# NCERT Solutions for Class 12 Maths Exercise 1.2

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## NCERT Solutions for Class 12 Maths Exercise 1.2

### Relations and Functions Solutions

#### Q 1) 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*?

Hence, function g is one-one but not onto.

#### Q 2) Check the injectivity and surjectivity of the following functions:

(i) f: N → N given by f(x) = x2
(ii) f: Z → Z given by f(x) = x
(iii) f: R → R given by f(x) = x2
(iv) f: N → N given by f(x) = x3
(v) f: Z → Z given by f(x) = x3

(i) f: N → N is given by,

f(x) = x2

It is seen that for x, y ∈N, f(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) f: Z → Z is given by,

f(x) = x2

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

∴ f is not injective.

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

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iii) f: R → R is given by,

f(x) = x2

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

∴ f is not injective.

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

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iv) f: N → N given by,

f(x) = x3

It is seen that for x, y ∈N, f(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) = x= 2.

∴ f is not surjective

Hence, function f is injective but not surjective.

(v) f: Z → Z is given by,

f(x) = x3

It is seen that for x, y ∈ Z, f(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.

#### Q 3) 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.

∴ 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.

#### Q 5) Show that the Signum Function f: R → R , given by

is neither one-one nor onto.

f: R → R

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.

#### Q 6) 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)}.
∴ 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.

#### Q 7) 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) 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

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.

#### Q 8) Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) 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.

#### Q 9) Let f: N → N be defined by

Answer) 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.

#### Q 10) 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 exists such that

Hence, function f is one-one and onto.

#### Q 11) 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

Answer) fR → R is defined as f(x) = x4

Let x∈ R such that f(x) = f(y).

∴ f (x1) = f (x2) does not imply that x1 = x2

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.

#### Q 12) 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

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.