Consider f : {1, 2, 3} → {a, b, c} given by f (1) = a, f (2) = b, f (3) = c. Find (f - 1)- 1 = f
Solution:
A function is a process or a relation that associates each element 'a' of a non-empty set A, to a single element 'b' of another non-empty set B.
According to the given problem,
Function f : {1, 2, 3} → {a, b, c} given by
f (1) = a, f (2) = b, f (3) = c.
If we define g : {a, b, c} → {1, 2, 3} as
g (a) = 1, g (b) = 2, g (c) = 3
(fog )(a) = f ( g (a )) = f (1) = a
(fog )(b) = f ( g (b)) = f (2) = b
(fog )(c) = f ( g (c)) = f (3) = c
And,
(gof )(1) = g ( f (1)) = g (a) = 1
(gof )(2) = g ( f (2)) = g (b) = 2
(gof )(3) = g ( f (3)) = g (c) = 3
Therefore,
gof = IX and fog = IY = {(1, 2, 3)} and Y = {a, b, c}
Thus, the inverse of f exists and f - 1 = g.
⇒ f - 1 : {a, b, c} → {1, 2, 3} is given by, f - 1 (a) = 1, f - 1 (b) = 2, f - 1 (c) = 3
We need to find the inverse of f -1 i.e., inverse of g.
If we define h : {1, 2, 3} → {a, b, c} as h (1) = a, h (2) = b, h (3) = c
(goh)(1) = g (h (1)) = g (a) = 1
(goh)(2) = g (h (2)) = g (b) = 2
(goh)(3) = g (h (3)) = g (c) = 3
And,
(hog )(a) = h (g (a)) = h (1) = a
(hog)(b) = h (g (b)) = h (2) = b
(hog)(c) = h (g (c)) = h (3) = c
⇒ goh = IX and hog = IY where X = {(1, 2, 3)} and Y = {a, b, c}
Thus, the inverse of g exists and g- 1 = h ⇒ (f - 1)- 1 = h.
It can be noted that h = f.
Hence, (f - 1)- 1 = f
NCERT Solutions for Class 12 Maths - Chapter 1 Exercise 1.3 Question 11
Consider f : {1, 2, 3} → {a, b, c} given by f (1) = a, f (2) = b, f (3) = c. Find (f - 1)- 1 = f.
Summary:
For the function f : {1, 2, 3} → {a, b, c} given by f (1) = a, f (2) = b, f (3) = c. It can be concluded that the inverse of g exists and g- 1 = h ⇒ (f - 1)- 1 = h. It can be noted that h = f. Hence, (f - 1)- 1 = f
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