QBManager

Sets, Relation and Function

117114 The number of elements in the set $S = {x \in R : 2 \cos (\frac{x^{2} + 6}{6}) = 4^{x} + 4^{- x}}$ is :

1 1

2 3

3 0

4 Infinite

Explanation:

Exp: (A) : $2 \cos (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}$
L.H.S $\leq 2 . &$ R.H.S. $= 2$
$2 \cos (\frac{x^{2} + x}{6}) = 2 4^{x} + 4^{- x} = 2$
It is possible only $x = 0$
Hence, it gives only one solution.

[JEE Main-29.07.2022

Sets, Relation and Function

117115 Let $f (x)$ be a function such that $f (x + y) = f (x)$ . $f (y)$ for all $x, y \in N$ . If $f (1) = 3$ and $\sum_{k = 1}^{n} f (k) =$ 3279 , then the value of $n$ is

1 8

2 6

3 7

4 9

Explanation:

Exp: (C) : Given $f (x + y) = f (x) \cdot f (y) \forall x, y \in n$ Given, $f (1) = 3$
$f (2) = f (1) . f (1) = 3 \times 3 = 3^{2}$
$f (3) = f (2) . f (1) = 3^{2} \times 3 = 3^{3}$
$∴ f (1) + f (2) + f (3) + \dots . . + f (n) = 3279$
L. H. S. is in G.P. which is
$3 + 3^{2} + 3^{3} + \dots \dots f (n) = 3279$
$\Rightarrow$ Here, $a = 3$
$r = 3$
$\frac{(r^{n} - 1)}{(r - 1)} = 3 \cdot \frac{(3^{n} - 1)}{(3 - 1)} = 3279$
$∴ \frac{3^{n} - 1}{2} = 1093$
$∴ 3^{n} - 1 = 2186$
$3^{n} = 2187$
$∴ 3^{n} = 3^{7}$
Hence, $n = 7$

[JEE Main-24.01.2023

Sets, Relation and Function

117116 If $f : R \to A$ defined by $f (x) = \frac{1}{x^{2} + 2 x + 2} \forall x \in R$ is surjective, then $A =$

1

[1, \infty]

2

(1, \infty)

3

[0, 1]

4

(0, 1]

Explanation:

Exp: (D) : Given,
$f (x) = \frac{1}{x^{2} + 2 x + 2}$
For quadratic expression,
$x^{2} + 2 x + 2 = (x + 1)^{2} + 1$
$x \in [1, \infty)$
Now, $f (x) = \frac{1}{(x + 1)^{2} + 1}$
$x \in R$ is surjective,
Then, $x \in (0, 1]$ .

[AP EAMCET-23.04.2018

Sets, Relation and Function

117117 Let $f : R - {n} \to R$ be a function defined by $f (x) = \frac{x - m}{x - n}$ such that $m \neq n$ , then

1

f

is one one into function

2

f

is one one onto function

3

f

is many one into function

4

f

is many one onto function

Explanation:

Exp: (A) : Given $f (x) = \frac{x - m}{x - n}$ where $m \neq n$ , $\forall x \in R - {n}$ .
Let $x_{1} x_{2} \in R$
$∴ f (x_{1}) = f (x_{2}) \Rightarrow \frac{x_{1} - m}{x_{1} - n} = \frac{x_{2} - m}{x_{2} - n} \Rightarrow x_{1} = x_{2}$
$∴ f$ is one-one.
Let, $λ \in R$ such that $f (x) = λ$
$∴ \frac{x - m}{x - n} = λ ∴ x = \frac{m - n λ}{1 - λ}$
$X$ is not defined for $λ = 1$ , also $x$ is not real.
$∴ f (x)$ is not an onto function.
If a function is not onto, then it is an into function.
Hence, $f$ is one-one into function.

COMEDK 2020

Sets, Relation and Function

117118 The function $f : [0, \infty) \to R$ given by $f (x) = \frac{x}{x + 1}$ is

1 one-one and onto

2 one-one but not onto

3 onto but not one-one

4 Neither one-one nor onto

Explanation:

Exp: (B) : Given,
The function $f : [0, \infty) \to R$ given by
$f (x) = \frac{x}{x + 1} = \frac{x + 1 - 1}{x + 1} = 1 - \frac{1}{x + 1}$
So, $f (x)$ is not contains the value 1 .
Now, then, $x \in [0, \infty)$ , it is a one-one function but not onto function.
Because every point in the co-domain is not a value of $f (x)$ .
Hence, function $f$ is one-one but not onto

[BCECE-2018

Sets, Relation and Function

117114 The number of elements in the set $S = {x \in R : 2 \cos (\frac{x^{2} + 6}{6}) = 4^{x} + 4^{- x}}$ is :

1 1

2 3

3 0

4 Infinite

Explanation:

Exp: (A) : $2 \cos (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}$
L.H.S $\leq 2 . &$ R.H.S. $= 2$
$2 \cos (\frac{x^{2} + x}{6}) = 2 4^{x} + 4^{- x} = 2$
It is possible only $x = 0$
Hence, it gives only one solution.

[JEE Main-29.07.2022

Sets, Relation and Function

117115 Let $f (x)$ be a function such that $f (x + y) = f (x)$ . $f (y)$ for all $x, y \in N$ . If $f (1) = 3$ and $\sum_{k = 1}^{n} f (k) =$ 3279 , then the value of $n$ is

1 8

2 6

3 7

4 9

Explanation:

Exp: (C) : Given $f (x + y) = f (x) \cdot f (y) \forall x, y \in n$ Given, $f (1) = 3$
$f (2) = f (1) . f (1) = 3 \times 3 = 3^{2}$
$f (3) = f (2) . f (1) = 3^{2} \times 3 = 3^{3}$
$∴ f (1) + f (2) + f (3) + \dots . . + f (n) = 3279$
L. H. S. is in G.P. which is
$3 + 3^{2} + 3^{3} + \dots \dots f (n) = 3279$
$\Rightarrow$ Here, $a = 3$
$r = 3$
$\frac{(r^{n} - 1)}{(r - 1)} = 3 \cdot \frac{(3^{n} - 1)}{(3 - 1)} = 3279$
$∴ \frac{3^{n} - 1}{2} = 1093$
$∴ 3^{n} - 1 = 2186$
$3^{n} = 2187$
$∴ 3^{n} = 3^{7}$
Hence, $n = 7$

[JEE Main-24.01.2023

Sets, Relation and Function

117116 If $f : R \to A$ defined by $f (x) = \frac{1}{x^{2} + 2 x + 2} \forall x \in R$ is surjective, then $A =$

1

[1, \infty]

2

(1, \infty)

3

[0, 1]

4

(0, 1]

Explanation:

Exp: (D) : Given,
$f (x) = \frac{1}{x^{2} + 2 x + 2}$
For quadratic expression,
$x^{2} + 2 x + 2 = (x + 1)^{2} + 1$
$x \in [1, \infty)$
Now, $f (x) = \frac{1}{(x + 1)^{2} + 1}$
$x \in R$ is surjective,
Then, $x \in (0, 1]$ .

[AP EAMCET-23.04.2018

Sets, Relation and Function

117117 Let $f : R - {n} \to R$ be a function defined by $f (x) = \frac{x - m}{x - n}$ such that $m \neq n$ , then

1

f

is one one into function

2

f

is one one onto function

3

f

is many one into function

4

f

is many one onto function

Explanation:

Exp: (A) : Given $f (x) = \frac{x - m}{x - n}$ where $m \neq n$ , $\forall x \in R - {n}$ .
Let $x_{1} x_{2} \in R$
$∴ f (x_{1}) = f (x_{2}) \Rightarrow \frac{x_{1} - m}{x_{1} - n} = \frac{x_{2} - m}{x_{2} - n} \Rightarrow x_{1} = x_{2}$
$∴ f$ is one-one.
Let, $λ \in R$ such that $f (x) = λ$
$∴ \frac{x - m}{x - n} = λ ∴ x = \frac{m - n λ}{1 - λ}$
$X$ is not defined for $λ = 1$ , also $x$ is not real.
$∴ f (x)$ is not an onto function.
If a function is not onto, then it is an into function.
Hence, $f$ is one-one into function.

COMEDK 2020

Sets, Relation and Function

117118 The function $f : [0, \infty) \to R$ given by $f (x) = \frac{x}{x + 1}$ is

1 one-one and onto

2 one-one but not onto

3 onto but not one-one

4 Neither one-one nor onto

Explanation:

Exp: (B) : Given,
The function $f : [0, \infty) \to R$ given by
$f (x) = \frac{x}{x + 1} = \frac{x + 1 - 1}{x + 1} = 1 - \frac{1}{x + 1}$
So, $f (x)$ is not contains the value 1 .
Now, then, $x \in [0, \infty)$ , it is a one-one function but not onto function.
Because every point in the co-domain is not a value of $f (x)$ .
Hence, function $f$ is one-one but not onto

[BCECE-2018

Sets, Relation and Function

117114 The number of elements in the set $S = {x \in R : 2 \cos (\frac{x^{2} + 6}{6}) = 4^{x} + 4^{- x}}$ is :

1 1

2 3

3 0

4 Infinite

Explanation:

Exp: (A) : $2 \cos (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}$
L.H.S $\leq 2 . &$ R.H.S. $= 2$
$2 \cos (\frac{x^{2} + x}{6}) = 2 4^{x} + 4^{- x} = 2$
It is possible only $x = 0$
Hence, it gives only one solution.

[JEE Main-29.07.2022

Sets, Relation and Function

117115 Let $f (x)$ be a function such that $f (x + y) = f (x)$ . $f (y)$ for all $x, y \in N$ . If $f (1) = 3$ and $\sum_{k = 1}^{n} f (k) =$ 3279 , then the value of $n$ is

1 8

2 6

3 7

4 9

Explanation:

Exp: (C) : Given $f (x + y) = f (x) \cdot f (y) \forall x, y \in n$ Given, $f (1) = 3$
$f (2) = f (1) . f (1) = 3 \times 3 = 3^{2}$
$f (3) = f (2) . f (1) = 3^{2} \times 3 = 3^{3}$
$∴ f (1) + f (2) + f (3) + \dots . . + f (n) = 3279$
L. H. S. is in G.P. which is
$3 + 3^{2} + 3^{3} + \dots \dots f (n) = 3279$
$\Rightarrow$ Here, $a = 3$
$r = 3$
$\frac{(r^{n} - 1)}{(r - 1)} = 3 \cdot \frac{(3^{n} - 1)}{(3 - 1)} = 3279$
$∴ \frac{3^{n} - 1}{2} = 1093$
$∴ 3^{n} - 1 = 2186$
$3^{n} = 2187$
$∴ 3^{n} = 3^{7}$
Hence, $n = 7$

[JEE Main-24.01.2023

Sets, Relation and Function

117116 If $f : R \to A$ defined by $f (x) = \frac{1}{x^{2} + 2 x + 2} \forall x \in R$ is surjective, then $A =$

1

[1, \infty]

2

(1, \infty)

3

[0, 1]

4

(0, 1]

Explanation:

Exp: (D) : Given,
$f (x) = \frac{1}{x^{2} + 2 x + 2}$
For quadratic expression,
$x^{2} + 2 x + 2 = (x + 1)^{2} + 1$
$x \in [1, \infty)$
Now, $f (x) = \frac{1}{(x + 1)^{2} + 1}$
$x \in R$ is surjective,
Then, $x \in (0, 1]$ .

[AP EAMCET-23.04.2018

Sets, Relation and Function

117117 Let $f : R - {n} \to R$ be a function defined by $f (x) = \frac{x - m}{x - n}$ such that $m \neq n$ , then

1

f

is one one into function

2

f

is one one onto function

3

f

is many one into function

4

f

is many one onto function

Explanation:

Exp: (A) : Given $f (x) = \frac{x - m}{x - n}$ where $m \neq n$ , $\forall x \in R - {n}$ .
Let $x_{1} x_{2} \in R$
$∴ f (x_{1}) = f (x_{2}) \Rightarrow \frac{x_{1} - m}{x_{1} - n} = \frac{x_{2} - m}{x_{2} - n} \Rightarrow x_{1} = x_{2}$
$∴ f$ is one-one.
Let, $λ \in R$ such that $f (x) = λ$
$∴ \frac{x - m}{x - n} = λ ∴ x = \frac{m - n λ}{1 - λ}$
$X$ is not defined for $λ = 1$ , also $x$ is not real.
$∴ f (x)$ is not an onto function.
If a function is not onto, then it is an into function.
Hence, $f$ is one-one into function.

COMEDK 2020

Sets, Relation and Function

117118 The function $f : [0, \infty) \to R$ given by $f (x) = \frac{x}{x + 1}$ is

1 one-one and onto

2 one-one but not onto

3 onto but not one-one

4 Neither one-one nor onto

Explanation:

Exp: (B) : Given,
The function $f : [0, \infty) \to R$ given by
$f (x) = \frac{x}{x + 1} = \frac{x + 1 - 1}{x + 1} = 1 - \frac{1}{x + 1}$
So, $f (x)$ is not contains the value 1 .
Now, then, $x \in [0, \infty)$ , it is a one-one function but not onto function.
Because every point in the co-domain is not a value of $f (x)$ .
Hence, function $f$ is one-one but not onto

[BCECE-2018

Sets, Relation and Function

117114 The number of elements in the set $S = {x \in R : 2 \cos (\frac{x^{2} + 6}{6}) = 4^{x} + 4^{- x}}$ is :

1 1

2 3

3 0

4 Infinite

Explanation:

Exp: (A) : $2 \cos (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}$
L.H.S $\leq 2 . &$ R.H.S. $= 2$
$2 \cos (\frac{x^{2} + x}{6}) = 2 4^{x} + 4^{- x} = 2$
It is possible only $x = 0$
Hence, it gives only one solution.

[JEE Main-29.07.2022

Sets, Relation and Function

117115 Let $f (x)$ be a function such that $f (x + y) = f (x)$ . $f (y)$ for all $x, y \in N$ . If $f (1) = 3$ and $\sum_{k = 1}^{n} f (k) =$ 3279 , then the value of $n$ is

1 8

2 6

3 7

4 9

Explanation:

Exp: (C) : Given $f (x + y) = f (x) \cdot f (y) \forall x, y \in n$ Given, $f (1) = 3$
$f (2) = f (1) . f (1) = 3 \times 3 = 3^{2}$
$f (3) = f (2) . f (1) = 3^{2} \times 3 = 3^{3}$
$∴ f (1) + f (2) + f (3) + \dots . . + f (n) = 3279$
L. H. S. is in G.P. which is
$3 + 3^{2} + 3^{3} + \dots \dots f (n) = 3279$
$\Rightarrow$ Here, $a = 3$
$r = 3$
$\frac{(r^{n} - 1)}{(r - 1)} = 3 \cdot \frac{(3^{n} - 1)}{(3 - 1)} = 3279$
$∴ \frac{3^{n} - 1}{2} = 1093$
$∴ 3^{n} - 1 = 2186$
$3^{n} = 2187$
$∴ 3^{n} = 3^{7}$
Hence, $n = 7$

[JEE Main-24.01.2023

Sets, Relation and Function

117116 If $f : R \to A$ defined by $f (x) = \frac{1}{x^{2} + 2 x + 2} \forall x \in R$ is surjective, then $A =$

1

[1, \infty]

2

(1, \infty)

3

[0, 1]

4

(0, 1]

Explanation:

Exp: (D) : Given,
$f (x) = \frac{1}{x^{2} + 2 x + 2}$
For quadratic expression,
$x^{2} + 2 x + 2 = (x + 1)^{2} + 1$
$x \in [1, \infty)$
Now, $f (x) = \frac{1}{(x + 1)^{2} + 1}$
$x \in R$ is surjective,
Then, $x \in (0, 1]$ .

[AP EAMCET-23.04.2018

Sets, Relation and Function

117117 Let $f : R - {n} \to R$ be a function defined by $f (x) = \frac{x - m}{x - n}$ such that $m \neq n$ , then

1

f

is one one into function

2

f

is one one onto function

3

f

is many one into function

4

f

is many one onto function

Explanation:

Exp: (A) : Given $f (x) = \frac{x - m}{x - n}$ where $m \neq n$ , $\forall x \in R - {n}$ .
Let $x_{1} x_{2} \in R$
$∴ f (x_{1}) = f (x_{2}) \Rightarrow \frac{x_{1} - m}{x_{1} - n} = \frac{x_{2} - m}{x_{2} - n} \Rightarrow x_{1} = x_{2}$
$∴ f$ is one-one.
Let, $λ \in R$ such that $f (x) = λ$
$∴ \frac{x - m}{x - n} = λ ∴ x = \frac{m - n λ}{1 - λ}$
$X$ is not defined for $λ = 1$ , also $x$ is not real.
$∴ f (x)$ is not an onto function.
If a function is not onto, then it is an into function.
Hence, $f$ is one-one into function.

COMEDK 2020

Sets, Relation and Function

117118 The function $f : [0, \infty) \to R$ given by $f (x) = \frac{x}{x + 1}$ is

1 one-one and onto

2 one-one but not onto

3 onto but not one-one

4 Neither one-one nor onto

Explanation:

Exp: (B) : Given,
The function $f : [0, \infty) \to R$ given by
$f (x) = \frac{x}{x + 1} = \frac{x + 1 - 1}{x + 1} = 1 - \frac{1}{x + 1}$
So, $f (x)$ is not contains the value 1 .
Now, then, $x \in [0, \infty)$ , it is a one-one function but not onto function.
Because every point in the co-domain is not a value of $f (x)$ .
Hence, function $f$ is one-one but not onto

[BCECE-2018

Sets, Relation and Function

117114 The number of elements in the set $S = {x \in R : 2 \cos (\frac{x^{2} + 6}{6}) = 4^{x} + 4^{- x}}$ is :

1 1

2 3

3 0

4 Infinite

Explanation:

Exp: (A) : $2 \cos (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}$
L.H.S $\leq 2 . &$ R.H.S. $= 2$
$2 \cos (\frac{x^{2} + x}{6}) = 2 4^{x} + 4^{- x} = 2$
It is possible only $x = 0$
Hence, it gives only one solution.

[JEE Main-29.07.2022

Sets, Relation and Function

117115 Let $f (x)$ be a function such that $f (x + y) = f (x)$ . $f (y)$ for all $x, y \in N$ . If $f (1) = 3$ and $\sum_{k = 1}^{n} f (k) =$ 3279 , then the value of $n$ is

1 8

2 6

3 7

4 9

Explanation:

Exp: (C) : Given $f (x + y) = f (x) \cdot f (y) \forall x, y \in n$ Given, $f (1) = 3$
$f (2) = f (1) . f (1) = 3 \times 3 = 3^{2}$
$f (3) = f (2) . f (1) = 3^{2} \times 3 = 3^{3}$
$∴ f (1) + f (2) + f (3) + \dots . . + f (n) = 3279$
L. H. S. is in G.P. which is
$3 + 3^{2} + 3^{3} + \dots \dots f (n) = 3279$
$\Rightarrow$ Here, $a = 3$
$r = 3$
$\frac{(r^{n} - 1)}{(r - 1)} = 3 \cdot \frac{(3^{n} - 1)}{(3 - 1)} = 3279$
$∴ \frac{3^{n} - 1}{2} = 1093$
$∴ 3^{n} - 1 = 2186$
$3^{n} = 2187$
$∴ 3^{n} = 3^{7}$
Hence, $n = 7$

[JEE Main-24.01.2023

Sets, Relation and Function

117116 If $f : R \to A$ defined by $f (x) = \frac{1}{x^{2} + 2 x + 2} \forall x \in R$ is surjective, then $A =$

1

[1, \infty]

2

(1, \infty)

3

[0, 1]

4

(0, 1]

Explanation:

Exp: (D) : Given,
$f (x) = \frac{1}{x^{2} + 2 x + 2}$
For quadratic expression,
$x^{2} + 2 x + 2 = (x + 1)^{2} + 1$
$x \in [1, \infty)$
Now, $f (x) = \frac{1}{(x + 1)^{2} + 1}$
$x \in R$ is surjective,
Then, $x \in (0, 1]$ .

[AP EAMCET-23.04.2018

Sets, Relation and Function

117117 Let $f : R - {n} \to R$ be a function defined by $f (x) = \frac{x - m}{x - n}$ such that $m \neq n$ , then

1

f

is one one into function

2

f

is one one onto function

3

f

is many one into function

4

f

is many one onto function

Explanation:

Exp: (A) : Given $f (x) = \frac{x - m}{x - n}$ where $m \neq n$ , $\forall x \in R - {n}$ .
Let $x_{1} x_{2} \in R$
$∴ f (x_{1}) = f (x_{2}) \Rightarrow \frac{x_{1} - m}{x_{1} - n} = \frac{x_{2} - m}{x_{2} - n} \Rightarrow x_{1} = x_{2}$
$∴ f$ is one-one.
Let, $λ \in R$ such that $f (x) = λ$
$∴ \frac{x - m}{x - n} = λ ∴ x = \frac{m - n λ}{1 - λ}$
$X$ is not defined for $λ = 1$ , also $x$ is not real.
$∴ f (x)$ is not an onto function.
If a function is not onto, then it is an into function.
Hence, $f$ is one-one into function.

COMEDK 2020

Sets, Relation and Function

117118 The function $f : [0, \infty) \to R$ given by $f (x) = \frac{x}{x + 1}$ is

1 one-one and onto

2 one-one but not onto

3 onto but not one-one

4 Neither one-one nor onto

Explanation:

Exp: (B) : Given,
The function $f : [0, \infty) \to R$ given by
$f (x) = \frac{x}{x + 1} = \frac{x + 1 - 1}{x + 1} = 1 - \frac{1}{x + 1}$
So, $f (x)$ is not contains the value 1 .
Now, then, $x \in [0, \infty)$ , it is a one-one function but not onto function.
Because every point in the co-domain is not a value of $f (x)$ .
Hence, function $f$ is one-one but not onto

[BCECE-2018

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