If f : A → B and g : B → C are two functions such that g o f : A → C is a bijection, then which one of the following is always true?

f is an injection and g is a surjection

f is a surjection and g is an injection

f is a bijection but g is not a bijection

EASY

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If $f (x) = \frac{1}{\sqrt{- x}},$ then domain of $f o f (x)$ is

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Important Questions on Relations and Functions

HARD

Mathematics>Algebra>Relations and Functions>Composition of Functions

For a suitably chosen real constant

a

, let a function,

f : R - \{- a\} \to R

be defined by

f (x) = \frac{a - x}{a + x}

. Further supposed that for any real number

x \neq - a,

and

f (x) \neq - a, (fof) (x) = x

. Then

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

is equal to :

HARD

Mathematics>Algebra>Relations and Functions>Composition of Functions

g (x) = x^{2} + x - 1

and

(g o f) (x) = 4 x^{2} - 10 x + 5,

then

f (\frac{5}{4})

is equal to

MEDIUM

Mathematics>Algebra>Relations and Functions>Composition of Functions

Let $f (x) = - 2 x^{2} + 1$ and $g (x) = 4 x - 3$ , then $(g o f) (- 1)$ is equal to

HARD

Mathematics>Algebra>Relations and Functions>Composition of Functions

Let

f : R \to R, g : R \to R

be differentiable functions such that

(f o g) (x) = x

. If

f (x) = 2 x + \cos x + \sin^{2} x

, then the value of

\sum_{n = 1}^{99} g (1 + (2 n - 1) π)

MEDIUM

Mathematics>Algebra>Relations and Functions>Composition of Functions

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

then

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

is equal to

HARD

Mathematics>Algebra>Relations and Functions>Composition of Functions

f : R \to R

is defined as

f (x) = {(2020 - x^{2019})}^{1 / 2019}, \forall x \in R,

find

(f \circ f \circ f \circ f) (\frac{2019}{2020})

MEDIUM

Mathematics>Algebra>Relations and Functions>Composition of Functions

f : R \to R

and

g : [0, \infty) \to R

is defined by

f (x) = x^{2}

and

g (x) = \sqrt{x} .

Which one of the following is not true?

MEDIUM

Mathematics>Algebra>Relations and Functions>Composition of Functions

Let

f, g : R \to R

be two functions defined as

f (x) = |x| + x

and

g (x) = |x| - x \forall x \in R

. Then

(f o g) (x)

for

x < 0

EASY

Mathematics>Algebra>Relations and Functions>Composition of Functions

For

x ϵ (0, \frac{3}{2}),

let

f (x) = \sqrt{x}, g (x) = t a n x

and

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

. If

ϕ (x) = (h o f) o g) (x),

then

ϕ (\frac{π}{3})

is equal to:

EASY

Mathematics>Algebra>Relations and Functions>Composition of Functions

Let

f (x) = x^{2}, x \in R

. For any

A \subseteq R,

define

g (A) = {x \in R : f (x) \in A}

. If

S = [0, 4]

, then which one of the following statements is not true?

MEDIUM

Mathematics>Algebra>Relations and Functions>Composition of Functions

f : A \to B

and

g : B \to C

are two functions such that

g o f : A \to C

is a bijection, then which one of the following is always true?

HARD

Mathematics>Algebra>Relations and Functions>Composition of Functions

Let

f (x) = \sqrt{2 - x - x^{2}} a n d g (x) = \cos x

. Which of the following statements are true?

(I) Domain of

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

Domain of

f (g (x))

(II) Domain of

f (g (x)) + g (f (x)) =

Domain of

g (f (x))

(III) Domain of

f (g (x)) =

Domain of

f (g (x))

(IV) Domain of

g ({(f (x))}^{3}) =

Domain of

f (g (x))

EASY

Mathematics>Algebra>Relations and Functions>Composition of Functions

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

and the composite function

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

, then the function

g (x)

is equal to

MEDIUM

Mathematics>Algebra>Relations and Functions>Composition of Functions

Let

f (x) = \{\begin{cases} | x - 3 |, & if x < - 1 \\ 3 x + 4, & if x ⩾ - 1' \end{cases}

g (x) = x^{2} - b x - 2 (x \in R)

and

b

is a real constant. If

g

of is continuous at

x = - 1

, then

b

is equal to______.

HARD

Mathematics>Algebra>Relations and Functions>Composition of Functions

g (x) = x^{2} + x - 2

and

\frac{1}{2} (g o f) x = 2 x^{2} - 5 x + 2,

then

f (x)

is equal to

EASY

Mathematics>Algebra>Relations and Functions>Composition of Functions

f (x) = e^{x}

and

g (x) = \log_{e} e^{x}

, then which of the following is True?

EASY

Mathematics>Algebra>Relations and Functions>Composition of Functions

Let

f (x) = x^{2}

and

g (x) = 2^{x}

then the solution set of

fog (x) = gof (x)

EASY

Mathematics>Algebra>Relations and Functions>Composition of Functions

f (x) = 3 x + 6, g (x) = 4 x + k

and

f o g (x) = g o f (x)

then

k =

HARD

Mathematics>Algebra>Relations and Functions>Composition of Functions

For

x \in R, x \neq 0, x \neq 1,

let

f_{0} (x) = \frac{1}{1 - x}

and

f_{n + 1} (x) = f_{0} (f_{n} (x)), n = 0, 1, 2, \dots . .

Then the value of

f_{100} (3) + f_{1} (\frac{2}{3}) + f_{2} (\frac{3}{2})

is equal to :

HARD

Mathematics>Algebra>Relations and Functions>Composition of Functions

Let

N

be the set of natural numbers and two functions

f

and

g

be defined as

f, g : N \to N

such that

f (n) = \{\begin{matrix} \frac{n + 1}{2}, i f n i s o d d \\ \frac{n}{2}, i f n i s e v e n \end{matrix}

and

g (n) = n - {(- 1)}^{n} .

Then

f o g

is:

If fx=1-x, then domain of fof(x) is

Important Questions on Relations and Functions

Learn and Prepare for any exam you want !

If $f (x) = \frac{1}{\sqrt{- x}},$ then domain of $f o f (x)$ is