Looking for NCERT Solutions for Class 12 Maths Exercise 1.3? If yes then you are in the right place. Here we have provided you with the Question and Answers of Exercise 1.3. These solutions are written by expert teachers and are so accurate to rely on.
| Chapter | 1. Relations and Functions |
| Exercise | 1.3 |
| Class | Twelve |
| Subject | Maths |
| Category | NCERT Solutions for Class 12 |
NCERT Solutions for Class 12 Maths are available online for students to use and refer to while studying for their exams. The solutions provide step-by-step explanations for each problem in the textbook, making it easy for students to understand the concepts. With Class 12 Maths NCERT Solutions, students can be sure that they are getting the most accurate and up-to-date information available. The solutions are regularly updated by experts in the field, so students can be confident that they are using the most current information.
NCERT Solutions for Class 12 Maths Exercise 1.3
Relations and Functions Solutions
Q 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.
Answer)
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)}.
Q 2) Let f, g and h be functions from R to R. Show that
Answer)
To prove:
Q 3) Find gof and fog, if
(i) f(x) = | x | and g(x) = | 5x – 2 |
Answer)
Q 4) If 
, show that f o f(x) = x, for all 
. What is the inverse of f?
Answer) It is given that
Hence, the given function f is invertible and the inverse of f is f itself.
Q 5) 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)}
Answer)
(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.
Q 6) Show that f : [–1, 1] → R, given by f (x) = 
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., 
Answer) 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
Q 7) Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Answer)
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
Q 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.
Answer)
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
Q 9) Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with 
Answer)
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, ∞).
We now have:
Q 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).
Answer)
Let f: X → Y be an invertible function.
Also, suppose f has two inverses (say g1 and g2)
Then, for all y ∈Y, we have:
Hence, f has a unique inverse.
Q 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.
Answer)
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:
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:
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.
Q 12) Let f: X → Y be an invertible function. Show that the inverse of f−1 is f, i.e., (f−1)−1 = f.
Answer)
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.
Q 13) If f: R → R be given by 
, then fof(x) is
(A) 
(B) x3 (C) x (D) (3 − x3)
Answer) f: R → R is given as 
The correct answer is C.
Q 14) Let 
be a function defined as 
. The inverse of f is map g: Range Given by
Answer) It is given that
Let y be an arbitrary element of Range f.
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.
That’s it. These were the solutions of NCERT Class 12 Maths Exercise 1.3. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.