Advertisements
Advertisements
प्रश्न
A function f from the set of natural numbers to integers defined by
`{([n-1]/2," when n is odd" is ),(-n/2,when n is even ) :}`
विकल्प
neither one-one nor onto
one-one but not onto
onto but not one-one
one-one and onto both
Advertisements
उत्तर
one-one and onto both
Injectivity:
Let x and y be any two elements in the domain (N).
\[\text{Case}-1: \text{Bothxandyare even}.\]
\[\text{Let}f\left( x \right) = f\left( y \right)\]
\[ \Rightarrow \frac{- x}{2} = \frac{- y}{2}\]
\[ \Rightarrow - x = - y\]
\[ \Rightarrow x = y\]
\[\text{Case}-2: \text{Bothxandyare odd}.\]
\[Letf\left( x \right) = f\left( y \right)\]
\[ \Rightarrow \frac{x - 1}{2} = \frac{y - 1}{2}\]
\[ \Rightarrow x - 1 = y - 1\]
\[ \Rightarrow x = y\]
\[Case-3:\text{Let x be even andybe odd}.\]
\[\text{Then},f\left( x \right) = \frac{- x}{2}\text{and}f\left( y \right) = \frac{y - 1}{2}\]
\[\text{Then, clearly}\]
\[x \neq y \]
\[ \Rightarrow f\left( x \right) \neq f\left( y \right)\]
\[\text{From all the cases,f is one-one}.\]
Surjectivity:
\[\text{Co-domain of f} = Z = \left\{ . . . , - 3, - 2, - 1, 0, 1, 2, 3, . . . . \right\}\]
\[\text{Range of f } = \left\{ . . . , \frac{- 3 - 1}{2}, \frac{- \left( - 2 \right)}{2}, \frac{- 1 - 1}{2}, \frac{0}{2}, \frac{1 - 1}{2}, \frac{- 2}{2}, \frac{3 - 1}{2}, . . . \right\}\]
\[ \Rightarrow \text{Range of f} = \left\{ . . . , - 2, 1, - 1, 0, 0, - 1, 1, . . . \right\}\]
\[ \Rightarrow \text{Range of f} = \left\{ . . . , - 2, - 1, 0, 1, 2, . . . . \right\}\]
\[ \Rightarrow \text{Co-domain of f} = \text{Range of f}\]
⇒ f is onto.
APPEARS IN
संबंधित प्रश्न
Check the injectivity and surjectivity of the following function:
f : N → N given by f(x) = x2
Show that the modulus function f : R → R, given by f(x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is –x, if x is negative.
Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = `((x - 2)/(x - 3))`. Is f one-one and onto? Justify your answer.
Let f : R → R be defined as f(x) = x4. Choose the correct answer.
Find the number of all onto functions from the set {1, 2, 3, ..., n} to itself.
Let A = {−1, 0, 1} and f = {(x, x2) : x ∈ A}. Show that f : A → A is neither one-one nor onto.
Classify the following function as injection, surjection or bijection : f : N → N given by f(x) = x3
Classify the following function as injection, surjection or bijection :
f : Z → Z, defined by f(x) = x2 + x
Classify the following function as injection, surjection or bijection :
f : Z → Z, defined by f(x) = x − 5
Classify the following function as injection, surjection or bijection :
f : R → R, defined by f(x) = `x/(x^2 +1)`
Let A = [-1, 1]. Then, discuss whether the following function from A to itself is one-one, onto or bijective : `f (x) = x/2`
If f : R → R be the function defined by f(x) = 4x3 + 7, show that f is a bijection.
If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.
Let f : N → N be defined by
`f(n) = { (n+ 1, if n is odd),( n-1 , if n is even):}`
Show that f is a bijection.
[CBSE 2012, NCERT]
Find gof and fog when f : R → R and g : R → R is defined by f(x) = 2x + 3 and g(x) = x2 + 5 .
Find fog and gof if : f (x) = |x|, g (x) = sin x .
if `f (x) = sqrt(1-x)` and g(x) = `log_e` x are two real functions, then describe functions fog and gof.
State with reason whether the following functions have inverse :
g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
Consider the function f : R+ → [-9 , ∞ ]given by f(x) = 5x2 + 6x - 9. Prove that f is invertible with f -1 (y) = `(sqrt(54 + 5y) -3)/5` [CBSE 2015]
If f : A → A, g : A → A are two bijections, then prove that fog is an injection ?
If f : R → R is given by f(x) = x3, write f−1 (1).
Let C denote the set of all complex numbers. A function f : C → C is defined by f(x) = x3. Write f−1(1).
Let \[f : \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \to R\] be a function defined by f(x) = cos [x]. Write range (f).
Write the domain of the real function
`f (x) = sqrtx - [x] .`
If f(x) = 4 −( x - 7)3 then write f-1 (x).
The function \[f : [0, \infty ) \to \text {R given by } f\left( x \right) = \frac{x}{x + 1} is\]
Let
\[A = \left\{ x : - 1 \leq x \leq 1 \right\} \text{and} f : A \to \text{A such that f}\left( x \right) = x|x|\]
Let
\[f : R - \left\{ n \right\} \to R\]
\[f : Z \to Z\] be given by
` f (x) = {(x/2, ", if x is even" ) ,(0 , ", if x is odd "):}`
Then, f is
Let \[f\left(x\right) = x^3\] be a function with domain {0, 1, 2, 3}. Then domain of \[f^{-1}\] is ______.
Let
\[f : R \to R\] be given by \[f\left( x \right) = x^2 - 3\] Then, \[f^{- 1}\] is given by
Mark the correct alternative in the following question:
If the set A contains 7 elements and the set B contains 10 elements, then the number one-one functions from A to B is
Let A = R − (2) and B = R − (1). If f: A ⟶ B is a function defined by`"f(x)"=("x"-1)/("x"-2),` how that f is one-one and onto. Hence, find f−1.
Let R be the set of real numbers and f: R → R be the function defined by f(x) = 4x + 5. Show that f is invertible and find f–1.
Let A be a finite set. Then, each injective function from A into itself is not surjective.
Given a function If as f(x) = 5x + 4, x ∈ R. If g : R → R is inverse of function ‘f then
Sherlin and Danju are playing Ludo at home during Covid-19. While rolling the dice, Sherlin’s sister Raji observed and noted the possible outcomes of the throw every time belongs to set {1,2,3,4,5,6}. Let A be the set of players while B be the set of all possible outcomes.
A = {S, D}, B = {1,2,3,4,5,6}
- Raji wants to know the number of functions from A to B. How many number of functions are possible?
