In this page we have *Relations and Functions NCERT Solutions for Class 12 Maths Chapter 1* for
EXERCISE 1.2 . Hope you like them and do not forget to like , social share
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1) We have to prove that

if f(x) = f(y), then x = y

1) for every y ∈ B, there exists an element x in A where f(x)=y

To prove a function is bijective, you need to prove that it is injective and also surjective.

"Injective" means no two elements in the domain of the function gets mapped to the same image. The method has been already described above.

"Surjective" means that any element in the range of the function is hit by the function. The method has been already described above

Show that the function f: R∗ → R∗ defined by f(x) = 1/x

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 same as R∗?

It is given that f: R* → R* is defined by f(x) =1/x

Let x, y ∈ R*

such that f(x) = f(y)

1/x =1/y

⇒ x = y

Therefore, f is one – one.

For y ∈ R*

, there exists x =1/y ∈ R*

[as y ≠ 0] such that

f(x) = 1/ [1/y]

= y

Therefore, f is onto.

Thus, the given function f is one – one and onto.

Now, consider function g: N → R* defined by g(x) =1/x

We have,

g(x) = g(y)

1/x=1/y

x = y

Therefore, 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) =1/1.2

Check the injectivity and surjectivity of the following functions:

(i) f: N → N given by f(x) = x

(ii) f: Z → Z given by f(x) = x

(iii)f: R → R given by f(x) = x

(iv)f: N → N given by f(x) = x

(v) f: Z → Z given by f(x) = x

(i) f: N → N is given by f(x) = x

It is seen that for x, y ∈ N, f(x) = f(y) ⇒ x

⇒ x = y.

Therefore, f is injective.

Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x

Therefore, f is not surjective.

(ii) f: Z → Z is given by f(x) = x

It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.

Therefore, f is not injective.

Now, −2 ∈ Z. But, there does not exist any element x ∈ Z such that

f(x) = −2 or x

Therefore, f is not surjective.

(iii) f: R → R is given by f(x) = x

It is seen that f (−1) = f(1) = 1, but −1 ≠ 1.

Therefore, f is not injective.

Now, −2 ∈ R. But, there does not exist any element x ∈ R such that

f(x) = −2 or x

2 = −2.

Therefore, f is not surjective.

(iv) f: N → N given by f(x) = x

It is seen that for x, y ∈ N, f(x) = f(y) ⇒

x

Therefore, f is injective.

Now, 2 ∈ N. But, there does not exist any element x ∈ N such that

f(x) = 2 or x

Therefore, f is not surjective

(v) f: Z → Z is given by f(x) = x

It is seen that for x, y ∈ Z, f(x) = f(y) ⇒ x

Therefore, f is injective.

Now, 2 ∈ Z. But, there does not exist any element x ∈ Z such that

f(x) = 2 or x

3 = 2.

Therefore, f is not surjective.

Prove that the Greatest Integer Function f: R → R 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.

Therefore, f(1.2) = f(1.9), but 1.2 ≠ 1.9.

Therefore, 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.

Therefore, f is not onto.

Show that the Modulus Function f: R → R given by f(x) = |x|, is neither one – one nor onto, where |x| is x, if x is positive or 0 and |x| is − x, if x is negative.

f: R → R is given by f(x) = |x|

= {x, if x ≥ 0

-x, if x < 0}

It is clear that f (-1) = |-1| = 1 and f (1) = |1| = 1

Therefore, f (−1) = f (1), but −1 ≠ 1.

Therefore, f is not one – one.

Now, consider −1 ∈ R.

It is known that f(x) = |x| is always non-negative. Thus, there does not exist any

element x in domain R such that f(x) = |x| = −1.

Therefore, f is not onto.

Show that the Sig-num Function f: R → R, given by

f(x) = {1, if x > 0

0, if x = 0

−1, if x < 0

is neither one-one nor onto.

f: R → R is given by

f(x) = {1, if x > 0

0, if x = 0

-1, if x < 0

It is seen that f (1) = f(2) = 1, but 1 ≠ 2.

Therefore, 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.

Therefore, f is not 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.

It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.

f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.

Therefore, f (1) = 4, f (2) = 5, f (3) = 6

It is seen that the images of distinct elements of A under f are distinct.

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 + x

(i) f: R → R is defined as f(x) = 3 − 4x.

Let x, y ∈ R such that f(x) = f(y)

3-4x = 3-4xy

-4x = -4y

x = y

Therefore, f is one – one.

For any real number (y) in R, there exists (3-y)/4 in R such that

f [(3-y)/4] = 3-[4 (3-y)/4]

= y

Therefore, f is onto.

(ii) f: R → R is defined as f(x) = 1 + x

Let x, y ∈ R such that f(x) = f(y)

1+ x

⇒ x = ± y

Therefore, f(x) = f(y) does not imply that x1 = x2

For example f (1) = f(-1) = 2

Therefore, f is not one – one.

Consider an element −2 in co-domain R.

It is seen that f(x) = 1 + x

Thus, there does not exist any x in domain R such that f(x) = −2.

Therefore, f is not onto.

Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is bijective function.

f: A × B → B × A is defined as f(a, b) = (b, a).

Let (a

f (a

⇒ (b

⇒ b

⇒ (a

Therefore, 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 ]

Therefore, f is onto.

Let f: N → N be defined by

f(n) = {(n+1)/2 , if n is odd

n/2, if n is even}

for all n ∈ N

State whether the function f is bijective. Justify your answer.

It can be observed that:

f (1) = (1+1)/2

= 1

and f (2) =2/2

= 1 [By definition of f(n)]

f (1) = f (2), where 1 ≠ 2

Therefore, f is not one-one.

Consider a natural number (n) in co-domain N.

Therefore, n = 2r + 1 for some r ∈ N. Then, there exists 4r + 1∈ N such that

f (4r + 1) =(4r + 1 + 1)/2

= 2r + 1

Therefore, n = 2r for some r ∈ N. Then, there exists 4r ∈ N such that

f(4r) =4r/2

= 2r.

Therefore, f is onto.

Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by f(x) = (x−2)/(x−3) Is f one-one and onto? Justify your answer.

A = R − {3}, B = R − {1} and f: A → B defined by f(x) = (x-2)/(x-3)

Let x, y ∈ A such that f(x) = f(y)

(x-2)/(x-3) =(y-2)/(y-3)

(x – 2)(y – 3) = (y – 2)(x – 3)

xy – 3x – 2y + 6 = xy – 2x – 3y + 6

– 3x – 2y = – 2x – 3y ⇒ x = y

Therefore, 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, f(x) = y

⇒(x-2)/(x-3) = y

⇒ x – 2 = xy – 3y

⇒ x (1 – y) = – 3y + 2

⇒ x =(2-3y)/(1-y) ∈ A [y ≠ 1]

Thus, for any y ∈ B, there exists (2-3y) /(1-y) ∈ A such that

f [(2-3y)/(1-y)]

=[(2-3y)/(1-y) -2]/ [(2-3y)/(1-y) -3]

= (2-3y-2 + 2y)/(2-3y-3 + 3y) = y

Therefore, f is onto.

Let f: R → R be defined as f(x) = x

(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) = x

Let x, y ∈ R such that f(x) = f(y).

x

x = ± y

Therefore, f(x) = f(y) does not imply that x = y.

For example f (1) = f(–1) = 1

Therefore, 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.

Therefore, f is not onto.

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

Therefore, f is one-one.

Also, for any real number (y) in co-domain R, there exists y/3 in R such that

f (y/3) =3 (y/3) = y

Therefore, f is onto.

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