Find the best tutors and institutes for Class 12 Tuition
Search in
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
f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}.
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.
State with reason whether following functions have inverse
(i) f: {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(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
∴f is not one-one.
Hence, function f 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.
∴g 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.
Show that f: [−1, 1] → R, given by
is 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 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:
Now, let us define g: Range f → [−1, 1] as
∴gof =
and fog = 
f−1 = g
⇒ 
Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
f: R → 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 y ∈ R, there exists 
such that
∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define g: R→ R by
.
∴
Hence, f is invertible and the inverse of f is given by
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.
f: R+ → [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 y ∈ 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, ∞) → R+ by,
∴
Hence, f is invertible and the inverse of f is given by
Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with
.
f: R+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5.
Let y be an arbitrary element of [−5, ∞).
Let y = 9x2 + 6x − 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
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 f: X → 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.
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} → {a, b, c} is given by,
f(1) = a, f(2) = b, and f(3) = c
If we define g: {a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:
∴
and
, where X = {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
Let us now find the inverse of f−1 i.e., find the inverse of g.
If we define h: {1, 2, 3} → {a, b, c} as
h(1) = a, h(2) = b, h(3) = c, then we have:
∴
, 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.
Let f: X → Y be an invertible function. Show that the inverse of f−1 is f, i.e.,
(f−1)−1 = f.
Let f: X → Y be an invertible function.
Then, there exists a function g: Y → X such that gof = IXand fog = IY.
Here, f−1 = g.
Now, gof = IXand fog = IY
⇒ f−1of = IXand fof−1= IY
Hence, f−1: Y → X is invertible and f is the inverse of f−1
i.e., (f−1)−1 = f.
If f: R → R be given by
, then fof(x) is
(A) 
(B) x3 (C) x (D) (3 − x3)
f: R → R is given as
.
The correct answer is C.
Let
be a function defined as
. The inverse of f is map g: Range
(A) 
(B) 
(C) 
(D) 
It is given that
Let y be an arbitrary element of Range f.
Then, there exists x ∈
such that 
Let us define g: Range
as
Now,
∴
Thus, g is the inverse of f i.e., f−1 = g.
Hence, the inverse of f is the map g: Range, which is given by
The correct answer is B.
How helpful was it?
How can we Improve it?
Please tell us how it changed your life *
Please enter your feedback
UrbanPro.com helps you to connect with the best Class 12 Tuition in India. Post Your Requirement today and get connected.
Find best tutors for Class 12 Tuition Classes by posting a requirement.
Get started now, by booking a Free Demo Class
