QBManager

Sets, Relation and Function

117310 If $= 2 \leq 3 x - 4 \leq 5, x \in R$ , then $x$ belongs to theinterval

1

[0, 1]

2

[1, 3]

3

[\frac{1}{2}, 3]

4

[2, 3]

Explanation:

D Given,
$2 \leq 3 x - 4 \leq 5$
$2 + 4 \leq 3 x \leq 5 + 4$
$6 \leq 3 x \leq 9$
$2 \leq x \leq 3$ So, $x \in [2, 3]$

JCECE-2019

Sets, Relation and Function

117312 Let $f : (2, \infty) \to I R$ be the function defined by $f (x) = x^{2} - 4 x + 5$ . then, the range of $f$ is

1 IR

2

[1, \infty)

3

[4, \infty]

4

[5, \infty]

Explanation:

B Given,
$f : (2, \infty) \to IR$
$f (x) = x^{2} - 4 x + 5$
$f (x) = x^{2} - 4 x + 4 + 1$
$f (x) = (x - 2)^{2} + 1$ Range of $f (x) = [1, \infty)$

JCECE-2018

Sets, Relation and Function

117320 The domain of the real function $f (x) =$ $\frac{1}{\sqrt{4 - x^{2}}}$ is

1 the set of all real numbers

2 the set of all positive

3

(- 2, 2)

4

[- 2, 2]

Explanation:

C Given,
$f (x) = \frac{1}{\sqrt{4 - x^{2}}}$
$4 - x^{2} > 0$
$x^{2} < 4$
$- 2 < x < 2$ Domain $(- 2, 2)$ .

BCECE-2012

Sets, Relation and Function

117341 The domain of
$f (x) = \cos^{- 1} (\frac{x - 3}{2}) - \log_{10} (4 - x)$ is.....

1

(1, 4)

2

[1, 4)

3

(1, 4]

4

[1, 4]

Explanation:

B Let $f (x) = \cos^{- 1} (\frac{x - 3}{2}) - \log_{10} (4 - x)$
For domain of $\log_{10} (4 - x)$
$\Rightarrow 4 - x > 0 \Rightarrow x < 4$
For domain of $\cos^{- 1} (\frac{x - 3}{2}) \Rightarrow - 1 \leq \frac{x - 3}{2} \leq 1$
$\Rightarrow 1 \leq x \leq 5$ Now domain of $f (x)$ will be $x \in [1, 4)$

Shift-I

Sets, Relation and Function

117351 The set of all real values of $x$ for which the real valued function $f (x) = {(1 + \frac{1}{x})}^{x}$ is defined, is

1

(0, \infty)

2

R - {0}

3

(- \infty, - 1) \cup (0, \infty)

4

R - {0, - 1}

Explanation:

C Given,
$f (x) = {(1 + \frac{1}{x})}^{x}$
$\Rightarrow (1 + \frac{1}{x}) > 0$
$\Rightarrow \frac{x + 1}{x} > 0$ Hence, $x \in (- \infty, - 1) \cup (0, \infty)$

Shift-I

Sets, Relation and Function

117310 If $= 2 \leq 3 x - 4 \leq 5, x \in R$ , then $x$ belongs to theinterval

1

[0, 1]

2

[1, 3]

3

[\frac{1}{2}, 3]

4

[2, 3]

Explanation:

D Given,
$2 \leq 3 x - 4 \leq 5$
$2 + 4 \leq 3 x \leq 5 + 4$
$6 \leq 3 x \leq 9$
$2 \leq x \leq 3$ So, $x \in [2, 3]$

JCECE-2019

Sets, Relation and Function

117312 Let $f : (2, \infty) \to I R$ be the function defined by $f (x) = x^{2} - 4 x + 5$ . then, the range of $f$ is

1 IR

2

[1, \infty)

3

[4, \infty]

4

[5, \infty]

Explanation:

B Given,
$f : (2, \infty) \to IR$
$f (x) = x^{2} - 4 x + 5$
$f (x) = x^{2} - 4 x + 4 + 1$
$f (x) = (x - 2)^{2} + 1$ Range of $f (x) = [1, \infty)$

JCECE-2018

Sets, Relation and Function

117320 The domain of the real function $f (x) =$ $\frac{1}{\sqrt{4 - x^{2}}}$ is

1 the set of all real numbers

2 the set of all positive

3

(- 2, 2)

4

[- 2, 2]

Explanation:

C Given,
$f (x) = \frac{1}{\sqrt{4 - x^{2}}}$
$4 - x^{2} > 0$
$x^{2} < 4$
$- 2 < x < 2$ Domain $(- 2, 2)$ .

BCECE-2012

Sets, Relation and Function

117341 The domain of
$f (x) = \cos^{- 1} (\frac{x - 3}{2}) - \log_{10} (4 - x)$ is.....

1

(1, 4)

2

[1, 4)

3

(1, 4]

4

[1, 4]

Explanation:

B Let $f (x) = \cos^{- 1} (\frac{x - 3}{2}) - \log_{10} (4 - x)$
For domain of $\log_{10} (4 - x)$
$\Rightarrow 4 - x > 0 \Rightarrow x < 4$
For domain of $\cos^{- 1} (\frac{x - 3}{2}) \Rightarrow - 1 \leq \frac{x - 3}{2} \leq 1$
$\Rightarrow 1 \leq x \leq 5$ Now domain of $f (x)$ will be $x \in [1, 4)$

Shift-I

Sets, Relation and Function

117351 The set of all real values of $x$ for which the real valued function $f (x) = {(1 + \frac{1}{x})}^{x}$ is defined, is

1

(0, \infty)

2

R - {0}

3

(- \infty, - 1) \cup (0, \infty)

4

R - {0, - 1}

Explanation:

C Given,
$f (x) = {(1 + \frac{1}{x})}^{x}$
$\Rightarrow (1 + \frac{1}{x}) > 0$
$\Rightarrow \frac{x + 1}{x} > 0$ Hence, $x \in (- \infty, - 1) \cup (0, \infty)$

Shift-I

Sets, Relation and Function

117310 If $= 2 \leq 3 x - 4 \leq 5, x \in R$ , then $x$ belongs to theinterval

1

[0, 1]

2

[1, 3]

3

[\frac{1}{2}, 3]

4

[2, 3]

Explanation:

D Given,
$2 \leq 3 x - 4 \leq 5$
$2 + 4 \leq 3 x \leq 5 + 4$
$6 \leq 3 x \leq 9$
$2 \leq x \leq 3$ So, $x \in [2, 3]$

JCECE-2019

Sets, Relation and Function

117312 Let $f : (2, \infty) \to I R$ be the function defined by $f (x) = x^{2} - 4 x + 5$ . then, the range of $f$ is

1 IR

2

[1, \infty)

3

[4, \infty]

4

[5, \infty]

Explanation:

B Given,
$f : (2, \infty) \to IR$
$f (x) = x^{2} - 4 x + 5$
$f (x) = x^{2} - 4 x + 4 + 1$
$f (x) = (x - 2)^{2} + 1$ Range of $f (x) = [1, \infty)$

JCECE-2018

Sets, Relation and Function

117320 The domain of the real function $f (x) =$ $\frac{1}{\sqrt{4 - x^{2}}}$ is

1 the set of all real numbers

2 the set of all positive

3

(- 2, 2)

4

[- 2, 2]

Explanation:

C Given,
$f (x) = \frac{1}{\sqrt{4 - x^{2}}}$
$4 - x^{2} > 0$
$x^{2} < 4$
$- 2 < x < 2$ Domain $(- 2, 2)$ .

BCECE-2012

Sets, Relation and Function

117341 The domain of
$f (x) = \cos^{- 1} (\frac{x - 3}{2}) - \log_{10} (4 - x)$ is.....

1

(1, 4)

2

[1, 4)

3

(1, 4]

4

[1, 4]

Explanation:

B Let $f (x) = \cos^{- 1} (\frac{x - 3}{2}) - \log_{10} (4 - x)$
For domain of $\log_{10} (4 - x)$
$\Rightarrow 4 - x > 0 \Rightarrow x < 4$
For domain of $\cos^{- 1} (\frac{x - 3}{2}) \Rightarrow - 1 \leq \frac{x - 3}{2} \leq 1$
$\Rightarrow 1 \leq x \leq 5$ Now domain of $f (x)$ will be $x \in [1, 4)$

Shift-I

Sets, Relation and Function

117351 The set of all real values of $x$ for which the real valued function $f (x) = {(1 + \frac{1}{x})}^{x}$ is defined, is

1

(0, \infty)

2

R - {0}

3

(- \infty, - 1) \cup (0, \infty)

4

R - {0, - 1}

Explanation:

C Given,
$f (x) = {(1 + \frac{1}{x})}^{x}$
$\Rightarrow (1 + \frac{1}{x}) > 0$
$\Rightarrow \frac{x + 1}{x} > 0$ Hence, $x \in (- \infty, - 1) \cup (0, \infty)$

Shift-I

Sets, Relation and Function

117310 If $= 2 \leq 3 x - 4 \leq 5, x \in R$ , then $x$ belongs to theinterval

1

[0, 1]

2

[1, 3]

3

[\frac{1}{2}, 3]

4

[2, 3]

Explanation:

D Given,
$2 \leq 3 x - 4 \leq 5$
$2 + 4 \leq 3 x \leq 5 + 4$
$6 \leq 3 x \leq 9$
$2 \leq x \leq 3$ So, $x \in [2, 3]$

JCECE-2019

Sets, Relation and Function

117312 Let $f : (2, \infty) \to I R$ be the function defined by $f (x) = x^{2} - 4 x + 5$ . then, the range of $f$ is

1 IR

2

[1, \infty)

3

[4, \infty]

4

[5, \infty]

Explanation:

B Given,
$f : (2, \infty) \to IR$
$f (x) = x^{2} - 4 x + 5$
$f (x) = x^{2} - 4 x + 4 + 1$
$f (x) = (x - 2)^{2} + 1$ Range of $f (x) = [1, \infty)$

JCECE-2018

Sets, Relation and Function

117320 The domain of the real function $f (x) =$ $\frac{1}{\sqrt{4 - x^{2}}}$ is

1 the set of all real numbers

2 the set of all positive

3

(- 2, 2)

4

[- 2, 2]

Explanation:

C Given,
$f (x) = \frac{1}{\sqrt{4 - x^{2}}}$
$4 - x^{2} > 0$
$x^{2} < 4$
$- 2 < x < 2$ Domain $(- 2, 2)$ .

BCECE-2012

Sets, Relation and Function

117341 The domain of
$f (x) = \cos^{- 1} (\frac{x - 3}{2}) - \log_{10} (4 - x)$ is.....

1

(1, 4)

2

[1, 4)

3

(1, 4]

4

[1, 4]

Explanation:

B Let $f (x) = \cos^{- 1} (\frac{x - 3}{2}) - \log_{10} (4 - x)$
For domain of $\log_{10} (4 - x)$
$\Rightarrow 4 - x > 0 \Rightarrow x < 4$
For domain of $\cos^{- 1} (\frac{x - 3}{2}) \Rightarrow - 1 \leq \frac{x - 3}{2} \leq 1$
$\Rightarrow 1 \leq x \leq 5$ Now domain of $f (x)$ will be $x \in [1, 4)$

Shift-I

Sets, Relation and Function

117351 The set of all real values of $x$ for which the real valued function $f (x) = {(1 + \frac{1}{x})}^{x}$ is defined, is

1

(0, \infty)

2

R - {0}

3

(- \infty, - 1) \cup (0, \infty)

4

R - {0, - 1}

Explanation:

C Given,
$f (x) = {(1 + \frac{1}{x})}^{x}$
$\Rightarrow (1 + \frac{1}{x}) > 0$
$\Rightarrow \frac{x + 1}{x} > 0$ Hence, $x \in (- \infty, - 1) \cup (0, \infty)$

Shift-I

Sets, Relation and Function

117310 If $= 2 \leq 3 x - 4 \leq 5, x \in R$ , then $x$ belongs to theinterval

1

[0, 1]

2

[1, 3]

3

[\frac{1}{2}, 3]

4

[2, 3]

Explanation:

D Given,
$2 \leq 3 x - 4 \leq 5$
$2 + 4 \leq 3 x \leq 5 + 4$
$6 \leq 3 x \leq 9$
$2 \leq x \leq 3$ So, $x \in [2, 3]$

JCECE-2019

Sets, Relation and Function

117312 Let $f : (2, \infty) \to I R$ be the function defined by $f (x) = x^{2} - 4 x + 5$ . then, the range of $f$ is

1 IR

2

[1, \infty)

3

[4, \infty]

4

[5, \infty]

Explanation:

B Given,
$f : (2, \infty) \to IR$
$f (x) = x^{2} - 4 x + 5$
$f (x) = x^{2} - 4 x + 4 + 1$
$f (x) = (x - 2)^{2} + 1$ Range of $f (x) = [1, \infty)$

JCECE-2018

Sets, Relation and Function

117320 The domain of the real function $f (x) =$ $\frac{1}{\sqrt{4 - x^{2}}}$ is

1 the set of all real numbers

2 the set of all positive

3

(- 2, 2)

4

[- 2, 2]

Explanation:

C Given,
$f (x) = \frac{1}{\sqrt{4 - x^{2}}}$
$4 - x^{2} > 0$
$x^{2} < 4$
$- 2 < x < 2$ Domain $(- 2, 2)$ .

BCECE-2012

Sets, Relation and Function

117341 The domain of
$f (x) = \cos^{- 1} (\frac{x - 3}{2}) - \log_{10} (4 - x)$ is.....

1

(1, 4)

2

[1, 4)

3

(1, 4]

4

[1, 4]

Explanation:

B Let $f (x) = \cos^{- 1} (\frac{x - 3}{2}) - \log_{10} (4 - x)$
For domain of $\log_{10} (4 - x)$
$\Rightarrow 4 - x > 0 \Rightarrow x < 4$
For domain of $\cos^{- 1} (\frac{x - 3}{2}) \Rightarrow - 1 \leq \frac{x - 3}{2} \leq 1$
$\Rightarrow 1 \leq x \leq 5$ Now domain of $f (x)$ will be $x \in [1, 4)$

Shift-I

Sets, Relation and Function

117351 The set of all real values of $x$ for which the real valued function $f (x) = {(1 + \frac{1}{x})}^{x}$ is defined, is

1

(0, \infty)

2

R - {0}

3

(- \infty, - 1) \cup (0, \infty)

4

R - {0, - 1}

Explanation:

C Given,
$f (x) = {(1 + \frac{1}{x})}^{x}$
$\Rightarrow (1 + \frac{1}{x}) > 0$
$\Rightarrow \frac{x + 1}{x} > 0$ Hence, $x \in (- \infty, - 1) \cup (0, \infty)$

Shift-I

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