The set of all images of the elements of X under f is called the ‘range’ of f.
The range of a function is a subset of its co-domain.
Question 1 :
Let A, B, C ⊆ N and a function f : A -> B be defined by f(x) = 2x + 1 and g : B -> C be defined by g(x) = x2 . Find the range of f o g and g o f .
Solution :
f o g = f[g(x)]
= f[x2]
Now we apply x2 instead of x in f(x).
f o g = 2 x2 + 1
y = 2x2 + 1
Range :
{y | y = 2x2 + 1 and x ∊ N}
g o f = g[f(x)]
= g[2x + 1]
Now we apply 2x + 1 instead of x in g(x).
g o f = (2x + 1)2
Range :
{y | y = (2x + 1)2 and x ∊ N}
Question 2 :
Let f (x) = x2 −1 . Find (i) f o f (ii) f o f o f
Solution :
(i) f o f
= f[f(x)]
= f[x2 −1]
Now we apply x2 −1 instead of x in f(x).
= (x2 −1)2 - 1
= x4 - 2x2 + 1 - 1
f o f = x4 - 2x2
(ii) f o f o f
f o f = x4 - 2x2
f o f o f = f [f o f]
= f[x4 - 2x2]
Now we apply x4 - 2x2 instead of x in f(x).
= (x4 - 2x2)2 - 1
Question 3 :
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?
Solution :
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.
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