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**RELATIONS AND FUNCTIONS**

**Q.1. Show that the function f: R _{âˆ—} â†’ R_{âˆ—} defined by f(x) = 1/x**

For one â€“ one:

Let x, y âˆˆ

â‡’ 1/x = 1/y

â‡’ x = y

âˆ´ f is one â€“ one.

For onto:

It is clear that for y âˆˆ

Thus, the given function f is one â€“ one and onto.

Now, consider function g: N â†’ R

We have,

g(x

â‡’ 1/x

âˆ´ g is one â€“ one.

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

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

(i) f : N â†’ N given by f(x) = x

(ii) f : Z â†’ Z given by f(x) = x

(iii) f : R â†’ R given by f(x) = x

(iv) f : N â†’ N given by f(x) = x

(v) f : Z â†’ Z given by f(x) = x

It is seen that for x, y âˆˆ N, f(x) = f(y) â‡’ x

âˆ´ f is injective.

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

âˆ´ f is not surjective.

Hence, function f is injective but not surjective.

It is seen that f(âˆ’1) = f(1) = 1, but âˆ’1 â‰ 1.

âˆ´ f is not injective.

Now, âˆ’2 âˆˆ Z. But, there does not exist any element x âˆˆ Z such that

f(x) = âˆ’2 or x

âˆ´ f is not surjective.

Hence, function f is neither injective nor surjective.

It is seen that f(âˆ’1) = f(1) = 1, but âˆ’1 â‰ 1.

âˆ´ f is not injective.

Now, âˆ’2 âˆˆ R. But, there does not exist any element x âˆˆ R such that

f(x) = âˆ’2 or x

âˆ´ f is not surjective.

Hence, function f is neither injective nor surjective.

It is seen that for x, y âˆˆ N, f(x) = f(y) â‡’ x

âˆ´ f is injective.

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

f(x) = 2 or x

âˆ´ f is not surjective

Hence, function f is injective but not surjective.

It is seen that for x, y âˆˆ Z, f(x) = f(y) â‡’ x

âˆ´ f is injective.

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

f(x) = 2 or x

âˆ´ f is not surjective.

Hence, function f is injective but not surjective.

f: R â†’ R is given by, f(x) = [x]

It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.

âˆ´ f(1.2) = f(1.9), but 1.2 â‰ 1.9.

âˆ´ f is not one â€“ one.

Now, consider 0.7 âˆˆ R.

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

âˆ´ f is not onto.

Hence, the greatest integer function is neither one â€“ one nor onto.

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

âˆ´ f(âˆ’1) = f(1), but âˆ’1 â‰ 1.

âˆ´ f is not one â€“ one. Now, consider âˆ’1 âˆˆ 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| = âˆ’1.

âˆ´ 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 â‰ 2.

âˆ´ f is not one â€“ one.

Now, as f(x) takes only 3 values (1, 0, or âˆ’1) for the element âˆ’2 in co-domain R, there does not exist any x in domain R such that f(x) = âˆ’2.

âˆ´ f is not onto.

Hence, the Signum function is neither one â€“ one nor onto.

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

âˆ´ 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 â€“ one.

(i) f : R â†’ R defined by f(x) = 3 âˆ’ 4x

(ii) f : R â†’ R defined by f(x) = 1 + x

Let x

â‡’ 3 - 4x

â‡’ x

âˆ´ f is one â€“ one.

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

âˆ´ f is onto.

Hence, f is bijective.

Let x

â‡’ 1+ x

â‡’ x

âˆ´ f(x

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

âˆ´ f is not one â€“ one.

Consider an element âˆ’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) = âˆ’2.

âˆ´ f is not onto.

Hence, f is neither one â€“ one nor onto.

Let (a

â‡’ (b

â‡’ b

âˆ´ f is one â€“ one.

Now, let (b, a) âˆˆ B Ã— A be any element.

Then, there exists (a, b) âˆˆ A Ã— B such that f(a, b) = (b, a). [By definition of f]

âˆ´ 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 â‰ 2

âˆ´ f is not one-one.

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

**Case I:** n is odd

âˆ´ n = 2r + 1 for some r âˆˆ N. Then, there exists 4r + 1âˆˆ N such that f(4r + 1) = 4r + 1 + 1/2 = 2r + 1

**Case II:** n is even

âˆ´ n = 2r for some r âˆˆ N. Then, there exists 4r âˆˆ N such that f(4r) = 4r/2 = 2r

âˆ´ f is onto.

Hence, f is not a bijective function.**Q.10. Let A = R âˆ’ {3} and B = R âˆ’ {1}. Consider the function f: A â†’ B defined by (x) = (x-2/x-3)****. Is f one-one and onto? Justify your answer.****Ans.**

A = R âˆ’ {3}, B = R âˆ’ {1} and f: A â†’ B defined by

Let x, y âˆˆ A such that f(x) = f(y)

â‡’ (x â€“ 2)(y â€“ 3) = (y â€“ 2)(x â€“ 3)

â‡’ xy â€“ 3x â€“ 2y + 6 = xy â€“ 2x â€“ 3y + 6

â‡’ â€“ 3x â€“ 2y = â€“ 2x â€“ 3y â‡’ x = y

âˆ´ f is one-one.

Let y âˆˆ B = R âˆ’ {1}. Then, y â‰ 1.

The function f is onto if there exists x âˆˆ A such that f(x) = y.

Now, f(x) = y

y = (x - 2/ x - 3)

â‡’ x â€“ 2 = xy â€“ 3y â‡’ x(1 â€“ y) = â€“ 3y + 2

â‡’ x=(2-3y )/(1-y) âˆˆA [y â‰ 1]

âˆ´ f is onto.

Hence, function f is one â€“ one and onto.**Q.11. Let f: R â†’ 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 âˆˆ R such that f(x) = f(y).

â‡’ x

â‡’ x = Â± y

âˆ´ f(x) = f(y) does not imply that x = y.

For example f(1) = f(â€“1) = 1

âˆ´ 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.

âˆ´ f is not onto.

Hence, function f is neither one â€“ one nor onto.

(a) f is one â€“ one and onto

(b) f is many â€“ one and onto

(d) f is neither one â€“ one nor onto

f : R â†’ R is defined as f(x) = 3x.

Let x, y âˆˆ R such that f(x) = f(y).

â‡’ 3x = 3y

â‡’ x = y

âˆ´ 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

âˆ´ f is onto.

Hence, function f is one â€“ one and onto.

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