QBManager

Sets, Relation and Function

117328 Let $f : R \to R$ and $g : R \to R$ be two functions defined by $f (x) = \log_{e} (x^{2} + 1) - e^{- x} + 1$ and $g (x)$ $= \frac{1 - 2 e^{2 x}}{e^{x}}$ . Then, for which of the following range of $α$ , the inequality
$f (g (\frac{(α - 1)^{2}}{3})) > f (g (α - \frac{5}{3}))$ holds?

1

(2, 3)

2

(- 2, - 1)

3

(1, 2)

4

(- 1, 1)

Explanation:

A Given,
$f (x) = \log_{e} (x^{2} + 1) - e^{- x} + 1$
$\Rightarrow f^{'} (x) = \frac{2 x}{x^{2} + 1} + e^{- x} > 0 \forall x \in R$
$\Rightarrow f$ is strictly increasing
$g (x) = \frac{1 - 2 e^{2 x}}{e^{x}} = e^{- x} - 2 e^{x}$
$g^{'} (x) = - e^{- x} - 2 e^{x}$
$\Rightarrow g^{'} (x) = - (2 e^{x} + e^{- x}) < 0 \forall x \in R$
$\Rightarrow g$ is decreasing
Now $f (g (\frac{(α - 1)^{2}}{3})) > f (g (α - \frac{5}{3}))$
$\Rightarrow g \frac{(α - 1)^{2}}{3} > g (α - \frac{5}{3})$
$\Rightarrow α^{2} - 5 α + 6 < 0$
$\Rightarrow (α - 2) (α - 3) < 0$
$\Rightarrow α \in (2, 3)$

Shift-I

Sets, Relation and Function

117329 The domain of the function
$f (x) = \frac{\cos^{- 1} (\frac{x^{2} - 5 x + 6}{x^{2} - 9})}{\log_{e} (x^{2} - 3 x + 2)}$ is

1

(- \infty, 1) \cup (2, \infty)

2

(2, \infty)

3

[- \frac{1}{2}, 1) \cup (2, \infty)

4

[- \frac{1}{2}, 1) \cup (2, \infty) - {\frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}}

Explanation:

D The domain of the function,
$- 1 \leq \frac{x^{2} - 5 x + 6}{x^{2} - 9} \leq 1$
$\frac{x^{2} - 5 x + 6}{x^{2} - 9} - 1 \leq 0 and \frac{x^{2} - 5 x + 6}{x^{2} - 9} + 1 \geq 0$
$\Rightarrow \frac{(x - 2) (x - 3)}{(x - 3) (x + 3)} - 1 \leq 0 and \frac{(x - 2) (x - 3)}{(x - 3) (x + 3)} + 1 \geq 0$
$\Rightarrow \frac{x - 2}{x + 3} - 1 \leq 0 and \frac{(x - 2)}{x + 3} + 1 \geq 0$
$\frac{- 5}{x + 3} \leq 0 and \frac{2 x + 1}{x + 3} \geq 0$
$\Rightarrow x > - 3$
$and (2 x + 1) (\frac{1}{x + 3}) \geq 0$
$\Rightarrow x \in (- 3, \infty) and x \in (- \infty, - 3) \cup [- \frac{1}{2}, \infty)$
After taking intersection -
$x \in [- \frac{1}{2}, \infty)$
$And, x^{2} - 3 x + 2 > 0$
$(x - 1) (x - 2) > 0$
$x > 1 or x > 2$
$x \in (1, \infty) \cup (2, \infty)$
$x^{2} - 3 x + 2 \neq 1$
$x \neq \frac{3 \pm \sqrt{5}}{2}$
After taking intersection of each solution -
$[\frac{- 1}{2}, 1) \cup (2, \infty) - {\frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}} .$

Shift-I

Sets, Relation and Function

117330 What is the range the function $h (x) = \frac{x - 2}{x + 3}$ ?

1

(- \infty, 2) \cup (2, \infty)

2

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

3

(- \infty, - 3) \cup (- 3, \infty)

4

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

Explanation:

B Let, $y = \frac{x - 2}{x + 3}$
$y x + 3 y = x - 2$
$x = \frac{3 y + 2}{1 - y}$
...(ii) ${$ write $x$ in terms of $y}$
Domain of $y$ in equation (ii), will be $R - {1}$ or in other words-
$(- \infty, 1) \cup (1, \infty),$ So, range of $y = \frac{x - 2}{x + 3}$ will be $(- \infty, 1) \cup (1, \infty)$

Shift-I

Sets, Relation and Function

117331 If $f : R \to [- 1, 1]$ and $g : R \to A$ are two subjective mappings and $\sin (g (x) - \frac{π}{3}) = \frac{f (x)}{2} \sqrt{4 - f^{2} (x)}$ , then $A =$

1

[0, \frac{2 π}{3}]

2

[- 1, 1]

3

[\frac{- π}{2}, \frac{π}{2}]

4

(0, π)

Explanation:

A Let, $f (x) = y$
Then, $\sin (g (x) - \frac{π}{3}) = \frac{f (x)}{2} \sqrt{4 - f^{2} (x)}$ ,
Let,
$\Rightarrow \frac{y}{2} \sqrt{4 - y^{2}} = p$
$\Rightarrow \frac{y^{2}}{4} (4 - y^{2}) = p^{2}$
$\Rightarrow y^{2} - \frac{y^{4}}{4} = p^{2}$
$\Rightarrow 4 y^{2} - y^{4} = 4 p^{2}$
$\Rightarrow {(y^{2})}^{2} - 2 \times 2 \times y^{2} + 2^{2} - 2^{2} = - 4 p^{2}$
$\Rightarrow p^{2} = 1 - \frac{1}{4} {(y^{2} - 2)}^{2}$
$∴ f (x) = y \in [- 1, 1]$
$\Rightarrow y^{2} \in [0, 1]$
$∴ p^{2} \in [0, \frac{3}{4}]$
$p \in [- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$
So, $\sin (g (x) - \frac{π}{3}) \in [- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$
$\Rightarrow g (x) - \frac{π}{3} \in [\frac{- π}{3}, \frac{π}{3}]$
$\Rightarrow g (x) \in [0, \frac{2 π}{3}]$

Shift-II

Sets, Relation and Function

117332 The domain of the function $f (x) = \sqrt{\frac{4 - x^{2}}{[x] + 2}}$ , where $[x]$ denotes the greatest integer not more than $x$ , is

1

(- \infty, - 2) \cup (1, 2)

2

(- \infty, - 2) \cup (- 1, 2)

3

(- \infty, - 2), \cup [- 1, 2]

4

(- \infty, - 1) \cup (1, 2)

Explanation:

C For $f (x)$ to be defined,
$\frac{4 - x^{2}}{[x] + 2} \geq 0$
$[x] + 2 \neq 0 x \notin - 2$
$Case 1 : 4 - x^{2} < 0,$
$[x] + 2 < 0$
$x > \pm 2$
$\Rightarrow x^{2} > 4$
$x \in (- \infty, - 2) \cup (2, \infty)$
Case $2 : 4 - x^{2} \geq 0, [x] + 2 > 0 x$
$\Rightarrow x^{2} \leq 4$
$\Rightarrow x \leq \pm 2$
$\Rightarrow x \in [- 2, 2] x \in [- 1, \infty) x \notin - 2$
$x \in [- 1, 2]$
Form (i) and (ii), we get -
$x \in (- \infty, - 2) \cup [- 1, 2]$

Shift-II

Sets, Relation and Function

117328 Let $f : R \to R$ and $g : R \to R$ be two functions defined by $f (x) = \log_{e} (x^{2} + 1) - e^{- x} + 1$ and $g (x)$ $= \frac{1 - 2 e^{2 x}}{e^{x}}$ . Then, for which of the following range of $α$ , the inequality
$f (g (\frac{(α - 1)^{2}}{3})) > f (g (α - \frac{5}{3}))$ holds?

1

(2, 3)

2

(- 2, - 1)

3

(1, 2)

4

(- 1, 1)

Explanation:

A Given,
$f (x) = \log_{e} (x^{2} + 1) - e^{- x} + 1$
$\Rightarrow f^{'} (x) = \frac{2 x}{x^{2} + 1} + e^{- x} > 0 \forall x \in R$
$\Rightarrow f$ is strictly increasing
$g (x) = \frac{1 - 2 e^{2 x}}{e^{x}} = e^{- x} - 2 e^{x}$
$g^{'} (x) = - e^{- x} - 2 e^{x}$
$\Rightarrow g^{'} (x) = - (2 e^{x} + e^{- x}) < 0 \forall x \in R$
$\Rightarrow g$ is decreasing
Now $f (g (\frac{(α - 1)^{2}}{3})) > f (g (α - \frac{5}{3}))$
$\Rightarrow g \frac{(α - 1)^{2}}{3} > g (α - \frac{5}{3})$
$\Rightarrow α^{2} - 5 α + 6 < 0$
$\Rightarrow (α - 2) (α - 3) < 0$
$\Rightarrow α \in (2, 3)$

Shift-I

Sets, Relation and Function

117329 The domain of the function
$f (x) = \frac{\cos^{- 1} (\frac{x^{2} - 5 x + 6}{x^{2} - 9})}{\log_{e} (x^{2} - 3 x + 2)}$ is

1

(- \infty, 1) \cup (2, \infty)

2

(2, \infty)

3

[- \frac{1}{2}, 1) \cup (2, \infty)

4

[- \frac{1}{2}, 1) \cup (2, \infty) - {\frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}}

Explanation:

D The domain of the function,
$- 1 \leq \frac{x^{2} - 5 x + 6}{x^{2} - 9} \leq 1$
$\frac{x^{2} - 5 x + 6}{x^{2} - 9} - 1 \leq 0 and \frac{x^{2} - 5 x + 6}{x^{2} - 9} + 1 \geq 0$
$\Rightarrow \frac{(x - 2) (x - 3)}{(x - 3) (x + 3)} - 1 \leq 0 and \frac{(x - 2) (x - 3)}{(x - 3) (x + 3)} + 1 \geq 0$
$\Rightarrow \frac{x - 2}{x + 3} - 1 \leq 0 and \frac{(x - 2)}{x + 3} + 1 \geq 0$
$\frac{- 5}{x + 3} \leq 0 and \frac{2 x + 1}{x + 3} \geq 0$
$\Rightarrow x > - 3$
$and (2 x + 1) (\frac{1}{x + 3}) \geq 0$
$\Rightarrow x \in (- 3, \infty) and x \in (- \infty, - 3) \cup [- \frac{1}{2}, \infty)$
After taking intersection -
$x \in [- \frac{1}{2}, \infty)$
$And, x^{2} - 3 x + 2 > 0$
$(x - 1) (x - 2) > 0$
$x > 1 or x > 2$
$x \in (1, \infty) \cup (2, \infty)$
$x^{2} - 3 x + 2 \neq 1$
$x \neq \frac{3 \pm \sqrt{5}}{2}$
After taking intersection of each solution -
$[\frac{- 1}{2}, 1) \cup (2, \infty) - {\frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}} .$

Shift-I

Sets, Relation and Function

117330 What is the range the function $h (x) = \frac{x - 2}{x + 3}$ ?

1

(- \infty, 2) \cup (2, \infty)

2

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

3

(- \infty, - 3) \cup (- 3, \infty)

4

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

Explanation:

B Let, $y = \frac{x - 2}{x + 3}$
$y x + 3 y = x - 2$
$x = \frac{3 y + 2}{1 - y}$
...(ii) ${$ write $x$ in terms of $y}$
Domain of $y$ in equation (ii), will be $R - {1}$ or in other words-
$(- \infty, 1) \cup (1, \infty),$ So, range of $y = \frac{x - 2}{x + 3}$ will be $(- \infty, 1) \cup (1, \infty)$

Shift-I

Sets, Relation and Function

117331 If $f : R \to [- 1, 1]$ and $g : R \to A$ are two subjective mappings and $\sin (g (x) - \frac{π}{3}) = \frac{f (x)}{2} \sqrt{4 - f^{2} (x)}$ , then $A =$

1

[0, \frac{2 π}{3}]

2

[- 1, 1]

3

[\frac{- π}{2}, \frac{π}{2}]

4

(0, π)

Explanation:

A Let, $f (x) = y$
Then, $\sin (g (x) - \frac{π}{3}) = \frac{f (x)}{2} \sqrt{4 - f^{2} (x)}$ ,
Let,
$\Rightarrow \frac{y}{2} \sqrt{4 - y^{2}} = p$
$\Rightarrow \frac{y^{2}}{4} (4 - y^{2}) = p^{2}$
$\Rightarrow y^{2} - \frac{y^{4}}{4} = p^{2}$
$\Rightarrow 4 y^{2} - y^{4} = 4 p^{2}$
$\Rightarrow {(y^{2})}^{2} - 2 \times 2 \times y^{2} + 2^{2} - 2^{2} = - 4 p^{2}$
$\Rightarrow p^{2} = 1 - \frac{1}{4} {(y^{2} - 2)}^{2}$
$∴ f (x) = y \in [- 1, 1]$
$\Rightarrow y^{2} \in [0, 1]$
$∴ p^{2} \in [0, \frac{3}{4}]$
$p \in [- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$
So, $\sin (g (x) - \frac{π}{3}) \in [- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$
$\Rightarrow g (x) - \frac{π}{3} \in [\frac{- π}{3}, \frac{π}{3}]$
$\Rightarrow g (x) \in [0, \frac{2 π}{3}]$

Shift-II

Sets, Relation and Function

117332 The domain of the function $f (x) = \sqrt{\frac{4 - x^{2}}{[x] + 2}}$ , where $[x]$ denotes the greatest integer not more than $x$ , is

1

(- \infty, - 2) \cup (1, 2)

2

(- \infty, - 2) \cup (- 1, 2)

3

(- \infty, - 2), \cup [- 1, 2]

4

(- \infty, - 1) \cup (1, 2)

Explanation:

C For $f (x)$ to be defined,
$\frac{4 - x^{2}}{[x] + 2} \geq 0$
$[x] + 2 \neq 0 x \notin - 2$
$Case 1 : 4 - x^{2} < 0,$
$[x] + 2 < 0$
$x > \pm 2$
$\Rightarrow x^{2} > 4$
$x \in (- \infty, - 2) \cup (2, \infty)$
Case $2 : 4 - x^{2} \geq 0, [x] + 2 > 0 x$
$\Rightarrow x^{2} \leq 4$
$\Rightarrow x \leq \pm 2$
$\Rightarrow x \in [- 2, 2] x \in [- 1, \infty) x \notin - 2$
$x \in [- 1, 2]$
Form (i) and (ii), we get -
$x \in (- \infty, - 2) \cup [- 1, 2]$

Shift-II

Sets, Relation and Function

117328 Let $f : R \to R$ and $g : R \to R$ be two functions defined by $f (x) = \log_{e} (x^{2} + 1) - e^{- x} + 1$ and $g (x)$ $= \frac{1 - 2 e^{2 x}}{e^{x}}$ . Then, for which of the following range of $α$ , the inequality
$f (g (\frac{(α - 1)^{2}}{3})) > f (g (α - \frac{5}{3}))$ holds?

1

(2, 3)

2

(- 2, - 1)

3

(1, 2)

4

(- 1, 1)

Explanation:

A Given,
$f (x) = \log_{e} (x^{2} + 1) - e^{- x} + 1$
$\Rightarrow f^{'} (x) = \frac{2 x}{x^{2} + 1} + e^{- x} > 0 \forall x \in R$
$\Rightarrow f$ is strictly increasing
$g (x) = \frac{1 - 2 e^{2 x}}{e^{x}} = e^{- x} - 2 e^{x}$
$g^{'} (x) = - e^{- x} - 2 e^{x}$
$\Rightarrow g^{'} (x) = - (2 e^{x} + e^{- x}) < 0 \forall x \in R$
$\Rightarrow g$ is decreasing
Now $f (g (\frac{(α - 1)^{2}}{3})) > f (g (α - \frac{5}{3}))$
$\Rightarrow g \frac{(α - 1)^{2}}{3} > g (α - \frac{5}{3})$
$\Rightarrow α^{2} - 5 α + 6 < 0$
$\Rightarrow (α - 2) (α - 3) < 0$
$\Rightarrow α \in (2, 3)$

Shift-I

Sets, Relation and Function

117329 The domain of the function
$f (x) = \frac{\cos^{- 1} (\frac{x^{2} - 5 x + 6}{x^{2} - 9})}{\log_{e} (x^{2} - 3 x + 2)}$ is

1

(- \infty, 1) \cup (2, \infty)

2

(2, \infty)

3

[- \frac{1}{2}, 1) \cup (2, \infty)

4

[- \frac{1}{2}, 1) \cup (2, \infty) - {\frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}}

Explanation:

D The domain of the function,
$- 1 \leq \frac{x^{2} - 5 x + 6}{x^{2} - 9} \leq 1$
$\frac{x^{2} - 5 x + 6}{x^{2} - 9} - 1 \leq 0 and \frac{x^{2} - 5 x + 6}{x^{2} - 9} + 1 \geq 0$
$\Rightarrow \frac{(x - 2) (x - 3)}{(x - 3) (x + 3)} - 1 \leq 0 and \frac{(x - 2) (x - 3)}{(x - 3) (x + 3)} + 1 \geq 0$
$\Rightarrow \frac{x - 2}{x + 3} - 1 \leq 0 and \frac{(x - 2)}{x + 3} + 1 \geq 0$
$\frac{- 5}{x + 3} \leq 0 and \frac{2 x + 1}{x + 3} \geq 0$
$\Rightarrow x > - 3$
$and (2 x + 1) (\frac{1}{x + 3}) \geq 0$
$\Rightarrow x \in (- 3, \infty) and x \in (- \infty, - 3) \cup [- \frac{1}{2}, \infty)$
After taking intersection -
$x \in [- \frac{1}{2}, \infty)$
$And, x^{2} - 3 x + 2 > 0$
$(x - 1) (x - 2) > 0$
$x > 1 or x > 2$
$x \in (1, \infty) \cup (2, \infty)$
$x^{2} - 3 x + 2 \neq 1$
$x \neq \frac{3 \pm \sqrt{5}}{2}$
After taking intersection of each solution -
$[\frac{- 1}{2}, 1) \cup (2, \infty) - {\frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}} .$

Shift-I

Sets, Relation and Function

117330 What is the range the function $h (x) = \frac{x - 2}{x + 3}$ ?

1

(- \infty, 2) \cup (2, \infty)

2

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

3

(- \infty, - 3) \cup (- 3, \infty)

4

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

Explanation:

B Let, $y = \frac{x - 2}{x + 3}$
$y x + 3 y = x - 2$
$x = \frac{3 y + 2}{1 - y}$
...(ii) ${$ write $x$ in terms of $y}$
Domain of $y$ in equation (ii), will be $R - {1}$ or in other words-
$(- \infty, 1) \cup (1, \infty),$ So, range of $y = \frac{x - 2}{x + 3}$ will be $(- \infty, 1) \cup (1, \infty)$

Shift-I

Sets, Relation and Function

117331 If $f : R \to [- 1, 1]$ and $g : R \to A$ are two subjective mappings and $\sin (g (x) - \frac{π}{3}) = \frac{f (x)}{2} \sqrt{4 - f^{2} (x)}$ , then $A =$

1

[0, \frac{2 π}{3}]

2

[- 1, 1]

3

[\frac{- π}{2}, \frac{π}{2}]

4

(0, π)

Explanation:

A Let, $f (x) = y$
Then, $\sin (g (x) - \frac{π}{3}) = \frac{f (x)}{2} \sqrt{4 - f^{2} (x)}$ ,
Let,
$\Rightarrow \frac{y}{2} \sqrt{4 - y^{2}} = p$
$\Rightarrow \frac{y^{2}}{4} (4 - y^{2}) = p^{2}$
$\Rightarrow y^{2} - \frac{y^{4}}{4} = p^{2}$
$\Rightarrow 4 y^{2} - y^{4} = 4 p^{2}$
$\Rightarrow {(y^{2})}^{2} - 2 \times 2 \times y^{2} + 2^{2} - 2^{2} = - 4 p^{2}$
$\Rightarrow p^{2} = 1 - \frac{1}{4} {(y^{2} - 2)}^{2}$
$∴ f (x) = y \in [- 1, 1]$
$\Rightarrow y^{2} \in [0, 1]$
$∴ p^{2} \in [0, \frac{3}{4}]$
$p \in [- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$
So, $\sin (g (x) - \frac{π}{3}) \in [- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$
$\Rightarrow g (x) - \frac{π}{3} \in [\frac{- π}{3}, \frac{π}{3}]$
$\Rightarrow g (x) \in [0, \frac{2 π}{3}]$

Shift-II

Sets, Relation and Function

117332 The domain of the function $f (x) = \sqrt{\frac{4 - x^{2}}{[x] + 2}}$ , where $[x]$ denotes the greatest integer not more than $x$ , is

1

(- \infty, - 2) \cup (1, 2)

2

(- \infty, - 2) \cup (- 1, 2)

3

(- \infty, - 2), \cup [- 1, 2]

4

(- \infty, - 1) \cup (1, 2)

Explanation:

C For $f (x)$ to be defined,
$\frac{4 - x^{2}}{[x] + 2} \geq 0$
$[x] + 2 \neq 0 x \notin - 2$
$Case 1 : 4 - x^{2} < 0,$
$[x] + 2 < 0$
$x > \pm 2$
$\Rightarrow x^{2} > 4$
$x \in (- \infty, - 2) \cup (2, \infty)$
Case $2 : 4 - x^{2} \geq 0, [x] + 2 > 0 x$
$\Rightarrow x^{2} \leq 4$
$\Rightarrow x \leq \pm 2$
$\Rightarrow x \in [- 2, 2] x \in [- 1, \infty) x \notin - 2$
$x \in [- 1, 2]$
Form (i) and (ii), we get -
$x \in (- \infty, - 2) \cup [- 1, 2]$

Shift-II

Sets, Relation and Function

117328 Let $f : R \to R$ and $g : R \to R$ be two functions defined by $f (x) = \log_{e} (x^{2} + 1) - e^{- x} + 1$ and $g (x)$ $= \frac{1 - 2 e^{2 x}}{e^{x}}$ . Then, for which of the following range of $α$ , the inequality
$f (g (\frac{(α - 1)^{2}}{3})) > f (g (α - \frac{5}{3}))$ holds?

1

(2, 3)

2

(- 2, - 1)

3

(1, 2)

4

(- 1, 1)

Explanation:

A Given,
$f (x) = \log_{e} (x^{2} + 1) - e^{- x} + 1$
$\Rightarrow f^{'} (x) = \frac{2 x}{x^{2} + 1} + e^{- x} > 0 \forall x \in R$
$\Rightarrow f$ is strictly increasing
$g (x) = \frac{1 - 2 e^{2 x}}{e^{x}} = e^{- x} - 2 e^{x}$
$g^{'} (x) = - e^{- x} - 2 e^{x}$
$\Rightarrow g^{'} (x) = - (2 e^{x} + e^{- x}) < 0 \forall x \in R$
$\Rightarrow g$ is decreasing
Now $f (g (\frac{(α - 1)^{2}}{3})) > f (g (α - \frac{5}{3}))$
$\Rightarrow g \frac{(α - 1)^{2}}{3} > g (α - \frac{5}{3})$
$\Rightarrow α^{2} - 5 α + 6 < 0$
$\Rightarrow (α - 2) (α - 3) < 0$
$\Rightarrow α \in (2, 3)$

Shift-I

Sets, Relation and Function

117329 The domain of the function
$f (x) = \frac{\cos^{- 1} (\frac{x^{2} - 5 x + 6}{x^{2} - 9})}{\log_{e} (x^{2} - 3 x + 2)}$ is

1

(- \infty, 1) \cup (2, \infty)

2

(2, \infty)

3

[- \frac{1}{2}, 1) \cup (2, \infty)

4

[- \frac{1}{2}, 1) \cup (2, \infty) - {\frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}}

Explanation:

D The domain of the function,
$- 1 \leq \frac{x^{2} - 5 x + 6}{x^{2} - 9} \leq 1$
$\frac{x^{2} - 5 x + 6}{x^{2} - 9} - 1 \leq 0 and \frac{x^{2} - 5 x + 6}{x^{2} - 9} + 1 \geq 0$
$\Rightarrow \frac{(x - 2) (x - 3)}{(x - 3) (x + 3)} - 1 \leq 0 and \frac{(x - 2) (x - 3)}{(x - 3) (x + 3)} + 1 \geq 0$
$\Rightarrow \frac{x - 2}{x + 3} - 1 \leq 0 and \frac{(x - 2)}{x + 3} + 1 \geq 0$
$\frac{- 5}{x + 3} \leq 0 and \frac{2 x + 1}{x + 3} \geq 0$
$\Rightarrow x > - 3$
$and (2 x + 1) (\frac{1}{x + 3}) \geq 0$
$\Rightarrow x \in (- 3, \infty) and x \in (- \infty, - 3) \cup [- \frac{1}{2}, \infty)$
After taking intersection -
$x \in [- \frac{1}{2}, \infty)$
$And, x^{2} - 3 x + 2 > 0$
$(x - 1) (x - 2) > 0$
$x > 1 or x > 2$
$x \in (1, \infty) \cup (2, \infty)$
$x^{2} - 3 x + 2 \neq 1$
$x \neq \frac{3 \pm \sqrt{5}}{2}$
After taking intersection of each solution -
$[\frac{- 1}{2}, 1) \cup (2, \infty) - {\frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}} .$

Shift-I

Sets, Relation and Function

117330 What is the range the function $h (x) = \frac{x - 2}{x + 3}$ ?

1

(- \infty, 2) \cup (2, \infty)

2

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

3

(- \infty, - 3) \cup (- 3, \infty)

4

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

Explanation:

B Let, $y = \frac{x - 2}{x + 3}$
$y x + 3 y = x - 2$
$x = \frac{3 y + 2}{1 - y}$
...(ii) ${$ write $x$ in terms of $y}$
Domain of $y$ in equation (ii), will be $R - {1}$ or in other words-
$(- \infty, 1) \cup (1, \infty),$ So, range of $y = \frac{x - 2}{x + 3}$ will be $(- \infty, 1) \cup (1, \infty)$

Shift-I

Sets, Relation and Function

117331 If $f : R \to [- 1, 1]$ and $g : R \to A$ are two subjective mappings and $\sin (g (x) - \frac{π}{3}) = \frac{f (x)}{2} \sqrt{4 - f^{2} (x)}$ , then $A =$

1

[0, \frac{2 π}{3}]

2

[- 1, 1]

3

[\frac{- π}{2}, \frac{π}{2}]

4

(0, π)

Explanation:

A Let, $f (x) = y$
Then, $\sin (g (x) - \frac{π}{3}) = \frac{f (x)}{2} \sqrt{4 - f^{2} (x)}$ ,
Let,
$\Rightarrow \frac{y}{2} \sqrt{4 - y^{2}} = p$
$\Rightarrow \frac{y^{2}}{4} (4 - y^{2}) = p^{2}$
$\Rightarrow y^{2} - \frac{y^{4}}{4} = p^{2}$
$\Rightarrow 4 y^{2} - y^{4} = 4 p^{2}$
$\Rightarrow {(y^{2})}^{2} - 2 \times 2 \times y^{2} + 2^{2} - 2^{2} = - 4 p^{2}$
$\Rightarrow p^{2} = 1 - \frac{1}{4} {(y^{2} - 2)}^{2}$
$∴ f (x) = y \in [- 1, 1]$
$\Rightarrow y^{2} \in [0, 1]$
$∴ p^{2} \in [0, \frac{3}{4}]$
$p \in [- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$
So, $\sin (g (x) - \frac{π}{3}) \in [- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$
$\Rightarrow g (x) - \frac{π}{3} \in [\frac{- π}{3}, \frac{π}{3}]$
$\Rightarrow g (x) \in [0, \frac{2 π}{3}]$

Shift-II

Sets, Relation and Function

117332 The domain of the function $f (x) = \sqrt{\frac{4 - x^{2}}{[x] + 2}}$ , where $[x]$ denotes the greatest integer not more than $x$ , is

1

(- \infty, - 2) \cup (1, 2)

2

(- \infty, - 2) \cup (- 1, 2)

3

(- \infty, - 2), \cup [- 1, 2]

4

(- \infty, - 1) \cup (1, 2)

Explanation:

C For $f (x)$ to be defined,
$\frac{4 - x^{2}}{[x] + 2} \geq 0$
$[x] + 2 \neq 0 x \notin - 2$
$Case 1 : 4 - x^{2} < 0,$
$[x] + 2 < 0$
$x > \pm 2$
$\Rightarrow x^{2} > 4$
$x \in (- \infty, - 2) \cup (2, \infty)$
Case $2 : 4 - x^{2} \geq 0, [x] + 2 > 0 x$
$\Rightarrow x^{2} \leq 4$
$\Rightarrow x \leq \pm 2$
$\Rightarrow x \in [- 2, 2] x \in [- 1, \infty) x \notin - 2$
$x \in [- 1, 2]$
Form (i) and (ii), we get -
$x \in (- \infty, - 2) \cup [- 1, 2]$

Shift-II

Sets, Relation and Function

117328 Let $f : R \to R$ and $g : R \to R$ be two functions defined by $f (x) = \log_{e} (x^{2} + 1) - e^{- x} + 1$ and $g (x)$ $= \frac{1 - 2 e^{2 x}}{e^{x}}$ . Then, for which of the following range of $α$ , the inequality
$f (g (\frac{(α - 1)^{2}}{3})) > f (g (α - \frac{5}{3}))$ holds?

1

(2, 3)

2

(- 2, - 1)

3

(1, 2)

4

(- 1, 1)

Explanation:

A Given,
$f (x) = \log_{e} (x^{2} + 1) - e^{- x} + 1$
$\Rightarrow f^{'} (x) = \frac{2 x}{x^{2} + 1} + e^{- x} > 0 \forall x \in R$
$\Rightarrow f$ is strictly increasing
$g (x) = \frac{1 - 2 e^{2 x}}{e^{x}} = e^{- x} - 2 e^{x}$
$g^{'} (x) = - e^{- x} - 2 e^{x}$
$\Rightarrow g^{'} (x) = - (2 e^{x} + e^{- x}) < 0 \forall x \in R$
$\Rightarrow g$ is decreasing
Now $f (g (\frac{(α - 1)^{2}}{3})) > f (g (α - \frac{5}{3}))$
$\Rightarrow g \frac{(α - 1)^{2}}{3} > g (α - \frac{5}{3})$
$\Rightarrow α^{2} - 5 α + 6 < 0$
$\Rightarrow (α - 2) (α - 3) < 0$
$\Rightarrow α \in (2, 3)$

Shift-I

Sets, Relation and Function

117329 The domain of the function
$f (x) = \frac{\cos^{- 1} (\frac{x^{2} - 5 x + 6}{x^{2} - 9})}{\log_{e} (x^{2} - 3 x + 2)}$ is

1

(- \infty, 1) \cup (2, \infty)

2

(2, \infty)

3

[- \frac{1}{2}, 1) \cup (2, \infty)

4

[- \frac{1}{2}, 1) \cup (2, \infty) - {\frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}}

Explanation:

D The domain of the function,
$- 1 \leq \frac{x^{2} - 5 x + 6}{x^{2} - 9} \leq 1$
$\frac{x^{2} - 5 x + 6}{x^{2} - 9} - 1 \leq 0 and \frac{x^{2} - 5 x + 6}{x^{2} - 9} + 1 \geq 0$
$\Rightarrow \frac{(x - 2) (x - 3)}{(x - 3) (x + 3)} - 1 \leq 0 and \frac{(x - 2) (x - 3)}{(x - 3) (x + 3)} + 1 \geq 0$
$\Rightarrow \frac{x - 2}{x + 3} - 1 \leq 0 and \frac{(x - 2)}{x + 3} + 1 \geq 0$
$\frac{- 5}{x + 3} \leq 0 and \frac{2 x + 1}{x + 3} \geq 0$
$\Rightarrow x > - 3$
$and (2 x + 1) (\frac{1}{x + 3}) \geq 0$
$\Rightarrow x \in (- 3, \infty) and x \in (- \infty, - 3) \cup [- \frac{1}{2}, \infty)$
After taking intersection -
$x \in [- \frac{1}{2}, \infty)$
$And, x^{2} - 3 x + 2 > 0$
$(x - 1) (x - 2) > 0$
$x > 1 or x > 2$
$x \in (1, \infty) \cup (2, \infty)$
$x^{2} - 3 x + 2 \neq 1$
$x \neq \frac{3 \pm \sqrt{5}}{2}$
After taking intersection of each solution -
$[\frac{- 1}{2}, 1) \cup (2, \infty) - {\frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}} .$

Shift-I

Sets, Relation and Function

117330 What is the range the function $h (x) = \frac{x - 2}{x + 3}$ ?

1

(- \infty, 2) \cup (2, \infty)

2

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

3

(- \infty, - 3) \cup (- 3, \infty)

4

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

Explanation:

B Let, $y = \frac{x - 2}{x + 3}$
$y x + 3 y = x - 2$
$x = \frac{3 y + 2}{1 - y}$
...(ii) ${$ write $x$ in terms of $y}$
Domain of $y$ in equation (ii), will be $R - {1}$ or in other words-
$(- \infty, 1) \cup (1, \infty),$ So, range of $y = \frac{x - 2}{x + 3}$ will be $(- \infty, 1) \cup (1, \infty)$

Shift-I

Sets, Relation and Function

117331 If $f : R \to [- 1, 1]$ and $g : R \to A$ are two subjective mappings and $\sin (g (x) - \frac{π}{3}) = \frac{f (x)}{2} \sqrt{4 - f^{2} (x)}$ , then $A =$

1

[0, \frac{2 π}{3}]

2

[- 1, 1]

3

[\frac{- π}{2}, \frac{π}{2}]

4

(0, π)

Explanation:

A Let, $f (x) = y$
Then, $\sin (g (x) - \frac{π}{3}) = \frac{f (x)}{2} \sqrt{4 - f^{2} (x)}$ ,
Let,
$\Rightarrow \frac{y}{2} \sqrt{4 - y^{2}} = p$
$\Rightarrow \frac{y^{2}}{4} (4 - y^{2}) = p^{2}$
$\Rightarrow y^{2} - \frac{y^{4}}{4} = p^{2}$
$\Rightarrow 4 y^{2} - y^{4} = 4 p^{2}$
$\Rightarrow {(y^{2})}^{2} - 2 \times 2 \times y^{2} + 2^{2} - 2^{2} = - 4 p^{2}$
$\Rightarrow p^{2} = 1 - \frac{1}{4} {(y^{2} - 2)}^{2}$
$∴ f (x) = y \in [- 1, 1]$
$\Rightarrow y^{2} \in [0, 1]$
$∴ p^{2} \in [0, \frac{3}{4}]$
$p \in [- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$
So, $\sin (g (x) - \frac{π}{3}) \in [- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$
$\Rightarrow g (x) - \frac{π}{3} \in [\frac{- π}{3}, \frac{π}{3}]$
$\Rightarrow g (x) \in [0, \frac{2 π}{3}]$

Shift-II

Sets, Relation and Function

117332 The domain of the function $f (x) = \sqrt{\frac{4 - x^{2}}{[x] + 2}}$ , where $[x]$ denotes the greatest integer not more than $x$ , is

1

(- \infty, - 2) \cup (1, 2)

2

(- \infty, - 2) \cup (- 1, 2)

3

(- \infty, - 2), \cup [- 1, 2]

4

(- \infty, - 1) \cup (1, 2)

Explanation:

C For $f (x)$ to be defined,
$\frac{4 - x^{2}}{[x] + 2} \geq 0$
$[x] + 2 \neq 0 x \notin - 2$
$Case 1 : 4 - x^{2} < 0,$
$[x] + 2 < 0$
$x > \pm 2$
$\Rightarrow x^{2} > 4$
$x \in (- \infty, - 2) \cup (2, \infty)$
Case $2 : 4 - x^{2} \geq 0, [x] + 2 > 0 x$
$\Rightarrow x^{2} \leq 4$
$\Rightarrow x \leq \pm 2$
$\Rightarrow x \in [- 2, 2] x \in [- 1, \infty) x \notin - 2$
$x \in [- 1, 2]$
Form (i) and (ii), we get -
$x \in (- \infty, - 2) \cup [- 1, 2]$

Shift-II