QBManager

Sets, Relation and Function

117284 The range of the function $f (x) = \frac{x - 3}{5 - x}, x \neq 5$ is

1

R - {- 1}

2

R - {1}

3

R - {5}

4

R - {- 5}

Explanation:

A Given,
$f (x) = \frac{x - 3}{5 - x}$
Let, $y = f (x) = \frac{x - 3}{5 - x}$
$∴ 5 y - xy = x - 3 \Rightarrow x + xy = 5 y + 3$
$∴ x = \frac{5 y + 3}{1 + y}$
Hence, range $= R - {- 1}$

MHT-CET 20

Sets, Relation and Function

117298 If $\frac{x^{2} + 2 x + 7}{2 x + 3} < 6, x \in R$ , then

1

x > 11

or

x < - \frac{3}{2}

2

x > 11

or

x < - 1

3

- \frac{3}{2} < x < - 1

4

- 1 < x < 11

or

x < - \frac{3}{2}

Explanation:

D Given,
$\frac{x^{2} + 2 x + 7}{2 x + 3} < 6$
$\Rightarrow \frac{x^{2} + 2 x + 7}{2 x + 3} - 6 < 0$
$\Rightarrow \frac{x^{2} + 2 x + 7 - 12 x - 18}{(2 x + 3)} < 0$
$\Rightarrow \frac{x^{2} - 10 x - 11}{(2 x + 3)} < 0$
$\Rightarrow \frac{(x - 11) (x + 1) (2 x + 3)}{(2 x + 3)^{2}} < 0$
$\Rightarrow (x - 11) (x + 1) (2 x + 3) < 0$
$\Rightarrow - 1 < x < 11 or x < \frac{- 3}{2}$

VITEEE-2011

Sets, Relation and Function

117285 If $R = {(a, b) : b = a - 1, a \in Z, 5 < a < 9}$ , then the range of $R$ is

1

{5, 6, 7}

2

{5, 6, 7, 8, 9}

3

{7, 8, 9}

4

{6, 7, 8}

Explanation:

A Given,
$R = {(a, b) : b = a - 1, a \in Z, 5 < a < 9},$
$∵ a = 6, 7, 8$
$∴ a = 6, b = 6 - 1 = 5$
$a = 7, b = 7 - 1 = 6$
$a = 8, b = 8 - 1 = 7$
$R = (6, 5), (7, 6) (8, 7)$
So, range of $R = {5, 6, 7}$

MHT-CET 20

Sets, Relation and Function

117286 If $| 3 x - 2 | \leq \frac{1}{2}$ , then $x \in$

1

(\frac{1}{2}, \frac{5}{6})

2

[\frac{1}{2}, \frac{5}{6})

3

[\frac{1}{2}, \frac{5}{6}]

4

(\frac{1}{2}, \frac{5}{6}]

Explanation:

C Given,
$| 3 x - 2 | \leq \frac{1}{2}$
$∴ \frac{- 1}{2} \leq (3 x - 2) \leq \frac{1}{2}$
$∴ \frac{- 1}{2} \leq 3 x - 2 and 3 x - 2 \leq \frac{1}{2}$
$∴ \frac{3}{2} \leq 3 x and 3 x \leq \frac{5}{2}$
$\frac{1}{2} \leq x and x \leq \frac{5}{6}$
$∴ x \in [\frac{1}{2}, \frac{5}{6}]$

MHT-CET 20

Sets, Relation and Function

117284 The range of the function $f (x) = \frac{x - 3}{5 - x}, x \neq 5$ is

1

R - {- 1}

2

R - {1}

3

R - {5}

4

R - {- 5}

Explanation:

A Given,
$f (x) = \frac{x - 3}{5 - x}$
Let, $y = f (x) = \frac{x - 3}{5 - x}$
$∴ 5 y - xy = x - 3 \Rightarrow x + xy = 5 y + 3$
$∴ x = \frac{5 y + 3}{1 + y}$
Hence, range $= R - {- 1}$

MHT-CET 20

Sets, Relation and Function

117298 If $\frac{x^{2} + 2 x + 7}{2 x + 3} < 6, x \in R$ , then

1

x > 11

or

x < - \frac{3}{2}

2

x > 11

or

x < - 1

3

- \frac{3}{2} < x < - 1

4

- 1 < x < 11

or

x < - \frac{3}{2}

Explanation:

D Given,
$\frac{x^{2} + 2 x + 7}{2 x + 3} < 6$
$\Rightarrow \frac{x^{2} + 2 x + 7}{2 x + 3} - 6 < 0$
$\Rightarrow \frac{x^{2} + 2 x + 7 - 12 x - 18}{(2 x + 3)} < 0$
$\Rightarrow \frac{x^{2} - 10 x - 11}{(2 x + 3)} < 0$
$\Rightarrow \frac{(x - 11) (x + 1) (2 x + 3)}{(2 x + 3)^{2}} < 0$
$\Rightarrow (x - 11) (x + 1) (2 x + 3) < 0$
$\Rightarrow - 1 < x < 11 or x < \frac{- 3}{2}$

VITEEE-2011

Sets, Relation and Function

117285 If $R = {(a, b) : b = a - 1, a \in Z, 5 < a < 9}$ , then the range of $R$ is

1

{5, 6, 7}

2

{5, 6, 7, 8, 9}

3

{7, 8, 9}

4

{6, 7, 8}

Explanation:

A Given,
$R = {(a, b) : b = a - 1, a \in Z, 5 < a < 9},$
$∵ a = 6, 7, 8$
$∴ a = 6, b = 6 - 1 = 5$
$a = 7, b = 7 - 1 = 6$
$a = 8, b = 8 - 1 = 7$
$R = (6, 5), (7, 6) (8, 7)$
So, range of $R = {5, 6, 7}$

MHT-CET 20

Sets, Relation and Function

117286 If $| 3 x - 2 | \leq \frac{1}{2}$ , then $x \in$

1

(\frac{1}{2}, \frac{5}{6})

2

[\frac{1}{2}, \frac{5}{6})

3

[\frac{1}{2}, \frac{5}{6}]

4

(\frac{1}{2}, \frac{5}{6}]

Explanation:

C Given,
$| 3 x - 2 | \leq \frac{1}{2}$
$∴ \frac{- 1}{2} \leq (3 x - 2) \leq \frac{1}{2}$
$∴ \frac{- 1}{2} \leq 3 x - 2 and 3 x - 2 \leq \frac{1}{2}$
$∴ \frac{3}{2} \leq 3 x and 3 x \leq \frac{5}{2}$
$\frac{1}{2} \leq x and x \leq \frac{5}{6}$
$∴ x \in [\frac{1}{2}, \frac{5}{6}]$

MHT-CET 20

Sets, Relation and Function

117284 The range of the function $f (x) = \frac{x - 3}{5 - x}, x \neq 5$ is

1

R - {- 1}

2

R - {1}

3

R - {5}

4

R - {- 5}

Explanation:

A Given,
$f (x) = \frac{x - 3}{5 - x}$
Let, $y = f (x) = \frac{x - 3}{5 - x}$
$∴ 5 y - xy = x - 3 \Rightarrow x + xy = 5 y + 3$
$∴ x = \frac{5 y + 3}{1 + y}$
Hence, range $= R - {- 1}$

MHT-CET 20

Sets, Relation and Function

117298 If $\frac{x^{2} + 2 x + 7}{2 x + 3} < 6, x \in R$ , then

1

x > 11

or

x < - \frac{3}{2}

2

x > 11

or

x < - 1

3

- \frac{3}{2} < x < - 1

4

- 1 < x < 11

or

x < - \frac{3}{2}

Explanation:

D Given,
$\frac{x^{2} + 2 x + 7}{2 x + 3} < 6$
$\Rightarrow \frac{x^{2} + 2 x + 7}{2 x + 3} - 6 < 0$
$\Rightarrow \frac{x^{2} + 2 x + 7 - 12 x - 18}{(2 x + 3)} < 0$
$\Rightarrow \frac{x^{2} - 10 x - 11}{(2 x + 3)} < 0$
$\Rightarrow \frac{(x - 11) (x + 1) (2 x + 3)}{(2 x + 3)^{2}} < 0$
$\Rightarrow (x - 11) (x + 1) (2 x + 3) < 0$
$\Rightarrow - 1 < x < 11 or x < \frac{- 3}{2}$

VITEEE-2011

Sets, Relation and Function

117285 If $R = {(a, b) : b = a - 1, a \in Z, 5 < a < 9}$ , then the range of $R$ is

1

{5, 6, 7}

2

{5, 6, 7, 8, 9}

3

{7, 8, 9}

4

{6, 7, 8}

Explanation:

A Given,
$R = {(a, b) : b = a - 1, a \in Z, 5 < a < 9},$
$∵ a = 6, 7, 8$
$∴ a = 6, b = 6 - 1 = 5$
$a = 7, b = 7 - 1 = 6$
$a = 8, b = 8 - 1 = 7$
$R = (6, 5), (7, 6) (8, 7)$
So, range of $R = {5, 6, 7}$

MHT-CET 20

Sets, Relation and Function

117286 If $| 3 x - 2 | \leq \frac{1}{2}$ , then $x \in$

1

(\frac{1}{2}, \frac{5}{6})

2

[\frac{1}{2}, \frac{5}{6})

3

[\frac{1}{2}, \frac{5}{6}]

4

(\frac{1}{2}, \frac{5}{6}]

Explanation:

C Given,
$| 3 x - 2 | \leq \frac{1}{2}$
$∴ \frac{- 1}{2} \leq (3 x - 2) \leq \frac{1}{2}$
$∴ \frac{- 1}{2} \leq 3 x - 2 and 3 x - 2 \leq \frac{1}{2}$
$∴ \frac{3}{2} \leq 3 x and 3 x \leq \frac{5}{2}$
$\frac{1}{2} \leq x and x \leq \frac{5}{6}$
$∴ x \in [\frac{1}{2}, \frac{5}{6}]$

MHT-CET 20

Sets, Relation and Function

117284 The range of the function $f (x) = \frac{x - 3}{5 - x}, x \neq 5$ is

1

R - {- 1}

2

R - {1}

3

R - {5}

4

R - {- 5}

Explanation:

A Given,
$f (x) = \frac{x - 3}{5 - x}$
Let, $y = f (x) = \frac{x - 3}{5 - x}$
$∴ 5 y - xy = x - 3 \Rightarrow x + xy = 5 y + 3$
$∴ x = \frac{5 y + 3}{1 + y}$
Hence, range $= R - {- 1}$

MHT-CET 20

Sets, Relation and Function

117298 If $\frac{x^{2} + 2 x + 7}{2 x + 3} < 6, x \in R$ , then

1

x > 11

or

x < - \frac{3}{2}

2

x > 11

or

x < - 1

3

- \frac{3}{2} < x < - 1

4

- 1 < x < 11

or

x < - \frac{3}{2}

Explanation:

D Given,
$\frac{x^{2} + 2 x + 7}{2 x + 3} < 6$
$\Rightarrow \frac{x^{2} + 2 x + 7}{2 x + 3} - 6 < 0$
$\Rightarrow \frac{x^{2} + 2 x + 7 - 12 x - 18}{(2 x + 3)} < 0$
$\Rightarrow \frac{x^{2} - 10 x - 11}{(2 x + 3)} < 0$
$\Rightarrow \frac{(x - 11) (x + 1) (2 x + 3)}{(2 x + 3)^{2}} < 0$
$\Rightarrow (x - 11) (x + 1) (2 x + 3) < 0$
$\Rightarrow - 1 < x < 11 or x < \frac{- 3}{2}$

VITEEE-2011

Sets, Relation and Function

117285 If $R = {(a, b) : b = a - 1, a \in Z, 5 < a < 9}$ , then the range of $R$ is

1

{5, 6, 7}

2

{5, 6, 7, 8, 9}

3

{7, 8, 9}

4

{6, 7, 8}

Explanation:

A Given,
$R = {(a, b) : b = a - 1, a \in Z, 5 < a < 9},$
$∵ a = 6, 7, 8$
$∴ a = 6, b = 6 - 1 = 5$
$a = 7, b = 7 - 1 = 6$
$a = 8, b = 8 - 1 = 7$
$R = (6, 5), (7, 6) (8, 7)$
So, range of $R = {5, 6, 7}$

MHT-CET 20

Sets, Relation and Function

117286 If $| 3 x - 2 | \leq \frac{1}{2}$ , then $x \in$

1

(\frac{1}{2}, \frac{5}{6})

2

[\frac{1}{2}, \frac{5}{6})

3

[\frac{1}{2}, \frac{5}{6}]

4

(\frac{1}{2}, \frac{5}{6}]

Explanation:

C Given,
$| 3 x - 2 | \leq \frac{1}{2}$
$∴ \frac{- 1}{2} \leq (3 x - 2) \leq \frac{1}{2}$
$∴ \frac{- 1}{2} \leq 3 x - 2 and 3 x - 2 \leq \frac{1}{2}$
$∴ \frac{3}{2} \leq 3 x and 3 x \leq \frac{5}{2}$
$\frac{1}{2} \leq x and x \leq \frac{5}{6}$
$∴ x \in [\frac{1}{2}, \frac{5}{6}]$

MHT-CET 20

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Question Bank Series

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