Winglish Coaching Centre, Puduvayal 1 X std Mathematics Made Easy
1. CARTESIAN PRODUCT
1. Let A = {1, 2, 3} and B = {a, b}. Write AXB
and BXA
(b,3)} (b,2), (b,1),(a,3), (a,2), {(a,1),
{1,2,3} b}{a, AB
b)}(3,a),(3,b),(2,a),(2,b),(1,a),{(1,
b}{a,{1,2,3} BA
=
×=×
=
×=×
2. Find A x B , A x A and B x A
(i) A = {2, −2, 3} and B = {1, −4}
(ii) A = B = {p, q}
(iii) A = {m, n} ; B = f
n)} (n, m) (n, n) (m, m) {(m,
n} {m, n} {m, AA
Ø. A x B B x A Ø, B Since
Ø B ;n} {m, A iii)
q)} (q, q) (p, p) (q, p) {(p, A x B
q)} (q, p) (q, q) (p, p) {(p,
q} {p, q} {p,A x A
q)} (q, p) (q, q) (p, p) {(p, B x A
q} {p, B and q} {p, A ii)
3)} 3)(-4, 2)(1,- 2)(-4,- 2)(1, 2)(-4, {(1, A x B
3)} 2)(3,- 2)(3, (3,
3) 2)(-2,- 2)(-2, 3)(-2, 2)(2,- 2)(2, {(2, A x A
4)}- 1)(3, 4)(3,- 1)(-2, 4)(-2,- 1)(2, {(2, B x A
4}- {1, B
3} 2,- {2, A i)
=
×=×
===
==
=
=
×=
=
==
=
=
=
=
=
3. Let A = {1, 2, 3} and B = {x | x is a prime
number less than 10}. Find A x B and B x A.
} 3) (7, 2) (7, 1) (7, 3) (5, 2) (5,
1) (5, 3) (3, 2) (3, 1) 3)(3, (2, 2) (2, 1) (2, { A x B
} 7) (3, 5) (3, 3) (3, 2) 7)(3, (2, 5) (2,
3) (2, 2) 7)(2, (1, 5) (1, 3) (1, 2) (1, { B x A
7} 5, 3, {2, B
10}.than less number prime a is x | {x B
3} 2, {1, A
=
=
=
=
=
4. If A = {1,3,5} and B = {2,3} then
(i) find A×B and B × A.
(ii) Is A×B=B×A? If not why?
(iii) Show that n(A×B) = n(B×A) = n(A)× n(B)
6. (B)n n(A) A)(Bn B)(An
6 32 (A)n (B)n
6 2 3 (B)n (A)n
6 A)(Bn B)(An
2. (B)n 3;n(A) iii)
AB BA Thus
),(),(),( (1,2) ii)
(3,5)} (3,3), (3,1), (2,5), (2,3), {(2,1),
{1,3,5} {2,3} AB
(5,3)} (5,2), (3,3), (3,2), (1,3), {(1,2),
{2,3} {1,3,5} BA i)
=×=×=×
=×=×
=×=×
=×=×
==
×≠×
≠≠
=
×=×
=
×=×
133112 and
5. If A×B = {(3,2), (3,4), (5,2), (5,4)} find A and B.
{2,4} B
B}.A of scoordinate second all of {set B
{3,5} A
} BA of scoordinate first all of {set A
(5,4)} (5,2), (3,4), {(3,2), BA
=
×=
=
×=
=×
6. If B × A = {(−2, 3),(−2, 4),(0, 3),(0, 4),(3, 3),(3,
4)} find A and B.
4}{3, A
A}B of scoordinate second all of {set A
3} 0, {-2, B
} AB of scoordinate first all of {set B
=
×=
=
×=
7. If A = {5, 6} , B = {4, 5, 6} , C = {5, 6, 7} , Show
that A × A = (B × B) n (C × C) .
R.H.S L.H.S
(2) (1)
(2)----- 6)} 5)(6, 6)(6, 5)(5, {(5, C) (C B) (B
7)} 6)(7, (7,
5) 7)(7, 6)(6, 5)(6, 7)(6, 6)(5, 5)(5, {(5,
7} 6, {5, 7} 6, {5, CC
6)} (6, 5) (6,
4)6)(6, 5)(5, 4)(5,6)(5, (4, 5) (4, 4){(4,
6} 5, {4, 6} 5, {4, BB R.H.S
-(1)--- 6) (6, 5) (6, 6) (5, 5) {(5,
6} {5, 6} {5, AA L.H.S
=
=
=×∩×
=
×=×
=
×=×
=
×=×
8. Given A = {1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and
D = {1, 3, 5}, check if (A n C) × (B n D) = (A ×
B)n(C × D) is true?
Winglish Coaching Centre, Puduvayal 2 X std Mathematics Made Easy
true. is statementgiven the Hence
(2) (1)
-(2)----- 3)(3,5)} {(3, D) (CB) (A
5)} 3)(4, 1)(4, 5)(4, 3)(3, 1)(3, {(3, D x C
5)} 3)(3, 2)(3, (3,
5) 3)(2, 2)(2, 5)(2, 3)(1, 2)(1, {(1, B x A
: R.H.S
(1)----- } 5) 3)(3, (3, { D)(B C) (A
5} {3, D B
{3} C A
:H.S L.
=
=×∩×
=
=
=∩×∩
=∩
=∩
9. Let A = {x ∈ N | 1< x < 4} , B = {x ∈W|0 ≤
x ≤ 2} and C = {x∈ N |x <3} . Verify that
i) A × (B ∪ C) = (A × B) ∪ (A × C)
ii) A × (B ∩ C) = (A × B) ∩ (A × C)
i) A × (B ∪ C) = (A × B) ∪ (A × C)
(2)(1)
..(2)3,1)(3,2)}2,2)(3,0)(2,0)(2,1)({()CA(B)(A
)})(3,1)(3,2{(2,1)(2,2CA
)})(3,0)(3,1{(2,0)(2,1BA
...(1)3,1)(3,2)}2,2)(3,0)(2,0)(2,1)({()(
},,{
=
=×∪×
=×
=×
=∪×
=∪
CBA
CB 210
ii) A × (B ∩ C) = (A × B) ∩ (A × C)
(2)(1)
}..(2)(2,1)(3,1){()CA(B)(A
)})(3,1)(3,2{(2,1)(2,2CA
)})(3,0)(3,1{(2,0)(2,1BA
}...(1)(2,1)(3,1){()(
}{
=
=×∩×
=×
=×
=∩×
=∩
CBA
CB 1
10. Let A = {x ∈ W | x < 2} , B = {x ∈|1 < x ≤ 4}
and C = {3, 5} . Verify that
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
A = {x ∈ W| x < 2} ,
A = {0, 1}
B = {x ∈N|1 < x ≤ 4}
B = {2, 3, 4}
C = {3, 5} .
i) A × (B ∪ C) = (A × B) ∪ (A × C)
proved Hence
(2) (1)
-(2)- 5)} (1, 4)(1,
3) (1, 2) 5)(1, (0, 4)(0, 3) (0, 2) {(0, C) x (A B) x (A
5)} (1, 3) (1, 5) 3)(0, {(0, C) (A
4)}(1, 3) (1, 2) 4)(1,(0, 3) (0, 2) {(0, B) (A
:R.H.S
-(1)- 5)} (1, 4)(1, 3) (1,
2) 5)(1, (0, 4)(0, 3) (0, 2) {(0, C)(B A
5} 4,3, {2, C)(B
:L.H.S
=
=∪
=×
=×
=∪×
=∪
(ii) A × (B n C) = (A × B) n (A × C)
proved. Hence
(2) (1)
(2)--- 3)} (1, 3) {(0, C) (AB) (A
5)} (1, 3) (1, 5) 3)(0, {(0, C) (A
4)}(1, 3) (1, 2) 4)(1,(0, 3) (0, 2) {(0, B) (A
:R.H.S
(1)--- 3)} (1, 3) {(0, C)(B A
{3} C) (B
: L.H.S
=
=×∩×
=×
=×
=∩×
=∩
(iii) (A U B) × C = (A × C) U (B × C)
(2) (1)
(2)----- } 5) 3)(4, 5)(4, 3)(3, 5)(3, (2,
3) 5)(2, 3)(1, 5)(1, 3)(0, (0, { C) (BC) (A
5)} (4, 3) (4, 5) 3)(3, (3, 5) (2, 3) {(2, C) (B
4)}(1, 3) (1, 2) 4)(1,(0, 3) (0, 2) {(0, B) (A
(1)----- } 5) 3)(4, 5)(4, 3)(3, 5)(3, (2,
3) 5)(2, 3)(1, 5)(1, 3)(0, (0, { C B)(A
4}3, 2, 1, {0, BA
=
=×∪×
=×
=×
=×∪
=∪
11. A = Set of all natural numbers less than 8,
B = Set of all prime numbers less than 8,
C = Set of even prime number. Verify that
i) (A ∩ B) × C = (A × C) ∩ (B × C)
ii) A × (B −C) = (A × B) − (A × C)
A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2}
i) (A n B) × C = (A × C) n (B × C)
Winglish Coaching Centre, Puduvayal 3 X std Mathematics Made Easy
proved. Hence
(2) (1)
)2)}.....(2 (7, 2) (5, 2) (3, 2) {(2, C) (B C) (A
2)} 2)(7, 2)(5, 2)(3, {(2, C) (B
2)} 2)(7, (6,
2) (5, 2) 2)(4, 2)(3, 2)(2, {(1, C) (A
:R.H.S
....(1) 2)} (7, 2) (5, 2) (3, 2) {(2, C B)(A
7} 5, 3, {2, B) (A
: L.H.S
=
=×∩×
=×
=×
=×∩
=∩
ii) A × (B − C) = (A × B) − (A × C)
proved. Hence
(2) (1)
-(2)--- 7)} (7, 5) (7, 3) 7)(7, (6, 5) (6, 3) (6,
7) (5, 5) (5, 3) (5, 7) (4, 5) 3)(4, 7)(4, 5)(3, (3,
3) (3, 7) (2, 5) (2, 3) (2, 7) (1, 5) 3)(1, {(1, C) (A - B) (A
2)} 2)(7, 2)(6, (5, 2) 2)(4, 2)(3, 2)(2, {(1, C) (A
7) 5)(7, 3)(7, 2)(7, (7,
7) 5)(6, 3)(6, 2)(6, 7)(6, 5)(5, 3)(5, 2)(5, (5,
7) 5)(4, 3)(4, 2)(4, 7)(4, 5)(3, 3)(3, 2)(3, (3,
7) 5)(2, 3)(2, 2)(2, 7)(2, 5)(1, 3)(1, 2)(1, {(1, B) (A
-(1)--- 7)} (7, 5) (7,
3) 7)(7, (6, 5) (6, 3) (6, 7) (5, 5) (5, 3) (5,
7) (4, 5) (4, 3) (4, 7) 5)(3, (3, 3) (3,
7) (2, 5) (2, 3) (2, 7) (1, 5) (1, 3) {(1, C) - (B A
7} 5, {3, C) - (B
=
=××
=×
=×
=×
=
2. RELATIONS 1. Define a relation between heights of
corresponding students using i) pair of
heights and students ii) an arrow diagram
R = {(heights, students)}
i) pair of heights and students
R= { (4.5, S1) (4.5, S4) (4.7, S9) (4.9, S10) (5, S3) (5, S5)
(5, S8) (5.1, S6) (5.2, S2) (5.2, S7)}
ii) an arrow diagram
2. Let A = {3,4,7,8} and B = {1,7,10}. Which of
the sets are relations from A to B?
(i) R1 ={(3,7), (4,7), (7,10), (8,1)}
(ii) R2= {(3,1), (4,12)}
(iii)R3= (3,7)(4,10)(7,7)(7,8)(8,11)(8,7)(8,10)}
A× B = {(3,1)(3,7)(3,10)(4,1)(4,7)(4,10)
(7,1)(7,7)(7,10)(8,1) (8,7) (8,10)}
(i) R1 ⊆ A×B
Thus, R1 is a relation from A to B.
(ii) (4,12)∈ R2 but (4,12)∉ A×B
So, R2 is not a relation from A to B.
(iii) (7,8) ∈ R3 but (7,8) ∉ A×B
So, R3 is not a relation from A to B.
3. Let A = {1, 2, 3, 7} and B = {3, 0, –1, 7}, which of
the following are relation from A to B ?
(i) R1 = {(2, 1), (7, 1)}
(ii) R2 = {(–1,1)}
(iii) R3 = {(2, –1), (7, 7), (1, 3)}
(iv) R4= {(7,–1), (0,3), (3,3), (0,7)}
A x B = {(1,3)(1,0) (1, -1) (1,7)(2,3) (2,0) (2, -1)
(2,7)(3,3)(3,0) (3,-1)(3,7)(7,3)(7,0)(7, -1)(7, 7)}
i) (2, 1) ∈ R1 but (2, 1) ∉A x B.
Hence R1 is not a relation.
ii) (-1, 1) ∈ R2, but (1, -1) ∉ A×B
Hence R2 is not a relation.
iii) R3 ⊆A x B,
Hence it is a relation from A to B.
iv) (0,3), (3,3), (0,7) ∈ R4 but ∉ A x B
Hence it is not a relation.
4. Let A = {1,3,5,7}, B = {4,8} and R be the
relation defined as “is less than”.
i) Write R as a subset of A x B.
ii) Define a relation R from A to B
ii) Write in xRy form
7R8 5R8, 3R8, 3R4, 1R8,R Or
(7,8)} (5,8), (3,8), (3,4), (1,8), {(1,4), R
}(7,4)(7,8) (5,8) (5,4) (3,8) (3,4) (1,8) {(1,4) BA
=
=
=×
5. Let A = {1, 2, 3, 4,..., 45} and R be the relation
defined as “is square of ” on A.
Winglish Coaching Centre, Puduvayal 4 X std Mathematics Made Easy
i) Write R as a subset of A x A.
ii) Find the domain and range of R.
36} 25, 16, 9, 4,{1, Range
6} 5, 4,3, 2, {1, Domain
36)} (1, 25) (1, 16) 9)(1, 1)(1,4)(1, {(1, A x A
36} 25, 16, 9, 4,{1, A of Squares
6} 5, 4,3, 2, {1, A
45} 4,...,3, 2, {1, A
=
=
=
=
=
=
6. A Relation R is given by set {(x, y) /y = x + 3,
x ∈ {0, 1, 2, 3, 4, 5}}. Determine its domain
and range.
8 3 5 y5, x
7 3 4 y 4, x
6 3 3 y3, x
5 3 2 y2, x
43 1 y1, x
3 3 0 y0, x
3 x y
=+==
=+==
=+==
=+==
=+==
=+==
+=
8} 7, 6, 5, 4,{3, R of Range
5} 4,3, 2, 1, {0, R ofDomain
8)} (5, 7) (4, 6) (3, 5) (2, 4)(1, 3) {(0, R
=
=
=
7. The arrow diagram shows a relationship
between the sets P and Q. Write the relation
in (i) Set builder form (ii) Roster form (iii)
What is the domain and range of R.
(i) Set builder form
R= {(x,y) |y= x – 2 x ∈ P, y ∈Q}
(ii) Roster form R R = {(5,3), (6,4), (7,5)}
(iii) Domain of R = {5,6,7}
Range of R = {3,4,5}
8. Represent each of the given relations by
(a) an arrow diagram, (b) a graph and (c) a set in
roster form, wherever possible.
(i) {(x, y)| x = 2y, x ∈ {2, 3, 4, 5}, y ∈ {1, 2, 3, 4}
5/2 y5 x
2 4/2 y 4 x
3/2 y3 x
1 2/2 y 2, x
x/2 y2y x
==
===
==
===
=⇒=
Required relation = {(2, 1) (3, 3/2) (4, 2) (5, 5/2)}
(i) Arrow diagram
(ii) A graph
(iii) Roster form R = {(2, 1) (4, 2)}
(ii) {(x, y)|y = x + 3, x, y natural numbers < 10}
9 y6 x
8 y5 x
7 y 4 x
6 y3 x
5 y2 x
4 y1 x
3 x y
==
==
==
==
==
==
+=
R = {(1, 4) (2, 5) (3, 6) (4, 7) (5, 8) (6, 9)}
i) An arrow diagram :
ii) A graph
Winglish Coaching Centre, Puduvayal 5 X std Mathematics Made Easy
iii) In Roster form :
R = {(1, 4) (2, 5) (3, 6) (4, 7) (5, 8) (6, 9)}
9. 10 families A, B, C, D, E, F, G, H, I and J with
two children. Among these, families B, F, I
have two girls; D, G, J have one boy and one
girl; the remaining have two boys. Define a
relation R by xRy, where x denote number of
boys and y denote family with x number of
boys. Represent this situation as a relation
through ordered pairs and arrow diagram.
Domain of R = {0,1,2},
where 0, 1, 2 represent no boy, one boy, two boys
Families with two girls are the ones with no boys
R= {(0,B)(0,F)(0,I)(1,D)(1,G)(1,J)(2,A)(2,C)(2,E)(2,H)
10. A company has four categories of employees
given by Assistants (A), Clerks (C), Managers
(M) and an Executive Officer (E). The company
provide Rs.10,000, Rs.25,000, Rs.50,000 and
Rs.1,00,000 as salaries to the people who work in
the categories A, C, M and E. If A1, A2, A3, A4and
A5 were Assistants; C1, C2, C3, C4 were Clerks;
M1, M2, M3 were managers and E1, E2 were
Executive officers and if the relation R is
defined by xRy, where x is the salary given to
person y, express the relation R through an
ordered pair and an arrow diagram.
"x" be the salary given to person
"y" be the set of employees
{(10000, A1)(10000, A2)(10000, A3)(10000, A4)
(10000, A5) (25000, C1) (25000, C2) (25000, C3)
(25000, C4) (50000, M1)(50000, M2)(50000, M3)
(100000, E1) (100000, E1)}
3. FUNCTIONS
1. X={1,2,3,4},Y= {2,4,6,8,10}, R = {(1,2),(2,4),(3,6),
(4,8)} Show that R is a function and find its
domain, co-domain and range?
i) for each x ∈ X , there exists only one y ∈ Y.
All elements in X have only one image in Y.
Therefore R is a function.
ii) Domain X = {1,2,3,4}
Co-domain Y = {2,3,6,8,10}
Range of f = {2,4,6,8}.
2. A relation ‘f’ is defined by f (x) =x2 – 2
where, x ∈ {-2,-1,0,3} (i) List the elements of f
(ii) Is f a function?
}0,-2)(3,7)2)(-1,-1)(,{(
72)()(
22)()(
-2)()(
22)()(
2 )(
2
2
2
2
2
2
33
00
111
22
−=
=−=
−=−=
=−−=−
=−−=−
−=
f
f
f
f
f
xxf
ii) Each element in the domain of f has a unique
image. Therefore f is a function.
3. If X={–5,1,3,4} Y = {a,b,c}, which of following
relations are functions from X to Y ?
(i) R1 = {(–5,a), (1,a), (3,b)}
(ii) R 2 = {(–5,b), (1,b), (3,a),(4,c)}
(iii) R3 = {(–5,a), (1,a), (3,b),(4,c),(1,b)}
i) R1 is not a function as 4∈ X does not have an
image in Y.
(ii) R2 is a function as each element of X has an
unique image in Y.
Winglish Coaching Centre, Puduvayal 6 X std Mathematics Made Easy
iii) 1∈X has two images a ( Y and b ( Y.
Image of an element should always be unique
R3 is not a function
4. Let f = {(x, y) | x, y ∈ N and y = 2x} be a
relation on N. Find the domain, co-domain
and range. Is this relation a function?
Since x and y ∈ N,
8 2(4) y 4 x
6 2(3) y3 x
4 2(2) y2 x
2 2(1) y1 x
===
===
===
===
f = {(1, 2) (2, 4) (3, 6) (4, 8)................}
For each values of x, we get different values of y.
So the given relation is a function.
.....} 8, 6, 4,{2, Range
...}.......... 4,3, 2, {1, domain Co
..}.......... 4,3, 2, {1, Domain
=
=
=
5. Let X = {3, 4, 6, 8}. Determine whether the
relation R = {(x, f (x)) | x ∈ X, f (x) = x2 +
1} is a function from X to N ?
f (x) = x2 + 1
65 18 f(8)8 x if
37 16 f(6)6 x if
17 1 4 f(4) 4 x if
10 13 f(3)3 x if
2
2
2
2
=+==
=+==
=+==
=+==
R = { (3, 10) (4, 17) (6, 37) (8, 65) }
For each values of x, we get different values of
f(x). Hence it is a function
6. Given 22 )( xxxf −= ,
find (i)f (1) (ii) f (x+1) (iii) f (x) + f (1)
121
1
1222
112
11
2
2
2
+−=+
+−=
−−−+=
+−+=+
=−=
−=
2
2
2
)()(
)()( 1)f(x
)(1)(2 (1) f
2 )(
xxfxf
x
xxx
xx
xxxf
7. Given function f : x → x2 −5x + 6 , evaluate
(i) f (-1) (ii) f (2a) (iii) f (2) (iv) f (x −1)
12 7x - x
6 5 5x - 1 2x - x
6 1)-5(x- 1)-(x 1)-f(x (iv)
0
6 10 - 4
6 5(2)- (2) f(2) (iii)
6 10a - 4a
6 5(2a)- (2a) f(2a) (ii)
12
6 5 1
6 5(-1)- (-1) f(-1) (i)
6 5x- x f(x)
2
2
2
2
2
2
2
2
+=
+++=
+=
=
+=
+=
+=
+=
=
++=
+=
+=
8. A graph representing function f (x) is given
it is clear that f (9) = 2.
(i) Find the following values of the function
(a) f (0) (b) f (7) (c) f (2) (d) f (10)
(ii) For what value of x is f (x) = 1?
(iii) Describe following (i) Domain (ii) Range.
(iv) What is the image of 6 under f ?
(i) (a) f(0) = 9, (b) f(7) = 6,
(c) f (2) = 6, (d) f (10) = 0
(ii) For x = 9.5, we get 1.
(iii) Domain = { 0 ≤ x ≤ 10 }
Range = { 0 ≤ x ≤ 9 }
(iv)
Winglish Coaching Centre, Puduvayal 7 X std Mathematics Made Easy
9. Let f (x) = 2x + 5. If x ≠ 0 find x
f(2) - 2) f(x +
2
x
9 - 9 2x
x
f(2) - 2) f(x
9
5 2(2) f(2)
2x
5 2) 2(x 2) f(x
5 2x (x) f
=
+=
+
=
+=
+=
++=+
+=
9
10. A function f is defined by f (x) = 2x – 3
(i) find [f(0) + f(1)]/2
(ii) find x such that f (x) = 0.
(iii) find x such that f (x) = x .
(iv) find x such that f (x) = f (1−x) .
21
x
2 4x
1 - 2x- 3 - 2x
3 - x) -2(1 3 - 2x
. x)-(1 f (x) f (iv)
3 x
x 3 - 2x
. x (x) f (iii)
23
x
0 3 - 2x
0. (x) f (ii)
2-
(-1) 3-
2
f(1) f(0)
1- 3 - 2(1) f(1)
3- 3 - 2(0) f(0) (i)
3 2x f(x)
=
=
=
=
=
=
=
=
=
=
=
=
+=
+
==
==
−=
2
11. An open box is to be made from a square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown. Express the volume V of the box as a function of x.
Since the original shape is square, length of all
sides will be equal.
576x 96x - 4x (x) Vcuboid of volume
] 4x 96x - [576 x
](2x) (2x) 2(24) - [24 x
2x) - x(24
2x) - (24 2x) - (24
height width length cuboid of Volume
x height
2x - 24 width length
23
2
22
2
+=
+=
+=
=
=
××=
=
==
x
12. A function f is defined by f (x) = 3−2x . Find
x such that f (x2) = (f (x))2 .
1 x
0 1 - x
0 1) - (x
0 1 2x - x
0 6 12x - 6x
0 3 - 9 12x - 2x 4x
4x 12x - 9 2x - 3
given][(x)) (f )(x f
4x 12x - 9 (x)) (f
(2x) 2(3)(2x) - 3
2x) - (3 (x)) (f
2x - 3 )(x f
2x - 3 (x) f
2
2
2
22
22
22
22
22
22
22
=
=
=
=+
=+
=++
+=
=
+=
+=
=
=
=
13. A plane is flying at a speed of 500 km per
hour. Express the distance d travelled by the
plane as function of time t in hours.
t 500 distance Required
500t d500d
t
hoursin t taken time
d travelled Distance
hour per km 500 plane of Speed
speed / Distance Time
=
=
=
=
=
=
=
Winglish Coaching Centre, Puduvayal 8 X std Mathematics Made Easy
14. The data depicts length of a woman’s
forehand and her corresponding height. A
student finds a relationship between height
(y) and forehand length(x) as y = ax +b ,
where a, b are constants.
i) Check if this relation is a function.
ii) Find a and b.
iii) Find the height of a woman whose forehand
length is 40 cm.
iv) Find the length of forehand of a woman if
her height is 53.3 inches.
(i) For every values of x,we get different y values
Hence it is a function.
24.5 0.9x y
24.5 b
40.95- 65.5 b
65.5 b 45.5(0.90)
(1),in 0.90 a Substitute
0.90 a
9.5/10.5 a
9.5 10.5a
9.5 b - b 35a - a 45.5
56 - 65.5 b) (35a - b) a (45.5
(2) - (1)
-(2)- b a 35 56
b a(35) 56
56 y35, x
-(1)- b a 45.5 65.5
b a(45.5) 65.5
65.5 y 45.5, x
b x a y
+=
=
=
=+
=
=
=
=
=+
=++
+=
+=
==
+=
+=
==
+=
(iii) Find height when forehand length is 40 cm.
inches. 60.5 woman of Height
60.5 y
24.5 0.9(40) y
40 x if ? y
=
=
+=
==
(iv) length of forehand if her height 53.3 inches
32 x
.
28.8x
0.9x 24.5 - 53.3
24.5 0.9x 53.3
53.3 yif ? x
=
=
=
+=
==
90
15. Find the domain of given functions
i) 1
1+
=x
xf )( ii) 65
12 +−
=xx
xf )(
} {2,3 ofdomain So
2,3 x except numbers real all for defined is
defined. not are (3) f and (2) fthen ,2,3 x If
)(ii)
}{R ofdomain
1- x except numbers real all for defined is
defined not is (-1) fthen 1- x If
)(
)(i)
−=
=
=+−
=
−−=
=
=
=−
+=
Rf
f
xxxf
f
f
f
xxf
65
1
1
01
1
11
2
4. TYPES OF FUNCTIONS
1. Let A = {1,2,3,4} and B = {2,5,8,11,14} be two sets.
Let f: A →B be a function given by f (x) = 3x−1.
Represent this function (i) arrow diagram (ii)
table form (iii) set of ordered pairs (iv) in a
graphical form
(i) Arrow diagram
111434
81333
51232
21131
13
=−=
=−=
=−=
=−=
−=
)()(
)()(
)()(
)()(
)(
f
f
f
f
x xf
(ii) Table form
x 1 2 3 4
f(x) 2 5 8 11
(iii) Set of ordered pairs :
f = {(1,2),(2,5),(3,8),(4,11)}
(iv)Graphical form
Winglish Coaching Centre, Puduvayal 9 X std Mathematics Made Easy
2. Let f : A → B be a function defined by
1 - 2x
)( =xf where A = {2, 4, 6, 10, 12}, B = {0, 1, 2,
4, 5, 9} . Represent f by
(i) set of ordered pairs; (ii) table;
(iii) arrow diagram; (iv) a graph
51 - 2
12 f(12)
4 1-2
10 f(10)
2 1 - 26
f(6)
11 - 24
f(4)
01 - 22
(2)
1 - 2x
)(
==
==
==
==
==
=
f
xf
i) Set of ordered pairs = {(2,
0)(4,1)(6,2)(10,4)(12,5)}
ii) a table
iii) an arrow diagram;
iv) a graph
3. Represent the function f = {(1, 2),(2, 2),(3, 2),(4,
3),(5, 4)} through
(i) arrow diagram (ii) a table form (iii) a graph
(i) an arrow diagram
(ii) a table form
(iii) a graph
4. Determine whether graphs given below
represent functions. Give reason
i) Curve does not represent a function as the
vertical line meets curve in two points P and Q
ii) Curve represents a function as the vertical lines
meet the curve in at most one point
iii) Curve does not represent a function as vertical
lines meet the curves in two points P and Q
iv) Curve represents a function as the vertical lines
meet the curve in at most one point
Winglish Coaching Centre, Puduvayal 10 X std Mathematics Made Easy
v) Since the graph intersects the vertical line (y-axis)
at two points, it is not a function.
vi) given graph intersects the vertical line (y-axis) at
one point. It is a function.
vii) Graph intersects the y-axis at three points,
hence it is not a function.
viii) The graph intersects the vertical line at most
one point. Hence it is a function.
5. Determine type of function.Justify your Answer
i) One-One and Onto (bijection)
Distinct elements of A have distinct images in B
and every element in B has a pre-image in A.
ii) One to One
Distinct elements of A have distinct images in B
iii) Many to One
Two or more elements of A have same image in B
iv) Onto
Range of f = co-domain
Every element in B has a pre-image in A
v) Into
Range of f is a proper subset of co-domain
There exists at least one element in B which is
not image of any element of A
6. Using horizontal line test determine which of
the following functions are one – one.
i) Curve represents a one–one function as hori
zontal line meet curve in only one point P
Winglish Coaching Centre, Puduvayal 11 X std Mathematics Made Easy
ii) Curve does not represent one–one function, as
horizontal line meets curve in two points P ,Q.
iii) Curve represents one–one function as hori
zontal line meets curves in only one point P.
7. Check whether a function. If so mention the
type. Justify your answer
i) A= {1,2,3,4}, B= {a,b,c,d,e}
f = {(1,a), (2,b), (3,d), (4,c)}
ii) A= {1,2,3,4}, B= {a,b,c,d,e}
g = {(1,a), (2,b), (3,c), (4,e)}
iii) A = {1,2,3,4}, B= {a, b, c}
f = {(1,a), (2,a), (3,b), (4,c)}
iv) A= {x,y,z}, B= {l,m,n}
f= {(x,m)(y,n)(z,l)}
v) A = {1,2,3,} , B= {w,x,y,z},
f = {(1,w) (2,z) (3,x)}
i) For different elements in A, there are different
images in B.
Hence f is a one – one function.
ii) g is a function from A to B.
Two distinct elements 1 and 2 in first set A
have same image b in the second set in B.
Hence, g is not a one–one function.
iii) f is a function from A to B
Different elements 1 and 2 of A have same
image a in B.
Hence f is a many – one function.
iv) Range of f ={l,m,n} = B
Hence f is an onto function.
v) y ∈ B is not image of any element in A.
Here, range of f = {w,x,z} ⊂ B
Therefore, f is a into function.
8. Let A = {1,2,3} , B = {4,5,6,7} and f = {(1,4)} ,( 2,5),( 3,6)}
be a function from A to B. Show that f is one –
one but not onto function.
For different elements in A, there are different
images in B.
∴ f is one–one function.
Element 7 in the co-domain does not have any
pre-image in the domain.
∴ f is not onto
Therefore f is one–one but not an onto
function.
9. If A = {-2,-1,0,1,2} f: A → B is an onto function
defined by 12 ++= x xf (x) then find B.
71222
31111
11000
11111
31222
1
2
2
2
2
2
2
=++=
=++=
=++=
=+−+−=−
=+−+−=−
++=
)()()(
)()()(
)()()(
)()()(
)()()(
f
f
f
f
f
x xf (x)
Since, f is an onto function,
Winglish Coaching Centre, Puduvayal 12 X std Mathematics Made Easy
Range of = co-domain of f.
Therefore, B= {1,3,7}
10. Show that the function f : N → N defined by f
(x) = 2x – 1 is one-one but not onto
If for all a1, a2 ∈ A, f(a1) = f(a2) implies a1 = a2 then
f is one – one function.
y x
2y 2x
1 - 2y 1 - 2x
f(y) f(x)
N, yx, Let
=
=
=
=
∈
Hence the function is one to one.
It is not onto :
co-domain of the function = range of function
Even numbers in co-domain are not associated
with the elements of domain.
Hence it is not onto.
11. Show that the function f : N → N defined by f
(m) = m2 + m + 3 is one-one function.
y x
0 y- x
0 1) y (x y)- (x
0 y)- (x y)- (x y) (x
0 y- x y- x
y y x x
3 y y 3 x x
f(y) f(x)
3 m m (m) f
f(y) f(x) N, yx, Let
22
22
22
2
=
=
=++
=++
=+
+=+
++=++
=
++=
=∈
Hence it is one to one function.
12. Let f be a function Ν→Ν:f , defined by
,)( Ν∈+= xxxf 23
i) Find the images of 1, 2, 3
ii) Find the pre-images of 29, 53
ii) Identify the type of function
i) ,)( Ν∈+= xxxf 23
112333
72232
52131
=+=
=+=
=+=
)()(
)()(
)()(
f
f
f
Images of 1, 2, 3 are 5, 8, 11
(ii) If x is the pre-image of 29 or 53
9
273
2923
29
=
=
=+
=
x
x
x
xf )(
17
213
5323
53
=
=
=+
=
x
x
x
xf )(
Pre-images of 29 and 53 are 9 and 17
iii) Since different elements of have different
images in the co-domain, the function f is one
– one function.
But range of f = {5, 8, 11, 14, 17, ...} is a proper
subset of .
Therefore f is not an onto function.
That is, f is an into function.
Thus f is one – one and into function.
13. Let A = {1, 2, 3, 4} and B = N . Let f : A → B be
defined by f (x) = x3 then, (i) find the range of f
(ii) identify the type of function
64} 27, 8, {1, f of Range
64 4 (4) f 4 x
27 3 (3) f3 x
8 2 (2) f2 x
1 1 (1) f1, x
x (x) f
3
3
3
3
3
=
===
===
===
===
=
Every element in A has associated with different
elements of B. Hence it is one to one.
14. In each of the following cases state whether the
function is bijective or not. Justify your answer.
i) f : R → R defined by f (x) = 2x +1
ii) f : R → R defined by f (x) = 3 – 4x2
i) Testing whether it is one to one :
y x
2y 2x
1 2y 1 2x
f(y) f(x)
f(y) f(x) R, yx, Let
=
=
+=+
=
=∈
So, it is one to one.
Testing whether it is onto :
If f :A→ B is onto function then, range of f = B
y f(x)
1 1 - y f(x) 2
1) -2(y f(x)
2
1 - y x
1 2x y
1 2x f(x) y
B (A) f i.e
=
+=
+=
=
+=
+==
=
1
Winglish Coaching Centre, Puduvayal 13 X std Mathematics Made Easy
It is onto function. Hence it is bijective function.
(ii) f : R → R defined by f (x) = 3 – 4x2
Testing whether it is one to one : Let x, y ∈ R,
y- x (or) y x
0 y x (or) 0 y- x
0 y) y)(x- (x
0 y- x
4y 4x-
4y3 4x-3
f(y) f(x)
22
22
22
==
=+=
=+
=
=
−=
=
It is not one to one.
Hence it is not bijective function.
15. Let A = {−1, 1}and B = {0, 2} . If the function f : A
→ B defined by f(x) = ax + b is an onto function?
Find a and b.
1 a
1- a-
0 1 a- (1),in b Sub
1 b
2 2b(2) (1)
-(2)----- 2 b a
2 b a(1)
2 f(1)
-(1)----- 0 b a-
0 b a(-1)
0 f(-1)
b ax f(x)
=
=
=+
=
=+
=+
=+
=
=+
=+
=
+=
16. If the function f : R→ R is defined by
≥−
≤≤−−
−<+
=
323
322
2722
xifx
xifx
xifx
xf )( Find
values of (i) f (4) (ii) f (-2) (iii) f (4)+2 f(1) iv)
)(
)()(
3
431
−
−
f
ff
2222
2
2
102434
23
4
2
2
=−−=−
−=
−=
=−=
−=
=
)()(
)(
interval. second thein lie ii)
)()(
)(
interval. third thein lie i)
f
xxf
x
f
xxf
x
311
1031
3
431
17323
72
3
81210124
1211
2
1
2
2
−=−−
=−
−
=+−=−
+=
−=
=−+=+
−=−=
−=
=
)(
)(
)()(
)()(
)(
interval. first thein lie iv)
)()()(
)()(
)(
interval. second thein lie iii)
f
ff
f
xxf
x
ff
f
xxf
x
17. If the function f is defined by
−<<−−
≤≤−
>+
=
131
112
12
xifx
xif
xifx
xf )(
Find the values of
(i) f (3) (ii) f (0) (iii) f (−1.5) (iv) f (2)+ f (−2)
(i) f(3)
5 f(3)
2 3 f(3)
2 x f(x)
=
+=
+=
(ii) f(0) 0 lies between -1 and 1.
f (0) = 2.
(iii) f (−1.5)
2.5- f(-1.5)
1 - 1.5- f(-1.5)
1 - x f(x)
=
=
=
(iv) f (2)+ f (−2)
1
(-3) 4 (-2) f (2) f
3-
1 - 2- f(-2)
1 - x f(x) Thus,
1xin lies 2-
4
2 2 f(2)
2 x f(x)Thus,
1xin lies 2
=
+=+
=
=
=
<
=
+=
+=
>
18. A function f : [−5,9] → R is defined as follows:
Winglish Coaching Centre, Puduvayal 14 X std Mathematics Made Easy
<<−
<≤−
<≤−+
=
9643
6215
25162
xifx
xifx
xifx
xf )( Find
(i) f (−3) + f (2) (ii) f (7) - f (1) (iii) 2f (4) + f (8)
(iv) f(-2) f(4)
f(6) -2f(-2)
+
(i) f (−3) + f(2)
219 17 - f(2) (-3) f
191 - 5(2) f(2)
f(2) for 1 - 5x f(x)
17- 1 6(-3) f(-3)
f(-3) for 1 6x f(x)
2
2
=+=+
==
=
=+=
+=
(ii) f (7) - f (1)
10 7 -17 (1) f - (7) f
1 6(1) f(1)
f(1) for 1 6x f(x)
17 4- 3(7) f(7)
f(7) for 4- 3x f(x)
==
=+=
+=
==
=
7
(iii) 2f (4) + f (8)
178
20 2(79) (8) f (4) 2f
20 4- 3(8)
4- 3x f(8)
f(8) for 4- 3x f(x)
79 1 - 80
1 - 5(4) f(4)
f(4) for 1 - 5x f(x)2
2
=
+=+
==
=
=
==
=
=
(iv) f(-2) f(4)
f(6) - 2f(-2)
+
179-
(-11) 79
14 - 2(-11)
f(-2) f(4)
f(6) - 2f(-2)
14 4- 3(6) f(6)
f(6) for 4- 3x f(x)
791 - 5(16) f(4)
f(4) for 1 - 5x f(x)
11- 1 6(-2) f(-2)
f(-2) for 1 6x f(x)
2
2
=
+=
+
==
=
==
=
=+=
+=
19. Forensic scientists can determine height of a
person based on the length of their thigh bone.
They do so using function bh (b) 1054472 .. +=
where b is the length of the thigh bone.
(i) Check if the function h is one – one
(ii) Find height of a person if the length of his
thigh bone is 50 cms.
(iii) Find the length of the thigh bone if the
height of a person is 147.96 cms.
21
21
21
21
472472
10544721054472
1054472
bb
bb
bb
bhbh
bh (b)
=
=
+=+
=
+=
..
....
)()( Assume,
..
So, the function h is one – one.
ii) If the length of the thigh bone b = 50
6177
10545047250
.h Height
.)(.
=
+= )h (
iii) If the height of a person is 147.96 cms,
cmsLength ..
...
.)(
384728693
961471054472
96147
=
=
=+
=
b
b
b
bh
20. The distance S (in kms) travelled by a particle in
time ‘t’ hours is given by 2
2 ttts
+=)( . Find the
distance travelled by the particle after
(i) three and half hours.
(ii) eight hours and fifteen minutes.
(i) three and half hours i.e t = 3.5 hrs
kms.
.).().(
87572
535353
2
=
+=s
(ii) 8 hours and 15 minutes i.e t = 8.25 hrs
kms.
.).().(
15625382
258258258
2
=
+=s
21. Distance S an object travels under the influence
of gravity in time t seconds is given by S(t) =
(1/2) gt2 + at + b
where, g is acceleration due to gravity, a, b are
constants. Check if the function S(t) is one-one.
If for all a1, a2 ∈ A, f(a1) = f(a2) implies a1 = a2 then
f is called one – one function.
Winglish Coaching Centre, Puduvayal 15 X std Mathematics Made Easy
yx
x - y
yx x - y
x - y
baygybax gx
b ay gy baxgx
=
=
=++
=++
=++−++
++=++
=
=∈
0
0 a] )g( [ )(
0 b - b )a( ) y- g(x
S(y) S(x)
f(y) yx, Let
22
21
21
021
21
21
21
22
22
Hence it is one to one function.
22. Function ‘t’ which maps temperature in Celsius
(C) into temperature in Fahrenheit (F) is defined
by t(C) = F where 32 C 59
F += . Find,
(i) t(0) (ii) t(28) (iii) t(-10)
(iv) value of C when t (C) = 212
(v) temperature when Celsius value is equal to
Farenheit value.
C 100 C of value
32 - 212 C
212 32 C
212 (C) t (iv)
F 14
3218-
32 (-10) t(-10) (iii)
F 82.4
32 (28) t(28) (ii)
F 32
32 (0) t(0) (i)
)(
°=
=
=+
=
°=
+=
+=
°=
+=
°=
+=
+=
59
59
59
59
59
3259
C Ct
(v) temperature when Celsius =Farenheit value.
F = C
Hence answer is -40°
23. Identify the type of function
i)A= { a,b,c,d }, B = {1,2,3} , f = {(a,3)(b,3)(c,3)(d,3)}
ii) A = { a, b, c }, f = {(a, a) (b, b)( c, c)}
i) Since, f(x) = 3 for every x ∈ A , Range of f =
{3}
f is a constant function.
ii) Since f (x) = x for all x ∈ A
It is identity function on A.
24. f is a function from R to R defined by f (x) = 3x – 5
Find values of a and b given that (a, 4) and (1, b)
belong to f.
f (x) = 3x – 5 written as f = {( x, 3x – 5)|x ∈ R}
(a, 4) means image of a is 4.
3 a
4 5-3a
4 (a) f
=
=
=
(1, b) means the image of 1 is b.
2 b
b 5 -3(1)
b (1) f
−=
=
=
5. COMPOSITION OF FUNCTIONS
1. Find fog and gof if f (x) = 2x+1, g (x) = x2 – 2
144
212
12
32
122
2
2
2
2
2
2
−+=
−+=
+=
−=
+−=
−=
xx
x
xggof
x
x
xffog
)(
)(
)(
)(
2. If f (x) = 3x−2 , g(x) = 2x+k and if fo g =go f then find the value of k.
1
46236
232223
−=
+−=−+
+−=−+
=+
=
+==
k
kxkx
kxkx
g
)()(
2)-3x(k)f(2x
f gog fo
k 2x g(x) ,2-3x (x) f
3. f (x) = 2x+3 g (x) = 1 −2x, h (x) = 3x¸ Prove that (fo go h) = fo( goh)
Winglish Coaching Centre, Puduvayal 16 X std Mathematics Made Easy
x
x
xfgohfo
x
x
xggoh
x
x
xfoghfog
x
x
xffog
125
3612
61
61
321
3
1125
345
3
45
3212
21
−=
+−=
−=
−=
−=
=
−=
−=
=
−=
+−−=
−=
)(
)()(
)(
)(
).....(
)(
)()(
)(
)(
4. Find x if gff(x) = fgg(x), given f x (x) = (3x +1) and gx (x) = x + 3
.
2x
126x
19 + 3x 7 + 9x
] 1 + 6) + 3(x [ 3] + 4)+ (9x [
6) + (x f 4)+ (9xg
3] + 3) + (x [ f 1]+1)+ 3(3x [g
3)] + (xg [ f 1)]+ (3x f [g
(x)}]{g [g f (x)}] {f [fg
fgg(x)= gff(x)
=
=
=
=
=
=
=
=
5. Using the functions f and g, find f o g and g o f . Check whether f o g = g o f . (i) f (x) = x −6, g(x) = x2
(x) f og (x)g o f
6) - (x
g[f(x)] (x) f og
6 - x
f[g(x)] (x)g o f
2
2
≠
=
=
=
=
(ii) f (x) = 2/x, g(x) = 2x2 - 1
(x) f og (x)g o f
1-x
8
1 - 2
2
g[f(x)] (x) f og
1 - 2x
2
1] - f[2x (x)g o f
2
2
2
2
≠
=
=
=
=
=
x
iii) f (x) = (x + 6)/3, g(x) = 3 - x
3
x-9
3
6x-3
x] - f[3 (x)g o f
=
+=
=
(x) f og (x)g o f3
x-3
3
6 x - 3
3
6 xg (x) f og
≠
=
+=
+=
(iv) f (x) = 3 + x, g(x) = x - 4
(x) f og (x)g o f
(2)--- 1 - x
4- x 3
x] [3g (x) f og
1 - x
4- x 3
4]- f[x (x)g o f
=
=
+=
+=
=
+=
=
(v) f (x) = 4x2 − 1, g(x) = 1 + x
(x) f og (x)g o f
4x
1 - 4x 1
1] - [4xg (x) f og
3 8x 4x
1 - 2x) x 4(1
1 - x) 4(1
x] f[1 (x)g o f
2
2
2
2
2
2
≠
=
+=
=
++=
++=
+=
+=
6. Find the value of k, such that f o g = g o f
(i) f (x) = 3x +2, g(x) = 6x −k
5- k
10 2k -
12 2- k 3k -
k - 12 18x 2 3k - 18x
k - 2) 6(3x 2 k) - 3(6x
2] g[3x k] - f[6x
g[f(x)] f[g(x)]
f og g o f
=
=
+=+
+=+
+=+
+=
=
=
(ii) f (x) = 2x −k, g(x) = 4x + 5
35
- k
5- 3k
10 - 5 4k k -
5 4k - 8x k - 10 8x
5 k) - 4(2xk - 5) 2(4x
k] - g[2x 5] f[4x
g[f(x)] f[g(x)]
f og g o f
=
=
=+
+=+
+=+
=+
=
=
Winglish Coaching Centre, Puduvayal 17 X std Mathematics Made Easy
7. f (x) = 2x −1,g (x) = 2
1 x +, show that fog=gof = x
-(2)- x 2
1) 1 - (2x
1] - g[2x f og
-(1)- x
1 - 2
1 x2
f g o f
=
+=
=
=
+
=
+
=2
1 x
(i) If f (x) = x2 −1, g(x) = x −2 find a, if g o f (a) = 1
2 a
4 a
1 3- a
1 (a) f og
: thatGiven
3- a (a) f og
2 - 1)- (a
1]- g[a (a) f og
2- a g(a)
1- a f(a)
2
2
2
2
2
2
±=
=
=
=
=
=
=
=
=
(ii) Find k, if f (k) = 2k −1 and f o f (k) = 5.
. 2 k
8 4k
3 5 4k
5 3 -4k
5 (k) f o f
: thatGiven
3 - 4k
1 - 1) -2(2k
1] -f[2k (k) f o f
=
=
+=
=
=
=
=
=
8. Let A, B, C ∈N and function f:A → B be defined
by f(x) = 2x + 1 and g:B → C be defined by g(x) = x2 . Find range of f o g and g o f.
N} x and 1) (2x y| {y
: Range
1) (2x
1] g[2x f og
N} x and 1 2x y| {y
: Range
1 2x y
1 x 2
]f[x g o f
2
2
2
2
2
2
∈+=
+=
+=
∈+=
+=
+=
=
9. Let f (x) = x2 −1 . Find (i) f o f (ii) f o f o f
1 - )2x - (x
]2x - f[x
f] o [f f f o f o f (ii)
2x - x f o f
1 - 1 2x - x
1 - 1)- (x
1]- f[x f o f (i)
224
24
24
24
22
2
=
=
=
=
+=
=
=
10. If f : R → R and g : R → R are defined by f(x) = x5
and g(x) = x4 then check if f, g are one-one and f
o g is one-one?
f(x) = x5
For every positive and negative values of x, we get positive and negative values of y.
Every element in x is associated with different elements of y. Hence it is one to one function.
g(x) = x4
For every positive and negative values of x, we get only positive values of y.
Negative values of y is not associated with any elements of x. Hence it is not one to one function.
fog(x) = f[g(x)]
= f[x4]
now, we apply x4 instead of x in f(x)
f[x4] = (x5)4
fog(x) = x20
fog is not one to one function.
11. Consider functions f (x), g(x), h(x) given below.
Show that (f o g) o h = f o (g o h) in each case.
(i) f(x) = x −1, g(x) = 3x +1 and h(x) = x2
h) o(g o f h o g) o (f
(2) (1)
(2)------- 3x h) o(g o f
1 - 1 3x
1] f[3x h) o(g o f
1 3x
]g[x h) o(g
:h) o(g o f
-(1)------- 3x
][x g) o (f h o g) o (f
3x
1 - 1 3x
1] f[3x fog(x)
:h o g) o (f
2
2
2
2
2
2
2
=
=
=
+=
+=
+=
=
=
=
=
+=
+=
Winglish Coaching Centre, Puduvayal 18 X std Mathematics Made Easy
(ii) f (x) = x2, g(x) = 2x and h(x) = x + 4
(2)-------64 x 32 4x
8 2(2x)(8) (2x)
8) (2x
8] f[2x h) o(g o f
8 2x 4) 2(x
4] g[x h) o(g
-(1)------- 64 32x 4x
16) 8x 4(x
4) 4(x
4] [x g) o (f h o g) o (f
4x (2x)
f[2x] fog(x)
2
22
2
2
2
2
22
++=
++=
+=
+=
+=+=
+=
++=
++=
+=
+=
==
=
(iii) f (x) = x −4, g(x) = x2 and h(x) = 3x −5
(2) (1)
-(2)------- 21 30x - 9x
25] 30x - f[9x h) o(g o f
25 30x - 9x
5)- (3x
5]- g[3x h) o(g
-(1)------- 21 30x - 9x
4- 5 (3x)(5) 2 - (3x)
4- 5) - (3x
5]- [3x g) o (f h o g) o (f
4- x
]f[x fog(x)
2
2
2
2
2
22
2
2
2
=
+=
+=
+=
=
=
+=
+=
=
=
=
=
12. Let f = {(−1, 3),(0,−1),(2,−9)} be a linear function from Z into Z . Find f (x).
1. - 4x- equation Rqd
and of valuetheApplying
4- a
4 a-
1 3 a-
3 (-1) a-
(1)in of valueApplying
1- b
b 1-
b a(0) 1-
1- then y 0, x if
(1)------- 3 b a-
b a- 3
b a(-1) 3
3 then y 1,- x if
function Linear
=
=
=
+=
=+
=
=
+=
==
=+
+=
+=
==
+=
y
ba
b
b ax y
13. In electrical circuit theory, a circuit C(t) is called
a linear circuit if it satisfies the superposition
principle given by C(at1 + bt2) = aC(t1) + bC(t2),
where a,b are constants. Show that the circuit
C(t) = 3t is linear.
Take two points t1 and t2 from domain of C(t).
c(at1 + bt2) = 3(at1 + bt2)
c(at1) = 3at1
ac(t1) = 3at1
bc(t2) = 3bt2
3(at1 + bt2) = 3at1 + 3at2
(or) C(at1 + bt2) = aC(t1) + bC(t2),
Hence c(t) is linear.
5. EUCLID’S DIVISION LEMMA
1. Find all positive integers, when divided by 3
leaves remainder 2.
............ 14, 11, 8, 5, 2,
8 = a
2 + 3(2) = a
(given) 3 = b ,2 = r 2, = q let
5 = a
2 + 3(1) = a
(given) 3 = b ,2 = r 1, = q let
2 = a
2 + 3(0) = a
(given) 3 = b ,2 = r 0, = q let
r + bq = a algorithm,division By
2. A man has 532 flower pots. He wants to arrange
them in rows such that each row contains 21
flower pots. Find the number of completed rows
and how many flower pots are left over.
Total number of flower pots = 532
Number of flower pots in each row = 21
We may arrange the flower pots in 25 rows with
each row consists of 21 pots.
The remaining number of flower pots = 5.
3. Prove that the product of two consecutive
positive integers is divisible by 2.
Let "x" be the positive integer
consecutive number be "x + 1"
When a positive integer is divided by 2 the
remainder is either 0 or 1. So, any positive integer
will of the form 2k, 2k+1 for some integer k.
Winglish Coaching Centre, Puduvayal 19 X std Mathematics Made Easy
1) +2k(2k 1) + (x x
2k =x
numbereven = x
: 1 Case
=
Hence it is divisible by 2.
1) +1)(2k +(k 2 =
2) +1)(2k +(2k =
1) + 1 +1)(2k +(2k = 1) + (x x
1 +2k = x
number odd = x If
: 2 Case
So, the product is divisible by 2.
4. When the positive integers a, b and c are divided by 13, the respective remainders are 9,7 and 10. Show that a+b+c is divisible by 13.
13 by divisible is c + b + a Hence
2] + )q + q + 13[(q =
13(2) + )q + q + 13(q =
26 +13q + 13q + 13q =
10 + 13q +7 + 13q + 9 + 13q = c + b + a
(3) + (2) + (1)
-(3)---10 + q 13 = c
-(2)---7 + q 13 = b
-(1)--- 9 + q 13 = a
remainder + quotient x divisor = divided
321
321
321
321
3
2
1
.
5. Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.
1. remainder leaves and 4by divisible isn
1 + m) + 4(m= n
1 + 4m+ 4m=
12 + 2(2m) + (2m) =
1) + (2m = n
1+2m =n let
odd... isn : 2 Case
0. remainder leaves and 4by divisible isn
4m= n
2m =n let
integereven isn : 1 Case
22
2
2
22
22
∴
∴
6. Use Euclid’s Division Algorithm to find the Highest Common Factor (HCF) of i) a = 273 and b = 119 .
7 55) ,(210 of HCF
0 +2 ×7 = 14
7+2× 14 = 35
114+3 ×35 = 119
35+ 2× 119 = 273
=
ii) a = 340 and b = 412
0 + 2(4) = 8
4+ 1(8) = 12
8 + 1(12) = 20
12 + 2(20) = 52
20 + 1(52) = 72
52 + 4(72)= 340
72 + (340) 1 = 412
340 > 412
Hence the required H.C.F is 4. iii) 867 and 255
0 + 2(51) = 102
51 + 2(102) = 255
102 + 3(255) =867
255 >867
255 = b and867 = a
Hence the H.C.F is 51.
iv) 10224 and 9648
0 + 3(144) = 432
144 + 1(432) = 576
432+ 16(576) = 9648
576 + 1(9648) = 10224
9648 > 10224
Hence the H.C.F is 144.
v) 84, 90 and 120
0 + 6(14) = 84
6 + 1(84) = 90
84 > 90
Hence the required H.C.F is 6. 7. Find the largest number which divides 1230 and
1926 leaving remainder 12 in each case.
0 + 174(3) = 522
174 + 522(1) = 696
522 + 696(1) = 1218
696 + 1218(1) = 1914
1914. and 1218 of H.C.F : find To
1914 = 12 - 1926
1218 = 12 - 1230
Rqd number is 174.
8. If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y .
4.is 32 and 60 of H.C.F
-(3)--- 0 + 4(7)= 28
-(2)--- 4+ 28(1) = 32
-(1)--- 28 + 32(1) = 60
32 > 60
∴
Winglish Coaching Centre, Puduvayal 20 X std Mathematics Made Easy
1.- = yand 2 = x Hence
(-1) 60 + 32(2) = 4
1 ? 60 - 1) + 32(1 = 4
1 ? 60 - (1) 32 + 32 = 4
1 ? (1) 32 + 1 ? 60 - 32 = 4
1 ? 32(1)) - (60 - 32 = 4
(4)in (5) from 28 of valueApply
(5)---32(1) - 60 = 28 (1), From
(4)--- 28(1) - 32 = 4(2), From
y60 + x 32 = 4
60y + 32x = d
4= d
9. A positive integer when divided by 88 gives the
remainder 61. What will be the remainder when
the same number is divided by 11?
11 by dividedwhen 6 remainder gives
11 by divisible is 55
11. by divisible is 6)-55)-((n
11. by divisible is 61)-(n
88. by divisible is 61)-(n
number. integer that be n"" Let
n
∴
∴
∴
10. Prove that two consecutive positive integers are always coprime.
prime.-co are 1 + x & x :ioncontradict by So
1 =n or 1 dividesn
x - 1 + x dividesn
1. + x and x dividesn
n, = 1) + x (x, of H.C.F
1 >n assume
1 + x x, : numbers econsecutiv
∴
∴
11. Find quotient,remainder when a is divided byb [Find q and r for a and b satisfying a=bq+r]
(i)a =−12 , b = 5
3
3
33512
=
−=
+−=−
+=
r
q
rbqa
Remainder
Quotient
)(
:lemmadivision Euclid By
(ii)a = 17 , b =−3
2
5
25317
=
−=
+−−=
+=
r
q
rbqa
Remainder
Quotient
))((
:lemmadivision Euclid By
(iii)a =−19 , b =−4
1
5
15419
=
=
+−=−
+=
r
q
rbqa
Remainder
Quotient
))((
:lemmadivision Euclid By
12. Show that the square of an odd integer is of the form 4q +1 , for some integer q.
)( where,
)(
1) +2k(x
k. integer some for 1, +2k= x
integer,even than more one integer odd
integer. odd any x
22
1
14
114
144 2
+=
+=
++=
++=
=
=
=
kkq
q
kk
kk
13. If the Highest Common Factor of 210 and 55 is expressible in the form 55 x -325 , find x.
6 x
5325-55x
5 55) ,(210 of HCF
0 +2 × 5 = 10
5+4× 10 = 45
10+1 × 45= 55
45+ 3 × 55 = 210
=
=
=
14. Find the greatest number that will divide 445 and 572 leaving remainders 4 and 5 respectively.
63. is number required
63 = 441,567 of HCF
02× 63= 126
633 ×126 = 441
1261× 441=567
567. and 441of HCF Find
7 556- ,5724414- 454
∴
∴
+
+
+
∴
==
15. Find the HCF of 396, 504, 636.
12. 636 and 504 396, HCF
12 = 636,36 of HCF Thus
02× 12 = 24
121× 24 = 36
2417×36 = 636
:36 ,636 of HCF find To
= 36,396,504 of HCF hus
02× 36 = 72
361× 72 = 108
723 ×108 = 396
1081×396 = 504
504 and 396: of HCF find To
=
+
+
+
=
+
+
+
+
T
Winglish Coaching Centre, Puduvayal 21 X std Mathematics Made Easy
6. FUNDAMENTAL THEOREM OF ARITHMETIC
1. In the given factor tree, find numbers m and n.
50 =n 300, = m are numbers Rqd
300=150×2 = m of Value
50=501×3 = bottom from box 2
50=10×5 =n of Value
10= 2×5 = bottom from box 1
nd
st
2. Is 7x5 x3x2 +3 a composite number? Justify your
answer.
number composite a is it
primes two toin factorized is numberGiven
1)+ x3x2 7x5( 3+ x3x2 7x5 ×= 3
3. Can the number 6n , n is a natural number end
with the digit 5? Give reason for your answer.
5. digit thewith end cannot 6 Hence,
odd. always is 5 is digit last whose number Any
even. always is6 So,
6 of factor a is 2
3 23)2(6
n
n
n
nnnn ×=×=
4. For what values of natural number n, 4n can end
with the digit 6?
6.with ends 4even, isn When
4096= then 4 6, =n if
1024 = then 4 5, =n if
256 = then 4 4,=n if
64 = then 4 3, =n if
16 = then 4 2, =n if
4= then 4 1, =n if
n
6
5
4
3
2
1
∴
5. Find the HCF of 252525 and 363636.
10101 =
37 137 3 = H.C.F
37137 3 2 = 363636
7 3 137 3 5 = 25252532
2
×××
××××
××××
6. If 13824 = 2a ×3b then find a and b.
3. b ,9 a Hence
3 x 2 = 13824 39
==
7. a,b are two positive integers such that ab ×ba=
800. Find ‘a’ and ‘b’.
5b2, a thus
b× a
5×5×2 ×2×2×2×2
800 =b× a
ab
ab
==
×=
=
25 52
8. If m, n are natural numbers, for what values of
m, does 2n x 5m ends in 5?
• For any value of n, 2n will become even.
• For any value m, 5m ends with 5.
• Product of even number and a number ends
with 5, we get a number ends with 0.
• We should not apply 0 for n and m, because
n and m are natural numbers.
9. If 113400 1 p ,p ,p ,p 4321 x
4x
3x
2x
1 = find the
value of 4321 p ,p ,p ,p ( primes in ascending
order) 432 x ,x ,x ,x1 ( integers, )
1 ,2 4,3, x ,x ,x ,x
7 5, 3, 2, p ,p ,p ,p
7 x 5 x 3 x 2 = 113400
4321
4321
1243
=
= .
10. Find the greatest number consisting of 6 digits
which is exactly divisible by 24,15,36?
99972036 and 24,15 by divisible
number digit-6 greatest
999720. = 279 - 999999 i.e
279. remainder i.e
360
999999.
36 ,24,15 of LCM
number digit-6 Greatest
36036 ,24,15 of LCM
999999. number digit-6 Greatest
=
=
+×=
=
=
=
2793602777
11. Find LCM and HCF of 408 and 170 by applying
the fundamental theorem of arithmetic.
2040 = L.C.M
17 x 5 x 3 x 23 = L.C.M
34 = H.C.F
17 ,2 factorsCommon
17 x 5 x 2 = 170
17 x 3 x 23 = 408
=
12. What is the smallest number that when divided
by three numbers such as 35, 56 and 91 leaves
remainder 7 in each case?
3647 =number Smallest Hence,
7 + 3640 = caseeach in 7 remainder a leaves
3640 =
x13 x2 2 x x2 5 7x = LCM
13 x7 = 91
7 x 2 x 2 x 2 = 56
7 x 5 = 35
,9156 35, of LCM = 91 56, 35, by divided no. Smallest
13. Find the least number that is divisible by the
first ten natural numbers.
2520 number smallest The
2520 =
7 × 5 × 3 × 3 × 2 × 2 × 2 numbers these of L.C.M
10 9, 8, 7, 6, 5, 4,3, 2, 1, :numbers natural 10
=
=
Winglish Coaching Centre, Puduvayal 22 X std Mathematics Made Easy
7. MODULAR ARITHMETIC
1. Find the least positive value of x such that
(i) 71 ≡ x (mod 8)
7x of valueleast
8 by divisible is 8
7 - 71 n
8
x - 71 n
8n = x - 71
=
=
=
(ii) 78 + x ≡ 3 (mod 5)
0x of valueleast
5 by divisible is,5
075
5
x75n
5n = x + 75
5n = 3 - x + 78
=
+=
+=
(iii) 89 ≡ (x + 3) (mod 4)
2x of valueleast
4by divisible is2 - 86
n
x - 86
n
4n= 3 - x - 89
=
=
=
4
4
(iv) 5) (mod 7
x 96 ≡
7 x of valueleast
7
x-96
7
x-96
=
=
=
5
5
n
n
.
(v) 5x ≡ 4 (mod 6)
2=
=
=
x of valueleast
6 by eisdivisibl6
4- 5(2)n
6
4- 5xn
6n = 4- 5x
vi) 67 + x≡1 (mod 4)
2x of valueleast
4by divsible is4
2674
x66n
4n x + 66
4n 1-x +67
4)(mod 1x +67
=
+=
+=
=
=
≡
vii) 98≡ (x+4) (mod 5)
4x of valueleast
5 by divsible is4
4675
x9n
5n x - 94
5n4-x - 98
=
−=
−=
=
=
4
2. Solve the following congruent equations
i) 5x ≡ 4 (mod 6)
.2,8,14,...
6,.......62 6,2 2, x of values
2 x of valuesleast
6 by divisible is6
4- 5(2) =n
6
4- 5x =n
6n = 4- 5x
=
+++=
=
ii) 3x −2 ≡ 0 (mod 11)
8,19,30
.11),......11(8 11),(8 8, x of values
8 x of valueLeast
11 by divisible is 11
2 - 3(8) =n
11
2 - 3x =n
11n = 2 - 3x
=
+++=
=
iii) 8x ≡ 1 (mod 11)
..7,18,29,..
.11),......11(7 11),(7 7, x of values
7 x of valueLeast
11 by divisible is 11
1 - 8(7) =n
11
1 - 8x =n
11n = 1 - 8x
=
+++=
=
iv) 3x ≡ 1 (mod 15)
solution integer no is there So,
integeran be tcan' 15
1-
5
x =n
15
1 - 3x =n
15n = 1 - 3x
3. Compute x, such that 104 ≡ x (mod 19)
( )
6
19610
192510
19510
19510
4
4
222
2
=
≡
≡
≡
≡
xThus
)(mod
)(mod
)(mod
)(mod
Winglish Coaching Centre, Puduvayal 23 X std Mathematics Made Easy
4. Prove that 2n + 6×9n is always divisible by 7 for any positive integer n.
7 by divisible is it Hence
)(
)(2
63m+9)- 2 ( 2
9 2-63m+ 2 2
9 )2-(7m+ 2 2 =
9 9× 6 + 2 2 =
9×6 + 2 = 1)P(k
1 +k =n for true is p(x) : prove To
2-7m = 9×6
7m = 9×6 + 2
7. by divisible is 9×6 + 2 =p(k)
k =n for true is p(x):Assume
7. by divisible is 56
56 = 54 + 2 =
9×6 + 2 =
1 =n for true is p(x) : prove To
k
k
kk
kk
kk
1+k1+k
kk
kk
kk
11
km
m
297
797
−=
×+−=
=
××=
××
××
+
5. Find remainders if70004 and 778 is divided by 7.
1. Remainder
7) (mod 1 778
7) 1(mod 1777
7) (mod 0 777
:7 by divisible is777 ii)
4. Remainder
7) (mod 4 70004
7) (mod 40 470000
7) (mod 0 70000
:7 by divided is 70004i)
=
≡
+≡+
≡
=
≡
+≡+
≡
0
6. Find the remainder when 281 is divided by 17.
2 =
2 1 =2 (-1) = Thus,
-1(mod17)16 but,
2 (16) =
2)(2 =
2 2 =2
)(mod
20
20
204
18081
××
≡
×
×
×
≡
81
81
2
1702
7. .Flight travel from Chennai to London through British Airlines is 11 hours. The airplane begins its journey on Sunday at 23:30 hours. If the time at Chennai is 41/2 hours ahead to that of London’s time, then find the time at London, when will the flight lands at London Airport. As journey time is 11 hrs, flight will reach London airport at 23:30 + 11 hrs = 10:30 Chennai time.
Chennai time is 4:30 hrs ahead of London time, so at the time of landing time will be 10:30 - 4:30 hrs = 6:30 London time.
8. Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?
Associate 0, 1, 2, 3, 4, 5, 6 to represent the weekdays from Sunday to Saturday
Friday. 5 number for Day
7) (mod 5
7) (mod 5 7) (mod47
after days 45 come will uncle
2. Tuesday for Number
=
≡
≡
=
=
9. Today is Monday . Vani celebrated her birthday 75
days ago Find when Vani celebrated her birthday.
Associate 0, 1, 2, 3, 4, 5, 6 to represent the weekdays from Sunday to Saturday
Wednesday. birthday VanisThus,
Wednesday. 3 number for Day
7) (mod 3 75-1 Thus,
7) (mod 3
7) (mod 4-7
7) (mod 4- 7) (mod 74-
ago days 75birthday Vanis
1. Monday for Number
=
=
≡
≡
≡
≡
=
=
10. A man starts his journey from Chennai to Delhi by
train. He starts at 22.30 hours on Wednesday. If it
takes 32 hours of travelling time and assuming that
the train is not late, when will he reach Delhi?
hrs 6.30 at Fridayon Reach
Friday Thursday
8 24)(1 32
(mod24) 6.30
.(mod24) 54.30 24) (mod 3222.30
:timeReaching
24. modulo use we Here
hours. 32 time Travelling
22.30 timeStarting
+×=
≡
≡+
=
=
11. i)What is the time 100 hours after 7 a.m.?
am 11 47 24) (mod 1.7
4Reminder
4 24)(4 100
24) (mod 1.7:24 modulousing
=+=+
=
+×=
+
0000
0000
ii) What is the time 15 hours before 11 p.m.?
am 8
3-1112) (mod 1.1
Reminder
3 )1(1 1
12) (mod 1.1:12 modulousing
=
=−
=
+×=
−
5001
3
25
5001
Winglish Coaching Centre, Puduvayal 24 X std Mathematics Made Easy
8.Sequences
1. Find next three terms of the following sequence. (i) 8, 24, 72, …
1944 648, 216,: terms 3 Next
1944 = 648(3) = a
648 = (3) 216 = a
216 = 72(3) = a
3. by multiplied is termEach
6
5
4
.
(ii) 5, 1,-3,…
15- = 4- 11- = a
11- = 4-7 - = a
7- = 4- 3- = a
4.by decreases termEach
6
5
4
iii) ,...,,14
1
6
1
2
1
26
1
422
122
1
418
118
1
414
1
6
5
=+
=
=+
=
=+
=
a
a
4a
4.by increased is rdenominato
same are numerators
iv) 4125 −− ,,,
13310
1037
734
7
6
−=−−=
−=−−=
−=−−=
a
a5a
3. by decreased is termEach
v) ,.....,., 010101
00001010
00010
0001010
0010
001010
010
..
a
..
a
..
a
10. by divided is termEach
6
5
4
==
==
==
vi) 1/4, 2/9, 3/16,…
......,36
5, : terms 3 Next
)(
6 = a
36
5
)(
5 = a
)(
4 = a
)(
n = a
6
5
4
n
49
6
25
4
49
6
16
15
25
4
14
1
2
2
2
2
=+
=+
=+
+n
2. Find the first four terms of the sequences whose nth terms are given by
(i) an = n3 −2
62. 25, 6, 1,- : terms four 1st
62 = 2- 4= a
25 2-3 = a
6 = 2-2 = a
1 = 2-1 = a
2-n = a
34
33
32
31
3n
=
(ii) an = (−1)n+1 n(n + 1)
20.- 12, 6,- 2, : terms four 1st
-201) + 4(4(-1) = a
121) + 3(3 (-1) = a
-61) + 2(2 (-1) = a
2 1) + 1(1 (-1) = a
1) +n(n (-1) = a
1+44
1+33
1+22
11+1
1+nn
=
=
=
=
(iii) an = 2n2 - 6
26 12, 2, 4,- : terms four 1st
266 - 2(4) = a
126 - 2(3) = a
26 - 2(2) = a
-46 - 2(1) = a
6 - 2n = a
24
23
22
21
2n
=
=
=
=
3. Find nth term /general term for the sequences i) 3, 6, 9
n3=na
3. of multiples are terms
ii)2, 5, 10, 17,...
1 + n = a
1),... + (4 1), + (3 1), + (2 1), + (1
1),... + (16 1), + (9 1), + (4 1), + (1
2n
2222
iii) ,.....,2
1
3
2..........
1
13
3
12
2
11
1
+=
+++=
n
nna
Nrthan more 1 is Dr n, is termnth of Nr.
,,
iv) ,.....,2
1 0,
3
2..........
n
1 -n a
,.....,2
1-2 ,
1
1-1 =
r.denominatothan less 1 is Numerator
n, is termnth of r.Denominato
n =
−
3
13
v) 3, 8, 13, 18,...
2 -5n a
2),... - (20 2), - (15 2), - (10 2), - (5 =
18,... 13, 8, 3, =
n =
Winglish Coaching Centre, Puduvayal 25 X std Mathematics Made Easy
vi) 5,-25,125,……
5 )(a
5. of powersin are they
ealternativsign - and
.. ,5 ,5- ,5 =
nn
321
11 +−=
+
n
4. Find the indicated terms of the sequences whose nth terms are given by
(i) an = 5n / (n + 2); a6 and a13
3
138
30
=
=2 + 13
5(13) = a
4
15 =
2 + 6
5(6) = a
2 +n
5n = a
136
n
(ii) an = -(n2 - 4); a4 and a11
117- 4)- (11- = a
12- 4)- (4- = a
4)- (n- = a
211
24
2n
=
=
5. Find a11 and a18 whose general term is
∈+
∈+=
Nnodd, is);(
Nneven, is;
nnn
nnan
3
12
154
31111
3
325
118
12
2
=
+=
+=
=
+=
+=
)(a
odd )(
a
even)(
1118
nnnannann
6. Find a8 and a15 whose nth term is
∈+
∈+
−
=
Nnodd, is;
Nneven, is;
nn
n
nn
n
an
12
3
1
2
2
31
225
1152
15
12
11
6338
18
3
1
2
2
2
2
=
+=
+=
=
+
−=
+
−=
)(a
odd is;
a
even is if
158
nn
nan
n
na nn
7. If a1 = 1, a2 = 1 and an = 2an - 1 + an - 2 n ≥ 3, n ∈ N, then find the first six terms of the sequence.
17 4
72(17)
a + 2a
a + 2a = a
1
32(7)
a + 2a
a + 2a = a
7
12(3)
a + 2a
a + 2a = a
3
12(1)
a + 2a
a + 2a = a
a + 2a = a
1 = a 1, = a
45
2 - 61 - 66
34
2 -51 - 55
23
2 - 41 - 44
12
2 - 31 - 33
2 -n 1 -n n
21
=
+=
=
=
+=
=
=
+=
=
=
+=
=
8. a1 = 1, a2 = 1 fin first 6 terms of
3≥+
n ,3 a
a = a
2 -n
1 -n n
52
1
34
116
1
3
16
131
3
4
131
1
3
3
3
4
41
2
3
1
2
=
+
=+
=
+
=
+=
+=
+
=
+=
+=
+
≥+
a
a
a
a
a
a
3 a
a = a
3 a
a = a
3 a
a = a
n ,3 a
a = a
1 = a 1, = a
2 - 5
1 - 55
2 - 4
1 - 44
2 - 3
1 - 33
2 -n
1 -n n
21
9. Arithmetic Progression
1. Check whether following sequences are in A.P. i) a - 3, a - 5, a -7,...
A.Pin is sequencegiven
t - tt - t
2- =
5 + a -7 - a =
5) - (a - 7) - (a =t - t
2- =
3 + a - 5 - a =
3) - (a - 5) - (a = t - t
2312
23
12
∴
=
=
ii) 1/2, 1/3, 1/4, 1/5........
A.Pin not is sequenceGiven
t - tt - t
t - t
6
1 =
3
1 = t - t
2312
2312
∴
≠
−=
−=−
12
13
1
4
1
12
1
iii) 9, 13, 17, 21, 25,...
A.Pin is sequenceGiven
t - tt - t
13-7 t - t
4=
9-3 = t - t
2312
23
12
∴
=
=
=
4
1
1
iv) -1/3, 0, 1/3, 2/3.........
Winglish Coaching Centre, Puduvayal 26 X std Mathematics Made Easy
A.Pin is sequenceGiven
t - tt - t
3
1 t - t
3
1 =
3
1-- = t - t
2312
2312
∴
=
=
−=
3
1
00
v) 1, -1, 1, -1, 1, -1,............
A.Pin not is sequenceGiven
t - tt - t
)(1 t - t
2- = 1-1 = t - t
2312
23
12
∴
≠
=−−=
−
21
vi) ,....,, 43322 +++ xxx
A.Pin is sequenceGiven
t - tt - t
)()( t - t
)()( = t - t
2312
23
12
∴
=
−=+−+=
−=+−+
13243
1232
xxx
xxx
vii) 2, 4, 8, 16,…..
A.Pin not is sequenceGiven
t - tt - t
4- t - t
2 = 2- = t - t
2312
23
12
∴
≠
== 48
4
viii) ,....,,, 29272523
A.Pin is sequenceGiven
t - tt - t
t - t
= - = t - t
2312
23
12
∴
=
=−= 222527
222325
2. Write an A.P. whose first term is 20 and
common difference is 8.
44,36, 28, 20,
,...838,202 20 ,820 20,
3d,... + a 2d,+ a ,da a,
:n Progressio Arithmetic
8 =d ,differencecommon
20 a term, First
×+×++=
+=
=
3. Find first term and common difference of Arithmetic Progressions whose nth terms are given (i) tn = −3 +2n
2differencecommon
1- term First
2 = d
(-1) - 1
t - td
2(2) + 3 t
1- = 2(1) + 3- = t
2n 3-t
12
2
1
n
=
=
=
=
=−=
+=
1
(ii) tn = 4 - 7n
7-differencecommon
1- term First
-
(-3) - 10- =d
7(2) - 4t
3- =
7(1) - 4= t
7n - 4t
2
1
n
=
=
=
=
−=
=
=
7
10
4. Find 15th , 24th and nth term (general term) of
an A.P. given by 3, 15, 27, 39,…
912
12123
1213
279
12243
121243
171
12143
121153
1
24
24
15
15
−=
−+=
−+=
=
×+=
−+=
=
×+=
−+=
−+=
==
=
nt
n
nt
t
t
t
t
dna
n
n )(term, n
)(
)(
)(t, term General
. 123- 15 d ,difference com.
3 a ,term first
th
n
5. Find the 19th term of an A.P. -11,-15,-19,...
83- = t
72 - 11 - =
18(-4) + 11- =
1)(-4) - (19 + 11- = t
1)d -(n + a = t term n
4- = 11 + 15- =
(-11) - 15- = d
,11- = a
19
19
nth
6. Determine the general term of an A.P. whose 7th
term is −1 and 16th term is 17
152
2213
2113
13126
218921
21715
17116
17
116
117
1
1
167
−=
−+−=
−+−=
−=⇒−=+
=
=⇒−=−−
=+
=−+
=
−=+
−=−+
−=
−+=
n
n
n
aa
dd
da
da
t
da
da
t
dna
n
n
nth
t
)(t
)(
(1)in 2 d Put
)()(
)......(
)(
)......(
)(
)(t, term n
7. Find number of terms in 3, 6, 9, 12,…, 111
Winglish Coaching Centre, Puduvayal 27 X std Mathematics Made Easy
37 =n terms, of No.
1 + 36 =
1 +3
3 - 111 =
= terms, of No.
33 - 6 = .difference com.
111 = term last
3 = a term First
+
−
=
1d
aln
l
8. Which term of an A.P. 16, 11, 6, 1,... is -54 ?
15 =n terms, of No.
1 + 14 =
1 +5-
16 - 54- =
= terms, of No.
-16 - 11 = .difference com.
54- = term last
16 = a term First
+
−
=
1
5
d
aln
l
9. Find middle terms of A.P. 9, 15, 21, 27,…,183.
99 =
15(6) + 9 =
15d + a = term16th
93 =
14(6) + 9 =
14d + a = term15th
term 2
nand term
2
n : terms middle
30(even) =n
1 +6
9 - 183 =n
=n terms of No.
69 - 15 = ddifference com.
183 = lterm last
9, = a term First
thth
+
+
−
=
1
1d
al
10. If nine times ninth term is equal to the fifteen times fifteenth term, show that six times twenty fourth term is zero.
0 = t
0 = 23d + a
0 = 23d) + 6(a
0 = 138d + 6a
0 = 72d - 210d + 9a - 15a
d 210 + 15a = 72d + 9a
14d) + 15(a = 8d) + 9(a
15t = 9t
24
159
.
11. If 3 + k, 18 - k, 5k + 1 are in A.P. then find k
4=k
16 =4k
k +3k = 2 - 18
2 +3k =k - 18
4+6k = k) - 2(18
1 +5k +k + 3 = k) - 2(18
c + a = 2b
:A.Pin are c b, a,
1 +5k = c k , - 18 = b k, + 3 = a
12. Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.
3 = x20 =17 + x
20 = y+ xin yApplying
31 = z 48= z +17
48= z + y
24 + 24 = z + y
24 - z = y- 24
t - tt - t
17 = y34 = 2y
10 + 24 = y+ y
y- 24 = 10 - y
t - tt - t
(1)--- 20 = y+ x
10 - y x - 10
t - t= t - t
:A.Pin is sequenceGiven
4534
3423
2312
⇒
⇒
=
⇒
=
=
Hence the values of x, y and z are 3, 17, 31 respectively.
13. In a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?
78. row lastin seats of No.
78 = l
2 1) - (3020 -
t
30 = (n) rows of No. Total
30 =n and 2 = d 20, = a
24 = 2 + 22 = row 3in seats of No.
22 = 2 + 20 = row 2in seats of No.
20 = row 1in seats of No.
n
rd
nd
st
=
=
=+
−
=+
−
=
l
l
d
al
3012
20
301
30
14. In an A.P., sum of 4 consecutive terms is 28
sum of their squares is 276. Find the 4 numbers.
Winglish Coaching Centre, Puduvayal 28 X std Mathematics Made Easy
2 ± = d )(
4a
3d) + (ad)a (d)-a (3d)-(a
276, squares their of Ssum
7 = a 28 4a
28 3d + a da d-a3d-a
28, terms 4of Sum
terms Four3d + a d,a ,d-a 3d,-a
2
2222
⇒=+
=+
=
++++++
−++−+
=++++
=
⇒=
=++++
=
→+
2762074
27620
276269
269
276
22
2
2222
2222
d
d
addaadda
addaadda
13 and 9 5, 1,
3(2)+7 ,2 ,72-7 3(2),-7 : Numbers
2 = d7,a
+
=
15. Sum of three consecutive terms that are in A.P. is 27 Their product is 288. Find the three terms.
16. 9, 2, terms three First
16 =7 + 9 = d + a = term 3rd
9 = a = term 2nd
2 =7 - 9 = d - a = term 1st
7 = d 49- = d -
81 - 32 = d -
32 = )d - (9
288 = )d - 9(9
288 = )d - a(a
288 = d) + (a a d) - (a
288 = terms 3 of Product
9 = a27 = 3a
27 = d + a + a + d - a
27 = terms 3 of Sum
d. + a a, d, - a terms 3
2
2
22
22
22
=
⇒
⇒
=
16. A mother devides 207 into three parts such that the amount are in A.P. and gives it to her three children. The product of the two least amounts
that the children had 4623. Find the amount received by each child.
,71.69 67,
2+69 69, 2,-69 :given Amount
2 = d
4623d)69-(69
4623= )(
4623. amounts least 2 of product
69 = a
207 3a
207 =
207amount the of sum Since,
,, :children 3 by received Amount
=
−
=
=
+++−
=
+−
ada
daada
daada
17. The ratio of 6th and 8th term of an A.P. is 7:9. Find the ratio of 9th term to 13th term.
7. : 5 term 13 : term 9
14d
10d =
12d + 2d
8d + 2d =
12d + a
8d + a =
t
t
term 13 : term 9
2d = a
0 = 2d - a
0 = 4d- 2a
0 = 49d- 45d+ 7a - 9a
49d+ 7a = 45d+ 9a
7d) + 7(a = 5d) + 9(a9
7 =
7d + a
5d + a
9
7
t
t
thth
13
9
thth
8
6
=
=
18. Temperature of Ooty from Monday to Friday to
be in A.P. Sum of temperatures from Monday to
Wednesday is 0° C. Sum of the temperatures
from Wednesday to Friday is 18° C. Find the
temperature on each of the five days.
C 9° = 2(3) + 3 = 2d + a
C 6° = 3 + 3 = d + a
C 3° = a
C 0° = 3 - 3 = d - a
C 3°- = 2(3) - 3 = 2d - a
eTemperatur
Friday
Thursday
Wednesday
Tuesday
Monday
Days
3 = d
3 - 6 = d
6 = d + 3 (1),in a of valueApply
3 = a
6 = 2a
6 = a + a (2) + (1)
(2)--- 6 = d + a
18 = 3d + 3a
18 = 2d + a + d + a + a
1friday to Wednesday from sum
(1)--- 0 = d - a
0 = 3d - 3a
0 = a + d - a + 2d - a
0Wednesday to Monday from sum
2d + a ,d + a a, d, - a 2d, - a
: Friday to Monday from temp.
8=
=
Winglish Coaching Centre, Puduvayal 29 X std Mathematics Made Easy
19. Priya earned 15,000 in the first month.
Thereafter her salary increased by 1500 per
year. Her expenses are 13,000 during the first
year and the expenses increases by 900 per
year. How long will it take for her to save
20,000 per month.
years31in 20000 save will she
30.8 = 12
361 yearsof number
361 =n months, of number
=
=n months, of number
-52000-2050d
200003000, a
APan is It
0002100,..,20 2050, 2000,:montheach ofSaving
13150,.... 13075, 13000,:sequence a as Expenses
75 = 12
900 = increment monthly
900 = earexpenses/y increasing
13000 = expensemonth 1st
15250,.. 15125, 15000, :sequence a as Earnings
125 = 12
1500 = increment monthly
1500 = year/ salary of Increment
=
=
+
−
+
−
==
==
150
200020000
1d
al
l
20. If lth, mth and nth terms of an A.P. are x, y, z
respectively, then show that
i) 0=−+−+− )()()( mlzlnynmx
0
00
111
1
1
111
=
+=
−−+−−+−+−−+
−+−+−=
−−++
−−++−−+=
−+−+−
=−+
=
=−+
=
=−+
=
)()(
))(())(()())([(
)]()()[(
)]()([
))](([)]()([
)()()(
)()()(
da
nmlmlnnmlnmd
lnnmmla
mldna
lnlnyanmdla
mlzlnynmx
zdna
zt
ydma
yt
xdla
xt nml
ii) 0=−+−+− mxzlzynyx )()()(
0=
−+−+−=
−+−+−=
−+−+−
−=−
−=−
−=−
)(
])()()[(
)()()(
)(
)(
)(
lmnmnl lmmnlnd
mlnlnmnmld
mxzlzynyx
dlnxz
dnmzy
dmlyx
10. Arithmatic Series
1. Find the sum of the following (i) 3, 7, 11,… up to 40 terms.
3240 =
156] + [6 20 =
1)4] - (40 + [2(3) 2
40 =
1)d] -(n + [2a 2
n = S
4= 3 -7 = (d) differenceCommon
3 = (a) term First
40= (n) terms of Number
n
(ii) 102, 97, 92,… up to 27 terms.
999 =
(74) 2
27 =
130] - [204 2
27 =
26(-5)] + [204 2
27 =
1)(-5)] -(27 + [2(102) 2
27 =
1)d] -(n + [2a =Sn
5- = 102 -97 = (d) differenceCommon
102 = (a) term First
27 = (n) terms of Number
2
n
(iii) 6 + 13 + 20 +...........+ 97
721 =
7[103] =
97] + [62
14 =
][ 2
n =Sn
14 = 1 +7
6-97 =
1 + =n terms of No.
7 = 6 - 13 = d 6, = a 97, = l
a + l
d
al
−
.
iv) 1460430400 ++++ ........ .
14.7 =
1] + [0.402
21 =
][ 2
n =Sn
21 = 1 +0.03
0.4-1 =
1 + =n terms of No.
7 = 6 - 13 = d 6, = a 97, = l
a + l
d
al
−
Winglish Coaching Centre, Puduvayal 30 X std Mathematics Made Easy
vi) ,...,,,4
35
2
16
4
178 to 15 terms
4
165
2
2116
2
15
4
311582
2
15
122
4
38
15
15
=
−=
−−+=
−+=
−=−
S
S
dnan
S
d
n
)()(
])([
4
17 =,differenceCommon
8 = a term First
15 =n terms, of Number
2. How many consecutive odd integers beginning with 5 will sum to 480?
20 =n or 24- =n
0 = 20) -(n 24) +(n
0 = 480- 4n + n)(
960 =8n + 2n
480= 8] +[2n
480= 2] -2n + [10
480= 1)(2)] -(n + [2(5)
480= 1)d] -(n + [2a
480=Sn
2 = 5 -7 = d 5, = a
480= ............. + 11 + 9 +7 + 5
2
2
2
2
2
2
2
÷
n
n
n
n
3. How many terms of the series 1+5 +9 +... must
be taken so that their sum is 190?
×
+
2
19- =n or 10=n
0 = 19)(2n 10)-(n
0 = 190 -n -n
380 = 2]-n[4n
190 = 4]- 4n +[2
190 = 1)(4)] -(n + [2(1)
190 = 1)d] -(n + [2a
190 =Sn
4= 2 - 5 = d 1, = a
190 = ...........+ 9 + 5 + 1
22
2
2
2
n
n
n
4. Find the sum of first 28 terms of an A.P. whose nth term is 4n -3.
1540 =
[110] 14 =
27(4)] + 14[2 =
1)4] - (28 + [2(1) 2
28 = S
1)d] -(n + [2a = S
4= 1 - 5 = d
5 =
3 - 4(2)= t
1 =a
3 - 4=
3 - 4(1)= t
3 - 4n =tn
28
n
21
2
n
5. The sum of first n terms of a certain series is given as 2n2 -3n . Show that the series is an A.P.
4is d whose A.Pan is It
7,.... 3, 1,-:series
t2
9 =ttt
9 = 3(3) - 2(3) = S
t1-
2 =tt i.e,
2 = 3(2) - 2(2) = S
1- = t i.e,
3 - 2 = 3(1) - 2(1) = S
3n - 2n = S
3
321
23
2
21
22
1
21
2n
729
9
312
2
3
2
=−=
=+
++
=+=
=+
+
t
t
6. In an A.P. the sum of first n terms is 2
3
2
5 2 nn+ .
Find the 17th term.
48466-748
SST
2
1280
)()( = S
S2
1445
)()( = S
= S
161717
16
17
17
n
==
−=
=
+=
+
=
+=
+
+
6642
482
163
2
165
7482
512
173
2
175
2
3
2
5
2
2
2 nn
7. The 104th term and 4th term of an A.P. are 125 and 0. Find the sum of first 35 terms.
Winglish Coaching Centre, Puduvayal 31 X std Mathematics Made Easy
612.5 = S
2
35 =
4
51) - (35 +
4
15-2
2
35 =
1)d] -(n + [2a 2
n =Sn
4
15- = a 0 =
4
53 + a (2),in sub
4
5 =d
100
125 = d
125 = 100d (2), - (1)
(2)----- 0 = 3d + a
0 = t
(1)----- 125 = 103d + a
125 = t
35
4104
⇒
⇒
2
70
d
8. Find sum of all odd positive integers less than 450
50625 = S
4502
225 =
449]+ [12
225 =Sn
l] + [a2
n =Sn
225 =n
1 + 24 =
1 + 2
1-449 =n
1 + =n
2, = 1 - 3 = d 449,= l 1, = a
449+ 447 + .............+7 + 5 + 3 + 1
225
×
2
d
l-a
9. Find the sum of all natural numbers between 602 and 902 which are not divisible by 4.
224848 =
299(752) =
[1504]2
299 =
901] + [603 2
299 = S
l] + [a2
n =Sn
299 =n
1 + 298 =n
1+1
603-901 =n
1 +d
a-l =n
901 + ........+ 604 + 603
902 and 602 b/w no. of Sum
301
902 and 602 b/w 4by divisible no. of Sum
56400 =
75(752) =
[1504]2
75 =
900] + [604 2
75 = S75
l] + [a2
n =Sn
75 =n
1 + 4
604 - 900 =
d
a-l =n
900 + ..... + 612 + 608 + 604 =
+
1
168448 =
56400 - 224848 =
902 &602 b/w
4by divisible
no. of Sum
-
902 and 602
between
no. of Sum
=
902 and 602 b/w
by4 divisible
not no. of Sum
10. Find the sum of all natural numbers between
300 and 600 which are divisible by 7.
19264 =
595] + [301 2
43 = S
l] + [a2
n =Sn
43=n
1+7
301-595 =n
1 +d
a-l =n
595. ,… 315, 308, 301,
are7 by Divisible
2,.....600300,301,30
:600 and 300 b/w numbers Natural
301
11. Raghu can buy a laptop by paying 40,000 cash or by giving it in 10 installments as 4800 in the first month, 4750 in the second month, 4700 in the third month and so on. Find (i) total amount paid in 10 installments. (ii) how much extra amount that he has to pay than the cost?
5750 =
40000- 45750= amount Extra
45750=
9(-50)] + 5[9600 =
(-50)] 1) - (10 + [2(4800)2
10 = S
d] 1) -(n + [2a2
n = S
50- = 4800- 4750= d 4800,= a 10, =n
tsinstallmen 10 ...to+ 4700+ 4750+ 4800
10
n
12. A man repays a loan of 65,000 by paying 400 in the first month and then increasing the payment by 300 every month. How long will it take for him to clear the loan?
Winglish Coaching Centre, Puduvayal 32 X std Mathematics Made Easy
months 20 :in clear amountloan
(x) 3
65- =n 20, =n 0 = 65) +(3n 20) -(n
0 = 20) -65(n + 20) -3n(n
0 = 1300 -65n +60n - 3n
0 = 1300 -5n + 3n
1300 = 3n2 +5n
1300 = 3n] + n[5
650 = 3n] + [5
65000 = 300] -300n + [800
65000 = 1)300] -(n + [2(400)
65000 = 1)d] -(n + [2a
65000 = amountloan
................ 1000, + 700 + 400
1000 = 300 + 700 =month 3rd For
700 = 300 + 400=month 2nd For
400=month 1st For
2
2
⇒
2
2
2
2
n
n
n
n
13. A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two bricks less than the previous step. How many bricks are required for top most step? How many bricks are required to build stair case?
2130 = 15(142) =
42]+ [1002
n =
l] + [a2
n =Sn (ii)
42= 58 - 100 =
29(-2) + 100 =
29d + a = t (i)
30 = (n) steps of number
2- = 100 - 98 = d
98 = 2 - 100 = step 2ndin bricks of No.
100 = (a) step 1stin bricks of No.
30
14. Find sum terms 12 ....to++
−+
+
−+
+
−
ba
ba
ba
ba
ba
ba 3523
+
−=
+
−−+
+
−=
−+=
+
−=
+
−−
+
−=
+
−=
ba
ba
ba
ba
ba
baS
dnAn
S
ba
ba
ba
ba
ba
baba
ba
n
13246
21122
2
12
122
223
12 )(
])([
enceCom.differ
A Term, First
15. If S1, S2, S3,....Sm are the sums of n terms of m A.P.’s whose first terms are 1,2, 3,...m and whose common differences are 1, 3, 5,..., (2m -1) Show
that 2
1321
)(....
mnmnSSSS m
+=+++
2
1
124
211222
12
2
1122
12151312
122
12122
12
152
51322
53
132
31222
32
12
11122
11
122
321
321
3
3
2
1
)(
][
])()([.....
Terms of No.
1) - 2n(m
1 )()(
Terms of No.
]...[....
)()])(()([
,)
)(])()([
,)
)(])()([
,)
)(])()([
,,)
])([
mnmn
mnmn
nmnmn
SSSS
m
n
n
nnmn
nmnnnnn
SSSS
nmnn
mnmn
S
mdmaiv
nn
nn
S
daiii
nn
nn
S
daii
nn
nn
s
dai
dnan
S
m
m
n
+=
+×=
−++=+++
=
+=
+
+−+−=
+−+++++=+++
+−=−−+=
−==
+=−+=
==
+=−+=
==
+=−+=
==
−+=
16. The sum of first n, 2n and 3n terms of an A.P.
are S1S2 , and S3. Prove that S3= 3(S2 – S1)
( )
312
12
12
3
2
1
3
1322
33
1322
1212242
122
1222
2
1322
3
1222
2
122
SSS
dnan
SS
dnan
dnadnan
dnan
dnan
SS
dnan
Sn
dnan
Sn
dnan
Sn
=−
−+=−
−+=
−+−−+=
−+−−+=−
−+=
−+=
−+=
)(
])([)(
])([
])([])([
])([])([
])([ terms 3 first of Sum
])([ terms 2 first of Sum
])([ terms first of Sum
Winglish Coaching Centre, Puduvayal 33 X std Mathematics Made Easy
11. Geometric Progression
1. Which of the following sequences are in G.P.? (i) 3, 9, 27, 81,…
GP a is It
t
t
t
t
t
t
t
t
2
3
1
2
2
3
1
2
∴
=
==== 39
273
3
9
(ii) 4,44,444,4444,...
GP not is It
t
t
t
t
44
444
t
t
t
t
2
3
1
2
2
3
1
2
∴
≠
=== 114
44
(iii) 0.5, 0.05, 0.005,…
GP a is It
t
t
t
t
0.1 =0.05
0.005
t
t0.1 =
0.5
0.05
t
t
2
3
1
2
2
3
1
2
∴
=
==
. c
(iv) 1/3, 1/6, 1/12,.......
GP a is It
t
t
t
t
t
t
t
t
2
3
1
2
2
3
1
2
∴
=
=×==×=2
1
1
6
12
1
2
1
1
3
6
1
(v) 1, −5, 25, −125,…
GP a is It
t
t
t
t
t
t
t
t
2
3
1
2
2
3
1
2
∴
=
=−
==−
= 55
255
1
5
(vi) 120,60,30,18,…
GP a is It
t
t
t
t
60
30
t
t
120
60
t
t
2
3
1
2
2
3
1
2
∴
=
====2
1
2
1
(vii) 16, 4, 1, 1/4,..........
GP a is It
t
t
t
t
t
t
16
4
t
t
2
3
1
2
2
3
1
2
∴
=
===4
1
4
1
viii) 7, 14, 21, 28, …
GP a not is It
t
t
t
t
t
t
7
14
t
t
2
3
1
2
2
3
1
2
∴
≠
====2
3
14
212
ix) ½,1,2 4, ...
GP a is It
t
t
t
t
t
t
2
1
1
t
t
2
3
1
2
2
3
1
2
∴
=
==== 21
22
x) 5, 25, 50, 75
GP not is It
t
t
t
t
t
t
5
25
t
t
2
3
1
2
2
3
1
2
∴
≠
=== 225
505
2. Write first three terms of the G.P. whose first term and the common ratio are given below. (i) a = 6, r = 3
54. 18, 6, : terms three 1st
54 = 6(3) = ar ,term Third
18 = 6(3) = ar ,term Second
6 = (a) term First
22
(ii) a = √2, r = √2
2,,2 : terms three First
22 = 222 = ar ,term Third
2 = 2 = ar term, Second
2 = (a) term First
2
22
(iii) a = 1000, r = 2/5
160. 400,1000, : terms three First
160 =5
2 1000 = ar term, Third
400= 5
2 1000 = ar term, Second
1000 = (a) term First
22
×
(iv) a =−7 , r = 6
252.42,--7,- : terms three First
252- =(6)7 - = ar term, Third
42- = 67- = ar term, Second
7- = a ,term First
22
×
(v) a = 256 , r = 0.5
,..6256,128, : terms three First
64 = 256 = ar term, Third
128 = 256 = ar term, Second
10
5r256, = a ,term First
22
4
2
1
2
12
1
×
==
3. i)In a G.P. 729, 243, 81,… find t7 . ii) Find 8th term of the G.P. 9, 3, 1,…
Winglish Coaching Centre, Puduvayal 34 X std Mathematics Made Easy
243
1 =
9 =
9 = t
ar =tn
9
3 = r
9, = a ii)
1 =
729 =
729 = t
ar =tn
729
243 = r
729, = a i)
7
1- 8
7
1-n
6
1-7
7
1-n
=
=
3
1
3
1
3
1
3
1
3
1
3
1
4. Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression.
18- = x
54- = 3x
144 - 90 = 21x - 24x
90 + 21x = 24x + 144
90 + 6x + 15x + x = 2x(12) + 122 + x
12 + x
15 + x
6 + x
12 + x
t
t
t
tG.P a For
22
2
3
1
2
=
=
.
5. Find the number of terms in the following G.P. (i) 4, 8, 16,…,8192 ?
12 =n
2 = 2
8192 = (2)
8192 = 4(2)
8192 = ar
8192 = t
2 = 4
8 = r 4,= a
111-n
1-n
1-n
1-n
n
(ii) 1/3, 1/9, 1/27,................1/2187
7 =n
2187
1 =
2187
1 =
3
1
2187
1 = ar
2187
19
1 = r
,3
1 = a
61-n
1-n
1-n
1-n
=
×
=
=×
3
1
3
1
33
1
3
1
3
1
1
3
nt
6. In a G.P. the 9th term is 32805 and 6th term is 1215. Find the 12th term.
1112
11
5
3 3
3
8
5
5
6
8
9
5(3) = t
ar term12th
5 = a343
1215 = a
1215 = a(3)(2),in r Apply
3 = r3 = r
27
1 =
r
1
32805
1215 =
ar
ar ,
(1)
(2)
-(2)- 1215 = ar
1215 = t
(1)- 32805 = ar
32805 = t
=
⇒
⇒
7. In a Geometric progression, the 4th term is 8/9
and the 7th term is 64/243 . Find the G.P.
,...,,
..3
2,
3
23,3 =
,...arar,a, GP
3 = a 8
27
9
8 = a
9
8 =
3
2a(1),in r Apply
3
2 = r
3
2 = r
27
8 = r
243
64 =
ar
ar ,
(1)
(2)
-(2)- 243
64 = ar
243
64 = t
(1)- 9
8 = ar
9
8 = t
2
3
3 3
3
3
6
6
7
3
4
3
423
3
8
9
2
=
=
⇒×
⇒
×
8. Find the 10th term of a G.P. whose 8th term is 768 and the common ratio is 2.
3072 =
6(512) =
)6(2 =
ar = t
6 = a128
768 = a
2] = r[768 = )a(2
768 = ar
768 = t
9
910
7
7
8
⇒
Q
9. If a, b, c are in A.P. then show that 3a, 3b, 3c are in G.P.
Winglish Coaching Centre, Puduvayal 35 X std Mathematics Made Easy
G.P.in are 3 ,3 ,3
2
ca = b
3 = 3
)(3 = 3
333
3
3 =
3
3 G.P,in are 3 ,3 ,3 If
t GP, For
2
cab
b-c = a-b A.P,in are c b, a, If
t AP, For
cba
c+a b
2
1c + a b
cab
b
c
a
bcba
2
2
∴
+
=
=
+=
−=−
2
2
3
1
231
t
t
t
ttt
.
10. If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb−c × yc−a × za−b = 1 .
( )
1
22
0
2222
2222
222
22
2
12
==
×=
×××=
××=
××=××
=⇒=
+=⇒+=
−+
−−+
−
−−−−
−−−
+−
−−
+−−−
o
caacacca
caacacca
caacca
caa
acc
cabaaccb
yx
zx
zzxx
zxzx
zyxzyx
xzyxzy
PGzyx
cabcab
)(
.in are,,
A.Pin are c b, a,
11. A man joined a company as Assistant Manager. The company gave him a starting salary of
60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?
76577 =
)60000(1.05 =
5%) + 60000(1 = years5 After
5%) + 60000(1 =
5%) + 5%)(1 + 60000(1 =
5%) + 60000(1 of 5% + 60000 = salary year3rd
5%) + 60000(1 =
60000 of 5% + 60000 = salary year2nd
60,000 = salaryStarting
5
5
2
12. Sivamani is attending an interview for a job and the company gave two offers to him. Offer A:
20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years. Offer
B: 22,000 to start with followed by guaranteed annual increase of 3% for first 5 years.
24040 =
)22000(1.03 =
3%) + 22000(1 = salary 4th year
3%) + 22000(1 =
22000 of 3% + 22000 = salary year2nd
22,000 = salaryStarting
5 =n and 0.03 = r 22000, = a,)
22820 =
)20000(1.06 =
6%) + 20000(1 = salary 4th year
6%) + 20000(1 =
20000 of 6% + 20000 = salary year2nd
20000 = salaryStarting
5 =n and 0.06 = r 20000, = a,)
3
3
3
3
BOffer
Offer Ai
ii
13. In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 57/2 . Find the three terms.
2
9 = 3 = ar ,term Third
3 = a ,term Second
2 = 3 = r
a ,term First
3
2 = r ,
2
3 = r0 = 2) - 3)(3r - (2r
0 = 6 + 13r - 6r
0 = 18 + 39r - 18r
0 = 18 + 57r - 18r + 18r
r57 = 18 + 18r + 18r
r57 = 1) + r + 18(r
2
57 =
r
)r + r + 9(1
2
57 = 1) + r +
r
1(a
2
57 = a + ra + a
2
57 = ar
r
aar .a .
r
a
2
57 = time a attaken terms 2 of Sum
3 = a27 = a
27 = ar a r
a
27 = terms 3 of Product
ar a, ,r
a : terms three
2
2
2
2
2
2
2
222
3
2
3
3
2
×
×
⇒
++
⇒
××
a
Winglish Coaching Centre, Puduvayal 36 X std Mathematics Made Easy
14. The product of three consecutive terms of a
Geometric Progression is 343 and their sum is
3
91. Find the three terms.
,,,: Numbers
,3
1(ii)Case
,,: Numbers
,(i)Case3
1 = r ,3 = r0 = 3) - 3)(r - (3r
0 = 3 + 10r - 3r
r 13 = 3 + 3r + 3r
r 13 = 1) + r + 3(r
3
91 =
r
)r + r + 7(1
3
91 = 1) + r +
r
1a(
3
91 = aa
r
a
3
91 = terms 3 of Sum
7 = a7 = a
343 = ar a r
a
343 = terms 3 of Product
ar a, ,r
a : terms three
2
2
2
2
33
3
7721
7
2173
7
73
==
==
⇒
++
⇒
××
ar
ar
r
15. The present value of a machine is 40,000 and
its value depreciates each year by 10%. Find the
estimated value of the machine in the 6th year
6023619
6
5
16
6
1
32
2
.6th yearin machine of Value
100
90 000 40
100
90 000 40
,100
90r
, 000 40 a GP, a is It
,.100
90 000 40,
100
90 000 40,
100
90 000 40i.e,
100
90 000 40 year2nd after machine of Value
100
90 000 40 yearfirst after machine of Value
=
×=
×=
=
==
=
×
××
×=
×=
−
−
t
art
n
nn
12. Geometric Series
1. Find the sum of first n terms of the G.P
(i) 5, -3, 9/5, -27/25, .........(ii) 256, 64, 16,…
−=
−
−
−
=
−=
−
−
−
<
n
n
n
n
rr
4
1
4
1
4
1
256 = S
r-1a = S
256
64 = r 256, = a ii)
5
3-
5
3-
5
3-
5 = S
r-1a = S
5
3- = r 5, = a i)
n
n
n
n
n
n
13
1024
1
1
1
4
1
18
25
1
1
1
1
2. i)Find sum of first 6 terms of G.P. 5, 15, 45, … ii) Find sum of 8 terms of the G.P. 1,-3,9, -27, ..
1640- =4
6561-1 =
(-3)-1
)(-3)-1(1 =S
)r-a(1 = S
1 3 - = 1
3- = r
1, = a 8, =n ii)
1820 =2
1) - 5(729 =
1 - 3
1) - 5(3 =S
)a(r = S
1 3 = 5
15 = r
5, = a 6, =n i)
8
8
n
n
6
6
n
n rr −
<
−
−
>
11
1
3. Find the first term of the G.P. whose i)common ratio 5. sum to first 6 terms is 46872.
ii) S6=4095 ,r = 4
3 = a
4095
34095a
340951) - a(4
40951 - 4
1) - a(4
4095= 1 - r
1) - (ra
4095= S ii)
12 = a
15624
446872a
46872(4)1) - a(5
468721 - 5
1) - a(5
46872= 1 - r
1) - (ra
46872= S i)
6
6
n
6
6
6
n
6
×=
×=
=
×=
=
=
4. Find the sum to infinity of (i) 9 + 3 + 1 + ..... (ii) 21 + 14 + 28/3 + .............
63 =
321 =
-1
21 =
r-1
a = S
3
2
21
14 = r21,= a ii)
27/2 =
39 =
-1
9 =
r-1
a = S
3
1
9
3 = r 9, = a i)
32
31
12×
∞
=
×
∞
=
Winglish Coaching Centre, Puduvayal 37 X std Mathematics Made Easy
5. i)First term of an infinite G.P. is 8 Sum to
infinity is 3
32. Find the common ratio.
ii)Find the least positive integer n such that 1+6 +62+....+ 6n >5000
6n Thus,
7 6
6
25001 6
55000 6
1-r
)a(r
S
6r1, = a ii)
4
1
4
3 - 1 = r
r - 1 = 4
3
r - 1 = 32
24
= r-1
8
= r-1
a
S i)
5
5
n
n
n
=
>
>
>
×>−
>−
=
=
=
=∞
776
46656
1
50001
5000
3
323
323
32
n
6. Find the sum of the Geometric series 3 + 6 + 12 +............+ 1536
3069 = S
3(1023) =
1) - 3(1024 =
1 - 2
1 - 23S
1 - r
1 - ra = S
10 =n
2 = 2
512 = 2
1536 = 3(2)
1536 =
1536 = t
2 = 3
6 = r3, = a
10
10
10
n
n
91-n
1-n
1-n
n
1-
=
nar
7. How many terms of 1+4+16 .. make the sum 1365
6
44
140956
=
=
+=
×=
=
=++
n
n
n
n
n
n
n
4
31365 1 - 4
13651 - 4
1) - 1(4
1365= 1 - r
1) - (ra
1365 = S
1365 n...... 16+4+1
8. Find rational form of i) 0.6666 ii) 1230.
333
41 =
0.999
0.123 =
0.123-1
0.123 =
r-1
a = S
.0.123
0.123r0.6,a
G.P a forms This
.. 0.01230.123 1230.
3
2 =
0.9
0.6 =
0.1-1
0.6 =
r-1
a = S
.0.6
0.06r0.6,a
G.P a forms This
006 0.0.060.6 0.6666
∞
===
++=
∞
===
+++=
1010
9. Find the sum to n terms of the series
(i) 0.4 + 0.44 + 0.444 +........ to n terms
×
9
(0.1) - 1 -n
9
4 =
0.1-1
(0.1) - 1 0.1 -n
9
4 =
terms]...n + 0.13+0.12+[0.1 - terms)1...n +1+[(19
4 =
terms].......n + 0.13) - (1 + 0.12) - (1 + 0.1) - [(19
4 =
terms]..n ..........+ 0.999 + 0.99 + [0.9 9
4 =
terms)n to ...... + 0.111 + 0.11 + (0.1 9
9 4=
terms)n to ...... + 0.111 + 0.11 + 4(0.1=
termsn to ........+ 0.444 + 0.44 + 0.4
n
n
(ii) 3 + 33 + 333 + ........... to n terms
×
3
n-
27
1) - 10(10 =
n-9
1) - 10(10
3
1 =
n-1-10
1) - 10(10
3
1 =
terms)]....n +1+1+(1-terms).....n + 1000+100+[(103
1=
terms]..n ..........+ 1) - (1000 + 1) - (100 + 1)-[(103
1 =
terms]..n ..........+ 999 + 99 + [99
3 =
terms)n to ...... + 111 + 11 + (1 9
9 3 =
terms]n +........... + 111 + 11 + [1 3 =
termsn to ........... + 333 + 33 + 3
n
n
n
(iii) 5 + 55 + 555 + ........... to n terms
×
9
5n-
81
1) - 50(10 =
n-9
1) - 10(10
9
5 =
n-1-10
1) - 10(10
9
5 =
term)]....n +1+(1-terms).....n + 1000+100+[(109
5=
terms]..n ..........+ 1) - (1000 + 1) - (100 + 1)-[(109
5 =
terms]..n ..........+ 999 + 99 + [99
5 =
terms)n to ...... + 111 + 11 + (1 9
9 5 =
terms]n +........... + 111 + 11 + [1 5 =
n to ........... + 555 + 55 + 5
n
n
n
Winglish Coaching Centre, Puduvayal 38 X std Mathematics Made Easy
10. A person saved money every year, half as much
as he could in the previous year. If he had
totally saved 7875 in 6 years then how much
did he save in the first year?
( )[ ]
40007875
1
1
=⇒=×
=−
=−
=
a
a
r
32
63a
7875-1
7875]r-a[1
.2
1 = r 6, =n
7875 Ssaved, amount Total
21
6
21
n
6
11. Kumar writes a letter to four of his friends. He asks each one of them to copy the letter and mail to 4 different persons with the instruction that they continue the process similarly. It costs 2 to mail one letter, find amount spent on postage when 8th set of letters is mailed.
174760 =
(87380) 2 = Cost
87380 =
3
65535 4=
]4[4 = S
]a[r =Sn
4.= 16/4 = r and 4= a 8, =n
series geometric a forms It
................... + 64 + 16 + 4
8
8
n
×
−
−
−
−
14
1
1
1
r
12. nyyxxyxyxyxyxSn ....)()()( +++++++++= 322322
prove that
−
−−
−
−=−
1
1
1
1 22
y
yy
x
xxSyx
nn
n
)()()(
1
1
1
1 22
432
432
443322
3223
22
3223
22
322322
−
−−
−
−=−
++
−++=
+−+−+−=
++++−
+++−++−=
++++
+++++−=−
+
++++++++=
y
yy
x
xxyxS
nyyy
nxxx
nyxyxyx
nyyxxyxyx
yxyxyxyxyx
nyyxxyx
yxyxyxyxyxS
n
yyxxyxyxyxyxS
nn
n
n
n
)()()(
)]term....(
term)....[(
terms....)()()(
]terms....))((
))(())((
]terms....)(
)())[(()(
y)- (x by sidesboth Multiply
....
)()()(
13.Special Series
1. Find the sum of the following series
i) 1 + 2 + 3 +............+ 60
1830 =
30(61) = 2
) 1 +60(60 = 60 +............+ 3 + 2 + 1
2
1) +n(n =n +...............+ 3 + 2 + 1
:numbers natural of Sum
ii) 1 + 2 + 3 +............+ 50
1275 =
25(51) = 2
) 1 +50(50 = 50 +............+ 3 + 2 + 1
2
1) +n(n =n +...............+ 3 + 2 + 1
iii) 2+4+6+......+80
1640 =
(41) (20) 2 =
2
1)+40(40 2 =
40)+............ + 3 + 2 + 2(1 =
iv) 3 + 6 + 9 + ............+ 96
1584 =
(33) (16) 3 =
2
1)+32(32 3 =
32) +............ + 3 + 2 + 3(1 =
v) 51+ 52 + 53 +............+ 92
3003 =
1275 - 4278=
25(51) - 46(93)=
21
1) + 50(50 -
2
1) + 92(92 =
50) +3... + 2 + (1 - 92) +..+ 3 + 2 + (1 = 92 +...+ 53 + 52 +51
vi)16+17+18+......+75
2730 =120 - 2850 =
15(8) - 75(38) =
21
1) + 15(15 -
2
1) + 75(75 =
15) +... 2 + (1 - 75) +..+ 2 + (1 = 75+......+17+16
2. Find the sum of the following series
i) 12+22+ 32+.......+192
2470 =
(13) 19(10) = 6
19(20)(39) =
6
1) + (2(19) 1) + 19(19 = squares of Sum
Winglish Coaching Centre, Puduvayal 39 X std Mathematics Made Easy
ii) 1 + 4 + 9 + 16 +...............+ 225
1240 =
(31) 5(8) = 6
15(16)(31) =
6
1) + (2(15) 1) + 15(15 =
15 +............+ 4+ 3 + 2 + 1 22222
iii)152+162+ 172+.......+282
82775 6
22125 =
6
1) + (2(21) 1) + 21(21 5 =
)21 +............+ 4+ 3 + 2 + (15
2
222222
=
×××
×
=
432
iv) 62 + 72 + 82 +........+212
3256 =
55 - 3311 =
6
(11) 5(6) -
6
(43) 21(22) =
6
1) + (2(5) 1) + 5(5 -
6
1) + (2(21) 1) + 21(21 =
6
1) +1)(2n +n(n = squares of Sum
)5 +.. + 2 + (1 - )21 +..... 2 +(1 = 21 +...+7 +6 22222 222 2
v)152+162+ 172+.......+282
6699 =
1015 - 7714 =
6
(29) 14(15) -
6
(57) 28(29) =
6
1) + (2(14) 1) + 14(14 -
6
1) + (2(28) 1) + 28(28 =
)14 +.. + 2 + (1 - )28 +..... 2 +(1 = 22222 2
3. Find the sum of the following series
i) 13 + 23 + 33 +.........+ 163
18496
6
=
=
+++
+++
2
2
333
2333
136
2
1) + 16(16 = 1..21
2
1) +n(n = n..21
ii) 103 + 113 + 123 +.........+ 203
42075=
(165) 255 =
45)- (210 45)+ (210 =
45- 210 =
2
1) + 9(9
2
1) + 20(20 =
)9 +...2 + (1 - )20 +....2(1 = 20 +... 11 + 10
22
22
333333333
−
+
ii) 93 + 103 + 113 +.........+ 213
52065
36) - (231 36) + (231 =
36 - 231 =
2
1) + 8(8
2
1) + 21(21 =
)8 +...2 + (1 - )21+....2(1 = 21 +... 10 + 9
22
22
333333333
=
−
+
4. Find the sum of the following series
i) 1 + 3 + 5 +.............+ 71
1296 =
36 =
2
1 =
2
1 = numbers odd of Sum
2
2
2
71
+
+l
ii) 1 +3+5+...... to 40 terms
784 =
28 =
2
1 = ...31
2
1 = ...31
2
2
2
55555
5
+++++
+++++
ll
iii)1 +3+5+...... +55
784 =
28 =
2
1 = ...31
2
1 = ...31
2
2
2
55555
5
+++++
+++++
ll
5. If 1+2+ 3 +......+n=666 then find n
, number, natural isn As
,
))((
n
6662
1)+n(n
666 =n +........... + 3 + 2 + 1
2
36
3637
03637
01332
=
=−=
=−+
=−+
=
n
nn
nn
n
6. If 1 + 2 + 3 + ....+ k = 325 , find 13+23 +33 +......+ k3
105625 =
325 =
2
1)+k(k = k +......+ 3 + 2+ 1
2
1)+k(k
325 =k +........... + 3 + 2 + 1
2
23333
= 325
7. If 13 + 23 + 33 +.....+ k3 = 44100.Find 1+2+3 +......+k
Winglish Coaching Centre, Puduvayal 40 X std Mathematics Made Easy
210 =k + ......... + 3 + 2 + 1
44100 = 2
1)+k(k
44100= 2
1)+k(k
44100= k +............+ 3 +2 + 1
2
33 33
8. How many terms of the series 13 + 23 + 33 +......... should be taken to get the sum 14400?
15. :terms of number Rqd
15 16,- =n
0 = 15) -16)(n +(n
0 = 240 -n + n2
240 =n + n
2120 = 1) +n(n
14400 = 2
1)+n(n
14400 = 2
1)+n(n
14400 =Sn
2
2
×
.
9. Sum of squares of first n natural numbers is 285, Sum of their cubes is 2025. Find the value of n.
9 =n 90
285 = 1 +2n
6
1) +(2n 906
1) +1)(2n +n(n
285 = n + ............+ 3 + 2 + 1
901)+n(n2
1)+n(n
2025 = 2
1)+n(n
2025 = n + ............+ 3 +2 + 1
2222
2
33 33
6
285
285
45
×
=
=
=
=
10. Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm,…, 24 cm. How much area can be decorated with these colour papers?
2
22222222
2222
cm 4615=
285 - 4900= 6
9(10)(19) -
6
24(25)(49) =
6
1) + 1)(2(9) + 9(9 -
6
1 + 1)(2(24) + 24(24 =
)9 +..... + 3 + 2 + (1 - )24 +..... + 3 + 2 + (1 =
24 +............. + 12 + 11 + 10
area Required
11. Find sum of series (23 − 1) + (43 −33)+(63 −53)+.. to
(i) n terms (ii) 8 terms First numbers are even ; second numbers are odd.
2240 =
192 + 2048 =
3(8) + 4(8)=S
3n + 4ntermsn of Sum
3n + 4n=
n +3n - 3n -2n + 6n + 4n=
n +3n - 3n - 1) +2n +n + 2n(2n =
n + 1) +3n(n - 1) +1)(2n +2n(n =
n + 2
1) +n(n 6 -
6
1) +1)(2n +n(n 12 =
1 +6n - 12n =
1 +6n - 12n + 8n - 8n =
1] -6n + 12n - [8n - 8n =
]1 - 3(2n)(1) + (1) 3(2n) - [(2n) - 8n =
1) -(2n - (2n) = term General
238
23
23
223
22
2
233
233
3233
33
=
××
14.System of Linear Equations in Three
Variables
1. Solve the following system of linear equations
in three variables
a) 6532223 =++=−+=+− zyxzyxzyx ;;
3 = z ,2 = y1, = x
3 = z , 6 z 21
(3),in 2 = y ,1 = xng Substituti
2 = y7 y+ 5
(4),in 1 = xng Substituti
4y20x5-(4)4
(5)-------1 y3x)()(
(3)-------
(2)-------
(4)-------7 y5x)()(
(2)-------
(1)-------
-(3)-
-(2)-
-(1),-
∴
⇒=++
⇒=
=⇒=
=+
=+×
=++
=++
=−+
=++
=−+
=+−
=++
=−+
=+−
11717
1143
28
1432
6
532
21
532
223
6
532
223
xx
yx
zyx
zyx
zyx
zyx
zyx
zyx
zyx
b) 1632925 =+−=+−=++ zyxzyxzyx ;;
(3)-------
(2)-------
(1)-------
1632
92
5
=+−
=+−
=++
zyx
zyx
zyx