Mathematics>Differential Calculus>Functions>Inverse Functions

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Let $f (x) = [x - 1] + {\{x\}}^{\{x\}}$ , $x \in (1, 3)$ , then $f^{- 1} (x)$ is

Here, $[\cdot]$ is GIF and $\{\cdot\}$ is the fractional part function.

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Important Questions on Functions

EASY

Mathematics>Differential Calculus>Functions>Inverse Functions

The inverse of

y = 5^{\log x}

is:

EASY

Mathematics>Differential Calculus>Functions>Inverse Functions

If

R

is the set of real numbers and

f : R \to R

is a function defined by

f (x) = \sin x,

then

f^{- 1} ([- 1, 1])

is

HARD

Mathematics>Differential Calculus>Functions>Inverse Functions

If

f : [0, \infty) \to [2, \infty)

is given by

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

, then

f^{- 1} (x)

equals

MEDIUM

Mathematics>Differential Calculus>Functions>Inverse Functions

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

, then the set

\{x : f (x) = f^{- 1} (x)\}

is equal to

MEDIUM

Mathematics>Differential Calculus>Functions>Inverse Functions

If

f : [1, \infty) \to [1, \infty]

is defined by

f (x) = \frac{1 + \sqrt{1 + 4 \log_{2} x}}{2}

, then

f^{- 1} (3) =

EASY

Mathematics>Differential Calculus>Functions>Inverse Functions

If

f : R - \{\frac{3}{5}\} \to R - \{\frac{3}{5}\}; f (x) = \frac{3 x + 1}{5 x - 3},

then ________.

EASY

Mathematics>Differential Calculus>Functions>Inverse Functions

Let

f : N \to Y

(

N

is the set of natural numbers) defined as

f (x) = 4 x + 3

where

Y = {y \in N; y = 4 x + 3

for some

x \in N}

. Then the inverse of

f (x)

is equal to

EASY

Mathematics>Differential Calculus>Functions>Inverse Functions

Let

f : A \to B, g : B \to A

be defined as

f (x) = x^{2} \forall x \in A

and

g (x) = x^{1 / 2} \forall x \in B

.

f (x)

and

g (x)

are inverse functions to each other when

HARD

Mathematics>Differential Calculus>Functions>Inverse Functions

Let

f (x) = (x + 1)^{2} - 1, x \geq - 1 .

Then

\{x ∣ f (x) = f^{- 1} (x)\} =

MEDIUM

Mathematics>Differential Calculus>Functions>Inverse Functions

If

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

are two functions defined by

f (x) = 2 x - 3, g (x) = x^{3} + 5

, then

{(f o g)}^{- 1} (x) =

MEDIUM

Mathematics>Differential Calculus>Functions>Inverse Functions

Let

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

then the solution of the equation

f (x) = f^{- 1} (x)

is

MEDIUM

Mathematics>Differential Calculus>Functions>Inverse Functions

Let

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

be two functions such that

f (x) = 2 x - 3, g (x) = x^{3} + 5

. The function

{(f o g)}^{- 1} (x)

is equal to

EASY

Mathematics>Differential Calculus>Functions>Inverse Functions

If

f (x) = \frac{3 x + 2}{5 x - 3},

then

MEDIUM

Mathematics>Differential Calculus>Functions>Inverse Functions

Let

f : R - {3} \to R - {1}

be defined by

f (x) = \frac{x - 2}{x - 3}

. Let

g : R \to R

be given as

g (x) = 2 x - 3

. Then, the sum of all the values of

x

for which

f^{- 1} (x) + g^{- 1} (x) = \frac{13}{2}

is equal to

MEDIUM

Mathematics>Differential Calculus>Functions>Inverse Functions

Let

f : R - \{\frac{α}{6}\} \to R

be defined by

f (x) = (\frac{5 x + 3}{6 x - α})

. Then the value of

α

for which

(f o f) (x) = x

, for all

x \in R - \{\frac{α}{6}\},

is

MEDIUM

Mathematics>Differential Calculus>Functions>Inverse Functions

Let

f (x) = {(x + 2)}^{2} - 2, x \geq - 2

. Then

f^{- 1} (x) =

MEDIUM

Mathematics>Differential Calculus>Functions>Inverse Functions

If

f : R \to R

and

g : R \to R

are two functions defined by

f (x) = a x + b (a \neq 0) \forall x \in R

and

g (x) = c x^{3} + d (c \neq 0) \forall x \in R

, then

{(f o g)}^{- 1} (x) =

EASY

Mathematics>Differential Calculus>Functions>Inverse Functions

Let $f : X \to Y$ and $A, B$ are non-void subsets of $Y,$ then (where the symbols have their usual interpretation)

EASY

Mathematics>Differential Calculus>Functions>Inverse Functions

The inverse of the function

f (x) = \log_{a} (x + \sqrt{x^{2} + 1})

(where,

a < 0, a \neq 1

) is

MEDIUM

Mathematics>Differential Calculus>Functions>Inverse Functions

The inverse function of

f (x) = \frac{8^{2 x} - 8^{- 2 x}}{8^{2 x} + 8^{- 2 x}}, x \in (- 1, 1),

is __________.