Consider the identity function I N : N → N defined as I N x = x ∀ x ∈ N . Show that although I N is onto but I N + I N : N → N defined as I N + I N x = I N ( x ) + I N ( x ) = x + x = 2 x is not onto.

Clearly I N is onto. But I N + I N is not onto, as we can find an element 3 in the co-domain N such that there does not exist any x in the domain N with I N + I N x = 2 x = 3 .

Consider the identity function I N : N → N defined as I N x = x ∀ x ∈ N . Show that although I N is onto but I N + I N : N → N defined as I N + I N x = I N ( x ) + I N ( x ) = x + x = 2 x is not onto.

Clearly I N is onto. But I N + I N is not onto, as we can find an element 3 in the co-domain N such that there does not exist any x in the domain N with I N + I N x = 2 x = 3 .

Let A = { x ∈ R : x is not a positive integer} . Define a function f : A → R as f x = 2 x x - 1 , then f is:

Neither injective nor surjective

The function f : R → - 1 2 , 1 2 defined as f x = x 1 + x 2 , is:

Neither injective nor surjective

Let f : R → R be defined by f ( x ) = x 2 - x 2 1 + x 2 for all x ∈ R . Then,

f is one-one but not onto mapping

f is onto but not one-one mapping

Let S , T , U be three non-void sets and f : S → T , g : T → U be so that g o f : S → U is surjective. Then,

g is surjective, f may not be so

f is surjective, g may not be so

f and g both may not be surjective

Show that the function f : R → R defined by f x = x x 2 + 1 , ∀ x ∈ ℝ is neither one-one nor onto. Also, if g : R → R is defined as g ( x ) = 2 x - 1 , find fog ( x ) .

Given, y = x x 2 + 1 ⇒ y x 2 − x + y = 0 Here a = y , b = - 1 and c = y ∴ x = − ( − 1 ) ± 1 − 4 y 2 2 y Clearly, for every value of y , x will have two different values . So the function is many-one not one - one. Since 1 − 4 y 2 ≥ 0 ⇒ ( 1 + 2 y ) ( 1 − 2 y ) ≥ 0 ⇒ − 1 2 ≤ y ≤ 1 2 That means no matter what is x , y belongs to the interval − 1 2 , 1 2 . Hence, the function is not onto. Now, f o g x = f g x = f 2 x - 1 = 2 x - 1 2 x - 1 2 + 1 = 2 x + 1 4 x 2 - 4 x + 1 + 1 = 2 x + 1 2 2 x 2 - 2 x + 1

MEDIUM

Earn 100

Consider the identity function $I_{N} : N \to N$ defined as $I_{N} (x) = x \forall x \in N$ . Show that although $I_{N}$ is onto but $I_{N} + I_{N} : N \to N$ defined as
$(I_{N} + I_{N}) (x) = I_{N} (x) + I_{N} (x) = x + x = 2 x$ is not onto.

Important Questions on Functions

MEDIUM

Mathematics>Differential Calculus>Functions>Function and its Types

Let

A = {a, b, c}

and

B = {1, 2, 3, 4} .

Then the number of elements in the set

C = {f : A \to B ∣ 2 \in f (A)

and

f

is not one-one

}

\dots

HARD

Mathematics>Differential Calculus>Functions>Function and its Types

Let

f : (- \frac{π}{2}, \frac{π}{2}) \to R

be given by

f (x) = {(\log (\sec x + \tan x))}^{3} .

Then

MEDIUM

Mathematics>Differential Calculus>Functions>Function and its Types

A = {x | x \in N, x \leq 5}, B = \{x | x \in Z, x^{2} - 5 x + 6 = 0\},

then the number of onto functions from

A

B

MEDIUM

Mathematics>Differential Calculus>Functions>Function and its Types

Let

A = [- 1, 1]

and

f : A \to A

be defined as

f (x) = x |x|

for all

x \in A

, then

f (x)

EASY

Mathematics>Differential Calculus>Functions>Function and its Types

Let

A =

{

x \in R : x

is not a positive integer}

.

Define a function

f : A \to R

f (x) = \frac{2 x}{x - 1}

, then

f

is:

EASY

Mathematics>Differential Calculus>Functions>Function and its Types

A = {1, 2, 3, 4},

then the number of functions on the set

A,

which are not one - one, is

MEDIUM

Mathematics>Differential Calculus>Functions>Function and its Types

If the function

f : R - \{1, - 1\} \to A

defined by

f (x) = \frac{x^{2}}{1 - x^{2}},

is surjective, then

A

is equal to

MEDIUM

Mathematics>Differential Calculus>Functions>Function and its Types

Let

f : [0, 1] \to [- 1, 1]

and

g : [- 1, 1] \to [0, 2]

be two functions such that

g

is injective and

g o f : [0, 1] \to [0, 2]

is surjective. Then,

HARD

Mathematics>Differential Calculus>Functions>Function and its Types

Let

X

be a set with exactly

5

elements and

Y

be a set with exactly

7

elements. If

α

is the number of one-one functions from

X \to Y

and

β

is the number of onto functions from

Y \to X

, then the value of

\frac{1}{5!} (β - α)

is _____-.

MEDIUM

Mathematics>Differential Calculus>Functions>Function and its Types

The function

f : R \to [- \frac{1}{2}, \frac{1}{2}]

defined as

f (x) = \frac{x}{1 + x^{2}},

is:

MEDIUM

Mathematics>Differential Calculus>Functions>Function and its Types

Let

A

and

B

be finite sets and

P_{A}

and

P_{B}

, respectively denote their power sets. If

P_{B}

has

112

elements more than those in

P_{A}

, then the number of functions from

A

B

which are injective is

EASY

Mathematics>Differential Calculus>Functions>Function and its Types

Which of the following function from

z

into

z

is bijection? (

z

is set of integers)

MEDIUM

Mathematics>Differential Calculus>Functions>Function and its Types

Let

f : R \to R

be defined by

f (x) = x^{2} - \frac{x^{2}}{1 + x^{2}}

for all

x \in R .

Then,

HARD

Mathematics>Differential Calculus>Functions>Function and its Types

If the function

f : R \to R

is defined by

f (x) = |x| (x - \sin x)

, then which of the following statements is TRUE?

HARD

Mathematics>Differential Calculus>Functions>Function and its Types

Let

S, T, U

be three non-void sets and

f : S \to T, g : T \to U

be so that

g o f : S \to U

is surjective. Then,

MEDIUM

Mathematics>Differential Calculus>Functions>Function and its Types

Let

f : [0, 1] \to R

be an injective continuous function that satisifes the condition

- 1 < f (0) < f (1) < 1

Then, the number of functions

g : [- 1, 1] \to [0, 1]

such that

(g o f) (x) = x

for all

x \in [0, 1]

MEDIUM

Mathematics>Differential Calculus>Functions>Function and its Types

Let $f : R \to R$ be a continuous function such that $f (x^{2}) = f (x^{3})$ for all $x \in R$ . Consider the following statements.

I. $f$ is an odd function.

II. $f$ is an even function.

III. $f$ is differentiable everywhere.

Then,

HARD

Mathematics>Differential Calculus>Functions>Function and its Types

Let

f : R \to R

be defined by

f (x) = \frac{|x| - 1}{|x| + 1},

then

f

MEDIUM

Mathematics>Differential Calculus>Functions>Function and its Types

Show that the function

f : R \to R

defined by

f (x) = \frac{x}{x^{2} + 1}, \forall x \in ℝ

is neither one-one nor onto. Also, if

g : R \to R

is defined as

g (x) = 2 x - 1

, find

fog (x)

EASY

Mathematics>Differential Calculus>Functions>Function and its Types

Let

f : R \to R

be defined as

f (x) = 3 x .

Then

Consider the identity function IN:N→N defined as INx=x∀x∈N. Show that although IN is onto but IN+IN:N→N defined as IN+INx=IN(x)+IN(x)=x+x=2x is not onto.

Important Questions on Functions

Learn and Prepare for any exam you want !

Consider the identity function $I_{N} : N \to N$ defined as $I_{N} (x) = x \forall x \in N$ . Show that although $I_{N}$ is onto but $I_{N} + I_{N} : N \to N$ defined as
$(I_{N} + I_{N}) (x) = I_{N} (x) + I_{N} (x) = x + x = 2 x$ is not onto.