JEE Mathematics CENGAGE LEARNING
Short Description
This is a document containing sample mock test paper of Cengage Learning Publication....
Description
Appendix B
Mock Test 1 Section A: Only One Option Correct Type Ê pˆ 1. For q Œ Á 0, ˜ , the value of definite integral Ë 2¯ q
(a) q ln (sec q) q ln 2 (c) 2
(b) q ln(cosec q) (d) 2q ln sec q
1 1 2 - = . v u f If errors made in measuring u and v are a, then the relative error in f is 2 Ê 1 1ˆ (a) (b) a Á + ˜ Ë u v¯ a Ê 1 1ˆ (c) a Á - ˜ Ë u v¯
(a)
a 2 + b2 - c2 2bc
(b)
c2 + a 2 - b2 2bc
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7 (b) 3
5 (c) 3
5 (d) 4
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5. A parabola touches two given straight lines originating from a given point. The locus of the mid point of the portion of any tangent, which is intercepted between the given straight lines, is a/an (a) parabola (b) ellipse (c) straight line (d) hyperbola p p 6. If x - + sin x - 2 = | x2 – 2 | + | sin x | + , then 4 4 2
(a) x Œ(0, 2 )
(b) x Π(- 2 , 2 )
(c) x ŒR
(d) x Π(- 2 , 0)
a b c - cos cos r r r equals b c sin sin r r
Lim
rƕ
(d) None of these
3. If the function f (x) = 2 tan x + (2a +1) loge | sec x | + (a –2) x is increasing on R, then (a) a Œ (1/2, •) (b) a Œ (–1/2, 1/2) (c) a = 1/2 (d) a Œ R 4. 3x + 4y – 7 = 0 and 3x – 4y –7 = 0 are equations of asymptotes of a hyperbola H. From a point P(3, 4), pair of tangents are drawn to hyperbola H in such a way that both tangents touch the same branch of hyperbola H. Then its eccentricity is 4 (a) 3
cos
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2. The focal length of a mirror is given by
di
0
a
Pv t
Ú ln (1 + tan q tan x) dx is equal to
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7. If A and B are different matrices satisfying A3 = B3 and A2B = B2A, then (a) det (A2 + B2) must be zero. (b) det (A – B) must be zero. (c) det (A2 + B2) as well as det (A – B) must be zero (d) At least one of det (A2 + B2) or det (A – B) must be zero. 8. A fair dice is thrown 3 times. The probability that the product of the three outcomes is a prime number is 1 1 1 1 (a) (b) (c) (d) 24 36 32 8 9. The number of real solution of equation 16 sin–1 x tan–1 x cosec–1 x = p3 is/are (a) 0 (b) 1 (c) 2 (d) infinite 10. Let a, b, c are non-zero constant numbers. Then
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Paper 1
b2 + c2 - a 2 2bc (d) Independent of a, b and c (c)
Section B: One or More Options Correct Type 11. Given 2 functions f and g which are integrable on every interval and satisfy (i) f is odd, g is even (ii) g(x) = f (x + 5), then (a) f (x – 5) = g(x) (b) f (x – 5) = – g(x) 5
(c)
Ú 0
5
f (t ) dt =
Ú g (5 - t ) dt 0
5
(d)
Ú 0
5
f (t ) dt = -
Ú g (5 - t ) dt 0
B.2 Mathematics
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area enclosed by the curve, the x-axis and the line x = 4 is 8 sq. units, then (a) a2 + b2 = 6 (b) a/b = 1 (c) a – b = 4 (d) ab = –3
Section C: Integer Value Correct Type ( n + 1) 2
16. The value of Lim
nƕ
Â
k = n2
1 is k
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ax at the point (1, 1) b-x is 2, then the value of a + b is Given a = 3i + j + 2k, b = i – 2j – 4k are the position vectors of point A and B. Then the distance of point –i + j + k from the plane passing through B and perpendicular to AB is If z = x + iy (x, y Œ R, x π –1/2), then the number of values of z satisfying | z |n = z2 | z |n–2 + z | z |n–2 + 1, (n Œ N, n > 1) is If x, y, z are distinct positive numbers such that 1 1 1 x + = y + = z + , then the value of xyz is y z x
17. If the slope of the curve y =
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19.
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a
18.
20.
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12. A continuous function f (x) on R Æ R satisfies the relation f (x) + f (2x + y) + 5xy = f (3x – y) + 2x2 + 1 for " x, y Œ R, then which of the following hold(s) good? (a) f is many-one (b) f has no minima (c) f is neither odd nor even (d) f is bounded 13. The equation x2 – 4x + a sin a = 0 has real roots (a) for all values of a (b) for all values of a provided –p/4 < a < p/4 (c) for all values of a ≥ 4 provided p £ a £ 2p (d) for all a provided | a | £ 4 14. Which of the following statement(s) is/are not correct? (a) If the roots of a quadratic equation are imaginary, then these are conjugates of each other. (b) If a continuous function is strictly monotonic, then it is differentiable. (c) If f (x) is periodic, then | f (x) | is also periodic. (d) sin x/(2x – 2np – p) and cos x tan x/(2x – 2np – p), where n Œ Z, are identical functions. 15. If the curve y = ax1/2 + bx passes through the point (1, 2) and lies above the x-axis for 0 £ x £ 9 and the
Paper 2
Section A: One or More Options Correct Type
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1. For natural numbers m and n, if (1 Рy)m (1 + y)n = 1 + a1 y + a2 y2 + L, and a1 = a2 = 10, then (a) m < n (b) m > n (c) m + n = 80 (d) m Рn = 20 2. Let P(x) = x2 + bx + c, where b and c are integer. If P(x) is a factor of both x4 + 6x2 + 25 and 3x4 + 4x2 + 28x + 5, then (a) P(x) = 0 has imaginary roots (b) P(x) = 0 has roots of opposite sign (c) P(1) = 4 (d) P(1) = 6 3. Let tan x Рtan2 x > 0 and | 2 sin x | < 1. Then the intersection of which of the following two sets satisfies both inequalities? (a) x > np + p/6, n ΠZ (b) x > np Рp/6, n ΠZ (c) x < np Рp/4, n ΠZ (d) x < np + p/4, n ΠZ
4. A bag initially contains one red and two blue balls. An experiment consisting of selecting a ball at random, noting its colour and replacing it together with an additional ball of the same colour. If three such trials are made, then (a) probability that atleast one blue ball is drawn is 0.9. (b) probability that exactly one blue ball is drawn is 0.2. (c) probability that all the drawn balls are red given that all the drawn balls are of same colour is 0.2. (d) probability that atleast one red ball is drawn is 0.6. 5. If a, b, c are non-zero real numbers such that bc ca ab ca ab bc = 0, then ab bc ca
1 1 1 1 1 1 + + = 0 (b) + + =0 a b c a bw 2 cw 1 1 1 + + = 0 (d) None of these (c) aw bw 2 c (a)
6. 2007201 + 2019201 – 1982201 – 2044201 is divisible by (a) 74 (b) 50 (c) 1850 (d) 2013
Appendix B: Mock Tests B.3 1 1 2 (b) (c) 22 23 33
(d)
5 22
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Paragraph for Questions 13 and 14
a
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Ê x + sin x ˆ , Consider a function f defined by f (x) = sin–1 sin Á Ë 2 ˜¯ " x Œ [0, p], which satisfies f (x) + f (2p – x) = p, " x Œ [p, 2p] and f (x) = f (4p – x) for all x Œ [2p, 4p], then. 13. If a is the length of the largest interval on which f (x) is increasing, then a (a) p/2 (b) p (c) 2p (d) 4p 14. If f (x) is symmetric about x = b, then b = (a) a/2 (b) a (c) a/4 (d) 2a Paragraph for Questions 15 and 16 Straight line x + y = 18 and common tangents of the curve x2 y 2 + –1=0 64 36 form a triangle ABC in the first quadrant and S3 = 0, the circumcircle of DABC. 15. The coordinates of the center of circle inscribed in triangle ABC are S1 = x2 + y2 – 28x + 160 = 0 and S2 =
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This section contains 4 paragraphs each describing theory, experiment, date etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of paragraph has only one correct answer along the four choice (a), (b), (c) and (d). Paragraph for Questions 9 and 10 A perpendicular is drawn from a fixed point (3, 4 ) to a variable line having x-intercept unity. P (x, y) = 0 represents the locus of the foot of perpendicular drawn from (3, 4) to the variable line, which is a circle. Then 9. The equation of the circle is: (a) x2 + y2 = 4 (b) (x – 2)2 + (y – 2)2 = 5 (c) (x – 3)2 + (y – 3)2 = 4 (d) (x – 3)2 + (y + 3)2 = 5 10. A tangent is drawn to P(x, y) = 0 from the origin, then the length of tangent is
(a)
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Section B: Paragraph Type
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7. If the equation sin2 x – a sin x + b = 0 has only one solution in (0, p), then which of the following statements are correct? (a) a Œ (– •, 1) » (2, •) (b) b Œ (– •, 0) » (1, •) (c) a = 1 + b (d) None of these 8. Let a, b, and g be some angles in the 1st quadrant satisfying tan (a + b) = 15/8 and cosec g = 17/8, then which of the following holds good? (a) a + b + g = p (b) cot a cot b cot g = cot a + cot b + cot g (c) tan a + tan b + tan g = tan a tan b tan g (d) tan a tan b + tan b tan g + tan g tan a = 1
(a) [12 - 2 2 , 10 - 2 2 ]
(b) [12 - 2 , 10 - 2 ]
(c) [12 + 2 2 , 10 + 2 2 ]
(d) [12 + 2 , 10 + 2 ]
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(a)
2 (b) 1
(c)
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3 (d) 2
Paragraph for Questions 11 and 12
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Let a point P whose position vector is r = xi + y j + zk is called lattice point if x, y, z Œ N. If atleast two of x, y, z are equal then this lattice point is called isosceles lattice point. If all x, y, z are equal, then this lattice point is called equilateral lattice point. 11. The number of lattice points on the plane r .(i + j + k ) = 10 are (a) 36 (b) 45 (c) 84 (d) 120 12. If a lattice point is selected at random from lattice points which satisfy r .(i + j + k ) £ 11, then the probability that the selected lattice point is equilateral given that it is isosceles lattice point is
16.
Equation of S3 is (a) x2 + y2 – 16x – 20x + 156 = 0 (b) x2 + y2 – 10x – 8x + 156 = 0 (c) x2 + y2 – 8x – 10x + 156 = 0 (d) x2 + y2 – 20x – 16x + 156 = 0
Section 3: Matching List Type 17. The graphs of f and g are given in Fig. B.1. Use them to evaluate each limit. y y = f (x)
2 1
-2
-1
O -1
1 2
x
B.4 Mathematics
-2
-1
1 2
O
x
(b) If smallest side =
Fig. B.1
3 f ( x) - 2
Ê f ( x) ˆ (c) lim Á + f ( x) g ( x)˜ x Æ 0 Ë g ( x) ¯
(p) 1
(d) If D = ( 3 - 1)/ 4, then
(s) largest side = 2 + 2 3
(q) does not exist
20. A bag contains 14 balls which are either white or black balls. (all number of white and black balls are equally likely). Five balls are drawn at random from the bag without replacement
+
x Æ1
(r) 0
3 f ( x) - g ( x) f ( x) + g ( x)
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(d) lim
6+ 2
di
x Æ1
2 , then (q) largest side =
3 +1
(r) largest side = 1
Column II
(a) lim f ( g ( x))
(p) largest side =
(c) If s = 3 + 3 + 2 , then
Column I
xÆ2
Column II
(a) If smallest side = 2, then
-1
(b) lim
Column I
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1
Pv t
y = g(x)
Column II (p) Portions of a line (q) Point of intersection of hyperbola (r) Pair of open rays (s) line segment 19. The angles of a triangle are in the ratio 2 : 3 : 7.
2
a
y
(s) 2
Column I
Column II
(a) Probability that all the five balls are (p) 3/13 black is equal to
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18. Match the locus of z given by equation in Column I with Column II. Column I
z - 1 - sin -1
(d)
1 1 p + z + cos -1 =1 3 3 2
(c) If all the five balls are black then (r) 6/65 the probability that the bag contains 11 black and 3 white balls is equal to (d) Probability that three balls are black (s) 3/65 and two are white is equal to
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Ê z 2 - 1ˆ (a) arg Á 2 ˜ = 0; z π ± i, ± 1 Ë z + 1¯ (b) || z – cos–1 cos 12 | – | z – sin–1 sin 12 || = 8 (p – 3) (c) z2 + k1 = i | z1 |2 + k2 ; k1 π k2 Œ R – {0} and z1 is fixed π 0
(b) If the bag contains 11 black and 3 (q) 1/6 white balls, then the probability that all five balls are black is equal to
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Mock Test 2
Paper 1
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Section A: Only One Option Correct Type
C
1. If f (x) =
50
’ ( x - n)
n ( 51 - n )
, then
n =1
(a) 5050
(c)
1 5050
(b)
f (51) = f ¢ (51)
1 1275
(d) 1275
2. Through the focus of the parabola y2 = 2px (p > 0), a line is drawn which intersects the curve at A(x1, y1) yy and B(x2, y2). The ratio 1 2 equals x1 x2 (a) 2 (b) –1 (c) –4 (d) some function of p n Ê ( 1 + k 1) k (k + 1) (k + 2) ˆ 3. Lim cos -1 Á ˜ is xÆ• k (k + 1) Ë ¯ k=2 equal to p p p p (a) (b) (c) (d) 6 4 3 2
Â
Appendix B: Mock Tests B.5
100
sin 2 a x +
cos 2 a x is p/8, then the value of a is
(a) ± 4 (b) ± 2 (c) ± 8 x ◊ f ¢( x) - 2 f ( x) 9. dx equals x 4 ◊ f ( x)
Ú
(a) x2 f (x) + c
(c)
(b) | x | f (x) + c
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(d) ± 1
2 f ( x) +c x
(d)
2 f ( x) +c x
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10. In which one of the following intervals, the inequality sin x < cos x < tan x < cot x can hold good? (a) (0, p/8) (b) (3p/4, p) (c) (5p/4, 3p/2) (d) (7p/4, 2p)
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Section B: One or More Options Correct Type
C
11. Let f (x) =
x
Úe
t - [t ]
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1ˆ Ê 6. The number of terms in Á x3 + 1 + 3 ˜ is Ë x ¯ (a) 300 (b) 200 (c) 100 (d) 201 7. The set of all values of a for which ax2 + (a – 2)x – 2 is negative for exactly two integral x is (a) (0, 2) (b) (1, 2) (c) (1, 2) (d) (0, 2) 8. If the fundamental period of the function f (x) =
a
5. A continuous and differentiable function y = f (x) is such that its graph cuts line y = mx + c at n distinct points. Then the minimum number of points at which f ≤(x) = 0 is/are (a) n – 1 (b) n – 3 (c) n – 2 (d) Cannot say
di
Ê 3p ˆ (d) Á 0, Ë 4 ˜¯
Ê 3p ˆ (c) Á , p˜ ¯ Ë 4
n
Ê 1 ˆ 12. Consider the binomial expansion of Á x + 4 ˜ , 2 x¯ Ë n Œ N, where the terms of the expansion are written in decreasing powers of x. If the coefficients of the first three terms form an arithmetic progression, then which of the following is/are true? (a) Total number of terms in the expansion of the binomial is 8. (b) Number of terms in the expansion with integral power of x is 3. (c) There is no term in the expansion which is independent of x. (d) Fourth and fifth are middle terms of the expansion. x y x y 13. If + = 1 and + = 1 intersect the axes at four a b c d concyclic points and a2 + c2 = b2 + d2, then these lines can intersect at (a, b, c, d > 0) (a) (1, 1) (b) (1, –1) (c) (2, –2) (d) (3, 3) 14. A forecast is to be made of the results of five cricket matches, each of which can be a win or a draw or a loss for the Indian team. Let p = number of forecasts with exactly 1 error q = number of forecasts with exactly 3 errors r = number of forecasts with all 5 errors then the incorrect statement is (a) 2q = 5r (b) 8p = q (c) 8p = 5r (d) 2(p + r) > q 15. A tangent drawn to the curve y = f (x) at P(x, y) cuts the x-axis at A. Now a perpendicular drawn from P(x, y) to the x-axis meets the x-axis at B. If B is the midpoint of OA (O being the origin), and f (1) = 1, then (a) General point on the curve is (t, 1/t), where t is a real parameter. (b) y = f (x) is a circle. (c) Tangent at P(1, 1) is x + y = 2. (d) f ¢(2) = –1/4.
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4. If f ≤(x) > 0 and f ¢(1) = 0 such that g(x) = f (cot2 x + 2 cot x + 2), where 0 < x < p, then the interval in which g(x) is decreasing is Êp ˆ (a) (0, p) (b) Á , p ˜ Ë2 ¯
dt ( x > 0), where [x] denotes great-
0
est integer less than or equal to x is (a) continuous and differentiable " x Π(0, 3) (b) continuous but not differentiable " x Π(0, 3) (c) f (1) = e (d) f (2) = 2(e Р1)
Section C: Integer Type 2p
16. The value of Lim tÆ0
Ú 0
sin ( x + t ) - sin x t
dx is
B.6 Mathematics
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(b) sin q sin f = 2 / 14
(c) tan f = 2 (d) cos q cos f = 1/ 14 5. If the side AB of an equilateral triangle ABC lying in the x–y plane is 3i, then the side CB can be 3 3 (i - 3 j ) (b) (i - 3 j ) 2 2 3 3 (c) - (i + 3 j ) (d) (i + 3 j ) 2 2 6. In a precision bombing attack, there is a 50% chance that any one bomb will strike the target. Two direct hits are required to destroy the target completely. The number of bombs which should be dropped to give a 99% chance of better of completely destroying the target can be (a) 12 (b) 11 (c) 10 (d) 13 7. The value(s) of p for which the equations ax2 – px + ab = 0 and x2 – ax – bx + ab = 0 may have a common root, given a, b are non-zero real numbers, is(are) (a) a + b2 (b) a2 + b (c) a(1 + b) (d) b(1 + a)
(a) -
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1. Let z be a complex number satisfying equation zp = z q, where n, m Œ N, then (a) If p = q, then the number of solutions of equation will be infinite. (b) If p = q, then the number of solution of equation will be finite. (c) If p π q, then the number of solution of equation will be p + q + 1. (d) If p π q, then the number of solution of equation will be p + q. x4 + 1 2 2. If dx = tan -1 f ( x) = tan -1 g ( x) + c, then 3 x6 + 1
(a) tan q = 5 / 3
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Section A: One or More Options Correct Type
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Paper 2
È 2 -1˘ È-1 - 8 -10˘ Í ˙ Í ˙ 19. If Í 1 0˙ A = Í 1 - 2 - 5˙ , then the sum of ÍÎ- 3 4˙˚ ÍÎ 9 22 15˙˚ all the elements of matrix A is 1 20. Let f (x) = sin23 x – cos22 x and g(x) = 1 + tan–1 | x |, 2 then the number of values of x in interval [–10p, 8p] satisfying the equation f (x) = sgn(g(x)) is
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17. If a, b and c are distinct positive real numbers such that a + b + c = 1, then the least integral value of (1 + a )(1 + b)(1 + c) is (1 - a )(1 - b)(1 - c) 18. The number of elements in the domain of the funcÊ x2 - 2 x ˆ + [ x] + [- x] , where [◊] tion f (x) = sin -1 Á 3 ˜¯ Ë represents greatest integer function, is
Ú
Ú
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(a) both f (x) and g(x) are odd functions. (b) g(x) is monotonic function. (c) f (x) = g(x) has no real roots. f ( x) 1 3 (d) dx = + 3 + c g ( x) x x
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3. A tangent is drawn at any point (x1, y1) other than vertex on the parabola y2 = 4ax. If tangents are drawn from any point on this tangent to the circle x2 + y2 = a2 such that all the chords of contact pass through a fixed point (x2, y2), then (a) x1, a, x2 are in GP. (b) y1/2, a, y2 are in GP. (c) –4, y1/y2, x1/x2 are in GP. (d) x1x2 + y1 y2 = a2 4. Let PM be the perpendicular from the point P(1, 2, 3) to x–y plane. If OP makes an angle q with the positive direction of the z-axis and OM makes an angle f with the positive direction of x-axis, where O is the positive direction of x-axis, where O is the origin, then (q and f are acute angles)
Ï x (t - 1), x £ t 8. Let f (x, t) = Ì , where t is a continuÓt ( x - 1), t < x ous function of x in [0, 1].
Ú
Let g ( x) =
1
0
f (t ) f ( x, t ) dt. Then
(a) g(0) = 1 (c) g(1) = 1
(b) g(0) = 0 (d) g(x) = f (x)
Section B: Paragraph Type Paragraph for Questions 9 and 10 The curve y = (x – a)(x2 + bx + c) meets the x-axis at a point P(p, 0). A straight line through P meets the curve at two more points A and B. Tangents drawn to the curve at P meet the curves at C. (b2 – 4c < 0)
Appendix B: Mock Tests B.7
and f (0) = 4. Let g(x) = 16 - x 2 . Given f (x) < g(x) " x Œ (0, 4) and graph of y = f (x) and y = f –1(x) are symmetrical about the line x + y = 4.
C2 : arg ( z - 2i ) = -
C3 : z - 1 - i = 2
15. The number of points common to curves C1, C2 and C3 is/are (a) 0 (b) 1 (c) 2 (d) None of these 16. Curves C4, C5, C6 are the reflections of the curves C1, C2 and C3, respectively, about the real axis. Then (a) C4 : z -
(b) C5 : arg ( z - 3i ) =
Ú
f -1 ( x) dx is
(c) C6 : z - 1 - i = 2 (d) None of these
2
Section C: Matching List Type
-1
( x)) dx = 2,
0 4
Úf
-1
( x) dx = 4, then
ng
Ú ( f ( x) - f
17.
4
2
Ú g ( x) - max ( f ( x), f
-1
( x)) dx is
0
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(a) 4p – 12 (b) 8p – 1 (c) 4p – 10 (d) 8 Paragraph for Questions 13 and 14 A function f (x) having the following properties: (i) f (x) is continuous except at x = 3. (ii) f (x) is differentiable except at x = –2 and x = 3 (iii) f (x) = 0, lim f (x) Æ – •, lim f (x) = 3, lim f (x) = 0 x Æ -•
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xƕ
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(iv) f ¢(x) > 0 " x Œ (– •, – 2) » (3, •) and f ¢(x) £ 0 " x Œ (– 2, 3) (v) f ≤(x) > 0 " x Œ (– •, – 2) » (– 2, 0) and f ≤(x) < 0 " x Œ (0, 3) » (3, •) 13. Maximum possible number of solutions of f (x) = | x | is (a) 2 (b) 1 (c) 3 (d) 4 14. f (x) + 3x = 0 has five solution if (a) f (– 2) > 6 (b) f ¢(0) < – 3 and f (– 2) > 6 (c) f ¢(0) > – 3 (d) f ¢(0) > – 3 and f (– 2) > 6 Paragraph for Questions 15 and 16 There are three curve defined here in Argand plane: 9 1 C1 : z + z=2 5 5
Column I
Column II
(a) Sum of coefficients in expansion of (p) 0 (px + qy – 5rz + 6t)3
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(a) 2 (b) 1 (c) 5 (d) 7 12. If f (x) – f –1(x) > 0 for x Œ (0, 2) and 2
3p 4
a
0
4
then
Ú f ( x) dx = 5,
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11. If f (x) Рf (x) < 0 for x Π(0, 2) and
9 1 + z=2 5 5
2
–1
p 4
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The abscissa of C is (a) – a – b (b) – a + b (c) a – b (d) a + b Number of points on the curve where the slope of the tangent is same as slope of AB is (a) 0 (b) 1 (c) 2 (d) 3 Paragraph for Questions 11 and 12 Let f (x) be one-one and onto function, such that | f (x) – f –1(x) | > 0 for x Œ (0, 2) » (2, 4) given f (2) = 2, f (4) = 0
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9. 10.
(b) Coefficient of x103 in (1 + x + x2 + (q) 100 x3 + x4)199(x – 1)201 is (c) Remainder when 22003 is divided by (r) 27 17 is (d) Remainder when (11111 … 1001 times) is divided by 1001 is
18. A tangent having slope -
(s) 8
4 x2 touches the ellipse 18 3
y2 = 1 at point P and intersects the major and 32 minor axes at A & B respectively, O is the centre of the ellipse +
Column I
Column II
(a) Distance between the parallel tangents (p) 24 4 having slopes - , is 3 (b) Area of DAOB is
(q) 7/24
(c) If coordinates of p are (l, m), then the value (r) 48/5 of l ¥ m is (d) If equation of the tangent intersecting posi- (s) 12 tive axes is x + my = 1, then + m is equal to
B.8 Mathematics 20. Match the following: Column II
Column I
(a) The number of integer lying in the domain (p) 1 of the function f (x) =
Ê 5 - 2x ˆ log 0.5 Á Ë x ˜¯
Ú
•
Ú
p /2
(c)
Ú
1
(d)
Ú
1
greatest integer function
(a)
(b) Number of positive roots of the equation (q) 2 (x – 1) (x – 2) (x – 3) + (x – 1) (x – 2) (x – 4) + (x – 2) (x – 3) (x – 4) + (x – 1) (x – 3) (x – 4) = 0 is/are
(b)
(c) If f (x) = e " x Œ [0, 1] and f (1) – f (0) = f ¢(c). (r) 3 where c Œ (0, 1) then n (ec + 1) is equal to
(q) –2/5
x dx
-1 x3
+
(r) –1
1 + x6
[ x] dx , where [◊] represents
-1
(s) –4/5
In
Mock Test 3
5. P and Q are two points on the upper half of the ellx2 y 2 ipse 2 + 2 = 1. The center of the ellipse is at the a b origin O and PQ is parallel to the x-axis such that the triangle OPQ has the maximum possible area. A point is randomly selected from inside of the upper half of the ellipse. The probability that it lies outside the triangle is p -1 2p - 1 (a) (b) p 2p
ng
Paper 1
(p) 0
ln (tan ( x)) dx
di
(d) Number of values of x satisfying the equa- (s) 4 1 p Ê 2 x - 1ˆ tion tan - 1 Á + tan - 1 = Ë 10 ˜¯ 2x 4
0
dx
a
x
x ln x (1 + x 2 )2
0
Column II
Pv t
Column I
.L td .
19.
Section A: Only One Option Correct Type
f (x) = sin x +
x
Ú
Le
ar ni
1. A curve passes through the point (2a, a) and is such that the sum of subtangent and abscissa is a. Its equation is (a) (x – a)y2 = a3 (b) (x – a)2 y = a3 2 (c) (x – a) y = a (d) none of these 2. A function f(x) satisfies f ¢(t) (2 sin t – sin2t) dt. Then f (x) is/are
0
(c)
sin x (b) 1 - sin x
ge
x (a) 1 - sin x
ga
1 - cos x cos x
(d)
tan x 1 - sin x
C
en
3. The number 916238457 is an example of nine-digit number which contains each of the digit 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. Number of such numbers are (a) 2268 (b) 2520 (c) 2975 (d) 1560 4. Let z1, z2, z3 be complex numbers such that z1 + z2 + z3 = 0 and | z1 | = | z2 | = | z3 | = 1. Then z12 + z22 + z32 is (a) greater than zero (b) equal to 3 (c) equal to zero (d) equal to 1
p -1 p -1 (d) 2p 4p 6. If x, y Œ R and satisfy the equation xy(x2 – y2) = x2 + y2 where x π 0, then the minimum possible value of x2 + y2 is (a) 1 (b) 2 (c) 4 (d) 8
(c)
Ï[ x]2 + sin[ x] for [ x] π 0 Ô 7. If f (x) = Ì [ x] Ô 0 for [ x] = 0 Ó where [x] denotes the greatest integer less than or equal to x, then lim f (x) equals: xÆ0 (a) 1 (b) 0 (c) –1 (d) None of these 8. Matrices of order 3 ¥ 3 are formed by using the elements of the set A = {–3, –2, –1, 0, 1, 2, 3}. Then the
Appendix B: Mock Tests B.9
1 + ln(cos1) 2 1 + ln(sin 1) (b) 2 1 - ln(sin 1 + cos1) (c) 2 1 + ln(sin 1 + cos1) (d) 2 2 10. If f (x, y) = x + y2 Р2xy (x, y ΠR) and the matrix A is given by
(a)
(a) a = –2 (c) b = –1
(b) a = 2 (d) b = 1
x3 - 6 x 2 + 11x - 6 a + x3 + x 2 - 10 x + 8 30 = 0 does not have real solution is (a) –10 (b) 12 (c) 5 (d) –30 15. The value of a for which
Section C: Integer Value Correct Type
ng
È f ( x1 , y1 ) f ( x1 , y2 ) f ( x1 , y3 ) ˘ A = ÍÍ f ( x2 , y1 ) f ( x2 , y2 ) f ( x2 , y3 ) ˙˙ such that ÍÎ f ( x3 , y1 ) f ( x3 , y2 ) f ( x3 , y3 ) ˙˚ trace (A) = 0, then (a) det(A) ≥ 0 (b) det(A) £ 0 (c) det(A) = 0 (d) None of these
.L td .
to
Pv t
ˆ 1ˆ Ê n 1 Ê 9. Lim Á tan -1 ˜ Á  has the value equal n Æ• Ë ¯ n Ë k =1 1 + tan ( k n) ˜¯
a
di
È -1 4 ˘ (a) a Œ Í , ˙ 3˚ Î 3 (b) Largest possible value of a is 3 . (c) Number of possible integral values of a is 3. (d) Sum of all possible values of a is 0. 13. If planes r (i + j + k ) = q1 , r (i + 2a j + k ) = q2 and r (ai + a 2 j + k ) = q3 intersect in a line, then the value of a is (a) 1/4 (b) 1/2 (c) 1 (d) 2 È 1 2 2˘ 1Í ˙ 14. If A = Í 2 1 -2˙ and AT = A-1 , then 3 ÍÎa 2 b ˙˚
In
probability that matrix is either symmetric or skew symmetric is 1 1 1 1 1 (a) 6 + 3 (b) 9 + 3 - 6 7 7 7 7 7 1 1 1 1 1 (c) 3 + 9 (d) 3 + 6 - 9 7 7 7 7 7
ar ni
16. If f : (0, •) Æ (0, •) satisfies f (x f (y)) = x2ya(a Œ R), then find a. 17. If a, b, and c are real numbers such that a2 + 2b = 7, b2 + 4c = –7, and c2 + 6a = –14, then find the value of (a2 + b2 + c2)/2. 18. Let an = 16, 4, 1, ... be a geometric sequence. Define Pn as the product of the first n terms. Then find the
Le
Section B: One or More Options Correct Type
11. If f : R Æ R be a continuous function such that f (x)
Ú 2t f (t )dt , then which of the following does not 1
ge
=
x
en
ga
hold good? 2 (a) f (p) = ep (b) f (1) = e (c) f (0) = 1 (d) f (2) = 2 12. Let (a – 1)(x2 + 3x + 1)2 – (a + 1)(x4 – x2 + 1) £ 0 " x Œ R. Then which of the following is/are correct?
 value of
C
Type f ( x) 1. If g(x) = , where f (x) is a ( x - a ) ( x - b) ( x - c ) polynomial of degree < 3, then
n =1
n
pn
. 4 19. Find the distance of the point P(3, 8, 2) from the line 1 1 1 ( x - 1) = ( y - 3) = ( z - 2) measured parallel 2 4 3 to the plane 3x + 2y Р2z + 15 = 0. 20. Consider the equation x2 + 2x Рn = 0, where n ΠN and n Π[5, 100]. Find the total number of different values of n so that the given equation has integral roots.
Paper 2
Section A: One or More options correct
•
(a)
Ú
1 a g ( x) dx = 1 b 1 c
f (a) log x - a f (b) log x - b f (c) log x - c 1 a a2
∏ 1 b b2 + k 1 c
c2
B.10 Mathematics
(b)
1 a a2 1 a f (a ) ( x - a ) -2 dg ( x) -2 = 1 b f (b) ( x - b) ∏ 1 b b2 (c) dx 1 c c2 1 c f (c) ( x - c) -2
(d)
Section B: Paragraph Type
∏ b
2
b 1 +k
c
2
c 1
9. The largest angle of triangle is (a) 70° (b) 100° (c) 110° (d) 130° 10. The triangle is (a) acute-angled (b) right-angled (c) isosceles (d) scalene Paragraph for Questions 11 and 12 b cos x b + sin x = Let , b ŒR 2 2 cos 2 x - 1 (cos x - 3 sin 2 x) tan x
(a) e = 386 /12
(b) e = 386 /13
(d) LR = 121/3 dy y 4. y = ae + b is a solution of = . Then dx x 2 (a) a Œ R (b) b = 0 (c) b = 1 (d) A takes finite number of values 5. If AB = A and BA = B, then (a) A2B = A2 (b) B2A = B2 (c) ABA = A (d) BAB = B 6. If the parabola y = (x2 + 4(b – c)x + 4a2)/4 touches the x-axis, then the line ax + by + c = 0 always passes through the fixed point/points (a) (1, 1) (b) (–1, –1) (c) (–1, 1) (d) (1, –1)
C
en
ga
ge
–1/x
Le
(c) LR = 121/6
ar ni
ng
2. If A and B are two events such that P(A) = 3/4 and P(B) = 5/8, then (a) P(A » B) ≥ 3/4 (b) P(A¢ « B) £ 1/4 (c) 3/8 £ P(A « B) £ 5/8 (d) 1/8 £ P(A « B¢) £ 3/8 3. If (5, 12) and (24, 77) are the focii of a hyperbola passing through the origin, then
Êa -g ˆ Ê 3a ˆ 3 Êa - bˆ + sin Á ˜ = . + sin Á sin Á Ë 2 ˜¯ Ë 2¯ 2 Ë 2 ˜¯
a
a2 a 1
Paragraph for Questions 9 and 10 Let angles a, b, g of a triangle satisfy the relation,
di
Ú
1 a f (a ) log x - a g ( x) dx = 1 b f (b) log x - b 1 c f (c) log x - c
.L td .
dg ( x) = 1 b f (b) ( x - b) -2 ∏ b 2 b 1 dx c2 c 1 1 c f (c) ( x - c) -2
(c) f (x) + p/4 < tan–1 x, "x ≥ 1 (d) f (x) + p/4 > tan–1 x, "x ≥ 1 8. If 10! = 2p3q5r7s, then (a) 2q = p (b) pqrs = 64 (c) number of divisors of 10! is 270. (d) number of ways of putting 10! as a product of two natural numbers is 135.
Pv t
a2 a 1
In
1 a f (a ) ( x - a ) -2
7. Let f (x) =
Ú
x
1
3t dt , x > 0, then 1 + t2
(a) for 0 < a < b, f (a) < f (b) (b) for 0 < a < b, f (a) > f (b)
11. Equation has solutions if
1ˆ Ï 1¸ Ê (a) b Œ Á - •, ˜ - Ì-1, 0, ˝ Ë 2¯ Ó 3˛
Ï (b) b Œ ( - •, 1) - Ì-1, 0, Ó
1¸ ˝ 3˛
1¸ Ï (c) b Œ R – Ì-1, 0, ˝ 3˛ Ó (d) None of these 12. For any value of b for which the equation has solution, then the number of solutions for x Œ(0, 2p) are always (a) infinite (b) depends upon the value of b (c) 4 (d) none of these Paragraph for Questions 13 and 14 The base of pyramid is rectangular, three of its vertices of the base are A(2, 2, –1), B(3, 1, 2) and C(1,1,1) (points
Appendix B: Mock Tests B.11 Column I
17. Match the following: Column I
Column II
(p) x = a
(b) The locus of P point is, if tangents from P to the parabola y2 = 4ax intersects the co-ordinate axes in concyclic points
(q) x + y = 3a
Le
(a) Common normals to the parabola y2 = 4ax and x2 = 4ay is/are
ge
ga
(d) The chord of contact of a point P w.r.t. the parabola y2 + 4ax = 0 subtends right angle at the vertex. The locus of P is
(s) x = 4a
en C
Pv t
(d) If a , b , c are unit vectors 3 (s) a.b + b .c + c .a = and a + b + c = 0, then 2
19. Match the following:
Column I
Column II
(p) 2
a
(a) If Langrange’s mean value theorem is applicable in [–2, 2] for the function
ÏÔmx + c , x < 0 then the value of x , x≥0 ÓÔ e
f (x) = Ì
m + 3c is
di
In
(b) If the ends of latus rectum of parabola are (q) 8 (2, 6) and (6, 2) and the equation of the possible directrix is x + y = li where i = 1, 2. Then the value of l1 + l2 is (c) The maximum value of f(x) = 2x3 – 3x2 – 12x (r) 4 in [–2, 5/2] is
1 1 Ê + Á (2n - 12 ) ( 4n - 2 2 ) (d) If lim Á n Æ• Á L+ ÁË
equal to k.
ˆ ˜ ˜ is 1˜ ˜ n¯ +
Column I
Column II
(a) If 1, a, a2, L a19 are 20th roots of unity, then 1 + a10 + a20 + L + a190 is equal to
Ê 1ˆ f(x) = x2e–2x, x > 0, then Á ˜ Ë 2¯
(p) a.c = b .c = c .a
(d) Number of points on
(Contd.)
(p) 4
(b) Number of integral values of x satisfying the (q) 2 inequality (log2x)2 + log2 0.03125x + 3 < 0
Column II
(s) 16
p , then k is 4
(c) If l is the maximum value of the function
(c) If a ¥ b = c , and b ¥ c = a , (r) a ¥ b = b ¥ c = c ¥ a then
18. Match the following:
.L td .
three adjacent sides of regular tetrahedron, then
20. Match the following:
(c) The locus of P, if tangents from it to (r) x + 3a = 0 the parabolas y2 = 4a (x + a) and y2 = 8a (x + 2a) are perpendicular is
(a) If the vectors a , b , c form sides BA , CA , AB of DABC, then
(q) a.b = b .c = c .a = 0
ar ni
Section C: Matching list Type
Column I
Column II
(b) If a , b , c are forming
ng
may or may not be in order). Its vertex at the top is Ê -26 -10 ˆ P Á 4, , and fourth vertex of the base is D. Ë 3 3 ˜¯ 13. Coordinates of D are (a) (4, 0, 2) (b) (4, 2, 0) (c) (2, 0, 4) (d) (0, 2, 4) 14. Volume of the pyramid is (in cubic units) (a) 20 (b) 10 (c) 40 (d) 30 Paragraph for Questions 15 and 16 Consider two circles C1 : x2 + y2 = r12 and C2 : x2 + y2 = r22 (r1 > r2). Let A be a fixed point on circle C1 say (r1, 0) and B be a variable point on circle C2. The line BA meets circle C2 again at C. Then 15. The maximum value of BC2 is (a) 4r12 (b) 4r22 2 2 (c) 4r2 + 4r1 (d) None 16. The value of OA2 + OB2 + BC2 is (a) [5r22 – 3r12, 5r22 + r12] (b) [4r22 – 4r12, 4r22] (c) [4r12, 4r22] (d) None of these
(r) 0
ln ( l )
is
x2 – y2 = 1 from 9
which pair of perpendicular tangents can be drawn to parabola y2 = – 12x is
(s) 1
B.12 Mathematics
Solutions of Mock Test 1 Paper 1 Section A
3 3 , m2 = 4 4 Since tangents are drawn on same branch and point P(3, 4) lies in obtuse angle between the asymptotes, therefore
Ú ln (1 + tan q tan (q - x)) dx 0
0 q
ˆ
Ú Ú
ln (1 + tan 2 q ) dx - ln (1 + tan q tan x) dx
x ˜¯
\
dx
q
Ú
0
0
fi I = 2q ln sec q – I fi 2I = 2q ln sec q fi I = q ln sec q Ans (b): We have 1 1 2 - = v u f 2.
Le
1 1 2 dv + 2 du = - 2 df 2 v u f
ge
ga
2 Ê 1 1ˆ 2 fi a Á + ˜ ¥ = 2 df Ë v u¯ f f
[Q du = dv = a]
È 1 1 2˘ - = ˙ ÍQ v u f˚ Î
df Ê 1 1ˆ =aÁ + ˜ Ë u v¯ f
en
q 3 = 2 4
e2 - 1 =
b q 4 = cot = a 2 3
5 3 Ans (c): Let the parabola is y2 = 4ax. a = at1t2, b = a(t1 + t2), Q ∫ (at1t3, a(t1 + t3)), R ∫ (at2t3, a(t2 + t3)) at bt3 Let T ∫ (h, k). Then h = 3 (t1 + t2) = and 2 2 4ah 2k = a(t1 + t2) + 2at3 = b + b
ng
2 Ê 1 1ˆ Ê 1 1ˆ fi a Á + ˜ Á - ˜ = 2 df Ë v u¯ Ë v u¯ f
Ê 1 1ˆ Thus, relative error in f = a Á + ˜ . Ë u v¯ Hence, (b) is the correct answer. 3. Ans (c): f (x) is strictly increasing ⇒ f ¢(x) ≥ 0 fi 2 sec2 x + (2a +1) tan x + (a – 2) ≥ 0, " x Œ R fi 2 tan2 x + (2a +1) tan x + a ≥ 0, " x Œ R fi (2a +1)2 – 8a £ 0 fi (2a –1)2 £ 0
C
5.
1ˆ 2 Ê 1 fi Á 2 - 2 ˜ a = 2 df ¯ Ëv u f
fi
tan
\ e =
ar ni
Ê 2ˆ Ê 1 1ˆ fi d Á - ˜ = d Á ˜ Ë v u¯ Ë f¯ fi -
Ê 1 + tan 2 q ln Á Ë 1 + tan q tan
0 q
fi I =
tan q (tan q - tan x) ˆ dx 1 + tan q tan x ˜¯
a
=
Ê
Ú ln ÁË1 +
In
fi I =
di
q
Pv t
4. Ans (c): m1 = -
q
1. Ans (a): I =
.L td .
fi 2a –1 = 0 fi a = ½
Therefore, the locus is 4ax Р2by + b2 = 0, which is a straight line. 6. Ans (d): | x + y + z | = | x | + | y | + | z | fi x, y, z are of same sign. Thus, x2 Р2 < 0 and sin x < 0 fi x Π(- 2 , 0) 7. Ans (d): A3 = B3 and A2B = B2A (1) Р(2) gives A3 РA2B = B3 РB2A A2(A РB) = РB2(A РB) fi (A2 + B2) (A РB) = 0 det (A2 + B2) (A РB) = 0 det (A2 + B2) . det (A РB) = 0 fi (d) 8. Ans (a): Possible outcomes are 11 2 (C); 11 3 (C); 11 5 (C) p=2
p=3
...(1) ...(2)
p=5
9 1 = 216 24 9. Ans (b): Domain is x = ± 1. However, only x = 1 satisfies. Hence, P(A) =
Appendix B: Mock Tests B.13
Ú g (5 - t ) dt 0
ng
(Q f (t) = g(t – 5))
(Q g is even)
Le
ge
ga
.L td .
1
f(n) =
n
2
1
+
2
(n + 2)
1
+
2
(n + 2) +
Therefore, choice (c) is correct. 12. Ans (a, b): Let 2x + y = 3x – y fi 2y = x fi y = x/2 Put y = x/2 f (x) + f (5x/2) + 5x2/2 = f (5x/2) + 2x2 + 1 \ f (x) = 1 – (x2/2) 13. Ans (c, d): x2 – 4x + a sin a ∫ (x – 2)2 + 4 – a sin a The roots are real if a sin a £ 4. As | sin a | £ 1, the roots are real for all values of a if | a | £ 4. For p £ a £ 2p, sin a £ 0. Even if a ≥ 4, a sin p £ 4 and the roots are real. 14. Ans (a, b): If the coefficients are imaginary, then the imaginary roots may not be conjugate of each other. Hence, Statement (a) is false. The continuous monotonic function may not be differentiable. Hence, Statement (b) is false. Statement (c) is true cos x tan x sin x for all value of x π np = 2 x - 2np - p 2 x - 2np - p
en
ar ni
5
a
16. Ans (2):
0
0
(2)
Section C
5
Ú
2a + b =1 3 Solving (1) and (2), we get a = 3, b = –1. fi
In
from (i)
Ú f (t ) dt
g (t - 5) dt
2a b ◊ 8 + ◊ 16 = 8 3 2
di
from (ii)
5
Further I =
C
Ú
fi
11. Ans (b, c): g(x) = f (x + 5) fi g(–x) = f (–x + 5) fi g(–x) = – f (x – 5) Thus, choice (b) is true.
=
4
A = (ax1/ 2 + bx) dx = 8
0
Section B
\ I =
p (which are not in the domains of both the sides). 2 Hence, Statement (d) is true. 15. Ans (c, d): Since the curve y = ax1/2 + bx passes through the point (1, 2), 2 = a + b (1) By observation, the curve also passes through (0, 0). Therefore, the area enclosed by the curve, x-axis and x = 4 is given by +
Pv t
10. Ans (c): Let 1/r = x so that as r Æ •, x Æ 0. cos ax - cos bx .cos cx Lim x Æ 0 sin bx sin cx . .bc . x 2 bx cx 1 cos ax - cos bx .cos cx = Lim bc x Æ 0 x2 - a sin ax + b sin bx cos cx + c sin cx ◊ cos bx 1 = lim bc x Æ 0 2x 1 (b 2 + c 2 - a 2 ) = 2bx
+ ... 1
2
n + 2n + 1
The terms of the sequence are decreasing and the number of terms are (2n + 2).
2n + 2 2
n + 2n + 1
Now Lim
£ f ( n) £
2n + 2 n2
1ˆ Ê 2n Á 1 + ˜ Ë n¯ =2 = Lim 2 nÆ• 2 1 n + 2n + 1 n 1+ + 2 n n 2 (n + 1)
nƕ
Similarly Lim
2 (n + 1)
nƕ
n
2
= Lim
nƕ
2 (n + 1) =2 n
\ Lim f ( x) = 2 nƕ
17. Ans (1): We have y = ax/(b – x) fi
dy (b - x)a - ax(-1) ab = = dx (b - x) 2 (b - x) 2
ab È dy ˘ = = 2 (given) Í dx ˙ ( b - 1) 2 Î ˚ (1, 1)
(1)
B.14 Mathematics
Section A
.L td .
fi z = z = x as z + z + 1 π 0 ( x π -1/ 2)
Pv t
fi The given equation reduces to xn = xn + x | x |n – 2 + 1 fi x | x |n – 2 = –1 fi x = –1. So, the number of solution is 1. 20. Ans (1):
y-z z-x x- y ,y-z= ,z-x= yz zx xy
x- y=
\ ( x - y )( y - z )( z - x) =
( x - y )( y - z )( z - x) ( xyz ) 2
fi xyz = 1
In
= 5 units
Paper 2
fi ( z - z ) ( z + z + 1) = 0
or –p/6 < x < p/6 or –p/6 + np < x < p, n Œ Z (generalizing) Then the common values are np + p/6 < x < np + p/4. Ans (a, b, c, d): (a) P(E1) = 1 – P(RRR) 1 2 3 = 1 - ¥ ¥ = 0.9 3 4 5 2 1 2 (b) P(E2) = 3P(BRR) = 3 ¥ ¥ ¥ = 0.2 3 4 5 (c) P(E3) = P(RRR/(RRR » BBB)) P ( RRR ) = P ( RRR ) + P ( BBB ) 4.
ng
fi z 2 + z = z 2 + z
a
(b - 1)b = 2 fi b = 2 fi a = 2 – 1 = 1 (b - 1) 2 Hence, a = 1, b = 2 18. Ans (5): AB = b - a = - 2i - 3 j - 6k Equation of the plane passing through B and perpen dicular to AB is (r - OB ) ◊ AB = 0 fi r ◊ (2i + 3 j + 6k) + 28 = 0 Hence, the required distance from c = – i + j + k is c ◊ (2i + 3 j + 6k ) + 28 - 2 + 3 + 6 + 28 = 2i + 3 j + 6k 7
19. Ans (1): The given equation is | z |n = (z2 + z) | z |n – 2 + 1 fi z2 + z is real
di
Since the curve passes through the point (1, 1), therefore, 1 = a/(b – 1) fi a = b – 1 (2) On putting a = b – 1 in Eq. (1), we get
Le
ar ni
1. Ans (b, c): (1 – y)m (1 + y)n = (1 – mC1 y + mC2 y2 – L) (1 + nC1 y + nC2 y2 + L) Ï m (m - 1) n (n - 1) ¸ + - mn ˝ y 2 + L = 1 + (n - m) + Ì 2 2 Ó ˛ Given: a1 = 10 fi a1 = n – m = 10 (1) m 2 + n 2 - m - n - 2mn = 10 2 2 fi (m – n) – (m + n) = 20 fi m + n = 80 (2) 2. Ans (c): Since P(x) divides into both of them, hence P(x) also divides (3x4 + 4x2 + 28x + 5) – 3(x4 + 6x2 + 25) = – 14x2 + 28x – 70 = – 14(x2 – 2x + 5 which is a quadratic. Hence, P(x) = x2 – 2x + 5 fi P(1) = 4 3. Ans (a, d): tan x – tan2 ¥ > 0 fi tan x(tan x – 1) < 0 fi 0 < tan x < 1 or 0 < x < p/4 or np < x < np + p/4, n Œ Z (generalizing) 1 sin x < 2 1 1 fi - < sin x < 2 2 a2 =
C
en
ga
ge
=
=
0.1 2 3 4 0.1 + ¥ ¥ 3 4 5 0.1 = 0.2 0.1 + 0.4
(d) P(E4) = 1 – P(BBB) = 1 -
2 = 0.6 5
bc ca ab 5. Ans (a, b, c): We have ca ab bc = 0 ab bc ca fi (ab)3 + (bc)3 + (ca)3 – 3(ab)(bc)(ca) = 0 or ab + bc + ca(abw + bcw2 + ca) (abw2 + bcw + ca) = 0
Appendix B: Mock Tests B.15 Ans 11. (a), 12. (b): Let r = xi + y j + zk
1 4
8¸ Ô ˝ 12 cases 2 Ô˛
1 2 4
1 2 4
7¸ Ô 5 ˝ 9 cases 1 Ô˛
1 2 3
1 2 3
1 2 3
1 2 3
6¸ Ô 4 ˝ 9 cases 2 Ô˛ 5¸ Ô 3˝ 9 cases 1 Ô˛
1
1
4} 3 cases
1 2
1 2
3¸ ˝ 6 cases 1˛
di
In
ar ni
Le
ge
Section B
C
en
ga
Ans 9. (b), 10. (c): Let the variable line AC be x + ay – 1 = 0 which passes through the point A(1, 0). Let the foot of perpendicular on variable line from B(3, 4) is C(h, k). Now BC ^ AC \ (slope of BC) ¥ (slope of AC) = –1 Ê k - 4ˆ Ê k - 0ˆ or Á = -1 Ë h - 3 ˜¯ ÁË h - 1 ˜¯ or x2 + y2 – 4x – 4y + 3 = 0 or (x – 2)2 + (y – 2)2 = 5 Thus, the length of tangent from origin is
S1 = 3 .
1 4
1
a
5
9¸ Ô ˝ 15 cases 1 Ô˛
1 5
Pv t
.L td .
r .(i + j + k ) = 10 \ x + y + z = 10 fi x ≥ 1, y ≥ 1, z ≥ 1 Therefore, the number of latice points = 9C2 x + y + z £ 11 For equilateral latice points, x = y = z = 1, 2, 3 So, three cases are possible. For isosceles lattice point, x = y π z
ng
or ab + bc + ca = 0, abw + bcw2 + ca = 0, abw2 + bcw + ca = 0 1 1 1 1 1 1 + + = 0, + + = 0, or a b c cw a bw 2 1 1 1 + + =0 c aw bw 2 6. Ans (a, b, c, d): Odd + Odd – Even – Even = Even So the given number is divisible by 2. Also 2007201 – 1982201 is divisible by 2007 – 1982 = 25 2044201 – 2019201 is divisible by 2044 – 2019 = 25 Hence, the given number is divisible by 25. Also 2007201 – 2044201 is divisible by 2007 – 2044 = –37 and 2019201 – 1982201 is divisible by 2019 – 1982 = 37 Hence, the Given number is divisible by 37. So the given number is divisible by 2 ¥ 37, 2 ¥ 25, 2 ¥ 25 ¥ 37 = 74, 50, 1850. 2007201 + 2019201 is divisible by 2007 + 2019 = 4026 1982201 + 2044201 is divisible by 1982 + 2044 = 4026 So, the given number is divisible by 4026 = 2 ¥ 2013 [\ an + bn has a factor a + b if n is odd] 7. Ans (a, b, c): sin2 x – a sin x + b = 0 has only one solution in (0, p). So, sin x = 1 gives one solution and sin x = a gives other solution such that a > 1 or a £ 0. Hence, (sin x – 1)(sin x – a) is the same equation as sin2 x – a sin x + b = 0 fi 1 + a = a and a = b or 1 + b = a and b > 1 or b £ 0 or b Œ (–•, 0) » (1, •) and a Œ (–•, 1) » (2, •) 8. Ans (b, d): tan (a + b) = 15/8 and tan g = 8/15 \ a + b + g = p/2 fi (b) and (d)
1 1 2} 3 cases Total number of isosceles lattice points which are not equilateral points = 15 + 12 + 9 + 9 + 9 + 3 + 6 + 3 = 66 Therefore, total number of isosceles latice point = 66 + 3 = 69 3 1 Therefore, required probability, P = = 69 23 Ans 13. (c), 14. (b): x + sin x g(x) = increasing function of x 2 È p˘ range Í0, ˙ Î 2˚
B.16 Mathematics x + sin x , x Œ [0, p] 2 p £ t £ 2p, then f (t) + f (2p – t) = p 2p - t + sin (2p - t ) f (t) + =p 2 t sin t f (t) + p – =p 2 2 t + sin t f (t) = 2 x + sin x f (x) = p £ x £ 2p 2 x + sin x f (x) = for 0 £ x < 2p 2 f (x) = f (4p – x) for x Œ [2p, 4p] f (x) is symmetric about x = 2p.
16. Centre of circumcircle is (10, 8) and radius 2 2 .
\ f (x) =
.L td .
17. Ans: (a Æ q); (b Æ s); (c Æ r); (d Æ p) (a) lim f (g(x)) = f (2–) = 2 x Æ 1-
lim f (g(x)) = f (1–) = 1 x Æ 1+
Pv t
Section C
Therefore, limit does not exist.
(b) lim f (x) = 2 fi
(c) lim f (x) = 0 and lim g(x) = finite quantity
xÆ2 xÆ0
a
Therefore, equation of circle is x2 + y2 – 20x – 16y + 156 = 0
\
lim
lim
xÆ2
3 f ( x) - 2 = 2
xÆ0
Ê f ( x) ˆ + f ( x) g ( x)˜ = 0 g ( x) ¯
x Æ 0Á Ë
di
3 f ( x) - g ( x) 3(1) - 1 2 = = =1 f ( x) + g ( x) 1+1 2 18. Ans: (a Æ r), (b Æ r), (c Æ q), (d Æ s) (d) lim
In
2p
4p
z2 - 1 z 2 - 1 = fi z - z = 0, z + z = 0 z2 + 1 z 2 + 1 y = 0, x = 0 Locus of z is portion of pair of lines xy = 0.
ar ni
O
Ê z 2 - 1ˆ (a) arg Á 2 ˜ =0;zπ±i Ë z + 1¯
ng
x Æ 1+
Fig. B.2
Le
a = 2p – 0 = 2p from graph b = a Ans 15. (a), 16. (d): From the graph (Fig. B.3), y = 6 and x = 8 are the common tangents.
C(8,10) A(8,6)
en
ga
ge
y
B(12,6) x
Fig. B.3
C
15. Area of DABC = 8 sq. unit
2s = 4 + 4 + 4 2
D 8 2 Radius r = = = = 2 2 -1 s 4 + 4 2 1+ 2 Co-ordinate of centre are (7 + 2 2 , 5 + 2 2 )
È ˘ Ê z 2 - 1ˆ ÍQ Á 2 ˜ > 0˙ Ë z + 1¯ ÍÎ ˙˚ (b) Given || z – cos–1 cos 12 | – | z – sin–1 sin 12 || = 8 (p – 3) Since, | cos–1 cos 12 – sin–1 sin 12 | = 8 (p – 3) Therefore, locus of z is the portion of a line joining z1 and z2 except the segment between z1 and z2. (c) z2 – i | z1 |2 = k2 – k1 x2 – y2 + 2 ixy – il1 = l2 l x2 – y2 = l2 and xy = 1 2 The locus of z is the point of intersection of hyperbola. (d) Given, 1 1 p + z + cos -1 =1 3 3 2 1 1 p Since 1 + sin -1 + cos -1 - =1 3 3 2
z - 1 - sin -1
Appendix B: Mock Tests B.17 P(A/Ei) = 0 for i = 0, 1, 2, 3, 4
P(A/Ei) =
i
C5 for i ≥ 5 C5
14 14
.L td .
a b c fi = = 1/ 2 1/ 2 ( 3 + 1)/ 2 2 a b c fi = = = k (say) 2 2 3 +1
 P( E ) P( A / E )
(a) P(A) =
i
i
i=0
=
1 15
=
1 15
1 5 ( C5 + 6C5 + L + 14C5) C5
14 15
C6 1 = C5 6
Pv t
| z – z1 | + | z – z2 | = | z1 + z2 | Thus, the locus of z is line segment joining z1 and z2. 19. Ans: (a Æ q), (b Æ p), (c Æ s), (d Æ r) Angles of the triangle are 30º, 45°, 105°. a b c Then = = sin 30∞ sin 45∞ sin 105∞
14
11
C5 3 = C5 13
Smallest side, a = 2 k and largest side, c = ( 3 + 1) k
(b) Clearly P(A/E11) =
If a = 2, k = 2 , c = 6 + 2
(c) From Baye’s theorem,
3 -1 1 = ¥ 2 k ¥ ( 3+ 1) k sin 30∞ 4 2
3 -1 ( 3 - 1) 2 3 -1 = fi k= 4 2 2 ( 3 + 1) and hence c = 1 20. Ans: (a Æ q); (b Æ p); (c Æ r); (d Æ q) Let Ei denotes the event that the bag contains i black and (14 – i) white balls (i = 0, 1, 2, ... 14) and A denotes the event that five balls drawn are all black. Then 1 P(Ei) = (i = 0, 1, ... 14) 15
Le
ge
ga en
Paper 1 50
1. Ans (b): f (x) =
’ ( x - n)
14
 n (51 - n) ln ( x - n) n =1
 P( E ) P( B / E ) i
i
i=0
=
1 1 3 11 . [ C3. C2 + 4C3.10C2 + L + 12C3.2C2] 15 14 C5
=
5005 1 = 14 15. C5 6
Differentiating both sides with respect to x, we get
f ¢ ( x) = f ( x)
f ¢ (51) = f (51)
n ( 51 - n )
n =1
ln f (x) =
C3 14 - i C2 for i = 3, 4, ..., 12 14 C5
\ P(B) =
50
i
Solutions of Mock Test 2
Section A
C
P(B/Ei) =
ar ni
fi k 2 =
a
ng
3 -1 1 , then from D = bc sin A, we have 4 2
If D =
di
If s = 3 + 2 + 3 , k = 2, c = 2 + 2 3
1 3 . P( E11 ) P( A / E11 ) 15 13 6 P(A/E11) = = = 1 P( A) 65 6 (d) Let B denotes the probability of 3 black and 2 white balls, then P(B/Ei) = 0 if i = 0, 1, 2 or 13, 14
In
If a = 2 , k = 1, c = 3 + 1
14
50
n (51 - n) x-n n =1
 50
n (51 - n) 51 - n n =1
Â
=1 + 2 + 3 + L + 50 = 1275
B.18 Mathematics
2. Ans (c): y2 = 4ax, 4a = 2p > 0 and t1t2 = –1. Ratio =
4a 2t1t2 = – 4 a 2t12t22
100
È Ê 1 ˆ˘ Í1 + Á x3 + 3 ˜ ˙ Ë x ¯˚ Î
3. Ans (a):
2
1 1 Let x = and y = k +1 k
=
100
1 (k + 1) 2
(k + 1) 2 - 1 k +1
Tk is in the form of
=
k ( k + 2) k +1
)
(
Ê 1 ˆ Ê 1ˆ Tk = cos -1 Á - cos -1 Á ˜ Ë k¯ Ë k + 1˜¯ Substituting k = 2, 3, 4, ...
ar ni
Ê 1 ˆ Ê 1ˆ Sum = Lim cos -1 Á - cos -1 Á ˜ nÆ• Ë 2¯ Ë n + 1¯˜ Ê 1ˆ = cos -1 (0) - cos -1 Á ˜ Ë 2¯
Sum =
-1
p p p - = 2 3 6
ge
Le
ng
cos -1 xy + 1 - x 2 . 1 - y 2 = cos–1(y) – cos–1(x) (∵ y < x)
Ê 1ˆ + L + B100 Á 3 ˜ Ëx ¯ All other terms obtained by the combination of x3 and 1/x3 well get converted into a term involving x3 or 1/x3 and hence it will be present among above terms. So number of terms = 1 + 100 + 100 = 201 7. Ans (b): f(x) = ax2 + (a – 2) x – 2 = (ax – 2) (x + 1) f (0) = – 2 and f (–1) = 0 If a is negative, then the expression becomes negative for infinite values of x. Therefore, it must be positive. Expression to be negative for exactly two integral values of x
a
1 - y2 = 1 -
Pv t
(k - 1) k (k + 1)(k + 2) ˆ ˜ k (k + 1) ¯
100
1ˆ 1ˆ Ê Ê + 100C2 Á x3 + 3 ˜ + L + 100C100 Á x3 + 3 ˜ Ë Ë x ¯ x ¯ 1 = (1 + r) + A1x3 + A2x6 + L + A100(x3)100 + B1 3 x
di
Ê1 1 Tk = cos -1 Á . + Ë k (k + 1)
1ˆ Ê = 1 + 100C1 Á x3 + 3 ˜ Ë x ¯
In
.L td .
5. Ans (c): From LMVT, there exists atleast (n – 1) point where f ¢(x) = m. fi $ atleast (n – 2) points where f ≤(x) = 0 (using Rolle’s theorem) 6. Ans (d):
f (51) 1 = f ¢ (51) 1275
or
ga
en C
Ê 3p ˆ Ê 3p ˆ , p˜ » " x Œ Á 0, ¯ Ë 4 ˜¯ ÁË 4 Thus, equation (i) holds, if cot x + 1 > 0
1
2 a
2
Fig. B.4
2
4. Ans (d): g(x) = f (cot x + 2 cot x + 2) fi g¢(x) = f ¢(cot2 x + 2 cot x + 2) ◊ {–2 cot x cosec2 x – 2 cosec2 x} for g(x) to be decreasing, g¢(x) < 0 fi f ¢{(cot x + 1)2 + 1} ◊ (–2 cosec2 x) (cot x + 1) < 0 fi f ¢{(cot x + 1)2 + 1} ◊ (cot x + 1) > 0 …(i) {as f ≤(x) > 0 fi f ¢(x) is increasing, then f ≤(cot x + 1)2 + 1} > f ¢(1) = 0
O
2 £ 2 fi a ≥1 a 2 and > 1 fi a < 2 a \ a Œ [1, 2) Ans (a): f (x) = | sin a x | + | cos a x | p p Period of f (x) = (given) = 2a 8
So 8.
fi a = ± 4 9. Ans (d): Given integral
Ê 3p ˆ fi cot x > -1 " x Œ Á 0, Ë 4 ˜¯
=
Ú
xf ¢ ( x) - 2 f ( x) dx = x 2 f ( x)
Ú
( x 2 f ¢ ( x) - 2 xf ( x))/ x 4 f ( x)/ x 2
dx
Appendix B: Mock Tests B.19
10. Ans (a): In 2nd quadrant, sin x < cos x is false as sin x is positive and cos x in negative. In 4th quadrant, cos x < tan x is false as cos x is positive and tan x in negative.
Section B 11. Ans (b, d):
Ú
x
\ Tr + 1 = 8Cr x
Ú
en
Ïe x - 1 if x Œ (0, 1) ÔÔ x -1 ⇒ f (x) = Ì(e - 1) + (e - 1) if x Œ (1, 2) Ô x-2 - 1) if x Œ (2, 3) ÔÓ2 (e - 1) + (e
C
Clearly f (x) is continuous " x > 0 but not differentiable " x > N. Also f (2) = 2(e – 1) = 0 = 2(e – 1) 1 ˆ Ê 12. Ans (b, c): Á x1/ 2 + x -1/ 4 ˜ ¯ Ë 2
n
8-r 2
¥
1 -r ¥x 4 2r
1 (4 - 34r ) ¥x r 2 Terms of x with integer power occur when r = 0, 4, 8. Thus, three terms. Hence, (b) and (c) are correct. 13. Ans (a, b, c, d):
= 8Cr ¥
y D(0, d) B(0, b) A(a, 0)
Le
ge
ga
n(n - 1) = (n - 1) 8 fi n = 8 (as n π 1)
O(0, 0)
Ú
Ú
1 , 2
1 1 = 2 ¥ nC1 ¥ 4 2
n(n - 1) =n 8
fi
0
Í et dt if x Œ (0, 1) Í Í0 x Í1 if x Œ (1, 2) So f (x) = Í et dt + et -1 dt Í 0 1 Í Í1 x 2 Í t t -1 e dt + e dt + et - 2 dt if x Œ (2, 3) Í ÍÎ 0 2 1
Ú
C0 + nC2 ¥
fi 1 +
Ú
Èx
Ú
n
1 22
et - [t ] dt = e{t } dt ,
0
Ú
C2 ¥
\
ar ni
We have f (x) =
n
ng
Ê 5p 3p ˆ In 3rd quadrant, i.e., Á , if tan x < cot x, , Ë 4 2 ˜¯ fi tan2x < 1, which is false. Hence, (a) can be correct. Ê pˆ Now sin x < cos x is true in Á 0, ˜ and tan x < cot x Ë 4¯ is also true. Further, cos x < tan x as tan x = (sin x)/(cos x) and cos x < 1.
x
Coefficient of the first three terms are nC0, n C1 ¥
.L td .
2 f ( x) 1 +c dt = 2t1/ 2 + c = x t
-r 1 ¥x4 r 2
a
Ú
¥
di
=
n-r 2
In
Tr + 1 = nCr x
Pv t
f ( x) dt x 2 f ¢ ( x) - 2 x f ( x) fi = 2 dx x x4 Therefore, the required integral is Let t =
C(c, 0)
x
Fig. B.5
Points A, B, C and D are concyclic, then ac = bd. The co-ordinates of the points of intersection of lines are
Ê ac (b - d ) bd (c - a ) ˆ ÁË bc - ad , bc - ad ˜¯
Let the co-ordinates of the point of intersection be (h, k). Then ac (b - d ) bd (c - a ) h= ,k = bc - ad bc - ad
Given c2 + a2 = b2 + d2 (∵ ac = bd) fi (c – a)2 = (b – d)2 or (c – a) = ± (b – d) Then the locus of the points of intersection is y = ± x.
B.20 Mathematics 14. Ans (a, b, d): p = 5C4 ◊ 2C1 = 10 q = 5C2(2C1)3 = 80 r = 5C0(2C1)5 = 32 fi 2q = 5r, 8p = q, and 2(p + r) > q 15. Ans (a, c, d): From Fig. B.6,
[ x] + [- x]
dy ˆ Ê -y+ x Á dx , 0˜ A=Á ˜ dy Á ˜ Ë ¯ dx
.L td .
Ê x2 - 2 x ˆ sin -1 Á is defined for 3 ˜¯ Ë
x2 - 2 x £ 1 and [ x] + [- x] defined only 3 for integral values of x. fi x = – 1, 0, 1, 2, 3 Therefore, total number of element in domain is 5. Ans: Since the produce matrix is 3 ¥ 3 matrix and the premultiplier of A is a 3 ¥ 2 matrix, A is a 2 ¥ 3 matrix.
-1 £ 19.
a
P(x, y)
Pv t
Ê x2 - 2 x ˆ + 18. Ans (5): f (x) = sin -1 Á 3 ˜¯ Ë
di
O
È1 m n ˘ Let A = Í ˙ , then the given equation becomes Îx y z˚
A
B(x, 0)
È 2 -1˘ È-1 - 8 -10˘ Í ˙ È1 m n ˘ Í ˙ Í 1 0 ˙ Í x y z ˙ = Í 1 - 2 - 5˙ ˚ Í 9 22 ÍÎ- 3 4˙˚ Î 15˙˚ Î
In
Fig. B.6
As per the given condition, dy dy 2x =-y+ x dx dx dy fi x = -y dx dx dy fi =x y fi ln x = –ln y + ln c fi xy = c Also, f (1) = 1 fi c = 1 fi xy = 1
Section C
Le
ar ni
ng
3m - y 2n - z ˘ È 2l - x Í ˙ fi Í l m x ˙ ÍÎ- 3l + 4 x - 3m + 4 y - 3n + 4 z ˙˚
2p
16. Ans (4): I = Lim
2p
Ú
=
0
t
dx
Ê tˆ t ˆ Ê 2 cos Á x + ˜ ¥ sin ˜ Á Ë 2¯ 2 ˜ Á Lim dt t ÁtÆ0 ˜ Á ˜ Ë ¯
ga
sin ( x + t ) - sin x
ge
tÆ0
Ú
È-1 - 8 -10˘ Í ˙ = Í 1 - 2 - 5˙ ÍÎ 9 22 15˙˚ fi 2l – x = –1, 2m – y = –8, 2n – z = –10, l = 1, m = –2, n = –5 fi x = 3, y = 4, z = 0, 1 = 1, m = –2, n = –5
en
0
2p
=
Ú cos x
dx = 4
0
C
17. Ans (9): a = 1 – b – c fi 1 + a = (1 – b) + (1 – c) > 2 Similarly, 1 + b > 2 and 1 + c > 2
(1 - b)(1 - c)
(1 - c)(1 - a )
(1 - a )(1 - b)
Required expression > 8.
È 1 m n ˘ È 1 - 2 - 5˘ fi A = Í ˙ ˙=Í 4 0˚ Î x y z ˚ Î3 20. Ans (9): g(x) =
1 tan–1 | x | + 1 fi sgn (g(x)) = 1 2
\ sin23 x – cos22 x = 1 fi sin23 x = 1 + cos22 x, which is possible if sin x = 1 and cos x = 0. p fi x = 2np + 2 p Hence, -10p £ 2np + £ 8p 2 21 15 fi - £ n £ fi –5 £ n £ 3 4 4 Hence, the number of values of x = 9.
Appendix B: Mock Tests B.21
1. Ans (a, c): If p = q, then the equation becomes z = z and it has infinite solution because any z Œ real will satisfy it. If p π q, let p > q, then zp = z q fi | z |p = | z |q fi | z |p (| z |p – q – 1) = 0 fi | z | = 0 or | z | = 1 | z | = 0 fi z = 0 + i0 | z | = 1 fi z = eiq fi e(p + q)qi = 1 fi z = 11/(p + q) Hence, the number of solution is p + q + 1. 2. Ans (a, c, d):
=
Ú (x
Ú
1ˆ Ê ÁË1 + 2 ˜¯ dx x 2 dx x -2 1ˆ Ê 2 ( x3 )2 + 1 ÁË x - 1 + 2 ˜¯ x
Ú
Ú
Le
=
Ú
( x 2 + 1) dx x 2 dx 2 ( x 4 - x 2 + 1) ( x 6 + 1)
In first integral, put x -
1 =t x
ga
ge
1ˆ Ê Therefore, Á1 + 2 ˜ dx = dt and in second integral, Ë x ¯ 3 put x = u. du , then \ x 2 dx = 3 I =
en
dt
Ú1+ t
2
-
C
= tan -1 t -
2 3
du
Ú1+ u
2
2 tan -1 u + c 3
Ê = tan -1 Á x Ë
1ˆ 2 - tan -1 ( x3 ) + c x ˜¯ 3
1 and g ( x) = x3 x Both the functions are odd. Here f ( x) = x -
4
3
Ê h + at 2 ˆ Any point on this tangent is Á h, . t ˜¯ Ë Chord of contact of this point with respect to the circle Ê h + at 2 ˆ 2 x2 + y2 = a2 is hx + Á ˜ y=a Ë t ¯
yˆ Ê or (aty – a2) + h Á x + ˜ = 0 Ë t¯ which is a family of straight lines passing through y the point of intersection of ty – a = 0 and x + = 0 t Ê a aˆ So, the fixed point is Á - 2 , ˜ . Therefore, Ë t t¯ a a , y2 = t t2 Clearly, x1x2 = – a2, y1 y2 = 2a2 x y Also, 1 = - t 4 , 1 = 2t 2 x2 y2
ar ni
=
( x 2 + 1) 2 - 2 x 2 dx + 1)( x 4 - x 2 + 1)
2
2
3. Ans (b, c, d): Let (x1, y1) ∫ (at2, 2at) Tangent at this point is ty = x + at2.
ng
Ú
Ú 1ˆ 1 3 Ê 1 = Á Ú Ë x - x ˜¯ dx = - x + x + c
In
( x 4 + 1) dx ( x 6 + 1)
Let I =
Ú
.L td .
p
Pv t
p
a
Section A
Also, g(x) is monotonic. f ( x) x - 1/ x Also dx = dx g ( x) x3
di
Paper 2
x2 = -
2
x Ê y ˆ fi 4 1 + Á 1 ˜ = 0 x2 Ë y2 ¯ 4. Ans (a, b, c): Here, let P be (x, y, z). Then, x = r sin q cos f, y = r sin q sin f, z = r cos q fi 1 = r sin q cos f, 2 = r sin q sin f, 3 = r cos q (i) fi 12 + 22 + 32 = r2 sin2q cos2f + r2 sin2f sin2q + r2 cos2q = r2 sin2q (cos2f + sin2f) + r2 cos2q = r2 sin2q + r2 cos2q = r2 fi r = ± 14 fi From (i), we have 1 2 , sin q sin f = sin q cos f = + , 14 14 3 cos q = 14 (neglecting –ve sign assuming acute angles)
sin q sin f 2 sin q 5 = and tan q = = 3 sin q cos f 1 cos q
fi tan f = 2 and tan q =
5 /3
B.22 Mathematics fi {1 – P(X < 2)} ≥ 0.99 fi 1 – {P(X = 0) + P(X = 1)} ≥ 0.99
Z
Fig. B.7
g(0) =
f (1, t) = t (1 – 1) = 0 for t < 1; so g(1) = 0
ge
ga
en
r
C
Ê 1ˆ Ê 1ˆ Then P(X = r) = n Cr Á ˜ Á ˜ Ë 2¯ Ë 2¯ n-r
n
n-r
n
Ê 1ˆ = n Cr Á ˜ ; r Ë 2¯
Ê 1ˆ Ê 1ˆ Ê 1ˆ n Cr Á ˜ Á ˜ == nCr Á ˜ ; r = 0, 1, 2, ..., n Ë 2¯ Ë 2¯ Ë 2¯ We should have P(X ≥ 2) ≥ 0.99
Ú
x
f (t ) t ( x - 1) dt +
= ( x - 1)
Ú
x
0
Ú
t f (t ) dt + x
1
1
0
x
Ú t f (t ) dt - Ú
1 x
Ú
f (t ) x (t - 1) dt
1 x
f (t ) (t - 1) dt
f (t ) dt
g≤(x) = f (x)
Section B
\ b = 60° 6. Ans (a, b, d): We have P = probability that the bomb strikes the target = 1/2. Let n be the number of bombs which should be dropped to ensure 99% chance or better of completely destroying the target. Then the probability that out of n bombs at least two strike the target is greater than 0.99. Let X denote the number of bombs striking the target.
r
f (t )◊ 0 ◊ dt = 0
0
0
\ g¢(x) =
a ◊c - 9/ 2 1 cos y = = =a c 3.3 2 a ◊d 9/ 2 1 cos b = = = 3.3 2 a d
Le
Also g(x) =
ar ni
\ f = 60°
Ú
1
ng
5. Ans (b, d): 3 3 Let a = - (i - 3 j ), b = (i - 3 j ) 2 2 3 3 c = - (i + 3 j ), d = (i + 3 j ) 2 2 AB = 3i = a (say) Clearly, a = b = c = d = 3 If a makes angle q and f, y and b with a , b , c and d , respectively, then a ◊a - 9/ 2 1 cos q = = =a a 3.3 2 a ◊b 9/ 2 1 cos f = = = 3.3 2 a b
Pv t
M
x
a
f
di
y
O
1¸ Ï fi 1 - Ì(1 + n) n ˝ ≥ 0.99 2 ˛ Ó 1+ n fi 0.001 ≥ n 2 fi 2n > 100 + 100 n fi n ≥ 11 Thus, the minimum number of bombs is 11. 7. Ans (b, c): x2 – (a + b)x + ab = 0 or (x – a)(x – b) = 0 fi x = a or b If x = a is the root of other equation, then a3 – ap + ab = 0 fi p = a2 + b If x = b is the root of the other equation, then ab2 – pb + ab = 0 or p = a(1 + b) 8. Ans (b, d): f (0, t) = 0 for t ≥ 0, so
In
q
.L td .
P(1, 2, 3)
r
9. Ans (a) 10. Ans (c) y = (x – a) (x2 + bx + c) for y = 0, x = a \ p = a fi The point P is (a, 0). Let the slope of the line through P be m. Therefore, equation of the line is y = m (x – a). It meets the curve in A and B. Therefore, m (x – a) = (x – a) (x2 + bx + c) dy = x2 + bx + c + (x – a) (2x + b) dx = x2 + bx + c + 2x2 – 2ax + bx – ab = 3x2 + (2b – 2a) x + c – ab Therefore, slope of the tangent at P(a, 0) is a2 + ab + c. Therefore, equation of the tangent y – 0 = (a2 + ab + c) (x – a)
Appendix B: Mock Tests B.23
È ˘ b2 = 4 Ía 2 + + ab ˙ 4 ÍÎ ˙˚ 2 2 = 4a + b + 4ab = (2a + b)2 ≥ 0 \ D > 0 Hence, there are 2 points.
-1
fi
Ú g ( x) - max ( f ( x), f
A1
ge
R
= 4p – (A1 + A2 + 2A3 + 4) = 4p – (2 + 6 + 4) = 4p – 12. 13. Ans (c):
Pv t
(–2, f (–2))
a
3
di
–2
ng
(–2, f (–2))
ga en
Fig. B.10
M –1 y = f (x)
Five solution but f (–2) > 6 f ¢(0) < –3
4
C
x=3
y = f(x)
A2
= Area RMT =
Ú
f -1 ( x) dx = 1
2
2
0
y=3
x=–2
Area of PQR = 5 – Area QRST = 5 – 4 = 1
Ú ( f ( x) - f
-1
3
So three solutions 14. Ans (b):
Fig. B.8
O
Fig. B.9
A3
T
( x)) dx
A4
Q
S
-1
0
Le
P
A3 = 4 - 1 = 3
.L td .
( x) dx = 4 fi
2 4
Ú f ( x) dx = 5 = shaded region PSTR 0
12. Ans (a):
Úf
ar ni
2
11. Ans (b):
4
In
It meets the curve at C. \ (a2 + ab + c) (x – a) = (x – a) (x2 + bx + c) i.e., a2 + ab + c = x2 + bx + c i.e., x2 + bx – a2 – ab = 0 i.e., (x – a) (x + a) + b (x – a) = 0 i.e., (x – a) (x + a + b) = 0 Therefore, absicca of C is – a – b. Since the line through P intersects at two different points A and B, therefore, b2 – 4 (c – m) > 0 ...(i) Now if the slope of any tangent is m, then 3x2 + (2b – 2a) x + c – ab = m Its D = 4(b – a)2 – 12(c – ab – m) = 4[a2 + b2 – 2ab – 3c + 3ab + 3m] = 4[a2 + b2 + ab + 3m – 3c] ...(ii) From (i) and (ii), we get 3 ˘ È D > 4 Ía 2 + b 2 + ab - b 2 ˙ 4 ˚ Î
( x)) dx = 1 fi
A1 = A2 = 1
15. Ans (b): C1 : z Here
9 1 + z=2 5 5
9 1 8 - = 0 "x > 0 1 + x2
C
en
ga
ge
Le
fi ln y = -
=
È10 ˘ Exponent of 7 = Í ˙ = 1 Î7˚ Number of divisors of 10 is (8 + 1) (4 + 1) (2 + 1) (1 + 1) = 270. Number of ways of putting N as a product of two natural numbers is 270/2 = 135.
Ans 9. (b), 10. (c): a+b+g=p
ar ni
LR =
386
386
\ e =
fi f (x) > tan–1 x – tan–1 1 fi f (x) + p/4 > tan–1 x 8. Ans (a, b, c, d): È10 ˘ È 10 ˘ È 10 ˘ Exponent of 2 = Í ˙ + Í 2 ˙ + Í 3 ˙ = 5 + 2 + 1 = 8 Î 2 ˚ Î2 ˚ Î2 ˚
È10 ˘ Exponent of 5 = Í ˙ = 2 Î5˚
fi (24 - 0) 2 + (7 - 0) 2 - 122 + 52 = 12 \ a = 6 (24 - 5) 2 + (12 - 7) 2 =
dx
È10 ˘ È10 ˘ Exponent of 3 = Í ˙ + Í 2 ˙ = 3 + 1 = 4 Î 3 ˚ Î3 ˚
fi 0 £ P ( A¢ « B ) £
Also 2ae =
2
1
.L td .
1
1
Ú1+ x
Pv t
3 5 \ £ P( A « B) £ 8 8 Next, P(A « B¢) = P(A) – P(A « B) 3 5 3 3 fi - £ P( A « B¢) £ 4 8 4 8 fi P(A « B) = P(B) – P(A¢ « B) 3 5 \ £ P ( B ) - P ( A¢ « B ) £ 8 8
Ú
x
f ¢ ( x) dx >
a
x
fi
di
5 8
In
fi P(A « B) £ P(B) =
3x 1 , "x > 1 > 1 + x2 1 + x2
Êa -g ˆ Ê 3a ˆ 3 Êa - bˆ + sin Á ˜ = + sin Á sin Á Ë 2 ˜¯ Ë 2¯ 2 Ë 2 ˜¯ Ê (p - (b + g )) - b ˆ Ê (p - (g + b )) - g ˆ fi sin + sin Á ÁË ˜ ˜¯ 2 2 Ë ¯ Ê 3p - 3(b + g ) ˆ 3 + sin Á ˜¯ = Ë 2 2 2b + g ˆ 2g + b ˆ 3(b + g ) ˆ 3 Ê Ê Ê fi cos Á + cos Á + cos Á ˜= Ë Ë 2 ˜¯ Ë 2 ˜¯ 2 ¯ 2 3 b -g ˆ fi 2 cos ÊÁ (b + g )ˆ˜ cos ÊÁ Ë 4 ˜¯ Ë4 ¯ Ê3 ˆ 3 + 1 - 2 cos 2 Á (b + g )˜ = Ë4 ¯ 2 Êb -g ˆ Ê3 ˆ fi 2 cos Á (b + g )˜ cos Á Ë 4 ˜¯ Ë4 ¯ Ê3 ˆ 1 - 2 cos 2 Á (b + g )˜ = Ë4 ¯ 2 fi 4 cos 2 ÊÁ 3 (b + g )ˆ˜ Ë4 ¯ Êb -g ˆ Ê3 ˆ + 1 = 0 (1) - 4 cos Á (b + g )˜ cos Á Ë 4 ˜¯ Ë4 ¯
B.30 Mathematics Ê3 ˆ Since cos Á (b + g )˜ is a real number, D ≥ 0. Hence, Ë4 ¯
or
2b - 1 1 ≥ 0 and £0 b -1 b- 1
Êb -g ˆ - 16 ≥ 0 16 cos 2 Á Ë 4 ˜¯
1˘ Ê or b Œ Á - •, ˙ » [1, •) and b Œ (– •, 1) Ë 2˚
Êb -g ˆ ≥1 cos 2 Á Ë 4 ˜¯
when b =
Êb -g ˆ =1 cos 2 Á Ë 4 ˜¯
1ˆ Ï Ê fi b Œ Á - •, ˜ - Ì-1, 0, Ë 2¯ Ó (2)
2
di In
b cos x b + sin x = 2 cos 2 x - 1 (cos 2 x - 3 sin 2 x) tan x
Le
p (1) 2 cos 2x – 1 π 0 fi x π np ± 6 p 2
ge
(2) tan x π 0 fi x π ±
p 6 Also 2 cos 2x – 1 = 2 (cos2 x – sin2x) – (cos2 x + sin2 x) = cos2x – 3 sin2 x Now, the given equation reduces to b sin x = b + sin x b fi sin x = b -1
en
ga
(3) cos2 x – 3 sin2 x π 0 fi x π n p ±
C
\ – 1 £ sin x £ 1 b or – 1 £ £1 b -1 or
b b + 1 ≥ 0 and –1£0 b -1 b -1
P(4, –26/3, –10/3)
(2, 2, – 1) A
ar ni
3 Hence 2 cos (b + g ) = 1 4 3 1 cos (b + g ) = 4 2 3 (b + g ) = 60∞ 4 or b + g = 80° or a = 100° or b = 40°, g = 40°, and a = 100°. Ans 11. (a), 12. (c):
a
È Ê3 ˆ ˘ Í 2 cos ÁË 4 (b + g )˜¯ - 1˙ = 0 Î ˚
1¸ ˝ 3˛
For any other value b, sin x takes two values for x Œ(0, p). Hence, four solutions for x Œ(0, 2p). 13. Ans (b):
ng
1 fi sin x = 1 which is not possible. 2
Pv t
or b = g Hence, Expression (2) for b = g is
.L td .
C (1, 1, 1) E D(x, y, z)
B (3, 1, 2)
Fig. B.14
A (2, 2, – 1), B(3, 1, 2), C(1, 1, 1) AB = i - j + 3k , AC = - i - j + 2k , BC = - 2i - k \ AC ^ BC Mid point of AB = Mid point of CD x +1 5 y +1 3 = fix=4fi = fiy=2 2 2 2 2 z +1 1 = fiz=0 2 2 Therefore, coordinates of D are (4, 2, 0). 14. Ans (a): Equation of the base a(x – 1) + b(y – 1) + C(z – 1) = 0 which also passes through A(2,2,–1) and B(3,1,2). \ a + b – 2c = 0 (i) and 2a + c = 0 (ii) Therefore, c = –2a, b = –5a a (x – 1) – 5a (y – 1) – 2a (z – 1) = 0 \ x – 5y – 2z + 6 = 0 Foot of the normal from P 26 10 y+ z+ x-4 3 3 = = -5 1 -2
Appendix B: Mock Tests B.31
4 2 ,z= 3 3
Section C
Ê 4 2ˆ \ Á 2, , ˜ Ë 3 3¯
-1 (x + 2a) – 2am (2) m Subtracting from (1) to (2), we get x + 3 a = 0. (d) If the chord joining t1 and t2 subtends 90º at vertex, then t1t2 = –4. point of intersection of tangents is (–at1t2, –a(t1 + t2)). So the locus is x = 4a. 18. Ans: (a Æ r); (b Æ p); (c Æ q); (d Æ s) (a) | a ¥ b | = | b ¥ c | = | c ¥ a | = 2DABC Also directions of a ¥ b , b ¥ c and c ¥ a are the same. Hence, a ¥ b = b ¥ c = c ¥ a (b) For regular tetrahedron all sides are of equal length, hence, | a |=| b |=| c | . Also, all the faces are equilateral triangle. Therefore, angle between a and b is 60º, b and c is 60º, and between a and c is 60º. Hence, a .b = b . c = c . a (c) Since a ¥ b = c fi a ^ c and b ^ c and b ¥ c = a fi b ^ a and c ^ a . Therefore, a , b , c are mutually perpendicular. (d) Since a + b + c = 0 fi a 2 + b 2 + c 2 + 2(a.b + b .c + c .a ) = 0 3 fi a.b + b .c + c .a = 2 and y =
Y C
ar ni
ng
Volume of the pyramid 1 = (Base area) ¥ Height 3 1 = AB ¥ AC ¥ EP 3 1 = i - 5 j - 2k ¥ 120 3 1 = 30 120 = 20 cubic units 3 15. Ans (b): Coordinate of any point on line AB can be taken as h = r1 + r cos q
17. Ans: (a Æ q); (b Æ p); (c Æ r); (d Æ s) (a) we get common normal perpendicular to y = x. So, slope = – 1 fi x + y = 3a (b) Tangent to the parabola y = mx + a/m passes through the point P(h, k). fi m2h – mk + a = 0. If its roots are m1 and m2, then m1m2 = +1. Locus is x = a. (c) The tangents are y = m(x + a) + a/m (1)
a
fi x = 2, y =
Therefore, for max sin q = 0 BC2max = 4r22 16. Ans (a): Now OA2 + OB2 + BC2 = r12 + r22 + 4r22 – 4r12sin2q = 5r22 + r12 – 4r12sin2q \ 0 £ sin2q £ 1 So OA2 + OB2 + OC2 Œ [5r22 – 3r12, 5r22 + r12]
Le
B
O
X
A(r1, 0)
ga
ge
C2
C1
Fig. B.15
C
en
k = 0 + r sin q Therefore, it lies on C2. \ (r1 + r cos q)2 + r2 sin2q = r22 r2 + 2 r r1 cos q + r12 – r22 = 0 Let AB = rAB and AC = rAC. Therefore, rAB + rAC = –2r1cos q, rAB.rAC = r12 – r22 So (BC) = | AC – AB | = | rAC – rAB |
=
(rAC + rAB ) 2 - 4rAC rAB
=
4r12 cos 2 q - 4r12 + 4r22
.L td .
26 10 z+ 3 = 3 = -2 -5 -2
y+
-4r12 sin 2 q + 4r22
di
x-4 = 1
BC =
In
Pv t
Ê ˆ Ê -10 ˆ Ê 26 ˆ Á 4 - 5 ÁË - 3 ˜¯ - 2 ÁË 3 ˜¯ + 6 ˜ = -Á ˜ 1 + 25 + 4 Á ˜ ÁË ˜¯
B.32 Mathematics dx 2x - x2
0
=
1
dx
Ú
1 - ( x - 1) 2
0
1
= ÈÎsin -1 ( x - 1) ˘˚ 0
p Ê pˆ p = 0 - Á - ˜ = = k fi k = 2 Ë 2¯ 2 4 20. Ans: (a Æ r); (b Æ s); (c Æ p); (d Æ s) (a) 1 + a10 + a20 + L + a190 = 0 as 10 is not an integral multiple of 20 (b) (log2x)2 + log2(0.03125) + log2x + 3 < 0 fi (log2x)2 + log2(1/32) + log2x + 3 < 0 fi (log2x)2 + (log2x) – 2 < 0 fi (log2x + 2) (log2x – 1) < 0 fi –2 < log2x < 1
a
(4, 4)
Fig. B.16
Length of latus rectum is 4 2 = 4a
1 0 fi x = 2 is the point of minima. 33 Now f (–1) = 8, f (2) = –19, f(5/2) = - , 2 f (–2) = –5 \ maximum value of f(x) in [–2, 5/2] is 8. 1 1 Ê ˆ + + .... Á (2n - 12 ) ˜ ( 4n - 2 2 ) ˜ (d) lim Á n Æ• Á ˜ 1 + Á ˜ Ë 2n 2 - n 2 ¯
In
di
(6, 2)
1
Ú
Pv t
(2, 6)
=
.L td .
19. Ans: (a Æ r); (b Æ s); (c Æ q); (d Æ p) ÏÔmx + c x < 0 (a) f (x) = Ì x x≥0 ÓÔ e For application of LMV function must be continous and derivable in [–2, 2]. f (0+) = f (0–) fi c = 1 f ¢(0+) = f ¢(0–) fi m = 1 fi m + 3c = 4 (b)
n
= lim  n ƕ
r =1 n
= lim  n ƕ
r =1
1 2rn - r 2 1 r r2 n 2 - 2 n n
(3, 0)
2
y = – 12x
x
x=3
Fig. B.17
Perpendicular tangents can be drawn to the parabola from points which lies on directrix. There is only one point (3, 0) which lies on its directrix as well as on the hyperbola.
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