QBManager

Sets, Relation and Function

117423 If the range of the function $f (x) = - 3 x - 3$ is ${3$ , $- 6, - 9, - 18}$ , then which of the following elements is not in the domain of $f$ ?

1 -1

2 -2

3 1

4 2

Explanation:

A Given that,
$f (x) = - 3 x - 3$
And, range of function $= {3, - 6, - 9, - 18}$
Now, check all the number from the options put $x = - 1$
$f (- 1) = - 3 \times - 1 - 3$
$= 3 - 3 = 0$
Thus, -1 can not belong to domain of $f (x)$
Put, $x = - 2$
$f (- 2) = - 3 \times - 2 - 3$
$= + 6 - 3$
$= 3$
Put, $x = 1$
$f (1) = - 3 \times 1 - 3$
$= - 6$
Put $x = 2$
$f (2) = - 3 \times 2 - 3$
$= - 6 - 3$
$= - 9$ Hence, -1 can not be in the domain of $f (x)$ -

TS EAMCET-2017

Sets, Relation and Function

117424 $f : (- \infty, 0] \to [0, \infty)$ is defind as $f (x) = x^{2}$ . The domain and range of its inverse is

1 Domain

(f^{- 1}) = [0, \infty)

, Range of

(f^{- 1}) = (- \infty

,

0]

2 Domain of

(f^{- 1}) = [0, \infty)

Range of

(f^{- 1}) = (- \infty

,

\infty)

3 Domain of

(f^{- 1}) = [0, \infty)

Range of

(f^{- 1}) = (0

,

\infty)

4

f^{- 1}

does not exist

Explanation:

A We have a function : $(- \infty, 0] \to [0, \infty)$ Defined as $f (x) = x^{2}$
The graph of above function is.
original image
Since, each line parallel to $x$ - axis cuts that above curve at maximum one paint, therefore $f$ is one- one . Also from the graph, it is clear that range.
$f = [0, \infty)$
$∴ f$ is onto.
Thus, $f$ is invertible.
Hence, $f^{- 1} : [0, \infty) \to (- \infty, 0]$
Domain $(f^{- 1}) = [0, \infty)$ and range of
$(f^{- 1}) = (- \infty 0]$

TS EAMCET-2017

Sets, Relation and Function

117425 Let the sets $A$ and $B$ denote the domain and range respectively of the function
$f (x) = \frac{1}{\sqrt{[x] - x}}$ where $[x]$ denotes the smallest integer greater than or equal to $x$ . Then among the statements
$(S_{1}) : A \cap B = (1, \infty) - N$ and
$(S_{2}) : A \cup B = (1, \infty)$

1 only

(S_{1})

is true

2 both

(S_{1})

and

(S_{2})

are true

3 neither

(S_{1})

nor

(S_{2})

is true

4 only

(S_{2})

is true

Explanation:

A Given that,
$f (x) = \frac{1}{\sqrt{[x] - x}}$
If $x \in I [x] = [x]$ (greatest integer function )
If $x \notin I [x] = [x] + 1$
$\begin{aligned} f (x) = {\begin{cases} \frac{1}{\sqrt{[x] - x}}, & x \in I \\ \frac{1}{\sqrt{[x] + 1 - x}}, & x \notin I \end{cases} \\ f (x) = {\begin{cases} \frac{1}{\sqrt{- {x}}}, & x \in I \\ \frac{1}{\sqrt{1 - {x}}}, & x \notin I \end{cases} (does not exist) \end{aligned}$
Domain of function $= R - I$
Now,
$f (x) = \frac{1}{\sqrt{1 - {x}}}, x \notin I$
$0 < {x} < 1$
$0 < \sqrt{1 - {x}} < 1$
$\frac{1}{\sqrt{1 - {x}}} > 1$
Range $\in (1, \infty)$
$A = R - I$
$B = (1, \infty)$
So, $A \cap B = (1, \infty) - N$
$A \cup B \neq (1, \infty)$
Hence, $S_{1}$ is only correct.

JEE Main-06.04.2023

Sets, Relation and Function

117426 The domain of the function $f (x) = \sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7})$ is

1

[1, \infty]

2

(- 1, 2]

3

[- 1, \infty]

4

[- \infty, 2]

Explanation:

C Given, $f (x) = \sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7})$
Then, the domain is defined as -
$\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \geq - 1 and \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1$
$2 x^{2} - x + 9 \geq 0 and 5 x \geq - 5 \Rightarrow x \geq - 1$
$x \in R$ Hence, domain $x \in [- 1, \infty]$

JEE Main-29.07.2022

Sets, Relation and Function

117427 If $f (x) = 3 - x, - 4 \leq x \leq 4$ , then the domain of $\log_{e}$ (f(x)) is

1

[- 4, 4]

2

(- \infty, 3]

3

(- \infty, 3)

4

[- 4, 3)

Explanation:

D Given that,
$f (x) = 3 - x$
Taking log with base e on both the side,
$\log_{e} f (x) = \log_{e} (3 - x)$
This function will exist when,
$3 - x > 0$
$- x > - 3$
$x > 3$
$x \in (- \infty, 3)$
But $f (x)$ is defined for only $x \in [- 4, 4]$
Hence, the domain of $\log_{e} f (x)$ is $[- 4, 3)$

J and K CET-2011

Sets, Relation and Function

117423 If the range of the function $f (x) = - 3 x - 3$ is ${3$ , $- 6, - 9, - 18}$ , then which of the following elements is not in the domain of $f$ ?

1 -1

2 -2

3 1

4 2

Explanation:

A Given that,
$f (x) = - 3 x - 3$
And, range of function $= {3, - 6, - 9, - 18}$
Now, check all the number from the options put $x = - 1$
$f (- 1) = - 3 \times - 1 - 3$
$= 3 - 3 = 0$
Thus, -1 can not belong to domain of $f (x)$
Put, $x = - 2$
$f (- 2) = - 3 \times - 2 - 3$
$= + 6 - 3$
$= 3$
Put, $x = 1$
$f (1) = - 3 \times 1 - 3$
$= - 6$
Put $x = 2$
$f (2) = - 3 \times 2 - 3$
$= - 6 - 3$
$= - 9$ Hence, -1 can not be in the domain of $f (x)$ -

TS EAMCET-2017

Sets, Relation and Function

117424 $f : (- \infty, 0] \to [0, \infty)$ is defind as $f (x) = x^{2}$ . The domain and range of its inverse is

1 Domain

(f^{- 1}) = [0, \infty)

, Range of

(f^{- 1}) = (- \infty

,

0]

2 Domain of

(f^{- 1}) = [0, \infty)

Range of

(f^{- 1}) = (- \infty

,

\infty)

3 Domain of

(f^{- 1}) = [0, \infty)

Range of

(f^{- 1}) = (0

,

\infty)

4

f^{- 1}

does not exist

Explanation:

A We have a function : $(- \infty, 0] \to [0, \infty)$ Defined as $f (x) = x^{2}$
The graph of above function is.
original image
Since, each line parallel to $x$ - axis cuts that above curve at maximum one paint, therefore $f$ is one- one . Also from the graph, it is clear that range.
$f = [0, \infty)$
$∴ f$ is onto.
Thus, $f$ is invertible.
Hence, $f^{- 1} : [0, \infty) \to (- \infty, 0]$
Domain $(f^{- 1}) = [0, \infty)$ and range of
$(f^{- 1}) = (- \infty 0]$

TS EAMCET-2017

Sets, Relation and Function

117425 Let the sets $A$ and $B$ denote the domain and range respectively of the function
$f (x) = \frac{1}{\sqrt{[x] - x}}$ where $[x]$ denotes the smallest integer greater than or equal to $x$ . Then among the statements
$(S_{1}) : A \cap B = (1, \infty) - N$ and
$(S_{2}) : A \cup B = (1, \infty)$

1 only

(S_{1})

is true

2 both

(S_{1})

and

(S_{2})

are true

3 neither

(S_{1})

nor

(S_{2})

is true

4 only

(S_{2})

is true

Explanation:

A Given that,
$f (x) = \frac{1}{\sqrt{[x] - x}}$
If $x \in I [x] = [x]$ (greatest integer function )
If $x \notin I [x] = [x] + 1$
$\begin{aligned} f (x) = {\begin{cases} \frac{1}{\sqrt{[x] - x}}, & x \in I \\ \frac{1}{\sqrt{[x] + 1 - x}}, & x \notin I \end{cases} \\ f (x) = {\begin{cases} \frac{1}{\sqrt{- {x}}}, & x \in I \\ \frac{1}{\sqrt{1 - {x}}}, & x \notin I \end{cases} (does not exist) \end{aligned}$
Domain of function $= R - I$
Now,
$f (x) = \frac{1}{\sqrt{1 - {x}}}, x \notin I$
$0 < {x} < 1$
$0 < \sqrt{1 - {x}} < 1$
$\frac{1}{\sqrt{1 - {x}}} > 1$
Range $\in (1, \infty)$
$A = R - I$
$B = (1, \infty)$
So, $A \cap B = (1, \infty) - N$
$A \cup B \neq (1, \infty)$
Hence, $S_{1}$ is only correct.

JEE Main-06.04.2023

Sets, Relation and Function

117426 The domain of the function $f (x) = \sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7})$ is

1

[1, \infty]

2

(- 1, 2]

3

[- 1, \infty]

4

[- \infty, 2]

Explanation:

C Given, $f (x) = \sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7})$
Then, the domain is defined as -
$\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \geq - 1 and \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1$
$2 x^{2} - x + 9 \geq 0 and 5 x \geq - 5 \Rightarrow x \geq - 1$
$x \in R$ Hence, domain $x \in [- 1, \infty]$

JEE Main-29.07.2022

Sets, Relation and Function

117427 If $f (x) = 3 - x, - 4 \leq x \leq 4$ , then the domain of $\log_{e}$ (f(x)) is

1

[- 4, 4]

2

(- \infty, 3]

3

(- \infty, 3)

4

[- 4, 3)

Explanation:

D Given that,
$f (x) = 3 - x$
Taking log with base e on both the side,
$\log_{e} f (x) = \log_{e} (3 - x)$
This function will exist when,
$3 - x > 0$
$- x > - 3$
$x > 3$
$x \in (- \infty, 3)$
But $f (x)$ is defined for only $x \in [- 4, 4]$
Hence, the domain of $\log_{e} f (x)$ is $[- 4, 3)$

J and K CET-2011

Sets, Relation and Function

117423 If the range of the function $f (x) = - 3 x - 3$ is ${3$ , $- 6, - 9, - 18}$ , then which of the following elements is not in the domain of $f$ ?

1 -1

2 -2

3 1

4 2

Explanation:

A Given that,
$f (x) = - 3 x - 3$
And, range of function $= {3, - 6, - 9, - 18}$
Now, check all the number from the options put $x = - 1$
$f (- 1) = - 3 \times - 1 - 3$
$= 3 - 3 = 0$
Thus, -1 can not belong to domain of $f (x)$
Put, $x = - 2$
$f (- 2) = - 3 \times - 2 - 3$
$= + 6 - 3$
$= 3$
Put, $x = 1$
$f (1) = - 3 \times 1 - 3$
$= - 6$
Put $x = 2$
$f (2) = - 3 \times 2 - 3$
$= - 6 - 3$
$= - 9$ Hence, -1 can not be in the domain of $f (x)$ -

TS EAMCET-2017

Sets, Relation and Function

117424 $f : (- \infty, 0] \to [0, \infty)$ is defind as $f (x) = x^{2}$ . The domain and range of its inverse is

1 Domain

(f^{- 1}) = [0, \infty)

, Range of

(f^{- 1}) = (- \infty

,

0]

2 Domain of

(f^{- 1}) = [0, \infty)

Range of

(f^{- 1}) = (- \infty

,

\infty)

3 Domain of

(f^{- 1}) = [0, \infty)

Range of

(f^{- 1}) = (0

,

\infty)

4

f^{- 1}

does not exist

Explanation:

A We have a function : $(- \infty, 0] \to [0, \infty)$ Defined as $f (x) = x^{2}$
The graph of above function is.
original image
Since, each line parallel to $x$ - axis cuts that above curve at maximum one paint, therefore $f$ is one- one . Also from the graph, it is clear that range.
$f = [0, \infty)$
$∴ f$ is onto.
Thus, $f$ is invertible.
Hence, $f^{- 1} : [0, \infty) \to (- \infty, 0]$
Domain $(f^{- 1}) = [0, \infty)$ and range of
$(f^{- 1}) = (- \infty 0]$

TS EAMCET-2017

Sets, Relation and Function

117425 Let the sets $A$ and $B$ denote the domain and range respectively of the function
$f (x) = \frac{1}{\sqrt{[x] - x}}$ where $[x]$ denotes the smallest integer greater than or equal to $x$ . Then among the statements
$(S_{1}) : A \cap B = (1, \infty) - N$ and
$(S_{2}) : A \cup B = (1, \infty)$

1 only

(S_{1})

is true

2 both

(S_{1})

and

(S_{2})

are true

3 neither

(S_{1})

nor

(S_{2})

is true

4 only

(S_{2})

is true

Explanation:

A Given that,
$f (x) = \frac{1}{\sqrt{[x] - x}}$
If $x \in I [x] = [x]$ (greatest integer function )
If $x \notin I [x] = [x] + 1$
$\begin{aligned} f (x) = {\begin{cases} \frac{1}{\sqrt{[x] - x}}, & x \in I \\ \frac{1}{\sqrt{[x] + 1 - x}}, & x \notin I \end{cases} \\ f (x) = {\begin{cases} \frac{1}{\sqrt{- {x}}}, & x \in I \\ \frac{1}{\sqrt{1 - {x}}}, & x \notin I \end{cases} (does not exist) \end{aligned}$
Domain of function $= R - I$
Now,
$f (x) = \frac{1}{\sqrt{1 - {x}}}, x \notin I$
$0 < {x} < 1$
$0 < \sqrt{1 - {x}} < 1$
$\frac{1}{\sqrt{1 - {x}}} > 1$
Range $\in (1, \infty)$
$A = R - I$
$B = (1, \infty)$
So, $A \cap B = (1, \infty) - N$
$A \cup B \neq (1, \infty)$
Hence, $S_{1}$ is only correct.

JEE Main-06.04.2023

Sets, Relation and Function

117426 The domain of the function $f (x) = \sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7})$ is

1

[1, \infty]

2

(- 1, 2]

3

[- 1, \infty]

4

[- \infty, 2]

Explanation:

C Given, $f (x) = \sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7})$
Then, the domain is defined as -
$\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \geq - 1 and \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1$
$2 x^{2} - x + 9 \geq 0 and 5 x \geq - 5 \Rightarrow x \geq - 1$
$x \in R$ Hence, domain $x \in [- 1, \infty]$

JEE Main-29.07.2022

Sets, Relation and Function

117427 If $f (x) = 3 - x, - 4 \leq x \leq 4$ , then the domain of $\log_{e}$ (f(x)) is

1

[- 4, 4]

2

(- \infty, 3]

3

(- \infty, 3)

4

[- 4, 3)

Explanation:

D Given that,
$f (x) = 3 - x$
Taking log with base e on both the side,
$\log_{e} f (x) = \log_{e} (3 - x)$
This function will exist when,
$3 - x > 0$
$- x > - 3$
$x > 3$
$x \in (- \infty, 3)$
But $f (x)$ is defined for only $x \in [- 4, 4]$
Hence, the domain of $\log_{e} f (x)$ is $[- 4, 3)$

J and K CET-2011

Sets, Relation and Function

117423 If the range of the function $f (x) = - 3 x - 3$ is ${3$ , $- 6, - 9, - 18}$ , then which of the following elements is not in the domain of $f$ ?

1 -1

2 -2

3 1

4 2

Explanation:

A Given that,
$f (x) = - 3 x - 3$
And, range of function $= {3, - 6, - 9, - 18}$
Now, check all the number from the options put $x = - 1$
$f (- 1) = - 3 \times - 1 - 3$
$= 3 - 3 = 0$
Thus, -1 can not belong to domain of $f (x)$
Put, $x = - 2$
$f (- 2) = - 3 \times - 2 - 3$
$= + 6 - 3$
$= 3$
Put, $x = 1$
$f (1) = - 3 \times 1 - 3$
$= - 6$
Put $x = 2$
$f (2) = - 3 \times 2 - 3$
$= - 6 - 3$
$= - 9$ Hence, -1 can not be in the domain of $f (x)$ -

TS EAMCET-2017

Sets, Relation and Function

117424 $f : (- \infty, 0] \to [0, \infty)$ is defind as $f (x) = x^{2}$ . The domain and range of its inverse is

1 Domain

(f^{- 1}) = [0, \infty)

, Range of

(f^{- 1}) = (- \infty

,

0]

2 Domain of

(f^{- 1}) = [0, \infty)

Range of

(f^{- 1}) = (- \infty

,

\infty)

3 Domain of

(f^{- 1}) = [0, \infty)

Range of

(f^{- 1}) = (0

,

\infty)

4

f^{- 1}

does not exist

Explanation:

A We have a function : $(- \infty, 0] \to [0, \infty)$ Defined as $f (x) = x^{2}$
The graph of above function is.
original image
Since, each line parallel to $x$ - axis cuts that above curve at maximum one paint, therefore $f$ is one- one . Also from the graph, it is clear that range.
$f = [0, \infty)$
$∴ f$ is onto.
Thus, $f$ is invertible.
Hence, $f^{- 1} : [0, \infty) \to (- \infty, 0]$
Domain $(f^{- 1}) = [0, \infty)$ and range of
$(f^{- 1}) = (- \infty 0]$

TS EAMCET-2017

Sets, Relation and Function

117425 Let the sets $A$ and $B$ denote the domain and range respectively of the function
$f (x) = \frac{1}{\sqrt{[x] - x}}$ where $[x]$ denotes the smallest integer greater than or equal to $x$ . Then among the statements
$(S_{1}) : A \cap B = (1, \infty) - N$ and
$(S_{2}) : A \cup B = (1, \infty)$

1 only

(S_{1})

is true

2 both

(S_{1})

and

(S_{2})

are true

3 neither

(S_{1})

nor

(S_{2})

is true

4 only

(S_{2})

is true

Explanation:

A Given that,
$f (x) = \frac{1}{\sqrt{[x] - x}}$
If $x \in I [x] = [x]$ (greatest integer function )
If $x \notin I [x] = [x] + 1$
$\begin{aligned} f (x) = {\begin{cases} \frac{1}{\sqrt{[x] - x}}, & x \in I \\ \frac{1}{\sqrt{[x] + 1 - x}}, & x \notin I \end{cases} \\ f (x) = {\begin{cases} \frac{1}{\sqrt{- {x}}}, & x \in I \\ \frac{1}{\sqrt{1 - {x}}}, & x \notin I \end{cases} (does not exist) \end{aligned}$
Domain of function $= R - I$
Now,
$f (x) = \frac{1}{\sqrt{1 - {x}}}, x \notin I$
$0 < {x} < 1$
$0 < \sqrt{1 - {x}} < 1$
$\frac{1}{\sqrt{1 - {x}}} > 1$
Range $\in (1, \infty)$
$A = R - I$
$B = (1, \infty)$
So, $A \cap B = (1, \infty) - N$
$A \cup B \neq (1, \infty)$
Hence, $S_{1}$ is only correct.

JEE Main-06.04.2023

Sets, Relation and Function

117426 The domain of the function $f (x) = \sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7})$ is

1

[1, \infty]

2

(- 1, 2]

3

[- 1, \infty]

4

[- \infty, 2]

Explanation:

C Given, $f (x) = \sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7})$
Then, the domain is defined as -
$\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \geq - 1 and \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1$
$2 x^{2} - x + 9 \geq 0 and 5 x \geq - 5 \Rightarrow x \geq - 1$
$x \in R$ Hence, domain $x \in [- 1, \infty]$

JEE Main-29.07.2022

Sets, Relation and Function

117427 If $f (x) = 3 - x, - 4 \leq x \leq 4$ , then the domain of $\log_{e}$ (f(x)) is

1

[- 4, 4]

2

(- \infty, 3]

3

(- \infty, 3)

4

[- 4, 3)

Explanation:

D Given that,
$f (x) = 3 - x$
Taking log with base e on both the side,
$\log_{e} f (x) = \log_{e} (3 - x)$
This function will exist when,
$3 - x > 0$
$- x > - 3$
$x > 3$
$x \in (- \infty, 3)$
But $f (x)$ is defined for only $x \in [- 4, 4]$
Hence, the domain of $\log_{e} f (x)$ is $[- 4, 3)$

J and K CET-2011

Sets, Relation and Function

117423 If the range of the function $f (x) = - 3 x - 3$ is ${3$ , $- 6, - 9, - 18}$ , then which of the following elements is not in the domain of $f$ ?

1 -1

2 -2

3 1

4 2

Explanation:

A Given that,
$f (x) = - 3 x - 3$
And, range of function $= {3, - 6, - 9, - 18}$
Now, check all the number from the options put $x = - 1$
$f (- 1) = - 3 \times - 1 - 3$
$= 3 - 3 = 0$
Thus, -1 can not belong to domain of $f (x)$
Put, $x = - 2$
$f (- 2) = - 3 \times - 2 - 3$
$= + 6 - 3$
$= 3$
Put, $x = 1$
$f (1) = - 3 \times 1 - 3$
$= - 6$
Put $x = 2$
$f (2) = - 3 \times 2 - 3$
$= - 6 - 3$
$= - 9$ Hence, -1 can not be in the domain of $f (x)$ -

TS EAMCET-2017

Sets, Relation and Function

117424 $f : (- \infty, 0] \to [0, \infty)$ is defind as $f (x) = x^{2}$ . The domain and range of its inverse is

1 Domain

(f^{- 1}) = [0, \infty)

, Range of

(f^{- 1}) = (- \infty

,

0]

2 Domain of

(f^{- 1}) = [0, \infty)

Range of

(f^{- 1}) = (- \infty

,

\infty)

3 Domain of

(f^{- 1}) = [0, \infty)

Range of

(f^{- 1}) = (0

,

\infty)

4

f^{- 1}

does not exist

Explanation:

A We have a function : $(- \infty, 0] \to [0, \infty)$ Defined as $f (x) = x^{2}$
The graph of above function is.
original image
Since, each line parallel to $x$ - axis cuts that above curve at maximum one paint, therefore $f$ is one- one . Also from the graph, it is clear that range.
$f = [0, \infty)$
$∴ f$ is onto.
Thus, $f$ is invertible.
Hence, $f^{- 1} : [0, \infty) \to (- \infty, 0]$
Domain $(f^{- 1}) = [0, \infty)$ and range of
$(f^{- 1}) = (- \infty 0]$

TS EAMCET-2017

Sets, Relation and Function

117425 Let the sets $A$ and $B$ denote the domain and range respectively of the function
$f (x) = \frac{1}{\sqrt{[x] - x}}$ where $[x]$ denotes the smallest integer greater than or equal to $x$ . Then among the statements
$(S_{1}) : A \cap B = (1, \infty) - N$ and
$(S_{2}) : A \cup B = (1, \infty)$

1 only

(S_{1})

is true

2 both

(S_{1})

and

(S_{2})

are true

3 neither

(S_{1})

nor

(S_{2})

is true

4 only

(S_{2})

is true

Explanation:

A Given that,
$f (x) = \frac{1}{\sqrt{[x] - x}}$
If $x \in I [x] = [x]$ (greatest integer function )
If $x \notin I [x] = [x] + 1$
$\begin{aligned} f (x) = {\begin{cases} \frac{1}{\sqrt{[x] - x}}, & x \in I \\ \frac{1}{\sqrt{[x] + 1 - x}}, & x \notin I \end{cases} \\ f (x) = {\begin{cases} \frac{1}{\sqrt{- {x}}}, & x \in I \\ \frac{1}{\sqrt{1 - {x}}}, & x \notin I \end{cases} (does not exist) \end{aligned}$
Domain of function $= R - I$
Now,
$f (x) = \frac{1}{\sqrt{1 - {x}}}, x \notin I$
$0 < {x} < 1$
$0 < \sqrt{1 - {x}} < 1$
$\frac{1}{\sqrt{1 - {x}}} > 1$
Range $\in (1, \infty)$
$A = R - I$
$B = (1, \infty)$
So, $A \cap B = (1, \infty) - N$
$A \cup B \neq (1, \infty)$
Hence, $S_{1}$ is only correct.

JEE Main-06.04.2023

Sets, Relation and Function

117426 The domain of the function $f (x) = \sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7})$ is

1

[1, \infty]

2

(- 1, 2]

3

[- 1, \infty]

4

[- \infty, 2]

Explanation:

C Given, $f (x) = \sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7})$
Then, the domain is defined as -
$\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \geq - 1 and \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1$
$2 x^{2} - x + 9 \geq 0 and 5 x \geq - 5 \Rightarrow x \geq - 1$
$x \in R$ Hence, domain $x \in [- 1, \infty]$

JEE Main-29.07.2022

Sets, Relation and Function

117427 If $f (x) = 3 - x, - 4 \leq x \leq 4$ , then the domain of $\log_{e}$ (f(x)) is

1

[- 4, 4]

2

(- \infty, 3]

3

(- \infty, 3)

4

[- 4, 3)

Explanation:

D Given that,
$f (x) = 3 - x$
Taking log with base e on both the side,
$\log_{e} f (x) = \log_{e} (3 - x)$
This function will exist when,
$3 - x > 0$
$- x > - 3$
$x > 3$
$x \in (- \infty, 3)$
But $f (x)$ is defined for only $x \in [- 4, 4]$
Hence, the domain of $\log_{e} f (x)$ is $[- 4, 3)$

J and K CET-2011

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