SMM631101

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SOLUTIONS FOR

BASIC STUDY MATERIAL

Sol.SMM631101

MATHEMATICS


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


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%.


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


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


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


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.


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


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.


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


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


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


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


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 .


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

=

−

+

−

+


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


= 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


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


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


=

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


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


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 =


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


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


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


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 =∴



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


⇒ 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


= 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) +


(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


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|.


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−=


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


= 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


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)


= 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

+−+−=

−−


( ) ( )( ) ( )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 −+


⇒ 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 =++


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



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


= 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.


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 +=++++ .


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


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


∴ 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


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−

=


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


= 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 −=


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

−−

→


= 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


⇒ 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


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 =


1. (d) xa

x2lim

ax +→ is = 1

2. (c) 2222

2x22x

lim2x

==−

+→


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


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

+=

+−++=

++++−=

++++−=−

→


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

Documents

SMM631101