MOCK TESTFOR ByAlokkumar. B.Tech, IITKanpur The Indian Statistical Institute (ISI), Kolkata, is considered as one of the foremost centres in the world for training and research in statistics and the related sciences. The B.Stat (Hons) degree program, the flagship programme of the institute, offers comprehensive instruction in the theory, method and application of statistics, in addition to several areas of Mathematics and some basic areas of computer science. Each candidate applying for admission to this programme has to take a selection test comprising Objective type and Short-answer type questions in mathematics at the Higher Secondary level (10 + 2 year's programme). The selection tests consists of (1) A multiple choice type test having about 30 questions, and (2) A short-answer type test having about 10 questions. Questions will be set on the following and related topics. Algebra : Sets, operations on sets, prime numbers, factorization of integers and divisibility, rational and irrational numbers, permutations and combinations, binomial theorem, logarithms, theory of quadratic equations, polynomial and remainder theorem, arithmetic and geometric progressions, inequalities involving A.M., G.M., and H.M., complex numbers. Geometry : Plane geometry of class X level. Geometry of 2 dimensions with cartesian and polar co-ordinates. Concept of a locus, equation of a line, angle between two lines, distance from a point to a line. Areas of a triangle, equations of a circle, parabola, ellipse and hyperbola and equations of their tangents and normals, mensuration. Trigonometry : Measures of angles, trigonometric and inverse trigonometric functions, trigonometric identities including addition formulae, solutions of trigonometric equations. Properties of triangles, heights and distances. Calculus : Functions, one-one functions, onto functions, limits and continuity, derivatives and methods of differentiation, slope and curve, tangents and normals, maxima and minima, use of calculus in sketching graph of functions, methods of integration, definite and indefinite integrals, evaluation of area using integrals. Logical Reasoning : Consistency of statements. In response to growing demand from students preparing for the ISI, we bring to you the first Mock ISI paper, which closely simulates the real exam. There is more to follow in the coming months. MULTIPLE CHOICE TEST 1. (a) (c) How many real solutions does the equation x4 - 10x2 + x + 20 = 0 have ? exactly one (b) exactly two exactly three (d) exactly four The roots of the equation z" = (z + 1)" = lie on a circle of radius ~ lie on a straight line lie on a regular polygon of n sides lie on a regular polygon of 2n sides. 3. Let A and B be two complex numbers such that A B u + -7 = 1, then the triangle OAB, O being the originD A

2.(a) (b) (c)(d)

is (a) (b) (c) (d) 4.

equilateral isosceles but not right angled right angled isosceles right angled. Let A = 4 = + -]= + -n v3 + , 1 . Then [A], V10000

56

MATHEMATICS TODAY |FEBRUARY'07

where [] denotes the greatest integer function, equals (a) 297 (b) 187 (c) 197 5. Sudhanshu and Geetika start at the point P on the circle, but Sudhanshu moves along the bigger circle whereas Geetika moves along the smaller circle that touches the bigger one at P, and also touches two of its radii OA and OB. ZAOB = 60. If both travels at uniform speed, what is the rate of their speeds if both reach P at the same point after one round? (a) 6.111

2468 - sum of the digits appearing in the base 7 representation of 1234

6

^ ^

1234 - sum of the digits appearing in the base 7 representation of 1234 2468 - sum of the digits appearing in the base 6 representation of 12346

1234 sum of the digits appearing in the base 7 representation of 1234 12. and (a) (b) (c) (d) 13. has (a) (b) (c) (d) (a) Let the two numbers a and (3 be a = 4673 + 6734 p = 4929 + 929 4 then a and P are both composite. a and p are both prime. a is prime but p is composite. a is composite but p is prime. The equation (x2 + 3x + 4 f + 3(x2 + 3x + 4) + 4 = x all its solutions real but not all positive. only two of its solutions real. two of its solutions positive, two negative. none of its solutions real. V302 +31 2 + 830 2 = 931 V272 +28 2 +656 2 == 757 V322 +33 2 +1056 2 = 1057 /402 +41 2 +1340 2 = 1641 15. For a positive integer n, let Sn denote the sum of digits of n. The number of integers for which n - Sn= 1234, equals (a) infinite (b) 16 (c) 25 (d) no such integer is possible. 16. Consider the sequence of real numbers a,, a2, an. The sum of every 6 consecutive terms of this sequence is positive whereas the sum of every 13 consecutive terms is negative. Then the sequence can have maximum (a) 17 terms (b) 78 terms (c) 19 terms (d) 7 terms. 17. For how many positive integer values of n is 24 + 27 + 2" a perfect square?MATHEMATICS TODAY |FEBRUARY'07

73:211

(b) 2 : 1

(c)

73:1

(d) 3 : 1

What is the remainder when the 224 digit number

-' is divided by 41? 224 l'i(a) 18 (b) 23 (c) 4

(d) 0

7. Let f ( x ) be a function such that f ( x - 1) +/(* + 1)= V2 f ( x ) Then for what value of p is the relation fix + p) =/(*) necessarily true for every real xl (a) - 8 (b) - 6 (c) 4 (d) 2 8. 12 + 72

The sum to 168 terms of the series 1 273+372 1 6 + 473 475+10

14. Which of the following equations is true?

(b) (c)

1 5V6+6V5 12 (a) 13 (c) 9. n, n (a) (c) 741-740 2742 + 743

equals (b) (d) 13 12 74T + 740 2743-742

For how many values of n are the numbers + 4, n + 14 all primes? exactly on (b) finitely many infinitely many (d) none of these

10. Let av a2, , a ] 6 9 represent any arbitrary permutation of the numbers 1, 2, 3, ... 169. Then the product (a, - 1) (a2 - 2) (a ]69 - 169) (a) never even, whatever be the permutation (b) always even, whatever be the permutation (c) even only for some permutation, not all (d) none of these 11. The highest power of 7 in the number 1234! is given by 58

(a) (c)

infinitely many exactly one

(b) finitely many (d) zero.

18. How many times does 8 occur when we list all the numbers from 1 to 1000? (a) 297 (b) 300 (c) 271 (d) 273 19. The difference between the greatest and least values of the function f i x ) = (3 - V 4 - * 2 f + (l + V 4 - x 2 ) 3 is (a) 500 27 (b) 18 (c) 11 (d) 470 27

20. L e t / i x ) be a fourth degree polynomial with leading coefficient unity. A l s o / ( l ) = 2 , / ( 2 ) = 4 , / ( 3 ) = 6. Then the value o f / ( 1 0 ) + / ( - 6 ) . (a) is 8072 (b) is 8064 (c) is 2368 (d) cannot be determined 21. Consider the quadratic/^) = 90.x2 + 20jc + 1. The sum of the digits of the n u m b e r / ( l l 1111) is (a) 11 (b) 12 (c) 13 (d) 14 22. The cyclic octagon A]A2A3A4ASA6A7AS has sides a, a, a, a, 6, b, b, b respectively in that order. The radius of the circle that circumscribes A]A2A3A4ASA6 is (a) (c) -b2 + ab \la2+b2 + abj2 (b) (d) 1 a2 +b2+absf2 b +ab

numbers, say, a and b among them and change them to 0.8a - 0.66, and 0.6a + 0.86. Leave the third number unchanged. Starting with this new triplet again generate a triplet performing the same operation i.e. taking any two numbers, changing the 0.8a 0.66, 0.6a + 0.86 and leaving the third unchanged. Then of the following numbers which number can be arrived at after performing the operation described a finite number of times? (a) (3,4,5) (b) (5,12,3) (c) (3,4,13) (d) none of these. Directions for Q. 28 & 29 : Answer these questions on the basis of the information given below. Let P be an interior point of a triangle ABC and let BP and CP meet AC and AB in E and F respectively. Denote by [X] the area of figure X, whether triangle or quadrilateral, as the case may be. Area of the triangle BPF is 4 units, that of triangle BPC is 8 units. Also area of triangle CPE = 13 units. 28. What is the value in sq. units of [APF]? (a) 52 (b) 91 (c) 39 (d) 13 29. What is the value in sq. units, of [AEPF]1 (a) 52 (b) 39 (c) 137 (d) 143 30. How many ordered triplet solutions (jc, y, z) does the system x + y + z = 4, x2 + y2 + z2 = 14, jc3 + y3 + z3 = 34 have? (a) six (b) one (c) eighteen (d) thirty six Short Answer Type Test 31. Solve the system of equations for real x and y. 5* 1 + I ) . . . f, 2 2 = 12,5^1x + j/ 1 1 =4 +x2+y2

2

'

2

23. The number of positive integer that divide 10999 but not 10998 is (a) 1998 (b) 999 (c) 998 (d) 1999 24. How many solutions numbers does the system abed - 27 + ab + ac (a) exactly one (c) exactly 4(4!) (a, b, c, d) in positive real a + b + c + d= 12 + ad + be + bd + cd have? (b) exactly 4! (d) none of these

25. Given that x, y, z satisfy 1 7 1 1 x + = v + - = 4, z + = 1 the value of xyz is y 3 z x (a) 1 (b) - 1,35 z

32. In the triangle ABC, the altitude, angle bisector and median from C divide the angle C into four equal angles. Find the angles of the triangle. 33. Let {*} be a sequence satisfying the recurrence y/3x-1 r,n> 1 n+1 " Xn + yf3 ' Prove that the sequence is periodic having period 6.x

(c),21 z

^

(d)

^

26. Let a = 2 + 1,6 = 2 +1 , then the greatest common divisor of a and b i.e. gcd(a, b) is (a) 2 2 ? - 1 (b) if' +1 (c) 22'4 - 1 (d) 1 27. Start with the triplet (3, 4, 12), choose any twoMATHEMATICS TODAY |FEBRUARY'07

34. Let a and 6 be two of the roots of the equation x4 + x3 - 1 = 0 . Prove that ab is a root of the equation xb+x4+xixk--l= 0. 35. Find integer solutions for the equation 59

4x2 + 9 / = 72z 2 36. Find all real solutions to the system x3 - 3x = y f - 3 y =z z3 3z = x 37. Prove that 99899 is composite. 38. A four digit number has the following properties (a) It is a perfect square. (b) The first two digits are equal. (c) The last two digits are equal. Find all such numbers. 39. Given a triangle ABC. Let 1 be the centre of the inscribing circle. The internal bisectors of angles A, B and C meet the opposite sides in A', B' and C" respectively. Ptove that 1 AI BI CI < 8 4 AA'' BB'' CC' ~ 27 40. What is the remainder when 256 + 3-9" + 14 is divided by 41? SOLI I IONS 1. (d): The equation x4 - l(k 2 + * + 20 = 0 is a fourth degree one, so we can't factor it. Note that the equation doesn't have any integral root as no divisor of 20 satisfies it. Instead set a = 5 and rewrite the equation as x4 - 2ax2 + x + a2 - a = 0 The beautiful thing about this equation is that while it's a fourth degree one in x, it's a quadratic in ' a ' parameter - that is a2 - a(2x2 + 1) + x4 + x = 0 having discriminant = (2x2 + 1 )2 - 4(x4 + x) = 4;t4 + 4x2 + 1 - 4x" - 4x = 4x2-4x + 1 =(2jc - l ) 2 (a perfect square) The roots then are 2x2 +1 (2x -1) _ 2x2 + 2x 2x2 +1 - 2x +1 a=2 2 ' 2 = x2 + x, x2 - x + 1 Returning to a = 5, gives x2+X-5 = 0 and x 2. (b) -4 = 0 : 2 l V l + 16 2 1 lVl7 =>

, 1 2kn 1 + = cos z n

. 2kn + ;sin n

k = 0, 1,2, 3, But k = 0 => z=

n- 1 = 0 , impossible. ;k = 1,2, (-l)

. 2*71 . . 2kn 1 + cos - + sin n n ,, 2/nt - 1 + cos ncos

.

. . 2kn /sin n

"(_ 1 +

2^)2+(sin2 2^)

2kn L 2kn\ - ( 1 - c o s ) -isin n 2kn\ 2^1 - c o s n J . . 2k% ;sm

_

, ' 2 1 2

j

kn kn 2smcos n n_ siir kit .2 2-2 n / . kn 2 n

Thus, Re(z) = -= orx = - i , showing that the roots of the equation (z +1)" = z" lie on the line x = - \ B A A2 - AB + B2 = 0 5 V B 2 - 4 B2 A==

(a)

Bij3B2

( z + l)" = z" i+=(i)n z

KF60

_ l i S = -I/V3 D 2d - ^B 2 B = -(aB-(o B Then \A\ = |S| Now A-B = -(oB-B = -(co + 1)5 = -to 2B 2 2 or -co - B = - (co + l)B = coB \A-B\ = |co25| or |co5| = \B\ implying |A - B\ = \A\ = |5|

MATHEMATICS TODAY |FEBRUARY'07

(a) (c)

infinitely many exactly one

(b) (d)

finitely many zero.

18. How many times does 8 occur when we list all the numbers from 1 to 1000? (a) 297 (b) 300 (c) 271 (d) 273

19. The difference between the greatest and least values of the function / ( * ) = ( 3 - V 4 - J C 2 ) 2 + ( i + V 4 - j c 2 ) 3 is (a) 500 27 (b) 18 (c) 11 (d) 470 27

20. L e t / ( x ) be a fourth degree polynomial with leading coefficient unity. A l s o / ( l ) = 2 , / ( 2 ) = 4 , / ( 3 ) = 6. Then the value o f / ( 1 0 ) + / ( - 6 ) . (a) is 8072 (b) is 8064 (c) is 2368 (d) cannot be determined 21. Consider the q u a d r a t i c / ^ ) = 90.x2 + 20x + 1. The sum of the digits of the n u m b e r / ( l l 1111) is (a) 11 (b) 12 (c) 13 (d) 14 22. The cyclic octagon A!A2A3A4ASA6A7AS has sides a, a, a, a, b, b, b, b respectively in that order. The radius of the circle that circumscribes AlA2A3A4AsA6 is (a) (c) a +b + ab yla2+b22

numbers, say, a and b among them and change them to 0.8a - 0.66, and 0.6a + 0.86. Leave the third number unchanged. Starting with this new triplet again generate a triplet performing the same operation i.e. taking any two numbers, changing the 0.8a 0.66, 0.6a + 0.86 and leaving the third unchanged. Then of the following numbers which number can be arrived at after performing the operation described a finite number of times? (a) ( 3 , 4 , 5 ) (b) (5,12,3) (c) (3,4,13) (d) none of these. Directions for Q. 28 & 29 : Answer these questions on the basis of the information given below. Let P be an interior point of a triangle ABC and let BP and CP meet AC and AB in E and F respectively. Denote by [X] the area of figure X, whether triangle or quadrilateral, as the case may be. Area of the triangle BPF is 4 units, that of triangle BPC is 8 units. Also area of triangle CPE = 13 units. 28. What is the value in sq. units of [APF\1 (a) 52 (b) 91 (c) 39 (d) 13 29. What is the value in sq. units, of [AEPF\? (a) 52 (b) 39 (c) 137 (d) 143 30. How many ordered triplet solutions (x, y, z) does the system x + y + z = 4, x2 + y2 + z2 = 14, x3 +y3 +z3 = 34 have? (a) six (b) one (c) eighteen (d) thirty six Short Answer Type Test 31. Solve the system of equations for real x and y. 5x1 + 1 2 x+y )2

(b) (d)

a2+b2+absl2 yja2 +b2 +ab

+ abj2

23. The number of positive integer that divide 10999 but not 10998 is (a) 1998 (b) 999 (c) 998 (d) 1999 24. How many solutions numbers does the system abed = 27 + ab + ac (a) exactly one (c) exactly 4(4!) (a, b, c, d) in positive real a + b + c + d= 12 + ad + be + bd + cd have? (b) exactly 4! (d) none of these

. . . f, = 12,5^1-

1 =4 1 + jc2 + y 2

25. Given that x, y, z satisfy 1 7 1 1 x + =, _y + = 4, z + = 1 the value of xyz is y J z x (a) 1 (b) - 1,35

32. In the triangle ABC, the altitude, angle bisector and median from C divide the angle C into four equal angles. Find the angles of the triangle. 33. Let {*} be a sequence satisfying the recurrence \l3x n - 1 ',> 1XN + YF3

(c),21

^

(d)

Prove that the sequence is periodic having period 6. 34. Let a and 6 be two of the roots of the equation x4 + x3 - 1 = 0 . Prove that ab is a root of the equation xb. + x4 + x3 - xi - 1 = 0. 35. Find integer solutions for the equation 59

26. Let a = 2Z +1,6 = 2 + 1 , then the greatest common divisor of a and b i.e. gcd(a, b) is (a) 2 2 ? - l (b) 2 2 ? + l (C) 2 2 ' 4 - l (d) 1 27. Start with the triplet (3, 4, 12), choose any twoMATHEMATICS TODAY|FEBRUARY'07

4x2 + 9f = 72z2 36. Find all real solutions to the system x3 - 3x - y y3-3y = z z3 -3z - x 37. Prove that 99899 is composite. 38. (a) (b) (c) Find A four digit number has the following properties It is a perfect square. The first two digits are equal. The last two digits are equal. all such numbers. =>

, , 1 2ku 1 + = cos z n

. . 2kn + /sin n

k = 0, 1, 2, 3,

n- 1

But k = 0 => j = 0 , impossible. ,k = 1,2, (-l)

, , n2kn , . 2kn -1 + cos f sin h n n ,, 2kn + cos 1 t i , 2kn\2 J

. 2kn /sin n 2 2kn\

39. Given a triangle ABC. Let I be the centre of the inscribing circle. The internal bisectors of angles A, B and C meet the opposite sides in A', B' and C" respectively. Ptove that 1 AI BJ CI < 8 4 AA'' BB'' CC' ~ 27 40. What is the remainder when 256 + 3-9" + 14 is divided by 41? s o i l HONS 1. (d): The equation x4 - I0x2 + x + 20 = 0 is a fourth degree one, so we can't factor it. Note that the equation doesn't have any integral root as no divisor of 20 satisfies it. Instead set a = 5 and rewrite the equation as x4 - lax2 + x +a2- a = 0 The beautiful thing about this equation is that while it's a fourth degree one in x, it's a quadratic in ' a ' parameter - that is a2 - a(2x2 + I) + x4 + x = 0 having discriminant = (2x2 + 1 )2 - 4(x4 + x) = 4x4 + 4x2 + 1 - 4x4 - 4x = 4x2-4x+ 1 = (2x - l ) 2 (a perfect square) The roots then are _ 2x2 +1 (2x - 1 ) _ 2x2 + 2x 2x2 +1 - 2x +1 a 1 2 2 2 = X2 + X, X2 - X + 1 Returning to a = 5, gives X*+x-5=0 and 2.x

=

-(l-cos)-/sin n_l _

2kn /sm-

, 2 1 2'

i

kn kn 2 sincos n n_ ^^ .2kn 2-2 sin' n /' .kn 2 n?

Thus, Re(z) = - i orx = - ^ , showing that the roots 1 of the equation (z +1)" = z" lie on the line x = - (a) A2-AB A==

^ B

A

=1

+ B2 = 0 45"

B

=>x.- f= 0 =

l

2_x_4

,

x =

l M H i

=

lVl7

(b)

(z + l)" = z n i => I+=(i) n

Bisl3B 2 _ iiS p - 1 /'V3 D 2b - ^# = 2 5 = -co6-CO5 Then = |6| Now A - B = 5 - 5 = -(co + 1 )5 = - 2 5 or -co 2 5 - 5 = - (co2 + 1)6 = co5 M - 5| = |cn25| or |co6| = |5| implying M - 5| = = \B\

60

MATHEMATICS TODAY |FEBRUARY'07

i.e. \A - B\ = \A - 0| = \B - 0| Showing that the origin point A and point B form an equilateral triangle. 4. (c) All we need is to bound the sum+ T

11111..1 = 1111.224 L'J 220 L'J

1,0000+1111 = 104 x l l l l l . . 1 + 1111220 r.v

..(A)

J = V2 V3 VlOOOO from above and from below by numbers that differ by unity. Note that 1 \[k y[k +y[k y[k 10000 successively and adding 1 .+(Vioooo-V9999)} => A A < 2 x 99 => A < 198 Similarly

J Now H Ur.v '220 "

= L L y i x 10000100001. ' 5 f' V ' .

.00001

= 41 x 271 x a natural number ... (B) From (A) and (B) when n is divided by 41, the remainder is the remainder obtained upon division of 1111 by 41, that is, remainder = 4. Note that we have made use of the prime factorisation of 11111, i.e. 11111 = 4 1 x 271. 7. (a) : f ( x - 1) +f(x + 1) = V 2 / W (A) Change the lowest argument, x + I to x. our equation reads/(x) +f(x + 2) = V 2 / ( * + 1) O ) As there is a difference of 2 in the arguments of function on the left side, we change x to (x + 2) to obtain fix + 2) + / ( x + 4) = fix + 3)..... (2) Adding (1) and (2) fix) + 2/(x + 2) + / ( x + 4) = V2 {/(* + 1) + / ( * + 3)} = V2 V 2 / ( x + 2) [from the parent relation (A)] =

Putting k = 2,3,

d)

Putting k = 2, 3,

10000 successively and adding1

V2

+

V 3 + " ' + V10000

-2{(V3-V2)+(V4-V3)

+

....+(Vi 0001-V10000)}=> A>2 [VlOOOl - y/2] >197(2)

comparing (1) and (2) we get, 197 < A < 198 /. [/(] = 197 5. (d) : Let the radius of smaller circle be r and that of the bigger one be R, OP = OO' + O'P r R=:+ r " ~ sin 30 (In triangle ONA, O'N = OO' x sin 30) => R = 2r + r = 3r R r = ^ l .. . , 2nR/t R 3 Ratio of their speeds = ^ n r f t ~ ~ ~ JD

^ / ( x ) + / ( x + 4) = 0 Now once again instead of increasing the argument by 1, we recast the relation as f ( x ) = - f ( x + 4) We would have immediately found the period but for the negative sign ! Change x to (x + 4) to obtain fix + 4) = -fix + 8) =>/(x + 8) = -fix + 4) = -{-/(x)}(from the above relation) =>/(x + 8) = / ( x ) i.e./(x) - fix + 8) Thus the period of function is 8. But the problem doesn't say ip) has to be positive, so p can be of the form p = 8k, kel 8.u

(a)

: Write the n'h term as

"

1 Vw+7+(+i)VM (w+l)Vw-V + 1

{ ( + i ) V + W + 1 } {(+1) V - W + i } _ (/+i)Vw-Vw+i nin+\)2 -n2in+ \) 61

(c)

Write n =

MATHEMATICS TODAY |FEBRUARY'07

(n +1 )4n - %/ +1 _ (n +1 )yjn - V +1 (h + 1){/I + 1 - 1 } 1 1 Vrt Vh +1 Putting = 1, 2, 3.... 168 successively in the above relation and adding 1 W 1 1 1 + "2 + + "168 = 1 f 1 V Vl68 1 Vl69 J

1234 - sum of the digits appears in the js^Qyy base 7 representation of 1234 1234- (3 + 4 +1 + 2) = 1234-10 ^ 1224 204 6 6 6 In general we have the highest exponent of prime p in ! n - sum of digits of n in base p representation P~ 1 12. (a) : Rewrite a as a = 4673 + 673 4 = 4-4672 + 673 4 = 4(4 168 ) 4 + 673 4 - 4a 4 + bA (say) Where a = 4 168 , b = 673 Observe that 4a 4 + b4 can be factorised 4a 4 + b4 = (2a2)2 + (b2)2 + 2(2a 2 )(b 2 ) - 4a2b2 = (2a 2 + b2)2 - (2ab)2 = (2a2 + 2ab + b2) (2a2 - lab + b2) implying that a is composite, viz, a has a factor (say) b2 + 2ab + 2a 2 (i.e.) (673) 2 + 2-4168 673 + 2 (4 168 ) 2 (3 can also be written in the formp = 4929 +929

6

1 = 1_J_=12 = 1Vl69 13 13 9. (a) : For n = 3, the numbers , + 4, w + 14 yield 3, 7, 17 and they are all prime. We claim that for no other value of n are n, n + 4, n + 14 all prime. Recall that any prime p > 5 can be written in the form 6k l,k>l. When n = 6k + 1 n, n + 4, n + 14 = 6k + 1, 6k + 5, 6k + 15, of which 6k + 15 = 3(2k + 5) is composite. When n = 6k - 1 n, n + 4, n + 14 = 6k- 1, 6k- 3, 6k- 13, of which 6k - 3 = 3(2k - 1) is composite. Hence only for n = 3 are the numbers n, n + 4, n + 14 all prime. 10. (b) : The idea behind the solution is to prove that at least one factor in the product is even. Of the numbers 1, 2, 3, 169, exactly 85 are odd. Again as the ajs consist of the numbers from 1 to 169 because a's are simply a permutation of 1,2, 3,.... 169, we have among a!s also 85 odd numbers. Both the sets {1,2, , 169} and { a v a 2 , . . . , a 168 } have together 170 odd numbers. Thus among the 169 factors (a, - 1), (a 2 - 2), (a 3 - 3), ( a m - 169) We have 170 odd numbers. Accordingly some factor say ak - k has both the numbers odd implying ak - k is even and so is the entire product. 11. (d) : Write 1234 in base 7 1234 = 3 x 343 + 4 x 49 + 7 + 2= 3X73 + 4x72 + 7 + 2

4 = 4.(4232)4 +

929

4

Showing that p is composite too. 13. (d) : Look at the equation (x2 + 3x + 4)2 + 3(x2 + 3x + 4) + 4=x for a while and your powers of observation tell you that it can be written as f ( f (x)) = x ! where f ( x ) = x2 + 3x + 4 consider the equation f ( x ) = x, i.e. x2 + 3x + 4 = x => x2 + 2x +4 = 0 D= 22-4X4 0 v x e R. i.e. f ( x ) > x \/ x e R. Setting/(x) in place of x, as f ( x ) eR, we have / i f (*)) > f(x)> x, from the above equation establishing that fif(x))>x Thus when f ( x ) = x has no real solution, / ( f (x)) = x doesn't have either. 14. (c) : We embed the problem in a more general one and look at the expression. f(h) = h2 + (h+ 1)2 + {h(h+ l)}2 = h2 + h2 + 2h + 1 + h2 (h2 + 2h + 1)MATHEMATICS TODAY | MARCH '07

= (3412), Highest power of 7 in j 1234 is 1234 1234 1234 = 176 + 25 + 3 = 204 49 L 34362

i.e. \A - B\ = \A - 0| = \B - 0| Showing that the origin point A and point B form an equilateral triangle. 4. (c) All we need is to bound the sum+

11111..1-1111. 10000+1111 224 L'J 220 L'J

= 104 x 11111. .1 + 1111 220 r.vNow

(A)

V2 V3 VlOOOO from above and from below by numbers that differ by unity. Note that TTvFTvT Putting k = 2,3, 1 2 { ( V 3 - V 2 ) + ( V 4 - V 3 ) V3 Vioooo....+(Vi A>2 [VlOOOl - V2} >197

oooi-Vioooo)}(2)

comparing (1) and (2) we get, 197 < A < 198 = 197 5. (d) : Let the radius of smaller circle be r and that of the bigger one be R. OP = OO' + O'P r R-. :+ r sin 30 (In triangle ONA, O'N = OO' x sin 30) => R = 2r + r = 3r R => r = TD f U ' A Ratio of .their speeds =2 7 l R / t

(a)

: Write the nth term as

"

1 W + i + (+i)V jn + \)Jn - V + 1 {( +1 )4n + V + 1} {( +1) Vn - V +1}

= = j

R

3

_ jn + l)Vw-wVw+1 ~m( + 1) 2 -W 2 ( + 1) 61

6.

(c)

Write n =

MATHEMATICS TODAY|FEBRUARY'07

= 2h2 + 2h + 1 + h4 + 2h3 + h2 = h4 + 2 h3 + 3h2 + 2h+\ = h4 + h2 + 1 + 2-h2 h + 2-h-l + 2-h2-l = (h2 + h+ l) 2 = {h(h+ 1 ) + l} 2 (a2 + b2 + c2 + 2 ab + 2 be + 2ca = (a + b + c)2) Now 40 x 41 = 1640, 27 x 28 = 756, 3 2 x 3 3 = 1056, 30 x 31 = 9 3 0 Thus the true equality is A/322 +33 2 +(32x33) 2 = 3 2 x 3 3 + 1 = 1057. 15. (d) : Let n = (a m am_x a,-a0) 10 where a0, av am are digits of positive integer n be the decimal representation of n. n-Sn = (am 10 + a m _, (10' + .... + 10 + a0) -iam+am-, + .... + a , + a 0 ) = am (10"' - 1) + ra_, (10 - 1)+..+7,(10-1).... (1) As every number of the form 10* - 1 is divisible by 10 - 1 (=9), we conclude from (1) that n - Sn is divisible by 9. But n - Sn is given to be 1234 which is not divisible by 9. Hence no such n exists. 16. (a) : We establish that the sequence can't have more than 17 terms. Assume to the contrary that the sequence has more than 17 terms. Consider the first 18 terms. al + a2 + a3 + aA + a5 + a6 > 0 a2 + a3 + + a7 > 0 ,3+ai4+ + I8>0 Adding coloumnwise (a, + a2 + ... + a | 3 ) + (a2 + ... + a, 4 ) + ... .... (a6 + ... + a 18 ) > 0 But since the sum of every 13 consecutive terms is negative, the sum on the left can't be positive. Impossible. Hence the sequence can have at the most 17 terms. 17. (c) : Let 2 4 + 2 7 + 2" = m2, m being an integer => 16+ 128 + 2" = m2 => 2" = m2 - 144 = (/w 12) (m + 12) From the above it follows that m - 12 and m + 12 are both powers of 2. A1SO(/M+ 1 2 ) - ( M - 12) = 24. So the problem amounts to finding two powers of 2 whose difference is 24. Observe 2' = 2, 2 2 = 4, 23 = 8, 2 4 = 16, 2 s = 32, 2 6 = 64, 27 = 128 We have 2 5 - 2 3 = 24 and as 2" grows rapidly, we can'tMATHEMATICS TODAY |FEBRUARY'07a

have a pair other than (2 5 /2 3 ) having a difference of 24. Consequently / - 12 = 8 and m + 12 = 32 giving w = 20 and 2" = 32 x 8 = 250 = 2 s n = 8. Accordingly there is just one value of n that makes 24 + 27 + 2" a perfect square. 18. (b) : As 8 doesn't appear in 1000, we have to count the number of times 8 occurs when we list the integers from 1 to 999. Now any number from 1 to 999 can be thought of as axa2a3 where 0 < a,, a 2 , a3 < 9. For example the numbers 81 and 4 will be written as 081 and 004 respectively in this system of representation. Let us first count the numbers in which 8 occurs exactly once. Since 8 can occur in any of the places, we have three ways for it corresponding to every occurrence of 8, other two places can be filled with the remaining numbers in 8 ways each. So number of such type of numbers = 3 x 9 x 9 = 243. Again 8 can occur in exactly two places in 3 c 2 = 3 ways. The remaining place can be filled in 9 ways. So there are 27 (= 9 x 3) such numbers. Finally, 8 can occur in all three places in just one way, viz 888. Number of times 8 occurs in all the numbers from 1 to 1000=243 + 2 x 27 + 3 x l = 243 + 54 + 3 = 300 19. (a) : First of all observe that the function is defined only when 4 - x2 > 0 => x2 < 4 => \x\ < 2. Denoting V 4 - x by t, we have t e [0, 2] Note that t has to be positive. so

The problem of finding the extreme values of f ( x ) amounts to finding the extreme values of g(l) where Sit) = (3 - t)2 + (1 + t)3 for the closed interval [0, 2] g(t) = -2(3 - f ) + 3(1 + 2t + t2) = 3/2 + 8/ - 3 = (/ + 3 ) ( 3 / . - l ) For extreme values g'{t) = 0 => (/ + 3) (3/ - 1) = 0 I 3 Also g"(t) = 6t + 8 = 2(3/ + 4) g"(-3) = - 10 < 0 = - 3 is a point of maxima. = 10 > 0 / = -j is a point of minima. g(t). 63 But / = - 3 is not in the domain of /=-3

At the end points g(0) = 3 2 + 1 3 = 10 g{2) = (3 - 2)2 + (1 + 2)3 = 1 + 27 = 28

1013 - 1 0 + 9

1 |3" - 1 0

= 11111...1 = Rx 313 r.v

Hence the greatest value on the closed interval [0, 2] is g(2) = 28 256 and the least value is g 27 256 _500 Difference between extreme values = 28 - : 27 27 20. (a) Our first reaction on seeing the problem is to w r i t e / ( x ) a s / ( x ) = x4 + atx3 + a^c2 + a3x + a4 and then attempt to find the constants with the help of f (1),/(2) etc. This idea, beside being unattractive, can't be put to use when the number of unknowns increases. Instead we construct an auxiliary function. g(x) =f(x) - 2x So that g(l) = / ( l ) - 2 = 0 Similarly g(2) = g( 3) = 0 As g(x) is a polynomial of degree 4 having 1, 2, 3 as its three zeros. Assume the fourth zero to be a . Then we can write g(x) = (x - 1) (x - 2) (x - 3) (x - a) =>/(*) - 2x = (x - 1) (x - 2) (x - 3) (x - ct) Let x = 10 and - 6 successively to obtain /(10)-20 = 9 x 8 x 7(10-a) / ( - 6 ) + 12 = (7)(8)(9)(6 - a ) = 9 x 8 x 7 (6 + a ) Adding/(10) + f(-6) - 8 = 9 x 8 x 7 x. 16 = 8064 . - . / ( 1 0 ) + / ( - 6 ) = 8072. 21. (c) : R6 = 111111, the number made up only of Is, called a repunit /? 6 = 111111 = ^ (99999) = - | ( 1 0 6 - 1 ) /(111111) = 90 x 10(10 - 1 ) 9106 6 2

- M -S-S-L^'-W G H

n digits Sum of digits of the n u m b e r / ( l l 1111)= 13 x l = 13 22. (b) : We have only shown the part of octagon that interests us. Let r be the radius of the circumcircle and 0, the j, angles subtended by the A4' sides of length a and b respectively at the centre of the circle. Then 40 + 4 0 + < = j > | Using cosine rule in triangles OAlA2 and OA^Az 2 a2 = r2 + r2_ 2r cos0 = 2f- (1 - COS0) b2 = ^ + ^-2r 2 QOsi> = 2 r 2 { \ -sinG) (v = - | - e j Upon division q z _ 1 cos 0 l-sin0 a2 2sin.290 . 9 , :20 cos 20 + sin 7r-2sin^cos^r

2 tan 2 0 l + tan2|-2tan It0 , , t = ttan -

(1-0

2

a _ yflt b 1 -/

a-at=bj2t Giving t = l-cosG = a + -Jib 2r 1 + /2 a2+b2+absf2

10 96

6

2 2 Thus r = a + abyfl + b

+ 1

23. (cf)

:

10999 = (2 x 5)999 = 2 999 x 5999 has been

( 1 0 - 1) (10 - 1 + 2 ) + 1

written in prime factorisation form. Number of divisors of 10999 is (999 + 1) (999 + 1) =10002. Again 10998 = 2998 x 5998 So the number of divisors of 10 998 in

= y ( 1 0 f ' - l ) ( 1 0 " + l) + l (10 1) + 1 6412

MATHEMATICS TODAY |FEBRUARY'07

(998 + 1) (998 + 1) = 9992. All the divisors of 10 10999 = 10002 - 999 2 = (1000+ 999) ( 1 0 0 0 - 9 9 9 ) = 1999 x l = 1999 24. (a) : Whenever you see the system is in complete, i.e. the number of unknowns exceed the number of equations it should suggest you to look for the possibility of an inequality turning into an equality. As positive numbers are involved, the stage is all set for an application of AM - GM theorem. ab + ac + ad + bc + bd + cd ^ ab + ac + ad + bc + bd + cd => ab + ac + ad + be+ bd + cd> 6j{ab){ac){ad){bc){bd){cd)998

also divide 10 .

999

Hence the number of divisors that divide 10999 but not

eliminating two variables out of three and then arrive at an equation in just one variable. But the symmetry of the system can be exploited to observe some elegant decomposition.

= xyz+y

i + x+-+z+-+y

i

i i + xyz

Plugging the values,

^ 3 _j3 abed

3

f

3

+U V

+ J xyz

The equation abed = 27 + ab + ac + ad + be + bd + cd can now be written as abed > 27 + 6 sjabed => ( J a t e d ) 1 ~ Seabed - 27 > 0 => (-Jabed + 3){jabcd As Jabcd > >we have

=>xyz + -2 xyz => (xyz)2 - 2 (xyz) (xyz - l) 2 = 0 xyz = 1 26. (d) form a -2= 2,35

+1=0

: Let a-2 + 1 - 2 = 2",35

,35

+ l a n d = 2z,22,132

,21

+1, ,22 X21:

-1 = 2

-1=(22

)

-9)>0 -Jabed - 9 > 0 ....(A)

= ( 2 2 2 2 f - l , A = 2 13 But (2 2 2 2 )* - 1 is divisible by 2 2 " - 1 , so we can write a2= (t?22 - 1 ) , m e N,22

=> -fated * 9 The first equation is a + b + c + d ^12 Using AM - GM inequality we havea + b+ c + A d

= (22-22' - l ) / = { ( 2 2 2 ! ) 2 - l } m = (2 221 - l ) ( 2 2 2 ' +1> = (222' - l ) bm = bl, where / = m (2 2 2 ' -1) .(B) => a - W = 2 ... (A) The d be the gcd of a and b, then d divides a - bl, so from (A) d must divide 2. i.e. d = 1 or 2. But a and b are both odd, so gcd can't be even. Accordingly gcd (a, b) = 1. 27. (d) : This problem has at its heart the idea of invariance when things change, look for something that doesn't change. So understand the problem better, 65

4

>Vated

4

=> 3 > $labcd => 9 > Vabed (A) and (B) give J abed = 9 Equality holding when a=b=c=d= 3 Thus AM - GM inequality turns into an equality giving (3, 3, 3, 3) as the only solution. 25. (a) : A solution could be given consisting in

MATHEMATICS TODAY |FEBRUARY'07

let's play by the rule. We choose 3, 4 as our a and b in the first move. Then the original set {3, 4, 12} transforms to {0.8 * 3 - 0.6 x 4, 0.6 x 3 + 0.8 x 4, 12} But we needn't evaluate them exactly. Instead observe that (0.8a - 0.6bf + (0.6a + 0.8b) 2 = (0.8 2 + 0.6 2 ) (a 2 + b2) =a +b Thus at any stage of operation, sum of square of numbers of the triplet = constant = 3 2 + 4 2 + 122 = 5 2 + 122 = 132 = 169 That is, no matter how many times we perform the operation on the set {3, 4, 12}, the triplets generated will always have their sum of squares as 169. But 3 2 + 4 2 + 5 2 * 169 5 2 + 122 + 3 2 * 169 3 2 + 4 2 + 1 3 2 * 169 So none of these number can be reached by the operation described. 28. (a) 29. (d) Join AP. Let [AFP} = x and [AEP] = y Form the triangles AFC and BFC, : Let [BPF\ = u, [BPC] = v and [CPE] = w2 2

x _ y _ 4x13 4 + 8 _ 8+13 _ g 2 4x13 x _ y _ 52 12 21 6 4 - 5 2 .'. x = 52, y = 9\.=

(putting the

values)

52 12

Area of triangle APF, i.e. [APF] = x = 52 sq. units. Area of quadrilateral AEPF, i.e. [AEPF] = jc+3^= 52 + 91 = 143 sq. units. 30. (a) : Consider the monic polynomial P(t) = t3 + at2 + bt + c with roots x, y, z x+y+z = -a=>4 = -a i.e. a = -4 Again x2 + y2 + z2 = (jc + y + z)2 -2(xy + yz + zx) => 14 = 4 2 - 2 {xy + yz + zx) 16-14 > xy + yz + zx = -=1 =b The numbers x, y, z are roots of p implying x 3 - 4x2 + x + c = 0 f-ty+y+c^O z 3 - 4z2 + z + c = 0 Adding these equalities and using the equations of system (x3 + y3 + z3) - 4(x2 + y2 + z2) + (x + y + z) + 3c = 0 3 4 - 4 - 1 4 + 4 + 3c = 0 => - 1 8 + 3c = 0 c = 6 Thus P{t) = t3 - 4/ 2 + t + 6 />(_!) = _ 1 - 4 - 1 + 6 = 0 Thus P factors as P(t) = (t + 1) (t2 - 5t + 6) = (/:+ 1) (/ - 2) (t - 3) giving / = - 1 , 2, 3 Thus the solution of the system are (-1, 2, 3) and its permutations. So there are six solutions.

B

[.APF] _ x FP ^ [BPF] = u [APC] y + w PC [BPC] v x _ u _ vx - ny = uw (1) y +w v Again from the triangles AEB and CEB, [APE] _ y _ PE _ [CPE] _ w [APB] x + u PB [CPB] v y =w. wx - vy = uwX+ U V

-(2)

solving the simultaneous linear equations vx uy - uw = 0 wx vy + uw = 0 by cross multiplication=

31.

1 5xM + 2 2 = 12 2 X +}T 5J12 x+y 2

(1)

(2)

y- W - V

=

ltw_V + ICW

From the two equations on division 1+1

u+v 66

y v+w

x2 + y2J 1 ^

12 4

v2 - uwMATHEMATICS TODAY | MARCH '07

(998 + 1) (998 + 1) = 9992. All the divisors of 10 10 999 = 1000 2 -999 2 = (1000+ 999) ( 1 0 0 0 - 9 9 9 ) = 1999 x 1 = 1999 24. (a) : Whenever you see the system is in complete, i.e. the number of unknowns exceed the number of equations it should suggest you to look for the possibility of an inequality turning into an equality. As positive numbers are involved, the stage is all set for an application of AM - GM theorem. ab + ac + ad + bc + bd+cd> ab + ac + ad + be+ bd + cd ej{ab){ac){ad){bc)(bd){cd)998

also divide 10 .

999

Hence the number of divisors that divide 10999 but not

eliminating two variables out of three and then arrive at an equation in just one variable. But the symmetry of the system can be exploited to observe some elegant decomposition. x+ = xyz+y i + x + -+z+

i i i + - + y + xyz+

= (x+y

+ z) + ( l

7

7)+

+

xyz

1

J_xyz

Plugging the values,

f4-i=!+4+i+(^+ir28 22 , I = + xyz + 3 3 v xyz > xyz + -^- = 2 xyz => (xyz)2 - 2 (xyz) 2 => (xyz - l) = 0 xyz = 1 26. (d)

>FA

3

bW

1

=> ab + ac + ad + be + bd + cd> 6-Jabcd The equation abed = 27 + ab + ac + ad + be + bd + cd can now be written as abed >21 + 6Jabed => {yfatedf - bJabcd - 27 > 0

+1=0

,35

,21

: Let a = 21

+\mdb

= 2A

+1

form ,35 ,35 , ,22 ^2''' ,22,13 a -2= 2 +1-2 = 2 - 1 = 2 2 - 1 = (2 2 ) -1

=> (yjabed + 3)(y/abcd -9) 2:0 As sjabed > 0. we have ^abcd - 9> 0 ....(A) But ( 2 2 2 2 ) * - l is divisible by 2 2 2 2 - l , so we can write a - 2 = ( 2 2 ^ - 1 ) , me N = = (222,-l _ l)m = j(2 22 ' f -1) m = ^ \bm -1) +1 }m

=> -Jabed * 9 The first equation is a + b + e + d=?\2 Using AM - GM inequality we havea+b+ c+d >y^d A

4

4

3 > \jabcd 9 > jabed (A) and (B) give Jabcd = 9 Equality holding when a=b=c=d3 Thus AM - GM inequality turns into an equality giving (3, 3, 3, 3) as the only solution. 25. (a) : A solution could be given consisting in (B)

= bl, where I = m (2 221 -1) =>a-bl=2 ... (A) The d be the gcd of a and b, then d divides a - bl, so from (A) d must divide 2. i.e. d = 1 or 2. But a and b are both odd, so gcd can't be even. Accordingly gcd (a, b) - 1. 27. (d) : This problem has at its heart the idea of invariance when things change, look for something that doesn't change. So understand the problem better, 14

MATHEMATICS TODAY |FEBRUARY'07

Set = t to obtain y X2+y2 _ 3 1 1- 2 x+y 2 Using componendo and dividendo 1 _ 3-t x2+y2 3+/

2 = a c a-b b 2 a+6 .... . , a+6 which is equivalent to => b2 + 2ab(A) a2 = 0

a

)2 + 2 - 1 = 0 ,al a From which we have = >/2 1 a lab cos 2 a Again b = CE =

Plugging this into 2nd equation

..(B)

_ a+6 1 , 1 6 1 1 / r0 =>cos2a = = + = _ + _ ( 2a 2 la 2 => 2 a 4 a = f = ZC .'. 33. Z C = 90,=

li) f

x

>/2 1 = jl_1 ^

_ 4 .(3 + t)2 3 f 25 3+/ x +y From (A) and (B) 1 9-r z 25 1+t 25 r2 = 4(1 + t2) (9 - r2) => 25t 2 = 4(9 + 8t 2 - t4) => 4/ 4 - It2 - 36 = 0 => (/2 - 4) (4/ 2 + 9) = 0 Since t e R, t2 = 4 Giving t = 22 1 2 z

y2

= 67.5, Z 5 = 22.5

V3*-l JT X + yJ 3 73 |

Changing n to n + 1

x

n+]

+

V3 x = ty = 2 If / - - 2 , x = ty- ( y2/3-2 5 1 -2 Changing to + 1 V3*-l ^ "+3=

t

y^

2\[3xn + 2 y/3x +1

x + ^3

/2 1\ Thus the solutions are (2, 1) and I - j , - j l 32. AE EB>

' -2>H)-f

y^=

x

n + l

S

=

s + >/3

-4 4*

=

1 *

is the altitude, the angular bisector and the median from the angular bisector theorem.:

j-2Sxn-\ ^ Xn + yfi Changing n to n + 3 -*7I+6 1 *n+3 1

V3x +1+ 1

AC CB

b_ a fa a+b be a+b

C VVers.

Showing that {xn} is periodic having period 6. 1 \ \ D E F \ 34. Let c and d be two other roots of x4 + x3- 1 = 0 From the relation between roots and coefficients a+b+c+d=-1 ab + ac + ad + be + bd + cd = 0 abc + abd + acd + bed = 0 abed = -1 We write these equalities in terms of 15

S0

FB FE

BC CE

=

a b =

But FB = ^-,FE = AF-AE 2 So that we obtain

2

~-=C}a a + b 2 (a + b)

MATHEMATICS TODAY |FEBRUARY'07

S = a + b, S" = c + Q, p = ab and p' = cd to obtain S + S' = - \ p + p' + SS' = 0 pS' + p'S = 0 pp' = -1 Substituting p'=~, equalities yields p = - - - 5 2 - 5 =0 P and p ( - l - S ) - | = 0 It follows from the second equality that P2+I Substituting this into the first equality gives P

x = 3k(2mn + m2 - n2), y = 2k(2mn -m2 z = k(m + n )2 2

+ n2),

36. The presence of x3 - 3x brings to find the triple angle formula for cosine. Let's look for the solutions in the interval [ - 2 , 2], Set x = 2cos,y = 2cosv, z = 2cosw with u, v, we[0, n] S* = - 1 - 5 in the second and third The system becomes 8cos 3 u - 6cosu = 2cosv =>=>

2{4cos3w - 3cos} = 2cosvCOS3M

= cosv

....(i) ..(ii) ....(iii)

Similarly we have cos3v = cosve cos3iv = cos u cos3u = cosv Gives 4COS33M - 3cosu = 4cos 3 v - 3cosv=>

COS9M

= cos3v = cosw

1

P*

~P~{p2+1)2

0

and as beforeCOS27M = c o s 3 w = c o s u

=> p1 (p 2 + l) 2 - (p2 + l) 2 - / 7 s +p2(p2+ 1 )p = 0 which reduces to p6 + p4 + p3 - p2 - 1=0 Showing that p = ab is a root of the equation x6 + x* + x3 - x2 - \ = 0 35. 4JT + 5 / = 72z 2 We first observe that x must be divisible by 3 and that is an even integer. Setting x = 3u and y = 2v yields 36 u2 + 36v 2 = 72z2 => u 2

The equality holds if and only if 27w = 2kn u for some integer k. The solutions in the interval [0, n] are u = k= 0, 1 , 2 , . . . 14 and " = k= 1, 2, .... 12 kiz ,

Consequently kn 3kn _ 5kn , , , . * = 2cos-j^j-, j = 2 c o s ^ j - , z = 2 c o s - ^ - , k = A 0,1, ...14 andJ: = 2 C O S ^ - _ = 2 C O S - ^ - , z = 2 C O S - j y , ^ = 1 , 2 , . . . 12 V

kit

3kn

_

5 kn ,

,

+ V

2

= 2z

2

and v are either both odd or both even, so that

are the solutions to the given system of equations. Since there are at most 3 x 3 x 3 = 27 solutions, which is obvious from the degree of the system, and we have already found 27 distinct ones, these are all the solutions. 37. 99899 is a palindromic number. Write 99899 = 9x4 + 9x3 + 8x2 + 9x + 9, x = 10 Above is a reciprocal equation that can be factorised by pairing off terms equidistant from the beginning and the end. 9x4 + 9x3+4

u + v and u - v are both even integers. The equation now reads_2

This is the well-known Pythagorean equation, whose solutions are z= k(m +n ), '2 2 2 }i

Last two equations yield2

2

2kmn, '

V

2

- = k(m

2

-H )2 2

2

u = k{2mn + m - n ) i.e. x = 3k(2mn + m - n ) v = k (2mn - m2 + n2), y = 2k(2mn - m2 + 2) Thus the solution are 16

8x2 + 9x + 9

= 9(x + 1) + 9(x3 + x) i 8-rMATHEMATICS TODAY|FEBRUARY'07

= = = =

9{(x2 + I)2 - 2x2} + 9x (x2 + 1) + 8x2 9(x2 + L ) 2 - 18A: 2 + 9x (x2 + 1 ) + Sx2 9(x2 + 1)2+ 9x(x2 + 1)- 10x2 9(x2 + l)2 + 15x(x2 + 1) - 6x(x2 + 1) - lOjt21) {3(JC2 + I ) + 5 x } - 2 * { 3 ( x 2 + 1) + 5x}2

x,

Now

AL ^

I L c / ... BB' CC~

Qa + b\b

{a+b+cf

+ c)(c + a)

= 3(JC2 t

= (3JC - 2x + 3 ) ( 3 x 2 + 5 x + 3 )

Consider 4(a + b)(b + c)(c + a) - (a + b + c)3 = 2bca + a2(b + c - a) + b2(c + a - b) + c\a + b - c) As sum of two sides is greater than third, we have R.H.S. > 0 => 4(a + b) (b + c) (c + a) > (a + b + cf _ {a + b){b + c\c + a) {a+b + cf 1 4 ..(A)

Returning to x = 10 we obtain 99899 = (3-102 - 2 10 + 3) (3 102 + 5 10 + 3) = 283 x 353 Showing that 99899 is composite. 38. Let N = aabb be the base 10 representation of such a number, 1 < a +c

Again by the AM - GM inequality (b + c)(c + a) + { a + b ) > ^ a+ b)(b + c){c + a )

=>2ia+3b

+ c)

>l](a

+ b)(b + c)(c + a)

Cubing both sides we get 8 (a + b + cY>{a + b)(b + c)(c + a) 27 giving (a + b)(b + c)(c + a) ^ 8 -27 {a + b + c f ..(B)

combining (A) and (B) we get 1 4 AI BI AA'' BB''CC'

CI ^ 8 ~ 27

ABI bisects angle ABA' BA' = A'l C A1ac b +c C=

40. Let m = 2 s6 + 3-9" + 14 (41) = 40, whose )>() is Euler's coefficient function giving the number of positive integer less than 41 and prime to it. By Euler's theorem 240= 1 (mod 41) i.e. 2 40 when divided by 41 leaves a remainder 1. 2 s = 2256 = 10 (mod 41) => 216 = 100 s 18 (mod 41) 256 _ 2 . 216 256 + 3 9 " + 14 = 5 (mod 41) Thus the remainder is 5.

17

Practice Paper forTime: 2 hrs.-By : Vidyalankar

IIT-JEE 2007Part I Max. Marks : 96 (a) (2,4) (c) ( 1 8 , - 1 2 ) 6. 7 Vl3 7. If f(x)= 1sinx ("TC2x)2 K

Max. Marks : 182Mumbai

Institute*,

Time : 1 hr. (One correct option)

(b) ( 2 , - 4 ) (d) (8, 8)

Section I : (Q.l to 12) 1. The distance of point (3, 5) from the line 2x + 3y - 14 = 0 measured parallel to line x 2y - 1 (a) (c) 7 Vl5 (b) (d)

Let A(x0, y 0 ) be any point on t h e curve a y = (ex/a + e-"'a). Let P denote the length of the normal to the curve at A. Then a, y(l, P are in (a) AP (b) GP (c) HP (d) none of these. log (sin x)IX X

2. The number of points (x, y, z) in space, whose each coordinate is a negative integer such that x+ y + z + 12 = 0, is (a) 385 (b) 55 (c) 110 (d) none of these If x" = e"\ log x2 (a) (c) 4. (log x-1) 2 xy-2 ,, ,,T (log x - 1 ) g'(x),

(b)

(-,-2)u(-l,-jJ(d)

= Q j 521+1 and g(x) = 5' + 4x In 5, is (Ib) (0, 1) (d) (0, oo) y/3-i , then (/101 + z101)103 equals (b) z (d) none of theseMATHEMATICS TODAY | FEBRUARY '07

17. The number of ways in which 10 candidates At, A2, , Aw can be ranked so that ,4, is always above A2 is (a) (c)10!10

(a) (l,oo) (c) (0,oo) 24. If z = (a) iz (c) z

2

(b) 8! x "C2 (d),U

P

C2

18. There are three boxes each containing 4 white 72

Part II Time : 1 hr. Passage-1 Let ABC be a triangle. R be the circumradius of the triangle. Also given R2 = - ( a 2 + b2 + c2).8

Passage-3 Max. Marks : 86 Given a function 'g' which has a derivative g'(x) for every real 'x' and which satisfy g ' ( 0 ) = 2 and g(x + y) = eyg(x) + e'g(y) for all x and y. 30. The function g(x) is (a) x (2 + xex) (b) x(e" + 1) (c) 2x e* (d) x + ln(x + 1) 31. The range of the function g(x) is (a) R (c) -i,oo e (b) (d) [0,oo)

Section III : (Q. 25 to 34) (Comprehensions)

Then,

25. Which of the following is/are true? (a) (c) cos2A =

-1

(b> S (d)

c

?s2A

=

1

sin 2 ^ = 1

sin 2A = - 1

26. Hence the triangle ABC can be (a) equilateral (b) isosceles (c) scalene (d) none of these 27. Further, we have (a) r + 2R = 2s (c) / + 2R = s (b) r - 2R = j (d) none of these

32. The value of X>00 g(x) j s I'm (a) 0 (c) -oo (b) 1 (d) does not exist

Passage-2 A system of vectors a,, a 2 .... is said to be linearly d e p e n d e n t , if there exists a system of scalars cv c2, cn (not all zero) such that c,5,+c2a2+ + cfl=0 That means a t , a 2 a n are linearly dependent if one can be expressed as the linear combination of others. Again a,,a 2 a n is said to be linearly independent, if there exists scalars cv c2 cn such that c,a, +c2a2+ c. = c,= + ca = 0= c =0

Passage-4 Sometimes differential equations which are not linear can be reduced to the linear form by suitable transformations. For example an equation of the form cfy + Py-Qy where P and O are functions of x only ax or constants and and 1) is a constant. ... ~f+Py ax y~"+

= Qy"=

Q (1) [dividing by y both sides] (1 -n)y cfy dz ~T = ax ax

Lety- = z

Put the value of y dz

-n (b); (Q) -> (c); (R) -> (d); (S) (a) (P) (c); (Q) (a); (R) - (d); (S) - 4 (b) For detail answers visit : www.vidyalankar.org

MATHEMATICS TODAY | FEBRUARY '07

PRACTICE PAPER

IIT-JEE 2007PART -1S E C T I O N - I : (Q-l to 25) (One option correct)371

(c) ( 0 , )

(d) none of these

1.

If T < a < and z = 2 tan a + 2i then arg (z) is C (b)

(a) 3K - a -(c) a - 2 7 t

f-a

(d) none of these

7. In following series of three digit numbers, how many terms are common? 102,110,118,126, series first 108, 120, 132, 144, series second (a) zero (b) 37 (c) 41 (d) none of these 8. Value of b for which equation 2log j (bx + 28) + log 5 ( l 2 - 4 x - x 2 j = 0 has only 25 one solution is (a) 4 only (b) - 1 2 only (c) 4 o r - 1 2 (d) none of these 9. The sides of triangle are 3x + 4y = k, 4x + 3y = k and 5x + 5y = k units, where k > 0. The triangle is (a) right angled (b) equilateral (c) obtuse angled (d) none of these 10. ax + by + c = 0 is concurrent with x cos 0 + y sin 0 = 2 and x sin 0 - y cos 0 = 0 then for all values of 0 locus of their point of intersection is x2 +y2 = 4, then relation of a, b, c is (a) \c c '2 < 4 +bz yla2+b2 (b) ) c | < \la2 +b2

2. x,y, 0 e R, (x2- y2) sin 6 + 2xy cos 6 = x1 + f then tan 9 is equal to*2 -y2

(a) (c) 3. (a)

y 2 -* 2 (b) 2 xy

2 xy 2 xy 7 ^l

(d) none of these8 8 8

Value of sin' ( l 0 ) + s i n ( l 3 0 ) + s i n ( 2 5 0 ) is 93128

(b)

93 256

11 (c) 64 4. n e N the value of . ( n ) . (2n) sin sin I yin) \2n) (a) (c) V2 + l2"

(d) none of these

. (3n sin U n (b)

.sin v ^ZT

(n-l)n2

"

,

(c)

(d) none of these

(d) none of these 2" 5. Number of solutions of x + y + z = 20, 20x + 4y + z = 80, where x, y, z are non-negative integers is (a) one (b) two (c) zero (d) infinite6

Vw+T

11. ax + by + ab = 0, cx + dy + cd = 0, y = 0 these three lines intersect to make triangle ABC whose area is not 0 then necessary and sufficient condition is (a) ac (ad -be)* 0 (b) ac * 0 (c) ac (ad - be) (b - d) (d) none of these 12. 0.) represents integral part of X, then value of Lt j->0 sinx x is (b) 0 (d) none of these

neR+,(l

+ /)" = \ + nt+"-^^-t

2

+

(a) 1 (c) - 1

Expansion is convergent, largest interval for t is (a) [ - 1 , 1 ) (b)(-l,l)

13. kx+y=3,2x-y = 3, 4x + k y - 9k are concurrent lines then value of k

Contributed by Prof. S.S. Dahiya (Director), QOON Ace Education Pvt. Ltd. 66MATHEMATICS TODAY | FEBRUARY '07

(a) - 2 or 1 (c) - 2 14. ap a2, a,, a,1+2 L + A, + C 3

(b) 1 (d) none of these ak are non-negative integers such that + a = n where n > k then value of k Vtf,\al! a2\ ak\, (b) (ky (d) none of these.

22. The common root is (a) 1 or 2 only (b) 3 or 4 only (c) 2 or 4 only (d) 1 or 2 or 3 or 4 23. (a + b + c) is maximum when common root is (a) 3 (b) 4 (c) I (d) none of these 24. (a + b + c) is minimum when common root is (a) 2 (b) 1 (c) 4 (d) none of these 25. b + c is maximum when common root is (a) 3 only (b) 4 only (c) 3 or 4 (d) none of these SECTION - II : (Q-26 to 30) (More than one option correct) 26. a, b, c are sides of triangle if x y z then value of b + c-a c + a-b a + b-c + z) + y (z + x) + z(x + y) r ; is equal to ax + by + cz ^ 2 ( x + . y + z) a+b+c y +z (b) x+y c z+x

(a) ()* (c) n\ k\

15. ii', x, y, z are non-negative integers. l f w + jc + iy + z = 20 then from the solutions one s o l u t i o n is s e l e c t e d at r a n d o m , p r o b a b i l i t y that x w > y > x (b) w = x + y (c) 2y = z (d) z + x = w+y 28. a, b, c are non null vectors such that ax(b (c) xc) = ( a x i ) x c =0 then

47t

O'then

(a) cx(axb) which of the following is (are) true? (a) xy + yz + zx = 0 2 2 2 xy : 2 (b) xyz = cos (39) 2 2 ^ yz zx x y

(b) a x e = 5 (d) none of these

a, b, c are parallel

(d) none of these

29. n e N, cCx+ b"y + c" = 0, for three distinct values of n we get three distinct lines, which are concurrent then (a) c = 0, ab (b) b = c, a * 0 (c) a = c,b* 0 (d) none of these 30. An operation * is defined as x * y ~ Jx2 + y2 ~ xy where x & y are distinct real numbers, which of the following is(are) true (a) x * y > 0 (b) zero is identity of * operation (c) x * y = y * x (d) (x * y) * z = x * (y * z) 67

Direction - (Q.21 to 25) : a, b, c are natural numbers and equations x2-5x + a = 0,x2-6x + b = 0,x2-Sx + c = 0 have exactly one root common and the common root is natural number. 21. Which of the following holds? (a) 3a = 6 + 2 c (b) 3b = c + 2a (c) 3c = a + 2b (d) none of theseMATHEMATICS TODAY | FEBRUARY '07

PART - IISECTION-I : (Q-lto20) Comprehensions Passage - 1 a, b, c, d, e,f are non-zero distinct digits, number ab c d ef is six digit number and number d efabc is also six digit number in which first three digits and next three digits exchange their places preserving same order. Assume one of such numbers is NK so that (7 - k). Nk = k. N7_k fork- 1,2,3,4 V 5,6 using this rule we try to find numbers N{, NV NV A'4, JV,, and NS and name these as star numbers. 1. Which of the following is true (a) N6, NY N4 are possible and other star numbers are not possible (b) N2, NS, NF, N6 are possible and other star numbers are not possible (c) N N 6 are possible and other star numbers are not possible (d) NJ, N2, N3, NA, NS, N, all are possible. 2. For star numbers which are possible which" of the following is (are) true? (a) Nt From (i) and (ii) we get,P(ALNA2) = P(A,)-P(A2).

)

...(i) ...(ii)

2. ( c ) : Required probability P (getting 8) + P(9) + P(10) + P(11) + P{\2) 36 3. 4.5 5.

36

36

3626

36C3 x

3626

12

35. The mean of values 0, 1 , 2 , corresponding weight "C 0 , " C p "C2,2"2" '

( c ) : Required probability is

52

C3

(d) : Required probability is = 1 ^ = 8 oFO (c ) .PP(BIA) A ) ~ (BIP { A N B )

(a) (c)

n +1 n+I

(b) (d)

n(n +1) H

p(A)

-

J / 2

I ^ I -- 2 I.

2

6.

( d ) : Let P (fresh egg) =10 100\5 / , \0=

= ^

= P

36. In a series of 2 n observations, half of them equal to a and remaining half equal to -a. If the standard deviation of the observations is 2, then | a \ equals (a) JI/n 0>) V2 (d) Mn. (c) 2

P (rotten egg)

1 10/ -

q, n = 5, r 5n5

required probability that none egg is rotten f JJL)'10/ \io/

37. If variance of x = 9 and regression equations are 4x-5>' + 33 = 0 and 20x9y10 = 0, then the coefficient of correlation between A and y and the variance of y respectively are (a) 0.6; 16 (b) 0.16; 16 (c) 0.3; 4 (d) 0.6; 4 38. If bIA and b, are both positive v(where b xy are regression coefficients), then (a) V+

7. ( b ) : For a moderately skewed distribution, Mode = 3 median - 2 mean =>7 = 3 median-2>=4=> 15 = 3 median .'. mean = f 8. ( d ) : t f = ( Z / ) = 20A =A^ + l

observation = |N +L

observation = 3

and bxy

Similarly, Q3 = 3

observation

V

r

(f)" observation = 5Now Q.D. = 1 ( ^ - 2 , ) = 1 ( 5 - 3 ) =1. 9. ( a ) : Here ( x - x ) O > - J ) = 10, ^ ( x - x )2

(d) none of these. 39. If acute angle between the two regression line is 0,MATHEMATICS TODAY | FEBRUARY '07

^ 29

and E ( j - J ) 2 = 20 Hence the coefficient of correlation is given by r = = IX*-*)^-^)10 =

10

2x5! 1 10x9x8x7x6 126' 17. (d) : Ways of selection of two subset of A = (2")2 Ways of selection A(/!) = 1 -P(A) = 1-

18. (b) : P(A') = 0.3 P(A) = 0.7 .-. P(B) = 0.4 and P(A n B') = 0.5 P(A u B') = P(A) + P(B') - P(A n B') = 0.7 + 0 . 6 - 0 . 5 = 0.8(cV P(A.'B) P { A

19

P(B)

B )

=

P ( A

^ B) P{B)

1

"

n A

y P(B)

B )

P(B)

20. (b) : Since 4 has appeared on the first, so we are required 4 or 5 or 6 on second dice. Hence required probability = 3/6 = 1/2. 21. (d) : Required probability =10

C4|ij

7>(5) = 1 - / > ( 5 ) = 1 - = z J 3

|

_ 10^

(If6 6 12

22. (d) : We are given that n = 3, p = 1/6, q = 5/6 Mean = np - 3 x = 6 2 Variance = npq = 3x x = y 23. (a) : Let the mean of the remaining 4 observations be 5Ej _ nM - a Then, M = X] (-4) + 4 4 24. ( d ) : Since, root mean square > arithmetic mean

14. ( a ) : Current in the upper part will flow only if both the switches a and b are closed. Their probability = p x p= p1 Now current will flow in lower part of C, if C is closed, its probability is p. Thus current will flow from A to B if current flows either in upper part or flow in lower part, required probability = p 2 + p. 15. (d) : Let A occupy any seat at the round table. Then there are 14 seats available for B. If there are to be four persons between A and B. . Then B has only two ways to sit, as V J shown in the figure. J2_ = 1 Hence required probability 14 T 16. (b) : Let n = total number of ways = 10! m = favourable number of ways = 2 x 5! x 5! Since the boys and girls can sit alternately in 5! x 5! ways if we begin with a boy and similarly they can sit alternately in 5! x 51 ways if we begin with a girl. Hence, required probability:

i = 1

n

n

|400 > 80 v n

n> 16

Hence, possible value of n = 18. 25. ( a ) : Variance = (S.D) 2 = I

Sfn. . X = *

n

=

m _ 2x5!x5! n 10!

n( + l)(2w + l) 6

f h(h + 1) V 2n

n2-l12 '

26. (b) : Mean deviation is minimum when it isMATHEMATICS TODAY | FEBRUARY '07

48

considered about the item, equidistant from the beginning and the end i.e. the median. In this case median is101+1 th

- [(*/ -

) + (x, - x2) +..... + (x, - X, _ , )

^

i.e. 51 th item i.e. x5j. y=-jx+2 (.x,-x)2 ] 2

27. ( b ) : Here,

[v |x,-xy|=> J>,--*)2 /=1 < w S2

V / - \ 2 . AT2 7 I (*/"*) - 7 n

n-\

_

c o v ( x y) >ar(x)->ar.O/)

_

8 V9VT6

S < r,

\n-Y

83x4

23 y = -|-*+

33. (d) : Required angle for food and clothing 400 x360 = 144. 1000 +13

29. (c) i Here 3 a + 2y 26 and6A

34. (b) : M

A1+A2+A3+...A

+ j> = 31 = > =

x =

1

31

n nM = x{ + x2 + .... xn _ j + xn nM-xn = A, + x2 n- J

30. (c): The angle between lines of regression is 6, then tanO = here 6 = 90 .\ r = 0

+ x' _ xx + x2 +... + xn_\ + x' n nM -x + x' New average = n 35. (d) : The required mean is _ _ 0 1 + 1 " Q +. 2 " C2 + 3 " C 3 +... n n C 1 +"Cl + "C2+...nCn

nM -xn

2 2 GX+GYJ

There is no linear correlation. 31. V b7) : L e t ^ = ^ (C

i.e. J = >

C

+ *C r =0 I r =0 "Cr r =a

i . e . y = Ax + B where A =, B = c c /. y = = ^ Z ( Y - y )1 => /? GY AY2

Ax+B

r =0

r =0

I

n

cr

=> CP-?)2 =2 n-G"X

=A2(x-x)2

V* - 1 M n 2 _ n_ 2" 2' 36. (c) : Let a, a .... n times and - a , - a , ... n times i.e mean = 0 and S.D. = n(a-0)2 + n(-a-0)2

A*(X-X?2 2 => GY = = 2 2 A GX

= A

= \A\

GX

=> A;,

InI na2 + na2 2~n=

Thus new S.D.0

Hi

=

,

32. (a) : We have r = m a x \ X j - X ; \

, , ^ TT Hence \a\ = 2.

37. (a) : var(x) = G2 = 9 (given) /=1 Now (x,-x)2= \ X , X I + X 2 +

The given lines of regression are 4JC-5^ + 33 = 0, 20x-9y10 = 0" "+ XH

i,.

y =j x

+

f

and

+

f

M A T H E M A T I C S TODAY

FEBRUARY '07

4(

F r o m here, w e obtain Now

b

y x

4 = ~ and b ^ =

9

39. (c) : t a n e = |

1=\

axOj, 2x

2 y j of the same

Since A . M . of t w o quantities > G M . quantities Hence => - f 9 - M= t a n 9 < ~ - ^ => t a n 2 0 < 'a

l

>' = y ' " Y y)

=

16

v a r ( ^ ) = 16 36 5x3x4v s i n 2

2r

2 \2 ~r~

2r

N o w p(x,

cov(s, j ) a ^ ay

e =

0.6

- J a ^ f 1 + tan 9 1 -r4

H e n c e coefficient of correlation = 0.6 and variance of

16.3 8 , ( b ) : W e k n o w t h a t byx+bxyb

V sin2 8 < > 2^jb~bxy

I

2r 2\

=> s i n 2 8 ^ 1+ r2

1 + 1 -r

2r

=> byx + bxy > b^+k

2 r

yx xy

+b

22r

since 1 - r 40. ( c ) :

2

< 1+ r

2

a n d s i n 2 9 < 1 => s i n 9 < . ^

1 - r2.

1 xy 16

cov(x,j;) axGy

0.6

.

16 4a,

^r

J _b

+

J _ >b

203

xy

yx

'

4x0.6

OO

Wl-fcG-

On the Latest Pattern

AIEEEEXPLORERB a s e d on L a t e s t P a t t e r n 10 Very Similar Practice Test PapersEE H H H fiii XPLORR : m 7wmv^km^f H P CB$fl?^MI;

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MATHEMATICS TODAY | FEBRUARY '07

PRACTICE PAPER

IIT-JEE 2007PAPER ISection - 1 (Q.No. 1 to 15) (Only one option correct) 1. If a, b, c are positive then minimum value of a+3 - + b+3 + -c +3 b+c c+a a+b (a) 6 (b) 3 (c) no conclusion can be drawn (d) none of these 2. ax + by + ab - 0, cx + dy + cd = 0, y = 0 intersect each other to make triangle whose area is non-zero then necessary and sufficient condition is (a) ac (ad - be) * 0 (b) ac* 0 (c) ac (ad-be) (b-d)*0 (d) none of these 3. a , (3, Y, 5 are the smallest positive angles in ascending order which have their sines equal to k where - 1 < k < 1 then 4sin^- + 3sin + 2sin^- + sin^- is equal to H 2 2 2 2 (a) 1 v r r * 2 (c) 2-fl-k (b) 2VT7I (a) 30 (c) 120 (b) 60 (d) None of these 7. Two series of three digit numbers are 102, 110, 118, 126, and 108, 120, 132, 144, number of common terms in these series is (a) 37 (b) 41 (c) zero (d) None of these 8. a and b are distinct natural numbers such that a2 + b2 is divisible by ab+ 1 then (a) either a = 63 or b-c? (b) either a - b3 or b = a3 and a > 1 (c) either a - b2 or b = a2 and a > 1 (d) none of these 9. (a) (c) In AABC, interval for a cos A + b cos B + c cos C a+b+c (b) (0,1]

i-'i

(d) none of these

(d) none of these

10. a, b, c are non-zero real number line ax + by + c = 0 is concurrent with lines x cos 6 + y sin 9 = 2 and x sin 0 -y cos 0 = 0 then locus oftheir point of intersection for all values of 0 is (a) x2+y2 = 4 and (b) x2+y* = 4 and (c) x +y2 2

4. Six persons have alotted seats while they take seats at random. Probability that exactly two persons are found at allotted seats is _3_ 3 (a) 16 (b) i 5 (d) None of these (c) 12 5. If k = "Cj0 then k is not divisible by which of the following (a) 8 (b) 29 (c) 31 (d) 23 6. Coefficient of a 4 bs cJ is expansion of (ab + be + caf is

< 2 ^ +b a i < AA +b | 0 for > 0 and fixy) = fix) . fiy) and for 1, fix) * 1 then which of the following hold fix) = x1 where k e real, k* 0 fix) is continuous for x > 0 fix) is bijective for x e (0, none of these 2 for xe\ 0 28. / ( * ) = then which of 2 2l + x

(c) local minima exists at two points in (d) local maxima & global maxima coincide ANSWERS (c) 3. (d) (b) 7. (c) (a) 11. (c) (d) 15. (c) (b) 19. (b) (a,b,c) 23. (a,b,c) (a,c) 27. (a,b,c)

^

tan

(x)

the following holds

13. 17. 21. 25.

(b) (d) (a) (b) (a) (a,b,c) (a,b,c)

2. 6.10.

14. 18. 22. 26.

4. 8. 12. 16. 20. 24. 28.

(a) (b) (a) (a,b,c) (c) (a,b,c) (d)

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MATHEMATICS TODAY | J N A Y '07 AUR

PRACTICE PAPER

IIT-JEE 2007PAPER - IISection - 1 Read the following passage carefully and give the answers. (Q.No. 1 to 20) Passage - 1 : S = ax2 + 2hxy + by2 + 2gx + 2fy + c is general expression of second degree is x and y and S = 0 is general second degree equation is x and y. Solving quadratic equation : ax2 + 2(hy + g)x + by2 + 2fy + c = 0, a * 0 _-{hy + g)yl(h2

4. x2 2xy + 4yi 4x Sy+l6 = 0 represents (a) pair of lines (b) point (c) ellipse (d) None of these 5. 3x2 - 6xy + 3 / - 1 Ox + 1 Oy + 15 = 0 represents (a) pair of parallel line (b) parabola (c) point (d) None of these 6. Pair of lines Lv Z,2 is ax2 + 2hxy + by2 = 0 and Lv Lt is (a + k) x2 + 2hxy + (b + k) y2 = 0 for k 6 R - {0} are (a) angle bisector of each other (b) angle between Lv L3 is same as angle between Lv L4 (c) angle between Lv L4 is same as angle between Lv L3 (d) none of these 7. x3 + kx2y + kxy2 = 0 represents three distinct lines passing through origin then value of k is given by interval (a) [ - 1 , 3 ] (b) R [1,3] (c) R {1, 3} (d) none of these

- ab)y2

- l{af - gh)y + g 2 ~ ac

a Discriminent of expression within the square root is -4a(abc + 2fgh - af - bg2 - ch2) Solving quadratic equation by2 + 2(hx +f)y + ax2 + 2gx + e = 0, b*0 -(hx + f)+y](h2-ab)x2-2{bg-hf)+ b f2-bc

8. fgh * 0, 2hxy + 2gx + 2fy + c = 0 represents a Discriminent of expression within the square root is hyperbola then 2 2 4b(abc + 2fgh - a f - b g - ch ). (a) 2fg = ch If abe + 2 f g h af bg2 ch2 = V and h2-ab = \ then (b) 2 \ f g * e h values of V and A, determine the geometry of (c) no conclusion can be drawn5 = 0

(d) none of these 9. fgh * 0,2hxy + 2gx + 2fy + e = 0 represent a hyperbola then equation of its conjugate hyperbola is 4Jg (a) 2hxy + 2gx + 2fy + - c = 0 h 2Jg_ (b) 2hxy + 2gx + 2fy + c = 0 h 2 fg (c) hxy + gx +fy + = 0 (d) none of these Passage - 2 : ABCD is a cyclic quadrilateral for ABCD, AACD, AABD, AABC the orthocentres are P, Q, R, S respectively and Gv Gv G}, G4 are their centroids respectively.

1. ax2 + 2hxy + by2 + 2gx + 2Jy + c = 0 represents pair of lines and its line members intersect at a point on j'-axis other then origin, then (a) (b) (c) (d) bg = hf a n d / 2 = be bg = hf* 0 a n d / 2 = be * 0 h2(iii), (b)->(iv), (c)->(ii), (d)^(i), (e)->(v) 22. (a)->(v), (b)->(iv), (c)->(i), (d)^(ii), (e)->(iii) 5. 9. 13. 17. (b) (d) 23. (a) 27. (a) 24. (a) 28. (c) 25. (c) 26. (c) 1.

MATHEMATICS TODAY |FEBRUARY'07

77

Solved Paper1. If a = 2i + 3j-k, b =i + 2j-5k, c=3i + 5j-k ,

then a vector perpendicular to a and in the plane containing b and c is (a) - 1 7 / + 2 1 j-91k (c) 17/ 21 j + 91k 2. (b) 17? + 21 j-\23k

8.

The conjugate of the complex number (b) 1 + / (d) - 1 - /

- is

(a) 1 - / (c) - I + I

(d) 17/ 21 ]-91k

9. ABC is a triangle with jA = 30 , 5 C = 10 cms. Tlie area of the circum-circle of the triangle is(a) IOOtt s q . c m s (b) 5 sq. c m s

OA and BO are two vectors of magnitudes 5 and (b) 15 (c) -15 (d) 15^3

6 respectively. If j BOA = 60, then OA OB is equal to (a) 0 3. A vector perpendicular to the plane containing the points A( 1, -1, 2), B(2, 0, -1), C(0, 2, 1) is (a) 4/ + 8 j - 4 k (c) 3 / + j + 2k4

(c) 25 sq. cms

... 100* (d) ---- - sq. cms

10. If sin 36 = sin 0, how many solutions exist such that -2n < 0 < 2n ? (a) 8 (b) 9 (c) 5 (d) 7 11. A graph G has 'in vertices of odd degree and '' vertices of even degree. Then which of the following statements is necessarily true ? (a) m + n is an odd number (b) m + n is an even number (c) n + 1 is an even number (d) m + 1 is an odd number 12. If P is any point on the ellipse - - + j ~ = 1, and Sx2 y2

(b) Si+4j + 4k (d) i+j-k I" ' ( 3 - 1 ) ( 3 k + 2)

-

2 5 + 5 8 + 8 11 +6-4 n

(a)

6n + 3 (d) 6T4 's6n + 4 5.1 The ninth term of the expansion \ ^ x " ' 2 x j -1 (b) 512x9 1 (d) 256x"1l (a) 512x9 -1 (c) 256-x8 6. The solution of tan 1 (c) -V3and S' are the foci, then PS + PS' = (a) 4 (b) 8 (c) 10 13. The value of sin 2 cos V5 (a)S(d) 12x + 2cot ' x I ( > 73b2n (b)x2 V22-JS(c)4V52V5 (d) ~(d) V37. sin 2 17.5 + sin2 72.5 is equal to (a) cos2 90 (b) tan2 45 2 (c) cos 30 (d) sin2 45MATHEMATICS TODAY |FEBRUARY'0714. If - - ~ = 1 is a hyperbola, then which of the 36 k~ following statements can be true ? (a) (-3, 1) lies on the hyperbola (b) (3, 1) lies on the hyperbola (c) (10, 4) lies on the hyperbola (d) (5, 2) lies on the hyperbola 6515. The focus of the parabola is (a) (c) -1 3 '2 J I 3 3'2 (1-3 (b) i 3' 2 W (d) 1 _1 3' 2 2 0 2 a and B is(d) pq (x - p) (x - q) 21. The imaginary part of i' is (a) 0 (b) 1 (c) 2 22. The amplitude of (1 + if is 3C T -3n -57t (b) (C) (a) T ~4~ ~4~ (d) - 1 5ttl -1 1 " "4 16. If A = 2 1 - 3 ,10J? = - 5 1 1 1 l(d)-2 323. ABC is a triangle. G is the centroid. D is the mid point of BC. If A = (2, 3) and G = (7, 5), then the point D is (a) f,4j (b) l y , 6 1 (c) (b)J9the inverse of A, then the value of a is (a) 2 (b)0 (c) 5 (d) 4011 u2'2(d)x16. . ,. t a n ( x ' - l ) . . 24. lim s -- is equal to Jt->i x - 1 (a) 2 (c) - 2 25. If.V = 2 logx17. If A= x 5 0 9 values of x are (a) 0, +12, -12 (c) 0 , 4 , - 4 1 -2 If A = 0 2 _3 - 2 "5 l (a) f1 is singular, then the possible x (b) 0, 1, - 1 (d) 0, 5, - 5 2 - 3 , then A- adj (A) 4 "5 0 0" (b)*(b) (d) , then ax is (b) 2'0g r-log2 (d)2 log* (a) fogs 2 ig* (c) 26. 721"875 is equal to (a) log735 (b)51 5 1 1 1 5 "8 0 0" 0 8 0 0 0 80 5 0 0 0 5 "0 0 0" 0 0 0 0 0 0(c) 25(d) log725(c)(d)27. In the group ( C , l 5 ) , where G = {3, 6, 9, 12}, 15 is multiplication modulo 15, the identity element is (a) 3 (b) 6 (c) 12 (d) 9 28. A group (G, *) has 10 elements. The minimum number of elements of G, which are their own inverses is (a) 2 (b) 1 (c) 9 (d) 0 29. If a and b are vectors such that19. I f f : R~>R is defined by f ( x ) = |x|, then, (a) f~](x) = - x (b) /""'(*) =(c) the function /"'(x) does not exist 1 (d) / " ' ( * ) = ix ppq\a + b\=\a-b\, then the angle between a and b is (a) 120 (b) 60 (c) 90 (d) 30 30. (a) ,3x2 + 1 is equal to x -6x+8320. The value of \ |p q x (a) x(x-p)(x-q) (b) (x -p)(xq) (x + p + q) (c) (p - q) (x- q) (x- p) 66l3-2(x - 4 ) 2 ( x - 2 ) 49 13_ b ( ) 2(x-4) 2(x-2) +MATHEMATICS TODAY |FEBRUARY'07(c) (d)-49 2(jc-4) 49 2(x-4)13 2(x-2) 13 2(x-2)(c) (-5,1)(d) (1,-5)31. The number of common tangents to the circles x2 + / = 4 and x2 + y2 - 6x - 8j> - 24 = 0 is, (a) 3 (b) 4 (c) 2 (d) 1 32. If 3x + y + k = 0 is a tangent to the circle x2 + y2 = 10, the values of k are, (a) 7 (b) 5 (c) 10 (d) 9 33. The negation of the proposition "If 2 is prime, then 3 is odd" is (a) if 2 is not prime then 3 is not odd (b) 2 is prime and 3 is not odd (c) 2 is not prime and 3 is odd (d) if 2 is not prime then 3 is odd 34. The equation of two circles which touch the F-axis at (0, 3) and make an intercept of 8 units on X-axis are (a) x2 + y 2 lOx - 6y + 9 = 0 (b) x2 + y2 6* - lOy + 9 = 0 (c) x2 + y2-8x )0y + 9 = 0 (d) x2 + y2 + lOx 6y + 9 = 0 35. The orthocentre of the triangle with vertices ,4(0,0), s f 0 , | ] , C ( - 5 , 0 ) i s (a)39. The area enclosed by the pair of lines xy = 0, the line x - 4 = 0 and y + 5 = 0 is (a) 20 sq. units (b) 10 sq. units 5. (c) . sq. units (d) 0 sq. units 40. If the area of the auxiliary circle of the ellipse x2 v2 - y + - j = K > b) is twice the area of the ellipse, then a' b the eccentricity of the ellipse is (a) ^ 1 (b) J3 f41. The range in which y = -x2+ 6x-3 is increasing is (a) x < 3 (b) x > 3 (c) 7 < x < 8 (d) 5 < x < 6 42. The value of the integral J (sin100 x - cos100 x)dx is J (a) 100 100! (b) ( 100 yoo (d) 0/ \ n ^ Too( I ' D ^ ( f ' ! ) < c > (- 5 >f) )36. x 2 - 6x - 6y + 4 = 0, x2 + y2 -2x -4y + 3 = 0, 2 x + f + 2kx + 2y + 1 = 0. If the radical centre of the above three circles exists, then which of the following cannot be the value of k ? (a) 2 (b) 1 (c) 5 (d) 4 37. If the circles x 2 + y2 - 2x - 2y - 7 = 0 and x2 + y2 + 4x + 2y + k = 0 cut orthogonally, then the length of the common chord of the circles is (a) j f i (b)2 (c) 5 (d)843. OA and OB are two roads enclosing an angleof 120. Zand y start from '?' at the same time, ^travels along OA with a speed of 4km/hour and Y travels along OB with a speed of 3km/hour. The rate at which the shortest distance between X and Y is increasing after 1 hour is(a) -Jyf km/h (c) 13 km/h 44. If k jx.f(3x)dx 0 (a) 9 (c) | 45. The value of(b) 37 km/h (d) ^ 3 km/h = jt f(t)dt, then the value o f * is 0 (b) 3 (d) |38. The co-ordinates of the foot of the perpendicular drawn from the point (3, 4) on the line 2x.+ y - 1 = 0 is (a) 12 5 , (b) (1,5)f n ^ g ^ i s 67MATHEMATICS TODAY | AUR '07 JNAY(a)M28*+ C(b) (d)ta 8 v" - +CO(c) tan4x + C 46. If sec"' |ta"i^ +c7-axis is (a) 16 sq. units , , 32 sq. units (c) y(b) 32 sq. units ,.. 16 (d) y s q . unitsj = a, then ^ is54. The differential equation of the family of straight lines whose slope is equal to ^-intercept is (a) ( x + l ) J 7(c)x-\ yi-fSt(d) y+Y=0(b) (x + l)~ x+i +1+y =0(c)dy _ x - 1 idx~y47. If y - cos 2 --sin 2 then is 2 2. dx~ (a) (c) -9y (b) (d) 9y 3,/Ty55. The order and degree of the differential equationdx2 (a) 1,5 (b) 2, (c) 2 , 52(d) 2, 31 cosx for .r * 0 48. If the function / ( * ) H x is [ k for x = 0 continuous at x = 0, then the value of k is (a) 1 (b) 0 (c) \256. The point on the curve j? = x, the tangent at which makes an angle 45 with X-axis is (a)J 14 '2(b) (d)1 L 2*4(d) - 149. If 1, (0, to are the cube roots of unity then (1 + C ) (1 + co2) (1 + C 4 (1 + C 8 is equal to O O) O) (a) 1 (b) 0 (c) co2 (d) co 50. If xx=yy, then ^ is (b) y 1 + logx (d) I + logy, 1 -1 (c) [ 2 ' 257. The length of the subtangent to the curve x2y2 = a4 at ( - a, a) is (a)2(b) 2a -LA(c) U (V, a(d) W2(a)458. The number of positive divisors of 252 is (a) 9 (b) 5 (c) 18 (d) 10 59. The remainder obtained when 5124 is divided by 124 is (a) 5 (b) 0 (c) 2 (d) 1 60. Which of the following is not a group with respect to the given operation ? (a) the set of even integers under addition (b) the set of odd integers under addition (c) {0} under addition (d) {1, -1} under multiplication.* ANSWERS(c) 1 + log51. The value of \e"(x5 + 5x4 +\)-dx is (a) e r x5 (b) e* x5 + e* + C x+] 5 (c) e x +C (d) 5x4 e* 52. The value of J]+ ' dx is x2-1w(0Iog(fri)+cx 1+ JCoiogfei)+c*+(d) log(x2 - 1) + c53. The area bounded by the curve x = 4 - y and the 681. (d) : Any vector perpendicular to a and in the plane containing ft and c is given by a x ( b x c )MATHEMATICS TODAY |FEBRUARY'07i bxcjkTr +1_ n(~r x "For (x + o f +1 256* 2n1 2 -5 = 2 3 / - 1 4 / - * 3 5 -1 ; k 3 - 1 -1 . -17/ - 2 1 j-91kTq = 8 Q(3x) 6. x(6 xc) = 2(d): tan ' x + 2cot ' x =123 - 1 4 :. 2.=> tanx + 2tan11=2n 2n "3/required vector is = -17/ -21 j-91k (b) : OA BO = jai||50jcose2(1/x) tan f x + tanV \X/(where 0 is the angle between OA and BO ) = OA OB cos 60 where OA and OB are magnitudes of the vectors = 5 x 6 x ^ = 15. 3. (b) : Vector perpendicular to plane containing AxB+BxC+CxA tan i .3(1,-1,2),fj(2,0,-l),C(0,2,1)is i2/ 2/x S o V x 2x x + 2x 2x x2-l j 2tc T271 T 2n T2n 3 x(x +1) (x +l) 2tc 32tc2j0k 2 = i+5j + 2k-1Now AxB = 1 - 1 i BxC = 2 j k0 - 1tan i| x + x2-x -Ttan2n 3= 2i-2]+ Ak. 2n => tan (x) = => tan_t_i - tan x tan -0 2 / Cx A = 011 | k 1 = 5i+j + (-A)k2271 x = - - =5j 2-1=> x = -tan(120) => x = -tan(7t-60) => x = tan60 => x = J l . 1. (b): sin2 17.5 +sin 2 72.5 = sin2 17.5 + sin2 ( 9 0 - 1 7 . 5 ) = sin2 17.5 + cos2 17.5= 1 = tan2 45 8.2 i+r+2/ 1-1+2/ (d): (1 + 0 i-/ i-z 1-/ 2/ 1 + i _ 2/(1 + /) _ 2/(1 + 0 1 - / 1+/ l_(lf l-(-l)AxB + BxC + CxA = Si+4j + 4k J_ < > 2-5c :J_ , JL , 5-8 8-11+1 (3-l)(3w + 2) . - L \3-l 3n + 2^i_I + I - I 2 5/ V5 1 2 1 3w + 2I--LU V8 117- U2/(1 + /) (3) 6(3n + 2)_1_=l+li- 1.3 + 2 - 2 2(3w + 2) 5. (d): 3x6n + 42x When we expand the given binomial, we get 9 terms, 9th term is the last term.MATHEMATICS TODAY |FEBRUARY'07Required conjugate is - / - 1 9. (a): In AABC, ZA = 30 therefore ZBOC = 60 OB = OC = radius of circle ZOBC = ZOCB = 60 AOBC is an equilateralB^69triangle i.e. having all sides equal therefore radius of circle = 10 cm < so, area of the circle = nr2 = n (10) 2 = 100 n sq.cms. 10. ( d ) : sin30 = sin9 . sin36 - sin0 = 0 ' 0x -36 Now for x 2 > 0 => x 2 > 36 This is true only for point (10, 4) (10, 4) lies on given hyperbola. 15. Question is incomplete. "1 - 1 1 " 16. (c) : A = 2 1 and B = A~ (Given)l2 cos(20) sin(0) = 0 => cos20 sin0 = 0 => cos20 = 0 or sin 9 = 0, n, 2n V cos20 = c o s ^ - J or 0 = 0, n, 2n => => 2 0 = | - or 0 = 0,71, 271 9 = or 9 = 0 , 7 1 , 271"4 12 0 -22" a 31 1- 3 , 105 = - 5 1Cn Adj A =^12-21 C22M3 -23 C13=lC23 =-2/. Total number of solutions = 711. fl>)12. (d) : The sum of the focal distances of any point on an ellipse is constant and equal to the length of the major axis of the ellipse (i.e. 2a) PS+PS' = 2a (i.e. length of major axis) = 2 x 6 (Here a = 6) = 12. 13. (c) : sin 2 c o s - ' ^Q l Q2 C 3 3 C!2 = - 5 C = 4 C 22 = 0 C21 = +2 C 32 = +5 C31=2 4 Adj/4 = -5 1 4 10 -5 1 2 0 -2 2 0 -2 2 0 -2 2" a 3_=C 33 = 32 5 3_ 2' 5 3 "4 -5 1 2 0 -2 2" 5 3_Ml = 1(1 + 3) + 1(2 + 3) + 1(2 - 1) = 4 + 5 + 1 = 10sin cos - lJl3-1 [ v 2 cos * x = cos >(2x2 - 1 ) ] "4 10JS = -5 1sm cos '( 2 - | - l sin m M l - f l sin sin sin -if x2 80 81-= sin cos - i f f [ v cos1x = sin ' V l - x 2 ~ \Alternative Solution : AB = I => 10/15 = 10/ "1 -1 1 1 1 " -3 1XJ 81 V0 _ 4^5 9 ~ 9 '0B) = 10/ ' 4 -5 1 2 0 -2 2" a 3_="10 0 00 10 00" 0 102 114. (c) : =>>-2 ~ y = 1 is a hyperbola 4 + a - 9 = 0 => a = 5 0 17. ( a ) : A = x 0 Therefore \A\ = 0MATHEMATICS TODAY | JULY '07x 5 916" 7 x is singular matrix.F>02 3 6 xt y 2 x2 1 Now = - 1 = * 36 3670Ml = 0(5x - 63) - x(x2 - 0) + 16(9x - 0) = 0 ~x3 + 144x = 0 => -x(x 2 - 144) = 0 => x(x2 - 144)= 0 => x = 0 , (x2 - 144) = 0 => x = 0, x 2 = 144 => x = 0, x = 12 => x = 0 , +12, -12. "1 - 2 18. (c) : A = 0 C = 8 - 6 C 2 1 = - ( - 8 + 4) C31 = +6 - 4 Qi Adj =C2" -3 423 - 2C 12 = -(0 + 9) C13 = 0 - 6 C 22 = 4 - 6 C 23 = - ( -2 + 6) C 32 = - { - 3 - 0) C 33 = 2 - 0 C,3-iT= (x-?)[(x2-jp2) + 9(x-p)] = (x - ?)[(x - />)(x + />) + ?(x - />)] = (x - q)(x-p)[x + p + q] = (x - q)(x - p)(x + p + q) 21. (a):A = t logA = log/' => logA = / log / log/f = / log(0 + 0 log/I = /[0 + m/2] log/! = -n/2 A = e"7"2 Therefore imaginary part is 0. 22.(d):(l + 05=(V2j ^ = ( ^ ) 5 ( c o s f + /sinJ = (V2)5(cos^+/smf 1.5C]221 C22 Q>3 C32 Q3 r '2 - 9 - 6 2 4 2 -2 -4 3 2- 2=+4 -2 -4 2 -9 -6 4 -2 -42 3 2 2 3 2=-9- 6A Adj A = 0 3"8 0 02- 2.*. Amplitude =^23. (b) : Here (x yx) = (2, 3), (p, q) = (7, 5) As G is centroid (P, q) fh+*2+x3 V 3 X] + X2 Xg yi+yi+y?} 3 J 7,s= 00 8 0 0 8 AiAd[A[)Alternative Solution: Adj AMlI M| = (AdjAA"Ml"8 0 0+^2 +>3 _ 3= > 2 + x2 +X3 = 2 1 ,53 + y2 + y3 = 15= > x 2 + x 3 = 19,i.fx\cNow Ml = 8 => ,4(Adjy-,+y3=l2= 8/ =08 0 tan(x -l) xl:0 0 8 19. ( c ) : / ( x ) = | x | f ( x ) = x if x > 0 = - x if x < 0 Therefore the function f~\x) x p q2-(I 4fo")124. (a) : Km does not exist. lim-2xsec 2_(x 2 - l ). lim 2xsec 2 (x 2 -1) = 2 1 sec 2 (0)X= 2 - 1 - 1 = 2. 25. (d) : y = 2ios* ax26. ( c ) :20. (b) : P x 1 'p q x = x(x2 - q2) - pipx - pq) + qipq - px) = x(x - q)(x + q) - p2(x - q) -pq(x - q) = (x - g)[x(x + 4) - p 2 - pq] = (x - >2 + 5x + 3j> + 4 = 0 Let C, and C2 intersect in the points A and D. Let a straight line BC pass through the points A and B be terminated by the circles C, and C2. 28. The length of the common chord is (a) 2 (b) 4 (c) 6 (d) None of these 29. The ZBDC is (a) always constant (c) always obtuse 30. (a) (b) (c) (b) always a right angle (d) None of theseSection IV : This section consists of 4 questions. Answers are to be given in between 0000 to 9999 in the f o r m of nearest integer. Each question carries +6 m a r k s and no negative marks. 33. From a point, common tangents are drawn to the circle x2 + y2 = 8 and parabola y2 = 16x. Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola. 34. Find the value of 100R if R is the radius of the smallest circle which touches the straight line 3x - y = 6 at (1, - 3 ) and also touches the line y = x. (Take =1.41 and S =2.24). 35. One of the diameter of the circle circumscribing the rectangle PQRS, 4y = x + 7. If P and Q are the points ( - 3 , 4 ) and (5,4) respectively, then find the area of rectangle.MATHEMATICS TODAY |DECEMBER'05The length of BC is a maximum when CD is perpendicular to AB CD makes an angle of 60 with AB CD passes through the centre of the circle through the points B, C and D (d) None of these Comprehension - 4(Question 31 & 32) Most of the formulae related to circle, parabola, ellipse and hyperbola are expressed in terms of three usual notations,7636. A triangle ABC, the length of whose sides are a (4 cm), b (8 cm), c (6 cm) is placed so that the middle points of the sides are on the axes. If equation of the plane is + + = T , then what will be the value of a (3 y 2a 2 ? Section V : This section consists of 4 questions has two columns with 4 entries in each column. Entries of column I a r e to be matched with column II. O n e entry of column I have exactly one m a t c h i n g in column II. Each question carries +6 m a r k s if all correct m a t c h i n g a r e indicated. Match of following: 37. Column I (P) The number of common tangents to the circles x + y - x = 0 and x2 + y2 + x = 0 is (Q) The length of the chord cut off by y = 2x + 1 from the circle x2 + y2 = 2 is (R) If a circle passes through the points of intersection of the co-ordinate axes with the lines he-y + 1 = 0 and * - 2y + 3 = 0, then the value of X is (S) For the circle x2 + y2 = 36, the length of the chord lying along the line 3x + Ay - 15 = 0 is 38. Column I (P) The normal chord at a point ' / ' on the parabola y2 = Ax subtends a right angle at the vertex. Then t2 is equal to (Q) The area of the triangle inscribed in the curve y2 = Ax, the parameter of coordinates of whose vertices are 1, 2 and 4 is: (R) The number of distinct normal possible from2 2 239.Column IC o l u m n II (A) 4(P) The radius of the circle passing through the foci of the ellipse x2 f + = 1 and having centre at 16 9 (0, 3) is (Q) If the length of the major axis of an ellipse is three times the length of its minor axis, then its eccentricity is (R) The eccentricity of hyperbola whose latusrectum is 8 and conjugate axis is half the distance between foci, is (S) The value of m for which y = mx + 6 is a tangent to the hyperbola x 10049(B)2>/2(C)>/l7 20Column II(D)4^(B) 36 IE= 1 is(C)240. Consider the general equation of second degree ax2 + by2 + 2hxy + 2gx + 2fy + c = 0. If this represents a pair of straight lines, map the two columns in the most accurate sense. Column I Column I I c (P) If ( j t , , ^ ) is the point of (A) 1 yj(a-b) +41r intersection of the two lines then (ox, + h{) (hxx + byt) = (Q) af + bg2 + ch2 = (R) The lines are parallel if h*= (S) Product of perpendiculars (B) (C) (D) ab gf abc + 2gh(D)6^3Column II (A) 4 1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 38. 39. 40.ANSWERS(a) 2. (b) 3. (a) 6. (b) 7. (b) 10. (a) 11. (a,b,d) 14. (b,c) 15. (a,c) 18. (b,d) 19. (b) 22. (a) 23. (d) 26. (a) 27. (a) 30. (c) 31. 34. 0147 35. 0060 (B); (Q) (P) (A); (R) (B); (Q) (P) (D); (R) (P) (D); (Q) (B); (R) (P) -> (C); (Q) (D); (R) (b) (c) 4. 8. 12. 16. 20. 24. 28. 32. 36. (S)- (S) -> (S) (S) (a) (a) (b) (a,b) (a,b) (a) (b) (c) 0021 (D) (A) (C) (A)(B)2(d)(b) (a,b,c) (a,c) (b) (d) 0032 -(C); ->(C); -(A); -(B);(C)3/IIU1'4.to the (D)parabola y = Ax is (S) If the normal at (a, 2a) on y2 = Aax meets the curve again at (at2, 2at), then the value of \t - 1| is :76 MATHEMATICS TODAY |DECEMBER'0575(a) (c)meso product laevo product(b) (d)dextro product racemic product114.! Q I ,H--7-4> O . What is O ?118. The ionisation constant of a weak electrolyte is 25 x 10"6 while equivalent conductance of its 0.01 M solution is 19.6 S cm 2 eq~'. The equivalent conductance at infinite dilution will be (a) 384 (b) 196 (c) 392 (d) 250 119. In the following reaction, FeS 2 + 0 2 - > F e 2 0 3 + S 0 2 if x equivalents of Oz are taken then find out the equivalents of F e 2 0 , and S O : with respect to FeS 2 respectively. , I Ox x (a) x; x (b) 11 II X , X 1 Ox (d) yy. (c) X, 2 120. The number of unpaired electrons in [Cr(CN) 6 ] 3 is (a) 5 (b) 7 (c) 6 (d) 3OilMATHEMATICS(d) none of these 121. If a system of linear equations x + ay + a2: = 0. x + by + b2z = 0, x + cy + c2: has a non zero solution, then (a) a + b + c = 0 (b) either a - b or b - c or a = c (c) a, b, c are in A.P. (d) none of these 122. If z, (/ = 1 to 6) represent the vertices of a regular hexagon, centered at origin with side 1 unit which has one of its vertices on +ve x-axis then zlz2z3z4zsz6 = (a) - 1 (b) 1 (c) 0 (d) oo .. log(2 + x 2 ) - l o g ( 2 - x 2 ) . , 123. hm , = k , the value of k is .v->0 x2 (a) - 1 (b) 2 (c) 1 (d) 0 124. Let T denote the number of triangles that can be formed using N vertices of a regular polygon of N sides. If Tn+lT = 21 then value of n is (a)OH115. NH, (a) (c)A nitrile on hydrolysis followed by treatment with and then on heating gives an amine (b) an amide a nitrile (d) a carboxylic acidr,116. Ethyl ester H,C 2 (a) H,C.,,CH.MgBr excess'> (/>. The product (P) will be CH, (b)C2H5OHII3COH OC,H,H,C2OC,H, (d)H3C7(b) 6(c)5(d) 4(c) H,COHH5C2117. H.C - CH - CH,C = CC0 2 H OH O i125. I f w is one of the roots of the equation ax2 + bx + c, where co is complex cube root of unity then the other root of the equation is (a) c/a (b) 0Mc/a) (c) -da (d) 1 o2(c/a). 126. If NCR denotes the number of combinations of N t h j n a s t a k e n r at a t i m e , t h e n t h e e x p r e s s i o n "L f (a) o+Reagents are (at Li/NH,; H , 0 ' (b) N a / E t O H R and i f / is not identically zero, then / (0) = (a) 0 (b) 1 (c) - 1 (d) not defined s i n ( a + l)x + sinx 130. If / ( * ) = (x + b x 2 f2x sin x * 0 137. I f / ( x ) = -! x , then / ( x ) is [ 0, x=0 (a) continuous as well as differentiable for all x (b) continuous for all x but not differentiable at x = 0 (c) (d) neither differentiable nor continuous at x = 0 discontinuous everywhere2for x < 0 for x = 01/2bx is continuous at x = 0 then a = (a) 3/2 (b) - 3/2 (c) 1/4 131. In a triangle PQR, (a) (c) p2 + q2 ,.2 q2 - r2 - p2 2pqsin(^(P (b) (d)312forx>0M J {/} dt is (i (where [ ] - greatest integer, { }- fractional part) (a) [x] (b) 2[x] (c) l/(2[x]) (d) [x]/2 138. The value of - 1/4 139. Let ^ F ( x ) = ex. If j sec2 x eanxdx then value of k is (a) 1/V3 (b) V3 (c) -V3 (d) - 1 / V 3 = F{k) - F(. 1) (d)+ O- R)j =r 2 + p 2 q2 r2 - p2 0, attains its maximum and minimum at I and m respectively such that / + m = 273, then a equals (a) - 1 (b) 1 (c) 0 (d) 2/3. 180. Equation of the locus of point z = x + iy such that(b) 1(c)- I(d) 0)". x2 2 (b) x 2 (c) x 2 (d) x (a) + + + += y2 y2 y2 y2is given by + 2x - 2x + 2y - 2y + + 1 1 1 1 =0 = 0 = 0 = 0.175. The value of (V2 + I) 5 (72 - 1 ) 5 is (a) 82 (b) 80 (c) 40 (d) 0.176. If x is a positive real number then the minimum value of X+ + J5 is .r (a) - 2 + V3 (c) i/3ANSWERS(b) (d) 2 + 73 0 is 1. 6. 11. 16. 21. 26. 31. 36. 41. 46. 51. 56. 61. 66. 71. 76. 81. 86. 91. 96. 101. 106. 111. 116. 121. 126. 131. 136. 141. 146. 151. 156. 161. 166. 171. 176. c) a) c) c) a) b) c) b) c) b) b) c) b) a) c) c) c) a) b) b) a) d) d) b) b) c) a) a) a) b) a) a) c) a) c) b) 2. (b) 7. (d) 12. (c) 17. (a) 22. (d) 27. (d) 32. (d) 37. (a) 42. (c) 47. (b) 52. (c) 57. (P.) 62. (c) 67. (b) 72. (c) 77. (c) 82. (c) 87. (d) 92. (c) 97. (c) 102 (d) 107.(b) '12.(c) 117.(d) 122.(a) 127.(b) 132.(a) 13 7. (a) 142.(d) 147.(c) 152.(a) 157.(c) 162. (a) 167.(c) 172.(c) 177.(b) 3. 8. 13. 18. 23. 28. 33. 38. 43. 48. 53. 58. 63. 68. 73. 78. 83. 88. 93. 98. 103. 108. 113. 118. 123. 128. 133. 138. 143. 148. 153. 158. 163. 168. 173. 178. c) b) b) d) c) b) a) a) a) b) d) b) b) b) c) a) d) d) c) c) b) c) a) c) c) a) a) d) c) a) d) a) b) b) c) b) 4. 9. 14. 19. 24. 29. 34. 39. 44. 49. 54. 59. 64. 69. 74. 79. 84. 89. 94. 99. 104. 109. 114. 119. 124. 129. 134. 139. 144. 149. 154. 159. 164. 169. 174. 179. c) c) c) d) a) a) b) a) b) c) b) c) b) c) c) d) c) d) a) d) a) b) d) d) a) b) a) b) b) b) d) c) a) a) a) b) 5. (b) 10. (c) 15. (b) 20. (b) 25. (b) 30. (d) 35. (b) 40. (d) 45. (c) 50. (a) 55. (c) 60. (c) 65. (b) 70. (c) 75. (c) 80. (a) 85. (a) 90. (b) 95. (a) 100.(a) 105.(c) 110 (c) I15.(c) 120.(d) 125.(d) 130.(b) 135.(a) 140.(a) 145.(a) 150.(c) 155.(d) 160.(d) 165.(b) 170.(a) 175.(a) 180.(c)177. Coefficient of x 4 in I 3x 2 (a) (c)7 72x7 7C(, C 4 3 4 2-1(b) (d)C 2 3 5 2~2 C2.,. (1 - c o s 2 x ) ( 2 + cos2x) . 178. lim is - < > > xtan .i.x (a) -172 (b) 2 (c) 0(d) - 1The 2.95% interest on credit card is actually 46.7%So far, though not for long, banks have been advertising a monthly interest rate for their cards like, say, 2.95 per cent. If you rolled over your clues, that's the interest you would pay in a month. If you didn't pay, you would think that you would pay an interest rate of 35.4 per cent (2.95 12). Well, it's about 11 percentage points m o r e for two reasons. ()ne. in what is a little-known fact, you also have to pay a 10.2 per cent service