Advertisements
Advertisements
प्रश्न
\[f : Z \to Z\] be given by
` f (x) = {(x/2, ", if x is even" ) ,(0 , ", if x is odd "):}`
Then, f is
पर्याय
onto but not one-one
one-one but not onto
one-one and onto
neither one-one nor onto
Advertisements
उत्तर
Injectivity:
Let x and y be two elements in the domain (Z), such that
LaTeX
\[f\left( x \right) = f\left( y \right)\]
\[Case-1: \text{Let both x andybe even}.\]
\[\text{Then},\]
\[f\left( x \right) = f\left( y \right)\]
\[ \Rightarrow \frac{x}{2} = \frac{y}{2}\]
\[ \Rightarrow x = y\]
\[Case-2: \text{Let bothx andybe odd}.\]
\[\text{Then},\]
\[f\left( x \right) = f\left( y \right)\]
\[ \Rightarrow 0 = 0\]
\[\text{Here, we cannot determine whether } x = y.\] So, f is not one-one.
Surjectivity:
Let y be an element in the co-domain (Z), such that
\[\text{Co-domain of f} = Z = \left\{ 0, \pm 1, \pm 2, \pm 3, \pm 4, . . . \right\} \]
\[\text{Range of f} = \left\{ 0, 0, \frac{\pm 2}{2}, 0, \frac{\pm 4}{2} , . . . \right\} = \left\{ 0, \pm 1, \pm 2, . . . \right\} \]
\[ \Rightarrow \text{Co-domain of f} = \text{Range of f}\]
\[\Rightarrow\] f is onto.
So, the answer is (a).
APPEARS IN
संबंधित प्रश्न
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
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.
Give examples of two functions f: N → Z and g: Z → Z such that g o f is injective but gis not injective.
(Hint: Consider f(x) = x and g(x) =|x|)
Given examples of two functions f: N → N and g: N → N such that gof is onto but f is not onto.
(Hint: Consider f(x) = x + 1 and `g(x) = {(x-1, ifx >1),(1, if x = 1):}`
Find the number of all onto functions from the set {1, 2, 3, ..., n} to itself.
Let f: R → R be the Signum Function defined as
f(x) = `{(1,x>0), (0, x =0),(-1, x< 0):}`
and g: R → R be the Greatest Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0, 1]?
Show that the function f: ℝ → ℝ defined by f(x) = `x/(x^2 + 1), ∀x in R`is neither one-one nor onto. Also, if g: ℝ → ℝ is defined as g(x) = 2x - 1. Find fog(x)
Give an example of a function which is not one-one but onto ?
Classify the following function as injection, surjection or bijection :
f : R → R, defined by f(x) = |x|
Classify the following function as injection, surjection or bijection :
f : R → R, defined by f(x) = 3 − 4x
Let A = {1, 2, 3}. Write all one-one from A to itself.
If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.
Find the number of all onto functions from the set A = {1, 2, 3, ..., n} to itself.
Find gof and fog when f : R → R and g : R → R is defined by f(x) = x and g(x) = |x| .
Let f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.
Let f : R → R and g : R → R be defined by f(x) = x + 1 and g (x) = x − 1. Show that fog = gof = IR.
Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and h(z) = sin z for all x, y, z ∈ N. Show that ho (gof) = (hog) of.
State with reason whether the following functions have inverse :
f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
If f : A → A, g : A → A are two bijections, then prove that fog is an injection ?
If f : A → A, g : A → A are two bijections, then prove that fog is a surjection ?
Which one the following relations on A = {1, 2, 3} is a function?
f = {(1, 3), (2, 3), (3, 2)}, g = {(1, 2), (1, 3), (3, 1)} [NCERT EXEMPLAR]
Let
\[A = \left\{ x \in R : - 1 \leq x \leq 1 \right\} = B\] Then, the mapping\[f : A \to \text{B given by} f\left( x \right) = x\left| x \right|\] is
Let M be the set of all 2 × 2 matrices with entries from the set R of real numbers. Then, the function f : M→ R defined by f(A) = |A| for every A ∈ M, is
Which of the following functions form Z to itself are bijections?
If a function\[f : [2, \infty )\text{ to B defined by f}\left( x \right) = x^2 - 4x + 5\] is a bijection, then B =
The function
\[f : R \to R\] defined by\[f\left( x \right) = \left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)\]
(a) one-one but not onto
(b) onto but not one-one
(c) both one and onto
(d) neither one-one nor onto
The inverse of the function
\[f : R \to \left\{ x \in R : x < 1 \right\}\] given by
\[f\left( x \right) = \frac{e^x - e^{- x}}{e^x + e^{- x}}\] is
If \[g\left( x \right) = x^2 + x - 2\text{ and} \frac{1}{2} gof\left( x \right) = 2 x^2 - 5x + 2\] is equal to
If \[f\left( x \right) = \sin^2 x\] and the composite function \[g\left( f\left( x \right) \right) = \left| \sin x \right|\] then g(x) is equal to
If \[f : R \to R\] is given by \[f\left( x \right) = x^3 + 3, \text{then} f^{- 1} \left( x \right)\] is equal to
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 the function f: R → R be defined by f(x) = 4x – 1, ∀ x ∈ R. Then, show that f is one-one.
For sets A, B and C, let f: A → B, g: B → C be functions such that g o f is surjective. Then g is surjective.
Which of the following functions from Z into Z are bijections?
The function f : A → B defined by f(x) = 4x + 7, x ∈ R 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 know among those relations, how many functions can be formed from B to G?
Let the function f: R → R be defined by f(x) = 4x – 1, ∀ x ∈ R then 'f' is
If f : R `rightarrow` R is defined by `f(x) = (2x - 7)/4`, show that f(x) is one-one and onto.
A function is called many-one if two or more different elements of the domain have:
In the function from \(X\) to \(Y\), if \(6\in Y\) is not the image of any element of \(X\), the function is:
