Plus Two Math's Solution Ex 1.3 Chapter1 Relations and Function

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Board	SCERT, Kerala
Text Book	NCERT Based
Class	Plus Two
Subject	Math's Textbook Solution
Chapter	Chapter 1
Exercise	Ex 1.3
Chapter Name	Relations and Function
Category	Plus Two Kerala

Kerala Syllabus Plus Two Math's Textbook Solution Chapter  1 Relation and Function Exercises 1.3

Chapter  1  Relations and Function Solution

• Exercises 1.1
• Exercises 1.2
• Exercises 1.3
• Exercises 1.4
• Miscellaneous Exercise Chapter 1

Kerala plus two maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

Chapter  1  Relations and Function Exercise   1.3

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.

Solution 1:

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)}.



Question 2:

Let f, g, and h be functions from R to R. Show that



Solution 2:

To prove:











Question 3:

Find gof and fog, if

(i) 

(ii) 

Solution 3:

(i) 



(ii) 



Question 4:

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

Solution 4:

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} 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)}

Solution 5:

(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.

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.,)

Solution 6:

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

⇒ 

Question 7:

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

Solution 7:

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



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.

Solution 8:

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



Question 9:

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

Solution 9:

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



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) = I_Y(y) = fog₂(y). Use one-one ness of f).

Solution 10:

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.

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.

Solution 11:

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.

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.

Solution 12:

Let f: X → Y be an invertible function.

Then, there exists a function g: Y → X such that gof = I_Xand fog = I_Y.

Here, f⁻¹ = g.

Now, gof = I_Xand fog = I_Y

⇒ f⁻¹of = I_Xand fof⁻¹= I_Y

Hence, f⁻¹: Y → 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)

Solution 13:

f: R → R is given as.



The correct answer is C.

Question 14:

Letbe a function defined as. The inverse of f is map g: Range

(A)  (B) 

(C)  (D) 

Solution 14:

It is given that

Let y be an arbitrary element of Range f.

Then, there exists x ∈such that 



Let us define g: Rangeas

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.

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Plus Two Math's Chapter Wise Textbook Solution PDF Download

            Plus Two Math's Part I

• Chapter 1: Relations and Functions
• Chapter 2: Inverse Trigonometric Functions
• Chapter 3: Matrices
• Chapter 4: Determinants
• Chapter 5: Continuity and Differentiability
• Chapter 6: Application of Derivatives



             Plus Two Math's Part II

• Chapter  7: Application of Integrals
• Chapter  8: Integrals
• Chapter  9: Differential Equations
• Chapter 10: Vector Algebra
• Chapter 11: Three Dimensional Geometry
• Chapter 12: Linear Programming
• Chapter 13: Probability

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