Relations and Functions

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New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

The binary operation? on the set {1, 2, 3, 4, 5} is defined as a? b = min {a, b} &mn For E; a, b? {1, 2, 3, 4, 5}.

Thus, the operation table for the given operation? can be given as:

?

1

2

3

4

5

1

1

1

1

1

1

2

1

2

2

2

2

3

1

2

3

3

3

4

1

2

3

4

4

5

1

2

3

4

5

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

 (i) On Z, * is defined by a * b = a − b.

It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1.

∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z

Hence, the operation * is not commutative.

Also we have:

(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4

1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z

Hence, the operation * is not associative.

(ii) On Q, * is defined by a * b = ab + 1.

It is known that:

ab = ba & mn For E; a, b ∈ Q

⇒ ab + 1 = ba&nb

...more

New answer posted

5 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

(i) On Z+, * is defined by a * b = a − b.

It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2= −1 ∉ Z+.

(ii) On Z+, * is defined by a * b = ab.

It is seen that for each a, b ∈ Z+, there is a unique element ab in Z+.

This means that * carries each pair (a, b) to a unique element a * b = ab in Z+.

Therefore, * is a binary operation.

(iii) On R, * is defined by a * b = ab2.

It is seen that for each a, b ∈ R, there is a unique element ab2 in R. 

This means that * carries each pair (a, b) to a unique element a * b = ab2 in R.

Therefore, * is a binary operation.

(iv) On Z+, * is defined by a * b = |a −

...more

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

It is given that f:R{43}R is defined as f(x)=4x3x4

Let y be an arbitrary element of Range f.

Then, there exists xR{43} such that y=f(x)

y=4x3x+4

3xy+4y=4x let us define g: Range fR{43}as,g(y)=4y43y

x(43y)=4yNow,(gof)(x)=g(f(x))=g(4x3x+4)x=4y43y=4(4x3x+4)43(4x3x+4)=16x12x+1612x=16x16=xAnd,(fog)(y)=f(g(y))=f(4y43y)=4(4y43y)3(4y43y)+4=16y12y+1612y=16y16=ygof=IR(43)and,fog=I_"Range,f"

Thus, g is the inverse of f i.e., f1=g.

Hence, the inverse of f is the map g: Range fR{43} which is given by

g(y)=4y43y

The correct answer is B.

New answer posted

5 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

f:RR is given as f (x)= (3x3) (1/3)

f (x)= (3x3)13fof (x)=f (f (x))=f ( (3x3)13)= [3 ( (3x3)13)3]13= [3 (3x3)]13= (x3)13=xfof (x)=x

The correct answer is C.

New answer posted

5 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Let f: X → Y be an invertible function. 

Then, there exists a function g: Y → X such that gof = IX and fog = IY. 

Here, f−1 = g. 

Now, gof = IX and fog = IY

⇒ f−1 of = IX and fof−1 = IY

Hence, f−1 : Y → X is invertible and f is the inverse of f−1 

i.e., (f−1)−1  = f.

New answer posted

5 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

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

If we define g:{a,b,c}{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)=cAnd(gof)(1)=g(f(1))=f(a)=1(gof)(2)=g(f(2))=f(b)=2(gof)(3)=g(f(3))=f(c)=3gof=IXand,fog=IYWhere,X={1,2,3},and,Y={a,b,c}.

Thus, the inverse of f exists and f1=g.

f1:{a,b,c}{1,2,3} is given by,

f1(a)=1,f1(b)=2,f1(c)=3

Let us now find the inverse of f1 i.e., find the inverse of g.

If we define h:{1,2,3}{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)=3And(hog)(a)=h(g(a))=h(1)=a(hog)(b)=h(g(b))=h(2)=b(hog)(c)=h(g(c))=h(3)=cgoh=IXand,hog=IYWhere,X={1,2,3},and,Y={a,b,c}.

Thus, the inverse of g exists and g1=h(f1)1=h.

It can be noted that h=f.

Hence, (f1)1=f.

New answer posted

5 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Let f:XY be an invertible function.

Also, suppose f has two inverses (say g1 and g2 ).

Then, for all y ∈ Y, we have:

fog1 (y)=Iy (y)=fog2 (y)f (g1 (y))=f (g2 (y))  [f is invertible => f is one-one]

g1=g2  [g is one-one]

Hence, f has a unique inverse.

New answer posted

5 months ago

0 Follower 19 Views

V
Vishal Baghel

Contributor-Level 10

f:R+ [4, ) is given as f (x)=x2+4 .

One-one:

Let, f (x)=f (y).x2+4=y2+4x2=y2x=y [as, x=yR]

f is a one-one function.

New answer posted

5 months ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

f:RR is given by,

f(x)=4x+3Oneone:Let,f(x)=f(y).4x+3=4y+34x=4yx=y

 f is a one-one function.

Onto:

For,yR,let,y=4x+3.x=y34R

Therefore, for any yR , there exists x=y34R such that

f(x)=f(y34)=4(y34)+3=y

 f is onto.

Thus, f is one-one and onto and therefore, f1 exists.

Let us define g:RR by g(x)=y34

Now,(gof)(x)=g(f(x))=g(4x+3)=(4x+3)34=x(fog)(y)=f(g(y))=f(y34)=4(y34)+3=y3+3=ygof=fog=IR

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

f1=g(y)=y34

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