If the function f is defined by f ( x ) = { x + 2 , x > 1 2 , - 1 ≤ x ≤ 1 x - 1 , - 3 < x < - 1 , then find the value of f ( 3 ) .

Given that the function f is defined as f ( x ) = { x + 2 , x > 1 2 , - 1 ≤ x ≤ 1 x - 1 , - 3 < x < - 1 The value of f ( x ) when x > 1 is x + 2 So, f ( 3 ) = 3 + 2 = 5 ⇒ f ( 3 ) = 5 ∴ The value of f ( 3 ) is 5 .

If the function f is defined by f ( x ) = { x + 2 , x > 1 2 , - 1 ≤ x ≤ 1 x - 1 , - 3 < x < - 1 , then find the value of f ( 3 ) .

Given that the function f is defined as f ( x ) = { x + 2 , x > 1 2 , - 1 ≤ x ≤ 1 x - 1 , - 3 < x < - 1 The value of f ( x ) when x > 1 is x + 2 So, f ( 3 ) = 3 + 2 = 5 ⇒ f ( 3 ) = 5 ∴ The value of f ( 3 ) is 5 .

Let f x = sin π 6 sin π 2 sin x , for all x ∈ R and g x = π 2 sin x , for all x ∈ R . Let f o g x denote f g x and g o f x denote g f x . Then which of the following is (are) true?

lim x → 0 f ( x ) g ( x ) = π 6

There is an x ∈ R such that g o f x = 1

The domain of the definition of the function f x = 1 4 - x 2 + l o g 10 x 3 - x is:

- 2 , - 1 ∪ - 1 , 0 ∪ 2 , ∞

The domain of definition of f ( x ) = 1 - | x | 2 - | x | is

( - ∞ , - 2 ) ∪ [ - 1 , 1 ] ∪ ( 2 , ∞ )

EASY

Earn 100

If the function $f$ is defined by $f (x) = {\begin{matrix} x + 2, x > 1 \\ 2, - 1 \leq x \leq 1 \\ x - 1, - 3 < x < - 1 \end{matrix}$ , then find the value of $f (3)$ .

Important Questions on Relations and Functions

EASY

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

f

is a continuous real-valued function, then the range of the function is _______

MEDIUM

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

The domain of the function $f : R \to R$ defined by $f (x) = \sqrt{x^{2} - 7 x + 12}$ is

EASY

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

The domain of the function

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

HARD

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

The number of non-constant functions

f

from

X = {0, 1, 2}

Y = {1, 2, 3, 4, 5, 6, 7, 8}

such that

f (i) \leq f (j)

for

i, j \in X

and

i < j

HARD

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

Let

f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))

, for all

x \in R

and

g (x) = \frac{π}{2} \sin x

, for all

x \in R

. Let

(f o g) (x)

denote

f (g (x))

and

(g o f) (x)

denote

g (f (x))

. Then which of the following is (are) true?

HARD

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

[x]

denotes the greatest integer $\leq x$, then the domain of the function

f (x) = \sqrt{\frac{4 - x^{2}}{[x] + 2}}

EASY

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

Let

f (x) = \{\begin{cases} 1 - 5 x if x < - 2 \\ x^{2} - 2 x if - 2 \leq x \leq 1 \\ - 1 + 2 x if x > 1 . \end{cases}

. Then the value of

f (- 1)

is equal to

MEDIUM

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

The domain of

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

EASY

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

The domain of the definition of the function

f (x) = \frac{1}{4 - x^{2}} + {l o g}_{10} (x^{3} - x)

is:

HARD

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

Let

f : (1, 3) \to R

be a function defined by

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

where

[x]

denotes the greatest integer

\leq x .

Then the range of

f

HARD

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

\frac{x - P}{x^{2} - 3 x + 2}

takes all real values for

x \in ℝ,

then the range of

P

HARD

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

The range of the polynomial

P (x) = 4 x^{3} - 3 x

x

varies over the interval

(- \frac{1}{2}, \frac{1}{2})

HARD

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

Let

D = \{x \in ℝ : f (x) = \sqrt{\frac{x - | x |}{x - [x]}}

is defined

}

and

C

be the range of the real function

g (x) = \frac{2 x}{4 + x^{2}} .

Then

D \cap C =

EASY

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

In the following statements -
(i) Relation is a special case of a function.
(ii) Function is a special case of a relation.
(iii) Both relation and function are same.

HARD

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

f : ℝ \to ℝ

is such that

f (x + y) = f (x) + f (y)

for all

x, y \in ℝ, f (1) = 7

and

\sum_{r = 1}^{n} f (r) = 14112,

then

MEDIUM

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

Let

f (x) = \sqrt{x^{2} - 3 x + 2}

and

g (x) = \sqrt{x}

be two given functions. If

S

be the domain of

f o g

and

T

be the domain of

g o f

, then

EASY

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

The domain of definition of

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

EASY

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

Let

f : [2, \infty) \to R

be the function defined by

f (x) = x^{2} - 4 x + 5

, then the range

EASY

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

Let

f : (- 1, 1) \to (- 1, 1)

be continuous,

f (x) = f (x^{2})

for all

x \in (- 1, 1)

and

f (0) = \frac{1}{2}

,

then the value of

4 f (\frac{1}{4})

HARD

Mathematics>Algebra>Relations and Functions>Domain and Range of Functions

Let

f

be a continuous function defined on

[0, 1]

such that

\int_{0}^{1} f^{2} (x) d x = {(\int_{0}^{1} f (x) d x)}^{2}

. Then the range of

f

If the function f is defined by f(x) = {x+2, x>1 2,-1≤x≤1x-1, -3<x<-1, then find the value of f(3).

Important Questions on Relations and Functions

Learn and Prepare for any exam you want !

If the function $f$ is defined by $f (x) = {\begin{matrix} x + 2, x > 1 \\ 2, - 1 \leq x \leq 1 \\ x - 1, - 3 < x < - 1 \end{matrix}$ , then find the value of $f (3)$ .