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Vol. XXXIV
No. 1
January 2016
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contents 79
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Maths Musing Problem Set - 157
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10 Practice Paper - Jee Main
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mathematics today | JANUARY ‘16
7
M
aths Musing was started in January 2003 issue of Mathematics Today with the suggestion of Shri Mahabir Singh. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material. During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India. Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced.
Set 157 jee main 1. Let f be twice differentiable such that f ″(x) = –f (x)
and f ′(x) = g(x). If h(x) = {f(x)}2 + {g(x)}2, where h(5) = 11, then h(10) = (a) 11 (b) 10 (c) 9 (d) 8 2. The real values of a, the point (–2a, a + 1) will be an interior point of the smaller region bounded by the circle x2 + y2 = 4 and the parabola y2 = 4x is (a) [–1, 0] (b) (−1, − 5 + 2 6 ) (c) [0, 1] (d) [−1, − 5 − 2 6 ] 3. A plane whose equation is 2x – y + 3z + 5 = 0 is rotated through 90° about its line of intersection with the plane 5x – 4y – 2z + 1 = 0. Find the equation of the plane in the new position. (a) 27x – 24y – 26z = 13 (b) 27x + 24y + 26z = 13 (c) 27x – 24y + 26z = 13 (d) none of these 4. The coefficient of x50 in the series S=
101
∑ rx
r −1
r =1
101 − r
(1 + x )
is
100 101 102 103 (a) 50 (b) 50 (c) 50 (d) 50 5. A line through A(–5, –4) meets the lines x + 3y + 2 = 0, 2x + y + 4 = 0 and x – y – 5 = 0 at the points B, C, D respectively. 2
2
2
15 10 6 + = If , then the equation AB AC AD of the line is (a) 2x + 3y + 11 = 0 (b) 2x + 3y + 22 = 0 (c) x + 3y + 11 = 0 (d) 2x + y + 22 = 0 jee advanced 6. Two integers x and y are chosen (without random) at random from the set {x : 0 ≤ x ≤ 10, x is an integer} then the probability for|x – y| ≤ 5 is 80 91 85 90 (a) (b) (c) (d) 121 121 121 121 8
mathematics today | JANUARY ‘16
comprehension Let n ∈N. The A.M., G.M., H.M. and R.M.S. (root-mean square) of the n numbers n + 1, n + 2, n + 3, ...., n + n are An, Gn, Hn, Rn respectively. Then G 7. lim n = n→ ∞ n 1 2 3 4 (a) (b) (c) (d) e e e e Rn = 8. lim n→ ∞ n 5 7 (a) 3 (b) (c) (d) 3 3 3 integer match
9. If
n n
∑ ∑ nC j jCi = mn − 1, then the value of m2 is
i =0 j =1
matching list 10.
Column-I Column-II log e a log e b log e c 37 P. If 1. = = , then b−c c−a a −b 29 ab+c⋅ bc+a⋅ca+b = Q. In an equilateral triangle, 3 coins of 2. 2 radii 1 are kept so that they touch each other and also the sides of the triangle. The area of the triangle is R. The remainder when 82n – (62)2n+1 is 3. 1 divided by 9 is S. If ∫e–2x(3cos5x – 4sin5x)dx 4. 6 + 4 3 = e–2x(a cos5x + b sin5x) + c, then a+b= P Q R S (a) 3 4 1 2 (b) 3 4 2 1 (c) 4 3 1 2 (d) 4 1 2 3 See Solution set of Maths Musing 156 on page no. 78
1.
sin2 q
cos2 q
1 + 4 sin 4q
sin2 q
1 + cos2 q
4 sin 4q
1 + sin2 q
cos2 q
4 sin 4q
Let f(q) =
4 And the period of f(q) is T, then the value of × T is p (a) 2 (b) 3 (c) 4 (d) none of these 2. Let A = {x : x is a prime factor of 240} B = {x : x is the sum of any two prime factor of 240}. Then A ∪ B is (a) {2, 3, 5} (b) {2, 3, 5, 7, 8} (c) {2, 3, 5, 7} (d) none of these 3.
If x1 and x2 are the roots of the equation x +x 4x – 3(2x + 3) + 128 = 0 , then the value of 1 2 is | x1 − x2 | (a) 5 (b) 6 (c) 7 (d) 8 4. The number of values of ‘t’ for which there exist a non-zero 3 × 3 matrix B such that B′ = tB is (a) 0 (b) 1 (c) 2 (d) infinite 5. Let Tn be the number of possible quadrilaterals formed by joining vertices of an n sided regular polygon. If Tn + 1 – Tn = 20, then the number of triangles formed by joining the vertices of the polygon is (a) 15 (b) 16 (c) 20 (d) none of these 6.
Coefficient of x2 in the expansion of
E = (x
2 /3
(a) 20 (c) –18
+ 4x
1/3
+ 4) + 1 / 3 2 / 3 1 / 3 x −1 x + x +1 5
1
1
(b) 24 (d) none of these
−9
is
7. If t1, t2, t3, …….., tn, …… are in A.P. such that t4 – t7 + t10 = p, then sum of the first 13 terms of the A.P. is (a) 10p (b) 12p (c) 13p (d) 15p 8. If (20)19 + 2(21) (20)18 + 3(21)2 (20)17 + ……. + 20(21)19 = m(20)19, then the value of m + 41 is (a) 20 (b) 21 (c) 22 (d) none of these 2
9.
If lim
(a) 1 (c) 4
x →0
e x − cos x x2
= b, then value of 2b is
ex − e−x 10. If y = ln cos tan −1 2 Statement I : y′(0) = 0 ex − e−x Statement II : y′(x) = 1 + x2 (a) Statement I is True, Statement II is True; Statement II is a correct explanation for Statement I. (b) Statement I is True, Statement II is True; Statement II is not a correct explanation for Statement I. (c) Statement I is True, Statement II is False. (d) Statement I is False, Statement II is True. 11. The length of the rectangle of the greatest area, that can be inscribed in the ellipse x2 + 2y2 = 8, are given by (a) 2√6 (b) 4√6 (c) 3√6 (d) none of these 12. If the primitive of
mathematics today | JANUARY ‘16
1
is f(x) – ln | g(x) | + C (e − 1)2 then f(x) and g(x) are respectively (a) (1 – ex) and 1 – e–x (b) (1 – ex)–1 and 1 – ex (c) (1 – ex)–1 and 1 – e–x (d) none of these
Contributed by : Sanjay Singh, Mathematics Classes, Chandigarh
10
(b) 3 (d) none of these
x
∞
13. If ∫ 0
dx
(x 2 + 4)(x 2 + 9)
(a) ± 7 (c) ± 9
p
, then the value of k is = 2 k + 11 (b) ± 8 (d) none of these
dy 1 = is given by dx 2 x − y 2 x = cg(y) + ay2 + by + d where c, a, b, d are constants then a + b + d is 3 1 (a) (b) 4 4 5 (c) (d) none of these 4 15. If a, b, c are in A.P., a, x, b are in G.P. and b, y, c are in G.P., then the locus of the point (x, y) is (a) a straight line (b) a circle (c) an ellipse (d) a hyperbola 14. The solution of
16. If the circles x2 + y2 + 2ax + cy + a = 0 and x2 + y2 – 3ax + dy – 1 = 0 intersect at two distinct points P and Q, then the line 5x + by – a = 0 passes through P and Q for (a) infinitely many values of a (b) exactly two values of a (c) exactly one value of a (d) no value of a 17. AB is a chord of the parabola y2 = 4ax with the end A at the vertex of the given parabola. BC is drawn perpendicular to AB meeting the axis of the parabola at C. The projection of BC on the axis is 1 (a) (latus-rectum) (b) semi latus-rectum 4 (c) latus-rectum (d) twice latus-rectum 18. Consider the two curves C1 : y2 = 4x, C2 : x2 + y2 – 6x + 1 = 0. Statement I: C1 and C2 touch each other exactly at two points. Statement II: Equation of the tangent at (1, 2) to C 1 and C 2 both is x – y + 1 = 0 and at (1, –2) is x + y + 1 = 0. (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I. (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true. 12
mathematics today | JANUARY ‘16
19. Let L1 be the line 2x – 2y + 3z – 2 = 0 = x – y + z + 1 and L2 be the line x + 2y – z – 3 = 0 = 3x – y + 2z – 1. The vector along the normal of the plane containing the lines L1 and L2 is (a) 7 i^ − 7 ^j + 8 k^ (b) 7 i^ + 7 ^j + 8 k^ (c) 7 i^ − 7 ^j + 7 k^ (d) none of these 20. If a, b, γ and δ are unit vectors, then | a − b |2 + | b − γ |2 + | γ − δ |2 + | δ − a |2 + | γ − a |2 + | b − δ |2 does not exceed (a) 4 (c) 8
(b) 12 (d) 16
21. The distance between the line and the plane x + 5y + z = 5 is 3 10 (a) (b) 10 3 10 10 (c) (d) 9 3 3 22. If
18
18
i =1
i =1
x −2 y +2 z −3 = = 1 −1 4
∑ (xi − 8) = 9 and ∑ (xi − 8)2 = 45 , then the
standard deviation of x1, x2, ….., x18 is 4 9 (a) (b) 9 4 3 (c) (d) none of these 2 23. Let A = {x, y, z, u} and B = {a, b}. A function f : A → B is selected randomly. Probability that function is an onto function is 1 5 (a) (b) 8 8 3 (c) 7 (d) 8 8 24. If letters of the word “ASSASSIN” are written down at random in a row, the probability that no two S's occur together is 1 1 (a) (b) 17 14 1 1 (c) (d) 28 35 25. If cot a equals the integral solution of the inequality 4t2 – 16t + 15 < 0 and sinb equals to the slope of the bisector of the second quadrant, then sin (a + b) sin (a – b) is equal to
3 5 2
(a) − (c)
5
26. If sin a2
+ 1 is (a) 10
(b) −
Solutions
4 5 1. (a) : f (q) =
(d) 3 −1 a
5 + cosec
(b) 9
(d) 26
27. The value of sin–1 (sin 10) is (a) 10 (b) 3p – 10 (c) 10 – 3p (d) none of these 28. Statement I : ~(p ↔ ~ q) is equivalent to p ↔ q. Statement II : ~(p ↔ ~ q) is a tautology. (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I. (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true. tan −1 x 29. Statement I : lim = 0, where [x] x x →0 represents greatest integer ≤ x. −1 Statement II : tan x < 1 for all x ≠ 0. x (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I. (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true. 30. The natural numbers are divided into rows as follows: 1 2 3 4 5 6 7 8 9 ……………………………….. Statement I : Sum of the numbers in the 10th row is a number which can be written as sum of two cubes in two different ways. Statement II : Sum of the numbers in the rth row is 1 3 (r + (r + 1)3) if r ≥ 2. 3 (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I. (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true. 14
mathematics today | JANUARY ‘16
0 1
1 0
1 + sin2 q cos2 q 4 sin 4q
p 4 = 2 , then the value of
−1 5
(c) 11
−1 −1
(R1 → R1 – R3; R2 → R2 – R3) 0 −1
=
0 1
1 0
1 + sin2 q + 4 sin 4q cos2 q 4 sin 4q
(C1 → C1 + C3 )
= – (cos2q + 1 + sin2q + 4 sin4q) = – 2(1 + 2 sin4q) 2p p = which is periodic function with period 4 2 4 4 p Hence the value of × T = . = 2 p p 2 2. (b) : 240 = 24 × 3 × 5 So, A = {2, 3, 5}, B = {5, 7, 8} Clearly, A ∪ B = {2, 3, 5, 7, 8} 3. (c) : Put 2x = y, given equation becomes y2 – 3(8y) + 128 = 0 ⇒ y2 – 24y + 128 = 0 ⇒ (y – 8) (y – 16) = 0 ⇒ y = 16, 8 ⇒ 2x = 16, 8 ⇒ x = 4, 3 x +x 4+3 Hence 1 2 = =7 | x1 − x2 | | 4 − 3 | 4. (c) : We have, B = (B′)′ = (tB)′ = tB′ = t2B As B ≠ 0, t2 = 1 ⇒ t = ± 1 5. (c) : Tn = Number of ways of choosing four vertices out of n = nC4 \ 20 = Tn + 1 – Tn = n + 1C4 – nC4 = n C4 + n C3 – n C4 = n C3 [
nC r–1
n(n − 1) (n − 2) 3. 2. 1 ⇒ n(n – 1) (n – 2) = 120 = 6.5.4
+ nCr =
n + 1C ] r
\ 20 =
⇒ n=6 \ Required number of triangles = nC3 = 6C3 = 20. 6. (c) : E =
= (2 + x
((x1/3
+
2)2)5
x 2/3 + x1/3 + 1 + x1/3 − 1 x −1
x 2/3 + 2 x1/3 ) x −1
1/3 10
−9
−9
(x − 1)9
= (2 + x1/3 )10
x1/3)x–3(x
1)9
= (2 + – (x1/3 )9 (x1/3 + 2)9 Coefficient of x2 = 2[coefficient of x5 in the expansion of x–3(x – 1)9] = 2(–1)9 [coefficient of x8 in (1 – x)9] = 2(–1) 9C8 (–1)8 = –18 7. (c) : Let d be the common difference of the A.P., then p = t4 – t7 + t10 = t4 + (t10 – t7) = t1 + 3d + 3d = t1 + 6d 13 Now, S13 = [2t1 + (13 – 1)d] = 13p 2 8. (b) : Dividing the given equation by (20)19, we get m = 1 + 2x + 3x2 + …… + 20x19, 21 where x = . 20 d x(x 20 − 1) d m = [x + x 2 + ...... + x 20 ] = dx x − 1 dx =
(x − 1)(21x
20
− 1) − x(x
20
− 1)
2
(x − 1)
21 But 20x21 – 21x20 = 20 20 1 \ m= = 400 2 21 − 1 20
21
=
20 x
21
− 21x
20
+1
21 − 21 20
=0
2
x →0
e x − 1 + 1 − cos x x2
x2 e − 1 2 sin2 x / 2 = 1+ 1 = 3 = lim + 2 x →0 x 2 2 2 ( / ) 4 x 2 Hence 2b = 2(3/2) = 3 ex − e−x sin tan −1 2 10. (c) : y ′(x ) = − ex − e−x cos tan −1 2 ex + e−x 1 × × 2 2 ex − e−x 1+ 2 x x − e −e 2 x −x = − tan tan −1 × x − x 2 × (e + e ) 2 e e ( + ) 16
mathematics today | JANUARY ‘16
11. (a) : Any point on the ellipse (2 2 cosq, 2 sinq). A = Area of the inscribed rectangle
x2 y2 + = 1 is 8 4
= 4(2 2 cosq) (2 sinq) = 8 2 sin 2q dA = 16 2 cos 2q = 0 ⇒ q = p / 4 dq d2 A p Also = − 32 2 sin 2q < 0 for q = . 2 4 dq Hence, the inscribed rectangle is of largest area if the sides are p p 4 2 cos and 4 sin i.e., 4 and 2 2 . 4 4 Hence the length of diagonal = 42 + (2 2 )2 = 16 + 8 = 24 = 2 6
(x − 1) 20
ex − e−x
ex + e−x y′(0) = 0
2
Hence, m + 41 = 441 = 21 9. (b) : b = lim
=−
12. (c) :
1
∫ (e x − 1)2
dx = ∫
x
e x dx x
2
e (e − 1)
=∫
dt t (t − 1)2
1 1 1 1 1 1 = − + − dt dt ∫ t − 1 t − 1 t (t − 1)2 t − 1 t 1 = – ln | t – 1 | + ln | t | + C 1− t = (1 – ex)–1 – ln | 1 – e–x | + C x –1 Hence, f(x) = (1 – e ) and g(x) = 1 – e–x
=∫
13. (a) :
∞
dx
∫ (x 2 + 4)(x 2 + 9) 0
=
∞
1 1 1 dx − 2 ∫ 2 5 x + 4 x + 9 0 ∞
x 1 x 1 1 = tan −1 − tan −1 5 2 2 3 3 0
1 p p p = − = , so k 2 + 11 = 60 5 4 6 60 ⇒ k2 = 49 ⇒ k = ± 7 dx = 2x – y2 dy This is a linear equation in x. 14. (c) : We have,
The integrating factor is e − ∫ 2dy = e −2 y d So (xe −2 y ) = − y 2e −2 y dy
Integrating, we have − y 2e −2 y − ∫ ye −2 y dy + Const −2 y 2 −2 y ye −2 y 1 −2 y = + − ∫ e dy + Const e 2 2 2 2 y −2 y y −2 y e −2 y = + e + +C e 2 2 4 y2 y 1 \ x = Ce 2 y + + + 2 2 4 On comparing with the above equation, we get xe −2 y =
1 1 1 a = ,b = ,d = 2 2 4
5 Hence a + b + d = . 4 15. (b) : We have 2b = a + c, x2 = ab, y2 = bc so that x2 + y2 = b(a + c) = 2b2 , which is a circle. 16. (d) : Equation of the line passing through the points P and Q is x2 + y2 + 2ax + cy + a – (x2 + y2 – 3ax + dy – 1) = 0 ⇒ 5ax + (c – d)y + a + 1 = 0 which coincides with 5x + by – a = 0 5a c − d a + 1 if = = −a 5 b or if a2 + a + 1 = 0 which does not give any real value of a. 17. (c) : Draw BD perpendicular to the axis of the parabola. Let the coordinates of B be (x, y) then slope y of AB is given by tanq = x Projection of BC on the axis of the parabola is DC = BD tanq y y 2 4ax = = 4a = y = x x x Y B(x, y)
A
D
C
X
18. (a) : Solving for the points of intersection, we have x2 + 4x – 6x + 1 = 0 ⇒ (x – 1)2 = 0 ⇒ x=1 ⇒ y=±2 18
mathematics today | JANUARY ‘16
Thus the two curves meet at (1, 2) and (1, –2). Tangent at (1, 2) to y2 = 4x is y(2) = 2(x + 1) ⇒ x – y + 1 = 0 Tangent at (1, 2) to the circle C2 is x + 2y – 3(x + 1) + 1 = 0 or x – y + 1 = 0 same as the tangent to the curve C1. Similarly the tangent at the point (1, –2) to the two curves is x + y + 1 = 0 ⇒ Statement II is true and hence statement I is also true. 19. (a) : The plane through L1 is 2x – 2y + 3z – 2 + l(x – y + z + 1) = 0 and the plane through L2 is x + 2y – z – 3 + m(3x – y + 2z – 1) = 0 Two planes are same 2+ l 2+ l 3+ l 2− l So = = = 1 + 3m m − 2 2m − 1 3 + m 3 ⇒ m=– and l = 5 and the equation of the plane 2 is 7x – 7y + 8z + 3 = 0. ^
^
^
Hence the vector along the normal is 7 i − 7 j + 8 k . 20. (d) : We have 0 ≤ | a + b + γ + δ |2 = | a |2 + | b |2 + | γ |2 + | δ |2 + 2(a ⋅b + b ⋅ γ + γ ⋅ a + a ⋅ δ + γ ⋅ δ + b ⋅ δ) = 4 + 2(a ⋅b + b ⋅ γ + γ ⋅ a + a ⋅ δ + γ ⋅ δ + b ⋅ δ) So, a ⋅b + b ⋅ γ + γ ⋅ a + a ⋅ δ + γ ⋅ δ + b ⋅ δ ≥ −2 Now | a − b |2 + | b − γ |2 + | γ − δ |2 + | δ − a |2 + | γ − a |2 + | b − δ |2 = 3(| a |2 + | b |2 + | γ |2 + | δ |2 ) −2(a ⋅ b + γ ⋅ δ + δ ⋅ a + a ⋅ γ + b ⋅ δ + b ⋅ γ ) ≤ 3(4) − 2(−2) = 16 21. (d) : Direction ratios of the line are 1, –1, 4 and that of the normal to the plane are 1, 5, 1. Since 1 × 1 + (–1) (5) + 4 × 1 = 0 ... Normal to the plane is perpendicular to the line and thus the line is parallel to the plane and the distance
between them is the distance of any point, let (2, –2, 3) on the line from the plane and is equal to 2(1) + (−2)(5) + 3 × 1 − 5 1 + 25 + 1 22. (c) : Let di = xi – 8 but 1 1 s 2x = s 2d = ∑ di2 − ∑ di 18 18 =
=
10 27
=
10 3 3
2
2
1 5 1 9 9 × 45 − = − = 18 18 2 4 4
Therefore sx = 3/2
23. (c) : Total number of functions from A to B is 24 = 16. There are exactly two functions which are not onto viz. one in which all the elements of A are mapped to a and other in which all the elements of A are mapped to b. Thus, there are 14 onto functions. 14 7 Therefore probability of required event is = 16 8 24. (b) : The number of ways of permuting the letters of the word “ASSASSIN” is 8! 8 × 7 × 6 × 5 = = 840 2!4! 2 To find the number of ways is which two S are not together, We first arrange A, A, I, N. This can be done in 4! 2! = 12 ways.
If in the following figure X positions of show one such arrangement, then 4S’s can be arranged at four boxes out of the five boxes. × × × × Thus, corresponding to each arrangement of A, A, I, N, there are 5C4 = 5 ways of arranging S’s. \ The number of favourable ways = (12) × (5) = 60 60 1 = Thus, required probability = 840 14 25. (b) : We have, 4t2 – 16t + 15 < 0 3 5 ⇒ 1 2 −1 −1 −1 x 1 − y sin x + sin y = p − sin + y 1 − x 2 if 0 < x , y ≤ 1 & x 2 + y 2 > 1 2 −1 x 1 − y p s in − − + y 1 − x 2 if − 1 ≤ x , y < 0 & x 2 + y 2 > 1
z
−1 2 2 sin x 1 − y − y 1 − x , 2 2 if − 1 ≤ x , y ≤ 1 & x + y ≤ 1 2 2 or if xy > 0 & x + y > 1 2 −1 x 1 − y p sin − 2 − y 1 − x sin −1 x − sin −1 y = if 0 < x ≤ 1, − 1 ≤ y < 0 & x2 + y2 > 1 2 −1 x 1 − y p sin − − 2 − 1 − y x if − 1 ≤ x < 0 , 0 < y ≤ 1 & x2 + y2 > 1
−1
= sec x " x ∈(−∞, −1] ∪ [1, ∞) x −1 cot x , x > 0 z tan −1 (1 / x ) = −1 − p + cot x , x < 0 Property iii p z sin −1x + cos −1x = , "x ∈[−1, 1] 2 p z sec −1x + cosec −1x = , "x ∈(−∞, −1] ∪ [1, ∞) 2 p − 1 − 1 z tan x + cot x = , "x ∈ R 2 Property iV −1 x + y tan 1 − xy , xy < 1 p + tan −1 x + y , x > 0, 1 − xy −1 −1 z tan x + tan y = y > 0, xy > 1 x+y − p + tan −1 , x < 0, 1 − xy y < 0, xy > 1 −1 x − y tan 1 + xy , xy > −1 p + tan −1 x − y , x > 0, 1 + xy −1 −1 z tan x − tan y = y < 0 & xy < −1 x−y − p + tan −1 , x < 0, 1 + xy y > 0 & xy < −1 z
remark : If x1, x2, x3,.....xn ∈ R, then
{
}
{
}
mathematics today | JANUARY ‘16
51
Property Vi
z
−1 2 2 cos xy − 1 − x 1 − y if − 1 ≤ x , y ≤ 1 & x + y ≥ 0 −1 −1 cos x + cos y = xy − 1 − x 2 2 p − cos −1 1 − y2 if − 1 ≤ x , y ≤ 1& x + y ≤ 0
−1 2 2 cos xy + 1 − x 1 − y if − 1 ≤ x , y ≤ 1, x ≤ y z cos −1 x − cos −1 y = − cos −1 xy + 1 − x 2 1 − y 2 if 0 < x ≤ 1, − 1 ≤ y ≤ 0 & x ≥ y Property Vii −1 1 −1 if ≤x≤ 2 sin x 2 2 1 z sin −1 (2 x 1 − x 2 ) p − 2 sin −1 x , if ≤ x ≤1 2 −1 −1 − p − 2 sin x if − 1 ≤ x ≤ 2 −1 2cos x if 0 ≤ x ≤ 1 z cos −1 (2 x 2 − 1) = −1 2 p − 2cos x if − 1 ≤ x ≤ 0
z
2 tan −1 x if − 1 ≤ x ≤ 1 2x sin −1 = p − 2 tan −1 x if x > 1 2 1+ x −1 − p − 2 tan x if x < −1
z
cos −1
z
2 tan −1 x , if − 1 ≤ x ≤ 1 2x tan −1 = p − 2 tan −1 x , if x < −1 2 1− x −1 − p + 2 tan x if x > 1
52
z
−1 −1 p + 3 tan x , x < 3 3 3x − x −1 1 tan −1 = 3 tan −1 x , 3
Property iX : Conversion z
z
mathematics today | JANUARY ‘16
1 −1 1 = sec −1 = cosec x 2 1− x
1 − x2 cos −1 x = sin −1 1 − x 2 = tan −1 x
=
1 x −1 1 −1 cot −1 = sec = cosec x 1 − x2 1 − x2 z
−1 2 tan x if x ≥ 0 = 1 + x 2 −2 tan −1 x if x ≤ 0
1 −1 −1 3 sin x ; 2 ≤ x ≤ 2 1 sin −1 (3x − 4 x 3 ) = p − 3 sin −1 x ; ≤ x ≤ 1 2 − p − 3 sin −1 x ; − 1 ≤ x ≤ −1 2
x sin −1 x = cos −1 1 − x 2 = tan −1 1 − x2 1 − x2 = cot −1 x
1 − x2
Property Viii
z
z
−1 −1 −2 p + 3cos x ; − 1 ≤ x ≤ 2 −1 1 ≤x≤ cos −1 (4 x 3 − 3x ) = 2 p − 3cos −1x ; 2 2 3cos −1x ; 1 ≤ x ≤ 1 2
1 −1 = cos 1 + x2 2 −1 1 −1 2 −1 1 + x = cot = sec 1 + x = cosec x x x tan −1 x = sin −1 1 + x2
Section-i Single correct Answer type
p If x = log cot + q , then sinh x = 4 (a) tan 2q (b) cot 2q (c) –tan 2q (d) –cot 2q 1.
2. AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60°. He moves away from the pole along the line BC to a point D such that CD = 7 m. From D the angle of
elevation of the point A is 45°. Then the height of the pole is
8.
If cos q =
a cos f + b , then tan q/2 is equal to a + b cos f
(a)
7 3 ( 3 + 1)m 2
(b)
7 3 ( 3 − 1)m 2
(a)
(c)
7 3 1 m 2 3 +1
(d)
7 3 1 m 2 3 −1
a −b a + b tan(f / 2) (b)
(c)
a −b a + b sin(f / 2) (d) none of these
3. The angle of elevation of a cloud from a point h m above the surface of a lake is q and the angle of depression of its reflection in the lake is f. The height of the cloud is hsin(f + q) hsin(f − q) (a) (b) sin(f − q) sin(f + q) hsin(q − f) sin(q + f)
(c)
hsin(q + f) sin(q − f)
4.
p y p x If tan + = tan3 + , then 4 2 4 2
(d)
3 + sin2 x sin x equals 1 + 3 sin2 x (a) cos y
(b) sin y
(c) sin 2y
(d) 0
5. If x1, x2, x3, ...., xn are in A.P, whose common difference is a, then the value of sin a [sec x1 sec x2 + sec x2 sec x3 + ..... + sec xn−1 sec xn ] = (a)
sin na cos xn cos x1
(b)
sin(n −1)a cos xn cos x1
(c)
sin(n +1)a cos xn cos x1
(d)
cos(n −1)a cos xn cos x1
6.
If a sin2x + b cos2x = c, b sin2y + a cos2y = d and 2
a a tanx = b tany, then 2 = b (a + d )(c + a) (a − d )(c − a) (a) (b) (b + c)(d + b) (b − c)(d − b) (c)
(a − d )(b − a) (a − c)(c − b)
7.
If cos x sin 2 x = ∑ar sin(rx ), "x ∈ R then 3
1 (a) n = 5, a1 = 2 1 (c) n = 5, a2 = 8
(d)
(d − a)(c − a) (b − c)(d − b)
n
r =0
1 4 1 (d) n = 5, a2 = 4
(b) n = 5, a1 =
9.
The value of
(a)
−9 2
(b)
10
a +b a − b cos(f / 2)
∑ cos3
pr is equal to 3
−7 2
(c)
r=0
−9 8
(d)
−1 8
10. The value of sin3100 + sin3500 – sin3700 is equal to 3 3 3 3 (a) − (b) (c) − (d) − 2 4 8 4 sin 2b 11. If tan(a − b) = , then 3 − cos 2b (a) tana = 2tanb (b) tanb = 2tana (c) 2tana = 3tanb (d) 3tana = 2tanb 12. In a triangle ABC, if angle C is obtuse and angles A and B are given by roots of the equation tan2x + p tanx + q = 0, then the value of q is (a) greater than 1 (b) less than 1 (c) equal to 1 (d) 0 13. If 2sinx – cos2x = 1, then cos2x + cos4 x is equal to (a) 1 (b) – 1 (c) − 5 (d) 5 14. If ABCD is a cyclic quadrilateral such that 13 cosA + 12 = 0 and 3tanB – 4 = 0, then the quadratic equation whose roots are tanC and cosD is (a) 15x2 + 60x – 11 = 0 (b) 60x2 + 11x – 15 = 0 (c) 11x2 + 60x – 15 = 0 (d) none of these 15. If A, B, C are the angles of a triangle such that A C cot = 3 tan , then sinA, sinB, sinC are in 2 2 (a) A.P. (b) G.P . (c) H.P. (d) none of these 16. If
2 tan a 2 tan a / 2 is equal to = l, then 1 + sec a + tan a 1 + tan a / 2
1 (b) l (c) 1 – l (d) 1 + l l 17. If 0 ≤ A, B, C ≤ p and A + B + C = p, then the minimum value of sin 3A + sin 3B + sin 3C is 3 3 (a) –2 (b) − 2 (c) 0 (d) none of these (a)
mathematics today | JANUARY ‘16
53
18. The value of x which satisfies the equation 4x is 2 tan −1 2 x = sin −1 1 + 4x2 1 1 (a) , ∞ (b) −∞, − 2 2 1 1 (d) − , 2 2
(c) [–1, 1]
19. Number of integral solutions of the equation 1 − 3x 2 3 tan −1 x + cos −1 = 0 is (1 + x 2 )3/2 (a) 1
(b) 2
(c) 0
(d) infinite
p 2p 4p 20. In a triangle ABC, with A = , B = ,C= , 7 7 7 then a2 + b2 + c2 is (R = circumradius of DABC) (a) 4R2
(b) 6R2
(c) 7R2
(d) 8R2
21. For what value of x, sin(cot–1(x + 1)) = cos(tan–1 x)? 1 1 (a) (b) 0 (c) 1 (d) − 2 2 22. If the equation x2 + 12 + 3 sin(a + bx) + 6x = 0 has atleast one real solution where a, b ∈ [0, 2p], then value of cosq, where q is least positive value of a + bx is p (a) p (b) 2p (c) 0 (d) 2 23. In any DABC, which is not right angled, ScosA cosecB cosecC is (a) constant (b) less than 1 (c) greater than 2 (d) none of these 24. If C = 90° in DABC, then a b + tan −1 tan −1 is equal to b +c c + a (a)
p 2
(b)
p 4
p 3 Section-ii (c)
(d) p
Multiple correct Answer type
25. In DPQR, if 3sinP + 4cosQ = 6 and 4sinQ + 3cosP = 1 then ∠R can be 3p p 5p p (a) (b) (c) (d) 4 6 6 4 2
sin q tan q 26. If = = 3, then sin f tan f 54
mathematics today | JANUARY ‘16
(a) tan f =
1 3
(c) tan q = 3
(b) tan f = −
1 3
(d) tan q = − 3
27. Which of the following quantities are rational? 11p 5p (a) sin sin 12 12 9p 4 p (b) cosec sec 10 5 4 p 4 p (c) sin + cos 8 8
2p 4p 8p (d) 1 + cos 1 + cos 1 + cos 9 9 9 28. A solution (x, y) of the system of equations 1 1 x − y = and cos2 (px ) − sin2 (py ) = is 3 2 7 5 2 1 (a) , (b) , 6 6 3 3 −5 −7 (c) , 6 6
(d) 13 , 11 6 6
2 y2 − y + 1 ≤ 2 is 29. If 0 ≤ x ≤ 2p then 2cosec x 2 (a) satisfied by exactly one value of y (b) satisfied by exactly two values of x (c) satisfied by x for which cos x = 0 (d) satisfied by x for which sin x = 0
30. If x cosa + y sina = x cosb + y sinb = 2a and a b 2 sin sin = 1 then 2 2 (a) y2 = 4a(a – x) (b) cos a + cos b = cos a cos b 4a 2 + y 2 (c) cos a. cos b = 2 x + y2 4ax (d) cos a + cos b = 2 x + y2 tan 3 A = k (k ≠ 1), then 31. If tan A cos A k − 1 = cos 3 A 2 1 (c) k < 3 (a)
(b)
sin 3 A 2k = sin A k − 1
(d) k > 3
Section-iV
q 32. Let fn (q) = tan (1 + sec q)(1 + sec 2q) 2
Matrix-Match type
n
(1 + sec 4q)....(1 + sec 2 q), then (a)
p f2 = 1 16
(b)
n
Column i
p f3 = 1 32
(A)
p =1 (d) f5 128
p (c) f 4 = 1 64 33. Sum of series
(B)
2r + 1
2 is 2 r (r + 1) r + 2r + r − 1
∑ sin−1
r =1
p 1 − sin −1 n + 1 2 1 (c) cos −1 n + 2 (a)
(C) (D)
1 (b) cos −1 n + 1
Paragraph for Question No. 35 to 37 p 3p 5p Let cos , cos , cos are the roots of equation 7 7 7 3 2 8x – 4x – 4x + 1 = 0 p 3p 5p 35. The value of sec + sec + sec is 7 7 7 (a) 2 (b) 4 (c) 8 (d) none of these
(b)
1 8
p 3p 5p sin sin is 14 14 14 (c)
7 4
(d)
7 8
p 3p 5p 37. The value of cos cos cos is 14 14 14 (a)
1 2
(b)
1 8
(c)
7 2
(d)
cot160 cot 440 + cot440 cot760 (r) – cot 760 cot 160 = 9 (s) rp ∑ sin2 18 =
3 5
Column i
Column ii
(A)
cos200 (p)
x3 – 3x2 – 3x + 1 = 0
(B)
sin100
32x5 – 40x3 + 10x – 1 = 0
(C)
tan150 (r)
8x3 – 6x – 1 = 0
(D)
sin60
8x3 – 6x + 1 = 0
(q) (s)
40. Match the following : Column i
comprehension type
1 4
If tanq is the G.M. between (p) 1 sinq and cosq, then 2 – 4 sin2q + 3sin4q – sin6q = (q) 0 3 cot 200 − 4 cos 200 =
39. Match the following trigonometric ratios with the equations whose one of the roots is given :
(d) none of these
Section-iii
(a)
Column ii
r =1
34. Minimum positive values of x and y such that p x + y = and sec x + sec y = 2 2 are 2 p p (a) x = (b) y = 4 4 p p (c) x = − (d) y = − 4 4
36. The value of sin
38. Match the following :
7 8
Column ii (A) If sinq = 3sin(q + 2a), then (p) 0 the value of tan(q + a) + 2 tana is (B) If p sinq + q cosq = a and (q) 1 p cosq – q sinq = b, then p+a q −b + +1is equal to q +b p−a (C) Then value of the expression (r) sec q p 2p 10 p cos cos cos 7 7 7 p 3p 5p − sin sin sin is 14 14 14 (D) If secq + tanq = 1, then one (s) 1 − root of the equation 4 (a – 2b + c)x2 + (b – 2c + a) x + (c – 2a + b) = 0 is (t) 1 − 2 mathematics today | JANUARY ‘16
55
Section-V
3. (a) : tan q =
integer Answer type
41. Let f(x) = 0 be an equation of degree six, having p integer coefficients and whose one root is 2 cos . 18 Then, the sum of all the roots of f ′(x) = 0, is 42. If cosq + cos2q + cos3q = 1 and sin6q = a + b sin2q + c sin4q, then a + b + c = sin q sin 3q sin 9q 1 + + = [tan Bq − tan Cq], cos 3q cos 9q cos 27q A then (27A – B – 27C) =
p , then the value of 14
(tana tan2a + tan2a tan 4a + tan 4a tan a) is 47. In a triangle ABC, if r1, r2, r3 are the ex-radius and bc ac ab abc s s s + + =k + + − 3 , then k is equal to 2D a b c r1 r2 r3 48. If a + b + g = p and a +b − g g + a −b g +b − a tan tan = 1, 4 4 4 then the value of 1 + cosa + cosb + cosg is K – 1, where K is
tan
SolutionS 1. (c) : x = log [cot(p/4 + q)]
h
⇒ h − h/ 3 = 7
56
7 3 3 −1
m
D
mathematics today | JANUARY ‘16
60° 45° 7m C x B
x B
h A
3
y
F h C x+h E
y x 1 + tan 2 = 2 4. (b) : y x 1 − tan 1 − tan 2 2 1 + sin y (1 + sin x )3 Squaring both sides, we get = 1 − sin y (1 − sin x )3 Using componendo and dividendo 2 sin y (3 + sin2 x ) = sin x 2 1 + 3 sin2 x 5. (b) : sina (secx1secx2 + secx2secx3 + .... + secxn–1secxn) sin(xn − xn−1 ) sin( x2 − x1 ) sin( x3 − x2 ) + + .... + cos xn−1 cos xn cos x1 cos x2 cos x2 cos x3 = tanx2 – tanx1 + tanx3 – tanx2 +....+ tanxn – tanxn–1 sin( xn − x1 ) sin(n − 1)a = tan xn − tan x1 = = cos xn cos x1 cos xn cos x1
=
6. (a) : a tan2 x + b = c(1 + tan2 x) c −b d −a ⇒ tan2 x = , tan2 y = a−c b − d a2
cos q − sin q cos q − sin q = log ⇒ ex = cos q + sin q cos q + sin q x −x 1 (cos q − sin q)2 − (cos q + sin q)2 e −e sinh x = = 2 2 (cos q + sin q)(cos q − sin q) 1 −4 cos q sin q − sin 2q = = − tan 2q = 2 cos2 q − sin2 q cos 2q A
⇒ h=
2h ⇒ x= cot q ⋅ tan f − 1
1 + tan
44. If tan x = tan y = tan z and x + y + z = p, 2 3 5 38 tan2x + tan2y + tan2z = , then K = K 45. If sinq + sin2q + sin3q = 1, then the value of cos6q – 4cos4q + 8cos2q must be
2. (a) : x = h cot 60° = h/ 3 x + 7 = hcot45°
2h + x y
h sin(f + q) CD = h + x = sin(f − q)
43. If
46. If a =
tan f =
D
x y
2
=
tan2 y 2
=
(a − d )(c − a) (b − c)(d − b)
tan x b 7. (b) : cos3x sin2x = cos2 x cosx sin2x 1 + cos 2 x 2 sin 2 x cos x = 2 2 1 = (1 + cos 2 x )(sin 3x + sin x ) 4 1 1 1 = [sin 3x + sin x + (2 sin 3x cos 2 x ) + (2 cos 2 x sin x )] 4 2 2 1 1 1 = [sin 3x + sin x + (sin 5x + sin x ) + (sin 3x − sin x )] 4 2 2 1 3 1 = [sin x + sin 3x + sin 5x] 4 2 2 1 By comparing, we get n = 5, a1 = 4
12. (b) : We have A + B = p – C ⇒ tan(A + B) = – tanC
1 − cos q 8. (a) : tan q / 2 = 1 + cos q
⇒
a cos f + b 1− a + b cos f (a − b)(1 − cos f) = = (a + b)(1 + cos f) a cos f + b 1+ a + b cos f =
=
= tanA⋅tanB < 1 ⇒ q 0 tan A > 0, tan B > 0, tan C < 0 1 − tan A ⋅ tan B
⇒
A+C A−C cos 2 2 = 2 ⇒ 2sinB = sinA + sinC A+C A+C 2 sin ⋅ cos 2 2 2 sin
mathematics today | JANUARY ‘16
57
16. (b) : We have =2
2 tan a 2 sin a = 1 + sec a + tan a 1 + cos a + sin a 2 tan a / 2
2
(1 + tan a / 2) + (1 − tan2 a / 2) + 2 tan a / 2 2 tan a / 2 = 1 + tan a / 2 17. (a) : Since A + B + C = p ⇒ All of sin3A, sin3B, sin3C can’t be negative. Let us take sin3A = –1 ⇒ A = p/2 ⇒ sin3A = –1, sin3B = –1 and sin 3C = 0 So minimum value is – 2.
1 and cos(tan −1 x ) = cos cos −1 1 + x2 By (i) and (ii), we get
cos A
∑ cos A cosec B cosec C = ∑ sin B sin C
23. (a) : Since,
So, number of integral solutions is 2. 20. (c) : a2 + b2 + c2 = 4R2 (sin2 A + sin2B + sin2C) = 2R2(1 – cos2A + 1 – cos2B + 1 – cos2C)
But in right angled DABC
2p 4p 8p = 2R2 3 − cos + cos + cos 7 7 7 2p 4p 6p = 2R2 3 − cos + cos + cos 7 7 7 2p 4p p p 2 sin ⋅ cos + 2 sin ⋅ cos 1 7 7 7 7 = 2 R 2 3 − p 6 p p +2 sin ⋅ cos 2 sin 7 7 7 5p 3p p 3p sin − sin + sin − sin 1 7 7 7 7 = 2 R 2 3 − 7p 5p 2 sin p + sin − sin 7 7 7 1 = 2 R 2 3 + = 7 R 2 2 1 −1 −1 21. (d) : sin(cot (x + 1)) = sin sin x 2 + 2 x + 2 1 ⇒ sin(cot −1 (x + 1)) = …(i) x 2 + 2x + 2 58
mathematics today | JANUARY ‘16
1
= x + 2x + 2 1 + x 2 22. (c) : (x + 3)2 + 3 + 3sin (a + bx) = 0 x = –3, sin(a + bx) = – 1 ⇒ sin(a – 3b) = –1 p a − 3b = (4n − 1) , n ∈ I 2 If n = 1 ⇒ a – 3b = 3p/2 ⇒ cos(a – 3b) = 0 2
1 1 p p ≤ 2 tan −1 2 x ≤ ⇒ − ≤ x ≤ 2 2 2 2 p p 19. (b) : Let tan −1 x = q, q ∈ − , 2 2 3q + cos–1 (cos 3q) = 0 cos–1 (cos 3q) = – 3q ⇒ –p ≤ 3q ≤ 0 p ⇒ − ≤q≤0 3 ⇒ x ∈ − 3 , 0 18. (d) : −
1 = 1 + x2 …(ii) 1
=
− ∑ cos( B + C ) sin B sin C
= ∑ (1 − cot B cot C ) = 3 − ∑ cot A cot B = 2 b a + ab 24. (b) : tan −1 b + c c + a as 3 3 1 − 3 tan A 2
n
32. (a, b, c, d) : fn (q) = tan(q / 2) ∏ (1 + sec 2r q) r =0
r 2 + 2r − r 2 − 1 33. (a, b) : Tr = sin −1 r (r + 1) 1 1 1 1 Tr = sin −1 1 − − 1− r r 2 (r + 1)2 r + 1 1 1 Tr = sin −1 − sin −1 r r + 1 ⇒ Sn =
p 1 1 − sin −1 = cos −1 n + 1 n + 1 2
x + y sec x + sec y 34. (a, b) : sec ≤ 2 2 p 3p 5p 35. (b) : sec , sec , sec are the roots of 7 7 7 x3 – 4x2 – 4x + 8 = 0 3p 5p p sec + sec + sec = 4 7 7 7 p 3p 8 x 3 − 4 x 2 − 4 x + 1 = 8 x − cos x − cos 7 7 5p x − cos 7 mathematics today | JANUARY ‘16
59
36. (b) : Put x = 1 p 3p 5p 1 ⇒ sin sin sin = 14 14 14 8 37. (d) : Put x = –1 7 p 3p 5p ⇒ cos cos cos = 14 14 14 8 38. A - p; B - p; C - r; d - s (A) tan2q = sinqcosq ⇒ sinq = cos3q \ (1 – sin2q) + (1 – 3sin2q) + 3sin4q – sin6q = cos2q + (1 – sin2q)3 = cos2q + cos6q = cos2q + sin2q = 1 (B) sin40° = sin(60° – 20°) 2 sin 200 cos 200 =
3 1 cos 200 − sin 200 2 2
4 cos 200 = 3 cot 200 − 1 (C)
= =
3 + cot 760 cot 160
=
cot 760 + cot 160
3 sin 760 sin160 + cos 760 cos160 siin(760 + 160 )
sin(760 + 160 ) 0
sin 92
=
1 − cos 920 0
sin 92
= tan 460
= cot 44° p 2p p (D) sin2 + sin2 + .... + sin2 = 5 18 18 2 39. A - r; B - s; C - p; d - q (A) Let A = 200 ⇒ 3 A = 600 ⇒ cos 3 A = ⇒ 4 cos3 A − 3 cos A = where x = cos 200
1 2
1 ⇒ 8x3 – 6x – 1 = 0, 2
1 (B) Let A = 100 ⇒ sin 3 A = ⇒ 8 x 3 − 6 x + 1 = 0, 2 where x = sin100 (C) Let A = 15° ⇒ 3A = 45° ⇒ tan 3A = 1 ⇒
3 tan A − tan3 A 2
1 − 3 tan A
=1
= x3 – 3x2 – 3x + 1 = 0, where x = tan15° 1 (D) Let A = 60 ⇒ sin 5 A = ⇒ 32 x 5 − 40 x 3 + 10 x − 1 = 0, 2 0 where x = sin6 60
p 3p 5p sin sin 14 14 14 p p p 3p p 5p = cos − cos − cos − 2 14 2 14 2 14
(C) Consider, sin
2 sin 760 sin 160 + cos(760 − 160 ) cos 600 − cos 920 + cos 600
40. A - p; B - q; C - s; d - q, r (A) Given, sinq = 3sin(q + 2a) ⇒ sin(q + a – a) = 3sin(q + a + a) ⇒ sin(q + a) cosa – cos(q + a)sina = 3sin(q + a) cosa + 3cos(q + a)sina ⇒ –2sin(q + a) cosa = 4cos(q + a) sina sin(q + a) 2 sin a ⇒ − = ⇒ –tan(q + a) = 2 tana cos(q + a) cos a ⇒ tan(q + a) + 2 tana = 0 (B) We have, psinq + qcosq = a ...(1) and pcosq – qsinq = b ...(2) Squaring (1) and (2) and then adding, we get (psinq + qcosq)2 + (pcosq – qsinq)2 = a2 + b2 ⇒ p2(1) + q2(1) – a2 – b2 = 0 ⇒ (p2 – a2) + (q2 – b2) = 0 ⇒ (p + a)(p – a) + (q + b)(q – b) = 0 p+a q −b ⇒ + =0 q +b p−a
mathematics today | JANUARY ‘16
3p 2p p = cos cos cos 7 7 7 p 2p 4p = − cos cos cos 7 7 7 10 p 4p = cos 7 7 p 2p 10 p p 3p 5p So, cos cos cos − sin sin sin 7 7 7 14 14 14 1 p 2p 4p = 2 cos cos cos = − 7 7 7 4 Also, cos
(D) Clearly, secq – tanq = 1 Also, 1 satisfy the given equation. So, the roots of the given equation are 1 & secq. p p 1 ⇒ 6q = ⇒ cos 6q = 18 3 2 1 ⇒ 4 cos3 2q − 3 cos 2q = 2 2 3 ⇒ 8(2 cos q − 1) − 6(2 cos2 q − 1) = 1. 41. (0) : Let q =
Let 2 cosq = x
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61
3
x2 x2 ⇒ 8 2. − 1 − 6 2. − 1 = 1 4 4
46. (1) : a + 2a + 4a = 7a =
p = tan(a + 2a + 4a) 2 1 tan a + tan 2a + tan 4a − tan a ⋅ tan 2a ⋅ tan 4a ⇒ = 0 1 − tan a ⋅ tan 2a − tan a ⋅ tan 4a − tan 2a ⋅ tan 4a ⇒ tan
⇒ (x2 – 2)3 – 3(x2 – 2) = 1 ⇒ x6 – 6x4 + 9x2 – 3 = 0 f ′(x) = 6x (x4 – 4x2 + 3) f ′(x ) = 0 ⇒ x = 0, ± 1, ± 3
⇒ 1 – tana ⋅ tan2a – tana ⋅ tan4a – tan2a ⋅ tan4a = 0 ⇒ tana ⋅ tan2a + tan2a ⋅ tan4a + tan4a ⋅ tana = 1 D D D , r2 = , r3 = 47. (2) : r1 = , Substitute this s−a s −b s−c
42. (0) : cosq (1 + cos2q) = sin2q (1 – sin2q) (2 – sin2q)2 = sin4q sin6q = 4 – 8sin2q + 4sin4q On comparing with the given equation, we get a = 4, b = –8, c = 4 \ a+b+c=0 43. (0) :
value and take abc common, we get abc s − a s − b s − c abc s L.H.S. = + + = − 3 ∑ D a b c D a
sin q 2 sin q cos q = cos 3q 2 cos 3q cos q
⇒ k=2 48. (1) : Let A =
sin 2q sin(3q − q) = = 2 cos 3q cos q 2 cos 3q cos q sin q 1 \ = [tan 3q − tan q] cos 3q 2 sin 3q 1 = [tan 9q − tan 3q] cos 9q 2 sin 9q 1 = [tan 27q − tan 9q] cos 27q 2 \ (1) + (2) + (3) ⇒ sin q sin 3q sin 9q 1 + + = [tan 27q − tan q] cos 3q cos 9q cos 27q 2 Now, on comparing, we get A = 2, B = 27, C = 1 \ 27A – B – 27 C = 0 44. (3) : Let tanx = 2t, tany = 3t, tanz = 5t 1 ∑ tan x = tan x ⋅ tan y ⋅ tan z ⇒ t 2 = 3 Now, tan2x + tan2y + tan2z = t2 (4 + 9 + 25) = 38t2 , K = 3 45. (4) : We have, sinq(1 + sin2q) = 1 – sin2q ⇒ sinq(2 – cos2q) = cos2q Squaring both sides, we get sin2q(2 – cos2q)2 = cos4q ⇒ (1 – cos2q)(4 – 4cos2q + cos4q) = cos4q ⇒ –cos6q + 5cos4q – 8cos2q + 4 = cos4q \ cos6q – 4cos4q + 8cos2q = 4 62
mathematics today | JANUARY ‘16
p 2
C= …(1) …(2) …(3)
b+g−a g +a −b ,B = and 4 4
a +b− g 4
⇒ tanA tanB tanC = 1 sin A sin B 1 = cos A cos B tan C sin A sin B − cos A cos B 1 − tan C = ⇒ sin A sin B + cos A cos B 1 + tan C [By Componendo and Dividendo] ⇒
p sin − C − cos( A + B) 4 ⇒ = cos( A − B) p cos − C 4 p p ⇒ 2 sin − C cos( A − B) + 2 cos − C cos( A + B) = 0 4 4
p p ⇒ sin − C + A − B + sin − C − A + B 4 4 p p + cos − C + A + B + cos − C − A − B = 0 ...(1) 4 4 A− B −C =
p p − a, B − A − C = − b, 4 4 p p C − A − B = − g and A + B + C = 4 4
\ (1) ⇒ cosa + cosb + cosg + 1 = 0 Thus, K = 1.
nn
8
LINEAR PROGRAMMING AND PROBABILITY
Linear Programming OptimiSatiOn prOblem A problem in which a linear function is to be optimised (maximise or minimise) satisfying certain linear inequalities is called optimisation problem. linear prOGramminG prOblem (l.p.p.) Linear programming (L.P.) is an optimisation technique in which a linear function is optimised (i.e., minimised or maximised) subject to certain constraints which are in the form of linear inequalities. mathematical fOrmulatiOn Of l.p.p. The procedure for mathematical formulation of a linear programming problem consists of the following steps : In every L.P.P. certain decisions are to be made. Step I These decisions are represented by decision variables. These decision variables are those quantities whose values are to be determined. Identify the variables and denoted them by x1, x2, x3, ..... Step II Identify the objective function and express it as a linear function of the variables introduced in step I. Step III In a L.P.P., the objective function may be in the form of maximising profits or minimising costs. So, after expressing the objective functions as a linear function of the decision variables, we must find the type of optimisation i.e., maximisation or minimisation. Identify the type of the objective function. Step IV Identify the set of the constraints, stated in terms of decision variables and express them as linear inequations or equations as the case may be.
uSeful termS in l.p.p. Objective function
The linear function Z = c1x1 + c2x2 + ...... + cnxn which is to be optimised (maximised or minimised) is called the objective function. Decision The variables x1, x2, x3, ....., xn whose values are to be decided, are called variables decision variables. Linear The linear inequations (inequalities) or constraints restrictions on the decision variables of a linear programming problem are called linear constraints. The conditions x ≥ 0, y ≥ 0 are called non-negative constraints. Convex A set S of points in a plane is said to be region convex, if the line segment joining any two points in it, lies in it completely. Feasible The set of points, whose co-ordinates region satisfy the constraints of a linear programming problem, is said to be the feasible region. Bounded A feasible region of a system of linear inequations is said to be bounded if it and unbounded can be enclosed within a circle. But if the feasible region extends indefinitely feasible in any direction, then the feasible region region is called unbounded. Points in the region which is the Corner intersection of two boundary lines are points of a feasible called corner points. region mathematics today | JANUARY ‘16
63
Feasible & infeasible solution
Optimal (feasible) solution
Points within and on the boundary of the feasible region of a L.P.P. represent feasible solutions. Any point outside the feasible region is called infeasible solution. The optimal solution of the feasible region is the optimal value (maximum or minimum) of the function at any point.
theOremS •
•
Let R be the feasible region (convex polygon) for a linear programming problem and let Z= ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at the corner point (vertex) of the feasible region. Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and minimum value on R and each of these occurs at a corner point (vertex) of R. Note : If R is unbounded, then a maximum or minimum value of the objective function may not exist. However, if it exists, it must occur at a corner point of R.
Graphical methOdS Of SOlvinG l.p.p. The graphical methods for solving L.P.P. is applicable to those problems which involve only two variables. Corner Point Method To solve a linear programming problem by corner point method, we follow the following steps : (i) Formulate the given L.P.P. in mathematical form. (ii) Convert all inequations into equations and draw their graphs.
(iii) Determine the region represented by each inequation. (iv) Obtain the region in xy-plane containing all points that simultaneously satisfy all constraints including non-negativity restrictions. The polygonal region so obtained is the feasible region and is known as the convex polygon of the set of all feasible solutions of the L.P.P. (v) Determine the coordinates of the vertices (corner points) of the convex polygon obtained in step (ii). These vertices are known as the extreme points of the set of all feasible solutions of the L.P.P. (vi) Obtain the values of the objective function at each of the vertices of the convex polygon. The point where the objective function attains its optimum (maximum or minimum) value is the optimal solution of the given L.P.P. different typeS Of l.p.p. A few important linear programming problems are listed below : • Manufacturing problems : In this problem, we determine the number of units of different products which should be produced and sold by a firm when each product requires a fixed manpower, machine hours, labour hour per unit of product, warehouse space per unit of the output etc., in order to make maximum profit. • Diet problems : In these problems, we determine the amount of different kinds of constituents/ nutrients which should be included in a diet so as to minimise the cost of the desired diet such that it contains a certain minimum amount of each constituent/nutrients. •
Transportation problems : In these problems, we determine a transportation schedule in order to find the cheapest way of transporting a product from plants/factories situated at different locations to different markets.
ProbabiLity cOnditiOnal prObability If two events A and B are associated with the same random experiment, the conditional probability P(A|B) of the occurrence of A knowing that B has occurred is given by P ( A | B) =
64
P ( A ∩ B) , P ( B) ≠ 0 P ( B)
mathematics today | JANUARY ‘16
Similarly, P (B | A) =
P ( A ∩ B) , P ( A) ≠ 0 . P ( A)
Properties of Conditional Probability • If E1 and E2 are any two events associated with an experiment and P(E2) ≠ 0, then 0 ≤ P(E1 | E2) ≤ 1. • If E is any event associated with an experiment, then P(E|E) = P(S|E) = 1, S being the sample space.
•
•
If E, E 1 and E 2 are events associated with an experiment and P(E) ≠ 0, then P((E1 ∪ E2)|E) = P(E1| E) + P(E2 |E) – P((E1 ∩ E2)|E) In particular, if E1 and E2 are mutually exclusive, then P((E1 ∪ E2)|E) = P(E1|E) + P(E2 |E). If E1 and E2 are any two events associated with an experiment, then P(E1c |E2) = 1 – P(E1 |E2).
multiplicatiOn theOrem On prObability If A and B are two events associated with a random experiment, then P(A ∩ B) = P(A) · P(B|A) ; P(A) ≠ 0 or P(A ∩ B) = P(B) · P(A|B) ; P(B) ≠ 0 If A1, A2, A3, ...., An are n events related to a random experiment, then P(A1 ∩ A2 ∩ .... ∩ An) = P(A1) × P(A2|A1) × P(A3|(A1 ∩ A2)) × .... × P(An|(A1 ∩ A2 ∩ ... ∩ An– 1)), where P(An|(A1 ∩ A2 ∩ ..... ∩ An–1)) represents the conditional probability of the event An, given that the events A1, A2, ....., An–1 have happened. independent eventS Two random experiments are said to be independent iff the probability of occurrence or non-occurrence of any event E2 associated with the second experiment is independent of the outcome of the first experiment. This means that if E1 is any event associated with the first experiment, then P(E1 ∩ E2) = P(E1) × P(E2) Note : • Two events E and F are said to be dependent if they are not independent, i.e., if P(E ∩ F) ≠ P(E) . P(F) • Three events A, B and C are said to be mutually independent, if P(A ∩ B) = P(A) . P(B) P(A ∩ C) = P(A) . P(C) P(B ∩ C) = P(B) . P(C) and P(A ∩ B ∩ C) = P(A) . P(B) . P(C) If at least one of the above is not true for three given events, we say that the events are not independent. theOrem Of tOtal prObability Partition of a sample space : A set of events E1, E2, ....., En is said to represent a partition of the sample space S if (i) Ei ∩ Ej = f, i ≠ j, i, j = 1, 2, 3 ....., n (ii) E1 ∪ E2 ∪ ..... ∪ En = S and (iii) P(Ei) > 0 for all i = 1, 2, ....., n.
where the events, E1, E2, ....., En represent a partition of the sample space S if they are pairwise disjoint, exhaustive and have non zero probabilities. For example any non empty event E and its complement E′ form a partition of the sample space S since they satisfy E ∩ E′ = f and E ∪ E′ = S. Law of total probability : Let S be the sample space and let E1, E2, ..., En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event of non-zero probability which occurs with E1 or E2 or ..... or En, then P(A) = P(E1)P(A|E1) + P(E2)P(A|E2) + ..... + P(En)P(A|En) n
= ∑ P (Ei )P ( A | Ei ) i =1
bayeS' theOrem Let S be the sample space and let E1, E2 , ....., En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event of non- zero probability which occurs with E1 or E2 or ..... or En, then P (E )P ( A | Ei ) P (Ei | A) = n i , i = 1, 2, ....., n
∑ P(Ei )P(A | Ei ) i =1
randOm variable and itS prObability diStributiOn • Random variable : Random variable is simply a variable whose values are determined by the outcomes of a random experiment; generally it is denoted by capital letters such as X, Y, Z etc. and their values are denoted by the corresponding small letters x, y, z etc. • Probability distribution : The system consisting of a random variable X along with P(X), is called the probability distribution of X. The probability distribution of a random variable X is the system of numbers X : x1 x2 ..... xn P(X) : p1 p2 ..... pn n
where •
pi > 0, ∑ pi = 1, i = 1, 2, ......, n i =1
Mean of a Random Variable The mean of a random variable X is also called the expectation of X. i.e., denoted by E(X). So, E( X ) = µ = p1x1 + p2 x2 + ..... + pn xn or mathematics today | JANUARY ‘16
n
∑ pi xi i =1
65
•
Variance of a Random Variable n
Var( X ) = ∑ (xi − µ)2 pi i =1
OR
2
n n Var( X ) = ∑ pi xi 2 − ∑ pi xi = E( X 2 ) − [E( X )]2 i =1 i =1
σ x = Var( X ) =
n
∑ (xi − µ)2 pi i =1
is called the Standard Deviation of the random variable X. bernOulli trialS and binOmial diStributiOn • Bernoulli trials : A sequence of independent trials which can result in one of the two mutually exclusive possibilities success or failure such that the probability of success or failure in each trial is constant, then such repeated independent trials are called Bernoullian trials. • Binomial distribution : A random variable X taking values 0, 1, 2, ...., n is said to have a binomial distribution with parameters n and p, if its probability distribution is given by P(X = r) = nCr prqn – r where p represents probability of success and q represents probability of non success, or failure and n is the number of trials. Note : While using above probability density functions of the binomial distribution in solving any problem we should first of all examine whether all the conditions, given below are satisfied : (i) There should be a finite number of trials. (ii) The trials are independent. (iii) Each trial has exactly two outcomes : success or failure. (iv) The probability of an outcome remains the same in each trial. • Mean, variance and standard deviation : (i) Mean = np (ii) Variance = npq
a chair at a profit of ` 15. Assume that he can sell all the items that he buys. Formulate this problem as an L.P.P., so that he can maximise the profit. 2. Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability of drawing two aces. 3. Six coins are tossed simultaneously. Find the probability of getting 3 heads. 4. If A and B are two independent events such that P(A ∪ B) = 0.7 and P(A) = 0.4. Find P(B). 5. A pair of dice is thrown 200 times. If getting a sum of 9 is considered a success, find the variance of the number of successes. short answer type 6. Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag I. 7. An instructor has a test bank consisting of 200 easy true/false questions, 100 difficult true/false questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from test bank, what is the probability that it will be a difficult question given that it is a multiple choice question? 8. Solve the following L.P.P. graphically : Minimize Z = 3x + 5y subject to –2x + y ≤ 4, x + y ≥ 3, x – 2y ≤ 2, x, y ≥ 0 9. A speaks truth in 60% of the cases and B in 90% of the cases. In what percentage of cases are they likely to contradict each other in stating the same fact. 10. A dice is thrown 6 times. If 'getting an even number' is a success, what is the probability of (i) 5 successes (ii) at most 5 successes Long answer type
Very short answer type
11. Find the probability distribution of the number of green balls drawn when 3 balls are drawn, one by one, without replacement from a bag containing 3 green and 5 white balls.
1. A furniture dealer deals in only two items viz., tables and chairs. He has ` 10, 000 to invest and a space to store at most 60 pieces. A table costs him ` 500 and chair ` 200. He can sell a table at a profit of ` 50 and
12. Anil wants to invest at most `12,000 in Bonds A and B. According to the rules, he has to invest at least `2,000 in Bond A and at least `4,000 in Bond B. If the rate of interest on Bond A is 8% per annum
(iii) Standard deviation = npq
66
mathematics today | JANUARY ‘16
and on Bond B is 10% per annum, how should he invest his money for maximum interest? Solve the problem graphically. 13. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both spades. Find the probability of the lost card being a spade. 14. There are 5 cards numbered 1 to 5, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Find the mean and variance of X. 15. A brick manufacturer has two depots, A and B, with stocks of 30,000 and 20,000 bricks respectively. He receives orders from three buildings, P, Q and R, for 15,000, 20,000 and 15,000 bricks respectively. The cost of transporting 1000 bricks to the builders from the depots (in rupees) are given below : From
To A B
P
Q
R
40 20
20 60
30 40
How should the manufacturer fulfil the orders so as to keep the cost of transportation minimum? soLUtions 1. Let x and y be the number of tables and chairs respectively. We have, 500x + 200y ≤ 10000 Thus, the mathematical formulation of the L.P.P. is Maximize Z = 50x + 15y subject to the constraints 5x + 2y ≤ 100, x + y ≤ 60, x ≥ 0, y ≥ 0 2. p = 4 = 1 52 13 \
Required probability =
1 1 1 × = 13 13 169
3. Let p be the probability of getting a head in the toss of a coin. 1 1 Then, p = ⇒ q = 1 − p = 2 2 Let X = No. of successes in the experiment. Then P(X = r) = nCr pr. qn–r \
3
1 1 P ( X = 3) = 6C3 ⋅ 2 2
6−3
=
20 20 5 = = 26 64 16 [... n = 6]
4. We know, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ⇒ P(A ∪ B) = P(A) + P(B) – P(A)P(B) [... A and B are independent events] ⇒ 0.7 = 0.4 + P(B) – 0.4 P(B) 0. 3 1 ⇒ P ( B) = = = 0. 5 0. 6 2 4 1 5. p = getting a sum of 9 on a pair of dice = = 36 9 1 8 ⇒ q = 1 − = . Here n = 200 9 9 Variance = npq = 200 × 1 × 8 = 1600 9 9 81 6. Let E1 be the event of choosing the bag I, E2 the event of choosing the bag II and A be the event of drawing a red ball. Then P (E1 ) = P (E2 ) = 1 2 3 Also P(A|E1) = P (drawing a red ball from bag I) = 7 and P(A|E2) = P(drawing a red ball from bag II) = 5 11 Now, the probability of drawing a ball from bag I, being given that it is red, is P(E1|A). By using Bayes' theorem, we have P (E1 )P ( A | E1 ) P (E1 | A) = P (E1 )P ( A | E1 ) + P (E2 )P ( A | E2 ) 1 3 × 33 2 7 = = 1 3 1 5 68 × + × 2 7 2 11 7. The given data may be tabulated as Easy Difficult Total True/False 200 100 300 Multiple choice 500 400 900 Total 700 500 1200 Let us denote E = Easy questions, D = Difficult questions, T = True/False questions and M = Multiple choice questions Number of difficult multiple choice questions = 400 Total number of questions = 1200 P(D ∩ M) = Probability of selecting a difficult and multiple choice question 400 = 1200 Total number of multiple choice questions = 500 + 400 = 900 mathematics today | JANUARY ‘16
67
P(M) = Probability of selecting one multiple choice question 900 = 1200 \
P(D | M ) =
\ A and B are likely to contradict each other in 42% cases.
P(D ∩ M ) 400 900 4 = ÷ = P(M ) 1200 1200 9
8. First we draw the lines – 2x + y = 4, x + y = 3 and x – 2y = 2 The feasible region has been shown shaded which is unbounded. The vertices of the feasible region are : 8 1 P , , D (0, 3), B (0, 4) 3 3 Given, Z = 3x + 5y 8 1 29 8 1 at P , , Z = 3 × + 5 × = = m (minimum) 3 3 3 3 3 at D(0, 3), Z = 0 + 5 × 3 = 15 at B(0, 4), Z = 0 + 5 × 4 = 20 29 8 1 at P , i.e., when Minimum value of Z = 3 3 3 8 1 x = and y = 3 3 Y
10. We have, the probability of success = the probability of even number on the die 3 1 = = 6 2 (i) The probability of getting 5 successes 5 6 6 3 1 1 1 P(5) = 6C5 ⋅ = 6 ⋅ = = 2 2 2 64 32 (ii) Probability of at most 5 successes = 1 – P(6) 6 1 63 1 = 1− = 1− = 2 64 64 11. Let X denote the total number of green balls drawn in three draws without replacement. Clearly, there may be all green, 2 green, 1 green or no green at all. Thus, X can assume values 0, 1, 2 and 3. Let Gi denote the event of getting a green ball in ith draw. Now, P(X = 0) = Probability of getting no green ball in three draws ⇒ P(X = 0) = P(G1 ∩ G2 ∩ G3 ) ⇒ P(X = 0) = P(G1)P(G2 | G1)P(G3 |(G1 ∩ G2 ))
x+
5 4 3 5 ⇒ P(X = 0) = × × = 8 7 6 28 P(X = 1) = Probability of getting one green ball in three draws ⇒ P(X = 1) = P ((G1 ∩ G2 ∩ G3 ) ∪ (G1 ∩ G2 ∩ G3 )
y=
3
4 B(0, 4)
3 D(0, 3) 2 X
1 O
1
P 2 3 E C (2, 0)(3, 0)
X
Y
9. Here the random experiment is "The stating of the fact by A and B". Let E = the event of A speaking the truth and F = the event of B speaking the truth 60 3 90 9 Then, P(E) = = and P(F ) = = 100 5 100 10 Required probability, P (A and B contradicting each other) = P(EF or EF ) = P(EF ) + P(EF ) = P(E) ⋅ P(F ) + P(E) ⋅ P(F ) = P(E) · [1 – P(F)] + [1 – P(E)] · P(F) 3 9 3 9 21 = 1 − + 1 − ⋅ = 5 10 5 10 50 68
mathematics today | JANUARY ‘16
∪ (G1 ∩ G2 ∩ G3 ))
⇒ P(X = 1) = P(G1 ∩ G2 ∩ G3 ) + P (G1 ∩ G2 ∩ G3 ) + P(G1 ∩ G2 ∩ G3 ) ⇒ P(X = 1) = P(G1)P(G2 | G1)P(G3 |(G1 ∩ G2 )) + P (G1)P(G2 | G1)P(G3 |(G1 ∩ G2 )) + P(G1)P(G2 | G1)P(G3 |(G1 ∩ G2 )) 3 5 4 5 3 4 5 4 3 15 ⇒ P(X = 1) = × × + × × + × × = 8 7 6 8 7 6 8 7 6 28 P(X = 2) = Probability of getting two green balls in the three draws. P(X = 2) = P ((G1 ∩ G2 ∩ G3 ) ∩ (G1 ∩ G2 ∩ G3 ) ∪(G1 ∩ G2 ∩ G3 ))
⇒ P(X = 2) = P(G1)P(G2 | G1)P(G3 |(G1 ∩ G2 ))
+ P (G1)P(G2 | G1)P(G3 |(G1 ∩ G2 )) + P(G1)P(G2 | G1)P(G3 |(G1 ∩ G2 ))
3 2 5 5 3 2 3 5 2 15 ⇒ P(X = 2) = × × + × × + × × = 8 7 6 8 7 6 8 7 6 56 and P(X = 3) = Probability of getting three green balls in the three draws. P(X = 3) = P (G1 ∩ G2 ∩ G3) = P(G1)P(G2 |G1)P(G3|G1 ∩ G2) 3 2 1 1 ⇒ P(X = 3) = × × = 8 7 6 56 Thus, the probability distribution of the number of green balls is given by X : 0 1 2 3
Now, we evaluate Z at the corner points A(2000, 10000), B(8000, 4000) and C(2000, 4000). 2x y Corner Point Z= + 25 10
5 15 15 1 28 28 56 56 12. Let Anil invest ` x in bond A and ` y in bond B, then 8 2x =` Interest on bond A = ` x × 100 25
13. Let E1, E2, E3 and E4 be the events of losing a card of spades, clubs, hearts and diamonds respectively. Let E be the event of drawing 2 spade cards from the remaining 51 cards. 13 1 Now P(E1) = P(E2) = P(E3) = P(E4) = = 52 4 P(E|E1) = Probability of drawing 2 spade cards given that a card of spade is missing.
P( X ) :
Y 12000
A(2000, 10000) B(8000, 4000) C(2000, 4000)
A(2000, 10000)
=
9000 8000
x = 2000
7000
x+
6000
y=
=
12
0
C
(2000, 4000)
y = 4000
B (8000, 4000)
x = 2000
1000
O
1000 2000 3000 4000 5000 6000 7000 8000 –1000
X
–2000 Y
10 y =` 100 10 2 x y His total annual interest = ` 25 + 10 Thus, our L.P.P. is to 2x y ... (i) + Maximise Z = 25 10 Subject to constraints x ≥ 2000 ... (ii) y ≥ 4000 ... (iii) and x + y ≤ 12000 ... (iv) On plotting inequalities (ii) to (iv), we have the required region shown shaded in the figure. and interest on bond B =` y ×
51
C2
C2
=
12 × 11 22 = 51 × 50 425
13
51
C2
=
13 × 12 26 = 51 × 50 425
\
2000 X –2000 –1000
12
C2 By Bayes' theorem, P(E ) ⋅ P(E | E1) P(E1 | E) = 4 1
5000
3000
Maximum
P(E|E2)= P(E|E3) = P(E|E4) 00
4000
←
\ Z is a maximum i.e., `1160 when x = 2000 and y = 10000. So, Anil should invest `2000 in Bond A and `10000 in Bond B to earn maximum profit of ` 1160.
11000 10000
1160 1040 560
∑ P(Ei ) ⋅ P(E | Ei ) i =1
1 22 × 4 425 = 1 22 1 26 1 26 1 26 × + × + × + × 4 425 4 425 4 425 4 425 22 22 = = = 0.22 22 + (3 × 26) 100 14. In this case the sample space contains 5 × 4 = 20 sample points of the type (x, y), x ≠ y, x, y ∈ {1, 2, 3, 4, 5} \ Sum of the numbers on the two cards can take values from 3 to 9. 2 1 P(X = 3) = P({(1, 2),(2, 1)}) = = , 20 10 2 1 P(X = 4) = P({(1, 3),(3, 1)}) = = , 20 10 4 1 P(X = 5) = P({(1, 4),(4, 1),(2, 3),(3, 2)}) = = , 20 5 mathematics today | JANUARY ‘16
69
P(X = 6) = P({(1, 5),(5, 1),(2, 4),(4, 2)}) =
4 1 = , 20 5
P(X = 7) = P({(2, 5),(5, 2),(3, 4),(4, 3)}) =
4 1 = , 20 5
2 1 = and 20 10 2 1 P(X = 9) = P({(4, 5),(5, 4)}) = = . 20 10 P(X = 8) = P({(3, 5),(5, 3)}) =
Thus, the probability distribution of X is as follows : X
3 1 P( X ) 10
4 1 10
5 1 5
6 1 5
7 8 1 1 5 10
9 1 10
Hence mean = SX P(X) 1 1 1 1 1 = 3× + 4× +5× +6× +7× 10 10 5 5 5 +8×
`40x `20
y 00 – `30 (x + y)
300
70
Q 20000
1 1 +9× 10 10
R 15000
`20 00 – x) `60 20000 – y 0 `40 – 1500 y) + (x
(150
mathematics today | JANUARY ‘16
+ 40[(x + y) – 15000]) 1 (1800000 + 30 x − 30 y ) = 1000 Now, we have to minimise the objective function 1 (1800000 + 30 x − 30 y ) (the cost of transZ= 1000 portation) which satisfies all the constraints (1) to (3). The graph of inequations of the constraints given in (1) –(3) gives the feasible region ABCDE, where the vertices are A(15000, 0) B(15000, 15000), C(10000, 20000), D(0, 20000) and E(0, 15000). x = 15000
15. Let x and y units be shipped to buildings P and Q respectively from depot A. Then, the units shipped to other buildings from two depots A and B are shown in the figure. Therefore, there is the following constraints to the problem : x ≥ 0, 15000 – x ≥ 0, i.e., 0 ≤ x ≤ 15000 ...(1) y ≥ 0, 20000 – y ≥ 0, i.e., 0 ≤ y ≤ 20000 ...(2) 30000 – (x + y) ≥ 0, (x + y) – 15000 ≥ 0 i.e. 15000 ≤ x + y ≤ 30000 ...(3) The total cost of transportation Z, for the distribution given in figure is
A 30000
1 (40 x + 20 y + 30[30000 − (x + y )] 1000 + 20(15000 – x) + 60(20000 – y)
Y
3 + 4 + 10 + 12 + 14 + 8 + 9 60 = = =6 10 10 2 and variance = SX P(X) – (mean)2 1 1 1 1 1 = 32 × + 4 2 × + 52 × + 6 2 × + 7 2 × 10 10 5 5 5 1 1 + 82 × + 92 × − (6)2 10 10 9 + 16 + 50 + 72 + 98 + 64 + 81 = − 36 10 390 = − 36 = 39 − 36 = 3 10
P 15000
Z=
B 20000
(0, 30000) D(0, 20000) E(0, 15000)
X
y = 20000
C(10000, 20000)
x+
0
B(15000, 15000)
y=
x + y = 30000
15
00
0
A(15000, 0)
(30000, 0)
X
Y
The values of the objective function at the corner points are given in the following table : Corner points Value of the objective (x, y) 1 function Z = × 1000 (1800000 + 30x – 30y) A(15000, 0) Z = 2250 B(15000, 15000) Z = 1800 C(10000, 20000) Z = 1500 D(0, 20000) Z = 1200 E(0, 15000) Z = 1350 We find that minimum cost of transportation is ` 1350, when x = 0 units and y = 15000 units. Thus, the units transported from depot A and B to three buildings P, Q and R will be as follows so as to minimize the cost. Depot A B
Bricks shipped to buildings Total bricks P Q R 0 20000 10000 30000 15000 0 5000 20000 nn
The entire syllabus of Mathematics of WB-JEE is being divided in to six units, on each unit there will be a Mock Test Paper (MTP) which will be published in the subsequent issues. The syllabus for module break-up is given bellow: Unit- I: Algebra, Unit-II: Trigonometry, Unit-III: Co-ordinate geometry of two dimension & three dimensions, Unit-IV: Calculus, Unit-V: Vector, Unit-VI: Statistics & probability.
cateGoRy 1 Category 1 shall have questions of 1 mark each. For incorrect response, 25% of full mark (1/4) would be deducted.
1. Given 5 line segments of the length 2, 3, 4, 5, 6 units. Then, the no. of triangles that can be formed by joining these segments is (a) 5C3 – 3 (b) 5C3 5 (c) C3 – 1 (d) 5C3 – 2 2. How many numbers greater than 10,00,000 be formed from 2, 3, 0, 3, 4, 2, 3? (a) 420 (b) 360 (c) 400 (d) 300 4 1 tan cos −1 − sin −1 is 5 2 17 3 29 3 29 (a) (b) (c) (d) 29 3 29 3 n 4. If in the expansion of (1 + px) , n ∈ N, then the coefficient of x and x2 are 8 and 24 respectively, then (a) n = 3, p = 2 (b) n = 5, p = 3 (c) n = 4, p = 3 (d) n = 4, p = 2 3.
5. If z =
( 3 + i)3 (3i + 4)2 (8 + 6i)2 (b) 3
(a) 1 2nd
5th
then | z | is equal to (c) 0
(d) 2
6. If the and terms of a G.P. are 24 and 3 respectively, then the sum of the first six terms is 2 189 179 (a) (b) (c) 189 (d) 189 5 2 2 7. If a and b are the roots of x2 – ax + b2 = 0, then a2 + b2 is equal to
(a) 2a2 – b2 (c) a2 – 2b2
(b) a2 + b2 (d) a2 – b2
8. If sin(q + f) = nsin(q – f), n ≠ 1, then the value of tan q = tan f n n +1 n n −1 (b) (c) (d) n −1 n −1 1− n n +1
(a)
9. If tanq = 5tanf then the maximum value of tan2(q – f) is 4 4 16 (a) (b) (c) (d) 1 9 5 25 2 − sin x − cos x = sin x − cos x
10.
x p (a) tan − 2 8
x p (b) tan + 2 8
x p (c) tan − 2 4
x p (d) tan + 2 4
(n !)1/n = n→∞ n (a) e (c) 1
(b) e–1 (d) none of these
11. lim
12. Given that, x 2dx
∞
∫
0
2
2
2
2
2
2
(x + a )(x + b )(x + c ) ∞
then ∫
0
dx 2
2
(x + 4 )(x 2 + 92 )
=
p , 2(a + b)(b + c)(c + a)
is
By : Sankar Ghosh, HOD(Math), Takshyashila. Mob : 09831244397 mathematics today | JANUARY ‘16
71
p 20 p (c) 80 (a)
(b)
p 40
(a)
(d) none of these
13. The length of the perpendicular drawn from x −6 y −7 z −7 (1, 2, 3) to the line is = = 3 2 −2 (a) 4 (b) 5 (c) 6 (d) 7 14. For any integer n, the integral p
∫e
cos2 x
0
cos3 (2n + 1)x dx = (b) p (d) none of these
(a) 1 (c) 2p
15. If [x] denotes the greatest integer ≤ x, then the value of lim | x |[cos x ] is x →0
(a) 0 (c) –1
16. If the angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the longest side and perimeter is (a) 3 : (2 + 3 ) (b) 1 : 6 (c) 1 : (2 + 3 ) (d) 2 : 6 x 3 + x 2 − 16 x + 20 , if x ≠ 2 17. If f (x ) = is continuous (x − 2)2 b, if x = 2 for all x, then b is equal to (a) 7 (b) 3 (c) 2
19. The area bounded by the curve y = f (x), x-axis and the ordinates x = l and x = b is (b – 1)sin(3b + 4). Then f (x) is (a) (x – 1)cos(3x + 4) (b) sin(3x + 4) (c) sin(3x + 4) + 3(x – 1)cos(3x + 4) (d) none of these x2 − y2 dy = log a, then = 20. If cos −1 2 2 dx x + y 72
(a) y
mathematics today | JANUARY ‘16
(c)
(b) xy
22. If I = ∫
x2
(d)
y2
y2 x2
dx (1 − x )(x − 2)
(c)
1 y 4
(c)
xy
, then I is equal to
(a) sin–1(2x – 3) + c (b) sin–1(2x + 5) + c (c) sin–1(3 – 2x) + c (d) sin–1(5 – 2x) + c 23. ∫
sin 2 x
(3 + 4 cos x )3
(c)
dx =
3 cos x + 8 2
(3 + 4 cos x ) 3 + cos x
2
(3 + 4 cos x )
+ c (b) + c (d)
3 + 8 cos x 16(3 + 4 cos x )2 3 − 8 cos x 16(3 + 4 cos x )2
+c +c
24. If the algebraic sum of the deviations of 20 observations from 30 is 20, then the mean of observations is (a) 30 (b) 30.1 (c) 29 (d) 31 25. The variance of first n natural numbers is (a)
(n2 − 1) 12
(b)
(n2 − 1) 6
(c)
(n2 + 1) 6
(d)
(n2 + 1) 12
(d) 5
18. An integrating factor of the differential equation dx y log y + x − log y = 0 is dy (a) log(logy) (b) logy 1 1 (c) (d) log y log(log y )
x y
(b)
d 2 y 1 dy x − x , then x + 21. If y = e + e = dx 2 2 dx
(a) (b) 1 (d) does not exists
y x
8 26. If 1, log 81 (3x + 48), log 9 3x − are in A.P., then 3 the value of x equals (a) 9 (b) 6 (c) 2 (d) 4 27. If the arithmetic mean of two positive numbers a and b (a > b) is twice their G.M., then a : b is (a) 6 + 7 : 6 − 7
(b) 2 + 3 : 2 − 3
(c) 5 + 6 : 5 − 6
(d) none of these
28. If S, P and R are the sum, product and reciprocal of n n k n terms of an increasing G.P. and S = R · P , then k is equal to (a) 1 (b) 2 (c) 3 (d) none of these
29. If the difference between the roots of x2 + 2p x + q = 0 is two times the difference between the roots of p x 2 + qx + = 0, where p ≠ q, then 4 (a) p – q + 1 = 0 (b) p – q – 1 = 0 (c) p + q – 1 = 0 (d) p + q + 1 = 0 30. The roots of the equation are (a) 2, 1, –9 (c) –1, 1, –9
1+ x 3 2 2+x 2 3
5 5 =0 x+4
(b) 1, 1, –9 (d) –2, 1, –8
0 3 2b 31. If 2 0 1 is singular, then the value of b is 4 −1 6 equal to (a) –3 (b) 3 (c) –6 (d) 6 32. If A and B are square matrices of the same order and if A = AT and B = BT, then (ABA)T = (a) BAB (b) ABA (c) ABAB (d) ABT 33. The distance between the point (1, 2) and the point of intersection of the lines 2x + y = 2 and x + 2y = 2 is 16 (c) 17 (d) 19 3 5 3 34. ABCD is a square with side a. If AB and AD are along the coordinate axes, then the equation of the circle passing through the vertices A, B and D is (a)
17 3
(b)
(a) x 2 + y 2 = 2a(x + y ) 2 2 (b) x + y =
a
(x + y ) 2 (c) x2 + y2 = a(x + y) (d) x2 + y2 = a2(x + y) 35. If the semi-major axis of an ellipse is 3 and the latus 16 rectum is , then the standard equation of the 9 ellipse is (a)
x2 y2 + =1 9 8
(b)
x2 y2 + =1 8 9
(c)
x2 3 y2 + =1 9 8
(d)
3x 2 y 2 + =1 8 9
36. The one end of the latus rectum of the parabola y2 – 4x – 2y – 3 = 0 is at (a) (0, –1) (b) (0, 1) (c) (0, –3) (d) (3, 0) x2
y2
5 4 a 2 b2 and 2x + 3y – 6 = 0 is a focal chord of the hyperbola, then the length of transverse axis is equal to
37. The eccentricity of the hyperbola
(a)
24 5
(b)
5 24
(c)
12 5
(d)
−
= 1 is
6 5
38. The solution of the differential equation y′(y2 – x) = y is (a) y3 – 3xy = C (b) y3 + 3xy = C (c) x3 – 3xy = C (d) x3 – xy = C ^ ^ ^ ^ ^ ^ 39. Let OB = i + 2 j + 2 k and OA = 4 i + 2 j + 2 k . The distance of the point B from straight line passing through A and parallel to the vector 2 i^ + 3 ^j + 6 k^ is (a)
7 5 9
(b)
5 7 9
(c)
3 7 7
(d)
9 5 7
40. The angle between the straight line ^ ^ ^ ^ ^ ^ r = ( i + 2 j + k ) + s( i − j + k ) and the plane ^ ^ ^ r × (2 i − j + k ) = 4 is −1 2 2 (a) sin 3
−1 2 (b) sin 6
2 (c) sin −1 3
−1 2 (d) sin 3
41. Let A and B be two events. Then 1 + P(A ∩ B) – P(B) – P(A) is equal to (a) P(Ac ∪ Bc) (b) P(A ∩ Bc) c (c) P(A ∩ B) (d) P(Ac ∩ Bc) 42. If a and b are two unit vectors inclined at an angle p/3 then the value of | a + b | is (a) equal to 1 (b) greater than 1 (c) equal to 0 (d) less than 1 43. If | a × b |2 + | a ⋅ b |2 = 144 and | a | = 4, then | b | is equal to (a) 12 (b) 3 (c) 8 (d) 4 mathematics today | JANUARY ‘16
73
44. Let S be the set of real numbers. A relation R has been defined on S by a R b ∀ |a – b| ≤ 1, then R is (a) symmetric and transitive but not reflexive (b) reflexive and transitive but not symmetric (c) reflexive and symmetric but not transitive (d) an equivalence relation n + 1 , if n is odd 45. Let f : N → N defined by f (n) = 2 n , if n is even 2 then f is (a) (b) (c) (d)
onto but not one-one one-one and onto neither one-one nor onto one-one but not onto
48. The set of values of x for which tan 3x − tan 2 x = 1 is 1 + tan 3x ⋅ tan 2 x p (a) f (b) 4 p (c) np + , n = 1, 2, 3 4 p (d) 2np + , n = 1, 2, 3 4 3 x −1 dx = 49. ∫ 3 x +1 (a) x – log x + log(x2 + 1) – tan–1x + c (b) x + logx + log(x2 + 1) – tan–1x + c 1 (c) x + log x + log(x 2 + 1) − tan −1 x + c 2 (d) none of these
(c) 0
(d) 2
(b)
ln x (ln x )2 + 1
+c
(d) none of these
54. The value of
3 1 2 log 3 6 (81) log5 9 + 3 log 25 7 log 25 6 − (125) is ( 7 ) 409
(b) 1
(c) 2
(d) 3
55. ABCD is a trapezium such that AB, DC are parallel p and BC is perpendicular to them. If ∠ADB = , 4 BC = 3, CD = 4 then AB = (a)
14 3
(b)
7
3
(c)
13 4
(d)
25 7
3 1 56. The value of tan −1 cos 2 tan −1 + sin 2 cot −1 4 2 is p p p p (a) (b) (c) > (d) < 4 4 4 3 57. Let z = cosq + isinq, then the value of 15
2m−1 ) at q = 2° is ∑ Im(z
2
mathematics today | JANUARY ‘16
(b) –1
2x 52. The function f (x ) = log(1 + x ) − is increasing 2+x in (a) (–1, ∞) (b) (–∞, 0) (c) (–∞, ∞) (d) none of these
(a) 0
dy 3 3 50. If x = a cos q, y = a sin q, then 1 + is dx
74
(a) 1
2
47. A fair die is rolled. Consider the events A = {1, 3, 5}, B = {2, 3} and C = {2, 3, 4, 5}. Then the conditional probability P((A ∪ B) | C) is (a) 1/4 (b) 5/4 (c) 1/2 (d) 3/4
(b) 1 (d) sec2q
51. If a, b, c, d > 0, x ∈ R and (a2 + b2 + c2)x2 – 2(ab + bc + cd)x + b2 + c2 + d2 ≤ 0 then 33 14 log a 65 27 log b is equal to 97 40 log c
ln x − 1 53. ∫ dx = (ln x )2 + 1 x (a) +c 2 x +1 x (c) +c (ln x )2 + 1
x −2 y −3 z −4 46. The lines and = = 1 1 −k x −1 y − 4 z − 5 are coplanar if = = k 2 1 (a) k = 2 (b) k = 0 (c) k = 3 (d) k = –1
(a) tan2q (c) tanq
cateGoRy 2 Category 2 shall have questions of 2 marks each. For incorrect response, 25% of full mark (1/2) would be deducted.
m=1
1 sin 2° 1 (c) 2 sin 2° (a)
1 3 sin 2° 1 (d) 4 sin 2°
(b)
58. The equation of the plane containing the lines 2x – 5y + z = 3, x + y + 4z = 5, and parallel to the plane, x + 3y + 6z = 1 is (a) x + 3y + 6z = 1 (b) 2x + 6y + 12z = –13 (c) 2x + 6y + 12z = 13 (d) x + 3y + 6z = –7 59. Locus of the image of the point (2, 3) in the line (2x – 3y + 4) + k(x – 2y + 3) = 0, k ∈ R is a (a) circle of radius 2 (b) circle of radius 3 (c) straight line parallel to x-axis (d) straight line parallel to y-axis. (1 − cos 2 x )(3 + cos x ) = x tan 4 x x →0 (a) 2 (b) 1/2 (c) 4
60. lim
(d) 3
61. The area (in square units) of the region described by {(x, y) : y2 ≤ 2x and y ≥ 4x – 1} is 7 9 5 15 (a) (b) (c) (d) 32 32 64 64 62. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is 11
12
1 (b) 22 3 10 55 2 (d) 3 3
1 (a) 220 3
11
55 2 3 3
(c)
k x + 1, 0 ≤ x ≤ 3 63. If the function g (x ) = is mx + 2, 3 ≤ x ≤ 5 differentiable, then the value of k + m is (a) 10/3 (b) 4 (c) 2 (d) 16/5 64. Let f ′(x ) = 1
192 x 3
1 for all x ∈ R with f = 0. 2 2 + sin px 4
If m ≤ ∫ f (x )dx ≤ M then the possible values of 1/2
m and M are (a) m = 13, M = 24 (c) m = –11, M = 0 x
du
ln 2
u
65. If ∫
(a) 4 (c) ln4
e −1
=
1 1 (b) m = , M = 4 2 (d) m = 1, M = 12
p , then the value of x is 6 (b) ln8 (d) none of these
cateGoRy 3 In category 3, questions may have more than one correct option. All correct answers only will yield two marks. Candidates must mark all the correct options in order to be awarded full marks
66. A line L is a tangent to the parabola y2 = 4x and a normal to the parabola x 2 = 2 y . The distance of the origin from L is 2 3 (a) 0 (b) (c) (d) 2 3 5 67. Let f (x) = ln|x| and g(x) = sinx. If A is the range of f (g(x)) and B is the range of g(f (x)), then (a) A ∪ B = (– ∞, 1] (b) A ∩ B = (– ∞, ∞) (c) A ∩ B = [–1, 0] (d) A ∩ B = [0, 1] 68. The general solution of the equation 3 – 2cosq – 4sinq – cos2q + sin2q = 0 is (a) np (b) 2np p p (c) 2np + (d) 2np + 2 4 69. A ship is fitted with three engines E1, E2 and E3. The engines function independently of each other 1 1 1 with respective probabilities , and . For the 2 4 4 ship to be operational at least two of its engines must function. Let X denote the event that the ship is operational and let X1, X2 and X3 denote respectively the events that the engines E1, E2 and E3 are functioning. Which of the following is(are) true? 3 c (a) P ( X1 | X ) = 16 (b) P(exactly two engines of the ship are 7 functioning |X) = 8 7 5 (c) P ( X | X2 ) = (d) P ( X | X1 ) = 16 16 70. Let DPQR be a triangle. Let a = QR, b = RP and c = PQ. If | a | = 12, | b | = 4 3 and b ⋅ c = 24 then which of the following is true? | c |2 | c |2 + | a | = 30 (a) (b) − | a | = 12 2 2 (c) | a × b + c × a | = 48 3 (d) a ⋅ b = −72 mathematics today | JANUARY ‘16
75
71. Let E1 and E2 be two ellipses whose centres are at the origin. The major axes of E1 and E2 lie along the x-axis and the y-axis respectively. Let S be the circle x2 + (y – 1)2 = 2. The straight line x + y = 3 touches the curves S, E1 and E2 at P, Q and R respectively. 2 2 Suppose that PQ = PR = . If e1 and e2 are the 3 eccentricities of E1 and E2 respectively, then the correct expression(s) is (or) 2 2 (a) e1 + e2 =
(c)
| e12
− e22 |
7
43 40
(b) e1e2 =
5 = 8
3 (d) e1e2 = 4
2 10
e|x|+[ x ] − 2 72. If f (x ) = x , then | x | +[x] (a) lim f (x ) = −1 (b) lim f (x ) = 0 x →0+
x →0−
(c) lim f (x ) = −1
(d) lim f (x ) = 0
x →0
x →0
73. The value of x satisfying the equation 2
p x − 2 x sin x + 1 = 0 is 2 (a) 1 (b) –1 (c) 0 (d) no value of x 4
74. Let f (x) be real valued function such that f (x + y) = f (x) f (a – x) + f (y) + f (a – x) ∀ x , y ∈ R then for some real a
76
mathematics today | JANUARY ‘16
(a) f (x) is a periodic function (b) f (x) is a constant function 1 1 (c) f (x ) = (d) f (x ) = cos x 2 2 75. System of equations x + 3y + 2z = 6, x + ly + 2z = 7 and x + 3y + 2z = m has (a) unique solution if l = 2, m = 6 (b) infinitely many solution, if l = 4, m = 6 (c) no solution if l = 5, m = 7 (d) no solution if l = 3, m = 5 ANSWER KEYS
1. 6. 11. 16. 21. 26. 31. 36. 41. 46. 51. 56. 61. 66. 69. 72. 75.
(a) 2. (c) 7. (b) 12. (a) 17. (c) 22. (c) 27. (c) 32. (a) 37. (d) 42. (b) 47. (c) 52. (c) 57. (b) 62. (a, b) (b, d) (a, b) (b, c, d)
(b) (c) (d) (a) (a) (b) (b) (a) (b) (d) (a) (d) (c)
3. 8. 13. 18. 23. 28. 33. 38. 43. 48. 53. 58. 63. 67. 70. 73.
(d) 4. (b) 9. (d) 14. (b) 19. (b) 24. (b) 29. (a) 34. (a) 39. (b) 44. (a) 49. (c) 54. (a) 59. (c) 64. (a, c) (a, c, d) (a, b)
(d) (a) (d) (c) (d) (d) (c) (d) (c) (d) (b) (a) (d)
5. (d) 10. (a) 15. (b) 20. (a) 25. (a) 30. (b) 35. (c) 40. (a) 45. (a) 50. (d) 55. (d) 60. (a) 65. (c) 68. (b, c) 71. (a, b) 74. (a, b, c) nn
and the other from row 3. 1 6 11 2 7 12 3 8 13 4 9 14 5 10 15
soLUtioN set-156
2ab = 2016 a +b \ (a – 1008)(b – 1008) = (1008)2 = 28⋅34⋅72. The 1 number of pairs = (The number of divisors – 1) 2 9⋅5⋅3 −1 = = 67 2 1 1 1 2. (c) : + = a b 2016 (a – 2016)(b – 2016) = (2016)2 = 210⋅34⋅72 1 Number of pairs = (11 ⋅ 5 ⋅ 3 − 1) = 82 2 2016 3. (b) : (5x – 2) = (5x)2016 – 2016(5x)2015 × 2 2016 ⋅ 2015 + (5x )2014 × 22 + ... = 0 2 2016 ⋅ 2015 (5)2014 × 22 The sum of the roots = 2 ⋅ 2016 (5)2015 × 2 = 403 1. (d) :
5 5 \ The number of pairs (a, b) = 5 + 2 2 1 = 100
\
5. (b) :
a 4
0
x 5 (2016 − x )6 dx
Put x = 2016t
1
I = 201612 ∫ t 5 (1 − t )6 dt = 201611 ⋅ 2016 ⋅ 0
5!6 ! 12 !
4 ⋅ 201611 11 π 6. (a, b, c) : In triangle ABC if C = 2 2 2 2r + c = a + b, (a + b – 2r) = a + b2 (a – 2r)(b – 2r) = 2r2 = 2(2016)2 = 211⋅34⋅72 1 Number of triangles is (number of divisors of 2r2) 2 = 12⋅5⋅3 = 22⋅32⋅5 =
7. (b) : a2 – b2 = (a – b)(a + b) is divisible by 5 if both a and b are from any of the five rows, or one from row 1 and the other from row 4, or one from row 2 78
mathematics today | JANUARY ‘16
2
b c 5 16 6 15 10 11
a b c a b c a b c 5 6 14 6 7 12 7 8 10 7 13 8 11 8 12 9 10 9 11 Total (a, b, c) are 6 + 4 + 3 + 1 = 14 and taking the permutations 14 × 3! = 84. a + b + c = 25, where a, b, c ≥ 4 ⇒ a + b + c = 16, where a, b, c ≥ 1. The number of 15 positive integer solutions = = 105 2 84 4 m Prob. = = = , n−m =1 105 5 n
(104030 − 1) (102015 − 1) +4 +1 (10 − 1) (10 − 1) = 444...4 + 444...4 + 1 S = (4 × 4030) + (4 × 2015) + 1 = 24181, with digit sum 16.
∫
2
5 4 5 \ The number of pairs (a, b) = 5 + 2 2 1 = 200 9. (1) : a < b < c, a + b + c = 25
N =4
2016
21 22 23 24 25
8. (b) : a4 – b4 = (a2 – b2)(a2 + b2) is divisible by 5 if both a and b over from way of the five rows or one each from any of the first 4 rows.
4. (a) : (102016 + 5)2 = 225 N 9N = (2⋅102015 + 1)2 = 4⋅104030 + 4⋅102015 + 1 \
16 17 18 19 20
10. (a) : (P) → 3, (Q) → 5, (R) → 4, (S) → 1 2 2 (P) 3(4 + 5 ) − 40∑ a ⋅ b ≤ 123 + 60 = 183 since −2∑ a ⋅ b ≤ 3 (Q) t1 + t2 + t3 = – 1 – 1 + 8 = 6. Sum = 33 × 6 – 1 = 197 (R) Using Leibnitz rule : y4(1)= 1 × 24 × (–1)(–8) = 192 (S) (a + b)(r + s) = – 15 ⇒ p = 15 Product of roots = 158 = q, \ p + q = 173 nn Solution Sender of Maths Musing set-155 1. 2.
Khokon Kumar Nandi (W.B.) Gajula Raninder (Karim Nagar (TS))
1. We are considering triangles ABC in space. (a) What conditions must be fulfilled by the angles a, b, g of triangle ABC in order that there exists a point P in space such that ∠APB, ∠BPC, ∠CPA are right angles? (b) Let d be the maximum distance among PA, PB, PC and let h be the longest altitude of triangle ABC. Show that
(
6 / 3) h ≤ d ≤ h.
2. Two circles intersect at A and B. P is any point on an arc AB of one circle. The lines PA, PB intersect the other circle at R and S, as shown below. If P′ is any other point on the same arc of the first circle and if R′, S′ are the points in which the lines P′A, P′B intersect the other circle, prove that the arcs RS and R′S′ are equal. P
S
S
3. For any positive integer n, evaluate an/bn, where k =1
n kp kp , bn = ∏ tan2 . 2n + 1 2n + 1 k =1
4. There are 9 regions inside the 5 rings of the Olympics. Put a different whole number from 1 to 9 in each so that the sum of the numbers in each ring is the same. What are the largest and the smallest values of this common sum?
∑ ai = 0. i =1
1 + a a + a + a i i +1 ) ( i i +1 i +2 ) ... (ai + ai +1 + ⋅⋅⋅ + ai +n−2 )
i =1 i
where an+1 = a1, an+2 = a2, etc., assuming that the denominators are non zero. solutions
1. (a) By Pythagoras theorem, we have AB2 = PA2 + PB2, AC2 = PA2 + PC2 and BC2 = PB2 + PC2. P
d A
R B
n
n
n
∑ a (a
R
A
P
an = ∑ tan2
5. Given are real numbers a1, a2, ..., an with Determine
2
h 2
2
B
MH
C
Hence AB + AC – BC = 2PA2 > 0, thus ∠BAC < 90° , i.e., a < 90°. Similarly, we get b < 90° and g < 90°. Conversely, if a, b and g are all acute angles, we may prove that there exists a point P such that ∠APB, ∠ABC, ∠CPA are right angles. (b) We may without loss of generality assume that PA ≥ PB ≥ PC, so PA = d, Because AB2 = AP2 + BP2 ≥ AP2 + CP2 = AC2 and AC2 = AP2 + CP2 ≥ BP2 + CP2 = BC2, we have AB ≥ AC ≥ BC. Let H be the foot of the perpendicular from A to BC, then AH is the longest altitude of DABC, so AH = h. As AP ^ BP and AP ^ CP, AP is perpendicular to the plane of BPC. Thus AP ^ BC and AP ^ PH so that AP < AH, i.e., d < h. ...(1) Because AP ^ BP and AH ^ BC, we get BC is perpendicular to the plane of APH. Thus, we have mathematics today | JANUARY ‘16
79
BC ^ PH. Let M be the midpoint of BC, then PH ≤ PM. As 1 ∠BPC = 90°, we have PM = BM = MC = BC. 2 Hence, 2PH ≤ 2PM = BC, so that 4PH2 ≤ BC2 = PB2 + PC2 ≤ 2PA2 As ∠APH = 90°, we get PH2 = AH2 – AP2 = h2 – d2 From (2) and (3), we have 4(h2 – d2) ≤ 2d2, i.e., 2h2 ≤ 3d2, from which, we have 6 h ≤ d. 3 From (1) and (4) we obtain
k =0
...(3)
So tan2
jp ,1 ≤ j ≤ n, are the roots of 2n + 1
n
...(4)
n n−2k −1 cos q sin2k +1 q 2 k 1 +
n k 2n + 1 2n−2k sin (2n + 1) q = ∑ ( −1) q sin2k +1 q cos 2 k 1 + k =0 n k 2n + 1 2k = tan q cos2n+1 q ∑ ( −1) tan q. 2 1 k + k =0
80
mathematics today | JANUARY ‘16
k
∑ (−1)
(−1)k
So
k =0
2n + 1 2k 2k + 1 tan q = 0
jp ,1 ≤ j ≤ n 2n + 1
3. Using De Moivre’s theorem cosnq + i sinnq = (cosq + i sinq)n, one finds easily that
∑
k
∑ (−1)
for q =
(Notice that this also gives the ‘power of the point’ result for P, PA ⋅ PR = PB ⋅ PS) Similarly P ′A = P ′B = AB P ′S P ′R ′ R ′S ′ Consider now traingles APS and AP′S′. We have ∠APS = ∠APB = ∠AP′B = ∠AP′S′, because P, P′ lie on the same arc of chord AB of the one circle. From the fact that S and S′ lie on the same arc of chord AB of the second circle ∠AS′P′ = ∠ASP. But then DAPS and DAP′S′ are similar. Thus PA P ′A AB AB = . So = PS P ′S ′ RS R ′S ′ Thus RS = R′S′ and the arcs are equal.
sin nq =
n
...(2)
6 h ≤ d < h, as required. 3 2. Because opposite angles of a cyclic quadrilateral are supplementary, we have that ∠PBA = p – ∠ABS = ∠ARS. Similarly ∠PAB = ∠BSR. Thus DPAB and DPSR are similar, from which PA PB AB = = PS PR RS
n−1 2
Thus
k =0 n
k
∑ (−1)
k =0
2n + 1 k 2k + 1 x = 0 and thus also of 2n + 1 n−k = 0. Since an and bn are the 2k x
sum and product of the roots, respectively, we have 2n + 1 2n + 1 ( ) an = = n n + b = 2 1 and n 2n = 2n + 1, 2 an =n and so bn
4. For the five rings, we have a+b=b+c+d=d+e+f=f+g+h = h + i = N. ...(1) Since we are dealing with the nine non-zero decimal digits, we have
9
∑ j = 9 (10) / 2 = 45. 1
The
five regions sum to a common N for 45/5 = 9 but then one pair must be 9 + 0 or one triplet 9 + 0 + 0, which isn’t allowed. So N > 9. Since a + b = h + i, there must be at least two pairs of decimal digits that sum to N. For 10 ≤ N ≤ 15, we have N = 9 + a = 8 + (1 + a) = ..., for 1 ≤ a ≤ 6 while N = 16 = 9 + 17 and N = 17 = 9 + 8 only. So N ≤ 15. From (1) a + b = b + c + d or a = c + d ...(2) and h + i = f + g + h or i = f + g ...(3) The five central digits must equal 45 – 2N (c + d) + e + (f + g) = a + e + i = 45 – 2N So, we have N
2N
45 – 2N
a, e, i
10
20
25
9, 8 – no digit available
11
22
23
9, 8, 6;...
12
24
21
9, 8, 4;...
13
26
19
9, 8, 2;...
14
28
17
9, 7, 1;...
15
30
15
9, 5, 1;...
So 11 ≤ N ≤ 15.
1
i =1 ai +1
(ai+1 + ai+2 )... (ai+1 + ai+2 + ⋅⋅⋅ + ai+n−1 ) i
be the given sum. Let bi = ∑ ar so bn = 0. Also, let n−1
p = ∏ br . Since for all j ≠ i
r =1
r =1
bj – bi = bn – bi + bj = ai+1 + ai+2 + ⋅⋅⋅ + an + a1 + a2 + ⋅⋅⋅ + aj = ai+1 + ai+2 + ⋅⋅⋅ + an + an+1 + ⋅⋅⋅ + an+j, We have 1 1 n−1 1 = +∑ ∏ p i =1 1≤ j≤n b j − bi i =1 1≤ j ≤n b j − bi n
S=∑
∏ j ≠i
=
n
1 S=∑ i =1 ai (ai + ai +1 ) ... (ai + ai +1 + ⋅⋅⋅ + ai +n−2 )
• • • • •
j ≠i
1 n−1 1 1 +∑ − . ∏ p i =1 bi 1≤ j≤n−1 b j − bi j ≠i
Using Lagrange’s interpolation, we let n−1 bj − x 1 F (x ) = ∑ − ∏ i =1 p 1≤ j ≤n−1 b j − bi j ≠i So that F(bk) = –1/p for all 1 ≤ k ≤ n – 1. Note that F(x) is a polynomial in x of degree at most n – 2, yet it is a constant at n – 1 distinct values. Hence F(x) is a constant, F(x) = –1/p. Since bj 1 1 1 − =− , ∏ ∏ p 1≤ j≤n−1 b j − bi bi 1≤ j≤n−1 b j − bi j ≠i
5. Let
Do you want yourself to be updated about
n
=∑
j ≠i
we get F(0) = S – 1/p and thus 1 S = F (0) + = 0. p
nn
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81
10 Best Proble
M
th rchives 10 Best Problems
Prof. Shyam Bhu
Math Archives, as the title itself suggests, is a collection of various challenging problems related to the topics of JEE (Main & Advanced) Syllabus. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for JEE (Main & Advanced). In every issue of MT, challenging problems are offered with detailed solution. The readers' comments and suggestions regarding the problems and solutions offered are always welcome.
1. If An is the area bounded by y = x and y = xn, n ∈ N, then A2 ⋅ A3 …… An = (a) (c)
1 n (n + 1) n−1
2
(b)
1 n (n + 1)
(d)
n
1
2 n (n + 1) n−2
2
1 n (n + 1)
cos–1(cos
2. The area bounded by y = x), x-axis and the lines x = 0, x = 2p is (a) p2 (b) 2p2 2 (c) 4p (d) Area is not bounded 3. The area bounded by y = x2, the lines y = 1 and y = 2, on the left of y–axis is 4 4 1− 2 2 2 2 −1 (a) (b) 3 3 2 2 1− 2 2 (c) (d) 2 2 −1 3 3
( (
( (
) )
) )
cos 4 x + 1 ∫ cot x − tan x dx = acos22x + bcos2x + c ln|cos2x| + l, l being a constant of integration, then 1 1 (a) a = − (b) b = − 8 8 5 (c) c = (d) a + b + c = 0 8 4.
5.
If
dx
∫ x (1 + x 4 ) =
(a) 3x2 + c
(b) ln
1 (c) ln | x | − ln(1 + x 4 ) + c 4 (d) None of these
x 1 + x2
+c
6. Let f(x) be a polynomial of degree three such that f(0) = 1, f(1) = 2 and 0 is a critical point of f having no f (x ) dx . local extreme. Find ∫ x2 + 7 7.
1 Evaluate ∫ sin −1 2 x 1 − x 2 dx , x ∈ , 1 . 2
8. Using derivatives up to order of f(x) = x1/3 + x2/3 and giving main points, draw the graph of y = f(x). Also find the area bounded by the curve y = f(x) and the x – axis. 9.
Solve
d3 y
d2 y
= 0 , given that dx 3 dx 2 1 y(0) = , y ′(0) = 0 and y ′′(0) = 1. 8 10. Draw the graph of the function f defined by f(x) = cot–1x, giving main points. Hence or otherwise evaluate
1
∫
−8
[cot −1 x]dx ([.] denotes the greatest inte-
− 3
ger function). soLUtioNs 1
1 x 2 x n+1 1. (d) : An = ∫ (x − x n ) dx = − 2 n + 1 0 0 1 1 n −1 y = − = 2 n + 1 2(n + 1) Thus A2 ⋅ A3 ⋅ A4 ⋅ ..... An
=
1 1 2 3 n −1 . . ....... + 3 4 5 n 1 2 1 n−1
O
= n−2 2 n(n + 1)
By : Prof. Shyam Bhushan, Director, Narayana IIT Academy, Jamshedpur. Mob. : 09334870021 82
mathematics today | JANUARY ‘16
1
x
2. (a) : Obviously area = p2
=∫
y
O
2
3. (c) : The required area 2
y
2
2 = − ∫ (− y ) dy = y 3/2 3 1 1
2
y = x2
2 = (2 2 − 1) sq. units 3
1 O
x
4 4. (b) : 1 (cos x + 1)sin 2 x dx 2∫ cos 2 x
(
)
Hence ∫ sin −1 (2 x 1 − x 2 ) dx = ∫ (p − 2 sin −1 x ) dx = px − 2 ∫ sin −1 x dx
2
1 1 (1 + t ) + 4 =− . ∫ dt t 8 2 1 5 = − ∫ t + 2 + dt 16 t
x = px − 2 x sin −1 x − ∫ dx 1 − x 2
1 1 = − . cos2 2 x + 2 cos 2 x + 5 ln| cos 2 x | + l 16 2 5. (c) :
x
1 1
1
Let I = ∫
∫ x 4 (1 + x 4 ) dx = 4 ∫ t − t + 1 dt
1 = ln | x | − ln(1 + x 4 ) + c 4 6. Let f(x) = ax3 + bx2 + cx + d. Since f(0)=1, so d=1. Moreover f ′ (0) = 0 i.e., c = 0. Also f ′′(x) = 6ax + 2b. Since f(x) doesn’t have extremum at 0, so f ′′(0) = 2b = 0 Hence f(x) = ax3 + 1 f(1) = 2 ⇒ a = 1 ⇒ f(x) = x3 + 1
∫
=∫
f (x ) x2 + 7 x3 2
x +7
dx = ∫ dx + ∫
x3 + 1 x2 + 7 dx 2
x +7
dx
2x
= px − 2 x sin −1 x + ∫
= −∫
1 t 1 x4 = ln + c = ln +c 4 t +1 4 x4 +1
Now,
(Put y2 = x2 + 7)
3
p = –sin–1(sin(2q – p)), − ≤ 2q − p ≤ 0 = p– 2q = p– 2 sin–1x 2
2 1 (1 + cos 2 x ) + 4 sin 2 x = ∫ dx 8 cos 2 x
3
y
+ log | x + x 2 + 7 | +C
y − 7 y + log| x + x 2 + 7 | + C 3 1 = (x 2 + 7)3/2 − 7(x 2 + 7)1/2 + log | x + x 2 + 7 | +C 3 p p 7. Let sin–1 x = q, then x = sinq and ≤ q ≤ 4 2 sin −1 2 x 1 − x 2 = sin–1(2sinq| cosq|) p p = sin–1(2 sinq cosq), as ≤q≤ 4 2 = sin–1(sin 2q) =
x
( y 2 − 7) ydy
2x
1− x
2
1 − x2
dx
dx
2t dt , Put 1 − x 2 = t 2 t
= −2 1 − x 2 + c Hence
∫ sin
−1
(2 x 1 − x 2 ) dx = px − 2 x sin −1 x − 2 1 − x 2 + c
8.
Let f (x ) = x1/3 + x 2/3
−2 x1/3 + 1 . 2 5 9 3 3⋅ x x3 Obviously f(x) = 0 at x = 0, f(x)>0 for x ∈(−∞, − 1) ∪ (0, ∞) and f(x) 0, ∀ x ∈ − , ∞ , where f increases and 8 ⇒ f ′(x ) =
1 + 2 ⋅ x1/3
⇒ f ′′(x ) =
1 f ′(x ) < 0, ∀ x ∈ −∞, − , where f decreases. 8 mathematics today | JANUARY ‘16
83
y
f ′(x ) < 0, ∀ x ⇒ f is decreasing f ′′(x ) < 0, ∀ x < 0 and f ′′(x ) > 0, ∀ x > 0 .
O
1
Thus x = 0 is a point of inflection and f is concave down for x < 0, f is concave up for x > 0.
x
f ′′(x ) < 0, ∀ (−∞, − 1) ∪ (0, ∞) , where f is concave down.
⇒
Lastly f (x ) → ∞ as | x | → ∞ .
⇒
The graph of y = f(x) is shown in adjacent figure. 2 1 3 3 dx = 3 sq.units = − x + x Required area ∫ 20 −1 0
⇒
9. Let yn = dny / dxn, then the given differential equation is y3 – 8y2 = 0 or y3/y2 = 8 which implies log|y2| = 8x + C1 (integrating both sides) Putting x = 0, we get C1 = log y2(0) = log 1 = 0. \ log|y2| = 8x or y2 = ± e8x, i.e., y1 = ± e8x/8 + C2. Again putting x = 0, we have C2 = ± 1/8. 1 8x 1 e8 x − x + C3 So y1 = ± (e − 1) ⇒ y = ± 8 8 8 Putting x = 0, we have
Thus
n
1 e 9 or y = − −x− 8 8 8
lim f (x ) = 0 and
n
84
n
n
lim f (x ) = p . Thus y = 0 and n
y = p are asymptotes. f(0) = p/2 ⇒ (0, p/2) is a point on the graph.
cot 3
2x
(1 + x 2 )2
x →−∞
y 3 2 /2 1 cot 2
O cot 1
n
n
y=
1
mathematics today | JANUARY ‘16
x
cot −1 x dx =
cot 2
∫
− 3
2dx +
cot 1
∫
1dx +
cot 2
1
∫
0 dx
cot 1
Your favourite MTG Books/Magazines available
8x
x →∞
∫
2, x ∈[− 3 , cot 2] x] = 1, x ∈(cot 2,cot 1] 0, x ∈(cot 1,1]
nn
n
1+ x
1
−1
= 2 (cot 2 + 3 ) + (cot 1 − cot 2) = cot 2 + cot 1 + 2 3
n
⇒ f ′′(x ) = 2
[cot
− 3
n
1 e8 x 7 Thus y = −x+ 8 8 8
−1
cot 3 < − 3 < cot 2 < 0 < cot 1 < 1.
in ASSAM
C3 = (1 / 8) + (1 / 64) = 7 / 64, 9 / 64.
10. f (x ) = cot −1 x ⇒ f ′(x ) =
p p 5p
