Critical Questions MATHS (English)Ans

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ICS

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CRITICAL

QUESTIONS

BANK

CRITICAL

QUESTIONS

BANK

TARGET: JEE (ADVANCED) 2014

MEDIUM : ENGLISH

Contents

1. Questions 1 - 20

2. Answer Key 21

3. Hints & Solutions 22 - 45

MEDIUM : ENGLISHSUBJECT : MATHEMATICS

A. Questions Format

In the booklet check that it contains all the 180 questions and corresponding answer choices are legible. Read carefully the

Instructions printed at the beginning of each section.

1. Section 1 contains 53 multiple choice questions. Each question has Four choices (A), (B), (C) and (D) out of which only ONE

is correct.

2. Section 2 contains 37 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or

MORE are correct.

3. Section 3 contains 2 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which Only ONE is

correct.

4. Section 4 contains 10 paragraphs each describing theory, experiment, data etc. Twenty questions related to Ten

paragraphs with one or two or three questions on each paragraph. Each question of a paragraph has ONLY ONE correct

answer among the four choices (A), (B), (C) and (D).

5. Section 5 contains 4 question. Each question contains statements given in two columns which have to be matched.

Statements in Column I are labelled as A,B,C and D whereas statements in Column II are labelled as p,q,r,s and t. The answers

to these questions have to be appropriately bubbled as illustrated in the following example.

6. Section 6 contains 1 multiple choice questions. Each questions has matching lists. The codes for the lists have coices (A),

(B), (C) and (D) out of which ONLY ONE is correct.

7. Section 7 contains 43 questions. The answer to each question is a single-digit integer, ranging from 0 to 9 (both inclusive).

8. Section 8 contains 17 questions. The answer to each question is a double-digit integer, ranging from 00 to 99 (both inclusive).

B. Marking Scheme

9. For each question in Section 1, you will be awarded 3 marks if you darken the bubble corresponding to only the correct answer

and zero mark if no bubbles are darkened. In all other cases, minus one (1) mark will be awarded.

10. For each question in Section 2, you will be awarded 4 marks if you darken the bubble(s) corresponding to only the correct

answer and zero mark if no bubbles are darkened. No negative marks will be awarded for incorrect answers in this section.

11. For each question in Section 3, you will be awarded 4 marks if you darken all the bubble(s) corresponding to only the correct

answer(s) and zero mark if no bubbles are darkened.

12. For each question in Section 4 contains 10 paragraphs each describing theory, experiment, data etc. Twenty questions

related to Ten paragraphs with one or two or three questions on each paragraph. Each question of a paragraph has ONLY ONE

correct answer among the four choices (A), (B), (C) and (D).

13. For each question in Section5, you will be awarded 2 marks for each row in which you have darkened the bubble

corresponding to the correct answer. Thus, each question in this section carries a maximum of 8 marks. There is no negative

marking for incorrect answer(s) in this section.


answer(s) and zero mark if no bubbles are darkened. In all other cases, minus one (1) mark will be awarded.

15. For each question in Section 7, you will be awarded 4 marks if you darken the bubble corresponding to only the correct answer

and zero mark if no bubbles are darkened. No negative marks will be awarded for incorrect answers in this section.


answer(s) and zero mark if no bubbles are darkened. No negative marks will be awarded for incorrect answers in this section.

QUESTION FORMAT & MARKING CRITERIA

SUBJECT : MATHEMATICS

RESONANCE Page - 1

MATHEMATICS

SECTION 1 : (Only One option correct Type)

This section contains 53 multiple choice questions. Each question has four choices (A), (B), (C) and (D)out of which ONLY ONE is correct.

1. Let f(x) = sin2(x + ) + sin2(x + ) 2cos( ) sin(x + ) sin(x + ). Which of the following is TRUE?(A) f(x) is strictly increasing in x (, )(B) f(x) is strictly decreasing in x (, )

(C) f(x) is strictly increasing in x

2

, and strictly decreasing in x

,

2(D) f(x) is a constant function

2. If the roots of the equation a(b c)x2 + b(c a)x + c(a b) = 0 (where a, b, c are unequal real numbers) arereal and equal and be the roots of equation ax2 + bx + c = 0, a 0 then harmonic mean of is(A) 1 (B) 1 + (C) 1 (D) 1

3. If A =

4321

and B =

dcba

are two matrices such that AB = BA and c 0, then value of cb3d3a3

is(A) 0 (B) 2 (C) 2 (D) 1

4.

x

+

(x -1) x

2xx 0

e - xlim(x) -1 =

(A) 1 (B) 81 (C) 1 (D) does not exist

5. If the equation 4y3 8a2yx2 3ay2x + 8x3 = 0 represent three straight lines, two of them are perpendicularthen sum of all possible values of a is

(A) 43 (B) 5 (C) 5

4 (D) 3

6. If th th thp , q , r terms of an AP are P, Q, R respectively , then which of the following must be CORRECT ?(A) p Q + q R + r P = p R + r Q + q P(B) (p + q + r)th term will be P + Q + R.(C) If p < q < r, then P < Q < R(D) If P,Q,R N then common difference of AP will be an integer..

7. The set of values of 'a' for which x2 + ax + sin1(x2 4x + 5) + cos1 (x2 4x + 5) = 0 has at least one solutionis(A) ( , 2 ] [ 2 , ) (B) ( , 2 ) ( 2 , )

(C) R (D)

4

8

8. Let = 10log 15 and = 10log 16 , a set A = A = 10 10 10{ log 1, log 2,.........log 50} the number of elements in setA which can be written in the form of a b c where a,b,c are rational numbers, is(A) 23 (B) 24 (C) 25 (D) none of these

9. If both the roots of the equation x2 + (3 2k) x 6k = 0 belongs to the interval (6, 10), then largest value ofk is(A) 1 (B) 3 (C) 5 (D) none of these

RESONANCE Page - 2

MATHEMATICS

10. The domain of the function f(x) = )xx3x2()1x2(

23

+ )x(logsin 21 is

(A)

,21 (B)

2,21 (C) [1, 2] (D) (1, )

11. Let G, S, I be respectively centroid, circumcentre, incentre of triangle ABC. If R, r are circumradius andinradius respectively then which of the following is INCORRECT ?(A) SI2 = R2 (1 cosA cosB cosC) ; A,B,C being angles of triangle.(B) SI2 = R2 2Rr

(C) SG2 = R2 91 (a2 + b2 + c2) ; a, b,c being sides of triangle.

(D) SG SI

12. Two circles are given as x + y + 14x 6y + 40 = 0 and x + y 2x + 6y + 7 = 0 with their centres as C1and C2. If equation of another circle whose centre C3 lies on the line 3x + 4y 16 = 0 and touches the circleC1 externally and also C1C2 + C2C3 + C3C1 is minimum, is x + y + ax + by + c = 0 then the value of( a + b + c) is(A) 2 (B) 0 (C) 16 (D) None of these

13. If g(x) = [f(x)], x 0, ,2 23, x / 2

where [x] denotes the greatest integer function and

f(x) = n n

n n

2(sin x sin x) | sin x sin x |,n R {1},

2(sin x sin x) | sin x sin x |

then

(A) g(x) is continuous and differentiable at x = /2, when 0 < n < 1(B) g(x) is continuous and differentiable at x = /2, when n > 1(C) g(x) is continuous but not differentiable at x = /2, when 0 < n < 1(D) g(x) is neither continuous nor differentiable, at x = /2, when n > 1

14. If

xsin2010xsec

2010

2dx = 2010)x(sin

)x(P+ C, then value of

3

P is

(A) 0 (B) 31 (C) 3 (D) 2

33

15. If ji bisects the angle between c and j k , then value of c.j is equal to

(A) 0 (B) 12 (C) 12

(D) 1

16. If A and B are two square matrices such that B = A1 BA, then (A + B)2 is equal to(A) O (B) A2 + B2 (C) A2 + 2AB + B2 (D) A + B

17. The value of where = sin1 4

32 + cos1

412

+ sec1 )2( is equal to

(A) 0 (B) 4 (C) 6

(D) 2

18. f(x) = QxQx

,

,

xcos

x

. The number of points of continuity of f(x) is(A) 0 (B) 1 (C) 2 (D) infinitely many

RESONANCE Page - 3

MATHEMATICS

19. Let f(x) =2

3x

dy1 y . The value of the integral

2

0

xf(x) dx is equal to

(A) 1 (B) 43 (C) 23 (D)

13

20. Let P be a point on ellipse 4x + y = 8 with eccentric angle 4

. If tangent at P intersects the x-axis at A and

y-axis at B and normal at P intersect the x-axis at A' and y-axis at B'. The ratio of area of triangle APA' toarea of triangleBPB' is(A) 1 : 1 (B) 2 : 1 (C) 3 : 1 (D) 4 : 1

21.

xcos1xsin1

ex/2 dx, x

2

9,

25

(A) ex/2 sin 2x

+ C (B) ex/2 sec 2x

+ C (C) ex/2 sin 2x

+ C (D) ex/2 cos 2x

+ C

22. |x|]x[2elim

|x|]x[

0x

= (where [.] denotes greatest integer function)

(A) 2 e

1 (B) 1 (C) 2 e (D) e

1 2

23. If 1 1 11 1tan (1) tan (2) tan (3)2 3

, 1 11 1 1tan (1) 2tan 3tan2 3

and

1p r cot (3)q s

,

where p, q, r, s N and are in their lowest form then which of the following is INCORRECT ?(A) p r 0 (B) q 4s (C) p q r s 42 (D) pr 1 q

24. The length of sub-normal at any point P(x,y) on the curve, which (curve) is passing through Q(0, 1) is unity.The area bounded by the curves satisfying this condition is equal to

(A)

23 (B)

43 (C)

13 (D)

83

25. Let f(x) = x + x + x4 + x8 + x16 + x32 + ......the coefficient of x10 in f(f(x)) is(A) 28 (B) 40 (C) 52 (D) none of these

26. If 1, 1, 2, 3,.....2008 are (2009)th roots of unity, then the value of 2008

r 2009rr 1

r( )

is equal to(A) 2009 (B) 2008 (C) 0 (D) 2009

27. Let

22

x

2x

)xsin(sec.)x(coslim , where , R

(A) = 2, = 0 (B) = 0, = 2 (C) = 37

, = 0 (D) = 35

, = 0

RESONANCE Page - 4

MATHEMATICS28. Let A(x1, y1), B(x2, y2) C(x3, y3) be vertices of triangle ABC with BC = a, AB = c, AC = b. If algebraic sum of

perpendicular distances from L

cba

ay3,

cbaax3 11

, M

cba

by3,

cbabx3 22

, N

cba

cy3,

cbacx3 33

to a

variable line is zero then all such lines passes through(A) orthocentre of ABC (B) centroid of ABC(C) circumcentre of ABC (D) incentre of ABC

29. If |f(x)| 1 x R and f(0) = 0 = f(0) then which of the following can be TRUE?

(A) 51

31f

(B) 41

31f

(C) 313f (D) none of these

30. If f(x) = cos 8{x} + sin 2x cosec 2x (where {.} represents fractional part function), then fundamental periodof f(x) is

(A) 41 (B) 2

1 (C) 1 (D) 87

31. The principle argument of z = x + iy, if it lies in second quadrant is equal to

(A) x

ytan2

1 (B)

x

ytan 1 (C) x

ytan 1 (D) x

ytan2

1

32. If the function f(x) = 2 1/ x

, x 0x {e }, x 0k

where {.} denotes fractional part function, is continuous at x = 0, then(A) k = 1 (B) f(x) is non-derivable at x = 0(C) f(x) is derivable at x = 0 (D) f(x) is continuous at every point in its domain.

33. f(x) = x2 + 1, g(x) = 3 x . Then (gofogofogogog)(x) is(A) odd function (B) even function (C) polynomial function (D) identity function

34. If a = 2 , then sum of seriescot1(2a1 + a) + cot1 (2a1 + 3a) + cot1(2a1 + 6a) + cot1 (2a1 + 10a) + .... upto infinite terms, is

(A) 4 (B) 2

(C) 3 (D) 6

35. The normal to the curve 5x5 10x3 + x + 2y + 6 = 0 at P (0, 3) meets the curve again at two points, then theequations of the tangents to the curve at these points is/are(A) 2x y 3 = 0 (B) 2x + y 3 = 0 (C) 2x + y + 3 = 0 (D) 2x y + 3 = 0

36. If f(x + y + 1) = 2)y(f)x(f and f(0) = 1, Ry,x then f(x) can be(A) 1 x2 (B) 1 x (C) (x + 1)2 (D) x2 1

37. f(x) = )xsine(cos 1x21 (A) domain of f is [1, 1] (B) f is injective(C) range of f is [0, 5] (D) f1(x) = f(x) have no solution.

38. A normal chord drawn to a parabola y2 = 4x at any point P intersects the parabola again at a point Q, then theminimum distance of Q from the vertex of the parabola is equal to(A) 4 2 (B) 4 3 (C) 4 5 (D) 4 6

39. Let p(x) be a real polynomial of least degree which has a local maximum at x = 2 and a local minimum atx = 4. If p(2) = 8, p(4) = 1 , then p(0) is

(A) 5168 (B) 42 (C) 43 (D) 45

RESONANCE Page - 5

MATHEMATICS

40. Let f(x) = (3x + 2)2 1, < x 23 . If g(x) is the function whose graph is the reflection of the graph of f(x)with respect to line y = x, then g(x) equals to

(A) 1 (2 x 1), x 13 (B) 1 (2 x 1), x 13

(C) 1 (1 x 2), x 23 (D) 1 (1 x 2), x 23

41. cot1

29

+ cot1

4

33 + cot1

8

129............ is

(A) 4 (B) 2

(C) 4

(D) none of these

42. The sum 10098

+ 9997

+ 9896

+.......+ 31

+ 20

is equal to, where n

r

= nC

r

(A) 10098

(B) 10199

(C) 10099

(D) 10198

43. A circle with centre (3, 3) and of variable radius cuts the hyperbola x2 y2 = 36 at the points P,Q,R and S.If the locus of centroid of PQR is (x 2)2 (y 2)2 = , then the value of is(A) 3 (B) 2 (C) 4 (D) 3

44. Let f(x) =

3x,18x

3x2,)3b3b(nxx4 23 . Find all the possible real values of b such that f(x) has the

smallest value at x = 3.(A) (, 2] [3, ) (B) (, 1] [2, ) (C) (1, 2] (D) (, 2]

45. The solution set of the inequality (cosec1x)2 2(cosec1x) 6

cosec1x 3

is (, a] [b, ), then(a + b) is equal to(A) 0 (B) 2 (C) 3 (D) 1

46. arg 72 318 9 10 14

1+ Z + Z + Z ......... + Z

1+ Z + Z + Z + ....... + Z

=

(A) 21

arg (Z1 Z2 .........Z7) (B) arg (Z1 Z2 .........Z7)

(C) 41

arg (Z1 Z2 .........Z7) (D) 81

arg (Z1 Z2 .........Z7)

47. Image of the curve xy = 1 in the curve

1xx 2

1yy 2

= 1 (on coordinate plane) is(A) xy = 1 (B) xy + 1 = 0 (C) xy = 0 (D) x2 + y2 = 1

48. |1 + Z1 + Z2 +.....+ Z7 | =

(A) sec

157 (B)

157

sec21 (C)

157

sec2 (D) None of these

49. Let O be centre, S, S be foci of hyperbola. If tangent at any point P on hyperbola cuts asymptotes at M andN then OM + ON =(A) |SP SP| (B) SP + SP (C) SS (D) distance between vertices

RESONANCE Page - 6

MATHEMATICS50. DABC be a tetrahedron such that AD is perpendicular to the base ABC and ABC = 30. The volume of

tetrahedron is 18. If value of AB + BC + AD is minimum then length of AC is

(A) 3 6 3 (B) 3 6 2 (C) 2 6 3 (D) 2 6 2

51. The solution set of values of x satisfying equation 2cot1x + cos1

2

2

x1x1

, is

(A) all real numbers (B) (, 0] (C) [0, ) (D) (, 1) (1, )

52. If g(x) = 2f (2x3 3x2) + f (6x2 4x3 3), x R and f (x) > 0, x R, then g(x) is increasing on theinterval

(A) 1, 0,12

(B) 1,0 1,2

(C) 0, (D) ,1

53. Let a circle with centre at C be made to pass through the point P(1,2), touching the straight lines 7x y = 5and x + y + 13 = 0 at A and B respectively. If tangents at A and B intersect at Q, then

(A) area of quadrilateral ACBQ can be 100 sq. units (B) radius of circle can be 40(C) area of quadrilateral ACBQ can be 200 sq. units (D) radius of circle can be10

SECTION 2 : (One or more options correct Type)This section contains 37 multiple choice questions. Each question has four choices (A), (B), (C) and (D)out of which ONE or MORE are correct.

54. Consider the locus of the complex number z in the Argand plane is given by Re(z) 2 = |z 7 + 2i|. Let P(z1)and Q (z2) be two complex number satisfying the given locus and also satisfying

arg 12

z - (2+ i)= ( R)

z - (2+ i) 2

then the minimum value of PQ is divisible by

(A) 3 (B) 5 (C) 7 (D) 2

55. Let f(x2 + y) = (f(x))2 + f(y) for all x, y R, then(A) f(x) is odd(B) f(x) is even f(x) = 0 x R(C) f(x) is continuous at x = 0 it is continuous everywhere(D) f(x) is differentiable at x = 0 f(x) = x f (0), x R

56. Let f(x) = x3 + px2 + qx + r, where p,q and r are integers, f(0) and f(1) are odd integers. Which of thefollowing is/are CORRECT ?(A) f(1) is an even integer (B) f(1) is an odd integer(C) f(x) = 0 has three distinct integer roots (D) f(x) = 0 cannot have three integer roots.

57. Let a,b,c R. If ax2 + bx + c = 0 have one root < 1 and other root > 1, then

(A) 1 b c 0a a (B) 1 + 0

a

c

a

b (C) 1 0a

c

a

b (D) 1 0a

c

a

b

58. The value(s) of x satisfying 1 log9(x + 1)2 = 21

3x5xlog 3 is/are

(A) 1 (B) 2 (C) 7 (D) 4

59. If g(x) = mlim 1x4x23)x(h)x(fx

m

m

when x 1 and g(1) = e3 such that f(x), g(x) and h(x) are continuous

functions at x = 1, then(A) f(1) = 2e3 (B) h(1) = 5e3 3 (C) f(1) + h(1) = 7e3 + 5 (D) f(1) h (1) = 7e3 + 5

RESONANCE Page - 7

MATHEMATICS60. A parabola C whose focus is S(0,0) and passing through P(3, 4). Equation of tangent at P to parabola is

3x + 4y 25 = 0. A chord through S parallel to tangent at P intersects the parabola at A and B. Which of thefollowing are CORRECT ?(A) Length of AB is 20 units(B) Latus rectum of parabola is 20 units(C) Only one real normal can be drawn from the point (3, 4)(D) Only one real normal can be drawn from the point (6, 8)

61. If 114 + 8 - 32 + 768 = a 2 cosb

, where a and b are natural numbers then a

b is divisible by

(A) 2 (B) 4 (C) 3 (D) 6

62. For the function f(x) = (x2 + bx + c) ex and g(x) = (x2 + bx + c) ex + ex (2x + b), which of the following holdsgood ?(A) f(x) > 0 for all real x g(x) > 0 (B) f(x) > 0 for all real x g(x) > 0(C) g(x) > 0 for all real x f(x) > 0 (D) g(x) > 0 for all real x f(x) > 0

63. If the equation 2

1 11 1 2cos x cos x1a a 0

22 2

has only one real solution then subsets of values

of 'a' are(A) (3, 1) (B) (, 3] (C) [1, ) (D) [3, )

64. f(x) = 23

x 2 , x 2x , 2 x 3

, x 3x 18

(A) f(x) is continuous in R. (B) f(x) is discontinuous at 'a' for some a R.(C) f(x) is continuous at infinitely many real x. (D) f(x) is discontinuous at infinitely many real x.

65. Which of the following must hold good for the expansion of the binomial 15

431

xx

?

(A) There exist a term which is independent of x.(B) 8th and 9th terms of the expansion have the greatest binomial coefficient(C) Coefficients of x32 and x17 are equal(D) If x = 2 , then number of rational terms in the expansion is 5

66. Which of the following is NOT in the solution set of the inequality |x2 1| + |2x2 3| < |2 x2| ?

(A) (1, 1) (B) (2, 2) (C)

21

,1 (D) (3, )

67. Let a function f(x), x 0 be such that 1 1f (x) f f (x) fx x

then f(x) can be :

(A) 1 x2013 (B) | x | 1

(C) 12 tan | x | (D) 21 k n | x | , k being arbitrary constant

68. TP and TQ are tangents to parabola y2 = 8x and normals at P and Q intersect at a point R on the parabola.The locus of circumcentre of TPQ is a parabola whose -

(A) Vertex is (2, 0) (B)foot of perpendicular from focus on the directrix is 7 , 04

(C) length of latus rectum is 1 (D) focus is 9 , 04

RESONANCE Page - 8

MATHEMATICS69. The function f(x) = 2|x| + |x + 1| ||x 3| 3|x|| has a local minimum or a local maximum at x =

(A) 0 (B) 23 (C) 4

3 (D) 3

70. If z1 , z2 ,z 3 , z4 are roots of the equation a0z4 + a1z3 + a2z2 + a3z + a4 = 0, where a0 , a1 , a2 , a3 and a4 are real,then(A)

1z , 2z , 3z , 4z are also roots of the equation

(B) z1 is equal to at least one of 1z , 2z , 3z , 4z(C) 1z , 2z , 3z , 4z are also roots of the equation(D) None of the above.

71.x3

X 3

sin(e 1)limn(x 2) = c, which of the following is/are CORRECT ?

(A) sin1c = 2 (B) c = 0

(C) c is a positive integer (D) c is neither prime nor composite

72. Let b,a

and c

be three unit vectors such that | cba | = 3 and )ba).(ac()ac).(cb()cb).(ba( . Which of the following are CORRECT ?

(A) The maximum value of is 0(B) If is maximum then the volume of parallelepiped determined by b,a and c is 3

(C) If is maximum then the value of )ac6cb5ba).(c4b3a2( is 32.(D) None of these

73. Let g be the inverse of the continuous function f, Let there be a point (, ), where , is such that itsatisfies each of y = f(x) and y = g(x) then(A) the equation f(x) = g(x) has infinitely many solutions(B) the equation f(x) = g(x) has atleast 3 solutions(C) f must be a decreasing function of x(D) g can be an increasing function of x

74. The maximum and minimum value of ab sin x + b 2a1 cos x + c (|a| 0), lie in interval(A) [c b, b + c] (B) (b c, b + c)(C) [c 2b, c + 2b] (D) [c, c]

75. If ( x 1)log 323x 2lim (log (ax 3x 1))

, where is a finite real number then

(A) a = 1 (B) 'a' can have more than one values(C) = e2/3 (D) = e1/3

76. Which of the following functions is an injective (one-one) function in their respective domain?(A) f(x) = 2x + sin 3x (B) f(x) = x. [x],(where[.] denotes the G.I.F.)

(C) f(x) =x

x

2 +14 -1

(D) None of these

RESONANCE Page - 9

MATHEMATICS77. The centre of a circle S = 0 lies on 2x 2y + 9 = 0 and S = 0 cuts orthogonally the circle x2 + y2 = 4. Then

the circle must pass through the point(A) (1, 1) (B) ( 1/2, 1/2) (C) (5, 5) (D) ( 4, 4)a

78. If 1 1 x2f(x)+ xf - 2f 2sin x + = 4cos + xcos , x R- {0}x 4 2 x

then which of the following

statements(s) is/are true?

(A) f(2) + f

21

= 1 (B) f(2) + f(1) = 0

(C) f(2) + f(1) = f

21 (D) f(1) + f

21

= 1

79. If 114 + 8 - 32 + 768 = a 2 cosb

, where a and b are natural numbers then (b a) is divisible by

(A) 2 (B) 23 (C) 69 (D) 46

80. Equation of the plane passing through the line of intersection of the two planes 1 1.r n q

and 2 2.r n q

and

parallel to the line of intersection of 3 3.r n q

and 4 4.r n q

is

(A) dependent on 1 3.n n (B) dependent on 3 4.n n

(C) independent of q1 and q2 (D) independent of q3 and q4

81. Let ABCD be a tetrahedron, where A = (2,0,0), B = (0,4,0). If edge CD lies on line x 1 y 2 z 31 2 3

, such

that centriod (,,) of tetrahedron satisfies b

za

y1

252

11

then which of the following are

CORRECT?

(A) a + b = 25 (B) y1 + z1 = 4

19 (C) y1 z1 = 41 (D) a + b + y1 = 5

82. The line 3x + 6y = k intersects the curve 2x2 + 2xy + 3y2 = 1 at points A and B. The circle on AB as diameterpasses through the origin. The possible value of k is -(A) 3 (B) 4 (C) 4 (D) 3

83. A rod of length 2 units whose one end is (1, 0, 1) and other end touches the plane x 2y + 2z + 4 = 0, isrotated on this plane, then(A) the rod sweeps a solid structure whose volume is cubic units(B) the area of the region which the rod traces on the plane is 2(C) the length of projection of the rod on the plane is 3 units

(D) the centre of the region which the rod traces on the plane is

35

,

32

,

32

84. Let f :

2

,0 [0, 1] be a differentiable function such that f(0) = 0, f

2 = 1, then

(A) f () = 2))(f(1 for all

2

,0 (B) f () = 2

for all

2

,0

(C) f() f () = 1

for at least one

2

,0 (D) f () = 28

for at least one

2

,0

RESONANCE Page - 10

MATHEMATICS85. Two equal sides of an isosceles triangle are given by the equation 7x y + 3 = 0 and x + y 3 = 0 and its third

side passes through the point (1, 10). The equation of the third side can be(A) x + 3y + 29 = 0 (B) x 3y = 31 (C) 3x + y = 3 (D) 3x + y + 7 = 0

86. Let f(x) = cos1 (cos2x) and g(x) = |cos x| then(A) number of solution of f(x) = g(x) in ]2,0[ is 4.(B) max {f(x), g(x)} is a periodic function(C) max {f(x), g(x)} is a non differentiable function for some x,(D) min {f(x), g(x)} is an even function

87. Consider f(x) = 2sinx + sin 2x , x

2

3,0

(A) Greatest value of f(x) is 3 (B) Greatest value of f(x) is 233

(C) Least value of f(x) is zero (D) Least value of f(x) is 2

88. For a Function f : A B such that n(A) a, n(B) b (a,b N) then which of the following statementssmust be CORRECT ?(A) If function is one - one, onto, then a>b(B) If function is one - one, into, then ab(D) If function is many - one, into, then a

RESONANCE Page - 11

MATHEMATICS

SECTION - 3 (True & False Statement Type)This Section Contains 2 questions.Each questions contains 4 statements S1, S2, S3 & S4. Eachstatement is either true (T) or false (F). Each questions has 4 choices (A), (B), (C) and (D) each of whichcontains whether S1, S2, S3 & S4 are true or false. Exactly one choice contains the correct order of truthnessor falseness of S1, S2, S3 & S4 respectively and is the correct choice.

92. Consider the following statements :S1 : Let ax2 + bx + c = 0 be a quadratic equation, if b2 4ac > 0 then roots are real and distinctS2 : Let a,b,c,d . Then the possible rational roots of the equation ax3 + bx2 + cx + d = 0 are of the

form qp

where p is an integral divisor of d and q is an integral divisor of a.

S3 : Let z = a + ib be a complex numbers where a,b C R. Then ibaz .

S4 : If 2)5ba( = 253

, then a,b Q.

State, in order, whether S1, S2, S3, S4 are true or false(A) FTTT (B) TFTT (C) FTTF (D) FFTF

93. Consider the following statements :S1 : Let a,b,c C and ax2 + bx + c = 0 be a quadratic equation. Then b2 4ac = 0 roots are real and equal.

S2 : Let b,c I and b2 4c be a perfect square. Then roots of the equation x2 + bx + c = 0 may not be integers.

S3 : If the quadratic equations a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0 have common root, then

1 1 1

2 2 2

a b ca b c

.

S4 : f(x) = 22

22

112

1

cxbxacxbxa

, g(x) = bax

mx

where a1, a2, , a are non zero real and other coefficients are

also real. Then range of f(x) range of g(x).State, in order, whether S1, S2, S3, S4 are true or false(A) TTFT (B) TTTF (C) FFTT (D) FFFF

SECTION 4 : (Paragraph Type)This section contains 10 paragraphs each describing theory, experiment, data etc. twenty questionsrelate to two paragraphs with one or two or three questions on each paragraph. Each question of aparagraph has only one correct answer among the four choices (A), (B), (C) and (D).

Paragraph for Question Nos. 94

Let 2 2x y

36 27 = 1 be an ellipse and P be any point on it and S and S be its foci.

94. The point of intersection of straight lines joining each focus to the foot of the perpendicular from the other

focus upon the tangent at a point P 3

, is

(A) 21 9,8 4

(B) 15 9,8 4

(C) 15 9,4 8

(D) 21 9,4 8

RESONANCE Page - 12

MATHEMATICSParagraph for Question Nos. 95 to 96

Consider f(x) = log{x}[x] where [.] is the greatest integer function and {x} is fractional part function.

95. f(x) is(A) a one-one function (B) a many-one function(C) a odd function (D) a periodic function

96. Consider the inequality f(x) < 2. Number of solutions of this inequality in x (1, 2) is/are(A) 0 (B) 1 (C) 3 (D) infinitely many

Paragraph for Question Nos. 97 to 98

Consider a polynomial y = P(x) of the least degree passing through A(1, 1) and whose graph has two pointsof inflection B(1, 1) and C with abscissa 0 at which the curve is inclined to the positive direction ofx-axis at an angle of sec1( 2 ).

97. The value of P(0) is

(A) 12 (B) 12 (C) 0 (D) 2

98. The equation P(x) = 0 has(A) three distinct real roots(B) one real root(C) three real roots such that one root is repeated.(D) none of these


If cos1 2x

+ cos1 3y

= then the value of 9x2 12xy cos + 4y2 is equal to N sin2

(where N is a positive integer) and complete set of values of x for which (cos1 x)2 (sin1 x)2 > 0 is satisfied,is [p, q).

99. Which of the following values lie in [p, q) ?(A) N 3 (B) N 4 (C) N 5 (D) N 6

100. Which of the following functions cannot be defined at x = N 6 ?(A) sin1 x (B) cos1 x (C) tan1 x (D) sec1 x


Base of a pyramid is rectangular, three of its vertices of the base are A(2,2, 1), B(3,1,2) and C(1,1,1) and its

vertex at the top is P26 104, ,3 3

.Fourth vertex of the base is D. Then answer the following questions.

101. If the coordinates of foot of normal drawn from P on the base of the pyramid are ( , m,n), then(A) + m + n = 4 (B) = 3m (C) n = 3m (D) < m < n

102. Volume (in cubic unit) of the pyramid is(A) 20 (B) 30 (C) 20 (D) 30

RESONANCE Page - 13


Consider f(x) = tan1 2

4 2( 12 2)xx 2x 3

and m and M are respectively minimum and maximum values of f(x) andx = a (a > 0) is the point in the domain of f(x), where f(x) attains its maximum value.

103. If cos1x + cos1y = 3 1 17M 3tan tan tan m tan2 8

, then (x + y) is equal to(A) 2 (B) 2 (C) 0 (D) 3/2

104. If and are roots of the equation x2 (tan(3sin1 (sinM))) x + a4 = 0, then ( + ) equals to(A) 1 (B) 4 (C) 3 (D) 2


Consider the quadratic equation ax2 + (2 a)x 2 = 0, where a R.

105. If exactly one root is negative, then the range of a2 + 2a + 5 is(A) [4, ) (B) [2, ) (C) (, 4] (D) (5, )

106. If , are roots of the quadratic equation and if there are at least four negative integers between and ,then the complete set of values of a is

(A)

3,27 (B)

21

,0 (C)

21

,23 (D)

3,27


Let M be the maximum value of c2 for which O(0, 0) and A(1, 1) does not lie on opposite side of the straightline (a + b)2x (ab + bc + ca + 1)y + 2 = 0 for all a,b R.Also lx + my + n = 0 be a variable line, where l, m,n are 1st, 3rd and 7th terms of an arithmetic progression andthe variable straight line always passes through a fixed point P(,).

107. If the circles x2 + y2 = M + 1 and x2 + y2 24x 10y + 2 = 0 have exactly two common tangents, then thenumber of possible integral values of , is(A) 11 (B) 13 (C) 12 (D) 10

108. The tangent of the circle x2 + y2 = M + 2 at the point ( 1, + 1) also touches the circlex2 + y2 4x 2y + 20 = 0 then its point of contact is(A) (3, 1) (B) (3,0) (C) (1, 1) (D) (2, 1)


Let f(x) = x2 + b1x + c1 , g(x) = x2 + b2x + c2 , , be real roots of f(x) = 0 and + , + be real roots

of g(x) = 0, minimum value of y = f(x) be 41

and minimum value of y = g(x) occurs at

x = 27

.

109. Minimum value of y = g(x) is

(A) 1 (B) 21 (C) 4

1 (D) 31

110. The value of b2 is(A) 6 (B) 7 (C) 8 (D) 7

RESONANCE Page - 14


There exists a matrix Q such that PQPT = N, where P =

100012021

.

Given N is a diagonal matrix of form N = diag.(n1, n2, n3) where n1, n2, n3 are three values of n satisfying theequation det.(P nI) = 0, n1< n2< n3.[Note : I is an identity matrix of order 33]

111. The value of det.(adjN) is equal to[Note : adj M denotes the adjoint of a square matrix M.]

(A) 4 (B) 41 (C) 9

1 (D) 9

112. If QT = Q + , then the value of is equal to

(A) 1 (B) 0 (C) 1 (D) 31

113. The trace of matrix P2012 is equal to(A) 32011 + 2 (B) 32012 (C) 32012 + 2 (D) 32011

SECTION - 5Matrix - Match Type

This section contains 4 questions. Each question contains statements given in twocolumns, which have to be matched. The statements in Column-I are labelled A,B, C and D, while the statements in Column-II are labelled p, q, r, s and t. Anygiven statement in Column-I can have correct matching with ONE OR MOREstatement(s) in Column-II. The appropriate bubbles corresponding to the answersto these questions have to be drakened as illustrated in the following example.If the correct matches are A-p, s and t ; B-q and r; C-p and q; and D-s and t; then the correct darkening of bubbles will look like the following :

114. Match the inequality in column-I with their complete solution set in column-II.

Column Column

(A) logsinx log3 log0.2 x < 0 (p) [1, 1]

(B) )1x(x)2x(sin)2xx)(3x2)(1e( 2x

0 (q) [3, 6)

(C) |2 | [x] 1| | 2, (r)

125

1,0

[.] represents greatest integer function.

(D) |sin1 (3x 4x3)| 2 (s) (, 1)

,23

(t)

RESONANCE Page - 15

MATHEMATICS

115. If f(x) = 2x1x2

and g(x) = x

1x .

Column Column

(A) Domain of f(g(x)) is (p) (, 1) (1, 2) (2, )

(B) Domain of g(f(x)) is (q) (, 2) (2, )

(C) Range of f(f(x)) is (r) (, 0) (0, 1) (1, )

(D) Range of g(g(x)) is (s) (, 1) (2, )

(t) (, 1/2) (1/2, 2) (2, )

116. Column-I Column-II

(A) If f : (1/2, 1) [0, ) and f(x) =

x1

xloge , then f(x) is (p) one-one function

(B) If f : [0, 1] [cos(sin1) + sin(cos1), 1 + sin1] and (q) many-one function

f(x) = cos(sinx) + sin(cosx), then f(x) is

(C) Let f : R R and f(x) = x 2m 2 , x 1

mx 4 , x 1

, then for (r) into function

all values of m (6, ). The function f is

(D) Let f(x) = )x)(x()x)(x(

, where < < < , then f(x) is (s) surjective function

(t) invertible function

117. Column-I Column-II

(A) an =

211 1n

+

211 1n

Then, the value of

20

1n na

1 is (p) 2

not divisible by(q) 5

(C) Number of solutions satisfying the equation xsin

1

x2sin1

=x4sin

2 (r) 7

in [0, 4] equals

(D) Integral multiple of solution of the equation cot4

12

r 1

1tan2r

= 2x3

4x3

is (s) 4

(t) 6

RESONANCE Page - 16

MATHEMATICS

SECTION 6 : (Matching list Type)This section contains 1 multiple choice questions. Each question has matching lists. The codes for the listshave choice (A), (B), (C) and (D) out of which ONLY ONE is correct.

118. Match List I with List II and select the correct answer using the code given below the lists :List - I List - II

P.2 3 9 10

sin sin sin ........sin sin21 21 21 21 21

= 1. 10211

Q. 2 3 9 10sin sin sin ........sin sin22 22 22 22 22

= 2. 1021

R.2 3 9 10

cos cos cos ........cos cos21 21 21 21 21

= 3. 5211

S. 2 3 9 10cos cos cos ........cos cos22 22 22 22 22

= 4. 10221

Codes:P Q R S

(A) 3 4 4 2(B) 4 1 2 1(C) 2 3 1 4(D) 4 3 1 3

SECTION 7 : (Integer value correct Type)This section contains 43 questions. The answer to each question is a single digit integer, ranging from0 to 9 (both inclusive).

119. I f f(x) =

0x0

n,nex

21

e

ex

)1xxlog(x 2I

(where [ . ] denotes the greatest integer function),

then find the value of f(0) + f(1947) + f(1947)

120. The equation of common tangent to the curve xy = 4 and x2 + y2 = 8, whose x and y intercepts are positiveis x + y = . Find the value of .

121. If function y = f(x) has only two points of discontinuities say x1, x2 , where x1, x2 < 0, then find the numberof discontinuities of y = f(|x|) .

122. Given equation |x2 5x + 4 + sinx| = |x2 5x + 4| + |sinx|, where 0 x 2 then find sum of all integral valuespresent in complete solution set of given equation.

123. If f(x) + f

x

11 = 1 + x x R {0, 1}. Then find the value of 4f(2).

124. If |z1| = 2, (1 i) z2 + (1 + i) 28z2 , (z1, z2 are complex variables) then the minimum value of |z1 z2| =

125. If 2

2 2x x k 0

x 5x 9 x 5x 9

for all real x, then find sum of modulus of all integral values of

k in [4, 7].

126. Find the number of integral values of 'a' for which the equation(3sinx 4cosx)2 (a2 + a + 5)|3sinx 4cosx| + a3 + 3a2 + 2a + 6 = 0 has a real solution.

RESONANCE Page - 17

MATHEMATICS127. In a sequence of circles c1, c2, c3, ....,cn , the centres lie along positive x-axis with abscissa forming an A.P.

of first term unity and common difference 3. The radius of these circles are in G.P. with first term unity andcommon ratio 2. If the tangents with slope m1 and m2 of c3 intersect each other at the centre of c5, then findthe value of 10|m1m2|.

128. In a meeting there are six ministers all speak exactly two languages. M1 speaks only L1 and L2, M2 speaksonly L2 and L3, M3 speaks only L3 and L4, M4 speaks only L4 and L2, M5 speaks only L4 and L1, M6 speaks onlyL1 and L3. If two ministers are chosen at random, the probability that they speak common language is , thenfind the value of 5.

129. If a,b and c are distinct real numbers, such that the quadratic expressions Q1(x) = ax2 + bx + c,Q2(x) = bx2 + cx + a and Q3(x) = cx2 + ax + b are always non-negative, then find the number of possible

integers in the range of the expression y = 2 2 2a b c

ab bc ca

.

130. Find sum of all real solutions of the equation xlog5 + log2 = x + log(2x 1).

131. Given f : RR; f(x) = 2x 3(k + 2)x + 12kx 7, 4 k 6, k I, then find the number of values of k forf(x) to be invertible.

132. The real value of a for which the integral 2a

2a

)2x( dxe2

attains its maximum value is . Find the value of ||.

133. If f(x) = 10xkx33x 22

3 is a many one function, find sum of all positive integral values of k.

134. Find the minimum value of xsin)1x2x2(22

e .

135. If a is the number of onto functions on A = {a,b,c,d} and b is the numbers of one one functions on A, find|a b|.

136. Find number of solutions of the equation sin3x = cos4x in the interval

4

,0.

137. If sum of the series cot1 (2.12) + cot1 (2.22) + cot1 (2.32) + .......... is equal to k, then find the valueof [k], where [.] denotes greatest integer function.

138. The function f(x) = x2 e2x, x > 0. If is maximum value of f(x) then find ][ 1 .(where [.] represents greater integer function)

139. Let A be any variable point on the x-axis and B be the point (2, 3). The perpendicular at A to the line AB meetsthe y-axis at C. Locus of mid-point of the segment AC as A moves is ax2 2x + by = 0. Find the value ofa + b.

140. x2 + (a b) x + (1 a b) = 0, a, b R. Find the least integral value of '

a

' for which both roots of theequation are real and unequal b R.

141. If a1, a2, a3, 5, 4, a6,a7,a8,a9 are in H.P. and

=

1 2 3

6

7 8 9

a a a

5 4 aa a a

then find the value of [] is, where [.] denotes G.I.F..

142. If f(x) = )cos()cos()cos()xcos()xcos()xcos()xsin()xsin()xsin(

and f(9) = (0), then find the value of

9

k 1f(k)

.

RESONANCE Page - 18

MATHEMATICS143. Find number of solutions of the equation 210 10 10log x log x 2log x 1 (where [.] denotes greatest

integer function)

144. If the value of the definite integral 1 1 2

20

sin x dxx x 1 n

(n N), then value of n

27

is

145. If barrK

, where K is a non zero scalar and b,a

are two given vectors. Then r will be

22 |a|K1

)ba(na

Kb.abKm

, find the value of nm .

146. Sum of all elements in range of f(x) = 16 xC2x 1 + 20 3xC4x 5 is then find the value of 469

.

147. If z = 21 )13i( , then find the value of (z z2 + 2z3)(2 z + z2).

148. If

sin3 1 1cos2 4 3

2 7 7

= 0, then find the number of values of in [0, 2].

149. If 1 lies between the roots of the quadratic equation 3x2 3(sin ) x 2 cos2 = 0 then find sum of all integralvalues of in [0, 2], where is in radians.

150. The volume of the parallelepiped whose coterminous edges are represented by the vectors 3(b c ),2( a b ) and 4( c a ) where

a

= (1 + sin) i + (cos)j + (sin2) k

b

= sin

i32

+ cos

j32

+ sin

k342

c

= sin

i32

+ cos

j32

+ sin

k342

is 18 cubic units then find number of values of in the interval

2

,0.

151. Let f(x) = sin xx (1 + xcosx. n x+ sinx)dx and f 2

=

2

4

then find the value |cos(f())|, where x > 0.

152. Find 2T

if T is the fundamental period of f(x) = sin

3

]x[2+cos

2]x[3

, where [.] denotes greatestinteger function.

153. f(x) = x3 + 4x2 + x + 2 is monotonically decreasing in the largest possible interval

32

,2. Then find

greatest value of .

154. If range of f(x) = n 3x2sinx tanx 1 x ,

6 3

is [a, b], then find the value of [a + b], where [.]denotes greatest integer function.

155. Find numbers of values of k if x2 + kx + 2 < 0 for exactly two integers.

RESONANCE Page - 19

MATHEMATICS156. Let z1, z2 and z3 be complex numbers such that |z1| = |z2| = |z3| = |z1 + z2+ z3| = 2. If |z1 z2| = |z1 z3| and

z2 z3. Then find the value of |z1 + z2| |z1 + z3|.

157. Following usual notations in a triangle ABC, if R(a + b) = c ab then r = a . Find the value of + .

158. Let a function f(x) be such that f''(x) = f'(x) + ex and f(0) = 0, f'(0) = 1, then find the value of

4))2(f(

n2

.

159. If the area bounded by |y| = sin1|x| and x = 1 is a( + b), then find the value of (a b).160. If z1, z2, z3 C satisfy the system of equations given by |z1| = |z2| = |z3| = 1, z1 + z2 + z3 = 1 and z1z2z3 = 1 such

that Im(z1) < Im(z2) < Im(z3), then find the value of [|z1 + 22z + 33z |], where [.] denotes the greatest integerfunction

161. If 2

2 2x x k 0

x 5x 9 x 5x 9

for all real x, then find sum of modulus of all integral values of

k in [4, 7].

162. Line L is a tangent to a unit circle S at a point P . Point A and the circle S are on the same side of L and thedistance from A to L is 3 unit. Two tangents from point A intersect line L at the point B and C. Find the valueof (PB)(PC)

SECTION 8 : (Integer value correct Type)This section contains 17 questions. The answer to each question is a Two digit integer, ranging from00 to 99 (both inclusive).

163. If x,y,z are non-negative integers such that 2(x3 + y3 + z3) = 3(x + y + z)2 then maximum value of x + y + zis

164. If ][ denotes the greatest integer function and let f be a function defined on the set of all non-negativeintegers and taking values in the same set. Given that -

(A) x f(x) = 20

20x

10

10

)x(f for all non negative integers

(B) 2008 < f(2010) < 2012then find the number of possible values of f(2010).

165. Eccentricity of an ellipse of minimum area, circumscribing two circles, of equal radius, touching externally,

is 22 . Find the value of .

166. Tangent at a point P (other than origin) on the curve y=x3 meets the curve again at P1 . The tangent at P1

meets the curve again at P2 and so on. then find the ratio a re a o f P P P1 2 3a re a o f P P P1 2

167. Let f : N [0, ) be a function satisfying the following conditions :(a) f(100) = 10,

(b) )1(f)0(f1 + )2(f)1(f

1 + )3(f)2(f

1 +......+ )1n(f)n(f

1 = f(n +1), for all non-negative integers n. Find

the value of f(9801).

RESONANCE Page - 20

MATHEMATICS

168. Complex number z satisfies arg (z (3 + 3i)) = and argz 2z 2i 11

then range of such that no z

exists is [a ,b], then find the value of [ 10 |a+b|] where [.] represents the greatest integer function and argmeans principal value of argument.

169. Range of f(x) = n

1x3xtanxsin2, x

3

,

6 is [a, b], then find the value of 11[a + b].[] represents greatest integer function.

170. A function g(x) is defined as g(x) = 41

f(2x2 1) + 21

f(1 x2) and f (x) is an strictly increasing function then

g(x) is strictly increasing on the interval p p0q q

, ,

then find the value of p2+q2.

(where p and q are coprime to each other)

171. If

e)xcos43(cxcosbadx)xcos43(

x2sind3 where a, b, c, d are positive integers and e is arbitrary constant

then minimum value of a + b + c + d equal to

172. If f(x) = x3 x2 + 100x + 1001, then find the number of correct statements among the following

(i) f(2000) > f(2001), (ii) f

1999

1 > f

2000

1, (iii) f(x + 1) > f(x 1), (iv) f(3x 5) > f(3x)

173. The numbers 2,3,4,5,6,7,8 are to be placed, one per square, in the diagram shown so that the sum of the fournumbers in the horizontal row equals 21 and the sum of the four numbers in the vertical column also equals21. In how many different ways can this be done ?

174. If the function f(x) = sin(n x) cos(n x), where n x , strictly increases in interval [e,e] and thelength of [e,e] is greatest possible, then find the value of 5 cos ().

175. If

1x3cos2x4cosx5cos

dx = a

x2sin sin x + C, find a.

176. If largest subset of (0, p ) at each point of which the function f(x) = 3 cos4 x + 10 cos3 x + 6 cos2 x 3 , is

decreasing is

,

r

2p

,0 , then find the value of (p + r).

177. Consider seven digit number x1 x2.......x7, where x1, x2.......,x7, 0 having the property that x4 is the great-est digit and all digits towards the left and right of x4 are in decreasing order. Then total number of suchnumber in which all digits are distinct is a, then find the sum of digits of a.

178. Let 1 1 1a tan tan30,b cos cos40,c sin sin(b a) then product of all integral values of x satisfying1cos cosx x a b c is

179. If S =

2n32

2

)1n(1n3

then S9

=

180. If

13n n

n2nn

n

ClimC

=

AB , where A & B are relatively prime numbers, then A+B is equal to-

RESONANCE Page - 21

MATHEMATICS

1. (D) 2. (D) 3. (D) 4. (B) 5. (A) 6. (A) 7. (D)

8. (B) 9. (D) 10. (C) 11. (A) 12. (B) 13. (B) 14. (C)

15. (A) 16. (B) 17. (D) 18. (B) 19. (C) 20. (D) 21. (B)

22. (A) 23. (C) 24. (B) 25. (B) 26. (D) 27. (C) 28. (D)

29. (C) 30. (B) 31. (B) 32. (C) 33. (B) 34. (A) 35. (A)

36. (C) 37. (B) 38. (D) 39. (B) 40. (B) 41. (D) 42. (D)

43. (C) 44. (B) 45. (D) 46. (C) 47. (A) 48. (B) 49. (B)

50. (B) 51. (C) 52. (B) 53. (A) 54. (BD) 55. (ABCD) 56. (BD)

57. (BD) 58. (ABC) 59. (AB) 60. (ABC) 61. (ABCD) 62. (AC) 63. (BC)

64. (AC) 65. (BC) 66. (ABCD) 67. (ABCD) 68. (ABCD) 69. (ACD) 70. (AB)

71. (CD) 72. (AC) 73. (BC) 74. (AC) 75. (AD) 76. (C) 77. (BD)

78. (ABC) 79. (ABD) 80. (ABD) 81. (ABCD) 82. (AD) 83. (ACD) 84. (CD)

85. (BD) 86. (ABCD) 87. (BD) 88. (BC) 89. (AB) 90. (ABC) 91. (ABCD)

92. (C) 93. (D) 94. (B) 95. (B) 96. (D) 97. (B) 98. (C)

99. (D) 100. (D) 101. (A) 102. (A) 103. (B) 104. (D) 105. (D)

106. (B) 107. (B) 108. (A) 109. (C) 110. (B) 111. (D) 112. (B)

113. (C) 114. (A) - (r), (B) - s, (C) - q, (D) - p 115. (A) - (r), (B) - (t), (C) - (q), (D) - (p)

116. (A) - (p,r), (B) - (p,s,t), (C) - (p,r), (D) - (q) 117. (A) -p,q,s,t ; (B) - p; (C) - s; (D) - p,s,t

118. (B) 119. 0 120. 4 121. 0 122. 5 123. 3 124. 2

125. 7 126. 6 127. 8 128. 4 129. 3 130. 1 131. 1

132. 2 133. 3 134. 1 135. 0 136. 1 137. 0 138. 7

139. 5 140. 2 141. 2 142. 9 143. 0 144. 4 145. 3

146. 5 147. 7 148. 5 149. 3 150. 1 151. 1 152. 6

153. 4 154. 1 155. 0 156. 8 157. 4 158. 4 159. 3

160. 2 161. 7 162. 3 163. 12 164. 00 165. 33 166. 16

167. 99 168. 47 169. 11 170. 13 171. 29 172. 02 173. 72

174. 05 175. 02 176. 05 177. 09 178. 24 179. 16 180. 43

ANSWER KEY

CRITICAL QUESTION BANK

RESONANCE Page - 22

MATHEMATICS

SOLUTIONS

CRITICAL QUESTION BANK

1. Method If '(x) = 2sin(x + ) cos(x + ) + 2sin(x + ) cos(x + ) 2cos( ) cos(x + ) sin(x + ) 2cos( ) sin(x + ) cos(x + )= sin(2x + 2) + sin(2x + 2) 2cos( ) sin(2x + + )= 2sin(2x + + ) cos( ) 2 cos( ) sin(2x + + )= 0 f(x) is a constant functionMethod IIf(x) = 1 cos2(x + ) + sin2 (x + ) 2cos( ) sin(x + )sin(x + )= 1 cos(2x + + ) cos( ) 2cos( ) sin(x + )sin(x + )= 1 cos( ) [cos(2x + + ) cos(2x + + ) +cos( )]= sin2 ( )a constant function

2. Product of roots

)cb(a)ba(c

= 1 b = )ca(ac2

a

b=

ca

c2

=

1c

a

2

(+) = 11

2

2= 1

3. AB =

4321

dcba

=

d4b3c4a3d2bc2a

BA =

dcba

4321

=

d4c2d3cb4a2b3a

if AB = BA, then a + 2c = a + 3b 2c = 3b b 0b + 2d = 2a + 4b 2a 2d = 3b

cb3d3a3

=

b23b3

b29

= 1

4.

x

x

x 0x

x 0

(x 1) x

2xx 0

x xx x

2x 2x 0

x x

2 2x xx 0

We know lim x 1

lim (x 1) 0

e xlim(x) 1

(x 1) (x 1)1 (x 1) ... 1 (x 1)2! 3!

lim(x ) 1

(x 1) (x 1)...... 1 1 12! 3!lim .

2 4 8(x ) 1 (x ) 1

5. Let 4m3 3am2 8a2m + 8 = 0 has three roots m1, m2, m3 m1m2m3 = 2

Given m1m2 = 1 m3 = 2 32 12a 16a2 + 8 = 0

16a2 + 12a 24 = 0 Sum of values of a = 1612

= 43

6. P (a d) pd Q (a d) qd R (a d) rd option A is correctFor option B

p q rT (a d) (p q r) d P Q R 3(a d) (p q r)d a d need not to be zero

p q rT P Q R For option C p q r If Common diff. is ve . then

pd qd qr P Q R For option D not necessarily true. if first term is not an integer.

7. Let y = x2 4x + 5y = (x 2)2 + 1

sin1 y, cos1 y defined only at y = 1 at which x = 2

So given equation

4 + 2a + 2

= 0

8a

4

8. 10log 15 = 10 101 log 3 log 2

= 104log 2

104log 3 = 4 4 we required those elements of A which can be written in termsof 10log 2 & 10log 3 .i.e. number of integers which are divisible 2 or 3, or 5 only.required numbers = 24

9. Roots are 2k, 3 6 < 2k < 10 k (3, 5)largest value of k is not defined.

10. )xx3x2()1x2(

23

0

x ( , 1)

0,21

1

,

2

for log2 x, x > 0

for )x(logsin 21 , 0 log2 x 1 x [1, 2] ........(2)

RESONANCE Page - 23

MATHEMATICSSo by (1) and (2) we get

x [1, 2]

11. R2 SG2 = 91

(a2 + b2 + c2) 93

(abc)2/3 .........(1)

abc = 4R = 4Rrs = 4Rr

2

cba 2Rr(abc)1/3.3

(abc)2/3 6Rr

R2 SG2 31

6Rr

SG2 R2 2RrSG2 SI2SG SI

12. Image of the centre C2 (1, 3) in the line 3x + 4y 16 = 0 is P(7, 5).Now for C1C2 + C2C3 + C3C1 to be minimum C1, C3 and P shouldbe on same line so C3 = (0, 4)C3 = (0, 4)

Distance between C3 and C150 5 2

radius of C1 = 3 2so radius of C3 = 2 2Equation of C3 (x 0) + (y 4) = 8x + y 8y + 8 = 0, a = 0, b = 8, c = 8

13. For 0 < n < 1, sin x 1, sin x > sinn xNow, for 0 < n

RESONANCE Page - 24

MATHEMATICS

19. Since f(x) = 0 31

1 x

2

0

xf(x) dx = 22

0

xf(x).2

2 2

0

xf (x) dx2

= 2f(2) + 12

2 2

30

x dx1 x f(2) = 0

= 0 + 16

2 2

30

3x dx1 x =

16

21/ 23

01 x

=

13 [3 1] =

23

20. Clearly %P (1,2)Equation of tangent at P is 4x (1) + y(2) = 8P ij Li'kZ js[kk dk lehdj.k 4x (1) + y(2) = 8 A (2,0) and B = (0, 4) A (2,0) vkSj B = (0, 4)Similarly normal at P is 2x 4y + 6 = 0blh izdkj P ij vfHkyEc 2x 4y + 6 = 0

A (3, 0) and B

23

,0 A (3, 0) rFkk

B

23

,0

Area BPB = 21

12/30121140

= 45

and area APAA = 5

Ratio = 4

21. I =

2x

cos2

dxe2x

sin2x

cos

2

2/x

Put x/2 = t

I = dt)ttantsect(sece t= et sec t + C

= ex/2 sec 2x

+ C

22. L =

1 x

x 0

e 2lim1 x

=

12e 1

= 2 e

1

23. 1 1 1tan 1 tan 2 tan 3

1 1tan 2 cot 34

..............................(1)

1 1 11 1tan (1) tan tan2 3 2

1 11tan co t 3

2 4

............................(2)

From 1

113 1cot 324 6

From 2

13 cot 34

13 13 11 cot 34 24 6

15 5 cot 324 6

p 5, q 24, r 5, s 6

24. dyy.dx = 1

ydy dx

2y

2= x + c

Q(0, 1) lies on curve 2y

2 = x +

12

y2 = 2x + 1 and y2 = 2x + 1

y2 = 21

x2

and y2 = 21

x 2

Required area = 4

= 4 23

12

1/ 23 / 20

1 2x

=

43 (0 1) =

43

25. f(f(x)) = f(x) + f2(x) + f4(x) + f8(x) + .....Coefficient of x10 in f(x) = 0Coefficient of x10 in f2(x) = 2Coefficient of x10 in f4(x) = 4C2 + 4C1 = 10Coefficient of x10 in f8(x) = 8C2 = 28

26. Let S = 2008

r 2009rr 1

r( )

S = 2009 (1 + 2 +.......2008)S = 2009

27. 2x

2

(cos x)limx

2

must be '0

iff > 228. Let variable line be x + my + n = 0

P1 =

1 1

2 2

3a x 3amyn

a b c a b cm

P2 =

2 2

2 2

3b x 3bmyn

a b c a b cm

RESONANCE Page - 25

MATHEMATICS

P3 =

3 3

2 2

3c x 3cmyn

a b c a b cm

Given P1 + P2 + P3 = 0

)cba()cxbxax(3 321

+ )cba()cybyay(m3 321

+ 3n = 0

)cba()cxbxax( 321

+ )cba()cybyay(m 321

+ n = 0

Hence line x + my + n = 0 passes through incentre

cbacybyay

,

cbacxbxax 321321

of triangle ABC.

29. According to LMVT

x

)0(f)x(f = 1)C(f 1

f (x) | x|

again by LMVT f (x) f (0)

x= 2f (C ) | x |

2f(x) x30. f(x) = cos 8 {x} = cos (8 x 8 [x]) = cos 8x. Its period is

41

.

Period of sin 2x cosec2x is 21

period of f(x) is 21

32. f(0) = k and x 0lim f(x) = x 0lim x2{e1/x} = 0

for f(x) to be continuous f(0) = x 0lim f(x) k = 0

f (0+) = h 0lim

2 1/ hh {e }h

= 0

f(x) is derivable at x = 0 and f (0) = h 0lim

2 1/ hh {e }h

= 0

due to {e1/x}, will be discontinuous at all value of x, where e1/xbecomes an integer.i.e. at x = log2e, log3e,.....etc.

33. f is even and g is odd gogog is odd gogog fogogog is even fogogog gofogofogogog is even. gofogofogogog

34. nth term of 1 + 3 + 6 + 10 + ....... is n = 2)1n(n

Tn = cot1

2

a)1n(na2 1

= cot1

a2

a)1n(n4 2

=tan1

2a)1n(n4

a2 = tan1

2a)1n(

2na1

2a

= tan-1

2na

.

2a)1n(1

2na

2a)1n(

= tan-1

2a)1n(

tan1

2

na

Put n = 1, 2, 3, .........., n we have

Sn = tan1 2a)1n( tan1 2

a; S = 2

tan1 2

a = cot-1

2a

a = 2 S = 4

35. 5x5 10x3 + x + 2y + 6 = 0 ......... (i)Differentiating w.r.t. x

25x4 30x2 + 1 + 2 dxdy

= 0

dxdy

= 21

(25x4 30x2 + 1)

)3,0(dxdy

= 2

1 ( 0 0 + 1) = 2

1

equation of the normal at P(0,3) is P(0,3) y + 3 = 2 (x0) y = 2x 3 .......... (ii)solving equation (i) and (ii)5x5 10x3 + x + 4x 6 + 6 = 05x (x42x2+1) = 05x (x21)2 = 0 x = 0 , 1 , 1y = 3 , 1 , 5 Normal at point P(0,3) meets the curve again at two points(1,1) and (1,5) .Now equation of tangent at point (1,1) is

y + 1 = )1,1(dxdy

(x1)

y + 1 = 2 ( x 1) 2x y 3 = 0 ........ (iii)Equation of tangent at point (1, 5) is

y + 5 = )5,1(dx

dy

(x+1)

y + 5 = 2 (x + 1) 2x y 3 = 0 ........ (iv)36. f(0) = 1

and f(x + y + 1) = 2)y(f)x(f substituting x = 0, y = 0, we get f(1) = (1 + 1)2 = 4

substituting y = 0, we get 21)x(f)1x(f f(0) = 1,f(1) = 22, f(2) = 32, f(3) = 42 and so onhence f(x) = (1 + x)2

37. Composition of one - one function is one - one

38. AQ = 2 4 2 22 2a t 4a t = at2 22t 4 =

22 2t t 4

( a = 1)

Since t2 = t1 1

2t = ( t1) + 1

2t

RESONANCE Page - 26

MATHEMATICS t1 < 0 (as shown is diagramnow apply A.M. G.M

1

1

2(t )t

22

(t1) + 1

2t

2 2

t2 2 2

(AQ)min = 2 2 8 4 = 4 6

39. p = (x 2) (x 4) = (x2 6x + 8)p(x) = (x3/3 3x2 + 8x) +

40. y = f(x) = (3x + 2)2 1, < x 23

It is clear from the graph that if x (, 2/3], then y [1,)y + 1 = (3x + 2)2 3x + 2 = y 1( x 2/3, negative sign)3x + 2 = y 1

x =2 y 1

3

= g(y) = f1(y)

g(x) = 2 x 13

41. 2tan4tan2.4124tan

92tan

29

cot 11111

4tan8tan8.41

48tan334tan

433

cot 11111

8tan16tan8.161

816tan129

8tan8

129cot 11111

Similarly Tn =

n11n1 2tan2tan on adding we get

sum = 2tan2tan 11n1

as n sum = 2

tan1 2 = cot1 2.

42. Let S = 2C0 + 3C1 + 4C2 + 5C3 +.......+ 99C97 + 100C98 S = 3C0 + 3C1 + 4C2 + 5C3 +.......+ 99C97 + 100C98 S = 4C1 + 4C2 + 5C3 + .......+ 99C97 + 100C98 S = 5C2 + 5C3 + .......+ 99C97 + 100C98 S = 101C98

43. Orthocentre lies on the rectangular hyperbola and

h = 12(3 ) x

3

and k = 12(3 ) y

3

x1 = 3(h 2) and y1 = 3(k 2) orthocentre (x1, y1) lies on x2 y2 = 36 9(h 2)2 9(k 2)2 = 36 (h 2)2 (k 2)2 = 4 locus of centroid G(h, k) is (x 2)2 (y 2)2 = 4 = 4

44. Here f (x) =

3x1

3x2x4 2,

f(x) is decreasing in [2, 3) and increasing in [3, ).Now, f(x) will have least value at x = 3 if= 3x

lim f(x) f(3) i.e. 15 + n(b2 3b + 3) 15

i.e. n (b2 3b + 3) 0 b2 3b + 3 1 i.e. b2 3b + 2 0i.e. b (, 1] [2, ).

45. Let cosec1x = t

t2 2t 6

(t 2) (t 2)(t /6) 0 t /6 or t 2 (not possible) t /6 0+ < t /6 or /2 t < 0 0+ < cosec1x /6 or /2 cosec1x < 0 2 x < or < x 1 x [2, ) ( , 1] a = 1, b = 2 a + b = 1

46. arg 1 2 7

8 9 10 14

1 Z Z Z1 Z Z Z Z

.........

.......

= arg 1 2 7

1 2 7

1 Z Z Z1 Z Z Z

.........

..........

= 2 arg (1+Z1 + Z2 + .........+Z7)

ZZ

arg = 2 arg (Z)

z7

z6

z5z4 z3

z2z1

1

arg (Z +Z )3 4

= 2 arg (Z3 + Z4)= 2 x 7 14

15 15

Also arg (Z1 Z2 Z3.......Z7)

=

2 4 6 1415 15 15 15

..........

=

215

(1+ 2 +3 +.......+7)5615

RESONANCE Page - 27

MATHEMATICS

47. x + 1x2 = 1yy1

2

take loge

1yy1

n)1xx(n2

2

= n

22

2

y1yy1y

1x2 + x = 1y2 y

x + y = 1x2 + 1y2 .....(1)

similarly y + 1y2 = 1xx1

2

1y2 + y = 1x2 x

x + y = 1x2 1y2 .....(2)adding (1) & (2) x + y = 0Image of xy = 1 in xy = 1 it self.

48. |1+ Z1 + Z2 + Z3 + ..........Z7|= | 1 + Z7| + |Z1 + Z6| + |Z2 + Z5| + |Z3 + Z4|

= 2cos 715

+ 2cos515

+ 2cos315

+ 2cos 15

= 2 4 3 42 215 15 15 15

cos cos cos cos

= 4 cos415

315 15

cos cos

= 8 cos 15

cos 215

cos 415

=

8 8 158 15sin( / )sin( / )

=

12 sec

715

49.

1by

a

x2

2

2

2

Tangent at P(asec, btan)

1tanby

seca

x

........(1)Asmptotes are

y = a

bx ..........(2)

y = a

bx .......(3)

M [a(sec tan), b(sec tan) ]and N [a(sec + tan), b(sec + tan) ] ON = 22 ba (sec + tan) = ae(sec + tan) OM + ON = ae(2sec) = 2ae(sec)

SP + SP = e(asec e

a) + e(asec +

e

a) = 2aesec

50.

D

A

B

C

90 30

V = 13 AD (

12

AB. BC sin 30) AD. AB. BC = 216Now, AB + BC + AD 3 (AD. AB. BC)1/3 AB + BC + AD 18and minimum value occurs when AB = BC = AD = 6

Hence AC = 2 2AB BC 2AB BC 30cos

= 3 6 252. Since f(x) > 0

f (x) is always increasing.g(x) = 2f (2x3 3x2) (6x2 6x) + f (6x2 4x3 3) (12x 12x2)= 12(x2 x) . (f (2x3 3x2) f (6x2 4x3 3))= 12x (x 1) [f (2x3 3x2) f (6x2 4x3 3)]For increasing g(x) > 0Case-I x < 0 or x > 1 f (2x3 3x2) > f (6x2 4x3 3) 2x3 3x2 > 6x2 4x3 3 { f(x) is increasing}

(x 1)2

21

x > 0 x >

21 x

0,21 (1, )

Case-II If 0 < x < 1f (2x3 3x2) < f (6x2 4x3 3)

(x 1)2

21

x < 0 x <

21

, so there

is no solution

Hence the values are x 1,0 1,2

RESONANCE Page - 28

MATHEMATICS53. Clearly, the point lies on 7x y = 5

Also, centre of the circle must lie on the bisectors of the lines x +y + 13 = 0 and 7x y 5 = 0 given by

x y 13 7x y 5x 3y 35 and 3x y 15 0

2 50

Let (h,k) be the centre of the circle, then h -3k = 35 ..... (1)and 3h + k +15 = 0..... (2)Clearly CA is perpendicular AQ

k 2 7 1 h 7k 15 0h 1

..... (3)

On solving, we get centres as C1 (29, 2) and C2 ( 6, 3) r21 = 800 and r21 = 50 Smaller circle has radius =

50

Therefore area of quadrilateral ACBQ = 50 200 sq.units = 100 sq. units. Hence (A)

54. z = x + iy(x 2) = (x 7) + ( y + 2)(x 2) - (x 7) = ( y + 2)

( y + 2) = 5(2x 9)

(y + 2) = 10 9

x2

min

5Y 4aX a2

PQ | L(LR) 10

55. (x2 + y) = (f(x))2 + f(y)f(0) = (f(0))2 + f(0) f(0) = 0for y = 0 : f(x2) = (f(x))2 ........(i)for y = x2 : 0 = f(0) = (f(x))2 + f(x2) (f(x))2 = f(x2) ........(ii)from (i) and (ii)f(x2) = f(x2) f(x) = f(x) ........(iii)thus f(x) is an odd functionif f(x) is even also, then f(x) = f(x) ........(iv) f(x) = 0 for all x {by (iii) and (iv)}since f(x) is continuous at x = 0, 0h

lim f(h) = 0

0h2lim

f(x + h2) = 0h2lim

{(f(h))2 + f(x)} = f(x) it is continuous everywhere

since 0hlim h

)0(f)h0(f = 0h

lim h

)h(f exists

0hlim 2

2

h)x(f)hx(f

= 0hlim 2

2

h)x(f)x(f))h(f(

= (f (0))2 f (x) = (f (0))2 f (0) = (f (0))2 f (0) = 0 or f (0) = 1 f(x) = (f (0))2 xNow (i) if f (0) = 0, then f(x) = 0 = x f (0)(ii) if f (0) = 1, then f(x) = x = x f (0)

56. f(0) = r is odd. Let r = 2n + 1, n I.f(1) = 1 + p q + 2n + 1 = p q + 2n is odd exactly one of p, q is oddf(1) = 1 + p + q + 2n + 1 = p + q + 2n + 2 is oddIf possible suppose , I be zeros of f(x). x3 + px2 + qx + r = x3 (+ + )x2 + (+ + )x r = are odd integers p = (+ + ) is oddand q = + + is also odd.It is a contradiction. Hence f(x) = 0cannot have three integer roots.

57. f(x) = x2 + a

bx +

a

c

f(1) < 0 & f(1) < 0

58. 1 23 )1x(log 2 = 21

3x5xlog 2/13

1 22

log3 |x + 1| = 22

log3

3x5x

log3

|1x|3

= log3

3x5x

|1x|3 = 3x

5x

Case- I x + 1 > 0 x > 1 3(x + 3) = (x + 1)(x + 5) x2 + 3x 4 = 0 x = 4 or x = 1x = 4 rejected ( x > 1) x = 1Case - II x + 1 < 0 x < 1 3(x + 3) = (x + 1)(x + 5) x2 + 9x + 14 = 0 x = 2 or x = 7 Set of value of x = {7, 2, 1}

59. 1xlim

g(x) = 1xlim

1x4x2

3)x(h)x(fxlimm

m

m = m

lim

1xlim

1x4x23)x(h)x(fx

m

m

= mlim

1xlim

m

m

x

1x42x

3)x(h)x(f = 2

)1(f

f(1) = 2e3

Similarly 1xlim

g(x) = 53)1(h

h(1) = 5e3 3

RESONANCE Page - 29

MATHEMATICS60. The tangent 3x + 4y 25 = 0 is tangent at vertex and axis is 4x

3y = 0So, PS = a = 5Latus rectum = AB = 20

61. Let 768 32cos

16 3 32cos3

cos2 6

So 4 8 32 32cos6

4 8 8cos12

114 4sin 4 4cos24 2411 11 114 1 cos 4.2cos 2 2 cos24 48 48

So a 2,b 48b 24a

62. f(x) = ex (x2 + bx + c) > 0 iff D = b2 4c < 0g(x)= ex (x2 + (b + 2)x + (b + c)) > 0 iff D=b2 4c + 4=D + 4 < 0

63.

2

1 1 2c o s x c o s x12 a 2 a 02

let 1cos x2

= t t 2

Now equation 12 2t a t a 02

has one root 2 or

greater than 2 and other root less than 2

(2) 0

21

a 2 a 02

4

22a 1 a 04 2 2a 3 0a

(a 3)(a 1) 0 a 3 or a 164. f is discontinuous at points outside its domain.

65. Tr+1

15Cr(x4)(15r)(x3)r = 15C

rx607r

(A) for the term independent of x, 60 7r = 0 r is not aninteger there is no term independent of x.(B) n = 15 is odd

nCr will be maximum if r =

n 12 or r =

n 12

i.e. r = 7 or r = 8 binomial coefficient of 8th and 9th terms will be greatest

(C) for the cofficient of x32; 60 7r = 32 r = 4 cofficient of x32 is = 15C4for the coeff. of x17; 60 7r = 17; r = 11 coeff. of x17 is = 15C11 = 15C4 (C) is correct

(D) If x = 2 , Tr+1 = 15Cr 607r

22

for rational terms r = 0, 2, 4, 6, ....... 1466. Solution set = 67. A & B are obvious

-1 -1

-1 -1 -1

1 1 1f -1 = i.e., f -1 isreciprocalof (f(x)-1)x (f(x)-1) x

/2 cot | x | 1 tan | x |Now,forf x = ,f(x)-1= ,f -1=xtan | x| tan | x | cot | x |

2 1-k n| x| 1Alsoforf(x) = ,f(x)-1= f -1+k n| x| 1+k n| x| x

1+k n| x|1=1-k n| x |

68. t1t2 = 2 and t1 + t2 + t3 = 0 and a = 2

R(t )3

Q(t )2

TP(t )1

Let circumcentre of TPQ be O(h,k) O(h, k) will be mid point of RT

h =2

1 2 3at t at2

h = 2 + t32 ........ (1)

and k = 1 2 3a(t t ) 2at

2

k = t3 .........

(2) h = 2 + k2 locus of O(h, k) is O(h, k)y2 = x 2y2 = 1(x 2)

69.

70. a0z4 + a1z3 + a2z2 + a3z + a4 = 0Taking conjugate on both sides.

a04)z( + a1 3)z( + a2 2)z( + a3 z + a4 = 0

4321 z,z,z,z are the roots of the equation if z1 is real,

then 1z is also real and if z1 is non real, then 1z is also rootbecause imaginary roots occur in conjugate pair.

71.x3

X 3

sin(e 1)limn(x 2)

put x = 3 + h

RESONANCE Page - 30

MATHEMATICSh

h 0

sin(e 1)limn(1 h) =

h h

hh 0

sin(e 1).(e 1) hlimn(1 h)(e 1)h

= 1 1 1 = 1

72.

a b . b c b.c a.b a.c

b c . c a a.c b.c a.b

c a . a b a.b a.c b.c

Given that

max

a b c 3 a b c . a b c 3 a.b b.c c.a 0

a.b . b.c b.c c.a c.a a.b a.b b.c c.a

a.b . b.c b.c c.a c.a a.b

0[since , x y z 0,xy yz zx 0]0only when a.b b.c a.c

0

a b,b cand c a

2a 3b 4c . a b 5b c 6c a

10a.(b c) 18b.(c a) 4c.(a b) 32

73. As the point (, ), lies on both f(x) and g(x) , the point ()will also lie on both the curves and as the functions arecontinuous they must cross (meet on) the line y = x in between.f must be on decreasing path, for all these to happen.

74. f (x)= ab cos x b 2a1 sin x = b cos (x + )

Where = cos1a. So f(x) = 0 x =(2m +1) 2

Also f(x) = b sin (x+). Thus f

2 < 0 and f

23

> 0

Hence f has maximum at 2

and minimum at

23

.

More over, max f = ab sin (/2 ) +b 2a1 cos ( 2 ) +

c = a2b + b (1 a2) + c = b + cand min f = b + c.

and min f = b + c.

75. 23x 2lim (log (ax 3x 1))

must be equal to 1

So 4a + 7 = 3 a = 1

Also

23

2x 2 x 23

log ( x 3x 1) 1 (x 1)(3 2x)lim limlog (x 1) ( x 3x 1) 1/ 3e e el

76.x

x x

2 1 1f (x)4 1 2 1

is one-one function.

77. Let S x2 + y2 + 2gx + 2fy + c = 0

it cuts x2 + y2 = 4 orthogonally c = 4Moreover 2g + 2f + 9 = 0( ( g, f) satisfy the given equation) S x2 + y2 + 2gx + 2fy + 4 = 0 x2 + y2 + (2f + 9)x + 2fy + 4 = 0 (x2 + y2 + 9x + 4) + 2f (x + y) = 0It is of the form S + P = 0 and hence passes through theintersection of S = 0 and P = 0 which when solved give ( 1/2,1/2), ( 4, 4).

78. Replace x by 2, 2f(2) + 2f12

-2f(1) = 4

f(2) + f12

=2 + f(1)-----(1)

Replace x by 1, f(1) = -1 ..............(2)

Replace x by 12

, 2f12

+12

f(2) + 2 = 52

..............(3)

Solve (1) and (3) => f 12

= 0 ; f(2) = 1

79. Let 768 32 cos

16 3 32 cos3

cos2 6

So 4 8 32 32 cos 4 8 8 cos6 12

114 4 sin 4 4 cos24 2411 11 114 1 cos 4.2 cos 2 2 cos24 48 48

So a 2, b 48(b a ) 46

80. Equation of the required plane is_ _ _ _ _ _ _

1 1 2 2 1 2 1 2(r.n q ) ( r.n q ) 0 r.(n n ) (q q )

Now , perpendicular to _ _

3 4n n

_ _ _ _

1 2 3 4(n n ).(n n ) 0 _ _ _

1 3 4_ _ _

2 3 4

[n n n ][n n n ]

So independent on q3 and q4 only.

RESONANCE Page - 31

MATHEMATICS81.

1 1 1

2 2 2

1 2 1 2 1 2

2 1 2 1

1

1

1

L e t C ( 1 , 2 2 , 3 3 ) a n dD ( 1 , 2 2 , 3 3 ) .C D 1 4

( ) 4 ( ) 9 ( ) 1 41 1

L e t G ( , , )4 5 24 1 0 44 9 6

4 x - 5 4 y - 1 0 4 z - 9L o c u s o f G i s = =2 4 6

5 5 92 x - y - z -2 2 4

= = 31 12

82. 3x 6y

k

= 1

........ (1)2x2 + 2xy + 3y2 1 = 0 ........ (2)Homogenizing (2) with the help of (1), we get

2x2 + 2xy + 3y2 23x 6y

k

= 0

k2(2x2 + 2xy + 3y2) (3x + 6y)2 = 0 ...... (3) circle described on AB as Diameter passes through (0, 0) AB will subtend right angle at (0, 0) coefficient of x2 + coefficient of y2 = 0 (2k2 9) + (3k2 36) = 0 k2 = 9 k = 3

83.

The rod sweeps out the figure which is a cone.The distance of point (A(1, 0, 1) from the plane

is .unit19|4211

The slant height l of the cone is 2 units.Then the radius of the base of the cone is

31412 l

Hence, the volume of the cone is 1.)3(3

2

cubic units.Area of the circle on the plane which the rodtraces is 3.Also, the centre of the circle is Q(x, y, z). Then

,

2)2(1)4201(

21z

20y

11x

222

or

35

,

32

,

32)z,y,x(Q

.

84. (A) Consider g(x) = sin1 f(x) x

since g(0) = 0, g

2 = 0

there is at least one value of

2

,0 such that g ()

=

2))(f(1)(f

1 = 0

i.e. f () = 2))(f(1 for atleast one value of but may

not be for all

2

,0 false

(B) Consider g(x) = f(x) x2

since g(0) = 0, g

2 = 0


2

,0 such that g ()

= f () 2

= 0

i.e. f () = 2

for atleast one value of but may not be for all

2

,0

false

(C) Consider g(x) = (f(x))2 x2

since g(0) = 0, g

2 = 0


2

,0 such that g ()

= 2f() f () 2

= 0

f() f () = 1

true

(D) Consider g(x) = f(x) 22x4

since g(0) = 0, g

2 = 0

RESONANCE Page - 32

MATHEMATICS


2

,0 such

that g () = f () 28

= 0

f () = 28

true

85.

third side will be parallel to bisectors of two given line

Bisectors are 7x y 3

5 2

= (x y 3)

2

Bisectors are 3x + y 1 = 0 and x 3y + 9 = 0Now required third side will be parallel to these bisectors 3x + y + 7 = 0 or x 3y 31 = 0

86. f(x) = cos1 (cos 2x), g(x) = |cos x|f(x), and g(x) both are even and periodic so max {f(x), g(x)}and min{f(x), g(x)} will also be periodic and

even.

but max {f(x), g(x)} will be non-differentiable when f(x) = g(x)no of points where f(x) = g(x) are four in [0, 2]

87. f(x) = 0 2cosx + 2cos2x = 0

x = 3

, are critical points

f(x) is increasing in

3

,0 and decreasing in

23

,

3

Hence f(0) = 0 , f

3 = 2

33, f() = 0 , f

2

3 = 2

Least value = 2, Greatest value = 233

88. (A) If function is one- one & onto then a = b since everyelement of set B should have exactly one pre-imagein A.(B) For one - one, into function, every element of set B shouldhave either one pre-image or no pre-image inset A. no. of elements in set B > no. of elements inA. b > a(C) For many - one, onto function, every element of set B shouldhave one or more than one pre-images inA. n(B) < n (A) b < a or a > b(D) For many - one, into function a R & b R .

89. (A) f(x) = 1 x6/5 f (x) = 65

15x

exist x (1, 1)and f(1) = 0 = f(1) Rolles theorem is applicable

(B) x 0lim

2x

t

0x

xe dt

1 x e

= x 0

lim

2x

t

0x

x e dt

1 x e

= x 0lim

2 2x

x t

0x

x(e ) e dt

1 e

00

= x 0lim =

2 2 2x x x

x

x(e .2x) e ee

=

0 1 11

= 2

(C) | a

+ b | = 1 + 1 + 2cos= 2(1 + cos ) = 4 cos2/2

and | a b

|2 = 1 + 1 2cos = 2(1 cos) = 4 sin2 /2

2

1

| a b | + 21

| a b |

=

14

2 2sec cosec2 2

=

14

2 21 tan 1 cot2 2

=

14

2 22 tan cot2 2

A.M. G.M

tan2 2

cot2 2 2 1

(D) M(1, 2) lies on the director circle of 7x2 12y2 = 84

90. m3 + (2a + 5)m2 6m 2a = 0

m1 + m2 + m3 = (2a + 5)m1 m2 + m2 m3 + m3 m1= 6m1 m2 m3 = 2aFor (A)

a +

3

1iim

= 1

a 2a 5 = 1 a = 4 m1 m2 m3 = 8 m1 = 1, m2 = 2, m3 = 4 m1 m2 m3 = 8for (B)a 2a 5 = 5 a = 0 m1 m2 m3 = 0 m1 = 1, m2 = 0, m3 = 6

a +

3

1iim = 0 + 0 = 0 (C is correct)

for (D)a + 2a = 32 not possibleHence (D) is incorrect.

91. fof(g(x)) = 1 + sin(fog(x)) f(1 x) = 1 + sin(1 x) f(x) = 1 + sinx

RESONANCE Page - 33

MATHEMATICS92. S1 : a, b, c must belongs to R.

S3 : z = a + bi = a biS4 : a, b R

93. S1 : D = 0 roots are real and equal if a, b, c R.S2 : Roots are integers

S3 : 1 1 1

2 2 2

a b ca b c

is true iff roots are both common

S4 : If Numerator and Denominator have common factor then Rfcan be equal to Rg.

94. GSS

F2F1P( )

Let I(h, k) and G(ae2cos, 0)

h =2e(acos ) ae cos

1 e

=

aecos (1 e)1 e

= aecos

and k =e(bsin ) 0

1 e

=

be1 e sin

h = aecos ; k =be

sin1 e

a = 6, b = 3 3 and e = 12

h = 3cos, k = 3 sin

locus of incentre I(h, k) is 2 2x y

9 3 = 1

and its eccentricity =319 =

23

we know that the straight lines joining SF1 and SF2 bisects thenormal PG. Required point of intersection is midpoint of PG.

P 3

= P93,2

and G(ae2cos,0) G3

, 04

Required point = 15 9

,

8 4

95 to 96f(1.1) = f(1.2) f is may onelog{x} [x] < 2 [x] > {x}2 [x] > (x [x])2 x2 2[x] x + [x]2 [x] < 0x [1, 0) x2 + 2x + 2 < 0,not possiblex [0, 1) x2 < 0, not possiblex (1, 2) 0 < 2 infinitely may solution.

97 to 98.two points of inflection occurs at x = 1 and x = 0 P(1) = P(0) = 0 P(x) = ax(x 1)

P(x) = 3ax

3 2ax

2 + b

x 0

dydx

= tan(sec1 2 ) = 1

P(0) = 1 b = 1

P(x) = 3ax

3 2ax

2 + 1

P(x) = 4ax

12

3ax

6 + x + c

P(1) = 1 a 1 1

12 6

1 + c = 1

a

4 + c = 2 .........(1)and P(1) = 1 a = 12c .........(2)solving (1) and (2), we get

c = 12 and a = 6

P(x) =4x

2 x3 + x +

12

P(0) = 12 P(x) = 0 2x3 3x2 + 1 = 0 (2x + 1)(x 1)2 = 0

99. to 100.

cos1 2x

+ cos1 3y

=

cos = 6xy

4x1

2

9y1

2

4x1

2

9y1

2

=

cos6xy

4x1

2

9y1

2

=

2

cos6xy

1 4x2

9y2

+ 36yx 22

= 36yx 22

+ cos2 3cos xy

1 cos2 = 4x2

+ 9y2

3cos xy

9x2 + 4y2 12 xycos = 36 sin2 N = 36and (cos1 x)2 (sin1 x)2 > 0

2

xsinxcos 11 > 0

x

21

,1 [p, q)

p = 1 , q = 21

RESONANCE Page - 34

MATHEMATICS

99. N 6 = 36 6 = 0

21

,1

100. sec1x is not defined at x = 0

101. to 102.

P26 104, ,3 3

AB = i j 3k , AC i j 2k

and BC 2i k

AC BC Fourth vertex D is (4, 2, 0) AB

AC

= i 5 j 2k Equation of base is x 5y 2z + 6 = 0Let E(x,y,z) be the foot of the perpendicular drawn from P to thebase

x 4

1 =y 26 / 35

=

z 10 / 32

= r (let)

(4 + r, 5r 263 , 2r

103 )

It lies on the base

4 + r 5265r 3

210 2r3

+ 6 = 0

r = 2

E4 22, ,3 3

+ m + n = 4

EP

= 2i 10 j 4k

height =EP

= 120

Volume =13 | AC

BC ||EP

| = 13 25 1 4 120

=

603 = 20 cubic unit

103. to 104

f(x) = tan1 22

2( 3 1)3

x 2x

x2 + 23x

2 3 x2 + 23x

+ 2

2( 3 1)

fmax

= tan12( 3 1)2( 3 1)

= 12

= M which occurs at x2 =

23x

i.e. at x = 31/4

= a

and fmin = 0 = m at x = 0

cos1x + cos1y =

31 17 3 7 3tan tan tan tan 3 2

24 8 24 8

x = y = 1 x + y = 2

3sin1(sinM) = 3sin1 sin12

= 4

x2 (tan(3sin1(sinM))) x + a4 = 0 x2 (tan/4) x + 3 = 0 x2 x + 3 = 0 + = 1, = 3 () (+) = 3 1 = 2

105. to 106.

Roots are 1, a

2

a

2 < 0 a > 0

a2 + 2a + 5 (5, )

a

2 < 4 a

21

,0

107. to 108Given, (a + b)2 x (ab + bc + ca + 1) y + 2 = 0Now, (a + b)2 (ab + bc + ca) + 1 0 for all a,b R a2 + b2 + ab bc ca + 1 0 a2 + (b c) a + (b2 bc + 1) 0 a R (b c)2 4(b2 bc + 1) 0 3b2 2bc + (4 c2) 0 b R 4c2 12(4 c2) 0 c2 12 + 3c2 0 c2 3 M = 3Also, lx + my + n = 0 (l, m, n are 1st, 3rd and 7th terms of anarithmetic progression) lx + (l + 2d)y + (l + 6d) = 0('d' be their common difference ) 'd' l (x + y + 1) + d(2y + 6) = 0 x + y + 1 = 0, 2y + 6 = 0 x = 2, y = 3Hence = 2, = 3

x2 + y2 = 4 ; x2 + y2 24x 10y + 2 = 0c1(0, 0) ; c2 = (12, 5)r1 = 2 ; r2 = 2169

169 2 > 0 48,48Since is integer

= 0, 1, 2,....,2 6 Number of possible integral values of is 13 Ans.x2 + y2 = M + 2 = 3 + 2 = 5tangent at (1, 2)

x 2y 5 = 0 .......(1)This line touches the circle x2 + y2 8x + 6y + 20 = 0

Solving 1,3x5y2x5yx2

Hence point of contact (3, 1) Ans.109 to 110

( ) = (( + ) ( + ))( + )2 4 = [( + )+ ( + )]2 4( + ) ( + )(b1)2 4c1 = (b2)2 4c2D1 = D2 ..........(i)

minimum of y = f(x) is 4D1

= 41

D1 = 1 D2 = 1

RESONANCE Page - 35

MATHEMATICS

minimum of y = g(x) is 4D2

= 41

minimum of y = g(x) occur at x = 2b2

= 27

b2 = 7b22 4c2 = D2

49 4c2 = 1 448

= c2

c2 = 12111 to 113

As det. (P nI) = 0

n1000n1202n1

= 0

(1 n)(n 3)(n + 1) = 0 n = 1,1,3

So N =

1 0 00 1 00 0 3

det.(N) = 3 det.(adj N) = (detN)2 = (3)2 = 9 Ans.112.Given PQPT = N PP 1 Q

1TT )P(P = P1N (PT)1 Q = P1N(P1)T [As (PT)1] = (P1)T]

Q

=

100

031

32

032

31

1 0 00 1 00 0 3

100

031

32

032

31

300

0310

0031

QT = Q + = 0 Ans.

113. Given P =

100012021

So P2 = PP =

100012021

100012021

=

100054045

Now P3 = P2P =

100054045

100012021

=

1000131401413

P4 = P3P =

1000131401413

100012021

=

1000414004041

Also P5 = P4P =

1000414004041

100012021

=

10001211220122121

Tr.(P) = 3, Tr. (P3) = 27 = 33Tr.(P5) = 243 = 35 Tr. (P2011) = 32011

Also Tr.(P2) = 11 = 32 + 2Tr.(P4) = 34 + 2Tr.(P2012) = 32012 + 2 Ans.

114. (A) logsinx (log3 (log0.2 x)) < 0 = logsinx1

log3 (log0.2x) > 1 log0.2x > 3 = log0.2(0.2)3

0 < x < (0.2)3 0 < x < 1251

(B) )1x(x)2x(sin)2xx)(3x2)(1e( 2x

0

)1x(x)2/3x)(1e( x

0 x < 1 or x 23

x (, 1)

,23

(C) |2 | [x] 1| | 2 ||[x] 1| 2| 2 0 |[x] 1| 4 3 [x] 5 x [3, 6)

(D) |sin1 (3x 4x3)| 2

2

sin1 (3x 4x3) 2

1 3x 4x3 1 1 x 1115. (A) x 0 and g(x) 2 x 1

Domain is x (, 0) (0, 1) (1, )(B) x 2 and f(x) 0 x 1/2

Domain is x (, 1/2) (1/2, 2) (2, )(C) f(f(x)) = x and x 2

Range is (, 2) (2, )

(D) g(g(x)) = 1x1x2

and x 0

Range is (, 1) (1, )

116. (A) Range of the function is (0, )(B) Function is decreasing(C) Range of function is not R and function is one-one.(D) The function is clearly many-one because the value offunction is zero at x = and .And range cannot be R.

RESONANCE Page - 36

MATHEMATICS

117. (A) an =

n

1n2n21n2n2 22

an =

1n2n21n2n2

422

na

1= 4

1

1n2n21n2n2 22

20

1n= 4

1 15

+ 41 513 ..................... 4

1 761841

= 41

(29 1) = 7

(C)xsin

1

x2sin1

=

x4sin2

x2sinxsinxsinx2sin

=

x2cosx2sin22

2/xcos2/xsin22/xsin2/x3cos2

=

x2cos1

cos 2x3

cos2x = cosx/2

cos 2x7

+ cos 2x

= 2 cos 2x

cos 2x7

= cos 2x

sin 2x sin 2x3

=0

sin 2x 0sin 2x3

=0 2x3

= n

x = 3n2

x

x = 32

,

34

,

38

,

310

(D) cot

4

1r2

1

r21tan

cot

4

1r2

1

r42tan

cot

4

1r

1)1r2)(1r2(1)1r2()1r2(tan

cot

4

1r

11 )1r2(tan)1r2(tan

cot (tan19 tan1(1))cot (tan1 4/5)

51x = 2x34x3

15x + 10 = 12x + 163x = 6x = 2

118. We have zn 1 = (z 1)i2 i2 (n 1)

n nz e ...... z e

1+ z +.....+ zn1 =

n

2i

ezi2 (n 1)

n...... z e

......(1)

Put z = 1 and take modulus n = 2n1 sinn

sin

2n

......sin (n1)

n

(P) If n = 21

21 = 220

220

10

2 9 10 20sin sin .......sin .sin ....sin

21 21 21 21 211021 sin ........sin .2

21 2110 21

sin .......sin21 21 2

(Q) If n = 222

21

2

21 20

2 1022 2 sin sin .........sin22 22 22

10 22 11sin .......sin

22 22 2 2

10211

2210

sin.....222

sin22

sin

(R) Again put z = 1 is...............(1) and take modulus

n 1

20

10

21 cos cos .......cos(n 1) 2n n n

n 212 201 cos cos .......cos 2

21 21 2110 1

cos .......cos21 21 2

RESONANCE Page - 37

MATHEMATICS

(S) 10

10

2 10 11sin .sin .....sin

22 22 22 210 9 11

cos cos ....cos22 22 22 2

sin cos(90 )

119. Let f(x) = )x(h)x(g.)x(p

where p(x) = x odd functiong(x) = log (x + 1x2 )

g(x) + g(x) = log (x + 1x2 ) + log (x + 1x2 ) = log (x2 + x2 + 1) = log (1) = 0 g(x) is an odd function.

h(x) = 21

e

ex

=

e

x + 2

1

h(x) + h(x) =

e

x + 2

1 +

e

x + 2

1

=

e

x +

e

x + 1

= 1 + 1 [x] + [x] = 1 if x R I = 0

h(x) is an odd function f(x) is an odd function f(1947) + f( 1947) + f(0) = 0.

120. Any point on the curve xy = 4 is

t2

,t2

slope of the tangent at

t2

,t2 is 2t

1

equation of tangent is x + t2y 4t = 0since it is tangent to x2 + y2 = 8 also

4t1

t4

= 22

i.e. t2 = 1 equation of the tangent is x + y = 4 Since the intercepts are positive the tangent is x + y = 4

121. y = f(|x|)

y =

0x,)x(f0x,)x(f

=

0t,)t(f0x,)x(f

where t = x and y = f(x) has no point

of discontinuity when x 0y = f(|x|) has no discontinuous points

122. |a + b| = |a| + |b| ab 0 (x2 5x + 4) sinx 0 x [0, 1] [, 4] {2}sum of all integral values is 0 + 1 + 4 = 5

123. Putting x = 2, 12 and 1 successively

1f ( 2 ) f 3 . . . . . (1 )2

1 3f f ( 1 ) . . . . . . . ( 2 )2 2

a n d f ( 1 ) f ( 2 ) 0 . . . . . . . ( 3 )3S o l v i n g , w e g e t f ( 2 )4

4f(2) = 3

124. |z1| = 2 and (1 i) z2 + (1 + i) 28z2 C : x2 + y2 = 4 L : x + y = 24

AB = OB rAB = 2.

125. Let f = 9x5xx

2 f

1,11

1

f2 + f + k < 0 f

1,11

1 f

111

< 0 & f(1) < 0

126. Let |3sinx 4cosx| = t t [0, 5] x RGiven equation can be written ast2 (a2 + a + 5) t + a2(a + 3) + 2(a + 3) = 0t2 ((a2 + 2) + (a + 3)) t + (a + 3)(a2 + 2) > 0 t = a + 3 or t = a2 + 2 0 a + 3 5 or 0 a2 + 2 5 3 a 2 or 2 a2 3

3 a 2 or 3 a 3 acceptable integer values of a are 3, 2, 1, 0, 1, 2 6 is the answer

127. Centre of circles c1, c2, c3 are in A.P.General term for abscissa of centres = 1 + (n 1).3 = 3n 2 centre of c5 is (13, 0)Radius of circles are in G.P. R

n = 1.2n 1 = 2n 1

R3 = 4 and centre of c3 is (7, 0)tangents of circle c3 intersect each other at (13, 0)equation of any line passing through (13, 0) is y 0 = m(x 13) mx y 13m = 0 now it will be required tangents if

2

7m 0 13m

m 1= 4

36m2 = 16m2 + 16 20m2 = 16 m = 25

RESONANCE Page - 38

MATHEMATICS

Let m1 = 25 , m2 =

25

10|m1m2| = 8

128.

4

3

2

1

654321

LLLL

MMMMMM

Required probability = 2

62

31

4

CCC

= 54

Alter : Required probability = 1 (no languages is common)

= 1 1 51

= 54

129. Q1(x) 0 a > 0 and b2 4ac 0Q2(x) 0 b > 0 and c2 4ab 0Q3(x) 0 c > 0 and a2 4bc 0 2a 4( ab) 0

2a 4( ab) a > 0, b > 0 and c > 0 ab 0

2a

ab 4 ........(1)

we know that (a b)2 + (b c)2 + (c a)2 > 0 2a ab 0

2a

ab

> 1 .........(2)

from (1) and (2), we can say that

1 < 2a

ab 4

possible of integers in the range are 2,3,4 Numb

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