# NCERT solution class 12 chapter 1 Relations and Functions exercise 1.2 mathematics part 1

## EXERCISE 1.2

#### Question 1:

Show that the function fR* → 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 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 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.

#### Question 2:

Check the injectivity and surjectivity of the following functions:

(i) fN → N given by f(x) = x2

(ii) fZ → Z given by f(x) = x2

(iii) fR → R given by f(x) = x2

(iv) f→ N given by f(x) = x3

(v) fZ → Z given by f(x) = x3

(i) fN → N is given by,

f(x) = x2

It is seen that for xy ∈Nf(x) = f(y) ⇒ x2 = y2 ⇒ x = y.

f is injective.

Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x2 = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

(ii) fZ → Z is given by,

f(x) = x2

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

∴ is not injective.

Now,−2 ∈ Z. But, there does not exist any element Z such that f(x) = x2 = −2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iii) fR → R is given by,

f(x) = x2

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

∴ is not injective.

Now,−2 ∈ R. But, there does not exist any element ∈ R such that f(x) = x2 = −2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iv) fN → N given by,

f(x) = x3

It is seen that for xy ∈Nf(x) = f(y) ⇒ x3 = y3 ⇒ x = y.

f is injective.

Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x3 = 2.

∴ f is not surjective

Hence, function f is injective but not surjective.

(v) f→ Z is given by,

f(x) = x3

It is seen that for xy ∈ Zf(x) = f(y) ⇒ x3 = y3 ⇒ x = y.

∴ f is injective.

Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x3 = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

#### Question 3:

Prove that the Greatest Integer Function f→ R given by f(x) = [x], is neither one-once 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.

#### Question 4:

Show that the Modulus Function f→ R given by , is neither one-one nor onto, where is x, if x is positive or 0 and is − 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.

#### Question 5:

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.

#### Question 6:

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.

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

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

∴ f (1) = 4, f (2) = 5, f (3) = 6

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

Hence, function f is one-one.

#### Question 7:

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) f→ R is defined as f(x) = 3 − 4x. . ∴ f is one-one.

For any real number (y) in R, there exists in R such that is onto.

Hence, is bijective.

(ii) fR → R is defined as . .  does not imply that For instance, ∴ f is not one-one.

Consider an element −2 in co-domain R.

It is seen that is positive for all x ∈ R.

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

∴ f is not onto.

Hence, f is neither one-one nor onto.

#### Question 8:

Let A and B be sets. Show that fA × B → × A such that (ab) = (ba) is 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.

#### Question 9:

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.

#### Question 10:

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

Now, Thus, for any ∈ B, there exists such that Hence, function f is one-one and onto.

#### Question 11:

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.

#### Question 12:

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. 