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(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}
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%.
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
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
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”.
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
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
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
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
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
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
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.
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
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
⇒ 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.
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
⇒ 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.
⇒ −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]
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
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 .
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
∴ 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)
∴ 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…..)