smm631101
DESCRIPTION
IIT PreparationTRANSCRIPT
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/1 ===================================================================================================
SOLUTIONS FOR
BASIC STUDY MATERIAL
Sol.SMM631101
MATHEMATICS
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/2 ===================================================================================================
2nd Floor, 95B, Siddamsetty Complex, Secunderabad – 500 003.
Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com HINTS/SOLUTIONS for M1101
(Sets)
Classroom Discussion Exercise 1. (c) A = {φ, x} Subsets of A are φ {φ}, {x} and A ⇒ P(A) = { φ , {φ}, {x}, A} 2. (d) 2m − 2n = 96 = 32 × 3 = 25 (4 − 1) = 25 (22 – 20) = 27 − 25 m = 7 and n = 5 3. (c) e x ≠ x for any x ∈ R So A ∩ B = φ. 4. (b) A = {x : x ≥ 3}; B = {x : x < 5} Clearly A ∩ B = {x : x ∈ R, 3 ≤ x < 5} 5. (d) X = {3, 5, 7, 9} 6. (d) The curves will intersect when x = 0 ∴ The curves meet at (0, 1) ∴ A ∩ B = {(0, 1)} 7. (c) The points of intersection are given by y2 = 4 ⇒ y = ± 2 ∴Points of intersection are (1, 2) and (1, –2) 8. (d) A – B = {x / x ∈ A, and x ∉ B} ∴ (A – B) ∩ B = φ. 9. (c) n (A) = 76, n (B) = 44, n (A ∪B) = 100 n (A ∩ B) = n (A) + n (B) – n (A ∪ B) = 76 + 44 – 100 = 20. 10. (a) The required number of subsets = 5C3 = 10 11. (d) All (i) , (ii) and (iii) represent the shaded region 12. (a) A = {7, 14, 21, 28,…,.105} B = {4, 8, 12, 16, …., 60} A ∩ B = {28, 56} ∴ n (A − B) = n(A) – 2 = 13. 13. (c) P ∩ (P ∪ Q)’ = P ∩ (P’ ∩ Q’) = (P ∩ P’) ∩ (P ∩ Q’) = φ ∩ (P ∩ Q’) = φ
14. (d) U = {1, 2, 3,…………, 10} A = {3, 4, 5, 6, 7, 8} B = {2, 3, 5, 7} A − B = {4, 6, 8} (A − B)’ = {1, 2, 3, 5 ,7, 9, 10}
C B
A
C −−−− B
B C
A
A −−−− C
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/3 ===================================================================================================
15. (d)
There is no region common to A − C and C − B. ∴ (A − C) ∩ (C − B) = φ 16. (a) People who are not teenagers is U – C = C’ People who have their weights less than 40 kg = U – D = D ’ ∴ Required set = C’ ∩ D’ 17. (c) Since k and m are relatively prime, the L.C.M.
of k and m is km. ∴∴∴∴ kN ∩∩∩∩ mN = (mk)N = nN ⇒⇒⇒⇒ mk = n 18. (d) n (A) = 300; n (B) = 500; n (A ∩ B) = 100 n (A ∪ B) = 300 + 500 – 100 = 700 n (A’ ∩ B’) = n [(A ∪ B)’] = 1000 − n (A ∪ B) = 1000 –
700 = 300
19. (c) Since A ∆ B = (A ∪ B) – (A ∩ B), therefore A ∩ B = φ 20. (c) A ∪ B = A ∪ C and A ∩ B = A ∩ C ⇒ B = C 21. (a) A – (A ∪ B)’ = A 22. (c) )BA(n)B(n)A(n)BA(n ∩−+=∪
)BA(n)B(n)A(n)BA(n ∪−+=∩⇒
)BA(n6465)BA(n ∪−+=∩⇒
But 100)BA(n ≤∪
1006465)BA(n −+≥∩⇒
%29)BA(n ≥∩⇒
)1(............%29x ≥⇒
and)A(n)BA(n ≤∩
)B(n)BA(n ≤∩
and%65)BA(n ≤∩⇒
%64)BA(n ≤∩
%64)BA(n ≤∩⇒
)2.....(..........%64x ≤⇒ Combining (1) and (2) 29% %64x ≤≤ 23. (c) Let A, B, C denote the set of all families
reading “The Hindu”, “Hindustan Times” and “The Chronicle” respectively.
n (A) = 10000 × 400010040 =
n (B) = 10000 × 200010020 =
n (C) = 10000 × 100010010 =
n (A ∩ B) = 10000 × 500100
5 =
n (B ∩ C) = 10000 × 300100
3 =
n (A ∩ C) = 10000 × 400100
4 =
n (A ∩ B ∩ C) = 10000 × 200100
2 =
n (A’ ∩ B’ ∩ C’) = 10000 – n (A ∪ B ∪ C) = 10000 – {n (A) +
n (B) + n (C) – [n (A ∩ B) + n (B ∩ C) ] + n (A ∩ C)] +
n (A ∩ B ∩ C)} i.e., n (A’ ∩ B’ ∩ C’) =10000 – {7000 – 1200 + 200} = 10000 – 6000 = 4000 = 40%.
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/4 ===================================================================================================
24. (d) A = { 2 } B = { − 2, 1 } A − B = { 2 } B − A = { − 2 , 1 }
[ ] .3)AB()BA(n}1,2,2{)AB()BA(
=−∪−∴−=−∪−
25. (c) )BA(n ∆
)BA(n2)B(n)A(n ∩−+= (1)
Now n(A ∩ B) = n(A) − n(A − B) ⇒ n(A ∩ B) = 5 − 4 = 1 Thus (1) gives 12 = 5 + n(B) − 2 ⇒ n(B) = 9
Regular Homework Exercise 1. (c) x 2 + 9 = 0 ⇒ x 2 = – 9 ⇒ x is imaginary. 2. (b) All equal sets are equivalent 3. (c) n(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y) is maximum
if n(X ∩ Y) = 0 ∴ maximum value of n(X ∩ Y) is n(X) + n(Y) = 12 4. (d) n(A∩B) = n(A) = 6 if A ⊂ B 5. (c) Number of elements of P(A) = Number of subsets of A = 2n(A)
= 25 6. (c) It represents the elements belonging to
exactly two of the sets A, B, C 7. (c) Let n(A) = m, n(B) = n Given 2m – 2n = 32 ⇒ 2n(2m − n − 1) = 25 ⇒ n = 5 (Θ 2m − n − 1 is odd) ⇒ m = 6 8. (a) A ∩ (A ∪ B) = (A ∩ A) ∪(A ∩ B) = A∪(A ∩ B) = A (since A ∩ B
⊂ A) 9. (c) P{A} = {φ , {φ} } ⇒ n [P(A)] = 2 10. (a) The number of subsets containing at least one
element = 25 − 1 = 31 11. (c) n(A ∪ B ∪C) = n(A) + n(B ) + n(C) – n(A ∩ B)
– n(B ∩ C) – n (C ∩ A) + n(A ∩ B ∩ C).
12. (b) )CA()BA()CB(A ∪∩∪=∩∪ 13. (b)
)ZX()YX()ZY(X −∪−=∩− 14. (a) x ≥ 100 – (20 + 15 + 30) = 35 ⇒ minimum value of x = 35 15. (d) A = {5, 6, 7} , B = {1, 3, 5, 7} ⇒ A ∪ B = {1, 3, 5, 6, 7} ⇒ A’ ∩ B’ = (A ∪ B)’ = {2, 4, 8} 16. (b) A ⊆ B ⇒ (A ∪ B) = B ∴ n(A ∪ B) = n(B) = 7 17. (b) Let A, B and C be the set of basketball players,
cricket players and general athletics players respectively.
n (A) = 21; n (B) = 26; n (C) = 29
n (A ∩ B) = 14; n (B ∩ C) = 15; n (A ∩ C) = 12; n (A ∩ B ∩ C) = 8 n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (B
∩ C) – n (A ∩ C) + n (A
∩ B ∩ C) = 21 + 26 + 29 – 14
– 15 – 12 + 8 = 43. 18. (d) .}..........,24,12,8,4{N4 =
.....},.........24,18,12,6{N6 =
..}..........,36,24,12{NN 64 =∩∴
= N12 19. (c) Number of students who like at least one juice
=125 + 118 + 117 − (60 + 60) + 20 = 260 ∴ total number of students = 260 + 70 = 330 20. (c) Clearly the shaded region represents B − (A ∪ C)
Assignment Exercise 1. (d) A = {18, 45, 108, …………} B = {18, 27, 36, 45,……….} ∴ B A ⊂
2. (a) n (A – B) + n (A ∩ B) = n (A). 3. (b)
A B
X
Y Z
A − B
A B A∩B
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/5 ===================================================================================================
We observe that A – (A – B) = A ∩ B. 4. (b) A ∆ B = {5, 6}. So, number of subsets of A ∆ B = 2 2 = 4.
5. (b) We have ex = .........!2
x!1
x1
2
+++ where x is real.
Hence, ex ≠ x for any real x. ∴ A ∩ B = φ 6. (c)
n (B – M) = n (M’)
= 40 − 30 = 10 7. (a) n (A’∩B’) = n (A∪B)’ (By De Morgan’s Law ) = n (U) – n (A∪B) = n (U) −
[ ])BA(n)B(n)A(n ∩−+
= 1000 – (400 + 300 − 100) = 400 8. (c) A ⊂ B ⇒ A’ ⊃ B’ ⇒ A’ ∩ B’ = B’.
9. (b) B only = 900500010018 =×
10. (b) Total number of subsets of two sets are 2m and 2n
324822 4nm ×==−∴
024n4m 22322 −==−⇒ −−
04nand24m =−=−⇒
4nand6m ==⇒ 11. (c) Since )BA(x ∪∉ ,
∴ x ∉ A and x ∉ B Option (c) is wrong. 12. (c) One half of the men belong to club A ⇒ 6 belong to club A One third of the men belong to club B ⇒ 4 belong to club B One forth of the men belong to both clubs ⇒ 3 belong to club A and club B ⇒ n (A∪B)=7 Thus 12 − 7 = 5 belong to neither clubs. 13. (b) n (A) = 4 and n ( B ) = 7 Minimum of n ( 7)B(n)BA ==∪ when A ⊆ B 14. (c) The shaded region represents A. C)BA(C =∪∩∴ . 15. (d) A = BC'BC −=∩ The shaded region represents A φ=∩∴ BA
A B
3
7
18
C
2
3 3 1
n(A)=6 n(B)=4 U = 12
5 M
B
8 22
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/6 ===================================================================================================
Additional Practice Exercise 1. (d) By definition. 2. (c) )BA(n2)B(n)A(n)BA(n Ι−+=∆
= 250 + 350 − 200 = 400 3. (d) n(A ∪ B) = n(A) + n(B) – n(A ∩ B) = 100 + 50 – 25 = 125. ∴ n(A’ ∩ B’) = 75. 4. (c) Obviously the smallest set is A = {1, 2, 5}.
5. (b) BA)BA(A ∩=−−
= )B'A(A ∪∩
6. (a) Let m be the number of elements in S. Then 9m = 3 × 45 ⇒ m = 15 Again 5n = 10m = 10 × 15 n = 30 7. (b) Let the number of news papers be x
Then the number of subscriptions = 30x
Every body subscribes 6 newspapers
∴ Number of people = 6
x30
⇒ 6
x30 = 240 ⇒ x = 48
8. (d) BA ∆ = )AB()BA( −∪−
= )BA()BA( ∩−∪
)BA(n ∆ = )BA(n)BA(n ∩−∪
= )BA(n2)B(n)A(n ∩−+ 9. (d) All of them are correct 10. (c) A = { x / x∈R, −2 < x < 2} B = { x / x∈R, 2 ≤ |x − 2|} = {x / x∈R, x∈(−∞, −2] ∪[4, ∞)} A ∪ B = (−∞, 2) ∪ [4, ∞) = R − {x/x ∈ R, 2 ≤ x < 4} 11. (c) {a, b} ∈ P({a, b} ) 12. (d) Number of students who passed with
distinction = 240 × 4810020 =
∴ By the given condition, number of girls who
passed with a distinction = 24248 =
∴ Percentage of girls who passed with
distinction = 10016024 × = 15%.
13. (a)
Number of persons belonging to at least two clubs = 53
14. (b) A−B, A∩B and B−A are mutually disjoint. 15. (b) (A ∩ B) − C = (A − C) ∩ (B − C) 16. (c) C −D = (A − B) − (B − A)
But there is no region common to both (A − B) and (B − A)
A − B B − A ∴ ( A − B) − (B − A) = (A − B) = C 17. (d) A ∪ B = C ∪ B and A ∩ B = C ∩ B ⇒ A = C 18. (b) n (m) = 120, n (p) = 90, n (c) = 70, n (m ∩ p) = 40, n (p ∩ c) = 30, n (m ∩ c) = 50, n (m’ ∩ p’ ∩ c’) = 20 n(m ∪ p ∩ c) = 200 – 20 = 180 ie n(m) + n(p) + n(c) – n (m ∩ p) – n (m ∩ c) – n(p ∩c) + n (m ∩ p ∩ c) = 180 ⇒ 120 + 90 + 70 – (40 + 30 + 50) + n(m ∩ p ∩ c) =
180 ⇒ n (m ∩ p ∩ c) = 20 19. (b) Delete both a and s from X. Then the resulting set
has 3 elements. Number of subsets of this set is 8. Now put “a” to each of these 8 sets. Thus there are 8 subsets of X containing “a” but not “s”.
20. (c) )CB(A ∩− )CA()BA( −∪−=
A
C B
24 12 12
6
11
15
B A
20
C
A B
2 −2 0 4
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/7 ===================================================================================================
21. (b) BABA ∩=− is an identity. 22. (c) A’− (A ∩ B) = A’ ≠ A’ (since A ∩ B ⊂ A) 23. (c) B and A − B are disjoint sets φ=−∩∴ )BA(B
24. (c) The shaded region represents A ∆ B .BA)BA(A ∩=∆−∴ 25. (c) Two sets A and B are said to be
equivalent if n (A) = n ( B ) 26. (d) A − (A ∩ B)C = A ∩ (A ∩ B) (Θ A − B = A ∩ BC) = A ∩ B 27. (d) n (m ) = 60 % n ( p) = 65 % n ( m ∪ p) = 100 − 20 = 80% Given that 80% of the students = 16
∴ Total number of students = 2080
10016 =×
∴ only 3.
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/8 ===================================================================================================
28. (b) A ∩ B = B − A’ 29. (c)
The shaded region represents P = (A∆B)’ Also Q = BA ∩ and R = BA ∪
'RQP ∪=∴ 30. (c)
HINTS/SOLUTIONS for M1102 (Relation & Functions)
Classroom Discussion Exercise 1. (a) (x, x + y) = (3, 7) ⇒ x = 3, y = 4
2. (c) n(B) = ( )
( ) 514
70
An
BAn==
×
3. (b) Since (B ∩ C) = φ, ∴ (A × B) ∩ (A × C) = φ. 4. (a) (A × B) ∩ (C × D) = (A ∩C) × (B ∩ D). 5. (d) Since A×B contains 18 elements, there are 218 relations from A to B 6. (d) Domain = {−1, 3, 4, 6} 7. (d) The relation “is perpendicular to” is not
reflexive and transitive. A line cannot be
perpendicular to itself and line l1 ⊥ to l2 and l2 ⊥ l3 does not imply that l1 ⊥ l3.
8. (b) Symmetric property is not satisfied. 9. (d) Total number of functions from A to B is (n(B))n(A) ∴ Number of functions on A is 33 = 27 10. (a) In choice (a), 3 it is related to more than one
element and hence is not a function. 11. (d) f1 = {(1, 1), (2, 3), (3, 5), (4, 7), (5, 9)} But 7,
9 ∉ A. f2 = {(1, 5), (2, 4), (2, 5), ….}, which is not a
function. f3 = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} But 6
∉ A. f4 = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} which
is a function from A to A.
12. (c) ( ) ( )yxfyxf −+
[ ] [ ]yxyxyxyx 3321
3321 +−−−−+ ++=
[ ]x2y2y2x2 333341 −− +++=
= ( ) ( )[ ]y2y2x2x2 333341 −− +++
= [ ])y2(f)x2(f21 +
13. (c) The graph shows that for all the positive
as well as negative values of x, y takes only positive
values. Hence the graph represents y = |x|. 14. (d) Rf = {−1, 0, 1}. 15. (a) x − [x] = {x}, the fractional part function
whose range is [0, 1) 16. (a) Range of cosx is [−1, 1] 17. (c) [x] is an integer
∴ The values of ]x[2
sinπ are − 1, 0 and 1.
18. (c) [π] = [3.14] = 3 and = [– π] = – 4 Θ f(x) = cos 3x – sin4x
∴∴∴∴ 23
134
sincos3
f +−=π−π=
π
= 2
23 −
19. (b) Since the function is real, we have 49 – x2 > 0 ⇒ x ∈ (−7, 7)
20. (c) The domain of ( )( ),
xbax
1
−− with a < b is
(a, b)
21. (a) Let y = 2x1
x
+⇒ yx2 − x + y = 0
b2 − 4ac ≥ 0 ⇒ 1 − 4y2 ≥ 0
⇒ 21
y21 ≤≤−
B A
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/7 ===================================================================================================
22. (a) f (x) = |x| + | x+ 4|
Minimum value of f(x) is 4 and it has no maximum.
∴ Range = [4,∞)
23. (d) f (x) + g (x) = 2
ee xx −+≥ 1 for all real x.
∴ Range of (f + g) (x) = [1, ∞) 24. (d) ( ) ( ) ( )yfxfyxf ⋅=+
( ) ( ) ( ) ( ) 23331f1f11f2f =×=⋅=+=
( ) ( ) ( ) 32 3331f2f3f =×=⋅=
( ) ( ) ( ) 43 3331f3f4f =×=⋅= . 25. (a) f (x) = 5x − |x| f (2x) = 10x −2 | x| f (−x) = − 5x − |−x| = − 5x − |x| f (2x)− f (−x) = 10x − 2 |x| + 5x + |x| = 15x − |x| = f(x) + 10x ∴ f(2x) − f(−x) − 10x = f(x)
Regular Homework Exercise
1. (b) n(A×B) = n(A) . n(B) = 0 2. (a) By definition of equality of ordered pairs, (A × B) ∩ (B × A) = φ. 3. (d) A × (B ∪ C) = (A × B)∪ (A × C). 4. (a) Co-domain is a superset of range Range = {6, 7, 8, 9} 5. (c) The first coordinate of the ordered pair
should be from B and second coordinate should be from A.
6. (d) Since R is reflexive (a, a)∈R ∀ a ∈ A, where
n is the given set having R elements. Since n(A) = n, R having at least n ordered pair. ∴ n(R) ≥ n
7. (a) By definition. 8. (d) By definition.
9. (a) f(x) + 2f x21x3x =
−+ ______(1)
let 1x3x
y−+= , ⇒
1y3y
x−+=
∴
−+=+
−+
1y3y
2)y(f21y3y
f
replacing y by x
( ))1x(3x2
)x(f21x3x
f−+=+
−+ _____(2)
2 × (2) − (1) ⇒ x2)1x()3x(4
)x(f3 −−+=
=
)1x(x2x212x4 2
−+−+
3f(x) = ( )1xx212x6 2
−−+
∴ )1x(3
x212x6)x(f
2
−−+=
10. (a) 2x
1x
x
1x
x
1xf
2
22 +
−=+=
−
∴ f (z) = z 2 + 2 ⇒ f (x) = x 2 + 2. 11. (b) There is only one element in the range of a
constant junction. 12. (d) The possible values of signum function are
±1 and 0. Hence range will contain 3 elements.
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/8 ===================================================================================================
13. (b) −1 ≤ cos 3x ≤ 1 3 ≥ 2 − cosx ≥ 1
⇒ 1x3cos2
131 ≤
−≤
∴ Range of f (x) =
1,
31
14. (a) f (x) is not defined when 16 – x 2 < 0 ⇒ 16 < x 2 or when x < –4 and x > 4 ∴ x ∈ [–4, 4]. 15. (c) By definition of f(x), f(x) is many–one, into. 16. (b) 0x1 2 >−
( ) 0x11 2 ≥−−
so 1− 0x11 2 ≥−− All these hold when 0x1 2 ≥− ⇒ x2 ≤ 1 ∴ The domain of f is [−1,1]
17. (d) The domain of bxax
−− when a < b is
( ] ( )∞∪−∞ ,ba,
18. (c) Since 2x is +ve the least value of 2
2
x2
x
+
is 0
also 1x2
xx2x
2
222 <
+∴+<
∴The range is [ )1,0 19. (d) By definition, 4x2 ≠ 0 ⇒ x2 ≠ 0⇒ x ≠ 0 ∴ x = R − {0}
20. (c) f (x) =5x
11x
−+−
x − 1 ≥ 0 ⇒ x ≥ 1, x −5 ≠ 0 ⇒ x ≠ 5 ∴ Domain is x ≥1, x ≠ 5
Assignment Exercise
1. (a) Number of subsets = 60125 22 =× 2. (a) A = {2, 4, 8}, B = {5, 6}. 3. (c) ( ) ( )ba2,54,2a +=+ ba24and52a +==+⇒ 2band3a −== . 4. (d) The number of relations from A to B = the number
subsets of A × B = 224
5. (b) By definition R is reflexive. 6. (d) By definition. 7. (d) R = {(1, 5), (2, 6), (3, 7), (4, 8), (5, 9), (6, 10)}
8. (d) ( )x1
xxf +=
( )[ ] ( )33 xfxf −
+−+++=3
33
3
x
1x
x
1x3
x3x
+=x1
x3 .
( )
==x1
f3xf3
9. (c) Obviously, the range is {1, –1}.
10. (c) Domain is R and range is [0, ∞)
11. (a) a x ≠ 0 and a x > 0 ∀ x ∈ R.
12. (c) 0x7 2 ≥−
⇒ x ∈ [ ]7,7− 13. (b) The domain of ( )( )xbaxlog −− for a < b is (a,
b). log(x − 2) (5 − x) = log(x − 2) + log(5 − x) ⇒ x > 2 and x < 5
14. (d) Let y = 9x5x
x2 +−
⇒ ( ) 0y91y5xyx2 =++−
0ac4b2 ≥− (as x is real)
⇒ ( ) ( )( ) 0y9y41y5 2 ≥−+
⇒ − 01y10y11 2 ≥++
01y10y11 2 ≤−−⇒
⇒ y ∈
−1,
111
∴ Range of the given function is
−1,
111
15. (d) The range of ( )7x45x3
xf−+= is same as the
domain of ( )3x45x7
xf 1
−+=− which is given by x
∈ R; x ≠ 43
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/9 ===================================================================================================
Additional Practice Exercise
1. (c) If A ⊆ B, then A × C ⊆ B × C. 2. (a) A ∩ B = φ ⇒ (A ∩ B) ∩ (A ∪ B) = φ ⇒ [(A ∩ B) ∩ (A ∪ B)] × A = φ 3. (c) n[(A × B) ∩ (B × A)] = 52 = 25 4. (b) A ∪ C = C ∴ (A ∪ C) × B contains 4 × 2 = 8 elements. 5. (c) n[(A × B)∩ (B × A)] = n[(A∩B) × (B ∩ A)] = n [(A ∩B) × (A ∩ B)] = n(A ∩ B) × n(A ∩ B) = 3 × 3 = 9. 6. (c) Domain = {x: | x | ≤ 4, x∈Z} = { −4, −3, −2, −1, 0, 1, 2, 3, 4} 7. (d) Clearly (a, b) ∗ (a, b) and (a, b) ∗ (c, d) ⇒ (c, d) ∗ (a, b) Let (a, b) ∗ (c, d) and (c, d) ∗ (e, f) ⇒ ad = bc & cf = de
⇒ cfbc
dead =
⇒ af = be ⇒ (a, b) ∗ (e, f) ∴ ∗ is transitive also. 8. (c) 2 1 = 2. 9. (d) Number of possible relations is a subset of A
× B A × B has p q elements ∴ number of subsets for A × B = 2pq. 10. (d) By definition, (a, b) ⊂ [a, b] 11. (d) R = { } }Nx;x,x ∈
xx ≠ for all x ∈ N except for x = 1 ∴(x, x) ∈ R ∀ x ∈ N (x, y) ∈ R ⇒ y = x ⇒ (y, x) ∉ R since x ≠
y
(x, y) ∈ R and (y, z) ∈ R ⇒ y2 = x and z2 = y ⇒ x = z4 ⇒ (x, z) ∉ R 12. (b) 2mn = 4096 = 212 where n (A) = m and n (B) = n ⇒ 26n = 212 ⇒ n = 2 ∴ B has 2 elements. 13. (c) n4 = 625 ⇒ n = 5 ∴∴∴∴ B has 5 elements. 14. (a) 25.n(B) = 1024 = 210
⇒ n(B) = 2 15. (c) 2mn 16. (c) Since n(B) < n(A), there is no one-one
function from A to B 17. (c) R = {(1,1) (1,2) (1,3) ,(2,1),
(2,2),(2,3),(3,1),(3,2)} 18. (d) By definition of f + g, f − g, αf and fg, all the
given statements in (a), (b), (c) are true. 19. (b) Either 11 − |x| ≥ 0 and 12 − |x| > 0 or 11 − |x| ≤ 0 and 12 − |x| < 0 ⇒ either |x| ≤ 11 or |x| > 12 ⇒ x ∈ (−∞, −12) ∪ [−11, 11] ∪ (12, ∞)
20. (b) 02x3x;2x3x
x 2
2>+−
+−
(x − 1) (x − 2) > 0 ⇒ x ∉ (1, 2) or x ∈ ( −∞, 1) ∪ (2, ∞)
21. (a) ( )
x1x
x1xx1x
x1x
x2
−−=
−−−=
−−
For 1x0x1,0x1
x2
>⇒<−≥
−−
and x = 0
∴∴∴∴ x ∈ (1, ∞) ∪ {0}
22. (b) 2222 baxcosbxsinaba +≤+≤+−
⇒ 2x3sinx3cos2 ≤+≤−
222x3sinx3cos0 +≤++≤⇒ 23. (a) Substitute x = − 1 and x = 3, image is (8, 72) 24. (c) f(x) is not defined when sin2 x = 0 ⇒ when x = nπ; n ∈ I.
25. (a) f (x) = 2
2
x1
x
+
2x0 ≤ < 1+ x2
⇒ 0 ≤ ( ) )1,0[xf1x1
x2
2
∈⇒<+
∴ Range is [0, 1) 26. (d) ∴ 10 – x ≥ 0 and 4 + x ≥ 0 10 ≥ x and x ≥ – 4 x ≤ 10 and –4 ≤ x – 4 ≤ x ≤ 10 D (f) = –4 ≤ x ≤ 10
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/10 ===================================================================================================
27. (d) g (x) = log
−+
x1x1
g(x1) + g (x2)
= log
−+
1
1
x1x1 + log
−+
2
2
x1x1
= log ( )
++−+++
2121
2121
xxxx1xxxx1
= log ( )( )
+−++++
2121
2121
xxxx1xxxx1
=
++−
+++
21
21
21
21
xx1xx
1
xx1xx
1log =
++
21
21
xx1xx
g
28. (c) Any polynomial satisfying the equation. f(x) .f(1/x) = f(x) + f(1/x) is of the form
1xn + or -xn +1 f(4) = 65 = 64 + 1 = 43 + 1 ∴ f(x) = x3 +1 f(3) = 33 +1 = 28
29. (d) 555 yx =+
⇒ xy 555 −=
⇒ y= ( )x5 55log −
∴ 1x055 x <⇒>−
30. (c) 2101x
x −<+
⇒ 100
11x
x100
1 <
+<−
− 100 > 100x
1x >+
− 100 > 1 + 100x1 >
i.e., − 101 > 99x1 >
991
x101
1 <<−
HINTS/SOLUTIONS for M1103 (Trigonometric Functions)
Classroom Discussion Exercise 1. (a) lr2P +=
θ= 2r21
A
θ+= rr2P
2r
A2.rr2P +=
A2r2Pr 2 +=
i.e., 0A2Prr2 2 =+− 2. (d) The ratio of the radii is the ratio of the angles in
radian measures. Hence required ratio is 5: 4. 3. (c) For 45°< x < 90°, cosx < sinx Hence y < 0
4. (d) tan θ = 1, cot θ = 1
==+ 1xiff2x1
xΘ
∴∴∴∴ tan3 θ + cot3 θ = 2 5. (a) m2 − n2 = (m + n) ( m − n) = 4 sinxtanx 6. (a) Let x = sec α + tan α
x1∴ = sec α – tan α
x1
x +∴ = 2 sec α
= 2a + a21
∴ x = 2a or a21
7. (c) sinx = cos2x ………….. (1) cos12x + 3 cos10x + 3 cos8x + cos6x = (cos4x + cos2x)3 = (sin2x + cos2x)3 = 1 by (1).
8. (a) A + B = C2
−π
tan (A + B) = tan
−πC
2
1Btan.AtanCtan.BtanCtan.Atan
Ctan1
CcotBtan.Atan1
BtanAtan
=++⇒
==−
+
9. (a)
+−=
+−
0
0
00
00
21tan1
21tan1
21sin21cos
21sin21cos
00
00
21tan45tan
21tan45tan
+−=
( )00 2145tan −= 024tan=
r
r
l θ
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/11 ===================================================================================================
10. (d) log tan 10 + log tan 20 + … + log tan 890 = log [tan 10 . tan 20 … tan 890] = log [tan 10 . tan 20 … tan 450 . cot 440
cot 2 0 . cot 1 0] = log 1 = 0. 11. (b) sin210 + sin220 + cos220 + cos210 = 2
12. (d) 4
B2
CA π==+
2
CAπ=+
tan(A +C) =CtanAtan1CtanAtan
−+
CtanAtan1CtanAtan
2tan
−+=π∴
tanA tanC = 1
But tan B = tan 4
π= 1
∴ tanA tanB tanC = 1 13. (a) tanA – tanB = x
yAtan
1Btan
1 =−
yx
Btan.Atan =⇒
cot(A – B) = BtanAtanBtan.Atan1
−+
=x
yx
1
+
y1
x1
xyyx +=+=
14. (a) xsin2x2sin
x2cos22−
=( )
xsin2xcosxsin2
xsinxcos22
22
−−
=( )( )
( )xsinxcosxsin2xsinxcosxsinxcos2
−−+
=1 + cotx. 15. (c) x2 + y2 = 1 Let x = cos θ, y = sin θ ∴∴∴∴ (4 cos3 θ − 3 cos θ)2 + (4 sin3 θ − 3 sin θ)2 = (cos 3θ)2 + (− sin 3 θ)2 = 1
16. (a) sin θ . cos θ = 2
2sin θ
Minimum of sin 2θ = –1
Minimum value = ( ) .2
11
2
1 −=−
17. (c) ( )βα−
β+α=β+α
tantan1tantan
tan
β−
β
=β
2tan1
2tan2
tan2
34
4
11
21
.2=
−=
∴∴∴∴ ( ) 3
3
4.
3
11
3
4
3
1
tan =−
+=β+α
18. (d) sin θ [sin 2 60 – sin 2 θ]
=
θ−θ 2sin43
sin
4
sin4sin3 3 θ−θ= θ= 3sin41
19. (b) cos20 cos40 cos80
= 20sin2
80cos40cos20cos20sin2
= 20sin4
80cos40cos40sin2
= 20sin8
160sin20sin8
80cos80sin2 = =81
20. (b) θ+θ=θ+θ 2sin8sin4sin6sin ⇒⇒⇒⇒ 2sin5 θθ=θθ 3cos5sin2cos
( ) 03coscos5sin =θ−θθ
⇒ sin5 0sin.2sin2. =θθθ
⇒ sin5 0sinor02sinor0 =θ=θ=θ
⇒ π=θπ=θπ=θ nor2
nor
5n
21. (d) 02sin3sin2 2 =−θ+θ
( ) ( ) 02sin1sin2 =+θ−θ
2sinor2
1sin −=θ=θ ( not possible )
6
sin21
sinπ==θ
∴∴∴∴ ( )6
1n n π−+π=θ .
22. (a) 1sin2
3cos
2
1 =θ+θ
1sin3
sincos3
cos =θπ+θπ
13
cos =
π−θ
π=π−θ n23
3
n2π+π=θ
In the interval [0, 2 π ], 3π=θ only
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/12 ===================================================================================================
23. (d) tan3x . tan7x = -1
tan7x = x3tan
1−
= − cot3x
=
+πx3
2tan
x32
nx7 +π+π=∴
π
+=∴2
1n2x4
π
+=⇒8
1n2x .
24. (b) The given equation is
( )222 xcosxsin +
xcosxsinxcosxsin2 22 =−
i.e. xcosxsinxcosxsin21 22 =−
i.e. 2
x2sin2
x2sin1
2
=−
i.e. x2sinx2sin2 2 =−
sin22x + sin2x − 2 = 0
sin2x = 2
811 +±− =
2
31±−= −2, 1
1x2sin =∴
2
n2x2π+π=
4
nxπ+π=∴ ( )1n4
4+π= .
25. (d) The given equation is
θ+θ 2tantan )2tantan1(3 θθ−=
i.e; 32tantan1
2tantan =θθ−
θ+θ
⇒⇒⇒⇒ 33tan =θ
3
n3π+π=θ
∴∴∴∴ ( )1n3993
n +π=π+π=θ .
Regular Homework Exercise
1. (b) We know that λ = rθ
⇒ 4
.r2
π=π
⇒ r = 2 cm 2. (c) Since the value of sinθ lies between −1 and 1, we
get sinθ1 = sinθ2 = sinθ3 = −1
⇒ θ1 = θ2 = θ3 = 2
3π
∴ cosθ1 + cosθ2 + cosθ3 = 0
3. (c) sec A + tan A = 23
⇒ sec A – tan A = 32
∴ 2 sec A = 23
+ 32
⇒ sec A = 1213
cos A = 1312
⇒ sin A = .135
4. (d) sin25 + sin210 +…+ sin290 = (sin25 + sin285) + (sin210 + sin280) +….+ sin245
+ sin290
= 8 + 21
+ 1 (since sin285 = sin2(90 − 5) = cos25)
= 921
5. (b) 37
cos6
sinπ+π
= 121
21 =+
6. (c)
π−π+π+π8
3cos
83
cos8
cos 222
π−π+8
cos2
= +
π+π8
sin8
cos 22
π+π8
3sin
83
cos 22 = 2
7. (d) cos36° = sin(90 − 36)° = sin54° and cos72° = sin(90 − 72) = sin18° ∴ sin18°.sin54° − cos36°.cos72° = 0 8. (d) Given expression = 3 (sin4 α + cos4 α)
– 2 (sin6 α + cos6 α) = 3 (1 – 2 sin2 α
cos2 α)
– 2(1 – 3 sin2 α cos2 α) = 1.
9. (c) Given expression =
π+π4
3cos
4cos2 44 = 1
10. (b) sin(18 + θ)cos(72 − θ) + cos(18 + θ)sin(72 − θ) = sin[(18 + θ) + (72 − θ)] = sin90 = 1
11. (b) Given expression = +−BsinAsin
BsinAcosBcosAsin
AsinCsinAsinCcosAcosCsin
CsinBsin
CsinBcosCcosBsin
−+
−
= cot B – cotA + cot C – cot B + cot A – cot C = 0
12. (a) Bcot
2
CAsin
2
CAsin2
2
CAsin
2
CAcos2
=
−
+
−
+
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/13 ===================================================================================================
B2CA
Bcot2
CAcot
=+⇒
=
+⇒
⇒ A, B, C are in A.P.
13. (d) 34cos
11sin11cos +
= 34cos
79cos11cos +
= 34cos
34cos45cos2 = 2
14. (b) ( )A16cos12222 ++++
= A8cos2222 +++
= A4cos222 ++ = A2cos22 +
= ( )A2cos12 +
= Acos.2.2 = 2 cos A
15. (c) cos4θ + sin4θ =1 − θ2sin21 2 which is maximum
when θ = 4π
or −4π
16. (b) 32π=θ ∴ General solution is
32
n2π+π=θ .
17. (d) tanx (tan2x – 3) = 0
tanx = 0 or tanx = 3±
∴ x = nπ or 3
nxπ±π=
18. (a) 2sincos =θ+θ
1sin2
1cos
2
1 =θ+θ
1sin4
sincos4
cos =θπ+θπ
cos 0cos4
=
π−θ
0n24
±π=π−θ
4
n2π+π=θ
19. (c) tanx = ± 1 and cot x = ± 1
x = 4
nπ+π .
20. (c) Multiplying by cos x and solving for sinx we
get θ = 10
)1(n n π−+π .
Assignment Exercise
1. (c) 6264
l ×π×= = 8π.
2. (a) Angle covered in 1second = 270 × 2π radians
Time taken to turn 6210π radians = ππ
5406210
= 11.5 sec.
3. (d) 2
2
2
2
2
2
c
z
b
y
a
x ++
= θ+φθ+φθ 22222222 cosrcossinrsinsinr
= r2 sin2 θ (sin2 φ + cos2 φ ) + r2 cos2 θ
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/14 ===================================================================================================
= r2 (sin2 θ + cos2 θ ) = r2
4. (b)
π+π+
π+π8
sin8
5sin
8sin
π+π+8
5sin
=8
5sin
8sin
85
sin8
sinπ−π−π+π
= 0
5. (c) cos (x + y) = cosx cosy – sinx siny = 25
1−
6. (a) 1m
BsinAcosBcosAsin
1m
BtanAtan =⇒=
Applying compendo dividendo, we get
1m1m
BsinAcosBcosAsinBsinAcosBcosAsin
+−=
+−
⇒ ( )( ) 1m
1mBAsinBAsin
+−=
+−
7. (a) 33
BcosAcosBsinAsin
BsinAsinBcosAcos
−++
−+
=
3
2BA
sin2
BAcos2
2BA
cos2
BAcos2
−+
−+
+
3
2BA
sin2
BAsin2
2BA
cos2
BAsin2
−+−
−+
= 3
3
2BA
cot2
BAcot
−−+− = 0.
8. (d) tan (A+B+C) = 0
⇒ AtanCtanCtanBtanBtanAtan1CtanBtanAtanCtanBtanAtan
−−−−++
= 0
∴ tanA + tanB + tanC = tanAtanBtanC 9. (d) cosxcos2xcos4xcos8x
= xsin2
12sinxcosxcos2xcos4xcos8x
= xsin2
1sin2xcos2xcos4xcos8x
= xsin2
12
sin4xcos4xcos8x
= xsin2
13
sin8xcos8x
= xsin2
14
sin16x
10. (c) 13tan.2tan =θθ
13cos2cos3sin2sin =
θθθθ
i.e, 03sin2sin3cos2cos =θθ−θθ
i.e, cos ( ) 032 =θ+θ
2
cos05cosπ==θ
⇒⇒⇒⇒ 2
n25π±π=θ
∴∴∴∴ 10
n52 π±π=θ .
11. (c) 0coteccos =θ+θ
0sincos
sin1 =
θθ+
θ
1 + cosθ = 0 ⇒ cosθ = −1 ⇒ θ = π 12. (d) Since sinx, sin2x, sin3x lies in the interval [-1,1],
the given equation is true only if sinx = 1, sin2x = 1 and sin3x = 1 which is impossible.
13. (b) x is in the third quadrant.
Particular solution is x = 4
5π
∴ General solution is x = 24
5n
π+π .
14. (b) secθ + tan θ = 3
secθ − tan θ = 3
1
Solving, secθ = 21
3
13
+ =
3
2
∴∴∴∴ θ = 6
n2π+π .
15. (d) Dividing by 2, we get θ = 3π
or 6π
Additional Practice Exercise
1. (c)
Arc length = r θ = π=π× 53
15
2. (d) Secant increases in the second quadrant and sine and cosine increases in the fourth quadrant
3. (b) cos(− 765°) = cos 765° = cos (720 + 45)°
= cos 45° =2
1
4. (a) tanθ being an increasing function, the value of
tanθ increases as θ increases 5. (b) secθ + tan θ = x
1
1
13π
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/15 ===================================================================================================
secθ − tan θ = x1
Solving, secθ = x2
1x2 +
⇒ cosθ = 1x
x22 +
∴ sinθ = xcos1 2− = 1x
1x2
2
+−
6. (a) )yxcos(22ba 22 −+=+ a2 – b2 = cos 2x + cos 2y + 2 cos (x + y) =
)yxcos(2)yxcos()yxcos(2 ++−+
= 2cos(x + y)[cos(x − y) + 1] = cos(x + y)(a2 + b2)
⇒ cos(x + y) = 22
22
ba
ba
+−
∴ sin (x + y) = 22 ba
ab2
+.
7. (d) )xsin()x(eccos
1−π=
−π=sinx
8. (b)
π−π
π−πππ5
sin.52
sin.52
sin.5
sin
= 5
sin.52
sin.52
sin.5
sinππππ
= 52
sin.5
sin 22 ππ
= sin236 × sin272
=
+
−16
521016
5210=
165
16
201002
=−
9. (c) cos x = 2p 2 – 1 = 2 cos 2 200 – 1 ⇒ cosx = cos40° = cos(360 − 40) = cos320° 10. (b) sin(45 + x) − cos(45 − x) = sin45cosx + cos45sinx − [cos45cosx +
sin45sinx]
= 2
1(cosx + sinx − cosx − sinx)
= 0 11. (c) (sinθ + cosθ)2 = 1 + 2sinθ cosθ
⇒ ac
21ab
2
+=
−
⇒ a2 − b2 + 2ac = 0
12. (c)
θ−π
θ+π4
sec4
sec
=
θ−π
θ+π4
cos4
cos
1
= θ−π 22 sin
4cos
1
= θ− 2sin21
2
= 2sec2θ 13. (b) sin(α+ 2β) = sin(α+ β) cosβ + cos(α+ β)sinβ = cosβ. 14. (c) Since A + B = 225, ∴ cot(A + b) = cot(225) = 1
⇒ BcotAcot1Bcotcot
+−
= 1
⇒ cotAcotB − 1 = cotA + cotB ⇒ cotAcotB = 1 + cotA + cotB ⇒ 2cotAcotB = (1 + cotA)(1 + cotB) (by adding cotAcotB to both LHS and
RHS)
⇒ 21
Bcot1Bcot
.Acot1
Acot =++
15. (a) cos74
cos.7
2cos.7
πππ
=
7sin8
78
sin
π
π
= 81−
16. (c) 4 sin 230 sin 370 sin 830 = 2 sin 230 (2sin 830 sin 370 ) = 2 sin 230 (− cos 1200 + cos 460) = sin 230 + 2 cos 460 sin 230 = sin 230 + sin 690 – sin 230 = sin 690 = cos 210
17. (c) cosA + cosB =
2C
sin4 2
2
Csin4
2
BAcos.
2
BAcos2 2=
−
+⇒
2C
sin22
BAcos =
−⇒
=
+2C
sin2
BAcoscesin
2C
cos.2
Csin.2.2
2C
cos.2
BAcos2 =
−
Csin22
BAsin
2BA
cos2 =
+
−
Csin2BsinAsin =+⇒ .
18. (d)
8cos
8sin
8tan
π
π
=π
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/16 ===================================================================================================
=
8cos2
8cos
8sin2
2 π
ππ
=
2
11
2
1
4cos1
4sin
+
=π+
π
= 1212
1 −=+
19. (d) 2524
491149
11
tan1
tan12cos
2
2=
+
−=
θ+θ−=θ
sinφ = 10
1 and cosφ =
10
3
∴ 4sinφ ≠ 2524
2sin2φ = 2524
56 ≠
sin3φ ≠ 2524
sin4φ = 2sin2ϕcos2ϕ
= 2524
φ+φ−=φ
φ+φ=φ
2
2
2
tan1
tan12cos
tan1
tan22sin
20. (a) sin2A + sin2B + sin2C = 2sin(A + B)cos(A − B) + sin2C = 2sinCcos(A − B) + 2sinCcosC = 2sinC[cos(A − B) + cosC] =
2sinC
−−+−2
CBAcos
2CBA
cos2
= 2sinC.2sinB.sinA = 4sinAsinBsinC
21. (a) x2sinx2cos
x2sinx6sin22 −
− =
4x cos2x sinx 4 2cos
= 2sin2x. 22. (a) cos6α + sin6α = 1 + msin22α(cos2α + sin2α)3 −
3cos2αsin2α(cos2α + sin2α) = 1 + msin22α i.e; 1 − 3sin2αcos2α = 1 + msin22α
∴ 4
3(sin2αcos2α) = msin22α
∴ m = −4
3
23. (a) tanA
−
+
+
−
Atan31
Atan3
Atan31
Atan3
=
−−
Atan31
Atan3Atan
2
2
= A3tanAtan31
AtanAtan32
3
=
−−
.
24. (a) cos4x = cos2x ∴ 4x = x2n2 ±π ⇒ 6x = 2nπ or 2x = 2nπ
⇒ x = nπ or 3
nπ
∴ 3n
xπ= which includes x = nπ also
25. (d) The given equation is 2sin4 θ cos3 θ + sin4 θ = 0 i.e;sin4 θ (2 cos3 θ +1) = 0
i.e; sin4 θ = 0 or cos 3 θ = 21−
sin4 θ = 0 ⇒ 0, 2
,4
ππ
cos3 θ = −2
1 ⇒ θ =
9
4,
9
2 ππ
26. (a) 06cos4cos2cos =θ+θ−θ
⇒ [ ] 04cos6cos2cos =θ−θ+θ
04cos2cos4cos2 =θ−θθ
i.e; ( ) 012cos24cos =−θθ
⇒ cos4 θ=θ 2cos2or0 =1
3
n22or2
n24π±π=θπ±π=θ∴
6
nor82
n π±π=θπ±π=θ∴ .
27. (c) sin3x + sinxcosx + cos3x = 1 ∴ sin3x + cos3x − (1 − sinxcosx) = 0 (sinx + cosx)(sin2x + cos2x − sinxcosx)
− (1 − sinxcosx) = 0 i.e; (1 − sinxcosx)(sinx + cosx − 1) = 0 i.e; sinxcosx = 1 (1) or sinx + cosx = 1 (2) (1) implies sin2x = 2 which is not possible From (2)
2
1xcos
2
1xsin
2
1 =+
2
1
4xcos =
π−
4
n24
xπ±π=π−∴
2
n2orn2xπ+ππ=∴ .
28. (b) Solving, cos x = 21
or cos x = 23
cosx = 2
1 ⇒ x =
35
,3
ππ
cosx = 2
3 is not possible
29. (d) Solving we get
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/17 ===================================================================================================
tanx = 3
1⇒ x =
6n
π+π .
30. (d) Both 2π
and 4π
does not satisfy the given
equation. Hence choice (d).
HINTS/SOLUTIONS for M1104 (Straight Lines)
Classroom Discussion Exercise 1. (d) Let the third vertex be (x, y). Then
23
12x =++ and 2
321y =+−
5y,3x ==∴ (3, 5) is the third vertex. 2. (a) Given (x − a1)
2 + (y − b1)2 = (x − a2)
2 + (y − b2)2
⇒ x2 + a12 − 2a1x + y2 + b1
2 − 2b1y = x2 + a2
2 − 2a2x + y2 + b22 −
2b2y ⇒ 2(a1 − a2)x + 2(b1 − b2)y + a2
2 + b2
2 − a1
2 − b12 = 0
⇒ (a1 − a2)x + (b1 − b2)y
+ ( ) 0baba21 2
121
22
22 =−−+
3. (a) Since y = mx + c passes through (3, −5) and
(2, 4), therefore −5 = 3m + c and 4 = 2m + c Solving m = −9, c = 22
4. (c) A(1, -2) and Q (2, -1) are points on QS
∴ 21
2y121x
+−+=
−−
⇒ x − y − 3 = 0
5. (a) Required equation is 2qy
px =+
⇒ 3y
2x + = 2
⇒ 3x + 2y − 12 = 0 6. (d) By the intercept form, equation is 5x + 7y + 35 = 0 7. (c) 3x + 4y – 5 – k(x + 2y – 3) = 0 ⇒ x(3 – k) + y(4 – 2k) + (3k – 5) = 0 This is parallel to x–axis. ⇒ Slope is zero
i.e; k24k3
−−
= 0 ⇒ k = 3
8. (a) If the required ratio is K:1, the point of division is
given by
++
+−
1K
1K8,
1K
1K7This being a point on
5yx2 =+ , gives 17
6K =
Thus the required ratio is 6 : 17 is 6 : 17. 9. (b) k2 + 13k + 5 + k2 + 1 + 14 = 0 ⇒ 2k2 + 13k + 20 = 0
⇒ k = 2
5−, −4
10. (a) The midpoint of the diagonal is (3, 2) which lie on
the line y = 2x + a Hence the value of a is – 4. 11. (b) The given line is
( ) ( ) 0b4a5y3x2by2xa =+−−++
i.e., a(x + 2y – 5) + b(2x − 3y + 4) = 0 This represents the family of lines through the
point of intersection of x + 2y – 5 = 0 and 2x − 3y + 4 = 0
Solving these two equations we get x = 1, y = 2. 12. (b) Equation of AB is y – 1 = −1 (x – 4) i.e., x + y – 5 = 0. B is the point of intersection of this line with
y = 3x.
∴ B is
415
,45
∴ AB2 = 22
14
1545
4
−+
− = 22
411
411
+
2
411
2
×=
∴ AB = 4
211
13. (c) Point of intersection of 4x + 3y = 1 and y = x + 5 is
(–2, 3) Substituting this in 5y + bx + 3 = 0 we get, b = 9.
14. (d) 2
cab
+=
0cy2
caax =+
++∴
ie ( ) ( ) 02ycyx2a =+++ This represents a family of lines passing through
the point of intersection of the lines 2x + y = 0 and y + 2 = 0.
A (4, 1)
135°
y = 3x
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/17 ===================================================================================================
15. (c) The equation of the required line is of the form 2x + y + 6 + λ (x – 2y+3) = 0. Since this passes through the origin, we get λ = − 2.
Substituting, the required line is y = 0. 16. (a) a 1 a 2 + b 1 b 2 = 0 ∴∴∴∴ θ = 900 17. (d) The lines 3x - 4y + 14 = 0 and 4x + 3y – 23 = 0 are perpendicular ∴ the orthocenter is the point of intersection of
these lines. Solving we get x = 2, y = 5. 18. (c) For parallel lines slopes are same. Hence k = 4.
19. (d) Dividing throughout with 211 =+ , we get
2
6y
2
1x
2
1 =−
⇒ cos α = 2
1; sin α =
2
1−
⇒ α = 315 0 = 4
7π
20. (d) The points (x1, y1) and (x2, y2) are on the same
side of the line ax + by + c = 0 if cbyaxcbyax
22
11
++++
is
positive
781
7+−−
< 0
(−7, 0) is a point on the line
761
7++
> 0
∴ (0, 0) and (1, 3) are on the same side of x + 2y + 7 = 0
21. (c) 13c1394
c0302=⇒=
+
+×+×
22. (a) The image (x2, y2) of the point (x1, y1) in the line
ax + by + c = 0 is given by
22111211
ba
)cbyax(2b
yya
xx
+++−=−=−
( )
( ) ( ).4,1y,x91
724323
8y1
3x
22
22
−−=⇒+
−+−=−=−⇒
23. (c) Distance = 22 43
205
+
+ = 5
24. (b) ( ) 2222 512
2y5x12
43
7y4x3
+
−+−=−+
+−
13
2y5x125
7y4x3 −+−=+−
39x – 52y + 91 = −60x – 25y + 10 99x – 27y + 81 = 0 11x – 3y + 9 = 0. 25. (c) Let the origin be shifted to (h ,k). Then the new equation is
( ) ( ) ( ) 02hx8ky4ky 2 =−+++++
Coefficient of y = 0 ⇒ 2k+ 4 = 0 ⇒ k = -2
Constant term = 0 ⇒ 02h8k4k2 =−++
⇒ 4 − 8 + 8h − 2 = 0 ⇒ 4
3h =
Regular Homework Exercise
1. (c)
+++−3
903,
3
512 = (2, 4)
2. (a) Let the point be P(x, y) (x + a)2 + y2 + (x − a)2 + y2 = (2a)2 ⇒ x2 + y2 = 2a2 3. (c) Slope of AB is 1 ∴AC is vertical where C is the new position of B
2222AB 22 =+=
∴ The required point is ( )22,2
4. (c) Given slope = 43 and c = –2
Equation of the line is y = mx + c
⇒ y = 43
x – 2
⇒ 3x – 4y – 8 = 0. 5. (c) Required equation is 1byax =+
6. (c) Let the line be by
ax + = 1
It passes through (1, 3) ⇒ 1b3
a1 =+
(1, 3) is the midpoint
⇒ 2a
= 1 and 2b
= 3
⇒ a = 2 and b = 6
∴ Required line is 6y
2x + = 1
⇒ 3x + y − 6 = 0
7. (a) Since a, b, c are in A.P, therefore b = 2
ca +
∴ ax + by + c = 0 ⇒ 2a
(2x + y) + 2c
(y + 2) = 0,
which always passed through (1, −2)
450
A (2, 0)
B (4, 2) C
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/18 ===================================================================================================
D
B
A
C
x + y – 6 = 0
x - 3y – 2 = 0
5x - 3y + 2 = 0
8. (c) Since the lines are concurrent,
312
32a
143
−−− = 0
⇒ a = 5 9. (a) Equation of the line is (4x + 3y – 7) + λ (8x + 5y – 1) = 0 (4 + 8λ) x + (3 + 5λ) y – 7 – λ = 0
Slope = 23−⇒
23
5384 −=
λ+λ+−
8 + 16λ = 9 + 15λ i.e., λ = 1. ∴ Equation of the line is (12x + 8y – 8 = 0) i.e., 3x + 2y – 2 = 0.
10. (d) 0
bac
acb
cba
=
i.e., a3 + b3 + c3 – 3abc = 0 11. (d) Equation of a line parallel to 4x + 10y − 8 = 0 is
0Ky10x4 =++ . It passes through ( )2,2 − .
Hence 12K0K21024 =⇒=+−×+× 01210y4xisequationrequired =++∴ 12. (a) 3m1 =
1m2 =
21
12
mm1
mmtan
+−
=α∴ 21
312 =+
=
113
m,32
m 43 ==
31132
1
113
32
tan
××+
−=β =
31
633922 =
+−
( )61
1
31
21
tan−
+=β+α∴ =1
4π=β+α∴ .
13. (d)
By solving the equations 5x – 3y + 2 = 0 and
x + y – 6 = 0 we get the point A (2, 4) Altitude AD is 3x + y + k = 0 It passes through (2, 4) ⇒ k = −10 ∴ equation of the altitude is 3x + y = 10.
14. (b) Any line perpendicular to the given line is x – 2y + k = 0. Since it passes through (1, 2), we get k = 3
Hence the required line is x – 2y + 3 = 0.
15. (b) Let the required ratio be λ : 1.
Then P is
+λ+λ
+λ+λ−
134
,12
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/19 ===================================================================================================
P lies on x + y + 1 = 0
∴ λ = −23
16. (d) With the given sides, the vertices are A(0, 0),
B(−1, 3), C(2, −1) Equation of the line through (−1, 3) and
perpendicular to x + 2y = 0 is 2x − y + 5 = 0 ……..(1) Equation of the line through (0, 0) and perpendicular to 4x + 3y = 5 is 3x − 4y = 0…………(2)
Point of intersection of (1) and (2) is the orthocentre
∴ Orthocentre is (−4, −3) 17. (a) The lines 3x – 4y – 2 = 0 and 4x + 3y – 11 = 0 are
perpendicular ∴ The orthocentre is the point of intersection of
these two lines. 3x – 4y – 2 = 0 4x + 3y – 11 = 0
169
1338
y644
x+
=+−
=+
⇒ x = 2, y = 1.
18. (a) If (x, y) is the image of (x1, y1) in the line ax + by + c = 0, then
( )
221111
ba
cbyaxb
yya
xx
+++−=−=−
( )
169520
45y
30x
+−−−=
−−=−
⇒ x = 3, y = 1 19. (c) If P (h, k) is the image of A (4,−13) in the lines
5x + y + 6 = 0, then
113k
54h +=−
1125
613202 −=
++−×−=
14k,1h −=−=∴ .
20. (b) 103
16a
125
2=
+
−
⇒ 103
16a2
252
=+
−
⇒ 516a2 =+
⇒ a2 + 16 = 25 ⇒ a2 = 9 ∴ a = ± 3.
Assignment Exercise
1. (c) [ ] [ ]22 )ab(y)ba(x −−++−
= [ ] [ ]22 )ba(y)ba(x +−+−−
bx − ay = 0 2. (b) P (h, k) lies on 3x + 2y − 13 = 0 ⇒ 3h + 2k − 13 = 0 Q (k, h) lies on 4x − y − 5 = 0 ⇒ 4k − h − 5 = 0 Solving them, we get h = 3, k = 2 ∴ Equation of PQ is y − 2 = −1 (x − 3) ⇒ x + y − 5 = 0
3. (b) The equation of the line is 1by
ax =+
where 2a = 3b
This passes through (3, −1) ⇒ 1b1
a3 =−
1a23
a3 =−⇒ ⇒
23
a =
∴ b = 1
∴ Required line 1y
23x =+
⇒ 2x + 3 y – 3 = 0.
4. (b) ax + by + 13 = 0 passes through (2, 5) and (−3, − 0)
⇒ 013b5a2 =++ and 013ba3 =+−−
Solving, a = 6, b = − 5
5. (d) c = 2
ba +
∴ ax + by + c = 0 ⇒ ,021
yb21
xa =
++
+
which always passes through
−−21
,21
6. (a) The equations of the sides are
Area = ( )
1221
2121
baba)dd(cc
−−−
= 28912 =
−×
7. (c) 0
bac
acb
cba
=
i.e; a3 + b3 + c3 − 3abc = 0
⇒ ( )( ) 0bcacabcbacba 222 =−−−++++
⇒ a + b + c = 0 (since a ≠ b ≠ c, ∴ a2 + b2 + c2 − ab − bc − ca ≠ 0) 8. (b) Solving the first and the second equations, the
point of intersection is (-1, 1). This lies on the third line. Substituting, we get k = 4
9. (c) 4x + 3y – 7 = 0 8x + 5y – 1 = 0 Solving, the point of intersection is (-8, 13).
∴ The required equation is ( )8x23
13y +−=−
⇒ 2y – 26 = − 3x – 24 ⇒ 3x + 2y – 2 = 0. 10. (b) x – y = 0 or x = y This line is equally inclined to the x and y axes ∴ The angle it makes with y = 0 (x – axis) is 450. 11. (c) The midpoint of (1, 1) and (3, 5) is (2, 3). Slope of the line joining (1, 1) and (3, 5) is 2
∴ Slope of its perpendicular is 21−
∴ Equation of the right bisector is
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/20 ===================================================================================================
y − 3 = ( )2x21 −−
⇒ x + 2y − 8 = 0 12. (b) The lines x = 7 and y = 5 are perpendicular ∴ the circumcentre lies on the third line 5x + 7y – 35 = 0. 13. (b) The required line can be taken as
kya
secx
btan =θ+θ
It passes through ( )θθ tanb,seca
ka
sectanbb
tanseca =θθ+θθ∴
θθ+=∴ tansecab
bak
22
Equation becomes
θθ+=θ+θ tansecab
basec
ay
tanbx 22
ie; 22 batanby
secax +=
θ+
θ
ie; 22 bacotbycosax +=θ+θ . 14. (a) Slopes of the three lines are
m1 = − 1, m2 = − 3, m3 = 31−
Angle between the lines x + y = 0 and 3x + y − 4 = 0 is given by
α =
++−−
3131
tan 1 =
−
21
tan 1
Angle between the lines x + y = 0 and x + 3y − 6 = 0 is given by
β =
+
+−−
31
1
131
tan 1 =
−
21
tan 1
Angle between the lines 3x + y − 4 = 0 and x + 3y − 6 = 0 is given by
γ =
−=
+
+−−−
34
tan113
13
tan 11
∴ The triangle is isosceles with the line x + y = 0 as the base.
15. (a) The lines are parallel.
Distance between the lines is 22 125
2
1725
+
−
26
33=
5233
radius =∴
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/21 ===================================================================================================
Additional Practice Exercise
1. (c) Area = ba1ac1cb1
21
= cbba0cabc0
cb1
21
−−−−
= [ ])ca()ba()cb()bc(21 −−−−−
=
[ ])bcabaca(bcbcbc21 222 +−−−+−−
= [ ]abcabccba21 222 −−−++−
=
[ ]ab2ca2bc2c2b2a241 222 −−−++−
= [ ]222 )ba()ac()cb(41 −+−+−−
2. (a) Required area is 4
49ab2c2
= .
3. (d) Area of triangle formed by the straight line
0cbyax =++ with the coordinate axes is
ab2c2
If a, c, b are in G.P; abc 2 =
⇒ Area of the triangle is 21
units.
4. (a) Solving the equations (0, -1) is the point of
intersection. 5. (d)
Orthocenter is O (0, 0) and circumcenter C is the
midpoint of AB. ie;
=23
,2C
49
4OC +=∴ =4
25=
25
= 2.5
6. (d) Solving 4x + 5y = 0 and 11x + 7y = 9 we get
−34
,35
solving 7x + 2y = 0 and 11x + 7y = 9 we
get
−34
,35
midpoint is
21
,21
.
The other diagonal should pass through (0, 0) and
21
,21
. Hence its equation is y = x
Aliter:
The given sides pass through the origin, while the given diagonal does not. Hence, the required
diagonal must pass through the origin. Thus, the only correct choice is (d).
7. (b) Consider P(a, b), Q(a, c), R(d, c)
Slope of PQ = 0
bc −
Slope of RQ = ad
0−
∴ PQ is perpendicular to QR ∴ Q(a, c) is the orthocentre. 8. (b) Equation of any line through ( )2,3 is
( )1...m3x2y =
−−
Now, slope of the line 21
is3y2x =−
12m
1m2
2
m1
2
1m
45tan ±=+−
⇒
+
−=
31
mor3m−==∴
Thus from (1) the required equations are 09y3x,07yx3 =−+=−−
9. (a) Any line perpendicular to 03yx3 =−+ is
0Ky3x =+− .
It passes through (2, 2) ⇒ k = 4. ∴ equation of the line is 04y3x =+− Putting x = 0, we get the intercept.
y = 3
4
10. (c) The sides are given by 5x + 2y + 5 = 0,
5x + 2y − 5 = 0, 5x − 2y + 5 = 0, 5x − 2y − 5 = 0. These lines form a rhombus.
Area = units.sq525
552 =×
××
11. (d) Slope of AC = 12
34
−−+
=37−
∴ slope of BD = 7
3
Midpoint of AC is
−21
,,2
1which is (the midpoint
of BD also) ∴ equation of BD is
+=−21
x73
21
y
ie; 6x − 14y + 10 = 0. ie; 3x − 7y + 5 = 0
12. (c) The angle θ is given by 1
23
1
21
3tan =
+
+−=θ
O A(4, 0)
B(0, 3)
23
,2C
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/22 ===================================================================================================
⇒ θ = 4π
13. (b) The triangle is right angled. So the circumcentre is the midpoint of the hypotenuse, Hence circumcentre is (3, 3). 14. (a) Equation of the line is 4x + 2y + c = 0
y-intercept = 2c− = 5
⇒ c = -10 ∴ equation is 4x + 2y – 10 = 0 i.e., 2x + y – 5 = 0.
15. (a) Any line perpendicular to 1sinby
cosax =θ+θ is
kcosay
sinbx =θ−θ
This passes through ( a cosθ, b sinθ)
∴ k = ab
ba 22 −sinθ cosθ
∴ required line is
θθ−=θ−θ cossinab
bacos
a4
sinbx 22
⇒ ax secθ − by cosec θ = a2 − b2
16. (a) Slope of PR = 25
Slope of PQ = m
tan θ = 1 1
2m5
1
m2
5
=+
−
m25
1m25 +=− or
2m5
1m25 +=+−
m = 73
of 37−
When m = 7
3, PQ is ( )3x
7
34y −=−
i.e., 7y – 28 = 3x – 9 ⇒ 3x – 7y + 19 = 0. 17. (c) The equation in normal form is
a60siny60cosx =+ οο
a2
y32x =+⇒
a2y3x =+⇒ 18. (b) 1 + 1 − 4 < 0
041a3a2 2 <−−+−∴
i.e. 08a2a2 <−+
ie; (a + 4) (a − 2) < 0 ∴ a lies in the interval (− 4, 2)
19. (b) θ+θ
=22 eccossec
ap
θ+
θ
=
22
22
sin
1
cos
1a
p
θθ= 222 cossina
⇒ θ= 2sinap4 222 (1)
Also θ+θ
θ=22 sincos
2cosap2
θ= 2cosap4 222 (2)
(1) + (2) → 22 ap8 =
⇒ 8a
p2
2 = .
20. (d) b
siny
a
cosx θ+θ = 1 ⇒ bxcosθ + aysinθ − ab = 0
Let α = 22 ba − ∴ Point is (α, 0) and (−α, 0)
∴ perpendicular distance from (α, 0) is
P1 = θ+θ
−θα2222 cosbsina
abcosb and
P2 = θ+θ
−θα−2222 cosbsina
abcosb
∴ P1P2 = θ+θθα−
2222
22222
cosbsina
cosbba
= ( )
θ+θθ−−
2222
222222
cosbsina
cosbabba
= [ ]
θ+θθ+θ−
2222
222222
cosbsina
cosbcosaab
= ( )
θ+θθ+θ
2222
22222
cosbsina
cosbsinab = b2
21. (b) p = θ+θ 22 eccossec
a
⇒ p2 =
θ+
θ 22
2
sin
1
cos
1a
= a2 sin2 θ cos2 θ
⇒ 4p2 = a2 sin2 2θ Similarly q2 = a2 cos2 2θ ∴ 4p2 + q2 = a2. 22. (c) The lines are 6x + 8y – 10 = 0 and 6x + 8y – 45 = 0 Distance between them
= 5.310
35
6436
4510
ba
cc
22
12 ==+
+−=
+
−
23. (d) Using
+
++−=
−=
−22
1111
ba
cbyax2
b
yy
a
xx
1
8y
1
6x
−−=−
( ) ( )6,8y,x2
862 =⇒
−−=
24. (c) The side of the square is equal to the distance between the parallel lines which is equal to
P (3, 4)
R (1, -1)
Q
S
O
θ
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/23 ===================================================================================================
= 2019
86
257
22=
+
+
∴ diameter of the circle is 20
219
∴ area =
2
202219
4
×π
=800361π
25. (a) Distance between the lines = 25
86
629
22=
+
−−
26. (a) Let the image be (h, k). Then
( )
191143
21
4k3
1h+
−+−−=−=+
⇒ h = 5 , k = 6 27. (b) Line joining (−1, 2) and (5, 4) is x − 3y + 7 = 0. If (h, k) is the foot of the perpendicular, then
( )
917031
130y
11h
++×−−=
−−=−
⇒ h = 5
12k,
51 =
28. (b) 2x + y = 7 passes through (2, 3) but it is parallel to
the given lines, so it will not make any intercepts with them. Point of intersection of x − 2 = 0 with 2x + y − 3 = 0 and 2x + y − 5 = 0 are (2, −1) and (2, 1) respectively. ∴ The intercepts made by x − 2 = 0 with them is 2 unit. x − 2y + 4 = 0 passes through (2, 3) but perpendicular to the given parallel lines
∴ The required intercept is the distance between
the parallel lines is 5
2
14
53 =+−− −
≠ 2
Finally, 2x + 3y − 4 = 0 does not pass through (2, 3)
∴ choice (b)
29. (b) Locus of P(x, y) such that PA = PB is the required equation
( ) ( ) ( ) ( )2222 3y2x1y1x −+−=−+−∴
011y4x2 =−+⇒ 30. (a) The other bisector is perpendicular to x + y – 2 = 0 Hence the equation is x – y + k = 0 This passes through (1, 1) ⇒ k = 0 ∴ Equation of the other bisector is x – y = 0
HINTS/SOLUTIONS for M1105 (Complex numbers and Quadratic Equations)
Classroom Discussion Exercise 1. (a) 5i.
2. (d) Let ibai247 +=− ⇒ 7 − 24i = a2 − b2 + 2iab. Equating the real and imaginary parts, we get
a2− b2 = 7 and ab = −12.
Now a2 + b2 = ( ) 22222 ba4ba +−
= 25 ⇒ 2a2 = 32 ⇒ a = ± 4 When a = 4, b = −3 When a = −4, b = 3.
∴ ( )i34i247 −±=− 3. (d) 5i5 + 4i4 – 3i3 + 2i2 – i = 5i + 4 + 3i – 2 – I = 2 + 7i
4. (d) tan−1
+
−
rs
tanqp 1
=
−
+−
rs
pq
1
rs
pq
tan 1 =
−+−
qsprpsqr
tan 1
= ( )4
n1tan 1 π+π=−
5. (d) (x + 1 + i) (x + 1 – i) (x – 1 + i) (x – 1 – i) =(x2 + 2x + 1 + 1) (x2 – 2x + 1 + 1) =[ (x2 + 2) + 2x ] [(x2 + 2) –2x] =(x2 + 2)2 – 4x2 = x4 + 4. 6. (c) The given equation is (1 + i) (3 – i) x – 2i (3 – i) +
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/24 ===================================================================================================
(2 – 3i) (3 + i) y + i (3 + i) = i (3 + i) (3 – i) ⇒ (4 + 2i) x – 6i – 2 + (9 – 7i) y + 3i – 1 = 10i ⇒ (4x + 9y – 3) + (2x – 7y) = 13i Equating real and imaginary parts, 4x + 9y – 3 = 0 2x – 7y = 13 Solving, we get x = 3, y = -1.
7. (c) 3423175675
5473322115
iiiii
iiiii
−+−−+−+−
( )32211554732
5473322115
iiiiii
iiiii
+−++−−+−+−
= 1i
12
=−
Aliter : Zkwhere
ii
1i
ii
1i
3k4
2k4
1k4
k4
∈
−=−=
==
+
+
+
Substituting for each term in the numerator and denominator, we get the answer.
8. (b) (5 + 2i) 2 = 21 + 20i ∴ Conjugate of (5 + 2i)2 is 21 – 20i.
9. (b) iyxyx 22 +=+
= 1i2i2 =
−+
10. (d) Complex conjugate of π+πcosi
2sin
π−π= cosi2
sin
2
sini2cosπ+π=
11. (c) ( ) ( ) ( )
( )i1i1i23i2
−+−−
= i1
i1i23i2
−+−−
=
.652
2135=
12. (a) Given 40 × (x 2 + y 2) × 25 × 10 = 2500
⇒ 41
yx 22 =+ .
13. (d) The given expression is
π+π
π+π
π+π
3sini
3cos
6sini
6cos
4sini
4cos
= sini364
cos +
π−π+π
π−π+π364
= 12
sini12
cosπ+π
14. (d) =−+
1z1z
2
2
k3ik1
k3ik3
−++
−++
= ( )( ) 22
22
k3k1
k3k3
−++−++
=
3k24k612 =
++
15. (a) AB = BC = CD = DA = 13 Also AC2 = | 17 – 7 i | = 338 AB2 + BC2 = 169 + 169 = 338 = AC2
∴∆ ABC is right angle ∴ABCD is a square.
16. (b) ( ) iba3i25 3940+=+
⇒ iba3i25 3940
+=+
⇒ 223940 ba.33 +=
⇒ 3ba 22 =+ ⇒ a2 + b2 =9. 17. (b) z = sinθ + icosθ
θ−π+
θ−π=2
sini2
cos
∴ z2 = cos (π − 2θ) + i sin (π − 2θ)
and 2z
1= cos(π − 2θ) − isin (π − 2θ)
( )θ−π=+∴ 2cos2z
1z
22
18. (a) The point P can be obtained by rotating the
complex number i3 + in anticlockwise and clockwise direction. Hence P can be either
( ) ( )i3iori3i +−+ 19. (d) θ+θ= sinicosxLet
θ−θ= sinicosx1
θ+θ= 6sini6cosx6
θ−θ= 6sini6cosx
16
θ=− 6sini2x
1x
66
20. (a) Putting x = 23− in the given equation,
2 x 49
- 23
p – 6 = 0
⇒ p = −1
21. (c) =α 10
i3i3
1 −=+
10
i3
+=β
106=β+α and
101
αβ =
Required equation is 0101
x1062x =+−
ie; 10x2 – 6x + 1 = 0 ∴ a = 10, b = −6 and c = 1
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/25 ===================================================================================================
22. (d) Since α, β satisfy x3 − 1 = 0 α3 = 1 and β3 = 1, α2 + α + 1 = 0 β2 + β + 1 = 0 ∴ (α2 − β2) + (α − β) = 0 α + β + 1 = 0, ⇒ Θ α ≠ β ⇒ α + β = −1 ∴ (α + β)27 + (α100 + β100)27 = α + β = −1 = (α + β)27 + (α + β)27 Θ α100 = α, β100 =
β = (−1)27 + (−1)27 = −2
23. (a) 01x2x4 2 =−+
∴41
- and21 =βα−=β+α
Since α is a root, therefore α+α 24 2 − 1 = 0
⇒ 24α = 1 − 2 α
⇒ 34α = 22 α−α
= 2
)21( α−−α
= 2
14 −α
21
2614
34 3 −α−=α−−α=α−α = β.
24. (a) α
−=+α⇒=+α+α rqp 0rqp 2 and
β
−=+β⇒=+β+β rqp 0rqp 2
∴ rrqpqp
αβ−αβ−=+β
α++α
β
pr
xr2−=
=p2−
25. (d) |z1 + z2| ≥≥≥≥ |z1| + |z2|.
Regular Homework Exercise
1. (b) Let z = i2i2
+−
z = i2i2
−+
= ( )
( ) ( )i2i2i2 2
+−+
5
i43 += .
2. (a) Let ibai6011 +=+−
⇒ −11 + 60i = a2 − b2 + 2iab ⇒ a2 − b2 = −11 and ab = 30
∴ a2 + b2 = ( ) 22222 ba4ba +− = 61
⇒ 2a2 = 50 ⇒ a = ± 5 When a = 5, b = 6 When a = −5, b = −6
∴ ( )i65i6011 +±=+− .
3. (c) ( )( )( )( )i2i2
i23i2i23i2
+−+−=
−−
2
2
i4
i36i2i4
−−−+= i
51
58 +−= .
4. (a) 3x + i(4x – 6y) = 2 – i Equating the real and imaginary parts, we get
3x = 2 and 4x − 6y = −1.
∴ x = 32
Thus 4x − 6y = −1 ⇒ y = 1811
5. ( ) ( )22i484i484 −−++− = ( ) ( )
+−22 i4842
= − 64
6. (d) Modulus is a non-negative real number.
7. (c) 2
zz.z = = 130.
8. (c) 0z0z0zz2 =⇒=⇒=
9. (b) Complex conjugate of π−πcosi
2sin
π+π= cosi2
sin
2
sini2cosπ−π=
10. (d) We cannot compare two complex numbers since there is no ordering in the set of complex numbers. [Note that the comparisons in (a), (b) and (c) are in R, the set of real numbers]
11. (b) Argument = 1243π=π−π
12. (b) 2.5.10 …… ( 1 + n2) = |1 + i| |1 + 2i| ….|1+ ni| = x2 + y2
13. (c) xx5858 = which is true only
when x = 0. Thus number of solutions = 1. 14. (d) The given complex number has modulus 1 and
argument
θ−π2
.
15. (b) [ ]20
1010
2i3
3iyx3
+=+
Now 2020
6sini
6cos
2i
23
π+π=
+
620
sini6
20cos
π+π=
21
x−=⇒ and
23
y−=
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/26 ===================================================================================================
21
431
yx 22 −=−=−
16. (d) Given that 91
1k5 2
=−
⇒ k2 = 2 Now, discriminant = 4k2− 4(5k2 − 1) = 8 − 4 (10 − 1) < 0 ∴ The roots are imaginary. 17. (c) Interchanging the constant term and the
coefficient of x we get the required equation as qx2 − px + 1 = 0
18. (c) If 3i2 + is a root then 3i2 − also a root.
Sum of the roots = 22 Product of the roots = 5
∴∴∴∴ Required equation is 05x22x2 =+−
19. (c) x2 + x + 1 = 01x,1x1x3
≠−−−
∴α, β are the roots of x2 + x + 1 = 0
⇒ 01
13
=−α−α
α ≠ 1 ⇒ α3 = 1
11011 3
3
=β⇒≠β⇒=−β−β
α400 = (α399. α) = α
β400 = (β399 . β) = β Again α2 + α + 1 = 0 and β2 + β + 1 = 0 ⇒ α + β = −1 ∴ α400 + β400 = α + β = −1 20. (c) We have
( )
23
i21
41i323
313i
3i
3i2
+=−+=+
+=+−
+
= cos3
sini3
π+π
3
200sini
3200
cos3i
3ii200
π+π=
+−+
Similarly 3
200sini
3200
cos3i
3i200
π−π=
+−
∴
200200
3i
3i
3i
3i
+−+
−+
121
23
200cos2 −=−×=π=
= (−ω2)200 + ω200 = ω400 + ω200 = −1
Assignment Exercise
1. (b) i 6 + i7 + i8 + i9 = –1 – i + 1 + i = 0.
2. (c) i21i21
+−
= ( )
5i21 2−
∴ Square root = ( )i215
1 −± .
3. (b) ( )
( ) ( ) i11
1i21i1i1
i1i1i1 2
=+
−+=−+
+=−+
which lies on
y–axis. 4. (d) –3 + ix2 y & x2 + y – 4i are conjugates ⇒ x2 + y = –3, x2 y = – 4. Solving, we get x = ± 1, y = −4
5. (a) (cos t + i sin t) (cos t – i sin t) = cos 2 t + sin 2 t =
1. 6. (d) |z1| + |z2| = |1 + 2i| + |2 + 3i|
= 9441 +++ = 135 +
7. (a) Modulus, r = 213 =+
tan θθθθ =3
1⇒⇒⇒⇒ θθθθ =
6π
+=+
2i
23
2i3
= 2
π+π6
sini6
cos
8. (a) Equating modulus on both sides, we get
ibaix1ix1 −=
+−
⇒ 1 = 22 ba + ⇒ 1ba 22 =+ 9. (a) Let z = x + iy.
∴ ( ) ( )2222 1yx1yx1iziz ++=−+⇒=
+−
⇒ y = 0, which represents the x - axis
10. (c) arg ( )3
3i1π=+
arg ( )6
i3π=+
∴663i3
3i1arg
π=π−π=
+
+
11. (d)
+=+ i
23
21
2i31 =
π+π3
sini3
cos2
∴ ( )
π+π=+3
5sini
35
cos2i31 55
and ( )
π−π=−35
sini3
5cos23i1 55
Thus, ( ) ( )553i13i1 −++ = 26 cos
35π
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/27 ===================================================================================================
= 25 = 32
12. (c) We have 2121 zzzz +≤+
∴∴∴∴ 2z2z +≤+
≤ 4 ( )4z ≤Θ
13. (b) Sita’s equation is (x - 2) (x - 5) = 0 i.e. x2 – 7x + 10 = 0 Mamata’s equation is (x + 6) (x + 1) = 0 i.e. x2 + 7x + 6 = 0 ∴ Correct equation is x2 – 7x + 6 = 0 ⇒ x = 1 or 6. 14. (b) Sum of the roots = − 6i ⇒ the other root = −3 − 10i Product of the roots = k
⇒ k = 31 − 42i
15. (c) ( )1003i1iyx −=+
= ( )1003i1+−
=2100
100
23
i21
+−
=2100
+−
23
i21
ω=ω=
+ 100
100
23
i1cesin
∴ x = −299, y = 299 3
Additional Practice Exercise
1. (c) The co-ordinates of the vertices of the triang le
are ( ) ( ) ( )37,3,7,,7,3,,3,7 +−−
Area = ( )( ))yy(x)yy(xyyx21
213132321 −+−+−
= ( )( ) ( )[ 3373377721 −+++−−
( )( )]7337 +−+
= [ ] 553721217721 =−=−++−−−
2. (b) =−+
i31i32 ( )( )
( )( )i31i31i31i32
+−++
= 10
i97 +−
∴∴∴∴ Im 109
i31i32 =
−+
3. (b) i32i21
+−
= ( ) ( )( ) ( )i32i32
i32i21
−+−−
= 13
i7
13
4
94
6i72 −−=+
−−.
4. (d) i 4 + i8 = 1 + 1 = 2.
5. (a) ( )( )iba
ibaz
−+=
( ) ( )( ) ( )ibaiba
ibaiba
+−++= =
22
22
ba
biba2a
+−+
Im z = 22 ba
ab2
+.
6. (c) z = ( ) ( )i1i1
i1
i1
1
−+−=
+ =
2
i1−
Re z = 2
1.
7. (c) The point 4i lies on positive y – axis.
8. (c) Multiplicative inverse of a + ib = iba
1
+
= 22 ba
iba
+−
9. (a) ( ) ( )2222 5yx5yx1i5z
i5z ++=−+⇒=+−
⇒ y = 0
10. (a) ( ) ( ) 01izi1izz2 =+++ (Θ i2 = −1)
( )( ) 01iziz2 =++⇒i1
zoriz2 −=−=⇒
In both cases |z| = 1
11. (c) The given points form a rectangle.
12. (a) (b + ia)5 = ( )[ ]5ibai −
= ( ) ( )β−α=− iiibai 55 = β+αi .
13. (b) ( ) qipyix 31
+=+
( )3iqpyix +=+⇒ = p3 – 3pq2 + i (3p2 q – q3) ⇒ x = p3 – 3pq2 and y = 3p2 q – q3
2222 qp3q3pqy
px −+−=+∴
= 4 (p2 – q2).
14. (a) Let iyxi125 +=−
∴ 5 − 12i = x2 − y2 + 2ixy ⇒ x2 − y2 = 5 and xy = − 6
Now x2 + y2 = ( ) 22222 yx4yx +−
(−5, 2) (5, 2)
(5, –2) (−5, −2)
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/28 ===================================================================================================
= 13 ∴ 2x2 = 18 ⇒ x = ± 3 When x = 3, y = −2 When x = −3, y = 2
∴ ( )i23i125 −±=−
15. (b) 1z7 −=
28917586 zzz ++
( ) ( ) ( ) 24172572127 zzzzz ⋅++⋅=
( ) ( ) ( ) 24125212 z.11z1 −+−+⋅−=
1z1z 22 −=−−= .
16. (c) x + 2 = 7i
⇒ x2 + 4 + 4x = −49
⇒ x2 + 4x + 53 = 0
∴ x3 + 4x2 + 53x + 5 = 5
17. (a) ( ) 2ttit1iyx 2 +++−=+
)t1(x −= 2tty 2 ++=
( ) ( ) 2x1x12tty 222 +−+−⇒++=
4x3x2 +−=
−+
−=49
423
x2
⇒
−−
47
23
x
47y
2
2
= 1, a hyperbola
18. (d) 212
1
2
1 zzzz
zz −≠=
19. (b) We have 6Z9
Z =+
∴ |Z| = Z9
Z9
Z −+
≤ Z9
Z9
Z ++
≤ |Z|
96 +
⇒ |Z|2 − 6 |Z| + 9 ≤ 18
(|Z| − 3)2 ≤ 18 ⇒ |Z| − 3 ≤ 18
⇒ |Z| ≤ 3 + 18
So maximum value of |Z| is 3 + 18
20. (d) 1 +
π+π=3
sini3
cos23i
∴ ( ) 33
3sini
3cos83i1
π+π=+
= 8 (cos π + i sin π) = – 8.
21. (c) ( )
i2
i1i1i1 2
=+=−+
∴arg
−+
i1i1
= 2π
22. (d) ( ) ( )
−=−−zz
argzargzarg
= arg (-1) = π.
23. (a) 2iz1iz =−+−
⇒ 2iziiz 2 =−++
⇒ ( ) 2izizi =−++
⇒ 2iziz =−++ , which represents a line
segment [Note: [|z − z1| + |z − z2| = k represent (i) a straight line if |z1 − z2| = k (ii) an ellipse if |z1 − z2| > k ]
24. (c) ( ) ( ) ( )4192i413i41322
−=−++
=− 64.
25. (b) 1 – sin α + i cos α
=
α−π+
α−π−2
sini2
cos1
=
α−π
α−π+
α−π24
cos24
sin2i24
sin2 2
=
α−π+
α−π
α−π24
cosi24
sin24
sin2
=
α+π+
α+π
α−π24
sini24
cos24
sin2
⇒ argument = 24α+π
.
26. (d) |ω| = 1
⇒ 1
2i3
z
z =−
⇒ 1
2i3
z
|z| =−
⇒ |z| = i23
z −
⇒ x2 + y2 = x2 + 2
23
y
− (where z = x + iy)
⇒ y = 43
, a straight line.
27. (b) Let z = x + iy
( )( )i54z
i23zzzzz
2
1
+−+−=
−−
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/29 ===================================================================================================
( ) ( )( ) ( )5yi4x
2yi3x−+−−+−=
( ) ( )[ ] ( ) ( )[ ]
( ) ( )22 5y4x
5yi4x2yi3x
−+−−−−−+−=
( )( ) ( )( )
( ) ( )22 5y4x
5y2y4x3x
−+−−−+−−=
( )( ) ( )( )
( ) ( )22 5y4x
5y3x4x2yi
−+−−−−−−+
( ) ( )22
22
5y4x
22y7yx7x
−+−+−+−=
( ) ( )22 5y4x
7yx3i
−+−−−+
arg 4zz
zz
2
1 π=
−−
122y7yx7x
7yx322
=+−+−
−−⇒
⇒ x2 − 10x + y2 − 6y + 29 = 0 ⇒ (x − 5)2 + (y − 3)2 = 5 ⇒ |(x − 5) + i (y − 3)|2 = 5
⇒ | (x + iy) − (5 + 3i) |2 = 5
⇒ | z − (5 + 3i) | = 5
28. (b) In a parallelogram diagonals bisect each other.
∴∴∴∴ 2
zz2
zz 4231 +=+
⇒ z1 + z3 = z2 + z4 29. (a) In a square, lengths of the diagonals are equal. ∴∴∴∴ |z1 − z3| = |z2 − z4|
30. (b)
π+π=+6
sini6
cos2i3
6sini
6cos
2i3 π+π=+
⇒ 6x
sini6x
cos2
i3x
π+π=
+
So from the given equation, we get
16
xsini
6
xcos =
π+
π
⇒ cos6xπ
= 1 and sin6xπ
= 0
sin6xπ
= 0 ⇒ 6xπ
= nπ; n ∈ Ι
⇒ x = 6n
Now cos6xπ
= 1 ⇒ cosnπ = 1
⇒ n is a multiple
of 2
⇒ x = 12,
24,………
HINTS/SOLUTIONS for M1106 (PMI, Sequence and Series)
Classroom Discussion Exercise
1. (b) t1 = 252; tn = 798; d = 7
n = 1d
tt 1n +− = 17
252798 +− = 79
2. (b) a = 5 , d = 5
t20 = a + 19d = 5 +19 5 = 520
3. (c) cb
1,
ac1
,ba
1+++
are in A.P
⇒ba
1
ac
1
ac
1
cb
1
+−
+=
+−
+
⇒ bacb
cbba
+−=
+−
⇒ 2222 cbba −=−
⇒ 222 b2ca =+
⇒ 222 c,b,a are in A.P
4. (b) a + 17d = 108 and a + 107d = 18 ⇒ d = −1 and a = 125 ∴ 126th term = 125 + (126 − 1) x – 1 = 0
5. (a) 1bc
)cb(a ++ , 1ca
)ac(b ++ , 1ab
)ba(c ++
are in A.P.
bc
cabcab ++ ,ca
cabcab ++ ,ab
cabcab ++ are in
A.P.
P.Ainareab1
,ca1
,bc1
⇒ abc
bc1 , abc
ca1 , abc
ab1 are in A.P.
⇒ a, b, c are in A.P. 6. (d) x2 = x1 + d d = 12 xx − = ( ) ( )1212 xxxx −+
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/30 ===================================================================================================
⇒ d
xx
xx
1 12
12
−=
+
Similarly, d
xx
xx
1 23
23
−=
+
d
xx
xx
1 34
34
−=
+
d
xx
xx
1 1nn
n1n
−
−
−=
+
n1n3221 xx
1.......
xx
1
xx
1
+++
++
+∴
−
= d
xx 1n −
But x n = x1 + (n −−−− 1)d
(n −−−− 1)d = xn – x1 = ( ) ( )1n1n xxxx −+
⇒⇒⇒⇒
+−=
−
1n
1n
xx
1nd
xx
n1n3221 xx
1.....
xx
1
xx
1
+++
++
+∴
−
=
1n xx
)1n(
+−
7. (d) a = 1; tn = 101; d = 2; n = 51
S = [ ]1011251 + = 2601
210251 =×
8. (b) 2nd term = S2 − S1 = 30 – 9 = 21 9. (b) a + 2d = 5 a + 6d = 3 (a + 2d) + 6 ⇒ a = −3 and d = 4 ∴ S32 = 16 [− 6 + 31 x 4] = 1888 10. (c) a + (n – 1) d = 164
( )[ ] n5n3d1na2.2n 2 +=−+
a + 164 = 3n + 5 Now, a = 3 × 1 + 5 × 1 = 8 Thus 3n + 5 = 86 ⇒ n = 27
11. (d) [ ]2
)1m(m1x)1m(2
2m
S1+=−+=
S2 = ( )[ ] ( )2
1m3m31m4
2m +=×−+
[ ])1p2)(1m(p22m
Sp −−+=
= [ ]1mp2pm2p22m +−−+
= [ ]1)1p2(m2m +−
∴∴∴∴2
mS.....SSS p321 =+++ [ ]{ }1)1p2(m +−∑
= [ ]pp.m2m 2 +
)1mp(2
mp +=
12. (d) 2b = a + c ∴ 72b = 7a + c ⇒ (7b)2 = 7a. 7c
13. (d) ar4 = 2 a . ar . ar2 … ar8 = a9 . r36 = (ar4)9 = 29 = 512 14. (b) Let the numbers be a, ar, ar2. Given a + ar + ar2 = 21 ⇒ a ( r + r + r2) = 21
⇒ a . 21r1r1 3
=−
−
(1) Also given a2 + (ar)2 + (ar2)2 = 189 ⇒ a2 (1 + r2 + r4) = 189
⇒ a2 189r1
r1 6
=−
− (2)
Solving these two equations, we get a and r. Aliter: The only numbers in the given choices
whose sum is 21 are 3, 6, 12 15. (d) 434 214 34 21
digitsndigitsn
2 )5...555()3...333( +
= (3 + 30 + 300 + ……..+ n terms )2 + (5 + 50 + 500 + ………+ n terms )
= ( )( )
( )110
1105
110
1103 n2n
−−+
−−
= ( )
+−−
59
11039
110 nn
= ( ) ( )
271410110 nn +−
16. (a) Let the terms be ra , a,ar.
a3 =21
a81 =⇒
( )16
21ar
r
aaraa
r
a =
×+×+
×
1621
arar
a 222
=++
1621
r1
r141 =
++
4
21
r
1r1 =++
( ) r211rr4 2 =++
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/31 ===================================================================================================
04r17r4 2 =+−
41
or4r =
Numbers are 81
,21
,2or2,21
,81 .
17. (b) Given arn – 1 = arn + arn + 1
⇒ rrrrr nnn
⋅+=
r2 + r − 1 = 0
⇒ r = 2
51±−
But r is positive
∴ r = 018sin22
15 =−.
18. (c) (2k + 3) – (k + 2) = (4k − 1) – (2k + 3) ⇒ k = 5
19. (b) Given ( ) 21
nn
1n1n
abba
ba =++ ++
21
n21
21
21
n1n1n b.ab.aba
++++ +=+⇒
( ) ( )babbaa 2
1n
2
1n
−=−⇒++
⇒ 1ba 2
1n
=
+
or 21
n021
n−=⇒=+
. 20. (d) b 2 = ac gives x = – 4 so that the numbers
are – 4, – 6, – 9
∴ 4th term = 227−
21. (d) a = 12; ar5 = 384
r5 = 3212384 =
∴ r = 2
22. (c) n th term nn
n
2
11
2
12 −=−=
Sum to n terms = ∑−
−n
1n
n
21
n
=
−
−
−
21
1
21
121
n
n
= n21n −+− .
23. (d) ( ) ( )( )6
1n21nn71
21nn ++×=+
⇒ 2n + 1 = 21 or n = 10
24. (d) tn = 2
1nn
n +=∑
∴ Sn = ( )2
1n∑ +
=
( )
2
n2
1nn ++
= ( )4
3nn +
25. (d) (12 − 22) + (32 − 42) + … + (492 − 502) + 512
= −1(3 + 7 + 11 + … + 99) + 512 = 1326
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/32 ===================================================================================================
Regular Homework Exercise
1. (d) 7 × t7 = 11 × t11 7 [a + 6d] = 11 [a + 10d] 7a + 42d = 11a + 110d 4a + 68d = 0 or a + 17d = 0 ∴ t`18 = 0 2. (b) Tn = a+(n-1)d 60th term = 3+59 × 5 = 298 3. (a) a = 20; d = 1; tn = 99
S = 2n (a + tn)
n = 11
2099 +− = 80
∴ S = ( ) =+ 99202
80 40 × 119 = 4760
4. (a) Tn = a + (n–1) d = 4 – 3i + (n – 1) (i – 2) = 6 – 2n + i (n – 4) Equating the imaginary part equal to
zero, we get n = 4. ∴ The 4th term of the sequence is purely real
. 4th term = T4
= – 2.
5. (c) n = 9, 61
d−= ;
2
1a =
( )23
61
1921
229
S9−=
−−+×=
6. (d) 6
2561
)1n(61 =−+ ⇒ n = 25
7. (a) a = 11, d = 2, a n = 99, an = 11 + (n – 1) 2 99 = 9 + 2n or n = 45. S = 2475. 8. (c) Let the numbers be a-d, a, a+d a = 5 5(5 − d) + 5(5 + d) + (52 − d2) = 71 d2 = 75 − 71 = 4 i.e. d = ± 2 The numbers are 3, 5, 7 9. (d) a = 1,x = t n = a + (n – 1) d where d = 5
⇒ 5
4xn
+=
∴ ( )[ ] 148d1na22n =−+ gives
( ) 148x110
4x =+
+
⇒ x = – 41, 36
But, the given A.P is increasing, and hence x = 36.
10. (d) 4k
4
4k2
4k
−=
+− Solving k = 16
11. (d) a =4; r = 3 ; tn = 36 × 34 arn – 1 = 36 × 34
( ) 61n3434 ×=×
−
62
1n
33 =−
⇒ 62
1n =− ⇒ n = 13
12. (d) a = 3, arn-1=192, Sn = 381 rn-1 = 64 i.e. rn = 64r
3811r
)1r(3 n
=−
−
i.e. rn – 1 = 127 (r-1) 64r – 1 = 127r-127 r = 2 2n-1 = 64 = 26 i.e. n = 7 13. (b) ar2 = 32 product of 1st 5 terms a.ar.ar2.ar3.ar4 = a5.r10 = (ar2) 5 = (32)5.
14. (d) 43
a = , r=2, nth term = 384
384)2(43 1n =− i.e 2n-1 = 384 x
34
= 29 n = 10
Sn = =−
−
12
)12(43 10
43069
1023x43 =
15. (b) x2 – 5x + 6 = 0 (x – 3).(x – 2) = 0 x = 3 or 2 ∴ roots are 3 and 2. α = 3 and β = 2 G.M. = 632 =×=αβ .
16. (c) There are now n + 2 terms in the A.P.
2 + 2d = 21 (7 – 2d). Solving, d =
21 .
Also, 2 + (n + 1)d = 7. Solving, n = 9. 17. (d) T1 = 2; T2 = 3 = 2+12 T3 = 7 = 2+12+22, --------- Tn = 2+12+22+--------+(n-1)2
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/33 ===================================================================================================
= 6
)1n2(n)1n(2
−−+
= 6
12)1n2)(1n(n +−−
18. (d) ( ) ( ) ( )nn.....2211 222 ++++++
∑ ∑= =
+=n
1i
n
1i
2 ii = ( )( ) ( )2
1nn6
1n21nn ++++
= ( )( ) ( )6
1nn31n21nn ++++
= ( ) ( )6
2n21nn ++ = ( )( )3
2n1nn ++ .
19. (c) 552
)1n(nn...321 =+=++++
( )
+=++++4
1nnn....321
233333
. = 55 × 55 = 3025
20. (b) n .1 + (n +1)2 +(n +2)3 +-----+ ( )n1nn −+ = [ n +2n +3n +---+n.n]+1× 2+2×3+---+(n−1)n = n (1+1+3+----+n) + ( )n1n −∑
= ( )nn
2
1nnn 2 ∑−∑+
+
= n ( ) ( )( ) ( )2
1nn
6
1n21nn
2
1nn +−+++
+
= (n−1) ( ) ( )3
1n2
2
1nn
2
1nn +×++
+
= ( )
++−+3
1n23n3
2
1nn
= ( )( )6
2n51nn −+ .
Assignment Exercise
1. (d) b1
c1
a1
b1 −=−
bc
cbab
ba −=−
ac
abbc
bacb ==
−−
2. (d) 2b = a + c
b2 – ac = ac2
ca2
−
+
4 (b2 – ac) = (a – c)2 3. (d) α , α + d, α + 2d be the angles α + α + d + α + 2d = 180 3α + 3d = 1800 Given α + 2d = 2α 2d = α
d = 2α
3α + 2
3α =1800
2
3α = 600
α = 400 ∴ largest angle is 800 4. (d) Sn = 16200, a = 100, d=20
[ ] 1620020)1n(2002n =−+
n(10+n-1) = 1620 n2 + 9n – 1620 = 0 (n + 45) (n-36) = 0 n = -45 or 36
5. (d) a + (p − 1) d = q and a + (p + q − 1) d = 0 a + ( p − 1) d + qd = 0 q + qd = 0 ⇒ d = −1 a + (p + q − 1) d = 0 a + (q − 1) d + pd = 0 ∴ a + (q − 1)d − p − 1 = 0 a + (q − 1) d = p 6. (d) 6a + 69d = 267 ⇒ 2a + 23d = 89
( ) 10688912d23a22
24ni
24
1
=×=+=∑
7. (c) G1 = {3}, n(G1) = 1 ⇒ Ist term of G1 = t1 of
A.P. G2 = {7, 11}, n (G2) = 2 ⇒ Ist term of G2 = t2
of A.P G3 = {15, 19, 23, 29} n (G3) = 4 = 22 ∴ Ist term of G3 = ( ) 15P.Aoft 22
=
G4 = {31, ….}, n(G4) = 8 = 23 ∴ Ist term of G4 = ( ) APoft 32
∴ n(G8) = 28−1 = 27 ∴ Ist term of G8 = ( ) P.Aoft 72
is t128 = 3 + (127) 4 = 511
8. (d) ( )[ ]
[ ] 3419410
111n182n
=×+
−+
n[11n +7] = (60)(84) This has no positive integral solution.
Thus n does not exist.
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/34 ===================================================================================================
9. (b) nth term of the first A.P. = −4 + (n − 1)7 = 7n − 11 nth term of the second A.P. = 61+(n-1)2 = 2n+59 ∴ 7n-11 = 2n + 59 5n = 70 i.e. n = 14 10. (a) −−+++ 777777
= [ ]terms n to99999997 −−+++
= [ ]−−−−+−+−+− )11000()1100()110(97
= ( )[ ]n10......101097 n2 −+++
= ( )10n910817 1n −−+ .
11. (a) 51n 3243)3.(3 ==− ⇒ n=10 12. (a) ar 2 = 8 a . ar . ar 2 . ar 3 . ar 4 = a 5 . r 10 = (ar 2) 5 = 8 5
13. (a) Let the numbers be ra , a, ar
6a216ar,a,ra =⇒=
22
r3636r
36 ++ =364
91r99r
9 22
=++
09r82r9 24 =+−
i.e. 0)9r()1r9( 22 =−−
r = 3 or 31 (+ve Nos.)
The numbers are 2, 6, 18
14. (a) c, a, b, d are in A.P. a – c = b – a = d – b
2
cbca
−=−⇒
15. (d) 1 + 8 + 27 + 64 + …. n terms
= ( )
4
1nnn...........321
223333 +=++++ .
Additional Practice Exercise
1. (d) It is only a sequence. 2. (c) p[a+(p-1)d] = q[a+(q-1)d] a(p-q) + [(p2-q2) – (p-q)]d = 0 i.e. a+(p+q-1)d = 0, Θ p ≠ q (p + q)th term = 0 3. (c) Let there be 2n terms in the A.P ∴ a + (a + 2d) +…… + [a + (2n – 2) d] = 72
⇒ ( ) 72]d21na2[2
n =−+
⇒ n [a + nd – d] = 72 (1) (a + d) + (a + 3d) + …… + (a+(2n–1)d) =
90
]d2)1n()da(2[2n −++ = 90
n [a+ (n – 1)d + d] = 90 i.e., n(a + (n − 1)d) + nd = 90 i.e., 72 + nd = 90, using (1) ∴ nd = 18 Moreover, [a+ (2n – 1)d] – a = 30 ∴ (2n − 1) d = 30 2nd − d = 30 2 × 18 − d = 30 ∴ d = 6 ∴ n = 3
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/35 ===================================================================================================
4. (c) a+(m-1)d = n1 and a+(n-1) =
m1
Solving, d = mn1 and a=
mn1
thmn term = a+(mn-1)d
= 1min
1mnmn1 =−+
5. (b) [ ]
[ ]D)1n(A22n
d)1n(a22
n
1n51n3
−+
−+=
+−
D
21n
A
d2
1na
−+
−+=
Since we are looking for the ratio of
the 4 th term, 32
1n =− i.e. n = 7
∴ Required ratio
= 95
3620
17x517x3 ==
+−
.
6. (c) A, B, C are in A.P ⇒ B = 60
2
3CsinBsin
2
3cb =⇒=
∴ sinC =
75A45C2
160sin
3
2 =∴=⇒=
7. (d) tp = 9p + 2 t1 = 9 + 2 = 11 tn = 9n + 2
∴ ( ) ( )2n9112n
tt2n
S n1n ++=+= = [ ]13n92n +
8. (c) Tn = a . rn-1 (G.P)
8th term = 3.729128
32
7
=
9. (a) 26r.a 24 = and 36.2r.a 511 =
737 )6(3.6.2r ==∴
i.e. 6r = and 2a =
∴ 3rd term = a . r2 = 26 10. (c) q – p, r – q, p are in G.P ⇒ (r – q)2 = p (q – p) (1) Since p, q, r are in A.P, r – q = q – p = d, the common difference. ∴ (1) ⇒ d2 = p.d ⇒ d = p, Θ d ≠ 0 ∴ q = p + d = 2p r = p+2d = 3p
∴ p : q : r = 1 : 2 : 3. 11. (d) r > 1 a + arn–1 = 516 2048arar 2n =× −
⇒ 1n2ra − = 2048
⇒ 02048a516a2 =+− ⇒ a = 512 or 4
⇒ ( ) 22
1n
4
2048or
512
2048r =−
⇒ 12816
2048r1r 1n ==∴> − and a = 4
Also given that ( )( )1r
1ra n
−− =1020
⇒ ( )1r
1r128a
−− = 1020
⇒ ( )1020
1r
1r1284 =−
−
⇒ 128r – 1 = 127r = 254 ⇒ r = 2. rn–1 = 128 ⇒ 2n–1 = 27 ⇒ n – 1 = 7 ⇒ n = 8.
12. (a) Let ar,a,r
a be the numbers in G.P.
∴∴∴∴ .P.areinAar,a2,r
a
∴ ( ) 01r4rrr1
aa22 2 =+−⇒
+=
32r ±=⇒ Given G.P is increasing.
32r +=∴ .
13. (d) a2 = 49
1
21
2
14
3=×
a = 71
14. (d) ar8 = 256 ar6 = 64 r2 = 4 r = ± 2 Since 9th term is > 7th term, r = +2 15. (b) t n = 2n(2n + 2) and n = 20 gives t 20 = 1680 16. (c) ar 9 = 9 ……….. (1) ar 3 = 4 …………(2)
Dividing (1) by (2); r 6 = 49 or r 3
= ± 23
⇒ a = ± 38 ⇒ t 7 = a r 6 = ± 6
49
38 ±=× .
17. (c) arp + q −1. arp − q − 1 = m n
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/36 ===================================================================================================
∴ a2. r2p − 2 = m n (arp − 1)2 = m n arp − 1 = mn 18. (b) ar4 = x; ar7 = y; ar10 = z i.e. x.z = a.r4.a.r10 = (a.r7)2 = y2 19. (c) 2p = a + b → (1) 2q = b + c → (2)
from (1) and (2); p + q = 2
cb2
ba +++
= 2
cb2a ++
and 4pq = (a + b) (b + c) = ab + b2 + ac + bc = ab + 2b2 + bc = b (a + 2b + c)
2pq = ( )cb2a2b ++
⇒=+
bqp
pq2 p, b, q are in H.P.
20. (c) r1
aS
−=∞
2
21
1
1S1 =
−= , 3
23
x2
31
1
2S2 ==
−=
−−−−==−
= 434
x3
41
1
3S3
1pp
1pxp
1p1
1
pSp +=+=
+−
=
p321 SSSS −−−−+++∴
=2+3+4+------+p+(p+1)
= 2
22p3p1
2)2p()1p( 2 −++=−++
= )3p(p21 +
21. (a) ar 2 = 6
a . ar . ar 2 . ar 3 . ar 4 = a 5 . r 10 = (ar 2) 5 = 6 5 22. (d) a, ar, ar2, -----, arn−1,---------
∴ =∑=
1000
1nn2a ar +ar3 +-------1000 terms
⇒ ar1
=α−
α ( 1+r2 +-----+r1998) ---(1)
∑=
−
1000
1n1n2a = a +ar2 +------1000 terms
⇒ =α+
α1
a ( 1+r2 +-----+r1998)----(2)
(1) ÷ (2) ⇒ r11 =
α−α+
⇒ 1 +α = r−rα ⇒ α +rα = r−1
⇒ α = 1r
1r
+− .
23. (c) xy22
yx =+
12
xy2
yx =+
13
xy2yx
xy2yx=
−+
++ ⇒
( )( ) 1
3
yx
yx2
2
=−
+
⇒ 13
yx
yx=
−
+ ⇒
13
13
y
x
−+=
⇒ 32
32yx
−+=
24. (a) From the data given, we have a, b, b + 3, b
+ 6 are the numbers so that a = b + 6 ∴ b + 6, b, b + 3, b + 6 are the numbers ∴ b2 = (b + 6) ( b + 3) b2 = b2 + 9b + 18 ⇒ b = − 2 ∴ numbers are 4, − 2, 1, 4 25. (c) a, x, b are in A.P i.e. 2x = a+b a, y, z, b are in GP y2 = az and z2 = by y3+z3 = yz (a+b) = 2xyz
26. (d) α + β = a
b− α β =
a
c
2b = a + c ⇒ 2 a
c1
a
b +=
∴ 2− (α + β) = 1 + α β
is 1 + α β + 2α + 2 β = 0
1 + 2 α = −β ( α + 2)
β2 −
+αα+2
21
27. (a) 4, a1, a2, ------a7, 52 are in A.P. ∴ 52 = 4+ (9−1) d 48 = 8d 6 = d ∴ a6 − a5 = d = 6 a1 + a7 = 4 + 52 = 56 (a2 + a6 = a3 + a5…..)
28. (c) S = 361002
20192
=
×
29. (a) S = sum of squares of first 20 numbers
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/37 ===================================================================================================
Sn = ( ) ( )6
1n21nn ++
S20 = 6
412120 ×× = 2870
30. (b) α + β = ab− αβ =
ac
ab11
22−=
β+
α
∴ ( ) a
b2
22−=
αββ+α
i.e., ( )
( ) ab2
2
2
−=αβ
αβ−β+α
i.e., 2
2
2
2
a
cab
ac
2a
b −=−
ac2
a
bc
a
b3
2
2
2
=+
Multiply by bca2
∴∴∴∴ ba
2ac
cb =+
∴ ac
andba
,cb
are in A.P
∴ bc
,ab
,ca
are in H.P.
HINTS/SOLUTIONS for M1107 (Limits & Derivatives)
Classroom Discussion Exercise
1. (a) [ ]
1x1x
lim1x +
+−→
=[ ]
1h11h1
lim0h +−
+−→
=
[ ]h2h2
lim0h −
−→
= 21
h21
lim0h
=−→
2. (c) 3xx3
lim3x
|3x|lim
3x3x −−
=−−
−− →→= −1
3. (d) ( )xflim1x −→
= 53x2lim1x
=+→
( ) ( ) 61x3limxflim1x1x
=+→→ +
⇒ ( ) ( )xflimxflim1x1x +→→
≠
⇒ ( )xflim1x→
does not exist
4. (a) 7xlim2x
14x5xlim
2x
2
2x+=
−−+
→→= 9
5. (c) x31x
2xlim
2x −−−−
→
=( ) ( )
−−−
−−−→ x31x
x31x21
lim2x
= ( )x31xlim21
2x−+−
→= 1
6. (d) LHL = ( ) ( )
]x[x5x4x
lim5x −
−−−→
( ) ( )4x
5x4xlim
5x −−−
→
= ( ) 05xlim5x
=−→
RHL = ( ) ( )5x
5x4xlim
5x −−−
+→
= ( ) 14xlim5x
=−→
LHL ≠ RHL ∴ Limit does not exist.
7. (a)
−
−=
−−
→−→
25
x
25
xlim
52x2
5x2lim
nn
25x1nn
nnn
25x
1n
25
n−
=
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/37 ===================================================================================================
8. (b)
22sin
lim
180x180
x2lim
limx
x2limlim
00x0x=
θθ=
π
π
=→θ→°
°
→
9. (a)
4x
cos
2xlim
2x π−
→
=
π+π→
4t
2cos
tlim
0t(put x−2 = t)
=
4t
sin
tlim
0t π−→ =
π−4
10. (d) ( ) ( )
xx10sinx10sin
lim0x
−−+→
=x
xsin10cos2lim
0x→
= 2 cos10
11. (a) ( )( )2
2
x x
2/xsinsinlim
−ππ
π→, put t = π − x
= 2
2
0t t
2t
2sinsin
lim
−ππ
→
= ( )
2
2
0t t
2/tcossinlim
π→
=( )
2
2
0t t
2/tsinsinlim
π−π→
=( )
2
2
0t t
2/tsinsinlim
π→
=
( )( ) 22
22
0t t2/tsin
2/tsin2/tsinsinlim
ππ×π
→=
4π
12. (c) Put θ = x − 3π
when x → 0,3
→θπ
∴
π+θ−
θ→θ
3cos21
sin2lim
0
= θ+θ−
θ→θ sin3cos1
sin2lim
0
=
2cos
2sin32
2sin2
2cos
2sin4
lim20 θθ+θ
θθ
→θ
=
θ+θ
θ
→θ
2cos3
2sin
2cos2
lim0
= 3
2
13. (a) [ ] 1x,0x1For −=<<−
[ ][ ]
[ ] 010sin
xx1sinLt
0x =−
=+∴ −→
14. (b) (2x+3)2 = 4x2 + 12x + 9
⇒ f’(x) = 8x + 12
15. (b) y = x+x1
2x
11
dxdy −=
dxdy
= 0 at x = 1
16. (d) f(x) = sin (π+x)
⇒ f(x) = −sinx
⇒ f’(x) = −cosx ⇒ f’
π3
= 2
1−
17. (a) LHL =( ) ( )
h
3fh3flim
0h
−+→
=
h
0|3h3|lim
0h
−−+→
=1
RHL=( ) ( )
h
3fh3flim
0h −−−
→ =
h
0|3h3|lim
0h −−−−
→= −1
⇒ f’(3) does not exist
18. (a)
−−−=
−−
→→ 2xx95
lim2x
)x(f)2(flim
2
2x2x
=
−+−
+−→ 2
2
2xx95)2x(
x95lim
=
−+−
−+→ 22x
x95)2x(
)2x()2x(lim
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/38 ===================================================================================================
= 5
2
55
4 =+
.
19. (a) ( )1xsecxcosdxdy
xcos 222 +=
= 1 + cos2 x = 1 + (1 − sin2x)
= 2 − sin2x.
20. (b) ( ) ( )
2xxf22xf
lim2x −
−→
= ( ) ( ) ( ) ( )( )
2xxf22f22f22xf
lim2x −
−+−→
=( ) ( ) ( ) ( )( )
−−+−
→ 2xxf2f22f2x
lim2x
= f(2) − ( ) ( )
−−
→ 2x2fxf
2lim2x
= f(2) −2 f’(2)
21. (d) f’(x) =
xsinx
dxd
= xsin
xcosxxsin2
−
= cosecx− x cotx cosecx
22. (a) f(x) = xsin1
xcos2
+
= xsin1xsin1 2
+−
= 1− sinx
⇒ f’(x) = −cosx
⇒ f’ 02
=
π
23. (c) f(x) = ( )
xsin1xcos3 2 −
= −3 sinx
f’(x) = −3 cosx
24. (b) xy = x+y
⇒ y = 1x
x−
⇒ ( )
( )21x
x1xdxdy
−−−=
= ( )21x
1
−−
25. (c) xy = c2 ⇒ y = x
c2
⇒ 2
2
x
cdxdy −=
Regular Homework Exercise
1. (c) 21
11111
1x1xx
lim37
1x=
−−+−−=
−++
−→
=
21
h21
lim0n
=−→
2. (a) ( ) 31xlimxflim 2
2x2x=−=
→→ +
3. (c) xx
xlim
0x −→
= 1x
1lim
0x −→
= −1
4. (c) 21x
1xlim
2
2
1x −+
−→
= 21xlim 2
1x++
→ = 22
5. (d) RHL = 4h4
4h4
0h
lim
−+−+
→
1h
h0h
lim=
→=
LHL 4h4
4h4
0h
lim
−−−−
→=
=h
h
0h
lim
−−
→
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/39 ===================================================================================================
= 1h
h0h
lim−=
−→
⇒ RHL ≠ LHL ⇒ the limit does not exist.
6. (c) xbb|x|
limbx −
−−→
= 0xxbbx
limbx
>−−−
−→Θ
= −1
7. (a) ( ) λ+=−→
6xflim2x
( ) 4xflim2x
=+→
⇒ λ = −2
8. (d) 216xx
8xlim
22
8x −−−
→
=2/32/3
22
8x 8x
8x8x8x
lim−−×
−−−
→
= −16×2/18
23
1
× =
328−
9. (a) ( )t
tsinlim
xxsin
lim0tx
−π=−π →π→
(put π−x = t)
= 1t
tsinlim
0t=
→
10. (a) ( )xx2sinxsin2
limx −π
−π→
= ( ) ( )
tt2sintsin2
lim0t
−π−−π→
(Put t = π −x)
= t
t2sintsin2lim
0t
+→
= t
t2sinlim
ttsin2
lim0x0t →→
+
= 4
11. (d) 5cos2x
xsin5cos2lim
0x=⋅
→.
12. (d) ( )2x x
2x2cos1lim
+π++
π−→
= ( )( )2x x
1xcos2lim
+π+
π−→
= ( )( )
20x t
1tcos2lim
+π−→
= ( )
20x t
tcos12lim
−→
=
44t
2t
sin22lim
2
2
0x×
×→
= 2
1
13. (c) 1
xxsin
xcoslim
0x=
→
14. (d) ( ) 5xflim
1x=
+→ so (a) is true
( ) 7xflim2x
=−→
( ) 7xflim2x
=+→
⇒ 2x
lim→
exists. So (b) is true
The domain of the function is [1, 3]
so that (d) is also true
15. (d) f(x) = x
2
f’(x) = 2x
2−
⇒ f’(−1) = −2
16. (a) f(x) = sina cosx + cosa sinx
⇒ f’(x) = −sina sinx + cosa cosx
= cos(x+a)
⇒ f’(−a) = 1
17. (d) Let f(x) = [x]
Left f’(0) = ( ) ( )
h
0fh0flim
0h −−−
→
⇒ Left f’(0) = h1
lim0h −
−=→
⇒ the left limit does not exist
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/40 ===================================================================================================
⇒ f’(0) here does not exist
18. (d) f(x) = tan2x, f’(x) = 2 tanx . sec2x
( )
hatanhatan
lim22
0h
−+→
= ( ) ( )
hafhaf
lim0h
−+→
= f’(a)
= 2 tan a sec2a
19. (c) Left Limit = h
)0(f)h0(flim
0h −−−
→
=
−−+
→ h0h1
lim3
0h= ∞
∴∴∴∴ f’(0) does not exist
∴∴∴∴ f(x) is not differentiable at x = 0
20. (d) f(x) = sin 2x = 2 sinx cosx
f’(x) = 2(cosx cosx − sinx sinx)
= 2(cos2x − sin2x)= 2 cos2x
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/41 ===================================================================================================
Assignment Exercise
1. (a) 32
39
1x
3xlim
2
3x=−=
−−
→
2. (d) 412x
4xlim
4x −+−
→
= 412x
1612xlim
4x −+−+
→ =
412xlim4x
++→
= 8
3. (d) 0x
lim→
( ) ( )
x
ax1ax1 −−+
=( ) ( )
( )
−++−−+
→ ax1ax1x
ax1ax10x
lim22
2
a20x
lim
→= = a
4. (c) LHL = ] 0xx2 = = 0,
RHL= ] 0xx2 = = 0
⇒ lim = 0.
5. (d) x
xlim
0x→ does not exist
6. (c) ( )
x11x
lim8
0x
−+→
( )
( ) 11x11x
lim88
11x −+−+
→+= 8
7. (a) 20x x
x5cosx3coslim
−→
=
20x x
xsinx4sin2lim→
= 8
8. (d) 0x
lim→ 3x
xcos.xsin2xsin2 −
= ( )
30x x
xcos1xsin2lim
−→
= 3
2
0x x2
xsin2.xsin2
lim→
=
2
0x x2x
sin.
xxsin
4lim
→
= 4.1. 121
2
=
9. (c)
−π
→ xcosxsin1
lim
2x
= 020
xsin1xcos
lim
2n
==
+π→
10. (a) f(x) = x9 −93
f’(x) = 9x8
⇒ f’ (1) = 9
11. (d) Standard result.
12. (b) 0100sin
1x
xsinlim
2
2
2
0x=
−=
−→
13. (a) f(x) = 2cosx+1
⇒ f’(x) = −2 sinx
14. (a) 3. 23
x.4
xsin22
2
→ 2
xxsin
= 23
1.23 2 =
15. (c) 1x
lim
→
( )
++
++
−
−+
23x
23xx
1x
23x
2
22
= ( )( )
++−
−
23x1x
1x
2
2
23x
1x2 ++
+= =2
1
4
2 =
Additional Practice Exercise
1. (d) xa
x2lim
ax +→ is = 1
2. (c) 2222
2x22x
lim2x
==−
+→
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/42 ===================================================================================================
3. (a) [ ]xxlim1x
−−→
is
= [ ]xlimxlim1x1x −− →→
−
= 1 − 0 = 1
4. (b) [ ] 1xlim4x
+−→
⇒ 4.
5. (d) By definition
6. (b) ( )
( )x3x33x
lim
31
x1
3xlim
3x3x −−=
−
−→→
= −9
7. (b) 1xsin
xlimecxcosxlim
0x0x==
→→
8. (b) x
xsin2lim
xx2cos1
lim0x0x →→
=−= 2
9. (b) ( )x2cos1x
xcos1lim
20x +−
→
=
x2cos11
limx
2x
sin2lim
0x2
2
0x +×
→→
= 241
4x
2x
sin2lim
2
2
0x××
→
= 41
10. (a) 30x x
xsin3x3sinlim
−−
→ =
3
3
0x x
xsin4lim
−−
→
= 4
11. (d) x
x8sinx5tanlim
0x
−→
= −3
12. (c) ( )
6xx
2xsinlim
22x −+−
→
=( )
( )( )3x2x2xsin
lim2x +−
−→
= 51
13. (a) x3sinx5x3x5tan
lim0x −
−→
13535
x
x3sin5
3x
x5tan
lim0x
=−−=
−
−
→
Since mxmxsin
lim0x
=→
and
mxmxtan
lim0x
=→
14. (b) xcos1
xsinxlim
0x λ−→=
2x
sin2
xsinxlim
20x λ→
= 2
1⇒ λ
= 2
15. (c) xtan1
1xsin2lim
4x +
+π−→
= ( )( )xtan1xsin21
xsin21lim
2
4x +−
−π−→
= ( )( )xtan1xsin21
x2coslim
4x +−π−
→
= ( )
( )( )xtan1xsin21
xtan1
xtan1
lim2
2
4x +−
+−
π−→
=
( )( )xsin21xtan1
xtan1lim
2
4x −+
−π−
→
=
( )
−−+
−−
2
1.2111
11 =
21
2.22 =
16. (b) 2x
.x
xsin
x
x
x2
xsin =
⇒ 1.0 = 0 as x → 0
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/43 ===================================================================================================
17. (c) 111
x18
x18
sinlim
1
xsin
xlim
0x
0
0
0x==
π
π=
→
→
18. (b) 0
lim→θ θθ
θθ.3.2
sin.3sin2 . 3 = 3.
19. (a) 1x
lim→
( )( )x
2cos
x1x1π
−+=
1x
lim
→( )( )
( )2
2
x12
sin
x1x1π
π
−π−+
1x
lim→
( )
( )( ) 1xas
42.22x1.
x12
sin
x12 →
π=
π→
π+
−π
−π
20. (b) 1x1x
lim2
1x −−
+→
( ) ( )
( )1x1x1x
lim1x −
−++→
→→+=
0h1x
h1x
( ) ( )
( )1h11h11h1
lim0h −+
−+++→
= 2.
21. (a)
θ−θθ−θ
=→θ 2cos1
2cos2sin
2sinlimitlim
0
= ( )θ−
θ−θθ
→θ 2cos12cos
12cos2sin
lim0
= 010
2cos2sin
lim0
=−=θθ−
→θ
22. (d)
+π→ 0
0x2cos1
x3coslim
2x
= x2sin2
x3sinlim
2x −
−π
→ (L–Hospitals’ rule)
x2sinx3sin
Lim23
2x
π→=
−∞=
=π−=π0
22sin,1
23
sinΘ
23. (b)
2
limπ
→θ θ−πθ
2
cot
11eccos
lim2
2
=−
θ−π
→θ.
24. (c) ( )xtan
nxlimxcotnxlim
nxnx π−=π−
→→
( ) π=
ππ→
1
xsec
1lim
2nx
25. (a) ( ) ( )( ) ( ) 3
1323
1
31
3231
xx8x8
xx8x8
++−+
−+−+
= 3
1323
1
31
3231
8xx
128x
12
8xx
128x
12
++−
+
−+−
+
(Expand using binomial series and neglect higher powers)
=
++−+
−+−+
24xx
124x
1
24xx
124x
1
32
32
=
24x
24x
24x
24x
24x
24x
32
32
−−
+−=
2
2
xx1
xx1
−−+−
∴ Limit = 1.
26. (d) ( ) ( )
( )
−−+++−
=+
→ 00
x12x2nnx1n1 1nn2
Lt1x
( ) ( )( )
( ) ( )( )
( ) ( )[ ]
( ).
21nn
21n2n1nn
21n2nn1nn
2x1n2nnx1nn1
2
n1n2Lt
1x
+=
+−++=
++++−=
++++−=−
→
===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003. Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631101/44 ===================================================================================================
27. (b) f(x) =
( )1x
n1x1x
xn
−
−−−
= ( ) ( )( )2
n
1x
1xn1xx
−−−−
= ( )2
1n
1x
nxnxx
−+−−+
Use L’ Hospital’s rule two times, Required limit
= ( )
( )1x21nx1n
limn
1x −−−+
→
= ( )
2x1nn
lim1n
1x
−
→
+
= ( )2
1nn + .
28. (d)
0xlim→
x log sin x =
0x
lim→
20x
x
1
xcos.xsin
1
lim
x
1xsinlog
−=
→
0xsec
x2lim
xtanx
lim20x
2
0x=−=
−=→→
By L Hospitals’ rule.
29. (d) f(x) = x2cos1+
= 2 cosx
⇒ f‘(x) = − 2 sinx
⇒ f’
π−2
= + 2
30. (b) f(x) = x2sin1+
= ( ) 2/122 xcossin2xsinxcos ++
= (cosx + sinx)
⇒ f’(x) = −sinx+cosx
⇒ f’(0) = 1