# Plus Two Math's Solution Ex 1.2 Chapter1 Relations and Function

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

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   1.2

Show that the function fR* → R* defined byis 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 fR* → R* is defined by

One-one:

f is one-one.

Onto:

It is clear that for y R*, there existssuch 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 is given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

fR → 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 is given by, is neither one-one nor onto, where is x, if x is positive or 0 andis − x if x is negative.

fR → 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 fR → R, given by

is neither one-one nor onto.

fR → 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}, = {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 $\rightarrow$ 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 defined by f(x) = 3 − 4x

(ii) f→ 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 fA × B → × A such that (ab) = (ba) is a bijective function.

fA × B → B × A is defined as f(ab) = (ba).

.

∴ f is one-one.

Now, let (ba) ∈ B × A be any element.

Then, there exists (ab) ∈A × B such that f(ab) = (ba). [By definition of f]

∴ f is onto.

Hence, f is bijective.

Let fN → N be defined by

fN → 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 4+ 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

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 is onto if there exists x ∈A such that f(x) = y.

Now,

Thus, for any ∈ B, there existssuch that

Hence, function f is one-one and onto.

Let fR → 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

fR → R is defined as

Let x∈ 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.

Let fR → 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.

fR → R is defined as f(x) = 3x.

Let x∈ R such that f(x) = f(y).

⇒ 3x = 3y

⇒ x = y

is one-one.

Also, for any real number (y) in co-domain R, there exists  in R such that.

is onto.

Hence, function f is one-one and onto.

#### Chapter 1: Relations and Function EX 1.2 Solution- Preview

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