MATHEMATICS
1. Let be a positive angle. If the number of degrees in is divided by the number of radians in , then an irrational
number 180results. If the number of degrees in is multiplied by the number of radians in , then an irrational
number 125
9 results. The angle must be equal to
(a) 30° (b) 45°
(c) 50° (d) 60°
2. In a triangle ABC, a (1 3) cm, b = 2 cm and angle C = 60°. Then the other two angles are
(a) 45° and 75° (b) 30° and 90° (c) 105° and 15° (d) 100° and 20°
DIRECTIONS: For the next two (2) items that follow.
Let be the root of the equation 25cos2 + 5cos – 12 = 0,
where .
2
3. What is tan equal to?
(a) 3 4 (b) 3 4 (c) 4 3 (d) 4 5
4. What is sin 2 equal to? (a) 24 25 (b) 24 25 (c) 5 1 2 (d) 21 25
5. The angle of elevation of the top of a tower from a point 20 m away from its base is 45°. What is the height of the tower?
(a) 10 m (b) 20 m
(c) 30 m (d) 40 m
6. The equation tan–1(1 + x) + tan–1 (1– x)
2 is satisfied by
(a) x = 1 (b) x = –1
(c) x = 0 (d) 1
2 x
7. The angles of elevation of the top of a tower standing on a horizontal plane from two points on a line passing through the foot of the tower at distances 49 m and 36 m are 43° and 47° respectively. What is the height of the tower?
(a) 40 m (b) 42 m
(c) 45 m (d) 47 m
8. (1 – sinA + cos A)2 is equal to (a) 2 (1 – cosA) (1 + sin A) (b) 2 (1 – sin A) (1 + cos A) (c) 2 (1 – cos A) (1 – sin A) (d) None of the above
9. What is cos sin
1 tan 1 cot equal to?
(a) sin – cos (b) sin + cos
(c) 2 sin (d) 2 cos
DIRECTIONS : For the next three (3) items that follow. Consider x = 4 tan 1 1 5 , 1 1 tan 70 y and tan 1 1 . 99 z
10. What is x equal to? (a) tan 1 60 119 (b) 1 120 tan 119 (c) tan 1 90 169 (d) 1 170 tan 169 11. What is x – y equal to?
(a) tan 1 828 845 (b) 1 8287 tan 8450 (c) tan 1 8281 8450 (d) 1 8287 tan 8471 12. What is x – y + z equal to?
(a)
2 (b) 3
(c)
6 (d) 4
DIRECTIONS: For the next three (3) items that follow. Consider the triangle ABC with vertices A (–2, 3), B (2, 1) and C (1, 2).
13. What is the circumcentre of the triangle ABC ?
(a) (–2, –2) (b) (2, 2) (c) (–2, 2) (d) (2, –2)
7
5
CHA
PT
ER
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14. What is the centroid of the tirnalge ABC ? (a) 1,1 3 (b) 1 , 2 3 (c) 1,2 3 (d) 1 ,3 2
15. What is the foot of the altitude from the vertex A of the triangle ABC?
(a) (1, 4) (b) (–1, 3)
(c) (–2, 4) (d) (–1, 4)
16. Let X be the set of all persons living in a city. Persons x, y in X are said to be related as x < y if y is at least 5 years older than x. Which one of the following is correct?
(a) The relation is an equivalence relation on X
(b) The relation is transitive but neither reflexive nor symmetric
(c) The relation is reflexive but neither transitive nor symmetric
(d) The relation is symmetric but neither transitive nor reflexive
17. Which one of the following matrices is an elementary matrix?
(a) 1 0 0 0 0 0 0 0 1 (b) 1 5 0 0 1 0 0 0 1 (c) 0 2 0 1 0 0 0 0 1 (d) 1 0 0 0 1 0 0 5 2
18. Consider the following statements in respect of the given equation :
(x2 + 2)2 + 8x2 = 6x (x2 + 2)
1. All the roots of the equation are complex. 2. The sum of all the roots of the equation is 6. Which of the above statements is/are correct?
(a) 1 only (b) 2 only
(c) Both 1 and 2 (d) Neither 1 nor 2
19. In solving a problem that reduces to a quadratic equation, one student makes a mistake in the constant term and obtains 8 and 2 for roots. Another student makes a mistake only in the coefficient of first-degree term and finds –9 and –1 for roots. The correct equation is
(a) x2 – 10x + 9 = 0 (b) x2 – 10x + 9 = 0 (c) x2 – 10x + 16 = 0 (d) x2 – 8x – 9 = 0 20. If
2 7 1 5
A then that is A + 3A–1 equal to?
(a) 3I (b) 5I
(c) 7I (d) None of these
where I is the identity matrix order 2.
21. In a class of 60 students, 45 students like music, 50 students like dancing, 5 students like neither. Then the number of students in the class who like both music and dancing is
(a) 35 (b) 40
(c) 50 (d) 55
22. If log10 2, log10 (2x – 1) and log10(2x + 3) are three consecutive terms of an A.P, then the value of x is
(a) 1 (b) log5 2 (c) log2 5 (d) log10 5 23. The matrix 0 4 4 0 i i is
(a) symmetric (b) skew-symmetric
(c) Hermitian (d) skew-Hermitian 24. Let Z be the set of integers and aRb, where a, b Z if and
only if (a – b) is divisible by 5. Consider the following statements:
1. The relation R partitions Z into five equivalent classes. 2. Any two equivalent classes are either equal or disjoint. Which of the above statements is/are correct?
(a) 1 only (b) 2 only
(c) Both 1 and 2 (d) Neither 1 nor 2
25. If 2 1 2 3 i z i wh er e i 1,th en ar gumen t of z is (a) 3 4 (b) 4 (c) 5 6 (d) 3 4
26. If m and n are the roots of the equation (x + p) (x + q) – k = 0, then the roots of the equation (x – m) (x – n) + k = 0 are
(a) P and q (b) 1 pand 1 q (c) –p and –q (d) p + q and p – q 27. What is the sum of the series 0.5 + 0.55 + 0.555 + ... to n
terms? (a) 59 92 1 1 10n n (b) 1 5 2 1 1 9 9 10n (c) 1 5 1 1 9 n 9 10n (d) 5 1 1 1 9 n 9 10n 28. If 1, , 2are the cube roots of unity, then the value
of(1 )(1 2)(1 4)(1 8) is
(a) –1 (b) 0
(c) 1 (d) 2
29. Let A = {1, 2 , 3, 4, 5, 6, 7, 8, 9, 10}. Then the number of subsets of A containing exactly two elements is
(a) 20 (b) 40
(c) 45 (d) 90
30. What is the square root of i, where i 1?
(a) 1 2 i (b) 1 2 i (c) 1 2 i (d) None of these
31. The point on the parabola y2 = 4ax nearest to the focus has its abscissa (a) x = 0 (b) x = a (c) 2 a x (d) x = 2a
32. A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is (a) 3 4 (b) 4 3 (c) 1 3 (d) 3 33. The hyperbola 2 2 2 2 1 x y
a b passes through the point
(3 5,1) and the length of its latus rectum is 4
3 units. The
length of the conjugate axis is
(a) 2 units (b) 3 units
(c) 4 units (d) 5 units
34. The Perpendicular distance between the straight lines 6x + 8y + 15 = 0 and 3x + 4y + 9 = 0 is (a) 3units 2 (b) 3 unit 10 (c) 3unit 4 (d) 2 unit 7
35. The area of a triangle, whose vertices are (3, 4), (5, 2) and the point of intersection of the lines x = a and y = 5, is 3 square units. What is the value of a?
(a) 2 (b) 3
(c) 4 (d) 5
36. The length of perpendicular from the origin to a line is 5 units and the line makes an angle 120° with the positive direction of x-axis.The equation of the line is
(a) x 3y 5 (b) 3x y 10
(c) 3x y 10 (d) None of these
37. The equation of the line joining the origin to the point of interesection of the lines x y 1
a b and 1 x y b a is
(a) x – y = 0 (b) x + y = 0
(c) x = 0 (d) y = 0
DIRECTIONS: For the next two (2) items that follow.
The projections of a directed line segment on the coordinate axes are 12, 4, 3 respectively.
38. What is the length of the line segment?
(a) 19 units (b) 17 units
(c) 15 units (d) 13 units
39. What are the direction cosines of the line segment? (a) 12 4, ,3 13 13 13 (b) 12 4 3 , , 13 13 13 (c) 12, 4, 3 13 13 13 (d) 12 4 3 , , 13 13 13 DIRECTIONS: For the next two (2) items that follow.
From the point P(3, – 1, 11), a perpendicular is drawn on the line
L given by the equation 2 3.
2 3 4
x y z
Let Q be the foot of the perpendicular.
40. What are the direction ratios of the line segment PQ?
(a) 1, 6, 4 (b) 1, 6, 4
(c) 1, 6, 4 (d) 2, 6, 4
41. What is the length of the line segment PQ?
(a) 47 units (b) 7 units
(c) 53 units (d) 8 units
DIRECTIONS: For the next two (2) items that follow.
A triangular plane ABC with centroid (1, 2, 3) cuts the coordinate axes at A, B, C respectively.
42. What are the intercepts made by the plane ABC on the axes?
(a) 3, 6, 9 (b) 1, 2, 3
(c) 1, 4, 9 (d) 2, 4, 6
43. What is the equation of plane ABC?
(a) x + 2y + 3z = 1 (b) 3x + 2y + z = 3 (c) 2x + 3y + 6z = 18 (d) 6x + 3y + 2z = 18 DIRECTIONS: For the next two (2) items that follow.
A point P (1, 2, 3) is one vertex of a cuboid formed by the coordinate planes and the planes passing through P and parallel to the coordinate planes.
44. What is the length of one of the diagonals of the cuboid?
(a) 10 units (b) 14 units
(c) 4 units (d) 5 units
45. What is the equation of the plane passing through P(1, 2, 3) and parallel to xy–plane?
(a) x + y = 3 (b) x – y = – 1 (c) z = 3 (d) x + 2y + 3z = 14 46. The decimal number (127. 25)10, when converted to binary
nunber, takes the form
(a) (1111111.11)2 (b) (1111110.01)2 (c) (1110111.11)2 (d) (1111111.01)2
47. Consider the following in respect of two non-singular matrices A and B of same order:
1. det (A + B) = det A + det B 2. (A + B)–1 = A–1 + B –1
Which of the above is/are correct?
(a) 1 only (b) 2 only
48. If 3 4 5 2 , 1 1 2 1 X B and A p q r s satisfy the equation AX = B, then the matrix A is equal to
(a) 7 26 1 5 (b) 7 26 4 17 (c) 7 4 26 13 (d) 7 26 6 23 49. What is 1 0 n r n r C equal to? (a) n + 2C 1 (b) n + 2Cn (c) n + 3C n (d) n + 2Cn + 1
50. How many words can be formed using all the letters of the word ‘NATION’ so that all the three vowels should never come together? (a) 354 (b) 348 (c) 288 (d) None of these 51. (x3 – 1) can be factorised as (a) (x – 1) (x – ) (x + 2) (b) (x – 1) (x – ) (x – 2) (c) (x – 1) (x + ) (x + 2) (d) (x – 1) (x + ) (x – 2)
where is one of the cube roots of unity.. 52. What is 3 sin 1 cos 6 6 sin 1 cos 6 6 i i
where i 1,equal to?
(a) 1 (b) –1 (c) i (d) –i 53. Let 2 , 2 1 x y y A B x x y and 3 2 C If AB = C, then what is A2 equal to? (a) 6 10 4 26 (b) 10 5 4 24 (c) 5 6 4 20 (d) 5 7 5 20 54. The value of 1 1 1 1 1 1 1 1 1 x y is (a) x + y (b) x – y (c) xy (d) 1 + x + y 55. If A = {x : x is a multiple of 3} and B = {x : x is a multiple of 4} and
C = {x : x is a multiple of 12}, then which one of the following is a null set?
(a) ( / )A B C (b) ( / ) /A B C
(c) (A B) C (d) (A B) /C
56. If (11101011)2 is converted decimal system, then the resulting number is
(a) 235 (b) 175
(c) 160 (d) 126
57. What is the real part of (sin x + i cos x)3 wherei 1 ?
(a) –cos 3x (b) –sin 3x
(c) sin 3x (d) cos 3 x 58. If cos sin sin cos E , then ( ) ( )E E is equal to (a) E( ) (b) E( ) (c) E( ) (d) E( )
59. Let A = {x, y, z}and B = {p, q, r, s}. What is the number of distinct relations from B to A?
(a) 4096 (b) 4094
(c) 128 (d) 126
60. If 2p + 3q = 18 and 4p2 + 4pq – 3q2 – 36 = 0, then what is (2p + q) equal to?
(a) 6 (b) 7
(c) 10 (d) 20
DIRECTIONS: For the next two (2) items that follow. Given that 2 4 2 1 . 1 d x x Ax B dx x x
61. What is the value of A?
(a) –1 (b) 1
(c) 2 (d) 4
62. What is the value of B?
(a) –1 (b) 1
(c) 2 (d) 4
DIRECTIONS: For the next two (2) items that follow. Given that 2 2 lim 3. 1 x x Ax B x
63. What is the value of A?
(a) –1 (b) 1
(c) 2 (d) 3
64. What is the value of B?
(a) –1 (b) 3
(c) –4 (d) –3
65. What is the solution of the differential equation
2 0 ?
ydx xdy y
(a) xy = c (b) y = cx
(c) x + y = c (d) x – y = c
66. What is the solution of the differential equation
sin dy a 0?
dx
(a) y = x sin–1 a + c (b) x = y sin–1 a + c (c) y = x + x sin–1 a + c (d) y = sin–1 a + c where c is an arbitrary constant.
67. What is the solution of the differential equation 2 0? dx x y dy y (a) xy = x4 + c (b) xy = y4 + c (c) 4xy = y4 + c (d) 3xy = y3 + c where c is an arbitrary constant.
68. What is 2 1 x xe dx x equal to? (a) (x + 1)2 ex + c (b) (x + 1)ex + c (c) 1 x e c x (d) 12 x e c x
where c is the constant integration.
69. The adjacent sides AB and AC of a triangle ABC are represented by the vectors 2i 3j 2kand 4i 5j 2k respectively. The area of the triangle ABC is
(a) 6 square units (b) 5 square units (c) 4 square units (d) 3 square units
70. A force F 3i 4j 3kis applied at the point P, whose position vector is r 2i 2j 3k. What is the magnitude of the moment of the force about the origin?
(a) 23 units (b) 19 units
(c) 18 units (d) 21 units
71. Given that the vectors and are non-collinear. The values of x and y for wh ich u v w h olds t r ue if
2 , 2 3
u x y v y x and w 2 5 ,are
(a) x = 2, y = 1 (b) x = 1, y = 2 (c) x = – 2, y = 1 (d) x = – 2, y = – 1
72. If | |a = 7, | |b = 11 and |a b| 10 3,then |a b is equal| to
(a) 40 (b) 10
(c) 4 10 (d) 2 10
73. Let , , be distinct real numbers. The points with position vectors i j k, i j kand i j k (a) are collinear
(b) form an equilateral triangle (c) form a scalene triangle (d) form a right-angled triangle
74. If a b c 0,then which of the following is/are correct? 1. a b c, , are coplanar..
2. a b b c c a
Select the correct answer using the code given below.
(a) 1 only (b) 2 only
(c) Both 1 and 2 (d) Neither 1 nor 2
75. If |a b| |a b then which one of the following is correct?|, (a) a b for some scalar
(b) a is parallel to b (c) a is perpendicular to b (d) a b 0 76. If G x 25 x2,then what is 1 1 lim 1 x G x G x equal to? (a) 1 2 6 (b) 1 5 (c) 1 6 (d) 1 6 77. Consider the following statements:
1. 2 x x e e y is an increasing function on 0, . 2. 2 x x e e y is an increasing function on , .
Which of the above statements is/are correct?
(a) 1 only (b) 2 only
(c) Both 1 and 2 (d) Neither 1 nor 2
78. For each non-zero real number x, let . | |
x f x
x The range of f is
(a) a null set
(b) a set consisting of only one element (c) a set consisting of two elements
(d) a set consisting of infinitely many elements 79. Consider the following statements:
1. f (x) = [x], where [.] is the greatest integer function, is discontinuous at x = n, where n Z.
2. f (x) = cot x is discontinuous at x = n , where n Z.
which of the above statements is/are correct?
(a) 1 only (b) 2 only
(c) Both 1 and 2 (d) Neither 1 or 2
80. What is the derivative of
2 1 1 1 tan x x with respect to tan–1 x? (a) 0 (b) 1 2 (c) 1 (d) x
81. If 3 2 1 3 log , 1 1 3 e x x x f x g x x x and g o f (t) =g(f(t)), then what is 1 1 e g f e equal to? (a) 2 (b) 1 (c) 0 (d) 1 2
DIRECTIONS: For the next two (2) items that follow. Given a function 1 If 0 If 0 1 1 If 1 x f x ax b x x
where a, b are constants. The function is continuous everywhere. 82. What is the value of a?
(a) –1 (b) 0
(c) 1 (d) 2
83. What is the value of b?
(a) –1 (b) 1
(c) 0 (d) 2
84. Consider the following functions: 1. f(x) = x3, x
2. f(x) = sin x, 0 < x < 2
3. f(x) = ex, x
Which of the above functions have inverse defined on their ranges?
(a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3 DIRECTIONS: For the next two (2) items that follow. The integral cos sin dx a x b x is of the form 1 In tan 2 x r .
85. What is r equal to?
(a) a2 + b2 (b) 2 2
a b
(c) a + b (d) a2 b2
86. What is equal to? (a) tan 1 a b (b) 1 tan b a (c) tan 1 a b a b (d) 1 tan a b a b DIRECTIONS: For the next two (2) items that follow. Consider the function
2 2 1 ( ) , 1 x f x x where x
87. At what value of x does f(x) attain minimum value?
(a) –1 (b) 0
(c) 1 (d) 2
88. What is the minimum value of f(x)?
(a) 0 (b) 1
2
(c) –1 (d) 2
DIRECTIONS: For the next two (2) items that follow. Consider the function
cos If 2 2 ( ) 3 If 2 x x x f x x Which is continuous at , 2 x where is a constant. 89. What is the value of ?
(a) 6 (b) 3
(c) 2 (d) 1
90. What is xlim0f x( )equal to?
(a) 0 (b) 3
(c) 3 (d) 6
91. The mean and the variance 10 observations are given to be 4 and 2 respectively. If every observation is multiplied by 2, the mean and the variance of the new series will be respectively
(a) 8 and 20 (b) 8 and 4
(c) 8 and 8 (d) 80 and 40
92. Which one of the following measures of central tendency is used in construction of index numbers?
(a) Harmonic mean (b) Geometric mean
(c) Median (d) Mode
93. The correlation coefficient between two variables X and Y is found to be 0 6.All the observations on X and Y are transformed using the transformations U = 2 – 3X and V = 4 Y + 1. The correlation coefficient between the transformed variables U and V will be
(a) 0 5 (b) 0 5
(c) 0 6 (d) 0 6
94. Which of the following statements is/are correct in respect of regression coefficients?
1. It measures the degree of linear relationship between two variables.
2. It gives the value by which one variable changes for a unit change in the other variable.
Select the correct answer using the code given below.
(a) 1 only (b) 2 only
(c) Both 1 and 2 (d) Neither 1 nor 2
95. A set of annual numerical data, comparable over the years, is given for the last 12 years.
1. The data is best represented by a broken line graph, each corner (turning point) representing the data of one year.
2. Such a graph depicts the chronological change and also enables one to make a short-term forecast. Which of the above statements is/are correct?
(a) 1 only (b) 2 only
96. Two men hit at a target with probabilities 1 2and
1 3
respectively. What is the probability that exactly one of them hits the target?
(a) 1 2 (b) 1 3 (c) 1 6 (d) 2 3
97. Two similar boxes Bi (i = 1, 2) contain (i + 1) red and (5 – i – 1) black balls. One box is chosen at random and two balls are drawn randomly. What is the probability that both the balls are of different colours?
(a) 1 2 (b) 3 10 (c) 2 5 (d) 3 5
98. In an examination, the probability of a candidate solving a question is 1
2. Out of given 5 questions in the examination,
what is the probability that the candidate was able to solve at least 2 questions? (a) 1 64 (b) 3 16 (c) 1 2 (d) 13 16
99. If A B,then which one of the following is not correct? (a) P A( B) 0 (b) ( | ) P A P A B P B (c) ( | ) P B P B A P A (d) ( | ( )) P A P A A B P B
100. The mean and the variance in a binomial distribution are found to be 2 and 1 respectively. The probability P(X = 0) is
(a) 1 2 (b) 1 4 (c) 1 8 (d) 1 16
101. The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is
(a) 28 (b) 30
(c) 35 (d) 38
102. If A and B are two events such that ( ) 3, 4 P A B 1 ( ) 4 P A B and ( ) 2, 3
P A then what is P(B) equal to?
(a) 1 3 (b) 2 3 (c) 1 8 (d) 2 9
103. The ‘less than’ ogive curve and the ‘more than’ ogive curve intersect at
(a) median (b) mode
(c) arithmetic mean (d) None of these
104. In throwing of two dice, the number of exhaustive events that ‘5’ will never appear on any one of the dice is
(a) 5 (b) 18
(c) 25 (d) 36
105. Two cards are drawn successively without replacement from a wellshuffled pack of 52 cards. The probability of drawing two aces is (a) 1 26 (b) 1 221 (c) 4 223 (d) 1 13
DIRECTIONS: For the next two (2) items that follows. Consider the line x 3y and the circle x2 + y2 = 4.
106. What is the area of the region in the first quadrant enclosed by the x-axis, the line x 3and the circle?
(a) 3 3 2 (b) 3 2 2 (c) 1 3 2 (d) None of these
107. What is the area of the region in the first quadrant enclosed by the x-axis, the line x 3y and the circle?
(a)
3 (b) 6
(c) 3
3 2 (d) None of these
DIRECTIONS: For the next two (2) items that follow. Consider the curves y = sin x and y = cos x.
108. What is the area of the region bounded by the above two curves and the lines x = 0 and ?
4
x
(a) 2 1` (b) 2 1
109. What is the area of the region bounded by the above two curves and the lines
x =
4 and x = 2?
(a) 2 – 1 (b) 2 1
(c) 2 2 (d) 2
DIRECTIONS: For the next two (2) items that follow. Consider the function
4 3 2
0 75 9 7
f x x x x
110. What is the maximum value of the function?
(a) 1 (b) 3
(c) 7 (d) 9
111. Consider the following statements:
1. The function attains local minima at x = – 2 and x = 3. 2. The function increases in the interval (–2, 0). Which of the above statements is/are correct?
(a) 1 only (b) 2 only
(c) Both 1 and 2 (d) Neither 1 nor 2 DIRECTIONS: For the next three (3) items that follow. Consider the parametric equation
2 2 2 1 2 , 1 1 a t at x y t t
112. What does the equation represent? (a) It represents a circle of diameter a (b) It represents a circle of radius a (c) It represents a parabola (d) None of the above
113. What is dy dx equal to? (a) y x (b) y x (c) x y (d) x y 114. What is 2 2 d y dx equal to? (a) 2 2 a y (b) 2 2 a x (c) 2 2 a x (d) 2 3 a y
115. Consider the following statements: 1. The general solution of dy f x( ) x
dx is of the form y = g (x) + c, where c is an arbitrary constant.
2. The degree of 2 ( ) dy f x dx is 2.
Which of the above statements is/are correct?
(a) 1 only (b) 2 only
(c) Both 1 and 2 (d) Neither 1 nor 2
116. What is 2 dx x a equal to? (a) 2 2 x+ x ln a c a (b) 2 2 ln x x a c a (c) 2 2 2 ln x x a c a (d) None of these
where c is the constant of integration.
DIRECTIONS: For the next four (4) items that follow. Consider the integral
0 sin 2 , sin m mx I dx x where m is a positive integer.
117. What is I1 equal to?
(a) 0 (b) 1
2
(c) 1 (d) 2
118. What is I2 + I3 equal to?
(a) 4 (b) 2
(c) 1 (d) 0
119. What is Im equal to?
(a) 0 (b) 1
(c) m (d) 2m
120. Consider the following: 1. Im – Im–1 is equal to 0. 2. I2m > Im
Which of the above is/are correct?
(a) 1 only (b) 2 only
Hints & Solutions
MATHEMATICS
1. (c) From going by the options, option (a), = 30°, as we know that 180° = radian
30° = 30
180 radian
Now according to question,
30 180 180
30
Now number of degree in is multiplied by number of radians in . 30° × 30 900 10 180 180 2 = 5 125 9 From option (b), = 45° 45° = 45 180 radian
Now according to question,
45 180 180
45
Now number of degree in is multiplied by number of radian in .
45° × 45 45 125
180 4 9
From option (c), = 50°
As we know that 180° = radian
50° = 50
180 radian
Now according to question
50 180 180
50
Now number of degree in ‘ ’ is multiplied by number of radian in . 50° × 50 2500 125 180 180 9 Option (c) is correct. 2. (a) 60° A B C b = 2 cm a (1 3 cm Now as a > b A > B
Now from Sine Rule, sin A sin B
a b
sin A sin B 2
1 3
From option (a),
sin 75 sin 45 2 1 3 6 2 1 2 2 4 3 2 12 + 4 = 4 + 4 3 4 + 4 3 = 4 + 4 3
Option (a) is correct. For (3-4):
3. (a) Here is the root of equation 25 cos2 + 5 cos – 12 = 0 25 cos2 + 5 cos – 12 = 0
25 cos2 + 20 cos – 15 cos – 12 = 0 5 cos (5 cos + 4) – 3(5 cos + 4) = 0 (5 cos – 3) (5 cos + 4) = 0 cos = 3 5 or cos = – 4 5 Here, 2 < < cos = – 4 5
( In 2nd quadrant, cos value is negative) Now, sin = 1 – cos2 1 –16
25
sin = 3 5
tan = sin 3 –5 –3
cos 5 4 4
Option (a) is correct. 4. (b) sin 2 = 2 sin . cos
= 2 3 – 4
5 5
= 6 – 4 –24
5 5 25
5. (b) In ABC, AB = h BC = 20 m C = 45° tan 45° = AB BC A B C h 45° 20 m 1 = h 20 [ tan 45° = 1] h = 20 m
Height of the tower = 20 m Option (b) is correct. 6. (c) tan–1(1 + x) + tan–1(1 – x) = 2 –1 1 x 1 – x tan 1 – 1 x 1 – x 2 1 x 1 – x tan 1 – 1 x 1 – x 2 2 1 1 – 1 x 1 – x 0 1 – (1 + x) (1 – x) = 0 (1 + x) (1 – x) = 1 1 – x2 = 1 x2 = 0 x = 0 Option (c) is correct. 7. (b) AB = h (height of the tower)
BD = 36 m BC = 49 m D = 47° C = 43° A B D 47° 36 m 43° h 49 m C Now, in ABD, tan 47° = h 36 m ... (i) and in ABC, tan 43° = h 49 m tan(90° – 47°) = h 49 cot 47° = h 49 ... (ii)
Multiplying equations (i) and (ii)
tan 47º. cot 47º h h 36 49 2 h 1 36 49 h = 6 × 7 = 42 m Option (b) is correct 8. (b) (1 – sin A + cos A)2
= 1 + sin2A + cos2A – 2 sin A
– 2 sin A . cos A + 2 cos A = 2 – 2 sin A – 2 sin A cos A + 2 cos A
= 2(1 – sin A) + 2 cos A(1 – sin A) = 2(1 + cos A) (1 – sin A) Option (b) is correct. 9. (b) cos sin 1 – tan 1 – cot = cos sin sin cos 1 – 1 – cos sin = 2 2 cos sin
cos – sin sin – cos
=
2 2
cos sin
–
cos – sin cos – sin
=
2 2 cos – sin cos sin
cos – sin
cos – sin cos – sin
= cos + sin Option (b) is correct. 10. (b) x = 4 tan–1 1 5 = –1 –1 2 2 1 5 2 2 tan 2 tan 5 1 1 – 5 = 2 tan–1 2 25 5 24 = 2 tan –1 10 24 = 2 tan –1 5 12 = tan–1 5 2 12 25 1 – 144 = tan–1 120. 119
11. (c) x – y = tan–1 120 – tan–1 1 119 70 = –1 120 1 – 119 70 tan 120 1 1 119 70 = –1 8400 – 119 8330 tan 120 1 8330 = –1 8281 8330 tan 8450 8330 –1 281 tan 8450 option (c) is correct. 12. (d) x – y = tan–18281 8450 –1 8281 –1 1 tan tan 8450 99 = –1 8281 1 8450 99 tan 8281 1 1 – 8450 99 = –1 828269 836550 tan 828269 836550 = 1 tan–1 = (1) = 4 Option (d) is correct.
13. (a) A circumcentre is a point at which perpendicular bisectors meet each other.
Here, ‘E’ represents circumcentre
A B C F D 3 3 , 2 2 1 5 – , 2 2 (1, 2) (2, 1) E (x, y) (–2, 3) Mid-point of BC = 2 1 1, 2 3, 3 2 2 2 2 Slope of BC = 2 – 1 1 – 2 = –1 Slope of DE = 1 Now, equation of ED is y – 3 2 = 3 1 x – 2 2y – 3 = 2x – 3 x = y ... (i) Now, mid-point of AC = –2 1 3, 2 –1 5, 2 2 2 2 Slope of AC = 3 – 2 –1 –2 – 1 3 Slope of EF = 3 Now, equation of EF is y –5 2 = 1 3 x 2 2y – 5 = 6x + 3 ... (ii)
From equations (i) and (ii), x = –2 and y = –2
Hence, circumcentre of ABC is (x, y) = (–2, –2) Option (a) is correct.
14. (b) Centroid of the triangle
= 1 2 3 1 2 3 x x x y y y , 3 3 A B C (–2, 3) (1, 2) (2, 1) (x y )2 1 (x y )2 2 (x y )3 3 = –2 2 1 3, 1 2 3 3 = 1 , 2 3 Option (b) is correct. 15. (d) Slope of BC = 2 – 1 1 – 2 = –1 A B C (–2, 3) (1, 2) (2, 1) (x, y) D Slope of AD = 1 Now, equation of BC is y – 2 = –1(x – 1)
y – 2 = –x + 1
x + y – 3 = 0 ... (i)
and equation of AD is (y – 3) = 1(x + 2)
x – y + 5 = 0 ... (ii)
From equations (i) and (ii), x = –1 and y = 4
Foot of altitude from the vertex A of the triangle ABC is (–1, 4)
Option (d) is correct. 16. (b) Given that x < y if y x + 5
For Reflexive: x / x
Hence, relation is not reflexive. For Symmetry:
if x < y, then y / x
Hence, relation is not symmetry. For Transitive:
if x < y and y < z, then x < z
Hence, relation is transitive. 17. (b)
1 5 0
0 1 0 0 0 1
An elementary matrix has each diagonal element 1. So, option (b) is correct answer.
18. (b) (x2 + 2)2 + 8x2 = 6x(x2 + 2) Let x2 + 2 = y y2 + 8x2 = 6xy y2 – 6xy + 8x2 = 0 y = 2 2 6x 36x – 32x 2 y = 6x 2x 2 = 3x ± x y = 4x, 2x At y = 4x, x2 + 2 = 4x x2 – 4x + 2 = 0 Discriminant, D = 16 – 8 = 8 > 0 Roots are real.
Sum of roots = –(–4) = 4 At x = 2x,
x2 + 2 = 2x x2 – 2x + 2 = 0 D = 4 – 8 = – 4 < 0 Roots are complex. Sum of roots = 2
Sum of all roots = 4 + 2 = 6 only statement 2 is correct. Correct option is (b)
19. (a) Let correct equation is ax2 + bx + c = 0 According to first student, equation is: ax2 + bx + c1 = 0 and roots are 8 and 2 8 + 2 = –b
a b a = –10
Quadratic equation according to second student ax2 + b1x + c = 0 and roots are –9 and –1 (–9) × (– 1) = c
a c a = 9
Putting value in original equation
x2 + bx a + c a = 0 x2 – 10x + 9 = 0. 20. (c) A = 2 7 1 5 Now, A–1 = 1 A adj (A) = 1 5 –7 –1 2 10 – 7 = 1 5 –7 –1 2 3 3A–1 = 5 –7 –1 2 Now, A + 3A–1 = 2 7 5 –7 1 5 –1 2 = 7 0 7 1 0
0 7 0 1 = 7I where I is Identity Matrix.
21. (b)
45 – x x 50 – x
Music Dancing
Let ‘x’ be the number of students who likes both music and dance.
5 students likes neither music nor dancing. Hence, total number of remaining students = 60 – 5 = 55
Now from Venn diagram, 45 – x + x + 50 – x = 55 95 – x = 55
x = 95 – 55 = 40.
22. (c) log102, log10(2x – 1) and log10(2x + 3) are in A.P. Hence, common difference will be same.
log10 x 2 – 1 2 = log10 x x 2 3 2 – 1 x x x 2 – 1 2 3 2 2 – 1 (2x – 1)2 = 2(2x + 3) 22x – 2x + 1 + 1= 2x + 1 + 6 22x – 2x + 2 = 5 Let 2x = y, then y2 – 4y – 5 = 0 y2 – 5y + y – 5 = 0 y(y – 5) + 1(y – 5) = 0 y = –1, y = 5 Therefore, 2x = 5 x = log25.
23. (d) The given matrix A = 0 – 4 i
4 i 0
Now, from options:
From option (a): For Symmetric matrix AT = A
Now, AT =
0 4 i
– 4 i 0 AA
The given matrix is not symmetric option (a) is wrong.
From option (b): For Skew-symmetric matrix AT = –A
= 0 4 i
–4 i 0 –A
Given matrix is not skew-symmetric option (b) is wrong.
From option (c): For Hermitian matrix AT = A, where A is conjugate of matrix AA
0 – 4 – i A
4 – i 0 AA
T option (c) is wrong.
From option (d): For Skew-Hermitian matrix
The diagonal element of a skew-hermitian matrix are pure imaginary or zero.
A = 0 – 4 i
4 i 0
Here, diagonal element indicates that the given matrix is skew-hermitian matrix. option (d) is correct. 24. (b) 25. (b) z = –2 1 2i 3 i = –2 – 4i –2 – 4i 3 – i 3 i 3 i 3 – i = 2 – 6 2i – 12i 4i 10 = – 6 – 10i – 4 –10 – 10i 10 10 = –1 – i z = – 1– i = r (cos + i sin )
On comparing real and imaginary part on both sides, we get
r cos = – 1 ...(i)
r sin = – 1 ...(ii)
On dividing eq. (ii) by (i), we get
r sin –1 r cos –1 tan = 1 = tan 4 4 = 4 Option (b) is correct.
26. (c) Here m and n are the roots of equation. (x + p) (x + q) – k = 0
x2 + x (p + q) + pq – k = 0 ... (i) If m and n are the roots of equation, then
(x – m) (x – n) = 0
x2 – (m + n) x + mn = 0 ... (ii) Now equation (i) should be equal to equation (ii), (m + n) = – (p + q) and mn = pq – k
Now, we have to find roots of (x – m) (x – n) + k = 0 x2 – (m + n) x + mn + k = 0 x2 + (p + q) x + (pq – k) + k = 0 x2 + (p + q) x + pq = 0 x2 + px + qx + pq = 0 x (x + p) + q (x + p) = 0 x + q = 0 or x + p = 0 x = – q and x = –p Option (c) is correct. 27. (d) Given 0.5 + 0.55 + 0.555 + ... to n = 5 [0.1 + 0.11 + 0.111 + ... to n terms] = 5 9 [0.9 + 0.99 + 0.999 + ... to n terms] = 5 9 99 999 ... to n terms 9 10 100 1000 5 9 1 1 1 1 – 1 – 1 – ... 10 100 1000 to n terms 5 2 3 9 n 1 1 1 1 – 1 – 1 – ... 10 10 10 1 1 – 10
= 2 n 5 1 1 1 n – .... 9 10 10 10 = n 1 1 – 10 5 1 n – 1 9 10 1 – 10 = 5 n – 1 1 – 1n 9 9 10
28. (c) Here cube root of unity is 1, , 2
Now as we know that 1 + + 2 = 0 and 3 = 1 =( 3)2. 2 = 2 4 = ( 3). = Now, (1 + ) (1 + 2) (1 + 4) (1 + 8) = (– 2) (– ) (1 + ) (1 + 2) = 3(1 + 2 + + 3) = 3( 3) = (1) (1) = 1 Option (c) is correct. 29. (c) A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Number of subsets of A containing two elements = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
= 45 9 9 1 90 45
2 2
Option (c) is correct Alternate Method
The number of subsets of A containing exactly two elements is: 10 C2 = 10 9 2 1 = 45 30. (c) Let i = x + iy i = (x + iy)2 x2 – y2 + 2xyi = 0 + i x2 – y2 = 0; 2xy = 1 Now, (x2 + y2)2 = (x2 – y2)2 + 4x2y2 (x2 + y2)2 = 0 + 1 x2 + y2 = 1 x2 – y2 = 0 ... (i) x2 + y2 = 1 ... (ii) 2x2 = 1 x2 = 1 2 x = ± 1 2 y2 = 1 2 y = ± 1 2 1 1 i i 2 2 = 1 2(1 + i) or –1 2(1 + i).
31. (a) Here, ‘S’ represents focus O(0, 0) is a point which is on parabola y2 = 4ax and nearest to focus (a, 0)
Y X X Y y = 4ax2 S(a, 0) (0, 0) O abscissa of O (0, 0) is x = 0 Option (a) is correct.
32. (b) A line passes through (2, 2) and is perpendicular to the line 3x + y = 3
Slope of line 3x + y = 3 is –3
Slope of line which passes through (2, 2) is 1
3
Equation of line passes through (2, 2) and having slope 1 3 is (y – 2) = 1 3(x – 2) 3y – 6 = x – 2 x – 3y + 4 = 0
In order to find y-intercept of the line Put x = 0 in x – 3y + 4 = 0 –3y = – 4 y = 4 3 Option (b) is correct. 33. (c) 2 2 2 2 x y – a b = 1
Hyperbola passes through 3 5, 1 2 2 2 3 5 1 – a b = 1 2 2 45 1 – a b = 1 ... (i)
Now length of latus rectum = 2 2b a 2 4 2b 3 a
2 2 b 3 a a = 2 3b 2 ... (ii)
Putting the value of ‘a’ from equation (ii) in equation (i), 4 2 45 4 1 – 9b b = 1 4 2 20 1 – b b = 1 20 – b2 = b4 b4 + b2 – 20 = 0 b4 + 5b2 – 4b2 – 20 = 0 b2(b2 + 5) – 4(b2 + 5) = 0 (b2 – 4) (b2 + 5) = 0 b2 = 4, b2 = –5 b2 = 4 b = 2
Now length of conjugate axis = 2b = 2(2) = 4 Option (c) is correct. 34. (b) 6x + 8y + 15 = 0 ... (i) and 3x + 4y + 9 = 0 ... (ii) d 6x + 8y + 15 = 0 6x + 8y + 18 = 0
Multiply equation (ii) by 2, we get 6x + 8y + 18 = 0
Distance between the straight lines
2 1 2 2 2 2 c – c 18 – 15 3 10 a b 6 8 unit Option (b) is correct. 35. (d) Area of ABC = 3 sq. unit
A B C (3, 4) ( , 5)a (5, 2) 3 4 1 1 5 2 1 2 a 5 1 = 3 3 4 1 5 2 1 a 5 1 = 6 3(2 – 5) – 4(5 – a) + 1(25 – 2a) = 6 – 9 – 20 + 4a + 25 – 2a = 6 2a = 10 a = 5 Option (d) is correct. 36. (b) A C 5 O B 60° 120° In OCB, sin 60° = 5 OB OB = 5 sin 60 OB = 5 2 10 3 3 In ACO, OAC = 30° sin 30° = 5 AO AO = 5 2 1 = 10 Normal form of line AB
X Y OB OA = 1 3 X Y 10 10 = 1 3X + Y = 10. 37. (a) x y a b = 1 ... (i) and x y b a = 1 ... (ii)
From solving equations (i) and (ii), we get the intersection point. bx + ay = ax + by (b – a)x = (b – a)y x = y ... (iii) x x a b = 1 x(a + b) = ab x = ab a b and y = ab
a b from equation (iii)
Now, equation of lin e joinin g (0, 0) and
ab ab
,
Here, slope of line = 1 y = x
x – y = 0
Option (a) is correct.
38. (d) The projection of a directed line segment on the co-ordinate axes are 12, 4, 3, respectively.
Length of the line segment = 122 42 32
= 144 16 9 169 = 13 unit Option (d) is correct.
39. (a) Direction cosine of line segment = 12, 4, 3 13 13 13 Option (a) is correct.
40. (b) Equation of line passing through P(3, –1, 11) and perpendicular to x y – 2 z – 3
2 3 4 is:
x – 3 y – 1 z – 1
–1 6 –4
The direction ratio are (–1, 6, – 4) Option (b) is correct. 41. (c) Now x2 – x1 = –1 x2 – 3 = –1 x2 = 2 Similarly, y2 – y1 = 6 y2 + 1 = 6 y2 = 5 and z2 – z1 = –4 z2 – 11 = –4 z2 = – 4 + 11 = 7 Co-ordinate of Q is (2, 5, 7) Length of segment PQ = 2 – 32 5 12 7 – 112 = 1 36 16 = 53 units. Option (c) is correct. 42. (a) Centroid = x31, y31, z31 A B C (x, 0, 0) (0, 0, z) (0, y, 0) C (1, 2, 3) = x31, y31, z31 x1 = 3, y1 = 6 and z1 = 9
Intercept made by plane on the axes are 3, 6 and 9, respectively.
Option (a) is correct.
43. (d) The plane passes through the point A (3, 0, 0), B(0, 6, 0) and C(0, 0, 9). So, it should satisfy the equation given in option for all the three points.
From option (a) For point A (3, 0, 0) x + 2y + 3z = 1 3 + 0 + 0 1 option (a) is wrong. From option (b) For point A(3, 0, 0) 3x + 2y + z = 3 3(3) + 0 + 0 3 option (b) is wrong. From option (c) For point A(3, 0, 0) 2x + 3y + 6z = 18 2(3) + 0 + 0 18 option (c) is wrong. From option (d) For point A (3, 0, 0) 6x + 3y + 2z = 18 6(3) + 0 + 0 = 18 For point B(0, 6, 0) 6x + 3y + 2z = 18 0 + 3(6) + 0 = 18 For point C (0, 0, 9) 6x + 3y + 2z = 18 0 + 0 + 2 × 9 = 18 Option (d) is correct.
44. (b) Length of one of the diagonal of cube = 12 22 32
= 1 4 9 14 units Option (b) is correct.
45. (c) Equation of plane passing through (1, 2, 3) and parallel to xy-plane is z = 3. Option (c) is correct. 46. (d) (127.25)10 Now, 2 127 2 63 – 1 2 31 – 1 2 15 – 1 2 7 – 1 2 3 – 1 1 – 1 127 = (1111111)2
Now,
0.25 × 2 = 0.5 0
0.5 × 2 = 1.0 1
(127.25)10 = (1111111.01)2 Option (d) is correct.
47. (d) Non-singular matrix is a matrix whose determinate Value is non-zero.
Let A = 1 1
1 2 and B = 2 1 1 2
Here, A and B are non-singular matrix Now from Statement 1:
A + B =
1 1 2 1 3 2
1 2 1 2 2 4
det (A + B) = 12 – 4 = 8
Now det (A) = 1 1
1 2 = 1
and det (B) = 2 1
1 2 = 3
Now, det (A + B) = 4 det (A + B) Statement 1 is wrong.
Now from Statement 2: (A + B)–1 = 1 A B adj (A + B) = 1 4 –2 –2 3 8 and A–1 = 1 A adj (A) = 2 –1 1 –1 1 1 and B–1 = 1 B adj (B) = 2 –1 1 –1 2 3 Now, A–1 + B–1 = 2 –1 2 –1 3 3 –1 1 –1 2 3 3 = 8 – 4 3 3 – 4 5 3 3 (A + B)–1 Statement 2 is wrong Option (d) is correct. 48. (a) X = 3 – 4 , 1 –1 B = 5 2 –2 1 and A = p q r s Now, AX = B p q 3 – 4 5 2 r s 1 –1 –2 1 3p q –4p – q 5 2 3r s –4r – s –2 1 3p + q = 5 ... (i) – 4p – q = 2 ... (ii) 3r – s = –2 ... (iii) –4r – s = 1 ... (iv)
From equations (i) and (ii), we get –p = 7
p = –7 q = 5 – 3(–7) q = 26
From equations (iii) and (iv), –r = –1 r = 1 s = –2 – 3 = –5 s = –5 Hence, A = p q –7 26 r s 1 –5
Option (a) is correct.
49. (d) 1 n r n r 0 C = n Cn + n + 1Cn = 1 + n 1 ! n 1 – n ! n! = 1 + n 1 n! n! = 1 + n + 1 = n + 2 n + 2 Cn + 1 = n 2 ! n 2 – n – 1 ! n 1 ! = n 2 n 1 ! n 1 ! = n + 2 Option (d) is correct.
50. (c) The given word is ‘NATION’.
Total number of words that can be formed from given word ‘NATION’
= 6! 6 5 4 3 2!
2! 2! = 360
Now numbers of words that can be formed from given word ‘NATION’, so that all vowels never come together.
= 360 – 4! 3! 2! = 360 – [24 × 3] = 360 – 72 = 288 Option (c) is correct.
51. (b) As we know that cube root of unity is 1, and 2 x3 – 1 = (x – 1) (x – ) (x – 2) Option (b) is correct. 52. (c) 3 sin i 1 – cos 6 6 sin – i 1 – cos 6 6 = 3 2 2
2 sin cos i 2 sin
12 12 12
2 sin cos – i 2 sin
12 12 12 = 3 3 i 12 –i 12 cos i sin e 12 12 cos – i sin e 12 12 = 3 i i 3 i 6 6 2 e e e = cos 2 + i sin 2 = i Option (c) is correct. 53. (a) A = x y y 2x x – y B = 2 –1 and C = 3 2 Here AB = C x y y 2 3 2x x – y –1 2 3 2 x y –y 2 4x –x y 2x + y = 3 ... (i) 3x + y = 2 ... (ii)
From equations (i) and (ii), we get x = –1 and y = 5 A = 4 5 –2 –6 Now, A2 = 4 5 4 5 –2 –6 –2 –6 = 16 – 10 20 – 30 6 –10 –8 12 –10 36 4 26
Option (a) is correct.
54. (c) 1 1 1 1 1 x 1 1 1 1 y = [(1 + x) (1 + y) – 1] – 1(1 + y – 1) + 1(1 – 1 – x) = 1 + x + y + xy – 1 – y – x = xy Option (c) is correct. 55. (d) A = {x : x is a multiple of 3} A = {3, 6, 9, 12, 15, 18, 24, ...} A B C B = {x : x is a multiple of 4} B = {4, 8, 12, 16, 20, 24, 28, 32, ...} C = {x : x is a multiple of 12} C = {12, 24, 36, 48, 60, 72, 84, 96, ...} A B = {12, 24, ...} = A B. C 56. (a) (11101011)2 = (1 × 27 + 1 × 26 + 1 × 25 + 0 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 1 × 20)10 = (128 + 64 + 32 + 8 + 2 + 1)10 = (235)10
Option (a) is correct. 57. (b) (sin x + i cos x)3
= sin3x + (i)3 cos3x + 3i (sin x) (cos x) (sin x + i cos x) = sin3x – i cos3x + 3i sin2x cos x – 3 sin x cos2x = sin3x – 3 sin x cos2x + i cos x (cos2x + sin2x) = sin x (sin2x – 3 cos2x) + i cos x
Real part of (sin x + i cos x)3 = sin x (sin2x – 3 cos2x) = sin x [sin2x – 3 (1 – sin2x)] = sin x [4 sin2 x – 3] = 4 sin3x – 3sin x = – (3 sin x – 4 sin3x) = –sin 3x Option (b) is correct. 58. (c) E ( ) = cos sin – sin cos
Now E ( ) = cos sin
– sin cos
and E ( ) = cos sin
– sin cos
Now E ( ) E( ) = cos sin cos sin
– sin cos – sin cos
=
cos cos cos sin
– sin sin sin cos
– sin cos – sin sin
= cos sin – sin cos = E( + ) Option (c) is correct. 59. (a) A = {x, y, z} B = (p, q, r, s) n(A) = 3 n(B) = 4
Number of distinct relations = 2n(A) × n(B) = 23 × 4 = 212 = 4096 option (a) is correct. 60. (c) 4p2 + 4pq – 3q2 – 36 = 0 (4p2 + 4pq + q2) – 4q2 – 36 = 0 (2p + q)2 = 4q2 + 36 2p + 3q = 18 p = 18 – 3q 2 2 18 – 3p 2 q 2 = 4q 2 + 36 (18 – 3q + q)2 (18 – 2q)2 = 4q2 + 36 324 + 4q2 – 72q = 4q2 + 36 72q = 324 – 36 = 288 q = 288 72 = 4
Putting the value of q in 2p + 3q = 18 2p + 3 × 4 = 18 2p = 18 – 12 p = 6 2 = 3 (2p + q) = 2 × 3 + 4 = 10 Option (c) is correct. For (61-62) Given 2 4 2 d 1 x x dx 1 x x = 2 4 2 d 1 x x x – x dx 1 x x = 4 2 d x – x 1 dx x x 1 = 3 2 x x – 1 d 1 dx x x 1 = d 1 x x – 1 dx = d 1 x – x2 dx = 2x – 1 ... (i)
Now comparing equation (i) with AX + B, we get A = 2 and B = –1. 61. (c) 62. (a) For (63-64): 2 x 2 x lim – Ax – B 1 x = 3 2 2 x 2 x – Ax – B – Ax – Bx lim 1 x = 3 2 x 1 – A x – A B x 2 – B lim 1 x = 3
Applying L’Hospital rule, 2x(1 – A) – (A + B) = 3 Comparing coefficients 2(1 – A) = 0 A = 1 and – (A + B) = 3 B = –3 – 1 = – 4 A + B = –3 A = 1, B = – 4 63. (b) 64. (c) 65. (b) y dx – x dy2 y = 0 y dx – x dy = 0 dy dx y x
Now integrating both sides,
dy dx
y x
log y = log x + log c log y = log c x y = cx Option (b) is correct. 66. (a) sin dy dx – a = 0 sin dy dx = a dy dx = sin –1 a dy = sin–1a dx
Now integrating both sides,
–1
y = x sin–1a + c Option (a) is correct.
67. (c) dx x dy y – y 2 = 0 dx x dy y = y 2
This is a linear differential equation of the form dx dy + P1x = Q1 Here, P = 1 y and Q = y 2 I.F. = 1 dy P dy y e e = elog y = y So, required solution is
x . y = y2 y dy c xy = y dy3 c xy = 4 y 4 + c 4xy = y4 + c Option (c) is correct. 68. (c) x 2 x e dx 1 x = x 2 1 1 e – dx 1 x 1 x = x e 1 x + C x x e f x f x dx e f x 69. (d) Area of ABC = 1 AB AC 2 = i j k 1 –2 3 2 2 –4 5 2 = 1 i 6 – 10 – j – 4 8 k –10 12 2 = 1 – 4i – 4j 2k 2 = 1 16 16 4 1 36 1 2 2 2 × 6 = 3 square units Option (d) is correct.
70. (a) Moment of force, m = r F
m = i j k 2 –2 –3 3 4 –3 = i (6 + 12) – j(– 6 + 9) + k(8 + 6) = 18i – 3j + 14k = 182 –32 142 = 529 = 23 units. 71. (a) u – v w 2x y – 2y 3x 2 – 5 2x – 2y y – 3x 2 – 5 2x – 2y = 2 ... (i) and 3x – y = –5 ... (ii)
Solving equations (i) and (ii), we get x = 2 and y = 1
Option (a) is correct.
72. (d) a = 7, b = 11 and a b 10 3 Now a b 2 a 2 b 2 2 a b cos 2 10 3 = 49 + 121 + 2 × 7 × 7 cos 300 = 170 + 154 cos 300 – 170 154 = cos 65 77 = cos Now, a – b 2 a 2 b 2– 2 a b cos = (7)2 + (11)2 – 2 × 7 × 11 × 65 77 = 49 + 121 – 2 × 65 = 170 – 130 = 40 a – b 40 2 10 Option (d) is correct.
73. (a) , and be distinct real numbers
i j k
i j k
i j k
a, b and c are collinear
If a = , b = , c = ( ˆ ˆi j k)ˆ Hence, the option (a) is correct.
74. (c) According to statements (1) and (2),
when a b c 0, then both statements are correct.
If a b b c c a = 0, then a, b and c are collinear.
Therefore, option (c) is correct.
75. (c) If a b a – b
We know that when a b a – b = 0, then a is
perpendicular to b. a.a – a.b b.a – b.b 0
ais perpendicular to b Option (c) is correct. 76. (a) G(x) = 25 – x2 Now, x 1 G x – G 1 lim x – 1 2 x 1 25 – x – 24 lim x – 1 (÷ form)
Applying L’Hospital rule,
2 2 x 1 –2x –2 lim 2 25 – x 2 25 – x = –2 –1 –1 2 24 24 2 6
Option (a) is correct. For (77-79) 77. (c) From statement -1 x – x e e y 2 x – x dy 1 [e – e ] dx 2 Now equating dy dx to zero we get, ex – e–x = 0 ex = e–x x – x e 1 e – + 0 e2x =1 2x = 0 x = 0
The given function is increasing on interval [0, ]
From Statement -2 x – x e – e y 2 x x dy 1 [e – e ] dx 2 Now equating dy dx to zero, we get. ex + e–x = 0 ex + x 1 0 e e2x + 1 = 0 e2x = –1
Hence, the given function is increacing from [– , ] Both statement are correct
Option (c) is correct. 78. (c) For a non-zero real number x
x f (x) | x | domain f (x) | x |x range –3 –2 –1 1 2 –1 +1
Here we can not take value of ‘x’ as zero as (x) is non zero real number
for any value of x in domain a set consisting of two elements i.e. (– 1, + 1)
Option (c) is correct. 79. (c) From statement-1
From the definition of greatest integer function f (x) = [x] is discontinuous at x = n for any value of n Z Statement1 is correct From statement-2 f (x) = cot x for x = = cot = – for x = 2 f (2 ) = cot 2 = –
Hence the function f (x) = cot x is discontinuous at x = n where n Z. Option (c) is correct. 80. (b) Let y = 2 –1 1 x – 1 tan x and u = tan –1 x
Put x = tan = tan–1x
Then, y = 2 –1 1 tan – 1 tan tan = 2 –1 sec – 1 tan tan
= –1 –1 1 – 1 sec – 1 cos tan tan sin tan cos = 2 –1 1 – cos –1 2 sin 2 tan tan sin 2 sin , cos 2 2 2
1 – cos 2 sin and 2
x x
sin x 2 sin cos
2 2 = tan–1 tan 2 y = 2 y = –1 tan x 2 [ = tan –1 x] y = u 2 dy 1 du 2 Option (b) is correct. 81. (b) For (82-83): F(x) = –1 , x 0 ax b , 0 x 1 1 , x 1 At x = 0, L.H.L. = R.H.L. – x 0 x 0 lim f x lim f x –1 = xlim ax0 b –1 = b
Since the given function is also continuous at x = 1 L.H.L. = R.H.L. – x 1 x 1 lim f x lim f x a + b = 1 a – 1 = 1 a = 1 + 1 = 2. 82. (d) Function is continuous at x = 0 – x 0 x 0 lim f x lim f x –1 = a(0) + b b = –1
Now, f(x) is also continuous at x = 1
– x 1 x 1 lim f x lim f x a(1) + b = 1 a = 1 – b = 1 + 1 a = 2.
83. (a) From solution 82, b = –1. 84. (c)
85. (b) Given that, dx 1 ln tan x
a cos x b sin x r 2
Let a = r sin , b = r cos
dx 1 1
r sin cos x r cos sin x r sin x
= 1 cosec x dx 1ln tan x
r r 2
a = r sin a2 = r2 sin2 ... (i) b = r cos b2 = r2 cos2 ... (ii) Adding (i) and (ii), we get
r2 = a2 + b2 r = a2 b .2
86. (a) a = r sin ... (i)
b = r cos ... (ii)
Dividing (i) from (ii),
a b = tan = tan–1 a . b 87. (b) Given f(x) = 2 2 x – 1 x 1
In order to find the value of ‘x’, where f(x) is maximum or minimum; equation f x equal to zero.
2 2 2 2 2 2 d d x 1 x – 1 – x – 1 x 1 dx dx f x x 1 2 2 2 2 x 1 2x – x – 1 2x f x x 1 = 2 2 2 2 2x x 1 – x 1 x 1 2 2 2x 2 f x x 1
Now equating f x to zero, we get 4x = 0
x = 0
Hence, f (x) attain minimum value at x = 0 Option (b) is correct. 88. (c) f (x) is minimum at x = 0 f (0) = –1 1 = –1 minimum value of f(x) is –1 Option (c) is correct. 89. (a) f (x) = cos x if x – 2x 2 3 if x 2 For continuity at x = 2 L.H.L. = – x 2 lim f(x) = – x 2 cos x lim – 2x Put x = 2 – h where x 2, then h 0 L.H.L. = h 0 h 0 cos – h sin h 2 lim lim – 2h – 2 – h 2 cos – sin 2 = h 0 h 0 2 sin h sin h lim lim 2h 2 h = 2 . 1 = 2 0 sin lim 1 and R.H.L. = x 2 lim f(x) x 2 cos x lim – 2 Put x = 2 + h when x 2, then h 0 R.H.L. = h 0 h 0 cos h – sin h 2 lim lim – – 2h – 2 h 2 = h 0 sin h lim – –2h cos 2 – sin = h 0 sin h lim 2 h 2 0 sin lim 1 Also, f 2 = 3 Since, f(x) is continuous at x = 2 L.H.L. = R.H.L. = f 2 2 2 = 3 = 6
Hence, for = 6, the given function (f) is continuous at
x = . 2 90. (d) x 0 cos x lim f x – 2x = cos 0 – 2 0 Option (d) is correct. 91. (c) Mean X = 4 Variance (X) = 2 Number of observations = 10 New average = 4 10 2 10 = 8
New variance = (2)2 × Variance (X) = 4 × 2 = 8
Option (c) is correct.
92. (b) Geometr ic mean is used in con str uction of index numbers.
option (b) is correct. 93. (a)
94. (a) Regression coefficients measures the degree of linear relationship between two variables and does not give the value by which one variable changes for a unit change in the other variable.
option (a) is correct.
95. (a) The annual numerical data for comparable for last 12 years is represented by broken line graph, where each turning point represent the data of a particular year, while such graph do not depict the chronological change.
Option (a) is correct.
= 1 1 –1 1 –1 1 2 3 2 3 = 1 2 1 1 2 3 2 3 = 1 1 1 3 6 2
Option (a) is correct. 97. (b) 98. (d) P = 5C2 2 3 1 1 2 2 + 5 C3 3 2 1 1 2 2 + 5C4 4 1 1 1 2 2 + 5 C5 5 0 1 1 2 2 = 5 5 5 5 5 2 3 4 5 1 C C C C 2 = 12 3 [10 + 10 + 5 + 1] = 12 3 × 26 = 13 . 16 99. (c) 100. (b) nP = 2 n Pq = 1 q = 1 2 P + q = 1 P = 1 – 1 2 = 1 2 1 n 2 = 2 n = 4 4 C0 = 4 – 0 1 1 . 2 16 101. (d) Mean of 5 numbers = 30
Total sum of 5 numbers = 30 × 5 = 150 After excluded one number
Mean of 4 numbers will be = 28 Total sum of 4 numbers = 4 × 28 = 112 Thus, excluded number
= (sum of 5 numbers – sum of 4 numbers) = 150 – 112 = 38 Option (d) is correct. 102. (b) P(A B) = 3 4 P(A B) = 1 4 2 P A , 3 P(A) = 1 3 P(B) = 3–1 1 9 – 4 3 8 4 3 4 12 2 12 3 Option (b) is correct.
103. (a) The ‘less than’ ogive curve and the ‘more than’ ogive curve intersect at median.
Option (a) is correct. 104. (c) n(S) = 6 × 6 = 36 1,1 1, 2 1, 3 1, 4 1, 5 1,6 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 3,1 3, 2 3, 3 3, 4 3, 5 3, 6 4,1 4, 2 4, 3 4, 4 4, 5 4, 6 5,1 5, 2 5, 3 5, 4 5, 5 5, 6 6,1 6, 2 6, 3 6, 4 6, 5 6, 6
Required number of exhausistance events = (6 – 1) × (6 – 1) = 5 × 5 = 25 Option (c) is correct. 105. (b) P(A) = 4 1 52 1 C 4 1 52 13 C P(B/A) = 3 1 51 1 C 3 1 51 17 C Required probability = 1 1 1 13 17 221 Option (b) is correct. For (106-107) x2 + y2 = 4 and x = 3y P 3, 1 = First quadrant Y X O x + y = 4 2 2 (0, 0) X Y x 3y P 3, 1 (2, 0) A B 3,0
The point of intersection of the line and the circle in the first quadrant is ( 3,1).
Area of OPA = Area of OPB + Area PAB 1 1 3 OB PB 3 1 2 2 2 Area of PAB 2 3 y dx 2 2 2 2 –1 3 3 x 4 x 4 – x dx 4 – x sin 2 2 2 –1 3 3 2 – [ 4 – 3] – 2sin 2 2 2 3 2 3 – – – 2 3 3 2 106. (a)
107. (a) Area enclosed by x-axis, the line 3y , and the circle x2 + y2 = 4 in the first quadrant
= 3 – 3 2 3 2 3 108. (a) y = sin x y = cos x 1 0
Area of shaded region = 4
0 cos x – sin x dx = sin x cos x 04 = 1 1 – 0 1 2 2 = 2 – 1 sq. units. 109. (a) 110. (c) f(x) = 0.75x4 – x3 – 9x2 + 7 f(0) = 7 f(–2) = 0.75(–2)4 – (–2)3 – 9(–2)2 + 7 = 2 + 8 – 36 + 7 = –9 f(3) = 0.75(3)4 – (3)3 – 9(3)2 + 7 = 60.75 – 27 – 81 + 7 = – 40.25 maximum value of f(x) is 7 Option (c) is correct. 111. (c) From Statement 1:
Function attain local minima at x = –2 and x = 3 As we have, f x = 3x3 – 3x2 – 18x Now, f x = 9x2 – 6x – 18 For x = –2, f –2 = 9(–2)2 – 6(–2) – 18 = 36 + 12 – 18 = 48 – 18 = 30 > 0 For x = 3, f 3 = 9(3)2 – 6(3) – 18 = 81 – 36 = 45 > 0 Statement 1 is correct From Statement 2: + –2 0 3 +
The function increases in the interval (–2, 0) Option (c) is correct. 112. (b) x2 + y2 = 2 2 2 2 2 a 1 – t 2at 1 t 1 t = 2 2 2 2 2 2 2 2 2 a 1 – t 4a t 1 t 1 t = 2 4 2 2 2 2 2 a 1 t – 2t 4a t 1 t = 2 2 4 2 2 2 2 2 2 2 a 1 t 2t a 1 t 1 t 1 t = a2
Hence, the given equation represent a circle of radius (a) Option (b) is correct. 113. (d) Here, x2 + y2 = a2 2x + 2ydy dx = 0 dy 2x x – dx –2y y Option (d) is correct. 114. (d) 2 2 2 d dy y –x – –x d y dx dx dx y = 2 2 – x dy –y x – y x y dx y y
= 2 2 2 2 2 3 3 3 – x y – y – x a – y y y Option (d) is correct. 115. (c) Statement 1:
The general solution of dy
dx = f(x) + x Now integrating on both side
dy f x x dx dx y = f(x) + C Statement 1 is correct. Statement 2: dy dx = f(x) + c
Squaring both sides,
2 2 dy f x c dx 2 2 2 dy f x 2c f x c dx
Hence, the differential equation is of order –1 and degree –2.
both Statements 1 and 2 are correct. Option (c) is correct. 116. (a) 2 2 dx x a Let x = a tan u dx = a sec2u du = 2 2 2 2 a sec u du a tan u a = 2 2 2 sec u du a a 1 tan u 2 a sec u du a sec u = 2 sec u du sec u du sec u
= ln[tan (u) + sec (u)] + c sec x dx tan x
= 2 2 2 x x ln 1 c a a = 2 2 a x x ln c a a = 2 2 x x a ln c a Option (a) is correct.
117. (a) Im = 0 sin 2 mx dx sin x I1 = 0 0
sin 2x 2 sin x cos x
dx dx
sin x sin x
= 0
0
2 cos x dx 2 sin x 2 sin – sin 0
= 2(0) = 0
Option (a) is correct.
118. (d) I2 =
0 0
sin 4x 2 sin 2x cos 2x
dx dx
sin x sin x
= 0
2 sin x cos x cos 2x
2 dx sin x = 2 0 4 cos x 1 – 2 sin x dx = 2 0 0
4 cos x dx – 8 cos x sin x dx 0
Let sin x = t cos x dx = dt = 4(sin – sin 0) – 0 = 0 I3 = 0 sin 6x dx sin x = 3 0
2 3 sin x – 4 sin x cos 3x dx sin x
= 2
02 3 – 4 sin x cos 3x dx
= 2
06 cos 3x dx – 08 sin x cos 3x dx
= 6 sin 3x 0 –8 1 0sin x sin 4x – sin 2x dx
3 2
= 6 – 0
8
sin x sin 4x – sin x sin 2x dx 2
= –8 1 0cos 3x – cos 5x – cos x cos 3x dx 2 2 = 0 2 sin 3x sin 5x –2 – – sin x 3 5 = 0 Hence, I2 + I3 = 0 + 0 = 0 Option (d) is correct.
119. (a) Im = 0 sin 2mx dx sin x = 0 sin 2m – x dx sin – x = 0 sin 2m – 2mx dx sin x = 0 – sin 2mx dx sin x Im = 0 sin 2mx – dx sin x 2Im = 0 Im = 0.
120. (a) I2m > Im is wrong statement Because, Im = Im – 1 = ... = In Thus,
Im – Im – 1 = 0 is the only correct statement. Option (a) is correct.
