Educative Commentary on Joint Entrance Examination 2015

8/17/2019 Educative Commentary on Joint Entrance Examination 2015

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EDUCATIVE COMMENTARY ON

JEE 2015 ADVANCED MATHEMATICS PAPERS(Revised on June 30, 2015)

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 positionsof 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 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 have

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

So, in effect, the JEE (Advanced) from 2013 took the role of the JEE in thepast except that the candidates appearing for it are selected by a procedureover which the IITs have no control. So, this arrangement is not quite the


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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 = 4 x withrespect to the line x+ y+4 = 0 . If A and B are the points of intersection

of C with the line y = −5, then the distance between A and B isAnswer 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 rst identify C which is given

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to be the reection of the parabola y2 = 4 x in the line x + y + 4 = 0.

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

2 = 4 x0∗2. The

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

A better approach is to realise that under reections 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 reection, 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 = 4 x. 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 = 4 x 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 reections 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 tonding 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 much

easier to nd 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 respectiveprobabilities are (

12

)n and n × 12 × (

12

)n−1 = n( 12

)n . Together, q =

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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 of reections 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 thequeue. Then the value of

mn

is

Answer and Comments: 5. To nd n think of the 5 girls linedtogether as a single object. Then the number of ways to arrange this

object 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 nd 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 of arrangements 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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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 direct

approach above is far better.

Q.45 Let f : IR −→IR be a function dened by f (x) = [x], x ≤ 20, x > 2

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

−1xf (x2)

2 + f (x + 1) dx,

then the value of (4 I −1) isAnswer 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 = √ 2and 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 themand add the four integrals. We do so one-by-one. Call the integrandas g(x) in all cases and denote the four integrals by I 1, I 2, I 3 and I 4respectively.

On both the intervals [ −1, 0] and [0, 1] 0 < x 2 < 1 except possiblyat the end-points and so f (x2) = [x2] = 0. Hence the rst 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 integrandg(x) simplies to

12

xf (x2). On [1, √ 2] this becomes x2

while on [√ 2, 2],it vanishes. So the whole integral I is merely

√ 21

x2

dx = x2

4√ 21

= 14

.Therefore 4I −1 = 0.

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A tedious problem which tests little more than some carefulness on

the part of the candidate. The integration part itself is trivial.Q.46 A cylindrical container is to be made from certain solid material with

the following constraints : It has a xed 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 containeris 10 mm, then the value of

V 250π

is

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 = πr 2h (1)

As V is xed, 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) −πr 2h= π[(r 2 + 4 r + 4)( h + 2) −r 2h]= π[h(4r + 4) + 2( r + 2) 2]

= V

r 2(4r + 4) + 2 π(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. So dW

dr must vanish

at r = 10. By a direct calculation,dW dr

= −4V r 2 −

8V r 3

+ 4π(r + 2) (3)

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As this vanishes for r = 10, we get

4V 100

+ 8V 1000

= 48π (4)

which simplies to 48V 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 simplies thework. If we were asked to nd where W is minimum, we would haveto solve a cubic equation and then hunt for the minimum among the

critical 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 those

who 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

x2cos2 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 of the region bounded by x = 0, y = 0, y = f (x) and x = a, then f (0) isAnswer 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 dened 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)

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where the R.H.S. needs to be calculated from the rst part of the

problem. (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 nd F ′′(0).To nd F ′′(x), we need to differentiate F (x) twice. The functionF (x) is dened by an integral in which both the upper and the lower

limits are functions of x. Therefore, by the generalised form of thesecond fundamental theorem of calculus,

F ′(x) = 2 cos2(x2 + π/ 6) ddx

(x2 + π/ 6) −2cos2 x ddx

(x)

= 4 x cos2(x2 + π/ 6) −2cos2 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 cos 2 x is one such term since itwill involve a factor sin x. As for the derivative of the rst term, viz.4x cos2(x2 + π/ 6), when we apply the product rule, the factor 4 x 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) =√ 32

.

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 +cos2t. But such a simplication has no role in the solution. In thisrespect, the problem is a bit tricky too.

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Note that the second term (viz. 2) on the R.H.S. of (1) plays no

role 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 of the 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 theproblem 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 to6

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 nd the value of f (0) + k. That would have forcedthe candidates to determine k as 2 from (1) and (3). The answer to

the problem would have been 5 instead of 3, still a single digit number.Q.48 The number of distinct solutions of the equation

54

cos2 2x + cos 4 x + sin 4 x + cos6 x + sin 6 x = 2

in the interval [0 , 2π] is

Answer and Comments: 8. Supercially, this is a problem of solvinga trigonometric equation. But there is no way to do this unless we rstsimplify the expression, say E on the L.H.S. The key idea is to notethat all terms are expressible in terms of sin 2 x and cos2 x. So the goodold identity sin 2 x +cos 2 x = 1 may be useful. Indeed, if we square this,we get

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

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

sin6 x + cos6 x = 1 −3sin4 x cos2 x −3sin2 x cos4 x= 1 −3sin2 x cos2 x(sin2 x + cos2 x)= 1 −3sin2 x cos2 x (2)

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

E = 54

cos2 2x + 2 −5sin2 x cos2 x (3)

and so the equation simplies to

cos2 2x −4sin2 x cos2 x = 0 (4)or equivalently,

cos2 2x −sin2 2x = 0 (5)and still further to

cos4x = 0 (6)

which has eight solutions in the interval [0 , 2π]. Specically, 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 of the basic identity for sin 2 x + cos2 x strikes. See Comment No. 14 of Chapter 7 for a problem where a similar trick is used.

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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 rst order linear differential equation. Normally,one would begin by recasting the equation in the standard form y′ + p(x)x = q (x) and nd 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 + x1 + ex

(2)

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

dydx

.

dydx

= (1 + ex ) −(4 + x)ex

(ex + 1) 2 (3)

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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 continuouseverywhere. Also g(−1) =

2e −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

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

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 ndingthe 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 = r 2 (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 with

the 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 of

y′′. But since these terms get cancelled, this ne thinking is not testedanyway.

Q.51 Let g : IR −→ IR be a differentiable function with g(0) = 0, g′(0) = 0and g′(1) = 0. Let f (x) =

x

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

and 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 dened) 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 toconsider whether limx→0

e|x |

−e0

x exists. For x = 0 we rewrite this ratioas

e|x| −e0x

= e|x | −1

|x| × |x|

x (1)

As x → 0, the rst 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

−1

as 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 > 0

0, x = 0

−g(x), x < 0(2)

Therefore, f ′+ (0), i.e. the right handed derivative of f at 0, will equalg′+ (0) which is 0 since it is given that g′(0) = 0. Similarly, f ′−(0) willequal −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) = 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 denition,(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 = 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 that g(x)

x → 0 as x → 0, it follows

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

Since sin x maps IR onto [−1, 1], π2 sin x maps it onto [−π2 , π2 ]. The sinefunction maps this interval onto [ −1, 1] and so

π6

sin(π2

sin x)) has range

[−π6

, π6

]. Under the sine function this interval goes to [ −1/ 2, 1/ 2] whichproves (A). In (B), we rst have to nd the range of g which is [−

π2

, π2

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

2π

. As the sine functionis 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’Hˆ opital’srule. But a better way out is to put u = g(x) =

π2

sin x and v = π6 sin u.

Then u →0 and v → 0 as x →0. Therefore,f (x)g(x)

= sin( π6 sin u)

u

= sin( π6 sin u)

π6 sin u ×

π6 sin u

u

= sin v

v × π6 ×

sin uu

(1)

Both the rst and the last factors tend to 1 and so the limit is π6

.

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 △P QR be a triangle. Let a = −→QR, b = −→RP and c = −→P Q. 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|2

2 +

|a

|= 30

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

b

Q

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 P QR is isosceles with P Q = P R.

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Let the angle between these sides be θ. Then θ is the angle between

the vectors − b and c. So,cos θ = − b · c

| b| |c|= −24

48 = −

12

(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|cos150◦ = −12 ×4√ 3 ×√ 32

= −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= |2c× 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 OP Q = OP R = 60◦ we have OP =

12

P Q =

2√ 3. This allows us to express all the three vectors a, b and c as linearcombinations 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)21


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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

α 2 + β 2 = (4 √ 3)2 = 48 (13)Similarly, b · c = 24 gives

−α(α + 12) −β 2 = 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 signicant 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 3Z 4 −Z 4Y 3 (B) X 44 + Y 44(C) X 4Z 3 −Z 3X 4 (D) X 23 + 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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Specically, 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 X 44, Y 44 and hencetheir sum symmetric and similarly X 23 + Y 23 skew-symmetric. Notethe analogy of these properties with the properties about the signs of real 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 of symmetric 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 = B T AT and (AT )T = A, for all A, B.)

With these rules in mind, we have

(Y 3Z

4

−Z 4Y

3)

T = ( Y

3Z

4)

T

−(Z 4Y

3)

T

= ( Z 4)T (Y 3)T −(Y 3)T (Z 4)T = −Z 4Y 3 + Y 3Z 4 (1)

because Y 3 is skew-symmetric and Z 4 is symmetric. So, Y 3Z 4 −Z 4Y 3is symmetric. If it were to be skew-symmetric too, it would have tovanish, which means that Y 3 and Z 4 must commute with each other.As this is not given, we discard (A).

By an analogous computation,

(X 4Z 3 −Z 3X 4)T = ( Z 3)T (X 4)T −(X 4)T (Z 3)T = Z 3X 4 −X 4Z 3since Z 3 and X 4 are both symmetric. This shows that X 4Z 3 −Z 3X 4is 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 a

possibility 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 Z 4 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) 4Answers 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 + α 2 1 + 4α + 4α 2 1 + 6α + 9α 2

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

2 2 2(2)

We now add the last row to the rst to get

D =α 2 4α 2 9α 2

3 + 2α 3 + 4α 3 + 6α2 2 2

(3)

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

D =α 2 3α 2 8α 2

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(12α 3 −16α 3) = −8α 3 (5)So the given equation, viz. D = −648α reduces to α3 = 81 α 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 ve 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. So

this is a good, classic type problem on determinants.There is an alternate, albeit a trickier solution to the problem.

We rst expand all the squares and then write the determinant as theproduct of two determinants, viz.

D =1 + 2α + α 2 1 + 4α + 4 α 2 1 + 6α + 9 α 24 + 4α + α 2 4 + 8α + 4 α 2 4 + 12α + 9α 29 + 6α + α 2 9 + 12α + 4α 2 9 + 18α + 9α 2

=1 1 14 2 19 3 1

1 1 12α 4α 6αα 2 4α 2 9α 2

= 2 α 31 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 rst and subtracting the rst 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 rst and the last column of the rst de-terminant and then take its transpose, we get the second determinant.)

So, we nally 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 ) = 2a2 a 1b2 b 1c2 c 1

1 1 1x y z x2 y2 z 2

(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 rst and the last columns of the rst 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

cos A sin A 0cos B sin B 0cos C sin C 0

cos P cosQ cosRsin P sin Q sin R

0 0 0(12)

Q.56 In IR 3, consider the planes P 1 : y = 0 and P 2 : x + z = 1. Let P 3 be aplane, different from P 1 and P 2 which passes through the intersectionof P 1 and P 2. If the distance of the point (0 , 1, 0) from P 3 is 1 and thedistance of a point ( α,β,γ ) from P 3 is 2, then which of the following

relations is (are) true ?(A) 2α + β + 2γ + 2 = 0 (B) 2 α −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 of the planes P 1 and P 2. Write the equations of P 1 and P 2 in the formE 1 = 0 and E 2 = 0 where E 1, E 2 are linear expressions in x,y,z .Then the equation of every plane passing through L is of the formλE 1 + µE 2 = 0 for some values of the parameters λ and µ. We can

dispense with one of the parameters, say µ and consider an equation of the form λE 1 + E 2 = 0. This will represent all possible planes throughL for various values of λ, except the plane P 1. (For the plane P 1,we need λ = 1 and µ = 0.) Similarly, the equation E 1 + µE 2 = 0will represent all planes through L except P 2. In the present case weare given that P 3 is different from both P 1 and P 2. So we are free totake either approach. (See Comment No. 13 of Chapter 9) for moreexamples of this technique.)

We take E 1 as y, E 2 as x + z −1. Then the equation of P 3 is of theformx + 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 simlies to

(λ −1)2 = λ2 + 2 (3)and determines λ as −

12

. So the equation of the plane P 3 is x + z −1 −

12

y = 0 or, equivalently,

2x −y + 2 z −2 = 0 (4)We are further given that the distance of the point ( α,β,γ ) from P 3 is1. This means

2α + β + 2γ −2√ 4 + 1 + 4 = 2 (5)This means

2α −β + 2 γ −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 P 1itself and is discarded by the data. Since we took the equation of P 3in the form λE 1 + E 2 = 0, which represents all planes containing Lexcept P 1, (2) degenerated into a linear equation in λ. Had we takenthe equation of P 3 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 + 2 µ = 0, having 0 and −2 asits roots. The root µ = 0 gives the plane P 1 and has to be discarded.28


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The second root, viz.

−2 will give the equation of the plane P 3 as

2x − 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 λE 1 + E 2 = 0 rather than with E 1 + µE 2 = 0. A sharp studentwill, however, not leave this choice to luck. He will observe that P 1 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 P 3 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 easier

option not so tempting, then this would have been a good problemwhich rewards the sharp candidates.

Q.57 In IR 3, let L be a straight line passing through the origin. Supposethat all the points on L are at a constant distance from the two planesP 1 : x + 2y −z + 1 = 0 and P 2 : 2x −y + z −1 = 0. Let M be thelocus of the feet of the perpendicular drawn from the points on L tothe plane P 1. 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 rst part of the data simplymeans that the line L is parallel to both P 1 and P 2 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 + 2 j

−k and

v = 2i − j + k are normals to the planes P 1, P 2 respectively.) Since Lpasses through the origin, its parametric equations arex = t, y = −3t, z = −5t (3)

where t is a parameter.

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As L is parallel to the plane P 1, the feet of the perpendiculars from

the points on L to the plane P 1 will form a line M parallel to L. Todetermine M , it suffices to know any one point P 0 = ( x0, y0, z 0) 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 P 1. To determine this point we needthree equations in x0, y0, z 0. one of them comes of course, from theequation of the plane P 1. That is

x0 + 2 y0 −z 0 = −1 (4)The other two equations come from the fact that the vector −→OP 0=x0i + y0 j + z 0k is perpendicular to P 1 and hence parallel to the normalvector u = i + 2 j −k . This gives

x0 = r, y 0 = 2r, z 0 = −r (5)for some real number r. Substituting this into (4), we get r = −1/ 6.Hence P 0 = (−1/ 6, −1/ 3, 1/ 6). Therefore the locus M is the line

x + 1 / 61

= 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/ 6

from x + 1/ 6 = λ. This value also satises 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. nding 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 = 2 x such thata circle with P Q as a diameter passes through the vertex O of the

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parabola. If P lies in the rst quadrant and the area of the triangle

△OP Q 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 P Q as a diameter. So instead of saying ‘a circle with P Q as a diameter’, the wording should have been ‘the circlewith P Q 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 are 2t1

and 2t2

respectively. As O lies

on the circle with P Q as a diameter, we have OP ⊥ OQ which means4t1t2

= −1 and hencet1t2 = −4 (3)

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

OP Q is right angled at O, we get

3√ 2 = 12 ×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 givest1 −t2 = ±3√ 2 (7)

We are given that P is in the rst 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 nd t1 by solving theequation

t1 + 4t1

= 3√ 2 (8)which becomes a quadratic

t21 −3√ 2t1 + 4 = 0 (9)

The roots are 3√ 2 ±√ 2

2 i.e. 2√ 2 and √ 2. The corresponding point

P = (t21

2 , 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 contains TWO questions. Each question contains two columnsColumn I and Column II . Column I has four entries (A), (B), (C) and(D). Column II has ve 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 IR 2, if the magnitude of the projection vectorof the vector αi + β j 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) = −3ax 2 −2, x < 1bx + a2, x ≥ 1 is differen-

tiable for all x ∈ IR , then the possible value(s)of a are(C) Let ω = 1 be a complex cube root of unity. If (3 −3ω + 2 ω2)4n +3 +(2+3 ω−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 havingve 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 √ 32 i + 12 j. Hence the projectionof αi + β j has magnitude |

√ 32

α + 12

β |. Equating this with √ 3 andsquaring gives

(√ 3 α + β )2 = 12 (1)33


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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 (4 n + 3)-th powers of threequadratic expressions in ω, specically,

E 1 = 3 −3ω + 2 ω2 (6)E 2 = 2 + 3 ω −3ω2 (7)

and E 3 = −3 + 2ω + 3 ω2 (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, nding theactual relationship is easy. Since ω3 = 1, ω4 = ω, ω5 = ω2 we see that

E 1 = ωE 3 and E 2 = ω2E 3 (9)

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So the given equation simplies to

(ω4n +3 + ( ω2)4n +3 + 1) E 4n +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 = 1.) So E 3 = −3 + 2 ω−3−3ω =−6 −ω which is non-zero. Hence E 4n +33 is also non-zero. So, only therst factor of the L.H.S. of (10) vanishes. Further, since ω3 = 1, ω3nand (ω2)3n also equal 1 each. So the equation simplies 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 rst 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 = 2 q (14)

Eliminating q ,

b = 2(10 −a) −5 = 15 −2a (15)Putting this into (12) we get

a(15 −2a) = 2 a + 30 −4a (16)which simplies to a quadratic in a, viz.

2a2 −17a + 30 = 0 (17)

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whose roots are 17±7

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

and 15/ 2. So |q −a| = 2 in the rst case and |q −a| = 5 in the second.(C) is the only interesting problem in the bunch. The others reduce

to 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, Y and Z respectively. If 2(a

2

− b2

) = c2

andλ =

sin(X −Y )sin Z

, then possible values of n forwhich 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, Y and Z , respectively. If 1 + cos2X − cos2Y =2sin X sin Y , then possible value(s) of

ab

is (are)

(C) In IR 2, let √ 3 i + j, i + √ 3 j and β i + (1−

β ) j be

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

3√ 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 = 4 x and y =

|αx −1|+ |αx −2|+ αx , where α ∈ {0, 1}. Thenthe value(s) of F (α) +

83√ 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(sin2 X −sin2 Y ) = sin 2 Z (1)By a well-known identity the L.H.S. factorises and we get

2 sin(X −Y ) sin(X + Y ) = sin 2 Z (2)which further simplies to

sin(X −Y )sin Z

= 12

(3)

since in any triangle △XY Z , sin(X + Y ) = sin( π −Z ) = sin Z . So weget λ = 12

and 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 nd the ratio ab

which equals sin X sin Y

. Let us rst recast the given

condition in terms of sin X and sin Y .1 + cos 2X −2cos2Y = 2 cos2 X −2(1 −2sin2 Y )

= (2 −2sin2 X ) −2 + 4 sin2 Y = 4sin 2 Y −2sin2 X (5)

Hence the condition given translates as

2sin2 Y −sin2 X = sin X sin Y (6)Dividing by sin2 Y and calling

sin X

sin Y as 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 supercially. The rst part

of 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 adiagram it is obvious that the points X 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 XOY .We are given that the distance of Z from this line y −x = 0 is

3√ 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 and

the other for α = 1. The rst one is easier because in this case y = 3and so, F (0) is the area bounded by x = 0, x = 2, y2 = 4 x and y = 3.It is the shaded area OABCO in the gure 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)38


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For x

∈ [0, 2],

|x

−2

|= 2

−x and so (10) simplies 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 gure 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 nding the area of aplane region, the rst half of the problem (where α = 0) would have

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served the purpose quite well. Adding one more part which hardly

tests 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 rst seven terms to thesum of the rst eleven terms is 6 : 11 and the seventh term lies between

130 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 some

specic 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 rst n terms(often denoted by S n ) is

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

2 d (1)

We are given that S 7S 11 = 611 . Because of (1) this means

7a + 21 d11a + 55 d

= 611

(2)

which yields 7a + 21 d = 6a + 30 d 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 supercially 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 100

terms of the form u1u2 . . . u 100 where each ui has two possibilities , either1 or xi , for i = 1, 2, . . . , 100. We can combine these two possibilitiesby saying that ui = xn i where the exponent ni is either 0 or i. So,u1u2 . . . u 100 will equal un 1 + n 2 + ... + n 100 . 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 nding 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 of terms in the summation. There is only one way to express 9 as a sumof a single integer, viz. 9 = 9. To nd 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 nally 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 of distinct 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 −1k 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 polynomial

1 + x + x2 + . . . + xk , let us take the entire innite 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 innite series

1+ x + x2 + . . . + xk + . . . can be identied as the geometric series 11 −x

.

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

in 1

1 −xn

. 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 where

the 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 be n(n + 1) . . . (n + k −1)

k! which is nothing but the binomial

coefficientn + 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 between

pure geometry and coordinate geometry. Pure geometry solutions areelegant but sometimes tricky. When coordinates are introduced thepowerful machinery of algebra makes the problem more amenable, if somewhat 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 algebraic

recasting of a combinatorial problem in terms of a suitable polynomialis sometimes useful when the answer is less demanding than nding thecoefficient of a specic 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 ndingit 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 common

sense 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 particular

value of n, but to nd the entire sum, say S 1 = a0+ a1+ a2+ . . .+ ark . We

could have written S 1 ostensibly as an innite sum S 1 =∞

n =0an because

all except nitely many terms of this series vanish. Since f (x) = a0 +a1x + a2x2 + . . . a rk xrk , we have S 1 = f (1). From the factorisation of f (x) as (1 + x + x2 + . . . + xk)r we immediately get S 1 = ( k + 1) r .Combinatorially, S 1 is the number of all possible ways of putting any

numbers 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 codication.

To do so, consider S 2 = a0 + a2 + + a4 + . . . + a2m + . . ., i.e. the sumof the coefficients of all even degree terms in f (x). Combinatorially, S 2is 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 balls

go 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 S 1. But the algebraic one is simple. If we addf (1) and f (−1), the even powers add up while the odd ones cancelout. In other words, S 2 = a0 + a2 + a4 + . . . =

f (1) + f (−1)2

. This is

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true for any polynomial. In our example, f (1) = S 1 = ( k + 1) r while

f (−1) = (1 −1 + 1 −. . . + ( −1)k+1 )r which equals 0 or 1 according ask is odd or even. Hence S 2 =

(k + 1) r

2 or

(k + 1) r + 12

depending uponwhether k is odd or even.

The essence of the calculations of S 1 and S 2 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 1 n + ωn + ( ω2)n equals 3 or 0 dependingupon whether n is a multiple of 3 or not. So, for any polynomial

f (x) = a0+ a1x+ a2x2

+ . . ., the sum, say S 3 = a0+ a3+ a6+ . . .+ a3m + . . .would equal

f (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 of this 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 nding thenumber of 6-digit numbers whose digits add to a number of the form4 p + 1. This is equivalent to counting the number of ways to putidentical balls into 6 boxes the rst of which can hold 1 to 9 balls andthe remaining 0 to 9 balls each, so that the total number of balls is of the form 4 p+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+ b2x2 + . . . = (1+ x+ x2 + . . . + x8)(1+ x + x2 + . . . + x9)5. By our

work, the count is g(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 have

g(1) = 900000, g(−1) = 0, g(i) = (1+ i)5 = −4−4i and g(−i) = −4+4 i.So the desired number is 9000000−4 −44

= 249998.

Q.43 Suppose that the foci of the ellipse x2

9 +

y2

5 = 1 are (f 1, 0) and (f 2, 0)

where f 1 > 0 and f 2 < 0. Let P 1 and P 2 be two parabolas with a com-

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mon vertex at (0 , 0) and with foci at ( f 1, 0) and (2f 2, 0), respectively.

Let T 1 be a tangent to P 1 which passes through (2 f 2, 0) and T 2 be atangent to P 2 which passes through ( f 1, 0). If m1 is the slope of T 1 and

m2 is the slope of T 2, then the value of 1m21

+ 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 − 59 = 23and so the foci are at (±3e, 0) i.e. f 1 = 2 and f 2 = −2.Since both the parabolas P 1 and P 2 have their vertices at (0 , 0), andtheir foci are at (2 , 0) and (−4, 0) respectively, their equations are

y2 = 8 x (1)y2 = −16x (2)

respectively. We are given that the line y = m1(x + 4) touches theparabola y2 = 8 x. So it intersects the parabola in two coincidentpoints. Therefore, the quadratic m1(x + 4) 2 −8x = 0 has discriminant0. On simplication, this gives (8 m21 −8)2 = 64 m41 and hence

m21 =

12 (3)

(Incidentally, this shows that there are two lines through ( −4, 0) thattouch the parabola P 1. The problem does not specify which of the twois to be chosen. But the answer does not depend on the choice since itinvolves only m21.)By an entirely analogous computation, the slope m2 of T 2 satises

m22 = 2 (4)

Hence the expression 1

m21+ m22 equals 4.

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

y = m1x + 2m1

(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 rst 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 given it atwist by rst 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( α n ) −eα m

= −e2

then the value of mn

is

Answer and Comments: 2. As α

→ 0 both the numerator and

the denominator tend to 0. So the limit, say L, in the question if of the indeterminate form

00

. It is, therefore, tempting to apply theL’Hôpital’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. Dene the function f (α ) = ecos( α n ) − 1.Then L’H ôpital’s rule implies that

L = limα→0

f ′(α)

1 (1)

provided this limit exists.

Now, by a direct computation, for α = 0 we have,f ′(α) = −sin(α n )ecos( α

n )nα n−1 (2)

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It is tempting to think that this is 0 at α = 0 because of the rst factor,

viz., sin(α n ). 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 xed number.

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

f ′(α) = −sin(α n )

α n ecos( α

n ) α 2n−1 (3)

Now the rst factor tends to 1 as α

→ 0. So the fate of the limit

depends 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 thelimit in question will exist. When it does, its value will be −

e2

. In theproblem, of course, n is given to be an integer. But the argument abovesuggests that if at all the ratio

mn

has some xed value, it is probably2.

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 = β 1/k .

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

L = limβ→0

ecos( β1 /k ) −eβ

(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 nite 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 because

sin uu

tends to a nite non-zero limit as

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

u2 tends

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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 −1α m

(5)

Then clearly, L = eL′ provided L′ exists. By what we just said, thenumerator is of the order of cos(α n ) −1, which, in turn, is of the orderof α 2n . The denominator, on the other hand, is of the order of αm . Sothe ratio will tend to a nite 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(α n ) −1 and then again by α 2n . 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

0

e9x+3tan− 1 x 12 + 9x2

x2

+ 1 dx where tan−1 x takes only princi-

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

4 is

Answer and Comments: 9. This is plainly a question about evalu-ating a denite 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 of the form ek+3 π/ 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 substitutionu = 9x + 3 tan −1 x (1)

which yieldsdudx

= 9 + 3

1 + x2 =

12 + 9x2

x2 + 1 (2)

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Hence we have

α = 9+3 π/ 4

0eu du

= e9+3 π/ 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)x12 + 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) =

12

. 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)

= 114

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

Answer and Analysis: 7. The problem is a combination of L’H ôpital’srule and the second Fundamental Theorem of Calculus (FTC). To ap-ply the former to the limit lim

x→1F (x)G(x)

we must rst ensure that it is

of the 00

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

−1

f (t) dt (1)

and G(1) = = 1−1 t|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. Hence

its 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 justied in applying the L’H ôpital’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)

= limx→1

F ′(x)G′(x)

Educative Commentary on Joint Entrance Examination 2015

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