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**RELATIONS AND FUNCTIONS**

**Q.1. Show that the function f: R _{∗} → R_{∗} defined by f(x) = 1/x**

For one – one:

Let x, y ∈

⇒ 1/x = 1/y

⇒ x = y

∴ f is one – one.

For onto:

It is clear that for y ∈

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

Now, consider function g: N → R

We have,

g(x

⇒ 1/x

∴ g is one – one.

Further, it is clear that g is not onto as for 1.2 ∈ R

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

(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

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

∴ f is injective.

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

∴ f is not surjective.

Hence, function f is injective but not surjective.

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) = −2 or x

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

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) = −2 or x

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

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

∴ f is injective.

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

f(x) = 2 or x

∴ f is not surjective

Hence, function f is injective but not surjective.

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

∴ f is injective.

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

f(x) = 2 or x

∴ f is not surjective.

Hence, function f is injective but not surjective.

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.

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

∴ f(−1) = f(1), but −1 ≠ 1.

∴ 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.

∴ f is not onto.

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

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.

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.

(i) f : R → R defined by f(x) = 3 − 4x

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

Let x

⇒ 3 - 4x

⇒ x

∴ f is one – one.

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

∴ f is onto.

Hence, f is bijective.

Let x

⇒ 1+ x

⇒ x

∴ f(x

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

∴ 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.

∴ f is not onto.

Hence, f is neither one – one nor onto.

Let (a

⇒ (b

⇒ b

∴ 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]

∴ f is onto.

Hence, f is bijective.

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

∴ 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 4r + 1∈ N such that f(4r + 1) = 4r + 1 + 1/2 = 2r + 1

**Case II:** n is even

∴ n = 2r for some r ∈ N. Then, there exists 4r ∈ N such that f(4r) = 4r/2 = 2r

∴ 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 (x) = (x-2/x-3)****. Is f one-one and onto? Justify your answer.****Ans.**

A = R − {3}, B = R − {1} and f: A → B defined by

Let x, y ∈ A such that f(x) = f(y)

⇒ (x – 2)(y – 3) = (y – 2)(x – 3)

⇒ xy – 3x – 2y + 6 = xy – 2x – 3y + 6

⇒ – 3x – 2y = – 2x – 3y ⇒ x = y

∴ 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

y = (x - 2/ x - 3)

⇒ x – 2 = xy – 3y ⇒ x(1 – y) = – 3y + 2

⇒ x=(2-3y )/(1-y) ∈A [y ≠ 1]

∴ f is onto.

Hence, function f is one – one and onto.**Q.11. Let f: R → R be defined as f(x) = x ^{4}. 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**

Let x, y ∈ R such that f(x) = f(y).

⇒ x

⇒ x = ± y

∴ f(x) = f(y) does not imply that x = y.

For example f(1) = f(–1) = 1

∴ 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.

(a) f is one – one and onto

(b) f is many – one and 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

∴ 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

∴ f is onto.

Hence, function f is one – one and onto.

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