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1 | 618-621 | Show that f : A ร B โ B ร A such that f (a, b) = (b, a) is
bijective function 9 Let f : N โ N be defined by f (n) =
n
n
n
n
+
๏ฃฑ
๏ฃฒ
๏ฃด๏ฃด
๏ฃด๏ฃณ
๏ฃด
1
2
2
,
,
if
is odd
if
is even
for all n โ N State whether the function f is bijective |
1 | 619-622 | 9 Let f : N โ N be defined by f (n) =
n
n
n
n
+
๏ฃฑ
๏ฃฒ
๏ฃด๏ฃด
๏ฃด๏ฃณ
๏ฃด
1
2
2
,
,
if
is odd
if
is even
for all n โ N State whether the function f is bijective Justify your answer |
1 | 620-623 | Let f : N โ N be defined by f (n) =
n
n
n
n
+
๏ฃฑ
๏ฃฒ
๏ฃด๏ฃด
๏ฃด๏ฃณ
๏ฃด
1
2
2
,
,
if
is odd
if
is even
for all n โ N State whether the function f is bijective Justify your answer 10 |
1 | 621-624 | State whether the function f is bijective Justify your answer 10 Let A = R โ {3} and B = R โ {1} |
1 | 622-625 | Justify your answer 10 Let A = R โ {3} and B = R โ {1} Consider the function f : A โ B defined by
f (x) =
2
3
xx
โ
๏ฃซ
๏ฃถ
๏ฃฌ
๏ฃท
โ
๏ฃญ
๏ฃธ |
1 | 623-626 | 10 Let A = R โ {3} and B = R โ {1} Consider the function f : A โ B defined by
f (x) =
2
3
xx
โ
๏ฃซ
๏ฃถ
๏ฃฌ
๏ฃท
โ
๏ฃญ
๏ฃธ Is f one-one and onto |
1 | 624-627 | Let A = R โ {3} and B = R โ {1} Consider the function f : A โ B defined by
f (x) =
2
3
xx
โ
๏ฃซ
๏ฃถ
๏ฃฌ
๏ฃท
โ
๏ฃญ
๏ฃธ Is f one-one and onto Justify your answer |
1 | 625-628 | Consider the function f : A โ B defined by
f (x) =
2
3
xx
โ
๏ฃซ
๏ฃถ
๏ฃฌ
๏ฃท
โ
๏ฃญ
๏ฃธ Is f one-one and onto Justify your answer 11 |
1 | 626-629 | Is f one-one and onto Justify your answer 11 Let f : R โ R be defined as f(x) = x4 |
1 | 627-630 | Justify your answer 11 Let f : R โ R be defined as f(x) = x4 Choose the correct answer |
1 | 628-631 | 11 Let f : R โ R be defined as f(x) = x4 Choose the correct answer (A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto |
1 | 629-632 | Let f : R โ R be defined as f(x) = x4 Choose the correct answer (A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto 12 |
1 | 630-633 | Choose the correct answer (A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto 12 Let f : R โ R be defined as f (x) = 3x |
1 | 631-634 | (A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto 12 Let f : R โ R be defined as f (x) = 3x Choose the correct answer |
1 | 632-635 | 12 Let f : R โ R be defined as f (x) = 3x Choose the correct answer (A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto |
1 | 633-636 | Let f : R โ R be defined as f (x) = 3x Choose the correct answer (A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto Rationalised 2023-24
MATHEMATICS
12
1 |
1 | 634-637 | Choose the correct answer (A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto Rationalised 2023-24
MATHEMATICS
12
1 4 Composition of Functions and Invertible Function
Definition 8 Let f : A โ B and g : B โ C be two functions |
1 | 635-638 | (A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto Rationalised 2023-24
MATHEMATICS
12
1 4 Composition of Functions and Invertible Function
Definition 8 Let f : A โ B and g : B โ C be two functions Then the composition of
f and g, denoted by gof, is defined as the function gof : A โ C given by
gof (x) = g(f (x)), โ x โ A |
1 | 636-639 | Rationalised 2023-24
MATHEMATICS
12
1 4 Composition of Functions and Invertible Function
Definition 8 Let f : A โ B and g : B โ C be two functions Then the composition of
f and g, denoted by gof, is defined as the function gof : A โ C given by
gof (x) = g(f (x)), โ x โ A Fig 1 |
1 | 637-640 | 4 Composition of Functions and Invertible Function
Definition 8 Let f : A โ B and g : B โ C be two functions Then the composition of
f and g, denoted by gof, is defined as the function gof : A โ C given by
gof (x) = g(f (x)), โ x โ A Fig 1 5
Example 15 Let f : {2, 3, 4, 5} โ {3, 4, 5, 9} and g : {3, 4, 5, 9} โ {7, 11, 15} be
functions defined as f (2) = 3, f(3) = 4, f(4) = f(5) = 5 and g (3) = g (4) = 7 and
g (5) = g(9) = 11 |
1 | 638-641 | Then the composition of
f and g, denoted by gof, is defined as the function gof : A โ C given by
gof (x) = g(f (x)), โ x โ A Fig 1 5
Example 15 Let f : {2, 3, 4, 5} โ {3, 4, 5, 9} and g : {3, 4, 5, 9} โ {7, 11, 15} be
functions defined as f (2) = 3, f(3) = 4, f(4) = f(5) = 5 and g (3) = g (4) = 7 and
g (5) = g(9) = 11 Find gof |
1 | 639-642 | Fig 1 5
Example 15 Let f : {2, 3, 4, 5} โ {3, 4, 5, 9} and g : {3, 4, 5, 9} โ {7, 11, 15} be
functions defined as f (2) = 3, f(3) = 4, f(4) = f(5) = 5 and g (3) = g (4) = 7 and
g (5) = g(9) = 11 Find gof Solution We have gof (2) = g (f (2)) = g (3) = 7, gof (3) = g (f (3)) = g (4) = 7,
gof(4) = g (f(4)) = g (5) = 11 and gof(5) = g (5) = 11 |
1 | 640-643 | 5
Example 15 Let f : {2, 3, 4, 5} โ {3, 4, 5, 9} and g : {3, 4, 5, 9} โ {7, 11, 15} be
functions defined as f (2) = 3, f(3) = 4, f(4) = f(5) = 5 and g (3) = g (4) = 7 and
g (5) = g(9) = 11 Find gof Solution We have gof (2) = g (f (2)) = g (3) = 7, gof (3) = g (f (3)) = g (4) = 7,
gof(4) = g (f(4)) = g (5) = 11 and gof(5) = g (5) = 11 Example 16 Find gof and fog, if f : R โ R and g : R โ R are given by f(x) = cos x
and g(x) = 3x2 |
1 | 641-644 | Find gof Solution We have gof (2) = g (f (2)) = g (3) = 7, gof (3) = g (f (3)) = g (4) = 7,
gof(4) = g (f(4)) = g (5) = 11 and gof(5) = g (5) = 11 Example 16 Find gof and fog, if f : R โ R and g : R โ R are given by f(x) = cos x
and g(x) = 3x2 Show that gof โ fog |
1 | 642-645 | Solution We have gof (2) = g (f (2)) = g (3) = 7, gof (3) = g (f (3)) = g (4) = 7,
gof(4) = g (f(4)) = g (5) = 11 and gof(5) = g (5) = 11 Example 16 Find gof and fog, if f : R โ R and g : R โ R are given by f(x) = cos x
and g(x) = 3x2 Show that gof โ fog Solution We have gof(x) = g (f(x)) = g(cos x) = 3 (cos x)2 = 3 cos2 x |
1 | 643-646 | Example 16 Find gof and fog, if f : R โ R and g : R โ R are given by f(x) = cos x
and g(x) = 3x2 Show that gof โ fog Solution We have gof(x) = g (f(x)) = g(cos x) = 3 (cos x)2 = 3 cos2 x Similarly,
fog(x) = f(g(x)) = f (3x2) = cos (3x2) |
1 | 644-647 | Show that gof โ fog Solution We have gof(x) = g (f(x)) = g(cos x) = 3 (cos x)2 = 3 cos2 x Similarly,
fog(x) = f(g(x)) = f (3x2) = cos (3x2) Note that 3cos2 x โ cos 3x2, for x = 0 |
1 | 645-648 | Solution We have gof(x) = g (f(x)) = g(cos x) = 3 (cos x)2 = 3 cos2 x Similarly,
fog(x) = f(g(x)) = f (3x2) = cos (3x2) Note that 3cos2 x โ cos 3x2, for x = 0 Hence,
gof โ fog |
1 | 646-649 | Similarly,
fog(x) = f(g(x)) = f (3x2) = cos (3x2) Note that 3cos2 x โ cos 3x2, for x = 0 Hence,
gof โ fog Definition 9 A function f : X โ Y is defined to be invertible, if there exists a function
g : Y โ X such that gof = IX and fog = IY |
1 | 647-650 | Note that 3cos2 x โ cos 3x2, for x = 0 Hence,
gof โ fog Definition 9 A function f : X โ Y is defined to be invertible, if there exists a function
g : Y โ X such that gof = IX and fog = IY The function g is called the inverse of f and
is denoted by f โ1 |
1 | 648-651 | Hence,
gof โ fog Definition 9 A function f : X โ Y is defined to be invertible, if there exists a function
g : Y โ X such that gof = IX and fog = IY The function g is called the inverse of f and
is denoted by f โ1 Thus, if f is invertible, then f must be one-one and onto and conversely, if f is
one-one and onto, then f must be invertible |
1 | 649-652 | Definition 9 A function f : X โ Y is defined to be invertible, if there exists a function
g : Y โ X such that gof = IX and fog = IY The function g is called the inverse of f and
is denoted by f โ1 Thus, if f is invertible, then f must be one-one and onto and conversely, if f is
one-one and onto, then f must be invertible This fact significantly helps for proving a
function f to be invertible by showing that f is one-one and onto, specially when the
actual inverse of f is not to be determined |
1 | 650-653 | The function g is called the inverse of f and
is denoted by f โ1 Thus, if f is invertible, then f must be one-one and onto and conversely, if f is
one-one and onto, then f must be invertible This fact significantly helps for proving a
function f to be invertible by showing that f is one-one and onto, specially when the
actual inverse of f is not to be determined Example 17 Let f : N โ Y be a function defined as f (x) = 4x + 3, where,
Y = {y โ N: y = 4x + 3 for some x โ N} |
1 | 651-654 | Thus, if f is invertible, then f must be one-one and onto and conversely, if f is
one-one and onto, then f must be invertible This fact significantly helps for proving a
function f to be invertible by showing that f is one-one and onto, specially when the
actual inverse of f is not to be determined Example 17 Let f : N โ Y be a function defined as f (x) = 4x + 3, where,
Y = {y โ N: y = 4x + 3 for some x โ N} Show that f is invertible |
1 | 652-655 | This fact significantly helps for proving a
function f to be invertible by showing that f is one-one and onto, specially when the
actual inverse of f is not to be determined Example 17 Let f : N โ Y be a function defined as f (x) = 4x + 3, where,
Y = {y โ N: y = 4x + 3 for some x โ N} Show that f is invertible Find the inverse |
1 | 653-656 | Example 17 Let f : N โ Y be a function defined as f (x) = 4x + 3, where,
Y = {y โ N: y = 4x + 3 for some x โ N} Show that f is invertible Find the inverse Solution Consider an arbitrary element y of Y |
1 | 654-657 | Show that f is invertible Find the inverse Solution Consider an arbitrary element y of Y By the definition of Y, y = 4x + 3,
for some x in the domain N |
1 | 655-658 | Find the inverse Solution Consider an arbitrary element y of Y By the definition of Y, y = 4x + 3,
for some x in the domain N This shows that
(
3)
4
y
x
โ
= |
1 | 656-659 | Solution Consider an arbitrary element y of Y By the definition of Y, y = 4x + 3,
for some x in the domain N This shows that
(
3)
4
y
x
โ
= Define g : Y โ N by
Rationalised 2023-24
RELATIONS AND FUNCTIONS
13
(
3)
( )
4
y
g y
โ
= |
1 | 657-660 | By the definition of Y, y = 4x + 3,
for some x in the domain N This shows that
(
3)
4
y
x
โ
= Define g : Y โ N by
Rationalised 2023-24
RELATIONS AND FUNCTIONS
13
(
3)
( )
4
y
g y
โ
= Now, gof(x) = g (f(x)) = g(4x + 3) = (4
3
3)
4
x
x
+
โ
=
and
fog(y) = f (g (y)) = f
(
3)
4 (
3)
3
4
4
y
y
โ
โ
๏ฃซ
๏ฃถ =
+
๏ฃฌ
๏ฃท
๏ฃญ
๏ฃธ
= y โ 3 + 3 = y |
1 | 658-661 | This shows that
(
3)
4
y
x
โ
= Define g : Y โ N by
Rationalised 2023-24
RELATIONS AND FUNCTIONS
13
(
3)
( )
4
y
g y
โ
= Now, gof(x) = g (f(x)) = g(4x + 3) = (4
3
3)
4
x
x
+
โ
=
and
fog(y) = f (g (y)) = f
(
3)
4 (
3)
3
4
4
y
y
โ
โ
๏ฃซ
๏ฃถ =
+
๏ฃฌ
๏ฃท
๏ฃญ
๏ฃธ
= y โ 3 + 3 = y This shows that gof = IN
and fog = IY, which implies that f is invertible and g is the inverse of f |
1 | 659-662 | Define g : Y โ N by
Rationalised 2023-24
RELATIONS AND FUNCTIONS
13
(
3)
( )
4
y
g y
โ
= Now, gof(x) = g (f(x)) = g(4x + 3) = (4
3
3)
4
x
x
+
โ
=
and
fog(y) = f (g (y)) = f
(
3)
4 (
3)
3
4
4
y
y
โ
โ
๏ฃซ
๏ฃถ =
+
๏ฃฌ
๏ฃท
๏ฃญ
๏ฃธ
= y โ 3 + 3 = y This shows that gof = IN
and fog = IY, which implies that f is invertible and g is the inverse of f Miscellaneous Examples
Example 18 If R1
and R2 are equivalence relations in a set A, show that R1 โฉ R2 is
also an equivalence relation |
1 | 660-663 | Now, gof(x) = g (f(x)) = g(4x + 3) = (4
3
3)
4
x
x
+
โ
=
and
fog(y) = f (g (y)) = f
(
3)
4 (
3)
3
4
4
y
y
โ
โ
๏ฃซ
๏ฃถ =
+
๏ฃฌ
๏ฃท
๏ฃญ
๏ฃธ
= y โ 3 + 3 = y This shows that gof = IN
and fog = IY, which implies that f is invertible and g is the inverse of f Miscellaneous Examples
Example 18 If R1
and R2 are equivalence relations in a set A, show that R1 โฉ R2 is
also an equivalence relation Solution Since R1
and R2 are equivalence relations, (a, a) โ R1, and (a, a) โ R2 โ a โ A |
1 | 661-664 | This shows that gof = IN
and fog = IY, which implies that f is invertible and g is the inverse of f Miscellaneous Examples
Example 18 If R1
and R2 are equivalence relations in a set A, show that R1 โฉ R2 is
also an equivalence relation Solution Since R1
and R2 are equivalence relations, (a, a) โ R1, and (a, a) โ R2 โ a โ A This implies that (a, a) โ R1 โฉ R2, โ a, showing R1 โฉ R2 is reflexive |
1 | 662-665 | Miscellaneous Examples
Example 18 If R1
and R2 are equivalence relations in a set A, show that R1 โฉ R2 is
also an equivalence relation Solution Since R1
and R2 are equivalence relations, (a, a) โ R1, and (a, a) โ R2 โ a โ A This implies that (a, a) โ R1 โฉ R2, โ a, showing R1 โฉ R2 is reflexive Further,
(a, b) โ R1 โฉ R2 โ (a, b) โ R1 and (a, b) โ R2 โ (b, a) โ R1 and (b, a) โ R2 โ
(b, a) โ R1 โฉ R2, hence, R1 โฉ R2 is symmetric |
1 | 663-666 | Solution Since R1
and R2 are equivalence relations, (a, a) โ R1, and (a, a) โ R2 โ a โ A This implies that (a, a) โ R1 โฉ R2, โ a, showing R1 โฉ R2 is reflexive Further,
(a, b) โ R1 โฉ R2 โ (a, b) โ R1 and (a, b) โ R2 โ (b, a) โ R1 and (b, a) โ R2 โ
(b, a) โ R1 โฉ R2, hence, R1 โฉ R2 is symmetric Similarly, (a, b) โ R1 โฉ R2 and
(b, c) โ R1 โฉ R2 โ (a, c) โ R1 and (a, c) โ R2 โ (a, c) โ R1 โฉ R2 |
1 | 664-667 | This implies that (a, a) โ R1 โฉ R2, โ a, showing R1 โฉ R2 is reflexive Further,
(a, b) โ R1 โฉ R2 โ (a, b) โ R1 and (a, b) โ R2 โ (b, a) โ R1 and (b, a) โ R2 โ
(b, a) โ R1 โฉ R2, hence, R1 โฉ R2 is symmetric Similarly, (a, b) โ R1 โฉ R2 and
(b, c) โ R1 โฉ R2 โ (a, c) โ R1 and (a, c) โ R2 โ (a, c) โ R1 โฉ R2 This shows that
R1 โฉ R2 is transitive |
1 | 665-668 | Further,
(a, b) โ R1 โฉ R2 โ (a, b) โ R1 and (a, b) โ R2 โ (b, a) โ R1 and (b, a) โ R2 โ
(b, a) โ R1 โฉ R2, hence, R1 โฉ R2 is symmetric Similarly, (a, b) โ R1 โฉ R2 and
(b, c) โ R1 โฉ R2 โ (a, c) โ R1 and (a, c) โ R2 โ (a, c) โ R1 โฉ R2 This shows that
R1 โฉ R2 is transitive Thus, R1 โฉ R2 is an equivalence relation |
1 | 666-669 | Similarly, (a, b) โ R1 โฉ R2 and
(b, c) โ R1 โฉ R2 โ (a, c) โ R1 and (a, c) โ R2 โ (a, c) โ R1 โฉ R2 This shows that
R1 โฉ R2 is transitive Thus, R1 โฉ R2 is an equivalence relation Example 19 Let R be a relation on the set A of ordered pairs of positive integers
defined by (x, y) R (u, v) if and only if xv = yu |
1 | 667-670 | This shows that
R1 โฉ R2 is transitive Thus, R1 โฉ R2 is an equivalence relation Example 19 Let R be a relation on the set A of ordered pairs of positive integers
defined by (x, y) R (u, v) if and only if xv = yu Show that R is an equivalence relation |
1 | 668-671 | Thus, R1 โฉ R2 is an equivalence relation Example 19 Let R be a relation on the set A of ordered pairs of positive integers
defined by (x, y) R (u, v) if and only if xv = yu Show that R is an equivalence relation Solution Clearly, (x, y) R (x, y), โ (x, y) โ A, since xy = yx |
1 | 669-672 | Example 19 Let R be a relation on the set A of ordered pairs of positive integers
defined by (x, y) R (u, v) if and only if xv = yu Show that R is an equivalence relation Solution Clearly, (x, y) R (x, y), โ (x, y) โ A, since xy = yx This shows that R is
reflexive |
1 | 670-673 | Show that R is an equivalence relation Solution Clearly, (x, y) R (x, y), โ (x, y) โ A, since xy = yx This shows that R is
reflexive Further, (x, y) R (u, v) โ xv = yu โ uy = vx and hence (u, v) R (x, y) |
1 | 671-674 | Solution Clearly, (x, y) R (x, y), โ (x, y) โ A, since xy = yx This shows that R is
reflexive Further, (x, y) R (u, v) โ xv = yu โ uy = vx and hence (u, v) R (x, y) This
shows that R is symmetric |
1 | 672-675 | This shows that R is
reflexive Further, (x, y) R (u, v) โ xv = yu โ uy = vx and hence (u, v) R (x, y) This
shows that R is symmetric Similarly, (x, y) R (u, v) and (u, v) R (a, b) โ xv = yu and
ub = va โ
a
a
xv
yu
u
u
=
โ
b
a
xv
yu
v
u
=
โ xb = ya and hence (x, y) R (a, b) |
1 | 673-676 | Further, (x, y) R (u, v) โ xv = yu โ uy = vx and hence (u, v) R (x, y) This
shows that R is symmetric Similarly, (x, y) R (u, v) and (u, v) R (a, b) โ xv = yu and
ub = va โ
a
a
xv
yu
u
u
=
โ
b
a
xv
yu
v
u
=
โ xb = ya and hence (x, y) R (a, b) Thus, R
is transitive |
1 | 674-677 | This
shows that R is symmetric Similarly, (x, y) R (u, v) and (u, v) R (a, b) โ xv = yu and
ub = va โ
a
a
xv
yu
u
u
=
โ
b
a
xv
yu
v
u
=
โ xb = ya and hence (x, y) R (a, b) Thus, R
is transitive Thus, R is an equivalence relation |
1 | 675-678 | Similarly, (x, y) R (u, v) and (u, v) R (a, b) โ xv = yu and
ub = va โ
a
a
xv
yu
u
u
=
โ
b
a
xv
yu
v
u
=
โ xb = ya and hence (x, y) R (a, b) Thus, R
is transitive Thus, R is an equivalence relation Example 20 Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9} |
1 | 676-679 | Thus, R
is transitive Thus, R is an equivalence relation Example 20 Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9} Let R1 be a relation in X given
by R1 = {(x, y) : x โ y is divisible by 3} and R2 be another relation on X given by
R2 = {(x, y): {x, y} โ {1, 4, 7}} or {x, y} โ {2, 5, 8} or {x, y} โ {3, 6, 9}} |
1 | 677-680 | Thus, R is an equivalence relation Example 20 Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9} Let R1 be a relation in X given
by R1 = {(x, y) : x โ y is divisible by 3} and R2 be another relation on X given by
R2 = {(x, y): {x, y} โ {1, 4, 7}} or {x, y} โ {2, 5, 8} or {x, y} โ {3, 6, 9}} Show that
R1 = R2 |
1 | 678-681 | Example 20 Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9} Let R1 be a relation in X given
by R1 = {(x, y) : x โ y is divisible by 3} and R2 be another relation on X given by
R2 = {(x, y): {x, y} โ {1, 4, 7}} or {x, y} โ {2, 5, 8} or {x, y} โ {3, 6, 9}} Show that
R1 = R2 Solution Note that the characteristic of sets {1, 4, 7}, {2, 5, 8} and {3, 6, 9} is
that difference between any two elements of these sets is a multiple of 3 |
1 | 679-682 | Let R1 be a relation in X given
by R1 = {(x, y) : x โ y is divisible by 3} and R2 be another relation on X given by
R2 = {(x, y): {x, y} โ {1, 4, 7}} or {x, y} โ {2, 5, 8} or {x, y} โ {3, 6, 9}} Show that
R1 = R2 Solution Note that the characteristic of sets {1, 4, 7}, {2, 5, 8} and {3, 6, 9} is
that difference between any two elements of these sets is a multiple of 3 Therefore,
(x, y) โ R1 โ x โ y is a multiple of 3 โ {x, y} โ {1, 4, 7} or {x, y} โ {2, 5, 8}
or {x, y} โ {3, 6, 9} โ (x, y) โ R2 |
1 | 680-683 | Show that
R1 = R2 Solution Note that the characteristic of sets {1, 4, 7}, {2, 5, 8} and {3, 6, 9} is
that difference between any two elements of these sets is a multiple of 3 Therefore,
(x, y) โ R1 โ x โ y is a multiple of 3 โ {x, y} โ {1, 4, 7} or {x, y} โ {2, 5, 8}
or {x, y} โ {3, 6, 9} โ (x, y) โ R2 Hence, R1 โ R2 |
1 | 681-684 | Solution Note that the characteristic of sets {1, 4, 7}, {2, 5, 8} and {3, 6, 9} is
that difference between any two elements of these sets is a multiple of 3 Therefore,
(x, y) โ R1 โ x โ y is a multiple of 3 โ {x, y} โ {1, 4, 7} or {x, y} โ {2, 5, 8}
or {x, y} โ {3, 6, 9} โ (x, y) โ R2 Hence, R1 โ R2 Similarly, {x, y} โ R2 โ {x, y}
Rationalised 2023-24
MATHEMATICS
14
โ {1, 4, 7} or {x, y} โ {2, 5, 8} or {x, y} โ {3, 6, 9} โ x โ y is divisible by
3 โ {x, y} โ R1 |
1 | 682-685 | Therefore,
(x, y) โ R1 โ x โ y is a multiple of 3 โ {x, y} โ {1, 4, 7} or {x, y} โ {2, 5, 8}
or {x, y} โ {3, 6, 9} โ (x, y) โ R2 Hence, R1 โ R2 Similarly, {x, y} โ R2 โ {x, y}
Rationalised 2023-24
MATHEMATICS
14
โ {1, 4, 7} or {x, y} โ {2, 5, 8} or {x, y} โ {3, 6, 9} โ x โ y is divisible by
3 โ {x, y} โ R1 This shows that R2 โ R1 |
1 | 683-686 | Hence, R1 โ R2 Similarly, {x, y} โ R2 โ {x, y}
Rationalised 2023-24
MATHEMATICS
14
โ {1, 4, 7} or {x, y} โ {2, 5, 8} or {x, y} โ {3, 6, 9} โ x โ y is divisible by
3 โ {x, y} โ R1 This shows that R2 โ R1 Hence, R1 = R2 |
1 | 684-687 | Similarly, {x, y} โ R2 โ {x, y}
Rationalised 2023-24
MATHEMATICS
14
โ {1, 4, 7} or {x, y} โ {2, 5, 8} or {x, y} โ {3, 6, 9} โ x โ y is divisible by
3 โ {x, y} โ R1 This shows that R2 โ R1 Hence, R1 = R2 Example 21 Let f : X โ Y be a function |
1 | 685-688 | This shows that R2 โ R1 Hence, R1 = R2 Example 21 Let f : X โ Y be a function Define a relation R in X given by
R = {(a, b): f(a) = f(b)} |
1 | 686-689 | Hence, R1 = R2 Example 21 Let f : X โ Y be a function Define a relation R in X given by
R = {(a, b): f(a) = f(b)} Examine whether R is an equivalence relation or not |
1 | 687-690 | Example 21 Let f : X โ Y be a function Define a relation R in X given by
R = {(a, b): f(a) = f(b)} Examine whether R is an equivalence relation or not Solution For every a โ X, (a, a) โ R, since f(a) = f (a), showing that R is reflexive |
1 | 688-691 | Define a relation R in X given by
R = {(a, b): f(a) = f(b)} Examine whether R is an equivalence relation or not Solution For every a โ X, (a, a) โ R, since f(a) = f (a), showing that R is reflexive Similarly, (a, b) โ R โ f (a) = f (b) โ f (b) = f (a) โ (b, a) โ R |
1 | 689-692 | Examine whether R is an equivalence relation or not Solution For every a โ X, (a, a) โ R, since f(a) = f (a), showing that R is reflexive Similarly, (a, b) โ R โ f (a) = f (b) โ f (b) = f (a) โ (b, a) โ R Therefore, R is
symmetric |
1 | 690-693 | Solution For every a โ X, (a, a) โ R, since f(a) = f (a), showing that R is reflexive Similarly, (a, b) โ R โ f (a) = f (b) โ f (b) = f (a) โ (b, a) โ R Therefore, R is
symmetric Further, (a, b) โ R and (b, c) โ R โ f (a) = f (b) and f(b) = f (c) โ f (a)
= f(c) โ (a, c) โ R, which implies that R is transitive |
1 | 691-694 | Similarly, (a, b) โ R โ f (a) = f (b) โ f (b) = f (a) โ (b, a) โ R Therefore, R is
symmetric Further, (a, b) โ R and (b, c) โ R โ f (a) = f (b) and f(b) = f (c) โ f (a)
= f(c) โ (a, c) โ R, which implies that R is transitive Hence, R is an equivalence
relation |
1 | 692-695 | Therefore, R is
symmetric Further, (a, b) โ R and (b, c) โ R โ f (a) = f (b) and f(b) = f (c) โ f (a)
= f(c) โ (a, c) โ R, which implies that R is transitive Hence, R is an equivalence
relation Example 22 Find the number of all one-one functions from set A = {1, 2, 3} to itself |
1 | 693-696 | Further, (a, b) โ R and (b, c) โ R โ f (a) = f (b) and f(b) = f (c) โ f (a)
= f(c) โ (a, c) โ R, which implies that R is transitive Hence, R is an equivalence
relation Example 22 Find the number of all one-one functions from set A = {1, 2, 3} to itself Solution One-one function from {1, 2, 3} to itself is simply a permutation on three
symbols 1, 2, 3 |
1 | 694-697 | Hence, R is an equivalence
relation Example 22 Find the number of all one-one functions from set A = {1, 2, 3} to itself Solution One-one function from {1, 2, 3} to itself is simply a permutation on three
symbols 1, 2, 3 Therefore, total number of one-one maps from {1, 2, 3} to itself is
same as total number of permutations on three symbols 1, 2, 3 which is 3 |
1 | 695-698 | Example 22 Find the number of all one-one functions from set A = {1, 2, 3} to itself Solution One-one function from {1, 2, 3} to itself is simply a permutation on three
symbols 1, 2, 3 Therefore, total number of one-one maps from {1, 2, 3} to itself is
same as total number of permutations on three symbols 1, 2, 3 which is 3 = 6 |
1 | 696-699 | Solution One-one function from {1, 2, 3} to itself is simply a permutation on three
symbols 1, 2, 3 Therefore, total number of one-one maps from {1, 2, 3} to itself is
same as total number of permutations on three symbols 1, 2, 3 which is 3 = 6 Example 23 Let A = {1, 2, 3} |
1 | 697-700 | Therefore, total number of one-one maps from {1, 2, 3} to itself is
same as total number of permutations on three symbols 1, 2, 3 which is 3 = 6 Example 23 Let A = {1, 2, 3} Then show that the number of relations containing (1, 2)
and (2, 3) which are reflexive and transitive but not symmetric is three |
1 | 698-701 | = 6 Example 23 Let A = {1, 2, 3} Then show that the number of relations containing (1, 2)
and (2, 3) which are reflexive and transitive but not symmetric is three Solution The smallest relation R1 containing (1, 2) and (2, 3) which is reflexive and
transitive but not symmetric is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} |
1 | 699-702 | Example 23 Let A = {1, 2, 3} Then show that the number of relations containing (1, 2)
and (2, 3) which are reflexive and transitive but not symmetric is three Solution The smallest relation R1 containing (1, 2) and (2, 3) which is reflexive and
transitive but not symmetric is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} Now, if we add
the pair (2, 1) to R1 to get R2, then the relation R2 will be reflexive, transitive but not
symmetric |
1 | 700-703 | Then show that the number of relations containing (1, 2)
and (2, 3) which are reflexive and transitive but not symmetric is three Solution The smallest relation R1 containing (1, 2) and (2, 3) which is reflexive and
transitive but not symmetric is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} Now, if we add
the pair (2, 1) to R1 to get R2, then the relation R2 will be reflexive, transitive but not
symmetric Similarly, we can obtain R3 by adding (3, 2) to R1 to get the desired relation |
1 | 701-704 | Solution The smallest relation R1 containing (1, 2) and (2, 3) which is reflexive and
transitive but not symmetric is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} Now, if we add
the pair (2, 1) to R1 to get R2, then the relation R2 will be reflexive, transitive but not
symmetric Similarly, we can obtain R3 by adding (3, 2) to R1 to get the desired relation However, we can not add two pairs (2, 1), (3, 2) or single pair (3, 1) to R1 at a time, as
by doing so, we will be forced to add the remaining pair in order to maintain transitivity
and in the process, the relation will become symmetric also which is not required |
1 | 702-705 | Now, if we add
the pair (2, 1) to R1 to get R2, then the relation R2 will be reflexive, transitive but not
symmetric Similarly, we can obtain R3 by adding (3, 2) to R1 to get the desired relation However, we can not add two pairs (2, 1), (3, 2) or single pair (3, 1) to R1 at a time, as
by doing so, we will be forced to add the remaining pair in order to maintain transitivity
and in the process, the relation will become symmetric also which is not required Thus,
the total number of desired relations is three |
1 | 703-706 | Similarly, we can obtain R3 by adding (3, 2) to R1 to get the desired relation However, we can not add two pairs (2, 1), (3, 2) or single pair (3, 1) to R1 at a time, as
by doing so, we will be forced to add the remaining pair in order to maintain transitivity
and in the process, the relation will become symmetric also which is not required Thus,
the total number of desired relations is three Example 24 Show that the number of equivalence relation in the set {1, 2, 3} containing
(1, 2) and (2, 1) is two |
1 | 704-707 | However, we can not add two pairs (2, 1), (3, 2) or single pair (3, 1) to R1 at a time, as
by doing so, we will be forced to add the remaining pair in order to maintain transitivity
and in the process, the relation will become symmetric also which is not required Thus,
the total number of desired relations is three Example 24 Show that the number of equivalence relation in the set {1, 2, 3} containing
(1, 2) and (2, 1) is two Solution The smallest equivalence relation R1 containing (1, 2) and (2, 1) is {(1, 1),
(2, 2), (3, 3), (1, 2), (2, 1)} |
1 | 705-708 | Thus,
the total number of desired relations is three Example 24 Show that the number of equivalence relation in the set {1, 2, 3} containing
(1, 2) and (2, 1) is two Solution The smallest equivalence relation R1 containing (1, 2) and (2, 1) is {(1, 1),
(2, 2), (3, 3), (1, 2), (2, 1)} Now we are left with only 4 pairs namely (2, 3), (3, 2),
(1, 3) and (3, 1) |
1 | 706-709 | Example 24 Show that the number of equivalence relation in the set {1, 2, 3} containing
(1, 2) and (2, 1) is two Solution The smallest equivalence relation R1 containing (1, 2) and (2, 1) is {(1, 1),
(2, 2), (3, 3), (1, 2), (2, 1)} Now we are left with only 4 pairs namely (2, 3), (3, 2),
(1, 3) and (3, 1) If we add any one, say (2, 3) to R1, then for symmetry we must add
(3, 2) also and now for transitivity we are forced to add (1, 3) and (3, 1) |
1 | 707-710 | Solution The smallest equivalence relation R1 containing (1, 2) and (2, 1) is {(1, 1),
(2, 2), (3, 3), (1, 2), (2, 1)} Now we are left with only 4 pairs namely (2, 3), (3, 2),
(1, 3) and (3, 1) If we add any one, say (2, 3) to R1, then for symmetry we must add
(3, 2) also and now for transitivity we are forced to add (1, 3) and (3, 1) Thus, the only
equivalence relation bigger than R1 is the universal relation |
1 | 708-711 | Now we are left with only 4 pairs namely (2, 3), (3, 2),
(1, 3) and (3, 1) If we add any one, say (2, 3) to R1, then for symmetry we must add
(3, 2) also and now for transitivity we are forced to add (1, 3) and (3, 1) Thus, the only
equivalence relation bigger than R1 is the universal relation This shows that the total
number of equivalence relations containing (1, 2) and (2, 1) is two |
1 | 709-712 | If we add any one, say (2, 3) to R1, then for symmetry we must add
(3, 2) also and now for transitivity we are forced to add (1, 3) and (3, 1) Thus, the only
equivalence relation bigger than R1 is the universal relation This shows that the total
number of equivalence relations containing (1, 2) and (2, 1) is two Example 25 Consider the identity function IN : N โ N defined as IN (x) = x โ x โ N |
1 | 710-713 | Thus, the only
equivalence relation bigger than R1 is the universal relation This shows that the total
number of equivalence relations containing (1, 2) and (2, 1) is two Example 25 Consider the identity function IN : N โ N defined as IN (x) = x โ x โ N Show that although IN is onto but IN + IN : N โ N defined as
(IN + IN) (x) = IN (x) + IN (x) = x + x = 2x is not onto |
1 | 711-714 | This shows that the total
number of equivalence relations containing (1, 2) and (2, 1) is two Example 25 Consider the identity function IN : N โ N defined as IN (x) = x โ x โ N Show that although IN is onto but IN + IN : N โ N defined as
(IN + IN) (x) = IN (x) + IN (x) = x + x = 2x is not onto Rationalised 2023-24
RELATIONS AND FUNCTIONS
15
Solution Clearly IN is onto |
1 | 712-715 | Example 25 Consider the identity function IN : N โ N defined as IN (x) = x โ x โ N Show that although IN is onto but IN + IN : N โ N defined as
(IN + IN) (x) = IN (x) + IN (x) = x + x = 2x is not onto Rationalised 2023-24
RELATIONS AND FUNCTIONS
15
Solution Clearly IN is onto But IN + IN is not onto, as we can find an element 3
in the co-domain N such that there does not exist any x in the domain N with
(IN + IN) (x) = 2x = 3 |
1 | 713-716 | Show that although IN is onto but IN + IN : N โ N defined as
(IN + IN) (x) = IN (x) + IN (x) = x + x = 2x is not onto Rationalised 2023-24
RELATIONS AND FUNCTIONS
15
Solution Clearly IN is onto But IN + IN is not onto, as we can find an element 3
in the co-domain N such that there does not exist any x in the domain N with
(IN + IN) (x) = 2x = 3 Example 26 Consider a function f : 0, 2
ฯ
๏ฃฎ
๏ฃนโ
๏ฃฏ
๏ฃบ
๏ฃฐ
๏ฃป
R given by f (x) = sin x and
g : 0, 2
ฯ
๏ฃฎ
๏ฃนโ
๏ฃฏ
๏ฃบ
๏ฃฐ
๏ฃป
R given by g(x) = cos x |
1 | 714-717 | Rationalised 2023-24
RELATIONS AND FUNCTIONS
15
Solution Clearly IN is onto But IN + IN is not onto, as we can find an element 3
in the co-domain N such that there does not exist any x in the domain N with
(IN + IN) (x) = 2x = 3 Example 26 Consider a function f : 0, 2
ฯ
๏ฃฎ
๏ฃนโ
๏ฃฏ
๏ฃบ
๏ฃฐ
๏ฃป
R given by f (x) = sin x and
g : 0, 2
ฯ
๏ฃฎ
๏ฃนโ
๏ฃฏ
๏ฃบ
๏ฃฐ
๏ฃป
R given by g(x) = cos x Show that f and g are one-one, but f + g is not
one-one |
1 | 715-718 | But IN + IN is not onto, as we can find an element 3
in the co-domain N such that there does not exist any x in the domain N with
(IN + IN) (x) = 2x = 3 Example 26 Consider a function f : 0, 2
ฯ
๏ฃฎ
๏ฃนโ
๏ฃฏ
๏ฃบ
๏ฃฐ
๏ฃป
R given by f (x) = sin x and
g : 0, 2
ฯ
๏ฃฎ
๏ฃนโ
๏ฃฏ
๏ฃบ
๏ฃฐ
๏ฃป
R given by g(x) = cos x Show that f and g are one-one, but f + g is not
one-one Solution Since for any two distinct elements x1 and x2 in 0, 2
ฯ
๏ฃฎ
๏ฃน
๏ฃฏ
๏ฃบ
๏ฃฐ
๏ฃป
, sin x1 โ sin x2 and
cos x1 โ cos x2, both f and g must be one-one |
1 | 716-719 | Example 26 Consider a function f : 0, 2
ฯ
๏ฃฎ
๏ฃนโ
๏ฃฏ
๏ฃบ
๏ฃฐ
๏ฃป
R given by f (x) = sin x and
g : 0, 2
ฯ
๏ฃฎ
๏ฃนโ
๏ฃฏ
๏ฃบ
๏ฃฐ
๏ฃป
R given by g(x) = cos x Show that f and g are one-one, but f + g is not
one-one Solution Since for any two distinct elements x1 and x2 in 0, 2
ฯ
๏ฃฎ
๏ฃน
๏ฃฏ
๏ฃบ
๏ฃฐ
๏ฃป
, sin x1 โ sin x2 and
cos x1 โ cos x2, both f and g must be one-one But (f + g) (0) = sin 0 + cos 0 = 1 and
(f + g)
๏ฃซฯ2
๏ฃถ
๏ฃฌ
๏ฃท
๏ฃญ
๏ฃธ
= sin
cos
1
2
2
ฯ
ฯ
+
= |
1 | 717-720 | Show that f and g are one-one, but f + g is not
one-one Solution Since for any two distinct elements x1 and x2 in 0, 2
ฯ
๏ฃฎ
๏ฃน
๏ฃฏ
๏ฃบ
๏ฃฐ
๏ฃป
, sin x1 โ sin x2 and
cos x1 โ cos x2, both f and g must be one-one But (f + g) (0) = sin 0 + cos 0 = 1 and
(f + g)
๏ฃซฯ2
๏ฃถ
๏ฃฌ
๏ฃท
๏ฃญ
๏ฃธ
= sin
cos
1
2
2
ฯ
ฯ
+
= Therefore, f + g is not one-one |
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