Relation and Functions are the basic building blocks of mathematics. The entire chapter of relation should preferably be covered in the initial phase since it helps to understand concepts like transitive, reflexive, etc. It is quite possible that students may miss so many questions because they do not know how to deal with these concepts and would have to depend on their memory which cannot be trusted always.
Chapter 1 Ncert Plus Two Maths's Chapter-wise Textbook Solution for Exercise 1.2 Here we provide detailed and complete study materials for each chapter and the answers provide here is detailed so that the students can easily understand
| Board | SCERT, Kerala |
| Text Book | NCERT Based |
| Class | Plus Two |
| Subject | Math's Textbook Solution |
| Chapter | Chapter 1 |
| Exercise | Ex 1.2 |
| Chapter Name | Relations and Function |
| Category | Plus Two Kerala |
Kerala Syllabus Plus Two Math's Textbook Solution Chapter 1 Relation and Function Exercises 1.2
Chapter 1 Relations and Function Solution
Chapter 1 Relations and Function Exercise 1.2
Show that the function f: R* → R* defined by
is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being the same as R*?
It is given that f: R* → R* is defined by
One-one:
∴f is one-one.
Onto:
It is clear that for y∈ R*, there exists
such that
∴f is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function g: N → R*defined by
We have,
∴g is one-one.
Further, it is clear that g is not onto as for 1.2 ∈R* there does not exit any x in N such that g(x) =
.
Hence, function g is one-one but not onto.
Prove that the Greatest Integer Function f: R → R is given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
f: R → R is given by,
f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.
∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.
∴ f is not one-one.
Now, consider 0.7 ∈ R.
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.
∴ f is not onto.
Hence, the greatest integer function is neither one-one nor onto.
Show that the Modulus Function f: R → R is given by
, is neither one-one nor onto, where
is x, if x is positive or 0 andis − x if x is negative.
f: R → R is given by,
It is seen that
.
∴f(−1) = f(1), but −1 ≠ 1.
∴ f is not one-one.
Now, consider −1 ∈ R.
It is known that f(x) = is always non-negative. Thus, there does not exist any element x in domain R such that f(x) = = −1.
∴ f is not onto.
Hence, the modulus function is neither one-one nor onto.
Show that the Signum Function f: R → R, given by
is neither one-one nor onto.
f: R → R is given by,
It is seen that f(1) = f(2) = 1, but 1 ≠ 2.
∴f is not one-one.
Now, as f(x) takes only 3 values (1, 0, or −1) for the element −2 in co-domain R, there does not exist any x in domain R such that f(x) = −2.
∴ f is not onto.
Hence, the signum function is neither one-one nor onto.
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
f: A B defined as f (1) =4,f (2) = 5, f (3) = 6 It is seen that each image of distinct elements of A under f has distinct elements, so according to the definition, f is a function.
In each of the following cases, state whether the function is one-one, onto, or bijective. Justify your answer.
(i) f: R → R defined by f(x) = 3 − 4x
(ii) f: R → R defined by f(x) = 1 + x2
(i) it is given that f : R → R defined by f (x) = 3 – 4x
.
⇒ f is one- one
we know one thing if a function f(x) is inversible then f(x) is definitely a bijective function. means, f(x) will be one - one and onto.
Let's try the inverse of f(x) = 3 - 4x
y = 3 - 4x
y - 3 = 4x => x = (y - 3)/4
f?¹(x) = (x - 3)/4
hence, f(x) is inversible .
so, f(x) is one - one and onto function.
hence, f(x) is bijective function [ if any function is one -one and onto then it is also known as bijective function.]
(ii) it is given that f :R→R defined by f(x) = 1 +x²
.
now, f(1) = f(-1) = 2
so, f is not one - one function.
also for all real value of x , f(x) is always greater than 1 . so, range of f(x) ∈ [1,∞)
but co-domain ∈ R
e.g., Co - domain ≠ range
so, f is not onto function.
also f is not bijective function.
Let A and B set. Show that f: A × B → B × A such that (a, b) = (b, a) is a bijective function.
f: A × B → B × A is defined as f(a, b) = (b, a).
.
∴ f is one-one.
Now, let (b, a) ∈ B × A be any element.
Then, there exists (a, b) ∈A × B such that f(a, b) = (b, a). [By definition of f]
∴ f is onto.
Hence, f is bijective.
Let f: N → N be defined by
State whether the function f is bijective. Justify your answer.
f: N → N is defined as
It can be observed that:
∴ f is not one-one.
Consider a natural number (n) in co-domain N.
Case I: n is odd
∴n = 2r + 1 for some r ∈ N. Then, there exists 4r + 1∈N such that
.
Case II: n is even
∴n = 2r for some r ∈ N. Then, there exists 4r ∈N such that
.
∴ f is onto.
Hence, f is not a bijective function.
Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by
. Is f one-one and onto? Justify your answer.
A = R − {3}, B = R − {1}
f: A → B is defined as.
.
∴ f is one-one.
Let y ∈B = R − {1}. Then, y ≠ 1.
The function f is onto if there exists x ∈A such that f(x) = y.
Now,
Thus, for any y ∈ B, there exists
such that
Hence, function f is one-one and onto.
Let f: R → R be defined as f(x) = x4. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto
f: R → R is defined as
Let x, y ∈ R such that f(x) = f(y).
∴
does not imply that
.
For instance,
∴ f is not one-one.
Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2.
∴ f is not onto.
Hence, function f is neither one-one nor onto.
The correct answer is D.
Let f: R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto.
f: R → R is defined as f(x) = 3x.
Let x, y ∈ R such that f(x) = f(y).
⇒ 3x = 3y
⇒ x = y
∴f is one-one.
Also, for any real number (y) in co-domain R, there exists 
in R such that
.
∴f is onto.
Hence, function f is one-one and onto.
The correct answer is A.
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Chapter 1: Relations and Function EX 1.2 Solution
Chapter 1: Relations and Function EX 1.2 Solution- Preview
Plus Two Math's Chapter Wise Textbook Solution PDF Download
- 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
- 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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