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**RELATIONS AND FUNCTIONS**

**Q.1. Let f : {1, 3, 4} â†’ {1, 2, 5} and g : {1, 2, 5} â†’ {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.****Ans.**

The functions f: {1, 3, 4} â†’ {1, 2, 5} and g: {1, 2, 5} â†’ {1, 3} are defined as

f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}.

gof(1) = g[f(1)] = g(2) = 3 , [as f(1) = 2 and g(2) = 3]

gof(3) = g[f(3)] = g(5) = 1, [as f(3) = 5 and g(5) = 1]

gof(4) = g[f(4)] = g(1) = 3 , [as f(4) = 1 and g(1) = 3]

âˆ´ gof = {(1, 3), (3, 1), (4, 3)}**Q.2. Let f, g and h be functions from R to R. Show that(f + g)oh = foh + goh(f.g)oh = (foh).(goh)**

LHS = [(f + g)oh](x)

= (f + g)[h(x)]

= f [h(x)] + g[h(x)]

= (foh)(x) + (goh)(x)

= {(foh)(x) + (goh)}(x) = RHS

âˆ´ {(f + g)oh}(x) = {(foh)(x) + (goh)}(x) for all x âˆˆR

Hence, (f + g)oh = foh + goh

(f.g)oh = (foh).(goh)

LHS = [(f.g)oh](x)

= (f.g)[h(x)]

= f[h(x)]. g[h(x)]

= (foh)(x). (goh)(x)

= {(foh).(goh)}(x) = RHS

âˆ´ [(f.g)oh](x) = {(foh).(goh)}(x) for all x âˆˆR

Hence, (f.g)oh = (foh).(goh)

âˆ´ gof(x) = g(f(x)) = g(|x|) = |5|x| - 2|

fog(x) = f(g(x)) = f(|5x-2|) = ||5x - 2|| = |5x - 2|

âˆ´ gof(x) = g(f(x)) = g(8x

fog(x) = f(g(x)) = f(x

It is given that f(x)

â‡’ fo f = I_{x}

Hence, the given function f is invertible and the inverse of f is f itself.**Q.5. State with reason whether following functions have inverse(i) f : {1, 2, 3, 4} â†’ {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}(ii) g: {5, 6, 7, 8} â†’ {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}(iii) h: {2, 3, 4, 5} â†’ {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}**

From the given definition of f, we can see that f is a many one function as

f(1) = f(2) = f(3) = f(4) = 10

âˆ´ f is not one â€“ one.

Hence, function f does not have an inverse.

**(ii)** g: {5, 6, 7, 8} â†’ {1, 2, 3, 4} defined as

g = {(5, 4), (6, 3), (7, 4), (8, 2)}

From the given definition of g, it is seen that g is a many one function as

g(5) = g(7) = 4.

âˆ´ g is not one â€“ one.

Hence, function g does not have an inverse.

**(iii)** h: {2, 3, 4, 5} â†’ {7, 9, 11, 13} defined as

h = {(2, 7), (3, 9), (4, 11), (5, 13)}

It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.

âˆ´ Function h is one â€“ one.

Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5}, such that h(x) = y.

Thus, h is a one â€“ one and onto function.

Hence, h has an inverse.**Q.6. Show that f: [âˆ’1, 1] â†’ R, given by f(x) = x/x + 2 is one â€“ one. Find the inverse of the function f: [âˆ’1, 1] â†’ Range f.(Hint: For y âˆˆ Range f, y = **

f : [âˆ’1, 1] â†’ R is given as f(x) = x/x + 2

For one â€“ one

Let f(x) = f(y)

â‡’ x/x + 2 = y/y + 2

â‡’ xy + 2x = xy + 2y

â‡’ 2x = 2y

â‡’ x = y

âˆ´ f is a one â€“ one function.

It is clear that f: [âˆ’1, 1] â†’ Range f is onto.

âˆ´ f: [âˆ’1, 1] â†’ Range f is one â€“ one and onto and therefore, the inverse of the function f: [âˆ’1, 1] â†’ Range f exists.

Let g: Range f â†’ [âˆ’1, 1] be the inverse of f.

Let y be an arbitrary element of range f.

Since f: [âˆ’1, 1] â†’ Range f is onto, we have

y = f(x) for some x âˆˆ [âˆ’1, 1]

â‡’ y = xx + 2

â‡’ xy + 2y = x

â‡’ x(1 - y) = 2y

â‡’ x = 2y/1 - y, yâ‰ 1

Now, let us define g: Range f â†’ [âˆ’1, 1] as

g(y) = 2y/1 - y, y â‰ 1

Now,

âˆ´ gof = x = I[-1, 1] and fog = y = I_{Range f}

âˆ´ f^{-1} = g

â‡’ f^{-1}(y) = 2y/1 - y, y â‰ 1**Q.7. Consider f: R â†’ R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.****Ans.**

f: R â†’ R is given by, f(x) = 4x + 3

For one â€“ one

Let f(x) = f(y)

â‡’ 4x + 3 = 4y + 3

â‡’ 4x = 4y

â‡’ x = y

âˆ´ f is a one â€“ one function.

For onto

For y âˆˆ R, let y = 4x + 3.

â‡’ x = y - 3/4 âˆˆ R

Therefore, for any y âˆˆ R, there exists x = y - 34 âˆˆ R, such that

f(x) = f(y - 3/4) = 4(y - 3/4) + 3 = y.

âˆ´ f is onto.

Thus, f is one â€“ one and onto and therefore, f âˆ’1 exists.

Let us define g: R â†’ R by g(x) = y-3/4

Now, (gof)(x) = g(f(x)) = g(4x + 3) = (4x + 3) - 3/4 = 4x/4 = x

and (fog)(y) = f(g(y)) = f(y - 3/4) = 4(y - 3/4) + 3 = y - 3 + 3 = y

âˆ´ gof = fog = I_{R}

Hence, f is invertible and the inverse of f is given by f - 1(y) = g(y) = y - 34.**Q.8. Consider f: R _{+} â†’ [4, âˆž] given by f(x) = x^{2} + 4. Show that f is invertible with the inverse f^{âˆ’1} of given f by **

f :

For one â€“ one

Let f(x) = f(y)

â‡’ x

â‡’ x

â‡’ x = y [x = y, âˆˆ

âˆ´ f is a one â€“ one function.

For onto For y âˆˆ [4, âˆž), let y = x

â‡’ x

â‡’ x = âˆšy - 4 â‰¥ 0

Therefore, for any y âˆˆ [4, âˆž], there exists x = âˆšy - 4 âˆˆ R

âˆ´ f is onto.

Thus, f is one â€“ one and onto and therefore, f âˆ’1 exists.

Let us define g: [4, âˆž] â†’ R

Now, (gof)(x) = g(f(x)) = g(x

and (fog)(y) = f(g(y)) = f(âˆšy - 4) = (âˆšy - 4)

âˆ´ gof = fog = I

Hence, f is invertible and the inverse of f is given by f - 1(y) = g(y) = âˆšy âˆ’ 4

.

Let y be an arbitrary element of [âˆ’5, âˆž).

Let y = 9x

â‡’ y = (3x + 1)

â‡’ (3x + 1)

âˆ´ f is onto, thereby range f = [âˆ’5, âˆž).

We now have:

(gof) (x) = g (f(x)) = g(9x

= g ((3x + 1)

âˆ´ gof = I

Hence, f is invertible and the inverse of f is given by

**Q.10.**** Let f: X â†’ Y be an invertible function. Show that f has unique inverse.****(Hint: suppose g _{1} and g_{2} are two inverses of f. Then for all y âˆˆ Y, **

Also, suppose f has two inverses (say g

Then, for all y âˆˆ Y, we have:

fog

â‡’ f(g

â‡’ g

â‡’ g

Hence, f has a unique inverse.

f(1) = a, f(2) = b, and f(3) = c

If we define g: {a, b, c} â†’ {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:

(fog)(a) = f (g(a)) = f(1) = a

(fog)(b) = f(g(b)) = f(2) = b

(fog)(c) = f (g(c)) = f(3) = c

And,

(gof)(1) = g(f(1)) = g(a)=l

(gof)(2) = g(f(2)) = g(b) = 2

(gof) (3) = g (f (3)) = g(c) = 3

âˆ´ gof = I

Thus, the inverse of f exists and f

âˆ´ f

f

Let us now find the inverse of f

If we define h: {1, 2, 3} â†’ {a, b, c} as

h(1) = a, h(2) = b, h(3) = c, then we have:

(goh)(1) = g(h(1)) = g(a) = 1

(goh)(2) = g(h(2)) = g(b) = 2

(goh) (3) = g (h(3)) = g(c) = 3

And,

(hog)(a) = h(g(a)) = h(1) = a

(hog)(b) = h(g(b)) = h(2) = b

(hog)(c) = h (g(c)) = h(3) = c

âˆ´ goh = I

Thus, the inverse of g exists and g

It can be noted that h = f.

Hence, (f

Then, there exists a function g: Y â†’ X such that gof = I

Here, f

Now, gof = I

â‡’ f

Hence, f

i.e., (f

(b) x

(c) x

(d) (3 âˆ’ x

f:

f(x) = (3 - x

âˆ´ fof (x) = f (f(x)) = f((3 - x

= [3 - (3 - x

âˆ´ fof (x) = x

It is given that

Let y be an arbitrary element of Range f.

âˆ´

Thus, g is the inverse of f i.e., f^{âˆ’1} = g.

Hence, the inverse of f is the map g:

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