Determine the relation R on the set of whole numbers ≤ 10 defined by R = { ( x , y ) ∣ 2 x + 3 y = 12 } .

Given, R = ( x , y ) ; 2 x + 3 y = 12 , { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 910 } x = 0 , 2 × 0 + 3 y = 12 ⇒ y = 4 ⇒ ( 0 , 4 ) x = 1 , 2 × 1 + 3 y = 12 ⇒ y = 10 3 x = 2 , 2 × 2 + 37 = 12 ⇒ y = 8 3 x = 3 , 2 × 3 + 3 y = 12 ⇒ y = 2 ⇒ ( 3 , 2 ) x = 4 , 2 × 4 + 3 y = 12 ⇒ y = 4 3 x = 5 , 2 × 5 + 3 y = 12 ⇒ y = 2 3 x = 6 , 2 × 6 + 3 y = 12 ⇒ y = 0 ⇒ ( 6 , 0 ) ∴ R = { ( 0 , 4 ) , ( 3 , 2 ) , ( 6 , 0 ) }

Determine the relation R on the set of whole numbers ≤ 10 defined by R = { ( x , y ) ∣ 2 x + 3 y = 12 } .

Given, R = ( x , y ) ; 2 x + 3 y = 12 , { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 910 } x = 0 , 2 × 0 + 3 y = 12 ⇒ y = 4 ⇒ ( 0 , 4 ) x = 1 , 2 × 1 + 3 y = 12 ⇒ y = 10 3 x = 2 , 2 × 2 + 37 = 12 ⇒ y = 8 3 x = 3 , 2 × 3 + 3 y = 12 ⇒ y = 2 ⇒ ( 3 , 2 ) x = 4 , 2 × 4 + 3 y = 12 ⇒ y = 4 3 x = 5 , 2 × 5 + 3 y = 12 ⇒ y = 2 3 x = 6 , 2 × 6 + 3 y = 12 ⇒ y = 0 ⇒ ( 6 , 0 ) ∴ R = { ( 0 , 4 ) , ( 3 , 2 ) , ( 6 , 0 ) }

Let f : ℝ - { 3 } → ℝ - { 1 } is defined by f ( x ) = x - 2 x - 3 . Show that $f$ is a bijective function.

Let A = R - { 3 } and B = R - { 1 } f : A → B Given, f ( x ) = x - 2 x - 3 f x 1 = f x 2 x 1 - 2 x 1 - 3 = x 2 - 2 x 2 - 3 ⇒ x 2 - 3 x 1 - 2 = x 2 - 2 x 1 - 3 ⇒ x 1 x 2 - 3 x 1 - 2 x 2 + 6 = x 1 x 2 - 3 x 2 - 2 x 1 + 6 ⇒ - 3 x 1 - 2 x 2 = - 3 x 2 - 2 x 1 ⇒ - x 1 = - x 2 ⇒ x 1 = x 2 So, f ( x ) is one-one f ( x ) = x - 2 x - 3 ⇒ y = x - 2 x - 3 ⇒ y ( x - 3 ) = x - 2 ⇒ y x - 3 y = x - 2 ⇒ y x - x = 3 y - 2 ⇒ x ( y - 1 ) = 3 y - 2 ⇒ x = 3 y - 2 ( y - 1 ) f ( x ) = x - 2 x - 3 = 3 y - 2 y - 1 - 2 3 y - 2 y - 1 - 3 = 3 y - 2 - 2 ( y - 1 ) y - 1 3 y - 2 - 3 ( y - 1 ) y - 1 = 3 y - 2 - 2 y + 2 3 y - 2 - 3 y + 3 = 3 y - 2 y - 2 + 3 = y f ( x ) = y f ( x ) is onto Hence, f ( x ) is bijective

If f : ℝ → ℝ is defined by f ( x ) = x 2 - 3 x + 2 , find f ( f ( x ) ) .

Given f : R → R f ( x ) = x 2 - 3 x + 2 f f ( x ) = f x 2 - 3 x + 2 = x 2 - 3 x + 2 2 - 3 x 2 - 3 x + 2 + 2 = x 2 - 3 x 2 + 2 x 2 x 2 - 34 + 4 - 3 x 2 + 9 x - 6 + 2 = x 4 - 6 x 3 + 9 x 2 + 4 x 2 - 12 x - 3 x 2 + 9 x = x 4 - 6 x 3 + 10 x 2 - 3 x f f ( x ) = x 4 - 6 x 3 + 10 x 2 - 3 x

Let f : ℝ - { 3 } → ℝ - { 1 } is defined by f ( x ) = x - 2 x - 3 . Show that $f$ is a bijective function.

Let A = R - { 3 } and B = R - { 1 } f : A → B Given, f ( x ) = x - 2 x - 3 f x 1 = f x 2 x 1 - 2 x 1 - 3 = x 2 - 2 x 2 - 3 ⇒ x 2 - 3 x 1 - 2 = x 2 - 2 x 1 - 3 ⇒ x 1 x 2 - 3 x 1 - 2 x 2 + 6 = x 1 x 2 - 3 x 2 - 2 x 1 + 6 ⇒ - 3 x 1 - 2 x 2 = - 3 x 2 - 2 x 1 ⇒ - x 1 = - x 2 ⇒ x 1 = x 2 So, f ( x ) is one-one f ( x ) = x - 2 x - 3 ⇒ y = x - 2 x - 3 ⇒ y ( x - 3 ) = x - 2 ⇒ y x - 3 y = x - 2 ⇒ y x - x = 3 y - 2 ⇒ x ( y - 1 ) = 3 y - 2 ⇒ x = 3 y - 2 ( y - 1 ) f ( x ) = x - 2 x - 3 = 3 y - 2 y - 1 - 2 3 y - 2 y - 1 - 3 = 3 y - 2 - 2 ( y - 1 ) y - 1 3 y - 2 - 3 ( y - 1 ) y - 1 = 3 y - 2 - 2 y + 2 3 y - 2 - 3 y + 3 = 3 y - 2 y - 2 + 3 = y f ( x ) = y f ( x ) is onto Hence, f ( x ) is bijective

Let f : R → R is defined by f ( x ) = 3 x - 2 and g : R → R is defined by g ( x ) = x + 2 3 . Show that f · g = g · f .

Given that, f ( x ) = 3 x - 2 & g ( x ) = x + 2 3 f · g = f ( x ) · g ( x ) = 3 x - 2 x + 2 3 = x 2 + 4 x 3 - 4 3 g · f = g ( x ) · f ( x ) = x + 2 3 3 x - 2 = x 2 + 4 x 3 - 4 3 Hence, proved.

Show that f : [ - 1 , 1 ] → R , given by f ( x ) = x ( x + 2 ) is one-one. Find the inverse of the function f : [ - 1 , 1 ] → Range f .

f : [ 1 , 1 ] → R is given as f ( x ) = x x + 2 . let f ( x ) = f ( y ) . ⇒ x x + 2 = y y + 2 ⇒ x y + 2 x = x y + 2 y ⇒ 2 x = 2 y ⇒ x = y ∴ f is a one-one function. It is clear that f : [ - 1 , 1 ] → range f is onto. ∴ f : [ - 1 , 1 ] → Range f is one-one and onto and therefore, the inverse of the function: f : [ - 1 , 1 ] → Range f exists. Let g : Range f → [ - 1 , 1 ] → be the inverse of f . Let y be an arbitrary element of range. f . Since f : [ - 1 , 1 ] → Range f is onto, we have: y = f ( x ) for same x ∈ [ - 1 , 1 ] ⇒ y = x x + 2 ⇒ x y + 2 y = x ⇒ x ( 1 - y ) = 2 y ⇒ x = 2 y 1 - y , y ≠ 1 Now, let us define g : Range f → [ - 1 , 1 ] as g ( y ) = 2 y 1 - y , y ≠ 1 . Now, ( g o f ) ( x ) = g ( f ( x ) ) = g ( x x + 2 ) = 2 ( x x + 2 ) 1 - x x + 2 = 2 x x + 2 - x = 2 x 2 = x ( f o g ) ( y ) = f ( g ( y ) ) = f ( 2 y 1 - y ) = 2 y 1 - y 2 y 1 - y + 2 = 2 y 2 y + 2 - 2 y = 2 y 2 = y ∴ g o f = I - 1 , 1 and f o g = I Range f ∴ f - 1 = g ⇒ f - 1 ( y ) = 2 y 1 - y , y ≠ 1

Let L be the set of all lines in X Y plane and R be the relation in L defined as R = { ( L 1 , L 2 ) : L 1 is parallel to L 2 } . Show that R is an equivalence relation. Find the set of all lines related to the line y = 2 x + 4 .

R = ( L 1 , L 2 ) : L 1 is parallel to L 2 R is reflexive as any line L 1 is parallel to itself i.e., ( L 1 , L 1 ) ∈ R . Now , Let ( L 1 , L 2 ) ∈ R . ⇒ L 1 is parallel to L 2 . ⇒ L 2 is parallel to L 1 . ( L 2 , L 1 ) ∈ R ∴ R is symmetric. Now, Let ( L 1 , L 2 ) , ( L 2 , L 3 ) ∈ R . ⇒ L 1 is parallel to L 2 . Also, L 2 is parallel to L 3 . ⇒ L 1 is parallel to L 3 ∴ R is transitive. Hence, R is an equivalence relation. The set of all lines related to the line y = 2 x + 4 is the set of all lines that are parallel to the line y = 2 x + 4 . Slope of line y = 2 x + 4 is m = 2 It is known that parallel lines have the same slopes. The line parallel to the given line is of the form y = 2 x + c , where c ∈ R . Hence, the set of all lines related to the given line is given by y = 2 x + c , where c ∈ R .

If f : ℝ → ℝ is defined by f ( x ) = x 2 - 3 x + 2 , find f ( f ( x ) ) .

Given f : R → R f ( x ) = x 2 - 3 x + 2 f f ( x ) = f x 2 - 3 x + 2 = x 2 - 3 x + 2 2 - 3 x 2 - 3 x + 2 + 2 = x 2 - 3 x 2 + 2 x 2 x 2 - 34 + 4 - 3 x 2 + 9 x - 6 + 2 = x 4 - 6 x 3 + 9 x 2 + 4 x 2 - 12 x - 3 x 2 + 9 x = x 4 - 6 x 3 + 10 x 2 - 3 x f f ( x ) = x 4 - 6 x 3 + 10 x 2 - 3 x