## Question 1:Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.

The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as
= {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}.

To prove:

(i)
(ii)

## Question 4:If, show that f o f(x) = x, for all. What is the inverse of f?

It is given that.
Hence, the given function f is invertible and the inverse of f is f itself.

## Question 5:State with reason whether following functions have inverse(i) f: {1, 2, 3, 4} → {10} withf = {(1, 10), (2, 10), (3, 10), (4, 10)}(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} withg = {(5, 4), (6, 3), (7, 4), (8, 2)}(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} withh = {(2, 7), (3, 9), (4, 11), (5, 13)}

(i) f: {1, 2, 3, 4} → {10}defined as:
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10
is not one-one.
Hence, function does not have an inverse.

(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as:
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) = 4.
is not one-one,
Hence, function g does not have an inverse.

(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h
∴Function h is one-one.
Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5}such that h(x) = y.
Thus, h is a one-one and onto function. Hence, h has an inverse.

## Question 6:Show that f: [−1, 1] → R, given byis one-one. Find the inverse of the function f: [−1, 1] → Range f.(Hint: For y ∈Range f, y =, for some x in [−1, 1], i.e.,)

f: [−1, 1] → R is given as
Let f(x) = f(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 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:
Now, let us define g: Range f → [−1, 1] as
gof = and fo
f−1 = g
⇒

## Question 7:Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

fR → R is given by,
f(x) = 4x + 3
One-one:
Let f(x) = f(y).
∴ f is a one-one function.
Onto:
For y ∈ R, let y = 4x + 3.
Therefore, for any ∈ R, there exists  such that
∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define gR→ R by.
Hence, f is invertible and the inverse of f is given by

## Question 8:Consider f: R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given f by, where R+ is the set of all non-negative real numbers.

fR+ → [4, ∞) is given as f(x) = x2 + 4.
One-one:
Let f(x) = f(y).
∴ f is a one-one function.
Onto:
For y ∈ [4, ∞), let y = x2 + 4.
Therefore, for any ∈ R, there exists  such that
.
∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define g: [4, ∞) → Rby,
Hence, f is invertible and the inverse of f is given by

## Question 9:Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with.

fR+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5.
Let y be an arbitrary element of [−5, ∞).
Let y = 9x2 + 6− 5.
f is onto, thereby range f = [−5, ∞).
Let us define g: [−5, ∞) → R+ as
We now have:
and
Hence, f is invertible and the inverse of f is given by

## Question 10:Let f: X → Y be an invertible function. Show that f has unique inverse.(Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y,fog1(y) = IY(y) = fog2(y). Use one-one ness of f).

Let fX → Y be an invertible function.
Also, suppose f has two inverses (say).
Then, for all y ∈Y, we have:
Hence, f has a unique inverse.

## Question 11:Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f−1 and show that (f−1)−1 = f.

Function f: {1, 2, 3} → {abc} is given by,
f(1) = af(2) = b, and f(3) = c
If we define g: {abc} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:
and, where X = {1, 2, 3} and Y= {abc}.
Thus, the inverse of exists and f−1 = g.
f−1: {abc} → {1, 2, 3} is given by,
f−1(a) = 1, f−1(b) = 2, f-1(c) = 3
Let us now find the inverse of f−1 i.e., find the inverse of g.
If we define h: {1, 2, 3} → {abc} as
h(1) = ah(2) = bh(3) = c, then we have:
, where X = {1, 2, 3} and Y = {abc}.
Thus, the inverse of exists and g−1 = h ⇒ (f−1)−1 = h.
It can be noted that h = f.
Hence, (f−1)−1 = f.

## Question 12:Let f: X → Y be an invertible function. Show that the inverse of f−1 is f, i.e.,(f−1)−1 = f.

Let fX → Y be an invertible function.
Then, there exists a function gY → X such that gof = IXand fo= IY
Here, f−1 = g
Now, gof = IXand fo= IY
⇒ f−1of = IXand fof−1= IY
Hence, f−1Y → X is invertible and f is the inverse of f−1
i.e., (f−1)−1 = f.

## Question 13:If f: R → R be given by, then fof(x) is(A) (B) x3(C) x(D) (3 − x3)

fR → R is given as.