English

Consider F : {1, 2, 3} → {A, B, C} And G : {A, B, C} → {Apple, Ball, Cat} Defined As F (1) = A, F (2) = B, F (3) = C, G (A) = Apple, G (B) = Ball And G (C) = Cat. Show That F, G And Gof

Advertisements
Advertisements

Question

Consider f : {1, 2, 3} → {abc} and g : {abc} → {apple, ball, cat} defined as f (1) = af (2) = bf (3) = cg (a) = apple, g (b) = ball and g (c) =  cat. Show that fg and gof are invertible. Find f−1g−1 and gof−1and show that (gof)−1 = f 1o g−1

Advertisements

Solution

 f = { ( 1, a ) . (2, b) , (c , 3 ) } and g = {(a , apple) , (b , ball) , (c , cat)} Clearly , f and g are bijections.

So, f and g are invertible. 

Now,

 f -1 = {(a ,1) , (b , 2) , (3,c)} and g-1 = {(apple, a ) , (ball ,b), (cat , c)}

So, f-1 o g-1= {apple , 1} , (ball,2), (cat , 3 )} ......... (1)

f : {1,2,3,} → {a,b,c} and g : {a,b,c} → {apple , ball , cat}

So, gof : {1.2.3} → {apple , ball, cat}

⇒ (gof) (1) =g (f(1)) = g (a) = apple

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

and (gof) (3) = g (f(3)) = g (c) cat 

∴ gof = {(1 . apple) ,(2, ball) , (3 , cat)}

Clearly , gof is a bijection.

So, gof is invertible. 

(gof)-1 = {(apple , 1), (ball,2),(cat , 3)}  ....... (2)

Form (1) and (2) , we get :

(gof)-1 = f-1 o g -1

shaalaa.com
  Is there an error in this question or solution?
Chapter 2: Functions - Exercise 2.4 [Page 68]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 2 Functions
Exercise 2.4 | Q 3 | Page 68

RELATED QUESTIONS

Check the injectivity and surjectivity of the following function:

f : R → R given by f(x) = x2


Let fR → be defined as f(x) = 10x + 7. Find the function gR → R such that g o f = f o = 1R.


Show that the function f : R → {x ∈ R : –1 < x < 1} defined by f(x) = `x/(1 + |x|)`, x ∈ R is one-one and onto function.


Let S = {abc} and T = {1, 2, 3}. Find F−1 of the following functions F from S to T, if it exists.

F = {(a, 2), (b, 1), (c, 1)}


Classify the following function as injection, surjection or bijection :

f : R → R, defined by f(x) = 3 − 4x


Find gof and fog when f : R → R and g : R → R is defined by  f(x) = x2 + 8 and g(x) = 3x3 + 1 .


Find gof and fog when f : R → R and g : R → R is defined by  f(x) = x2 + 2x − 3 and  g(x) = 3x − 4 .


Let f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}. Show that gof and fog are both defined. Also, find fog and gof.


Let f be a real function given by f (x)=`sqrt (x-2)`
Find each of the following:

(i) fof
(ii) fofof
(iii) (fofof) (38)
(iv) f2

Also, show that fof ≠ `f^2` .


 If f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x|- x, ∀x∈R" .Then find fog and gof. Hence find fog(–3), fog(5) and gof (–2).


If A = {1, 2, 3} and B = {ab}, write the total number of functions from A to B.


If f : R → R is defined by f(x) = x2, find f−1 (−25).


Let f : R → R+ be defined by f(x) = axa > 0 and a ≠ 1. Write f−1 (x).


Let f : R → Rg : R → R be two functions defined by f(x) = x2 + x + 1 and g(x) = 1 − x2. Write fog (−2).


Write the domain of the real function

`f (x) = 1/(sqrt([x] - x)`.


What is the range of the function

`f (x) = ([x - 1])/(x -1) ?`


The function 

f : A → B defined by 

f (x) = - x2 + 6x - 8 is a bijection if 

 

 

 

 


Let

\[A = \left\{ x : - 1 \leq x \leq 1 \right\} \text{and} f : A \to \text{A such that f}\left( x \right) = x|x|\]

 


\[f : R \to R\] is defined by

\[f\left( x \right) = \frac{e^{x^2} - e^{- x^2}}{e^{x^2 + e^{- x^2}}} is\]

 


\[f : Z \to Z\]  be given by

 ` f (x) = {(x/2, ", if  x is even" ) ,(0 , ", if  x  is  odd "):}`

Then,  f is


Mark the correct alternative in the following question:
Let A = {1, 2, ... , n} and B = {a, b}. Then the number of subjections from A into B is


Let A be a finite set. Then, each injective function from A into itself is not surjective.


If f: R → R is defined by f(x) = x2 – 3x + 2, write f(f (x))


Let X = {1, 2, 3}and Y = {4, 5}. Find whether the following subset of X ×Y are function from X to Y or not

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


Let A = R – {3}, B = R – {1}. Let f: A → B be defined by f(x) = `(x - 2)/(x - 3)` ∀ x ∈ A . Then show that f is bijective.


Let f : R `->` R be a function defined by f(x) = x3 + 4, then f is ______.


The function f: R → R defined as f(x) = x3 is:


An organization conducted a bike race under 2 different categories-boys and girls. Totally there were 250 participants. Among all of them finally, three from Category 1 and two from Category 2 were selected for the final race. Ravi forms two sets B and G with these participants for his college project. Let B = {b1,b2,b3} G={g1,g2} where B represents the set of boys selected and G the set of girls who were selected for the final race.

Ravi decides to explore these sets for various types of relations and functions.

  • Ravi wants to find the number of injective functions from B to G. How many numbers of injective functions are possible?

If f: R → R given by f(x) =(3 − x3)1/3, find f0f(x)


Let f: R → R defined by f(x) = 3x. Choose the correct answer


If `f : R -> R^+  U {0}` be defined by `f(x) = x^2, x ∈ R`. The mapping is


A function f: x → y is said to be one – one (or injective) if:


The domain of the function `cos^-1((2sin^-1(1/(4x^2-1)))/π)` is ______.


The solution set of the inequation log1/3(x2 + x + 1) + 1 > 0 is ______.


ASSERTION (A): The relation f : {1, 2, 3, 4} `rightarrow` {x, y, z, p} defined by f = {(1, x), (2, y), (3, z)} is a bijective function.

REASON (R): The function f : {1, 2, 3} `rightarrow` {x, y, z, p} such that f = {(1, x), (2, y), (3, z)} is one-one.


A function is called bijective if it is both:


A perfect one-to-one matching between users and unique IDs is similar to:


Which condition represents a one-one (injective) function?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×