NCERT Maths 12 Solutions of Relations and Functions – Chapter 1 Ex 1.2

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

For one-one, f(x) = f(y) → 1/x = 1/y → x = y

Therefore, f is one-one

For onto,

It is clear that for y ϵ R_* , there exists x =  ϵ R_* . R_*(Exists 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(x₁) = g(x₂) →  1/x₁ = 1/x₂→ x₁ = x₂

Therefore, g is one-one.

Further, it is clear that g is not onto as for 1.2 ϵ R_*, there does not exist any x in N such that g(x) =1/1.2

Hence, function g is one-one but not onto

Sol. (i) f : N→N is given by f(x) = x²

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

→ x² = y² → x = y (x and y are positive numbers)

Therefore, f is injective

Now, 2 ϵ N but there does not exist any x in N such that f(x) = x² = 2

It means there is some element in co-domain in which do not have any images. Therefore, 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

Therefore, f is not injective

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

Therefore, 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

Therefore, f is not injective

Now, -2 ϵ R. But there does not exist any element x ϵ R such that f(x) = x² = -2. Therefore, 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 ϵ N, f(x) = f(y) → x³ = y³ → x = y

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

Therefore, f is not surjective

Hence, function f is injective but not surjective

(v) f : Z→Z given by f(x) = x³

It is seen that for x, y ϵ Z, f(x) = f(y) → x³ = y³ → x = y

Therefore, f is injective.

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

Hence, function f is injective but not surjective

Sol. Here, 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 ϵ R. It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ϵ R such that f(x) = 0.7

Therefore, f is not onto

Hence, the greatest integer function is neither one-one nor onto

Sol. Here, f : R →R  is given by

It is seen that f(-1) = |-1| = 1, f(1) = |1| = 1

Therefore, f(-1) = f(1) but -1 ≠ 1

Therefore, f is not one-one

Now, Consider -1 ϵ 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

Hence, the modulus functions is neither one-one nor onto

Now, as f(x) takes only three 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.

Hence, the Signum function is neither one-one nor onto.

Sol 6.

Given that A = {1, 2, 3} and b = {4, 5, 6, 7}

Now, 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. Hence, function f is one-one

Sol. (i) Here, f : R → R is defined by f(x) = 3-4x

Let x₁, x₂ ϵ R such that f(x₁) =  f(x₂)

→ 3 – 4x₁ = 3 – 4x₂ → – 4x₁ = – 4x₂ → x₁= x₂

Therefore, f is one-one.

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

Therefore, f is onto. Hence, f is bijective.

(ii) Here, f : R →R is defined as

f(x) = 1 + x²

Let x₁, x₂ ϵ R such that  f(x₁) =  f(x₂)

→ 

For instance, f(1) = f(-1)= 2

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

Therefore, f is not one-one

Consider an element -2 in co-domain R

It is seen that f(x) = 1 + x² is positive for all x ϵ R

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

Therefore, f is not onto. Hence, f is neither one-one nor onto.

Sol. Here, f : A X B → B X A is defined as f(a, b) = (b, a)

Let                  (a₁, b₁), (a₂, b₂) ϵ A X B

Such that      f(a₁, b₁) = f(a₂, b₂) → (b₁, a₁) = (b₂, a₂)

→     b₁ = b₂ and a₁ = a₂  → (a₁, b₁) = (a₂, b₂)

Therefore, f is one-one. Now, let (b, a) ϵ B X A of any element

Then, there exist (a, b) ϵ A X B such that f(a, b) = (b, a) [definition of f]

Therefore, f is onto. Hence, f is bijective.

Sol. Here, f :N→N is defined as, f(n),

for all n ϵ N. It can be observed that

f(1)= 1+1/2 = 1 and f(2) = 2/2 = 1         [By definition of f]

f(1) = f(2), where 1 ≠ 2

Therefore, f is not one-one. Consider a natural number n in co-domain N.

Case 1 When n is odd.

Therefore, n = 2r +1 for some r ϵ N

Then, there exists 4r + 1 ϵ N such that f(4r +1) = 4r+1+1/2 = 2r + 1

Therefore, f is onto.

Case 2 When n is even

Therefore, n = 2r for some r ϵ N

Then, there exists 4r ϵ N such that f(4r) = 4r/2 = 2r

Therefore, f is onto. Hence, f is not a bijective function.

Note If f is one-one and onto, then we say that f is bijective functive.

Sol. Here, A = R – {3}, B = R – {1}

And f : A → B is defined as f(x) = (x-2/x-3)

Let x, y ϵ 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 – 3y – 2x + 6

→ -3x – 2y = -3y -2x

→ 3x – 2x = 3y – 2y → x = y

Therefore, f is one-one. Let y ϵ B = R – {1}. Then y ≠ 1

The function f is onto if there exists x ϵ 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 ϵ A [y ≠ 1]

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

Therefore, f is onto. Hence, function f is one-one and onto.

Sol. Function f : R →R is defined as f(x) = x⁴

Let x, y ϵ R such that f(x) =f(y)

→            x⁴ = y⁴

→            x = ±y

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

For instance, 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. Hence, function f is neither one-one nor onto.

The correct answer is (d).

Sol. f : R →R is defined as f(x) = 3x

Let x, y ϵ 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. Hence, function f is one-one and onto.

The correct answer is (a).