublesomepart is cancelled by some term which is also of the same order. This givesa = 1 almost instinctively. And once a is determined, the equation forb is obtained by equating the limit with 4. And this much can be easilymanaged in a minute.

The essential idea in this problem is that of asymptotic estimation

(see Exercise (6.53). Instead of saying thatx2

x+ 1is of the order of x as

x → ∞, we also say that asymptotically x2

x+ 1equals x.

Q.2 The point P is the intersection of the straight line joining the pointsQ(2, 3, 5) and R(1,−1, 4) and the plane 5x−4y−z = 1. If S is the foot ofthe perpendicular drawn from the point T (2, 1, 4) to QR, then the lengthof the line segment PS is

(A)1√2

(B)√2 (C) 2 (D) 2

√2

Answer and Comments:(A). This is a straightforward problem askingfor the distance between two points P and S. Neither point is givendirectly. Instead, it is defined by some property. So the first task is todetermine these two points P and S. As both these points lie on the lineQR, we can write them parametrically, since we are given the point Q andalso the vector QR, viz.

Q = (2, 3, 5) and−→QR= −~i− 4~j − ~k (1)

Therefore we have

P = (2− α, 3− 4α, 5− α) (2)and S = (2− β, 3− 4β, 5− β) (3)

for some values of α and β. To determine α we use the fact that P lies onthe plane 5x− 4y − z = 1. A direct substitution gives

10− 5α− 12 + 16α− 5 + α = 1 (4)

i.e. 12α− 7 = 1, which gives

α =2

3(5)

and hence

P = (4

3,1

3,13

3) (6)

4

Determination of β (and hence of the point S) requires slightly more work.We are given that the line TS is perpendicular to the line QR. Hence the

vectors−→TS and

−→QR are perpendicular to each other. The second vector

is already known to us as −~i− 4~j − ~k. The first vector is given by−→TS= −β~i+ (2− 4β)~j + (1− β)~k (7)

Taking dot product with−→QR, we get

β + 16β − 8 + β − 1 = 0 (8)

i.e. 18β = 9 which gives

β =1

2(9)

and hence

S = (3

2, 1,

9

2) (10)

Having determined both P and S, a straightforward calculation now gives

PS =

√

(4

3− 3

2)2 + (

1

3− 1)2 + (13

2− 9

2)2

=

√

(1

36+

4

9+

1

36)

=

√

18

36=

1√2

(11)

An intelligent candidate would notice that the distance PS could alsohave been expressed from (2) and (3) as

PS = |α− β|√1 + 16 + 1 = |α− β|3

√2 (12)

From (5) and (9), we have

|α− β| = |23− 1

2| = 1

6(13)

Putting this into (12) we get PS =1√2

with a lot less calculation than

before. The explanation is simple. The work done in calculating P andS from the values of α and β respectively gets undone when we take thedifferences of the coordinates of P and S later.

So, there is some room to apply a little brain in this problem. Still,it is a very routine computational problem in solid geometry. The timesaved by taking the short cut is not significant as compared to the timespent on the main body of the problem. The real advantage of the shortcut is that it reduces the chances of numerical mistakes.

5

Q.3 The integral

∫

sec2 x

(secx+ tanx)9/2dx equals (for some arbitrary constant

K)

(A)−1

(secx+ tanx)11/2

{

1

11− 1

7(secx+ tanx)2

}

+K

(B)1

(secx+ tanx)11/2

{

1

11− 1

7(secx+ tanx)2

}

+K

(C)−1

(secx+ tanx)11/2

{

1

11+

1

7(secx+ tanx)2

}

+K

(D)1

(secx+ tanx)11/2

{

1

11+

1

7(secx+ tanx)2

}

+K

Answer and Comments: (C). The expression sec2 x dx in the integrandmakes it tempting to try the substitution t = tanx. In that case theintegral, say I, would become

I =

∫

dt

(t+√1 + t2)9/2

(1)

Rationalising the denominator would convert this to

I =

∫

(√

t2 + 1− t)9/2dt (2)

As the exponent is not an integer, we cannot expand the integrand (asa finite sum) by the binomial theorem. So we have to abandon this line.The given integrand involves a fractional power of a sum of two functions.In such cases, the right substitution is often to replace the sum by a singlevariable. Let us try this approach. So we let

u = secx+ tanx (3)

We now have

du = (secx tanx+ sec2 x)dx = secx udx (4)

and hence

secxdx =du

u(5)

We are still left with a factor secx which we must express in terms of u.This can be done from the identity

secx− tanx = 1secx+ tanx

=1

u(6)

6

which, coupled with secx+ tanx = u gives

secx =1

2(u +

1

u) (7)

(we also get tanx =1

2(u − 1

u) but that is not needed here.) We can

now transform the given integral completely in terms of an integral of afunction of u, viz.

I =

∫

u+ 1u2uu9/2

du (8)

The integrand can now be expressed as a sum of various powers of u, eachof which can be integrated separately. Thus

I =1

2

∫

u−9/2 + u−13/2du

= −12

[

u−7/2

7/2+

u−11/2

11/2

]

= − 1u11/2

[

1

11+

u2

7

]

(9)

Putting back u = secx+tanx and adding an arbitrary constant gives theanswer.

In the conventional examination this would have been a reasonable fulllength question on finding antiderivatives. It would take some thinkingon the part of the candidate to come up with the right substitution, viz.u = secx + tanx. In the MCQ format all given alternatives involve theexpression (secx + tanx)2. And this gives an unwarranted clue to thecorrect substitution. But a sneaky solution need not even bother. Onecan simply differentiate each of the four alternatives, one by one, and seewhose derivative equals the given integrand. The work involved in all thesefour differentiations is essentially the same, except for the coefficients.

Had the paper-setters been a little careful, they could have givensecx tanx instead of secx + tanx in some of the fake alternatives. Buteven then the problem is already diluted by asking it as an MCQ. It wouldhave been better to ask, instead, the definite integral from, say, 0 to π/4and give the possible answers numerically. In that case the form of theanswers would not give away the right substitution so easily. But in thatcase, the work involved would be far more than justified by the creditallotted.

Q.4 The total number of ways in which 5 balls of different colours can bedistributed among three persons so that each person gets at least one ballis

(A) 75 (B) 150 (C) 210 (D) 243

7

Answer and Comments: (B). This problem is exactly the same as the1981 JEE problem in Chapter 1, Comment No. 3. Even the numbers arethe same! The only change is that instead of giving the balls to threepersons, there they were to be put into three distinct boxes so that nobox was empty. As the numbers involved are small, the best solution isby decomposing the possibilities into mutually disjoint cases. Call thethree persons as P1, P2, P3. Since everybody is to get at least one ball,the numbers of balls they get are either 2, 2, 1 or 3, 1, 1. In each case,there are three possibilities depending upon which of the three personsis the odd man out. The number of ways to give 2 balls to P1, 2 (ofthe remaining three) to P2 and the last one to P3 is

(

52

)(

32

)

= 30. So inall there are 90 distributions of this type. As for the second kind, thenumber of distributions where P1 gets 3 balls and the other two one eachis(

53

)(

21

)

= 20. So, there are 60 distributions of this type. In all the answeris 90 + 60 = 150.

There are also more sophisticated ways of getting the answer. One,based on Stirling numbers of the second kind, is pointed out in Exer-cise (1.29). Yet another solution, based on the principle of inclusion andexclusion (see Exercise (1.43)) can be given as follows. Note that everydistribution of the 5 (distinct) balls to 3 (distinct) persons amounts toa function from a 5-set (i.e. a set with 5 elements) to a 3-set. The re-quirement that every person is to get at least one ball amounts to sayingthat the corresponding function is surjective, i.e. onto. So the problem isequivalent to counting the number of surjective functions from a 5-set toa 3-set. More generally, we count the number of surjective functions froma set, say X , with m elements to a set, say Y , with n elements using theprinciple of inclusion and exclusion.

We follow the notation of Exercise (1.43). Denote the set of all functionsfrom X to Y by S. Clearly,

|S| = nm (1)

We want the number of those functions in S which are surjective. Thisrequirement can be paraphrased as follows. Let the (distinct) elements ofY be y1, y2, . . . , yn. For i = 1, 2, . . . , n, let Ai be the set of those functionsin S whose ranges do not contain the element yi. Obviously no suchfunction can be onto. In fact, a function f : X −→ Y is onto if and onlyif it is not in any of these Ai’s, or equivalently it is in the intersection of

their complements, viz.

n⋂

i=1

A′i. We now apply the principle of inclusion

and exclusion to get

∣

∣

∣

∣

∣

n⋂

i=1

A′i

∣

∣

∣

∣

∣

= s0 − s1 + s2 − s3 + . . .+ (−1)nsn =n∑

r=0

(−1)rsr (2)

8

where for r = 1, 2, . . . , n, sr is the sum of the cardinalities of the intersec-tions of these subsets taken r at a time. Thus,

s1 =

n∑

i=1

|Ai|

s2 =∑

i

9

Q.5 Let P = (aij) and Q = (bij) be two 3× 3 matrices with bij = 2i+jaij forall 1 ≤ i, j ≤ 3. If the determinant of P is 2, then the determinant of Q is

(A) 210 (B) 211 (C) 212 (D) 213

Answer and Comments: (D). An honest solution would be to begin bywriting Q out as

Q =

4a11 8a12 16a138a21 16a22 32a2316a31 32a32 64a33

(1)

Taking out the factors 4, 8 and 16 from the first, the second and the thirdrow respectively, we get

|Q| =

∣

∣

∣

∣

∣

∣

4a11 8a12 16a138a21 16a22 32a2316a31 32a32 64a33

∣

∣

∣

∣

∣

∣

= 4× 8× 16×

∣

∣

∣

∣

∣

∣

a11 2a12 4a13a21 2a22 4a23a31 2a32 4a33

∣

∣

∣

∣

∣

∣

(2)

We now take out the factors 2 and 4 from the second and the third columnrespectively. What is left is simply |P |. So,

|Q| = 4× 8× 16× 2× 4× |P |= 22+3+4+1+2+1 = 213 (3)

This is a good problem with a reasonably straightforward solution.But there is a sneak solution which makes a mockery of even this simplesolution. Note that it is already implicit in the problem that the ratio ofthe determinants of Q and P is a constant. So, we shall get a solutionas soon as this constant is determined. And for this purpose it sufficesto take P to be any matrix with determinant 2. As the simplest choicetake P to be a diagonal matrix with entries 2, 1 and 1. Then Q is also adiagonal matrix with entries 8, 16 and 64. So its determinant is 213. Ina conventional examination, this solution would get little credit since itis applicable only in a special case. But when the problem is posed as amultiple choice question, there is no way to know how the candidate gotthe answer. So a candidate who comes up with a sneak solution not onlygets full credit but also some bonus in terms of the time saved which is aprecious commodity.

Q.6 Let z be a complex number such that the imaginary part of z is non-zeroand a = z2 + z + 1 is real. Then a cannot take the value

(A) −1 (B) 13

(C)1

2(D)

3

4

10

Answer and Comments: (D). Before commenting on the problem it-self, its wording calls for a mild rebuttal. Although the phrase ‘such that’is used commonly (and often inevitably) in precisely stating some compli-cated definitions (a well known instance being the definition of a limit),for a layman and at the JEE level it often appears clumsy and should beavoided as far as possible. Thus, the present problem could have beenworded as ‘Let z be a complex number with a non-zero imaginary part.Let a = z2 + z + 1. Then ......’ which is a simpler diction.

Now, coming to the problem itself, assume z = x + iy where x, y arereal. We are given y 6= 0. A direct calculation gives

a = (x+ iy)2 + (x+ iy) + 1

= (x2 − y2 + x+ 1) + i(2xy + y) (1)

As we are given that a is real, we get 2xy + y = 0. As y 6= 0, we getx = −1

2. With this substitution in (1) we have

a =3

4− y2 (2)

As y is real and non-zero, y2 > 0. Hence a <3

4. So, a 6= 3

4. The other

values given are all less than3

4and can be assumed by a with x = −1

2and a suitable choice of y.

The problem can also be tackled by looking at z2 + z + 1 = a as aquadratic in z. The two possible solutions are

z =−1±

√4a− 3

2(3)

If a =3

4, then z would be purely real contradicting the hypothesis. This

solution is shorter but not so straightforward.

Since the number 1 is real, the reality of z2 + z + 1 is equivalentto that of z2 + z. It is not clear what purpose is served by giving thatz2 + z + 1 is real instead of giving that z2 + z is real, which is moredirect. one possibility is that z2 + z + 1 can be expressed as a ratio, viz.z3 − 1z − 1 . The ratio of two non-zero complex numbers, say z1 and z2, is realif and only if their arguments differ by a multiple of π. Geometrically thismeans that the points 0, z1 and z2 are collinear. In the present case thisamounts to saying that the points z3, 1andz are collinear. But that doesnot apparently lead to any elegant solution of the problem. Anyway, theproblem is so simple as it stands that it is only of an academic interest tosee if there is any elegant solution.

11

Q.7 The locus of the mid-point of the chord of contact of tangents drawn frompoints lying on the straight line 4x− 5y = 20 to the circle x2 + y2 = 9 is

(A) 20(x2 + y2)− 36x+ 45y = 0 (B) 20(x2 + y2) + 36x− 45y = 0(C) 36(x2 + y2)− 20x+ 45y = 0 (AD 36(x2 + y2) + 20x− 45y = 0

Answer and Comments: (A). Let P = (x1, y1) be a point on the givenline, say L, and suppose the tangents from to P to the given circle, sayC, touch it at the points A and B. Let Q = (h, k) be the midpoint of thechord AB. We have to find the locus of the point Q.

For the formula lovers, there are two formulas which are almost tailormade for this problem. One is the equation of the chord of contact of thepoint P w.r.t. the circle C. Popularly, it is remembered by T = 0. In thepresent case, since

C = x2 + y2 = 9 (1)

and P = (x1, y1), the equation of the chord AB comes out to be

xx1 + yy1 = 9 (2)

The second formula, which is less frequently used, is for the equation of achord in terms of its middle point M . In a compact form it is T = S1. Inthe present case, since Q = (h, k), this becomes

hx+ ky = h2 + k2 (3)

As (2) and (3) are equations of the same line AB, we get

x1h

=y1k

=9

h2 + k2(4)

These are two equations. In addition we are given that (x1, y1) lies on theline L. So,

4x1 − 5y1 = 20 (5)

The locus of (h, k) can be obtained by eliminating x0 and y0 in thesethree equations. The most straightforward way to do this is by gettingthe values of x1 and y1 from (4) and substitute them in (5). Thus we get

36h

h2 + k2− 45k

h2 + k2= 20 (6)

i.e.

20(h2 + k2)− 36h+ 45k = 0 (7)

12

To get the locus of Q, replace its current coordinates h and k by x and yrespectively. This gives (A) as the answer.

Formulas analogous to (2) and (3) are valid for any conic. So the samemethod would have worked if instead of the circle C we were given someother conic in the plane. In the present problem as the conic is a circlewhich is heavily studied in pure geometry, instead of using the formulas(2) and (3), a solution based on some standard properties of tangents toa circle from a point outside it can be given as follows. We follow thenotation in the solution above. Then it can be seen by symmetry that thepoints P , Q and O (the centre of the circle) are collinear and moreoverthat OP.OQ = 9, the radius square. (An easy way to see this is thesimilarity of the triangles OQB and OBP .)

O x

y

P

A

B

(x , y )1 1

Q (h,k)

With this approach, we have, by collinearity of O,Q and B,

x1 = λh and y1 = λk (8)

for some λ. Also the equality OQ.OP = 9 gives

√

h2 + k2√

x21 + y21 = 9 (9)

Substituting the values from (8) into (9) we get

λ(h2 + k2) = 9 (10)

Hence we have

x1 =9h

h2 + k2and y1 =

9k

h2 + k2(11)

which is exactly the same as (4). The rest of the work is the same asbefore.

The problem is straightforward once you know which formulas orproperties (of circles) to apply. But again, the work involved is more thancan be justified on the basis of the credit allotted.

13

Q.8 Let f(x) =

{

x2∣

∣cos πx∣

∣ , x 6= 00, x = 0

for x ∈ IR. Then f is

(A) differentiable both at x = 0 and x = 2

(B) differentiable at x = 0 but not differentiable at x = 2

(C) not differentiable at x = 0 but differentiable at x = 2

(D) differentiable neither at x = 0 nor at x = 2

Answer and Comments: (B). Here we have two separate problemsabout checking differentiability at the given points. A detailed analysiswould be to consider the right as well as the left handed derivatives of f(x)at the two points given. But instead, we give an argument which is basedon the known differentiability of two very commonly studied functions,viz. g(x) = x2 sin 1x for x 6= 0 (with g(0) = 0) and h(x) = |x|, at the point0. The first is differentiable at 0 while the second one is not. We count onthis familiarity and tackle the problem using the same reasoning.

The reason that the function g is differentiable at 0 is that it is of

the form x2 times a bounded function. So the ratiog(x)− g(0)

x− 0 becomesx times a bounded factor and hence tends to 0. Exactly the same reasonapplies to show that f is differentiable at 0. As for its differentiabilityat 2, note that the first factor x2 of f(x) is differentiable at 2 with anon-zero derivative 4. As f is continuous at all x 6= 0 and hence in aneighbourhood of x = 2, differentiability of f at 2 depends entirely onthe differentiability of its second factor, viz. | cos πx |. (To see this more

formally, write | cosπx| as f(x)x2

in a neighbourhood of x = 2 and apply

the quotient rule for derivatives.)

Thus the problem is reduced to checking the differentiability ofthe function, say |k(x)| (where k(x) = cos πx ) at x = 2. Here the crucialpoint is that the absolute value function is differentiable everywhere exceptwhere it takes the value 0. In the present case, in a small neighbourhoodof x = 2, the function k(x) vanishes only at x = 2. However, from this wemust not hastily conclude that |k(x)| is not differentiable at x = 2. Whatmatters is not just that |k(x)| → 0 as x → 2 but how rapidly. If it tendsto 0 more rapidly than |x−2| then |k(x)| would indeed be differentiable atx = 2 with a vanishing derivative. (For example, the functions |(x − 2)3|or | sin3/2(x − 2)| are differentiable at x = 2.) On the other hand, if|k(x)| tends to 0 at a rate slower than or comparable to that of |x − 2|,then |k(x)| is not differentiable at x = 2. (In the former case, the rightand left handed derivatives don’t exist while in the latter they exist butare unequal. The two cases are illustrated, for example, by

√

|x− 2| and| sin(x− 2)| respectively.)

14

So, now we have to see how fast cos πx tends to 0 as x → 2. For this

we consider and simplify the ratiocos πxx− 2 .

cos πxx− 2 =

sin(

π2 − πx

)

x− 2

=sin

(

π(x−2)2x

)

x− 2 (1)

Now we use the well-known fact that as θ → 0, sin θ tends to 0 at a ratecomparable to that of θ. So, we replace the numerator in the expression

above byπ(x− 2)

2x. So, we get that

limx→2

cos πxx− 2 = limx→2

π(x− 2)2x(x− 2) =

π

4(2)

which shows that as x → 2, | cos πx | tends to 0 at a rate comparable tothat of |x − 2|. We are now justified in saying that cos πx and hence f(x)is not differentiable at x = 2.

This is a good problem in which brute force computations can beavoided with some fine thinking. Unfortunately, a candidate who arguesthat | cos πx | is not differentiable at x = 2 on the superficial ground thatthe absolute value function is not differentiable at 0 also gets full creditundeservedly.

Q.9 The function f : [0, 3] −→ [1, 29] , defined by f(x) = 2x3 − 15x2 +36x+1

(A) one-one and onto (B) onto but not one-one(C) one-one but not onto (D) neither one-one nor onto

Answer and Comments: (B). A straightforward problem about thebehaviour of a polynomial function on an interval. The only way a con-tinuous function defined on an interval, say [a, b], can be one-one is thatit is either strictly increasing or decreasing. If the function is further dif-ferentiable, then the sign of its derivative provides a handy criterion. (Sohandy, if fact, that many people wrongly believe that a positive derivativethroughout an interval is the very meaning of its being strictly increas-ing, forgetting that the concept of increasing/decreasing is based purelyon inequalities and has nothing to do with derivatives!)

Returning to the problem, a straight calculation gives

f ′(x) = 6x2 − 30x+ 36= 6(x2 − 5x+ 6)= 6(x− 2)(x− 3) (1)

15

Both the roots of f ′(x) are simple and lie in the interval [0, 3]. So, f ′

is positive on [0, 2) and negative on (2, 3]. Hence f is neither strictlyincreasing nor strictly decreasing on [0, 3] and hence not one-one.

Next, to ascertain which values f takes, too the derivatives help. As fis strictly increasing on [0, 2] the image of [0, 2] is the interval [f(0), f(2)]i.e. [1, 29]. This alone is enough to conclude that f is onto. (Just for thesake of completeness, f is strictly decreasing on [2, 3] and hence maps itonto [f(3), f(2)] = [28, 29]. As this interval is already included in [0, 29],it does not add to the range. It remains [0, 29].)

Q.10 The ellipse E1 :x2

9+

y2

4= 1 is inscribed in a rectangle R whose sides are

parallel to the coordinate axes. Another ellipse E2, passing through thepoint (0, 4) circumscribes the rectangle R. The eccentricity of the ellipseE2 is

(A)

√2

2(B)

√3

2(C)

1

2(D)

3

4

Answer and Comments: (C). A straightforward problem once you un-derstand the vocabulary involved, specifically, what is meant by ‘inscribe’,‘circumscribe’ and ‘eccentricity’. (Incidentally, these three are highly tech-nical words and even with an otherwise good command over English, acandidate is not likely to have come across them unless he has studiedgeometry in the English medium. This is no reason to shed tears if thecandidate is weak in geometry. But if he is strong, then it means that heis being penalised merely for not having studied in the English medium.Availability of the Hindi version of the question paper alleviates the prob-lem to some extent but is of little help to a candidate who has studied inKannada or Tamil. One solution might be to make dictionaries of technicalterms available to the candidates, similarly to the log tables).

Now coming to the problem itself, the extreme ties of the major andthe minor axes of the ellipse E1 are (±3, 0) and (0,±2) respectively. Hencethe rectangle R has its vertices’s at (±3,±2). The second ellipse is givento pass these four points. However, this does not determine the ellipseE2 uniquely. You need five conditions to determine an ellipse (and moregenerally, any conic). This happens because the general second degreeequation

Ax2 + 2Hy +By2 + 2G+ 2FY + C = 0 (1)

has five independent unknowns. (Superficially there are six unknowns,A,B,C,D,E, F , but it is only their relative proportions that matter). If(1) is to represent an ellipse (and indeed any conic) passing through thefour points (±3,±2) we must have

9A+ 12H + 4B + 6G+ 4F + C = 0 (2)

16

9A− 12H + 4B + 6G− 4F + C = 0 (3)9A− 12H + 4B − 6G+ 4F + C = 0 (4)

and 9A+ 12H + 4B − 6G− 4F + C = 0 (5)

Subtracting (3), (4) and (5) from (2), we get three equations in threeunknowns, viz.,

24H + 8F = 0 (6)

24H + 12G = 0 (7)

8F + 12G = 0 (8)

which yield H = F = G = 0. We are now justified in assuming thatthe ellipse E2 has its axes parallel to the coordinate axes. Hence we canrepresent E2 by an equation of the form

x2

a2+

y2

b2= 1 (9)

where we need two equations to determine the parameters a and b. Oneof them is already given by any of the four points (±3,±2).

9

a2+

4

b2= 1 (10)

The other equation comes from the data that E2 passes through the point(0, 4). In fact, this directly gives b = 4. Putting this into (10) and solving,we have a2 = 12. Note that for this ellipse, the major axis is along the y-axis rather than on the x-axis, as is usually the case. So the usual formulafor the eccentricity has to be modified by switching a and b. That is, wehave

a2 = b2(1 − e2) (11)

Substituting the values of a and b we get

1− e2 = 1216

=3

4(12)

which gives e2 =1

4and hence e =

1

2.

SECTION II

Multiple Correct Choice Type

This section contains five multiple choice questions. Each question has fourchoices out of which ONE OR MORE may be correct. All correct choicesget 4 marks, otherwise, none.

17

The problem is easy once you get that the ellipse E2 has its axes parallelto the coordinate axes. This is not given in the statement of the problemand deriving this from the rest of the data takes some work as shown above.The trouble is that nearly in all problems at the JEW level, ellipses are inthe standard form. As a result, many students will simply assume that (10)is the equation of E2. Such students will get an unfair advantage over thescrupulous ones who derive it with some work (which is simple, but takes sometime nevertheless). Yet another sad casualty of the competitive MC based tests.

SECTION II

Multiple Correct Answer(s) Type

This section contains five questions with four answers to each, out of whichone or more may be correct. 4 points for identifying all correct answers, 0otherwise.

Q.11 A ship is fitted with three engines E1, E2 and E3. The engines function

independently of each other with respective probabilities1

4,1

2and

1

2. Let

X denote the event that the ship is operational and let X1, X2 and X3denote, respectively, the events that the engines E1, E2 and E3 are func-tioning. Which of the following is (are) true?

(A) P [Xc1 |X ] =3

16

(B) P [exactly two engines of the ship are functioning |X ] = 78

(C) P [X |X2] =5

16

(D) P [X |X1] =7

16

Answer and Comments: (B, D). A question of calculating the condi-tional probabilities. The ship is operational if and only if at least two ofE1, E2, E3 are operational. This resolves into four mutually disjoint events: either all are operating or exactly one of the three is not functioning andthe other two are. As a result,

P (X) = P (X1X2X3) + P (XC1 Z2X3) + P (X1X

c2X3) + P (X1X2X

c3) (1)

We are given the values of P (Xi) for i = 1, 2, 3. Further they are given tobe mutually independent. So, (1) gives

P (X) =1

2

1

4

1

4+

1

2

1

4

1

4+

1

2

3

4

1

4+

1

2

1

4

3

4

=1

32+

1

32+

3

32+

3

32

=1

4(2)

18

In all the statements except (B), the desired probability is already ex-pressed in terms of symbols. In (B), it is expressed in words. It is theconditional probability that exactly two of the events X1, X2 and X3 oc-cur, given that X occurs. Call the former event ay Y . Then Y is thedisjunction of the three mutually disjoint events Xc1X2X3, X1X

c2X3 and

X1X2Xc3 , already considered in computing (1). So, taking the last three

terms in (1), we get

P (Y ) =7

32(3)

Moreover, Y is a sub-event of X . So Y ∩X is the same as Y . So

P (Y |X) = P (Y )P (X)

=7/32

1/4=

7

8(4)

Hence statement (B) is true. The calculations in the remaining threestatements are more direct. In (A),

P (Xc1 |X) = P (Xc1 ∩X)/P (X) = P (Xc1X2X3)/P (X)

=121414

14

=1

8(5)

Hence (A) is incorrect. In (C), X ∩X2 equals (X1X2X3) ∪ (Xc1X2X3) ∪(X1X2X

c3). Hence

P (X |X2) =P (X ∩X2)

P (X2)

=P (X1X2X3) + P (X

c1X2X3)) + P (X1X2X

c3)

P (X2)

=121414 +

121414 +

121434

14

=5/32

1/4=

5

8(6)

So the statement (C) is also incorrect. Finally, for (D), the calculationsare similar to that in (C) except that we interchange X1 and X2. So,analogously to (6), we have

P (X |X1) =P (X ∩X1)

P (X1)

=P (X1X2X3) + P (X1X

c2X3)) + P (X1X2X

c3)

P (X2)

=121414 +

123414 +

121434

12

=7/32

1/2=

7

12(7)

19

Therefore the statement (D) is correct.

This is a simple but highly repetitious problem. The statement (B) byitself (without the answer) could have been asked as an MCQ with onlyone correct answer. (B) tests the ability to correctly translate a verballydescribed event into a mathematical formula as well as the knowledgeof the formula for conditional probability. Nothing more and a lot lessis tested in the other three statements. They only waste a candidate’sprecious time and increase the chances of numerical errors.

Q.12 If S is the area of the region enclosed by y = e−x2

, y = 0, x = 0 and x = 1,then

(A) S ≥ 1e

(B) S ≥ 1− 1e

(C) S ≤ 14

(

1 +1√e

)

(D) S ≤ 1√2+

1√e

(

1− 1√2

)

Answer and Comments: (A, B, D). Clearly,

S =

∫ 1

0

e−x2

dx (1)

If we could evaluate S exactly, then we can answer all alternatives by directcomparisons. The trouble here is that the integrand e−x

2

does not havean andtiderivative in an explicit form and so the fundamental theorem ofcalculus cannot be applied. Therefore, the present problem is not aboutthe evaluation of a definite integral. The best we can do is to find someupper and lower bounds on S. Parts (A) and (B) of the problem givepossible lower bounds for S while (C) and (D) give possible upper boundsand our job is to find which of these are correct.

For any definite integral I =

∫ b

a

f(x)dx the most immediate upper

and lower bounds (obtained from the very definition of a definite integral)are the upper and lower Riemann sums of f(x) for various partitions, say,

P = (x0, x1, . . . , xi, . . . xn) (2)

(where a = x0 < x1 < x2 < . . . < xn = b). In the present problem, theintegrand is strictly decreasing on [0, 1]. So, the upper and lower Riemannsums for a partition P like (2) are

U(P ; f) =

n∑

i=1

f(xi−1)(xi − xi−1) (3)

L(P ; f) =

n∑

i=1

f(xi)(xi − xi−1) (4)

20

We now take f(x) = e−x2

and [a, b] as [0, 1]. Let us first see which of thegiven four numbers occur either as lower or as upper Riemann sums for I.Trivially, if we take the partition P to be the one whose only nodes are 0and 1, then

U(P ; f) = f(0)× 1 = 1 (5)

and L(P : f) = f(1)× 1 = 1e

(6)

Thus we see that the the expression in (A) is a lower bound for S. Hence(A) is true.

To see which of the other three expressions arise as lower or upperbounds for I, we need to recognise some points in [0, 1] at which f assumessome values which equal some of the numbers occurring in these expres-

sions. We already know (and used) that 1 and1

eequal, respectively f(0)

and f(1). A less obvious functional value is1√e. Its natural logarithm is

−1/2 which equals −( 1√2)2. Hence f(

1√2) =

1√e. Let us now consider

the partition P = (0,1√2, 1) with three nodes. For this partition we have

U(P ; f) = f(0)(1√2− 0) + f( 1√

2)(1 − 1√

2) =

1√2+

1√e(1 − 1√

2) (7)

L(P ; f) = f(1√2)(

1√2− 0) + f(1)(1− 1√

2) =

1√2e

+1

e(1 − 1√

2) (8)

The R.H.S. of (7) is precisely the expression in (D). So we see that (D) iscorrect. (As it turns out, calculating (8) was essentially a waste. Unfor-tunately there is no easy way to fortell this.)

The expressions in (B) and (C) cannot be recognised as upper or lowerRiemann sums in an obvious way. So we try some other approaches. Oneof them is based on the simple fact that integrals preserve inequalities offunctions. Specifically, if we have two functions, say f1(x) and f2(x) suchthat

f1(x) ≤ f2(x) (9)

for all x ∈ [a, b]. Then∫ b

a

f1(x) dx ≤∫ b

a

f2(x) dx (10)

Actually, the inequalities based on upper and lower Riemann sums canbe viewed as special cases of this where one of the functions is the givenintegrand and the other function is a step function whose values are the

21

maxima/minima of the integrand on the subintervals of the partition.But we have already exhausted these possibilities. So we try some otherfunctions for comparison. In the present problem, we take f1(x) as f(x).We now look for a function f2(x) for which the inequality (9) (or itsopposite) will be true throughout [0, 1]. Since the exponential function ismonotonically increasing (i.e. preserves inequalities), we can take f2(x) ofthe form eg(x) where the exponent g(x) is comparable to −x2 throughout[0, 1]. One such choice is to take g(x) as −x. (Of course there are many

others. But we are guided by the requirement that

∫ 1

0

eg(x)dx should

coincide with the expression in (B) or (C).) With this choice, we get

−x2 ≥ −x (11)

for all x ∈ [0, 1] and hence

S ≥∫ 1

0

e−xdx (12)

This integral, unlike the original integral, can be easily evaluated in a

closed form. It equals −e−x∣

∣

∣

1

0= 1− 1

ewhich is precisely the expression

in (B). So we see that (B) is also true.

Finally, we check if (C) is correct. The method we are applyingso far requires us to find a suitable function h(x) which is comparable to

e−x2

throughout the interval [0, 1] and whose integral over [0, 1] equals thegiven expression. This approach did work in (B) by taking h(x) as e−x

(and also indirectly in (A) and (D) where we took h(x) to be a suitablestep function). Unfortunately, in (C), there is no obvious h(x) which willfit the prescription. So here a more delicate analysis is needed.

We can take some clue from

the sum 1 +1√e. We already recog-

nised 1 as f(0) and1√e

as f(1√2).

Therefore we get that the expression1

2

(

1 +1√e

)

is, geometrically, the mid-

dle height of the trapezium OAPCshown in the accompanying diagram ,where O,A, P and C are respectively,

the points (0, 0), (1√2, 0), (

1√2,1√e) and

(0, 1).

O

A

B

P

.MP

Therefore, the area, say ∆, of this trapezium is,

∆ =1

2

(

1 +1√e

)(

1√2− 0

)

=1

2√2

(

1 +1√e

)

(13)

22

Let us now compare this area with the area, say S1, under the portion of

the curve y = e−x2

between x = 0 and x =1√2. That is,

S1 =

∫ 1/√2

0

e−x2

dx (14)

It is not easy to evaluate S1 in a closed form, the difficulty being exactly thesame as that in evaluating S, viz. that the integrand has no antiderivativethat can be expressed in a closed form. But our interest is not so muchin evaluating S1 as in comparing it with ∆. Here we crucially need theconcept of concavity of a function. Since f(x) = e−x

2

a direct calculationgives

f ′(x) = −2xe−x2 (15)and f ′′(x) = (4x2 − 2)e−x2 (16)

for all x. The exponential factor is always positive while the other factor

(4x2 − 2) is negative for 0 ≤ x < 1√2

and positive for x >1√2. So we

get that f is concave downward on [0,1√2]. We also get that it is concave

upwards on [1√2,∞) and hence that the point P on the graph y = e−x2 is

a point of inflection, where the tangent crosses the curve as shown in thefigure. But that is not relevant here. What matters is only the first part,

viz., that f(x) is concave downward on [0,1√2]. Geometrically, this means

that the curve always lies above the chord. Since ∆ is the area below thischord, while S1 is the area below the curve we have

S1 > ∆ (17)

Further as the integrand e−x2

is positive throughout, we also have

S > S1 (18)

Further since 2√2 < 4 we also have

∆ >1

4

(

1 +1√e

)

(19)

Combining the last three inequalities, we finally have

S >1

4

(

1 +1√e

)

(20)

which shows that the statement (C) is false.

23

This is an excellent problem on the estimation of definite integrals. It

takes certain perceptivity on the part of the candidate to recognise1√eas

f

(

1√2

)

and thereby to realise that the point 1/√2 has a special role in

the problem. A further realisation that it is the point where f(x) changesits concavity gives the necessary hint for (C). Unfortunately, the workdemanded for various alternatives is of a different kind. The work for(C) is unusual since it is based on the concept of concavity downwards.The time needed far exceeds the time allowed. Note that the inequality(17) we have proved using concavity is much stronger than what is neededto disprove (C). It is possible that the paper-setters had some simplerargument in mind to disprove (C).

Q.13 Tangents are drawn to the hyperbolax2

9− y

2

4= 1, parallel to the straight

line 2x− y = 1. The points of contacts of the tangents on the hyperbolaare

(A)

(

9

2√2,1√2

)

(B)

(

− 92√2,− 1√

2

)

(C) (3√3,−2

√2) (D) (−3

√3, 2

√2)

Answer and Comments: (A, B). This is a straightforward problemabout finding the point of contact of a tangent to a conic given its slope.But apparently the paper-setters thought it too straightforward and sothey have given it a twist by not specifying the slope directly, but, insteadsaying that the tangent is parallel to a given line. This line has no otherrole in the whole problem. It would be naive to suppose that a candidatewho gets stuck by not being able to find the slope of the tangent from thispiece of data would have been able to solve the problem correctly had theslope be given directly as 2. So no testing purpose is served by such cornhusks. In a conventional examination, perhaps some partial credit couldbe allotted for merely finding the slope of the tangent. But in an MCQ,that is not possible either.

Now coming to the problem itself, for the formula lovers, there isa formula for the equation of a tangent to a conic in terms of its slope.Knowing the equation, one can find its point of contact by solving simul-taneously with the equation of the conic. But a better method is availableif we keep in mind that our interest is more in the point of contact thanwith the tangent itself. Suppose (x1, y1) is a point on the hyperbola. Thenthe equation to the tangent at (x1, y1) is

xx19

− yy14

= 1 (1)

If this is to be parallel to the line 2x − y = 1, then its slope must be 2.

24

This gives an equation in x1, y1, viz.

x1/9

y1/4= 2 (2)

which upon simplification becomes

x1 =9

2y1 (3)

Putting this into the equation of the hyperbola gives a quadratic in y1,viz.,

9y214

− y21

4= 1 (4)

i.e. y21 =1

2which gives y1 = ±

1√2. Putting these values in (3), the two

points of contact are those on (A) and (B).

This solution is so simple that it is relatively pointless to see if it canbe improved. But there is a way to improve. Our interest is in the sloperather than in the entire equation of the tangent. If we take parametricequations, there is a handy formula for the slope of the tangent at a pointusing the chain rule. So we let

x = 3 sec θ, y = 2 tan θ (5)

for θ ∈ [0, 2π]. Differentiating,dx

dθ= 3 sec θ tan θ,

dy

dθ= 2 sec2 θ (6)

So, by the chain rule,

dy

dx=

dy/dθ

dx/dθ

=2

3

sec θ

tan θ

=2

3 sin θ(7)

Equating this with 2 we get

sin θ =1

3(8)

which then gives cos θ = ±2√2

3and further tan θ = ± 1

2√2

and sec θ =

± 32√2. Putting these into (5) gives the desired points of contact. (Ac-

tually, it suffices to find only one of the points of contact, since points ofcontact of parallel tangents are always symmetric w.r.t. the centre.)

25

Q.14 If y(x) satisfies the differential equation y′ − y tanx = 2x secx and y(0) =0, then

(A) y(π

4

)

=π2

8√2

(B) y′((π

4

)

=π2

18

(C) y(π

3

)

=π2

9(D) y′(

(π

3

)

=4π

3+

2π2

3√3

Answer and Comments: (A, D). A straightforward problem on solvinga differential equation. The given equation is linear with integrating factor

e−∫

tan xdx = e− ln sec x = cosx (1)

The formula lovers can now also use the ready-made formula for the solu-tion. But it is often a good idea to multiply the equation throughout bythe integrating factor and cast the L.H.S. as the derivative of some func-tion. Indeed this is the very purpose of integrating factors. So, multiplyingboth the sides by cosx, the given D.E. becomes

y′ cosx− y sinx = 2x (2)

i.e.

d

dx(y cosx) = 2x (3)

whose general solution is

y cosx = x2 + c (4)

where c is an arbitrary constant. The initial condition y(0) = 0 determinesc as 0. Hence we have

y = x2 secx (5)

and hence

y′(x) = 2x secx+ x2 secx tanx (6)

To complete the solution we have to substitute x =π

4and x =

π

3in (5)

and (6). Since tanπ

4= 1 while sec

π

4=

√2, a direct calculation gives

y(π

4

)

=π2

14

√2 =

π2

8√2

(7)

y′(π

4

)

=π

2

√2 +

π2

16

√2 (8)

26

which shows that (A) is true while (B) is false. Similarly, from tan(π

3

)

=

tan 60◦ =√3 and sec

(π

3

)

= 2 we see, after substitution, that (C) is false

while (D) is true.

A very routine problem about differential equations with a purelyclerical trigonometrical appendage.

Q.15 Let θ, φ ∈ [0, 2π] be such that

2 cos θ(1− sinφ) = sin2 θ(

tanθ

2+ cot

θ

2

)

cosφ− 1,

tan(2π − θ) > 0 and −1 < sin θ < −√3

2. Then φ cannot satisfy

(A) 0 < φ <π

2(B)

π

2< φ <

4π

3

(C)4π

3< φ <

3π

2(D)

3π

2< φ < 2π

Answer and Comments: (A, C, D). Superficially, this is a problemabout trigonometric equations. But in reality it is not so. There are twovariables here θ and φ. But only one equation in them. So, the data isinsufficient to determine θ and φ. All we can do is to use this equationto express the variable φ in terms of θ. We are also given (indirectly)the interval(s) over which θ varies and our job is to find the set, say S,over which φ varies. Once this set S is identified, the problem asks youto determine which of the given four intervals is (are) disjoint from S.(The language used in framing the last part of the question is likely to beconfusing to some candidates. Generally, the paper-setters are restrainedfrom asking questions where the assertion is in the negative.)

Once again, the paper-setters have given a twist to the data. Astraight-forward simplification gives that for any angle θ,

tan θ + cot θ =sin θ

cos θ+

cos θ

sin θ

=1

sin θ cos θ

=2

sin 2θ(1)

Using this identity (with θ replaced by θ/2), the given equation becomes

2 cos θ(1− sinφ) = 2 sin θ cosφ− 1 (2)

which upon further simplification gives

2 sin(θ + φ) = 2 cos θ + 1 (3)

27

We are also given certain restrictions on the possible values θ can take.First θ ∈ [0, 2π]. Secondly, tan(2π−θ) > 0 which is a twisted way of sayingthat tan θ < 0. So, θ is further restricted to

(π

2, π

)

∪(

3π

2, 2π

)

. Finally we

are also given that −1 < sin θ < −√3

2which means θ ∈ (4π

3,5π

3)− {3π

2}

and further restricts θ to

θ ∈(

3π

2,5π

3

)

(4)

From this and (3) we have to determine what possible values φ can take.One way to do this would be to solve (3) for φ to get

φ = nπ + (−1)n sin−1(cos θ + 12)− θ (5)

where n is an integer. We shall then have to consider various values of nfor which the R.H.S. has at least some values between 0 and 2π and in eachcase determine which possible values φ can take. Moreover, the analysiswould be different for even n than for odd n. This would be complicated.So we try a simpler approach. We recast (3) as

sin(θ + φ) =1 + cos θ

2(6)

Because of (4) the R.H.S. lies between1

2and 1. Further since both θ, φ

lie in [0, 2π], θ + φ ∈ [0, 4π]. Therefore

θ + φ ∈(

π

6,5π

6

)

∪(

2π +π

6, 2π +

5π

6

)

(7)

Thus, we have two cases:

π

6< φ+ θ <

5π

6(8)

or13π

6< φ+ θ <

17π

6(9)

We add the inequality

−5π3

< −θ < −3π2

(10)

to each of these two inequalities to get

−3π2

< φ < −2π3

(11)

orπ

2< φ <

4π

3(12)

28

The first possibility is ruled out because φ ≥ 0. The second possibilitymeans that φ must lie in the interval

(

π

2,2π

3

)

. This is precisely the

interval in (B). As the intervals in the remaining three possibilities are alldisjoint from this interval we see that φ can never lie in any of them.

We were somewhat lucky that we got the answer relatively easily.From the data that θ and φ both lie in [0, 2π], we concluded that θ + φlies in [0, 4π]. Implicitly this was obtained by adding the two inequalities0 ≤ θ ≤ 2π and 0 ≤ φ ≤ 2π. This is a true but a crude estimate. Whenθ and φ are independent of each other, this is the best estimate we cangive. But when two quantities are related to each other, sharper estimatesare often possible for their sums. As an extreme example, both sin2 x andcos2 x vary over [0, 1]. But their sum does not vary all over [0, 2]. In fact,it is a constant. In the present problem, φ is related to θ through theequation (5). So the variation of θ + φ may be much smaller than 0 to4π. There was also some extravagance in concluding (11) from (8) and(10) (and similarly, in deriving (12) from (9) and (10)). Accordingly, theset, say S, of possible values of φ may be only a subset of the interval(

π

2,4π

3

)

. The intervals in (A), (C) and (D) are all disjoint from this

interval and hence also from S. But if they were bigger then it is possiblethat they are not disjoint from this interval but still disjoint from thesubset S. In that case our approach would fail. To identify the set Sexactly, we need to follow the other approach of a detailed analysis of (5)for various values of the integer n. In that case the problem would havebeen far more challenging and complicated. As it stands, the problemis simple for the naive candidate. A scrupulous candidate who goes foranalysing (5) is penalised in terms of the time he spends.

SECTION III

Integer Answer Type

This section contains five questions. Each has a single digit correct answer.Correct answer gets 4 points, incorrect one none.

Q.16 Let p(x) be a real polynomial of least degree which has a local maximumat x = 1 and a local minimum at x = 3. If p(1) = 6 and p(3) = 2, thenp′(0) is ....... .

Answer and Comments: 9. The data implies that p′(x) has 1 and 3among its roots. So, its degree is at least two. Since p(x) has the leastpossible degree among polynomials which satisfy this requirement, so doesp′(x). Hence p′(x) is of the form

p′(x) = α(x − 1)(x− 2) (1)

29

for some (real) constant α. Expanding and integrating,

p(x) = α(x3

3− 2x2 + 3x) + β (2)

where β is some other real constant. The values of α and β can be foundfrom the given conditions, viz, p(1) = 6 and p(3) = 2 which imply

α(1

3− 2 + 3) + β = 6 (3)

and α(9 − 18 + 9) + β = 2 (4)

respectively. The second equation immediately gives β = 2. Putting thisinto (3) we get

4

3α = 6− 2 = 4 (5)

which implies α = 3. This gives

p′(x) = 3(x2 − 4x+ 3) (6)

whence p′(0) = 9.

A very straightforward problem about determining a polynomial, beinggiven its points of local maxima and local minima. The computationsinvolved are reasonable and not prone to numerical errors. The values ofp(x) at any two distinct points would serve to determine α and β uniquely.No particular advantage is gained by specifying the values of p(x) at thepoints where it has a local maximum or a local minimum. However, withthese values, the same data could have been given a little more subtly, byspecifying that the graph of p(x) has a local maximum at the point (1, 6)and a local minimum at (3, 2).

The problem could have been made a little more interesting byspecifying that p(x) is a cubic polynomial which has a point of inflectionat x = 2 and a local minimum at x = 3. In that case a candidate wouldfirst have to recognise that the local maximum of p(x) must be at x = 1(because, for a cubic, the point of inflection is the arithmetic mean of thepoint of local maximum and the point of local minimum). The rest of thesolution would remain the same.

Q.17 Let S be the focus of the parabola y2 = 8x and let PQ be the commonchord of the circle x2 + y2 − 2x − 4y = 0 and the given parabola. Thenthe area of the triangle PQS is ..... .

Answer and Comments: 4 (square units). A straightforward problemof finding the area of a triangle from the coordinates of its vertices. Inthe present problem, none of the three vertices P,Q, and S is specifieddirectly. So the real task is to determine these first. As the focus of the

30

parabola is at S, we get S = (2, 0). To determine P and Q, we have tosolve the equations

x2 + y2 − 2x− 4y = 0 (1)and y2 = 8x (2)

simultaneously. Substituting (2) into (1) and then solving for y gives

4y = x2 + 6x (3)

Squaring and using (2) again, we get an equation of degree 4 in x, viz.

x4 + 12x3 + 36x2 − 128x = 0 (4)

x = 0 is a solution of this equation. It gives (0, 0) as one of common pointsof the circle and the parabola which is obvious anyway. Let us take it tobe P . The other solutions satisfy the cubic As soon as we determine theother end of the common chord, viz. Q, the problem is almost solved. Butit is not easy to find Q by mere inspection. The x-coordinate of Q mustsatisfy the cubic equation

x3 + 12x2 + 36x− 128 = 0 (5)

By trial we see that x = 2 is a solution of this. (Geometrically, we alreadyknow that there are no other real solutions. For an analytical proof wefactorise the cubic as (x− 2)(x2 + 14x+ 64) and note that the quadraticfactor never vanishes for any real x since 72 − 64 < 0.)Since x = 2, y2 = 8x = 16. So y = ±4. The point (2,−4) does not lieon the circle. So we must have Q = (2, 4). Now that we know all threevertices of the triangle PQS its area is

=1

2

∣

∣

∣

∣

∣

∣

0 0 12 0 12 4 1

∣

∣

∣

∣

∣

∣

= 4 (6)

Determination of Q is the only non-trivial part in the problem. Our solu-tion was based on eliminating one of the variables (viz. y) from (1) and(2) and trial. An alternate (but essentially equivalent) approach is to takeone of the two curves in a parametric form. The parametric equations ofthe parabola t2 = 8x are

x = 2t2, y = 4t (7)

where −∞ < t < ∞. Substitution into (1) gives

4t4 + 16t2 − 4t2 − 16t = 0 (8)

or, in a factorised form

t(t3 + 3t2 − 4) = 0 (9)

31

Once again, 0 is an obvious root (which corresponds to the root 0 of (4)and to the point P = (0, 0) of the common chord). The other roots arethe roots of the cubic factor. As the sum of the coefficients is 0, t = 1 is aroot. The cubic factors as (t− 1)(t2+4t+4) and the quadratic factor hasno roots. So, t = 0, 1 are the only real roots of (8) and the correspondingcommon points are (0, 0) and (2, 4). In this approach too there was someguesswork. But it was not as much a shot in the dark as in (5). Addingthe coefficients of the terms in a polynomial is usually the first thing onedoes to see if it has any obvious non-zero roots.

Instead of the parabola, we can represent the circle in a parametricform. Its center is at (1, 2) and radius

√5, we have

x = 1 +√5 cos θ, y = 2 +

√5 sin θ (10)

Substituting these into the equation of the parabola, we get

(2 +√5 sin θ)2 = 8(1 +

√5 cos θ) (11)

which, after simplification, becomes

5 sin2 θ + 4√5 sin θ − 8

√5 cos θ − 4 = 0 (12)

If the cosine term were absent this would be a quadratic in sin θ. To getrid of the cosine term we can take it on one side and square both the sidesand replace cos2 θ by 1− sin2 θ. But then the equation that results wouldbe a a fourth degree equation in sin θ. So again we would face the samedifficulty.

Yet another way a candidate can hit upon the point Q is if it strikeshim that the ends of the latus rectum of the parabola y2 = 8x are at(2,±4). If he also observes luckily that (1, 2) is the centre of the circle,then he would hardly fail to notice that (2, 4) is the end of the diameterpassing through (0, 0). In particular, it is a common point of the parabolaand the circle.

Summing up, no matter which approach is tried there is an elementof luck in determining the point (Q. The contribution of luck is least indoing it through (9). Once Q is determined, hardly anything is left in theproblem.

Q.18 The value of

6 + log3/2

1

3√2

√

√

√

√

√4− 13√2

√

√

√

√4− 13√2

√

4− 13√2. . .

is ......

32

Answer and Comments: 4. This is a somewhat unusual questionbecause the expression given involves an infinitistic process. But it isdifferent from the way infinite sums are defined. The sum of an infiniteseries is defined as the limit of the sequence of its partial sums. Here eachpartial sum is an ordinary finite sum. In the present problem, however,a number is defined by an infinitistic process in such a manner that apart of that number is defined by the same infinitistic process. Logically,such a definition suffers from what is called a vicious cycle. It is beyondthe scope of the JEE level to consider the question as to whether sucha number indeed exists. (A similar situation arises for what are calledcontinued fractions which were popularly studied at one time.)

We bypass these philosophical questions and proceed to find the value ofthe number, assuming that the expression does define a number uniquely.Denote the number

√

√

√

√

√4− 13√2

√

√

√

√4− 13√2

√

4− 13√2. . . (1)

by x. Then, we get the equation

x =

√

4− 13√2x (2)

Squaring both the sides we get a quadratic in x, viz.

x2 +1

3√2x− 4 = 0 (3)

Solving,

x =− 1

3√2±√

118 + 16

2

=−1± 3

√2√

28918

6√2

=−1± 176√2

(4)

The negative sign is discarded since obviously x is non-negative. So,

x =8

3√2. Completing the solution is now straightforward. The given ex-

pression equals 6+log3/2

(

1

3√2× 8

3√2

)

which simplifies to 6+log3/2(

49

)

.

Since4

9=

(

2

3

)2

= (3/2)−2, the expression equals 6− 2 = 4.

33

Problems like this are not asked frequently. For those who have seensimilar problems earlier, obtaining (2) and hence (3) is rather routine. (Afairly well-known simple problem of this type is to evaluate

√

2 +

√

2 +

√

2 +√2 + . . .

which comes out to be 2.) But in the present problem, even after getting(3), the subsequent work requires considerable numerical calculation in-volving radicals. The candidate has also to realise that 289 is a perfectsquare. A numerical slip can be fatal here.

Q.19 Let f : IR −→ IR be defined as f(x) = |x|+ |x2 − 1|. The total number ofpoints at which f attains either a local maximum or a local minimum is..... .

Answer and Comments: 5. The expression x changes sign at x = 0while x2 − 1 does so at the points ±1. Therefore because of the absolutevalue signs in the definition of f(x), to study its behaviour, we have toseparately consider the intervals (−∞,−1), (−1, 0), (0, 1) and (1,∞). Fur-ther, f(x) is evidently an even function of x. So, it suffices to study itsbehaviour only on the first two of these four intervals.

For x < −1, x is negative but x2 − 1 is positive. So,

f(x) = x2 − x− 1 (1)

This is a quadratic whose derivative vanishes only at x = 1/2 which isoutside the interval (−∞,−1). So, f(x) is strictly decreasing on (−∞,−1).Next we consider f on (−1, 0). Here f(x) = −x+1−x2 which is a quadraticin x with derivative vanishing at x = −1/2. The function is strictlyincreasing on (−1,−1/2) and strictly decreasing on (−1/2, 0). As f is aneven function, this also implies that f is strictly increasing on (0, 1/2) andstrictly decreasing on (1/2, 1). Similarly, it is strictly increasing on (1,∞).

Summing up, f changes its behaviour from increasing to decreasing orvice versa at the points x = −1,−1/2, 0, 1/2 and x = 1. Therefore it hasa local maximum or a local minimum at these five points.

A fairly simple problem. The reason-ing given need not be in words. A goodgraph will give the answer equally easily.In fact the graphic approach has the ad-vantage that one need not determine thepoints ±1/2 exactly. All that matters forthe question is that there are two local max-ima, somewhere in (−1, 0) and in (0, 1).The question is well suited to test the ability to reason correctly withouthaving to express the reasoning in words.

34

Q.20 If ~a,~b,~c are unit vectors satisfying |~a−~b|2 + |~b − ~c|2 + |~c − ~a|2 = 9, then|2~a+ 5~b+ 5~c| is ..... .

Answer and Comments: 3. We first express the lengths in terms ofthe dot products. Thus,

|~a−~b|2 = (~a−~b) · (~a−~b)= |~a|2 + |~b|2 − 2~a ·~b= 2− 2~a ·~b (1)

with similar expressions for |~b − ~c|2 and |~c − ~a|2. Adding these threeexpressions the data means

~a ·~b+~b · ~c+ ~c · ~a = −32

(2)

To use this information we need some other expression where the term~a ·~b+~b · ~c+ ~c · ~a occurs. One such expression is |~a+~b+ ~c|2 which, uponexpansion, becomes

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

The values of all the terms on the R.H.S. are known to us. Putting themwe get

|~a+~b+ ~c|2 = 3− 3 = 0 (4)

This means that the vector |~a+~b + ~c| has zero length. Hence it must bethe zero vector, i.e.

~a+~b+ ~c = ~0 (5)

Therefore, ~b+ ~c = −~a. Substituting this into the given vector gives

2~a+ 5~b+ 5~c = 2~a− 5~a = −3~a (6)

Therefore

|2~a+ 5~b+ 5~c| = | − 3~a| = 3|~a| = 3 (7)

since ~a is a unit vector.

The method applied here works only in an extreme case. In gen-eral, knowing the length of a vector does not determine it uniquely. Thishappens to be the case only when the length is 0. Even after getting(5), if, instead of 2~a+ 5~b+ 5~c we had some other linear combination, say

α~a + β~b + γ~c where the scalars α, β, γ are all distint, we would not be

35

able to find its length. In the present problem it works because two of thescalars are equal.

These two unusual features make the problem a little tricky. Ananalytically minded candidate is likely to give up when he realises thatthe data is, in general, insufficient to give the answer uniquely. If he hasthe perseverance to see that in the present problem, the method goesthrough, then only he gets the answer.

Problems where an equation can be solved only in an extreme caseare not uncommon. Indeed nearly all problems of trigonometric charac-terisations of equilateral triangles are of this kind. (See Chapter 14.) Asa typical example, if ABC is a triangle, then an equation like

cosA+ cosB + cosC = k (8)

does not determine the angles A,B,C uniquely. But the maximum value

k can take is3

2. And when k equals this extreme value, there is a unique

solution, viz. an equilateral triangle. The first part of the present problemcan be looked at as a characterisation of an equilateral triangle. If ABC isa triangle with circumcentreO and circumradius 1 (an essentially arbitrary

choice) and we let ~a =−→OA,~b =

−→OB and ~c =

−→OC, then it is not hard to show

that these vectors satisfy the data in the problem if and only if the triangleABC is equilateral. Questions involving characterisations of equilateraltriangles are not suitable for an MCQ format because here the answer istrivial to guess but not so trivial to prove. A candidate who has studiedsuch characterisations is likely to be prompted by the very symmetry ofthe data to think that the triangle formed by the end points of the threevectors (assuming they have a common origin) is equilateral. Unless he isa scrupulous candidate, he will not bother to prove this. He will just takeit for granted. However, even after getting this, the last part does requirethat two of the scalars in the given linear combination of ~a,~b,~c are equal.

36

PAPER 2

Contents

Section I (Single Correct Choice Type)

Section II (Paragraph Type)

Section III (Multiple Correct Answers)

Single Correct Choice Type

This section contains eight multiple choice questions. Each question has 4choices out of which ONLY ONE is correct. 4 points for a correct answer −1for an incorrect one.

Q.21 Let a1, a2, a3, . . . be in harmonic progression with a1 = 5 and a20 = 25.The least positive integer n for which an < 0 is

(A) 22 (B) 23 (C) 24 (D) 25

Answer and Comments: (D). Problems on H.P.’s are not as commonas those on A.P.’s or G.P.’s. One of the reasons is that unlike in the caseof an A.P. or G.P., there is no handy formula for the sum of an H.P. Theharmonic progressions are defined in terms of arithmetic ones. So thestandard approach to tackle any problem about numbers in an H.P. is toconsider the A.P. of their reciprocals.

So, we let

bi =1

ai(1)

for i = 1, 2, . . . , 25. Then the b’s are in an A.P. We are also given that

b1 =1

5and b20 =

1

25(2)

Let the common difference of this progression be d. Then we get

b20 = b1 + 19d (3)

Substituting the values from (1), this gives an equation for d, biz.

19d =1

25− 1

5= − 4

25(4)

which gives d = − 4475

. Therefore,

bn = −4(n− 1)

475+

1

5=

96− 4n475

(5)

37

Clearly, an and bn have the same signs. So, an and hence bn is negative ifand only if

96 < 4n (6)

The least n for which this happens is n = 25.

A very straightforward problem. The problem is more about A.P.’sthan H.P.’s. But had it been asked about A.P.’s it would be too easy. Sothe paper-setters have given it a slight twist.

Q.22 Let α(a) and β(a) be the roots of the equation

( 3√1 + a− 1)x2 + (

√1 + a− 1)x+ ( 6

√a+ 1− 1) = 0

where a > −1. Then lima→0+

α(a) and lima→0+

β(a) are

(A) −52and 1 (B) −1

2and −1 (C) −7

2and 2 (D) −9

2

Answer and Comments: (B). The problem involves a novel combina-tion of the roots of a family of quadratic equations, parametrised by aparameter (viz. a here) and their limits as this parameter tends to somevalue. Such combinations are often baffling to the average candidates. So,the problem is a good test of a candidate’s ability to correctly understandwhat the problem is all about.

An honest approach to solve the problem is to first express α(a)and β(a) as functions of a and then take the limits of these functions asa → 0+. The various fractional powers make the expression look ugly. Sowe make a suitable substitution to make the expression look a little moremanageable. We put u = 6

√a+ 1. Then the quadratic becomes

(u2 − 1)x2 + (u3 − 1)x+ (u− 1) = 0 (1)

whose roots are

x =(1− u3)±

√

(u3 − 1)2 − 4(u2 − 1)(u− 1)2(u2 − 1) (2)

If we want, we could express these in terms of the original parameter a.A better way out, instead, is to work with u instead of a while finding thelimits of these roots. As a → 0+, u → 1+. So, our task is to find the limitsof the two expressions in (2) as u → 1+. This can be done easily evenwithout using the all time favourite l’Hôpital’s rule. We simply cancel thevanishing factor (u− 1) and get

x =−(u2 + u+ 1)±

√

(u2 + u+ 1)2 − 4(u+ 1)2

(3)

38

Then the limits are−3±

√9− 8

2, i.e. −1

2and−1. (The data is insufficient

to decide which one is the limit of α(a) and which one β(a). Nor does theproblem ask it.)

This gives the limits of the roots of the quadratic (1) as the parameteru tends to 1+. But the operation of a limit commutes with many otheroperations in mathematics. That is why we have theorems which can beverbally expressed by statements like ‘A limit of a sum is the sum of thelimits’ or ‘The limit of a quotient is the quotient of the limits providedthe limit of the denominator is non-zero’. We also have theorems whichassert that the limit of the sine is the sine of the limit, or more precisely,if lim

x→cf(x) = L, then lim

x→csin(f(x)) = sinL. Indeed this is equivalent to

saying that the sine function is continuous (at L). More generally theresult holds if we replace the sine function by any continuous function.But the function has to be single valued. (Otherwise it is not a functionat all!)

If we allow an extension of such results even when we have multi-valued functions, then we can give a shorter solution of the problem. Herethe roots of the quadratic (1) form a bivalued function of the parameteru. We identified them separately and found their limits as u → 0+. Wecan as well interchange the two processes. That is, we first take the limitof the quadratic (1) as u → 0+ and then find the roots of this limitingquadratic. If we do this indiscriminately, the limiting quadratic comes outto be 0x2 + 0x + 0 = 0. To avoid this degeneracy, we replace (1) by anequivalent quadratic valid for u 6= 1, viz.

(u+ 1)x2 + (u2 + u+ 1)x+ 1 = 0 (4)

We now let u → 1. Then the limiting quadratic is

2x2 + 3x+ 1 = 0 (5)

whose roots are−2±

√9− 8

4, i.e. −1

2and −1.

This alternate approach is certainly a short cut. But a rigorous proofof its validity is beyond the JEE level. So, once again a candidate whojust assumes its validity gets rewarded.

Q.23 Four fair dice D1, D2, D3 and D4 each having six faces numbered 1,2,3,4,5and 6 are rolled simultaneously. The probability that D4 shows a numberappearing on one of D1, D2 and D3 is

(A)91

216(B)

108

216(C)

125

216(D)

127

216

Answer and Comments: (A). Superficially, this is a problem on proba-bility. But since the desired probability is merely the ratio of the number

39

of favourable cases to the total number of cases, in reality it is a countingproblem. As the figures shown by the four dice are independent of eachother the total number of cases is 64.

To find the number of favourable cases, let xi denote the figureshown by Di for i = 1, 2, 3, 4. We want the number of ordered quadruples(x1, x2, x3, x4) with 1 ≤ xi ≤ 6 (i = 1, 2, 3, 4) for which x4 equals at leastone of x1, x2, x3. We resort to complementary counting, i.e. we countthe number of (ordered) quadruples (x1, x2, x3, x4) in which x4 is differentfrom all x1, x2, x3 and then subtract this number from 6

4, the total numberof quadruples. Here x4 can take six possible distinct values. For any fixedvalue of x4, the remaining three x1, x2, x3 are free to take any of the fiveremaining values independently of each other. This they can do in 53

ways. So the number of unfavourable cases is 6×53. Hence the number offavourable cases is 64−6×53, i.e. 6× (63−53) = 6× (216−125) = 6×91.

Therefore the desired probability is6× 9164

=91

216.

Instead of using complementary courting, the number of favourablecases could also have been obtained directly. But this resolves into severalcases depending upon how many and which of x1, x2, x3 equals x4. Thecomplementary counting gives the answer is one shot. So, this is a simpleproblem which rewards those who get the idea of using complementarycounting.

The language used in the question can be confusing. Some candidatesare likely to interpret ‘one’ as exactly one. What is intended here is ‘atleast one’. Fortunately, in this problem the wrong interpretation would

give the answer as75

216which is not one of the alternatives. So that will

alert the candidate, but only after wasting some precious time. The bestthing is to word the question so as to leave no ambiguity.

Q.24 Let PQR be a triangle with area ∆ with a = 2, b =7

2and c =

5

2, where

a, b, c are the lengths of the sides of the triangle opposite to angles at P,Q

and R respectively. Then2 sinP − sin 2P2 sinP + sin 2P

equals

(A) =3

4∆(B) =

45

4∆(C) =

(

3

4∆

)2

(D) =

(

3

4∆

)2

Answer and Comments: (C). Before tackling the problem, it is wellto comment the entertaining part of the notations used in its statement.In the good old days, the triangle would invariably be taken as ABCinstead of PQR and then nobody needed to be told what a, b, c stood for.The notation ∆ for the area of the triangle is also standard. When theJEE became an MCQ test, this time honoured practice was changed, therationale apparently was that since the alternatives to the question are

40

also marked as (A), (B), (C), (D), the use of the same symbols to denoteanything else could confuse the candidates. It is doubtful if anybody wouldreally get confused for this reason!

Now coming to the problem itself, let us first simplify the expressionto be evaluated.

2 sinP − sin 2P2 sinP + sin 2P

=1− cosP1 + cosP

=2 sin2 P22 cos2 P2

= tan2P

2(1)

There are numerous formulas for the various trigonometrical functionsof the angles of a triangle in terms of its sides. In the present problem,there is an implicit hint that we should choose a formula which involvesthe area ∆. (The sides are given and they determine ∆ numerically. So,there must be some purpose in giving all the alternatives in terms of ∆rather than as numbers.) One such formula is

tanP

2=

√

(s− b)(s− c)s(s− a) (2)

where s is the semi-perimeter1

2(a+ b+ c). Combining this with another

well-known formula for the area, viz.

∆ =√

s(s− a)(s− b)(s− c) (3)

we get

tanP

2=

(s− b)(s− c)∆

(4)

Squaring and putting b =7

2, c =

5

2and s =

1

2

(

2 +7

2+

5

2

)

= 4, we get

tan2P

2=

(1/2)2(3/2)2

∆2=

(

3

4∆

)2

(5)

A straightforward problem once you get an idea which formula to use.The hint for this is given in a subtle and an imaginative manner.

Q.25 If P is a 3× 3 matrix such that PT = 2P + I, where PT is the transposeof P and I is the 3× 3 identity matrix, then there exists a column matrix

X =

xyz

6=

000

such that

(A) PX =

000

(B) PX = X (C) PX = 2X (D) PX = −X

41

Answer and Comments: (D). As P (and hence also PT ) is a 3 × 3matrix, the single matrix equation

PT = 2P + I (1)

is equivalent to a system of 9 equations obtained by matching the corre-sponding entries of the matrices on both the sides. But to try to deter-mine P this way would be cumbersome. Instead we use the fact that everysquare matrix can be expressed uniquely as a sum of a symmetric and askew symmetric matrix. So, suppose

P = A+B (2)

where A is symmetric and B is skew symmetric. Then

PT = AT +BT = A−B (3)

So, (1) becomes

A−B = 2A+ 2B + I (4)

We rewrite this as

−A− I = 3B (5)

The L.H.S. is a symmetric while the R.H.S. is a skew symmetric matrix.Hence both the sides equal the zero matrix of order 3. This gives

B = 0 and A = −I (6)

which combined with (2) gives

P = −I (7)

As a result, for every 3× 1 matrix X ,

PX = −X (8)

Therefore (D) is true. In fact what we have proved is much stronger than(D). (D) only requires some non-zero 3×1 matrix X for which PX = −X .Those familiar with eigenvalues will recognise (D) to mean that P has −1as an eigenvalue. There are methods for finding eigenvalues of a squarematrix using determinants. But in this problem this is of little help.

Q.26 If ~a,~b are vectors such that |~a+~b| =√29 and

~a× (2î+ 3ĵ + 4k̂) = (2î+ 3ĵ + 4k̂)×~b

then a possible value of (~a+~b) · (−7î+ 2ĵ + 3k̂) is

42

(A) 0 (B) 3 (C) 4 (D) 8

Answer and Comments: (C). For brevity, denote the vector 2î+3ĵ+4k̂by ~v. Then the given condition is

~a× ~v = ~v ×~b (1)

The R.H.S. equals −~b× ~v. So (1) gives

(~a+~b)× ~v = ~0 (2)

which is equivalent to saying that ~a+~b is a scalar multiple of ~v, say

~a+~b = λ~v (3)

Taking lengths of both the sides

|~a+~b| = |λ||~v| (4)

The L.H.S. is given as√29. By a direct calculation |~v| also comes out

to be√4 + 9 + 16 i.e.

√29. So (4) gives |λ| = 1 and hence λ = ±1.

Therefore

~a+~b = ±(2î+ 3ĵ + 4k̂) (5)

and hence

(~a+~b) · (−7î+ 2ĵ + 3k̂) = ±(2î+ 3ĵ + 4k̂) · (−7î+ 2ĵ + 3k̂)= ±(−14 + 6 + 12) = ±4 (6)

So there are two possible values of the given expression and one of themis 4.

The problem is a simple but good test of the knowledge of the basicproperties of the cross product, specifically, its anti-commutativity andthe condition for its vanishing. It is a pure coincidence that the valueof |~a + ~b| is given to be the same as |~v|. Apparently, the paper-setterswanted to keep the computations simple and thereby reward a candidatewho gets the conceptual part correctly. By using the words ’a possiblevalue’ instead of merely saying ‘a value’ the paper-setters have made itabundantly clear that there may be more than one correct answers butonly one of them is asked. This leaves no room for confusion in the mindof the candidate.

Q.27 The equation of a plane passing through the line of intersection of the

planes x+ 2y + 3z = 2 and x− y + z = 3 and at a distance 2√3from the

point (3, 1,−1) is

43

(A) 5x− 11y + z = 17 (B)√2x+ y = 3

√2− 1

(C) x+ y + z =√3 (D) x−

√2y = 1−

√2

Answer and Comments: (A). The family of planes passing throughthe line of intersection of two given planes is a 1-parameter family and theequation of any member of it can be written by combining the equationsof these two planes. Thus, in the present problem, the desired plane willhave an equation of the form

(1− λ)(x + 2y + 3z − 2) + λ(x − y + z − 3) = 0 (1)

for some real number λ. (See Chapter 9, Comment No. 13 for a discussionabout the analogous result in the plane for the equation of a line passingthrough the point of intersection of two given lines. Often the coefficient(1− λ) is dropped. But the danger in doing so is that we miss the secondplane and if it happens to be a right answer, we miss it too. So we workwith (1).) Rewriting it,

x+ (2− 3λ)y + (3− 2λ)z − 2− λ = 0 (2)

The distance of this plane from the point (3, 1,−1) is given to be 2√3.

This gives us an equation for λ, viz.∣

∣

∣

∣

∣

3 + 2− 3λ+ 2λ− 3− 2− λ√

1 + (2 − 3λ)2 + (3− 2λ)2

∣

∣

∣

∣

∣

=2√3

(3)

Squaring and simplifying

λ2

13λ2 − 24λ+ 14 =1

3(4)

This becomes a quadratic in λ, viz.,

10λ2 − 24λ+ 14 = 0 (5)

whose roots are

λ =12±

√144− 14010

(6)

Hence λ = 1 or7

5. The first possibility does give the second plane, viz.

x − y + z = 3 as a possible answer. The second value, viz., λ = 75 , afterputting into (2) gives the plane

x− 115y +

1

5z − 17

5= 0 (7)

44

which is the plane in (A). Technically, the problem is not wrong becauseit asks for the equation of a plane with certain properties and the planein (A) meets the prescription. But in reality there are two such planes. Itwould have been better to ask this question as an MCQ where more thanone alternatives may be correct and also give the equation of the secondplane as one of the alternatives.

What is really disturbing about this problem is that the majority ofthe candidates would begin the solution by taking the parametrised familyof planes as

(x + 2y + 3z − 2) + λ(x− y + z − 3) = 0 (8)

where λ is a real parameter. With this choice the resulting equation in λcomes out to be

3λ2 = 3λ2 + 4λ+ 14 (9)

which is a linear equation, with λ = −72as the only solution. This solution

gives the plane in (A) correctly. But an analytically minded candidatewould realise that unless d = 0, there are two planes in the given familywhich are at a distance d from a given point. He will then wonder whichis the other plane. It is only when he realises that the second of the giventwo planes is also a right answer, that he would identify that the villainwas to use (8) instead of (1) which is the right thing.

It is hard to tell whether this possibility eluded the paper-setterstoo. Probably it did, as otherwise, they would have asked the questiondifferently. That is, they would have made it abundantly clear that oneof the answers given is only one of the possible answers to the problem,as they did in framing the last question. As the problem stands, thosewho miss the subtle difference between (1) and (8) get unduly rewarded.Ignorance is bliss!

Q.28 The value of the integral

∫ π/2

−π/2

(

x2 + lnπ + x

π − x

)

cosx dx is

(A) 0 (B)π2

2− 4 (C) π

2

2+ 4 (D)

π2

2

Answer and Comments: (B). When the integrand of a function splitsas the sum of two functions, the integral also splits as the sum of twointegrals. In the present case we can write the integral, say I as

I = I1 + I2 (1)

where

I1 =

∫ π/2

−π/2x2 cosx dx (2)

45

and I2 =

∫ π/2

−π/2

(

lnπ + x

π − x

)

cosx dx (3)

Occasionally, in some tricky problems on integration, it is not a good ideato split the integrand. Instead, it is easier to evaluate the integral as awhole by some trick. See for example, the JEE 1983 problem in CommentNo. 24 of Chapter 17. But the present integral is not of this tricky type.Here I1 can be evaluated by a standard application of integration by parts.So, the evaluation of I depends more on that of I2. Here no matter whichof the two factors of the integrand is taken as the first function, integrationby parts will only lead to more complicated integrands. No substitutionsuggests itself either. So, it is futile to try to evaluate I2 by finding anantiderivative of the integrand. Nor is it necessary. The integral askedis a definite integral and as pointed out in Chapter 18, there are ways toevaluate a definite integral taking advantage of some special features ofthe problem such as the symmetry of the interval around 0 or some specialproperty of the integrand.

In the present problem, the interval of integration is symmetric

about the point 0. Further for any x in this interval,π + x

π − x andπ − xπ + x

are reciprocals of each other and so their logarithms are negatives of eachother. Therefore the first factor of the integrand in (3) is an odd function.The second factor, cosx, is an even function. So the integrand of I2 isan odd function over the interval [−π/2, π/2] which is symmetric about 0.Thus we get

I2 = 0 (4)

without any computation. As for I1, the integrand is an even function ofx. So again by the symmetry of the interval of integration, we have

I2 = 2

∫ π/2

0

x2 cosx dx (5)

The integral on the R.H.S. can be evaluated by two standard applicationsof integration