10th Class Math Notes 2tg2
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Taleem City
1
Exercise 5.5 For more educational resources visit
www.taleemcity.com
=* = *( = *( = *( (ii) =* = *( = *( = *(
+ * + )( )( )( )( )+ ) ( )+
Ta le
Q. 3: (i)
em
ci
Q. 2:
* +, * +, then find two binary relations of If and . + * + =* )( ) ( ) ( ) ( ) ( )+ = *( + =* + * )( ) ( )( ) ( ) ( )+ = *( ) ( ) ( )+ = *( ) ( ) ( )+ = *( )( )+ = *( = *( ) ( ) ( ) ( )+ * +, then make two binary relations If . Also find their domain and range. + * + =* )( )( )( )( )( )( ) ( ) ( )+ = *( )( ) ( ) ( )+ = *( + =* + =* )( )( )+ = *( + =* =* + * + and * +, then find two binary relations in each: If
ty
Q. 1:
+ * )( )( )( )+ )( )(
+ )(
)(
)(
)(
)(
)(
)(
)(
)(
)(
)(
)+
)(
)(
)(
)+
)+
(iii)
Q. 4:
+ * + =* = *( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )+ = *( )( )+ = *( ) ( If set has 5 elements, then find the number of binary relations in M. No. of Elements in = =5 No. of binary relations in = www.taleemcity.com
)(
)(
)(
)(
)+
Taleem City
2
= =
Ta le
em
ci
ty
* +, * +, then make the following relations from L to M. Q. 5: If Also write the domain and range of each relation. So, we have from the question * + * + + * + =* = *( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )+ ) + (i) = *( = *( ) ( ) ( ) ( ) ( )+ + =* =* + ) + (ii) = *( = *( ) ( ) ( )+ + =* + =* ) + (iii) = *( = *( ) ( ) ( )+ + =* + =* ) + (iv) = *( = *( ) ( ) ( )+ + =* =* + Q. 6: Indicate relations, into function, one-one function, onto function, and bijective function from the following. Also find their domain and range. (i) = *( ) ( ) ( ) ( )+ + =* + =* As, we know A relation becomes a function if and Every appears in one and only one ordered pair in . So, the given relation is function. As, All distinct elements of A have distinct images in B so, the given relation is One-One. Also, every element of set B is an image of at least one element of set A i.e. relation is also Onto function. As, the given relation is One-One as well as Onto function so, it is bijective function. (ii) = *( ) ( ) ( ) ( )+ www.taleemcity.com
. So, given
Taleem City
3
+ =* + =* As, we know A relation becomes a function if
ty
and Every appears in one and only one ordered pair in . As, we can clearly see the 3 is repeated in 3rd and 4th ordered pair so the given relation is not a function, its only a relation. )( )( )+ (iii) = *( + =* =* + As, we know A relation becomes a function if
em
ci
and Every appears in one and only one ordered pair in . So, the given relation is a function. As, it doesn’t fulfill any condition of One-One, Onto or into function so the relation is only a function. (iv) = *( ) ( ) ( ) ( ) ( )+ + =* + =* As, we know A relation becomes a function if
Ta le
and Every appears in one and only one ordered pair in . So, the given relation is a function. As, It doesn’t fulfill condition of One-One. Every element of set B is an image of at least one element of set A. So, the given relation is an onto function. )( )( )( )+ (v) = *( + =* + =* As, we know A relation becomes a function if and Every appears in one and only one ordered pair in . So, the given relation is a function. As, As, All distinct elements of A have distinct images in B so, the given relation is One-One. It doesn’t fulfill condition of Onto function. So, the given relation is a One-One function. (vi) = *( ) ( ) ( ) ( )+ www.taleemcity.com
4
Taleem City
+ =* + =* As, we know A relation becomes a function if
ty
and Every appears in one and only one ordered pair in . As, we can clearly see the 1 is repeated in 1st and 3rd ordered pair so the given relation is not a function, it’s only a relation. (vii)
ci
= *( ) ( ) ( )+ + =* + =* As, we know A relation becomes a function if
Ta le
em
and Every appears in one and only one ordered pair in . As, All distinct elements of A have distinct images in B so, the given relation is One-One. It doesn’t fulfill condition of Onto function. So, the given relation is a One-One function. (viii)
) ( )+ = *( ) ( + =* + =* As, we know A relation becomes a function if and Every But
appears in one and only one ordered pair in . , So the given relation is not a function. it’s only a relation.
www.taleemcity.com
1
Exercise 5.5 For more educational resources visit
www.taleemcity.com
=* = *( = *( = *( (ii) =* = *( = *( = *(
+ * + )( )( )( )( )+ ) ( )+
Ta le
Q. 3: (i)
em
ci
Q. 2:
* +, * +, then find two binary relations of If and . + * + =* )( ) ( ) ( ) ( ) ( )+ = *( + =* + * )( ) ( )( ) ( ) ( )+ = *( ) ( ) ( )+ = *( ) ( ) ( )+ = *( )( )+ = *( = *( ) ( ) ( ) ( )+ * +, then make two binary relations If . Also find their domain and range. + * + =* )( )( )( )( )( )( ) ( ) ( )+ = *( )( ) ( ) ( )+ = *( + =* + =* )( )( )+ = *( + =* =* + * + and * +, then find two binary relations in each: If
ty
Q. 1:
+ * )( )( )( )+ )( )(
+ )(
)(
)(
)(
)(
)(
)(
)(
)(
)(
)(
)+
)(
)(
)(
)+
)+
(iii)
Q. 4:
+ * + =* = *( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )+ = *( )( )+ = *( ) ( If set has 5 elements, then find the number of binary relations in M. No. of Elements in = =5 No. of binary relations in = www.taleemcity.com
)(
)(
)(
)(
)+
Taleem City
2
= =
Ta le
em
ci
ty
* +, * +, then make the following relations from L to M. Q. 5: If Also write the domain and range of each relation. So, we have from the question * + * + + * + =* = *( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )+ ) + (i) = *( = *( ) ( ) ( ) ( ) ( )+ + =* =* + ) + (ii) = *( = *( ) ( ) ( )+ + =* + =* ) + (iii) = *( = *( ) ( ) ( )+ + =* + =* ) + (iv) = *( = *( ) ( ) ( )+ + =* =* + Q. 6: Indicate relations, into function, one-one function, onto function, and bijective function from the following. Also find their domain and range. (i) = *( ) ( ) ( ) ( )+ + =* + =* As, we know A relation becomes a function if and Every appears in one and only one ordered pair in . So, the given relation is function. As, All distinct elements of A have distinct images in B so, the given relation is One-One. Also, every element of set B is an image of at least one element of set A i.e. relation is also Onto function. As, the given relation is One-One as well as Onto function so, it is bijective function. (ii) = *( ) ( ) ( ) ( )+ www.taleemcity.com
. So, given
Taleem City
3
+ =* + =* As, we know A relation becomes a function if
ty
and Every appears in one and only one ordered pair in . As, we can clearly see the 3 is repeated in 3rd and 4th ordered pair so the given relation is not a function, its only a relation. )( )( )+ (iii) = *( + =* =* + As, we know A relation becomes a function if
em
ci
and Every appears in one and only one ordered pair in . So, the given relation is a function. As, it doesn’t fulfill any condition of One-One, Onto or into function so the relation is only a function. (iv) = *( ) ( ) ( ) ( ) ( )+ + =* + =* As, we know A relation becomes a function if
Ta le
and Every appears in one and only one ordered pair in . So, the given relation is a function. As, It doesn’t fulfill condition of One-One. Every element of set B is an image of at least one element of set A. So, the given relation is an onto function. )( )( )( )+ (v) = *( + =* + =* As, we know A relation becomes a function if and Every appears in one and only one ordered pair in . So, the given relation is a function. As, As, All distinct elements of A have distinct images in B so, the given relation is One-One. It doesn’t fulfill condition of Onto function. So, the given relation is a One-One function. (vi) = *( ) ( ) ( ) ( )+ www.taleemcity.com
4
Taleem City
+ =* + =* As, we know A relation becomes a function if
ty
and Every appears in one and only one ordered pair in . As, we can clearly see the 1 is repeated in 1st and 3rd ordered pair so the given relation is not a function, it’s only a relation. (vii)
ci
= *( ) ( ) ( )+ + =* + =* As, we know A relation becomes a function if
Ta le
em
and Every appears in one and only one ordered pair in . As, All distinct elements of A have distinct images in B so, the given relation is One-One. It doesn’t fulfill condition of Onto function. So, the given relation is a One-One function. (viii)
) ( )+ = *( ) ( + =* + =* As, we know A relation becomes a function if and Every But
appears in one and only one ordered pair in . , So the given relation is not a function. it’s only a relation.
www.taleemcity.com
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