Let a functoin f : ( 0 , ∞ ) → [ 0 , ∞ ) be defined by f ( x ) = 1 - 1 x . Then f is: Note: In actual JEE paper co-domain was wrong. We have corrected it.

Not injective but it is surjective

Neither injective not surjective

Both injective as well as surjective

Let a functoin f : ( 0 , ∞ ) → [ 0 , ∞ ) be defined by f ( x ) = 1 - 1 x . Then f is: Note: In actual JEE paper co-domain was wrong. We have corrected it.

Not injective but it is surjective

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

Neither injective nor surjective

If [ x ] denotes the greatest integer ≤ x , then the system of linear equations [ s i n θ ] x + [ - c o s θ ] y = 0 , [ c o t θ ] x + y = 0

have infinitely many solutions if θ ∈ π 2 , 2 π 3 and has a unique solution if θ ∈ π , 7 π 6

has a unique solution if θ ∈ π 2 , 2 π 3 ∪ π , 7 π 6

have infinitely many solution if θ ∈ π 2 , 2 π 3 ∪ π , 7 π 6

has a unique if θ ∈ π 2 , 2 π 3 and have infinitely many solutions if θ ∈ π , 7 π 6

Show that the function f : N → N defined by f n = n + 1 is not onto, by drawing arrow diagrams.

Given, the function f : N → N defined by f n = n + 1 . N = 1 , 2 , 3 , 4 , 5 , 6 , . . . . . . . f 1 = 1 + 1 = 2 f 2 = 2 + 1 = 3 f 3 = 3 + 1 = 4 f 4 = 4 + 1 = 5 f 5 = 5 + 1 = 6 Thus, f = 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , . . . . . . . . . . . . Different elements have different images. Here, the range is not equal to the codomain. Therefore, the given function is not an onto function.

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

Let t denote the greatest integer ≤ t . Then the equation in x , x 2 + 2 x + 2 - 7 = 0 has :

exactly four integral solutions

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

Let a functoin $f : (0, \infty) \to [0, \infty)$ be defined by $f (x) = |1 - \frac{1}{x}|$ . Then $f$ is:

Note: In actual JEE paper co-domain was wrong. We have corrected it.

50% studentsanswered this correctly

Important Questions on Functions

HARD

Mathematics>Differential Calculus>Functions>Types of Functions

For a real number

r

let

[r]

denote the largest integer less than or equal to

r .

Let

a > 1

a real number which is not an integer and

k

be the smallest positive integer such that

[a^{k}] > {[a]}^{k}

Then which of the following statement is always true?

MEDIUM

Mathematics>Differential Calculus>Functions>Types of Functions

The function

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

defined as

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

is:

HARD

Mathematics>Differential Calculus>Functions>Types of Functions

For a real number

r

we denote by

[r]

the largest integer less than or equal to

r

. If

x, y

are real numbers with

x, y \geq 1

then which of the following statements is always true?

HARD

Mathematics>Differential Calculus>Functions>Types of Functions

[x]

denotes the greatest integer

\leq x,

then the system of linear equations

[s i n θ] x + [- c o s θ] y = 0

[c o t θ] x + y = 0

MEDIUM

Mathematics>Differential Calculus>Functions>Types of Functions

Show that the function

f : N \to N

defined by

f (n) = n + 1

is not onto, by drawing arrow diagrams.

MEDIUM

Mathematics>Differential Calculus>Functions>Types of Functions

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

HARD

Mathematics>Differential Calculus>Functions>Types of Functions

The function

f : N \to I

defined by

f (x) = x - 5 [\frac{x}{5}]

, where

N

is the set of natural numbers and

[x]

denotes the greatest integer less than or equal to

x

, is:

EASY

Mathematics>Differential Calculus>Functions>Types of Functions

| 3 x - 5 | \leq 2

then

EASY

Mathematics>Differential Calculus>Functions>Types of Functions

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:

HARD

Mathematics>Differential Calculus>Functions>Types of Functions

a \in R

and the equation

- 3 (x - [x])^{2} + 2 (x - [x]) + a^{2} = 0

(where

[x]

denotes the greatest integer

\leq

x

) has no integral solution, then all possible values of

a

lie in the interval

HARD

Mathematics>Differential Calculus>Functions>Types of Functions

For

x \in R,

Let

[x]

denotes the greatest integer

\leq x,

then the sum of the series

[- \frac{1}{3}] + [- \frac{1}{3} - \frac{1}{100}] + [- \frac{1}{3} - \frac{2}{100}] + . . . . . + [- \frac{1}{3} - \frac{99}{100}]

EASY

Mathematics>Differential Calculus>Functions>Types of Functions

Which of the following function from

z

into

z

is bijection? (

z

is set of integers)

MEDIUM

Mathematics>Differential Calculus>Functions>Types of Functions

Let

f

and

g

be differentiable functions on

R

such that

f o g

is the identity function. If for some

a, b \in R, g^{'} (a) = 5

and

g (a) = b,

then

f^{'} (b)

is equal to:

HARD

Mathematics>Differential Calculus>Functions>Types of Functions

Let

f : R \to R

be defined by

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

then

f

MEDIUM

Mathematics>Differential Calculus>Functions>Types of Functions

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>Types of Functions

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

MEDIUM

Mathematics>Differential Calculus>Functions>Types of Functions

Let

[t]

denote the greatest integer

\leq t

. Then the equation in

x, {[x]}^{2} + 2 [x + 2] - 7 = 0

has :

HARD

Mathematics>Differential Calculus>Functions>Types of Functions

Let $R$ be the set of real numbers and $f : R \to R$ be defined by $f (x) = \frac{\{x\}}{1 + {[x]}^{2}},$ where $[x]$ is the greatest integer less than or equal to $x,$ and $\{x\} = x - [x] .$ Which of the following statement are true?

I. The range of $f$ is closed interval
II. $f$ is continuous on $R$
III. $f$ is one-one on $R$

MEDIUM

Mathematics>Differential Calculus>Functions>Types of Functions

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>Types of Functions

Let

f : R \to R

be defined as

f (x) = 3 x .

Then

Let a functoin f:(0,∞)→[0, ∞) be defined by f(x)=1-1x . Then f is: Note: In actual JEE paper co-domain was wrong. We have corrected it.

Important Questions on Functions

Learn and Prepare for any exam you want !

Let a functoin $f : (0, \infty) \to [0, \infty)$ be defined by $f (x) = |1 - \frac{1}{x}|$ . Then $f$ is:

Note: In actual JEE paper co-domain was wrong. We have corrected it.