Page 13 - Relations and Functions 19.10.06.pmd

RELATIONS AND FUNCTIONS 13

⎧⎫ ⎧ 3⎫ 3x + 4
7
:
Example 17 Show that if f R − → ⎨⎬ R − ⎨ ⎬ is defined by ()fx = and
5
⎩⎭ ⎩ 5⎭ 5x − 7
⎧⎫ ⎧ 7⎫ 7x + 4
3
: g R − → ⎨⎬ R − ⎨ ⎬ is defined by ()gx = , then fog = I and gof = I , where,
5
⎩⎭ ⎩ 5⎭ 5x − 3 A B
⎧⎫ ⎧⎫
3
7
A = R – ⎨⎬ , B = R – ⎨⎬ ; I (x) = x, ∀ x ∈ A, I (x) = x, ∀ x ∈ B are called identity
⎩⎭ ⎩⎭ A B
5
5
functions on sets A and B, respectively.
Solution We have
⎛ (3x + 4) ⎞
7 ⎜ ⎟ + 4
+
⎛ 3x + ⎞ 4 ⎝ (5x − 7) ⎠ 21x + 28 20x − 28 41x
gof () x = g ⎜ ⎟ = = = = x
⎝ 5x − ⎠ 7 ⎛ (3x + 4) ⎞ 15x + 20 15x− + 21 41
5 ⎜ ⎟ − 3
⎝ (5x − 7) ⎠
⎛ (7x + 4) ⎞
3 ⎜ ⎟ + 4
+
⎛ 7x + ⎞ 4 ⎝ (5x − 3) ⎠ 21x + 12 20x − 12 41x
Similarly, fog x = () f ⎜ ⎟ = = = = x
−
⎝ 5x − ⎠ 3 ⎛ (7x + 4) ⎞ 35x + 20 35x + 21 41
5 ⎜ ⎟ − 7
⎝ (5x − 3) ⎠
Thus, gof (x) = x, ∀ x ∈ B and fog(x) = x, ∀ x ∈ A, which implies that gof = I B
and fog = I .
A
Example 18 Show that if f : A → B and g : B → C are one-one, then gof : A → C is
also one-one.
Solution Suppose gof (x ) = gof (x )
1 2
⇒ g (f(x )) = g(f (x ))
1 2
⇒ f (x ) = f (x ), as g is one-one
1 2
⇒ x = x , as f is one-one
1 2
Hence, gof is one-one.
Example 19 Show that if f : A → B and g : B → C are onto, then gof : A → C is
also onto.
Solution Given an arbitrary element z ∈ C, there exists a pre-image y of z under g
such that g (y) = z, since g is onto. Further, for y ∈ B, there exists an element x in A