QBManager

Sets, Relation and Function

117438 Let $[x]$ denote the greatest integer $\leq x$ , where $x$ $\in R$ . If the domain of the real valued function $f (x) = \sqrt{\frac{| [x] | - 2}{| [x] | - 3}}$ is $(- \infty$ , a) $\cup [b, c) \cup [4, \infty), a <$ $b < c$ , then the value of $a + b + c$ is

1 8

2 1

3 -2

4 2

Explanation:

C Given function,
$f (x) = \sqrt{\frac{| [x] | - 2}{| [x] | - 3}}$
Define the function $f (x)$ is ,
$\frac{| [x] | - 2}{| x | - 3} \geq 0$
$Case-I : when | [x] | - 2 \geq 0 and [x] - 3 > 0$
$x \in (- \infty, - 3) \cup [4, \infty)$
$Case-II : when | [x] | - 2 \leq 0 and | [x] | - 3 > 0$
$x \in [- 2, 3)$
$From case - (I) and case - (II), we get -$
$Domain of function f (x)$
$x \in (- \infty, 3) \cup [- 2, 3) \cup [4, \infty)$
$Hence, a + b + c = - 3 + (- 2) + 3$
$= - 5 + 3$
$= - 2$

JEE Main 20.07. 2021 Shift-I

Sets, Relation and Function

117439 The real valued function
$f (x) = \frac{{cosec}^{- 1} x}{\sqrt{x - [x]}}$ , where $[x]$ denotes the greatest integer less than or equal to $x$ , is defined for all $x$ belonging to

1 all reals except integers

2 all non-integer except the interval

[- 1, 1]

3 all integers except

0, - 1, 1

4 all reals except the interval

[- 1, 1]

Explanation:

B Given that,
$f (x) = \frac{{cosec}^{- 1} x}{\sqrt{x - [x]}}$
$f (x) = \frac{{cosec}^{- 1} x}{\sqrt{{x}}} ∵ x - [x] = {x}$
Hence,
Domain of $f (x)$ is, $x \in (- \infty, - 1) \cup (1, \infty)$
We know that, ${x} \neq 0$
So, $x \neq$ integers.
Hence, option (b) is correct.

Shift-I

Sets, Relation and Function

117440 If the function are defined as $f (x) = \sqrt{x}$ and $g (x) = \sqrt{1 - x}$ , then what is the common domain of the following functions?
$f + g, f - g, f / g, g / f, g - f$ , where $(f \pm g) (x) = f (x) \pm$ $g (x), (f / g) (x) = \frac{f (x)}{g (x)}$

1

0 \leq x \leq 1

2

0 \leq x < 1

3

0 < x < 1

4

0 < x \leq 1

Explanation:

C Given that, $f (x) = \sqrt{x}$
$g (x) = \sqrt{1 - x}$
Now,
$f + g = \sqrt{x} + \sqrt{1 - x}$
We know,
$x \geq 0 and 1 - x \geq 0$
$\Rightarrow x \in [0, 1]$
And,
$f - g = \sqrt{x} - \sqrt{1 - x}$
Hence,
$x \geq 0 and 1 - x \geq 0$
$\Rightarrow x \in [0, 1]$
And, $\frac{f}{g} = \frac{\sqrt{x}}{\sqrt{1 - x}}$
$x \geq 0 and 1 - x > 0$
$\Rightarrow x \in [0, 1)$
And, $\frac{g}{f} = \frac{\sqrt{1 - x}}{\sqrt{x}}$
$1 - x \geq 0 and x > 0$
$\Rightarrow x \in (0, 1)$
And,
$g - f = \sqrt{1 - x} - \sqrt{x}$
$1 - x \geq 0 and x \geq 0$
$x \in [0, 1]$
For all condition, we get -
$x \in (0, 1)$

JEE Main-19.03.2021 Shift-II

Sets, Relation and Function

117441 The domain of the function
$f (x) = \sin^{- 1} (\frac{3 x^{2} + x - 1}{(x - 1)^{2}}) + \cos^{- 1} (\frac{x - 1}{x + 1})$ is

1

[0, \frac{1}{4}]

2

[- 2, 0] \cup [\frac{1}{4}, \frac{1}{2}]

3

[\frac{1}{4}, \frac{1}{2}] \cup (0)

4

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

Explanation:

C Given function,
$f (x) = \sin^{- 1} (\frac{3 x^{2} + x - 1}{(x - 1)^{2}}) + \cos^{- 1} (\frac{x - 1}{x + 1})$
We know,
$\sin^{- 1} θ, \cos^{- 1} θ$
Then, $- 1 \leq θ \geq 1$
So,
$- 1 \leq \frac{3 x^{2} + x - 1}{(x - 1)^{2}} \leq 1$
$- (x - 1)^{2} \leq 3 x^{2} + x - 1 \geq (x - 1)^{2}$
Now,
$- (x - 1)^{2} \leq 3 x^{2} + x - 1$
$- (x^{2} + 1 - 2 x) \leq 3 x^{2} + x - 1$
$- x^{2} - 1 + 2 x \leq 3 x^{2} + x - 1$
$4 x^{2} - x \geq 0$
$x (4 x - 1) \geq 0$
$4 x - 1 \geq 0, x \geq 0$
$x \geq 0, x \geq \frac{1}{4}$
original image
$\begin{aligned} x \in (- \infty, 0] \cup [\frac{1}{4}, \infty) \\ And, 3 x^{2} + x - 1 \leq (x - 1)^{2} \\ 3 x^{2} + x - 1 \leq x^{2} - 2 x + 1 \\ 2 x^{2} + 3 x - 2 \leq 0 \\ 2 x^{2} + 4 x - x - 2 \leq 0 \\ 2 x (x + 2) - 1 (x + 2) \leq 0 \\ (x + 2) (2 x - 1) \leq 0 \\ x \leq \frac{1}{2}, x \leq - 2 \end{aligned}$
original image
$x \in [- 2, \frac{1}{2}]$
Now, $- 1 \leq \frac{x - 1}{x + 1} \leq 1$
$\begin{aligned} - (x + 1) \leq x - 1 \leq x + 1 \\ - (x + 1) \leq x - 1 \\ - x - 1 \leq x - 1 \\ 2 x \geq 0 \\ x \geq 0 \\ x \in [0, \infty) \end{aligned}$
True for all values
Hence, $x \in [- 2, 0] \cup [\frac{1}{4}, \frac{1}{2}]$
So, $x \in {0} \cup [\frac{1}{4}, \frac{1}{2}]$

JEE Main 31.08.2021 Shift-II

Sets, Relation and Function

117438 Let $[x]$ denote the greatest integer $\leq x$ , where $x$ $\in R$ . If the domain of the real valued function $f (x) = \sqrt{\frac{| [x] | - 2}{| [x] | - 3}}$ is $(- \infty$ , a) $\cup [b, c) \cup [4, \infty), a <$ $b < c$ , then the value of $a + b + c$ is

1 8

2 1

3 -2

4 2

Explanation:

C Given function,
$f (x) = \sqrt{\frac{| [x] | - 2}{| [x] | - 3}}$
Define the function $f (x)$ is ,
$\frac{| [x] | - 2}{| x | - 3} \geq 0$
$Case-I : when | [x] | - 2 \geq 0 and [x] - 3 > 0$
$x \in (- \infty, - 3) \cup [4, \infty)$
$Case-II : when | [x] | - 2 \leq 0 and | [x] | - 3 > 0$
$x \in [- 2, 3)$
$From case - (I) and case - (II), we get -$
$Domain of function f (x)$
$x \in (- \infty, 3) \cup [- 2, 3) \cup [4, \infty)$
$Hence, a + b + c = - 3 + (- 2) + 3$
$= - 5 + 3$
$= - 2$

JEE Main 20.07. 2021 Shift-I

Sets, Relation and Function

117439 The real valued function
$f (x) = \frac{{cosec}^{- 1} x}{\sqrt{x - [x]}}$ , where $[x]$ denotes the greatest integer less than or equal to $x$ , is defined for all $x$ belonging to

1 all reals except integers

2 all non-integer except the interval

[- 1, 1]

3 all integers except

0, - 1, 1

4 all reals except the interval

[- 1, 1]

Explanation:

B Given that,
$f (x) = \frac{{cosec}^{- 1} x}{\sqrt{x - [x]}}$
$f (x) = \frac{{cosec}^{- 1} x}{\sqrt{{x}}} ∵ x - [x] = {x}$
Hence,
Domain of $f (x)$ is, $x \in (- \infty, - 1) \cup (1, \infty)$
We know that, ${x} \neq 0$
So, $x \neq$ integers.
Hence, option (b) is correct.

Shift-I

Sets, Relation and Function

117440 If the function are defined as $f (x) = \sqrt{x}$ and $g (x) = \sqrt{1 - x}$ , then what is the common domain of the following functions?
$f + g, f - g, f / g, g / f, g - f$ , where $(f \pm g) (x) = f (x) \pm$ $g (x), (f / g) (x) = \frac{f (x)}{g (x)}$

1

0 \leq x \leq 1

2

0 \leq x < 1

3

0 < x < 1

4

0 < x \leq 1

Explanation:

C Given that, $f (x) = \sqrt{x}$
$g (x) = \sqrt{1 - x}$
Now,
$f + g = \sqrt{x} + \sqrt{1 - x}$
We know,
$x \geq 0 and 1 - x \geq 0$
$\Rightarrow x \in [0, 1]$
And,
$f - g = \sqrt{x} - \sqrt{1 - x}$
Hence,
$x \geq 0 and 1 - x \geq 0$
$\Rightarrow x \in [0, 1]$
And, $\frac{f}{g} = \frac{\sqrt{x}}{\sqrt{1 - x}}$
$x \geq 0 and 1 - x > 0$
$\Rightarrow x \in [0, 1)$
And, $\frac{g}{f} = \frac{\sqrt{1 - x}}{\sqrt{x}}$
$1 - x \geq 0 and x > 0$
$\Rightarrow x \in (0, 1)$
And,
$g - f = \sqrt{1 - x} - \sqrt{x}$
$1 - x \geq 0 and x \geq 0$
$x \in [0, 1]$
For all condition, we get -
$x \in (0, 1)$

JEE Main-19.03.2021 Shift-II

Sets, Relation and Function

117441 The domain of the function
$f (x) = \sin^{- 1} (\frac{3 x^{2} + x - 1}{(x - 1)^{2}}) + \cos^{- 1} (\frac{x - 1}{x + 1})$ is

1

[0, \frac{1}{4}]

2

[- 2, 0] \cup [\frac{1}{4}, \frac{1}{2}]

3

[\frac{1}{4}, \frac{1}{2}] \cup (0)

4

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

Explanation:

C Given function,
$f (x) = \sin^{- 1} (\frac{3 x^{2} + x - 1}{(x - 1)^{2}}) + \cos^{- 1} (\frac{x - 1}{x + 1})$
We know,
$\sin^{- 1} θ, \cos^{- 1} θ$
Then, $- 1 \leq θ \geq 1$
So,
$- 1 \leq \frac{3 x^{2} + x - 1}{(x - 1)^{2}} \leq 1$
$- (x - 1)^{2} \leq 3 x^{2} + x - 1 \geq (x - 1)^{2}$
Now,
$- (x - 1)^{2} \leq 3 x^{2} + x - 1$
$- (x^{2} + 1 - 2 x) \leq 3 x^{2} + x - 1$
$- x^{2} - 1 + 2 x \leq 3 x^{2} + x - 1$
$4 x^{2} - x \geq 0$
$x (4 x - 1) \geq 0$
$4 x - 1 \geq 0, x \geq 0$
$x \geq 0, x \geq \frac{1}{4}$
original image
$\begin{aligned} x \in (- \infty, 0] \cup [\frac{1}{4}, \infty) \\ And, 3 x^{2} + x - 1 \leq (x - 1)^{2} \\ 3 x^{2} + x - 1 \leq x^{2} - 2 x + 1 \\ 2 x^{2} + 3 x - 2 \leq 0 \\ 2 x^{2} + 4 x - x - 2 \leq 0 \\ 2 x (x + 2) - 1 (x + 2) \leq 0 \\ (x + 2) (2 x - 1) \leq 0 \\ x \leq \frac{1}{2}, x \leq - 2 \end{aligned}$
original image
$x \in [- 2, \frac{1}{2}]$
Now, $- 1 \leq \frac{x - 1}{x + 1} \leq 1$
$\begin{aligned} - (x + 1) \leq x - 1 \leq x + 1 \\ - (x + 1) \leq x - 1 \\ - x - 1 \leq x - 1 \\ 2 x \geq 0 \\ x \geq 0 \\ x \in [0, \infty) \end{aligned}$
True for all values
Hence, $x \in [- 2, 0] \cup [\frac{1}{4}, \frac{1}{2}]$
So, $x \in {0} \cup [\frac{1}{4}, \frac{1}{2}]$

JEE Main 31.08.2021 Shift-II

Sets, Relation and Function

117438 Let $[x]$ denote the greatest integer $\leq x$ , where $x$ $\in R$ . If the domain of the real valued function $f (x) = \sqrt{\frac{| [x] | - 2}{| [x] | - 3}}$ is $(- \infty$ , a) $\cup [b, c) \cup [4, \infty), a <$ $b < c$ , then the value of $a + b + c$ is

1 8

2 1

3 -2

4 2

Explanation:

C Given function,
$f (x) = \sqrt{\frac{| [x] | - 2}{| [x] | - 3}}$
Define the function $f (x)$ is ,
$\frac{| [x] | - 2}{| x | - 3} \geq 0$
$Case-I : when | [x] | - 2 \geq 0 and [x] - 3 > 0$
$x \in (- \infty, - 3) \cup [4, \infty)$
$Case-II : when | [x] | - 2 \leq 0 and | [x] | - 3 > 0$
$x \in [- 2, 3)$
$From case - (I) and case - (II), we get -$
$Domain of function f (x)$
$x \in (- \infty, 3) \cup [- 2, 3) \cup [4, \infty)$
$Hence, a + b + c = - 3 + (- 2) + 3$
$= - 5 + 3$
$= - 2$

JEE Main 20.07. 2021 Shift-I

Sets, Relation and Function

117439 The real valued function
$f (x) = \frac{{cosec}^{- 1} x}{\sqrt{x - [x]}}$ , where $[x]$ denotes the greatest integer less than or equal to $x$ , is defined for all $x$ belonging to

1 all reals except integers

2 all non-integer except the interval

[- 1, 1]

3 all integers except

0, - 1, 1

4 all reals except the interval

[- 1, 1]

Explanation:

B Given that,
$f (x) = \frac{{cosec}^{- 1} x}{\sqrt{x - [x]}}$
$f (x) = \frac{{cosec}^{- 1} x}{\sqrt{{x}}} ∵ x - [x] = {x}$
Hence,
Domain of $f (x)$ is, $x \in (- \infty, - 1) \cup (1, \infty)$
We know that, ${x} \neq 0$
So, $x \neq$ integers.
Hence, option (b) is correct.

Shift-I

Sets, Relation and Function

117440 If the function are defined as $f (x) = \sqrt{x}$ and $g (x) = \sqrt{1 - x}$ , then what is the common domain of the following functions?
$f + g, f - g, f / g, g / f, g - f$ , where $(f \pm g) (x) = f (x) \pm$ $g (x), (f / g) (x) = \frac{f (x)}{g (x)}$

1

0 \leq x \leq 1

2

0 \leq x < 1

3

0 < x < 1

4

0 < x \leq 1

Explanation:

C Given that, $f (x) = \sqrt{x}$
$g (x) = \sqrt{1 - x}$
Now,
$f + g = \sqrt{x} + \sqrt{1 - x}$
We know,
$x \geq 0 and 1 - x \geq 0$
$\Rightarrow x \in [0, 1]$
And,
$f - g = \sqrt{x} - \sqrt{1 - x}$
Hence,
$x \geq 0 and 1 - x \geq 0$
$\Rightarrow x \in [0, 1]$
And, $\frac{f}{g} = \frac{\sqrt{x}}{\sqrt{1 - x}}$
$x \geq 0 and 1 - x > 0$
$\Rightarrow x \in [0, 1)$
And, $\frac{g}{f} = \frac{\sqrt{1 - x}}{\sqrt{x}}$
$1 - x \geq 0 and x > 0$
$\Rightarrow x \in (0, 1)$
And,
$g - f = \sqrt{1 - x} - \sqrt{x}$
$1 - x \geq 0 and x \geq 0$
$x \in [0, 1]$
For all condition, we get -
$x \in (0, 1)$

JEE Main-19.03.2021 Shift-II

Sets, Relation and Function

117441 The domain of the function
$f (x) = \sin^{- 1} (\frac{3 x^{2} + x - 1}{(x - 1)^{2}}) + \cos^{- 1} (\frac{x - 1}{x + 1})$ is

1

[0, \frac{1}{4}]

2

[- 2, 0] \cup [\frac{1}{4}, \frac{1}{2}]

3

[\frac{1}{4}, \frac{1}{2}] \cup (0)

4

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

Explanation:

C Given function,
$f (x) = \sin^{- 1} (\frac{3 x^{2} + x - 1}{(x - 1)^{2}}) + \cos^{- 1} (\frac{x - 1}{x + 1})$
We know,
$\sin^{- 1} θ, \cos^{- 1} θ$
Then, $- 1 \leq θ \geq 1$
So,
$- 1 \leq \frac{3 x^{2} + x - 1}{(x - 1)^{2}} \leq 1$
$- (x - 1)^{2} \leq 3 x^{2} + x - 1 \geq (x - 1)^{2}$
Now,
$- (x - 1)^{2} \leq 3 x^{2} + x - 1$
$- (x^{2} + 1 - 2 x) \leq 3 x^{2} + x - 1$
$- x^{2} - 1 + 2 x \leq 3 x^{2} + x - 1$
$4 x^{2} - x \geq 0$
$x (4 x - 1) \geq 0$
$4 x - 1 \geq 0, x \geq 0$
$x \geq 0, x \geq \frac{1}{4}$
original image
$\begin{aligned} x \in (- \infty, 0] \cup [\frac{1}{4}, \infty) \\ And, 3 x^{2} + x - 1 \leq (x - 1)^{2} \\ 3 x^{2} + x - 1 \leq x^{2} - 2 x + 1 \\ 2 x^{2} + 3 x - 2 \leq 0 \\ 2 x^{2} + 4 x - x - 2 \leq 0 \\ 2 x (x + 2) - 1 (x + 2) \leq 0 \\ (x + 2) (2 x - 1) \leq 0 \\ x \leq \frac{1}{2}, x \leq - 2 \end{aligned}$
original image
$x \in [- 2, \frac{1}{2}]$
Now, $- 1 \leq \frac{x - 1}{x + 1} \leq 1$
$\begin{aligned} - (x + 1) \leq x - 1 \leq x + 1 \\ - (x + 1) \leq x - 1 \\ - x - 1 \leq x - 1 \\ 2 x \geq 0 \\ x \geq 0 \\ x \in [0, \infty) \end{aligned}$
True for all values
Hence, $x \in [- 2, 0] \cup [\frac{1}{4}, \frac{1}{2}]$
So, $x \in {0} \cup [\frac{1}{4}, \frac{1}{2}]$

JEE Main 31.08.2021 Shift-II

Sets, Relation and Function

117438 Let $[x]$ denote the greatest integer $\leq x$ , where $x$ $\in R$ . If the domain of the real valued function $f (x) = \sqrt{\frac{| [x] | - 2}{| [x] | - 3}}$ is $(- \infty$ , a) $\cup [b, c) \cup [4, \infty), a <$ $b < c$ , then the value of $a + b + c$ is

1 8

2 1

3 -2

4 2

Explanation:

C Given function,
$f (x) = \sqrt{\frac{| [x] | - 2}{| [x] | - 3}}$
Define the function $f (x)$ is ,
$\frac{| [x] | - 2}{| x | - 3} \geq 0$
$Case-I : when | [x] | - 2 \geq 0 and [x] - 3 > 0$
$x \in (- \infty, - 3) \cup [4, \infty)$
$Case-II : when | [x] | - 2 \leq 0 and | [x] | - 3 > 0$
$x \in [- 2, 3)$
$From case - (I) and case - (II), we get -$
$Domain of function f (x)$
$x \in (- \infty, 3) \cup [- 2, 3) \cup [4, \infty)$
$Hence, a + b + c = - 3 + (- 2) + 3$
$= - 5 + 3$
$= - 2$

JEE Main 20.07. 2021 Shift-I

Sets, Relation and Function

117439 The real valued function
$f (x) = \frac{{cosec}^{- 1} x}{\sqrt{x - [x]}}$ , where $[x]$ denotes the greatest integer less than or equal to $x$ , is defined for all $x$ belonging to

1 all reals except integers

2 all non-integer except the interval

[- 1, 1]

3 all integers except

0, - 1, 1

4 all reals except the interval

[- 1, 1]

Explanation:

B Given that,
$f (x) = \frac{{cosec}^{- 1} x}{\sqrt{x - [x]}}$
$f (x) = \frac{{cosec}^{- 1} x}{\sqrt{{x}}} ∵ x - [x] = {x}$
Hence,
Domain of $f (x)$ is, $x \in (- \infty, - 1) \cup (1, \infty)$
We know that, ${x} \neq 0$
So, $x \neq$ integers.
Hence, option (b) is correct.

Shift-I

Sets, Relation and Function

117440 If the function are defined as $f (x) = \sqrt{x}$ and $g (x) = \sqrt{1 - x}$ , then what is the common domain of the following functions?
$f + g, f - g, f / g, g / f, g - f$ , where $(f \pm g) (x) = f (x) \pm$ $g (x), (f / g) (x) = \frac{f (x)}{g (x)}$

1

0 \leq x \leq 1

2

0 \leq x < 1

3

0 < x < 1

4

0 < x \leq 1

Explanation:

C Given that, $f (x) = \sqrt{x}$
$g (x) = \sqrt{1 - x}$
Now,
$f + g = \sqrt{x} + \sqrt{1 - x}$
We know,
$x \geq 0 and 1 - x \geq 0$
$\Rightarrow x \in [0, 1]$
And,
$f - g = \sqrt{x} - \sqrt{1 - x}$
Hence,
$x \geq 0 and 1 - x \geq 0$
$\Rightarrow x \in [0, 1]$
And, $\frac{f}{g} = \frac{\sqrt{x}}{\sqrt{1 - x}}$
$x \geq 0 and 1 - x > 0$
$\Rightarrow x \in [0, 1)$
And, $\frac{g}{f} = \frac{\sqrt{1 - x}}{\sqrt{x}}$
$1 - x \geq 0 and x > 0$
$\Rightarrow x \in (0, 1)$
And,
$g - f = \sqrt{1 - x} - \sqrt{x}$
$1 - x \geq 0 and x \geq 0$
$x \in [0, 1]$
For all condition, we get -
$x \in (0, 1)$

JEE Main-19.03.2021 Shift-II

Sets, Relation and Function

117441 The domain of the function
$f (x) = \sin^{- 1} (\frac{3 x^{2} + x - 1}{(x - 1)^{2}}) + \cos^{- 1} (\frac{x - 1}{x + 1})$ is

1

[0, \frac{1}{4}]

2

[- 2, 0] \cup [\frac{1}{4}, \frac{1}{2}]

3

[\frac{1}{4}, \frac{1}{2}] \cup (0)

4

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

Explanation:

C Given function,
$f (x) = \sin^{- 1} (\frac{3 x^{2} + x - 1}{(x - 1)^{2}}) + \cos^{- 1} (\frac{x - 1}{x + 1})$
We know,
$\sin^{- 1} θ, \cos^{- 1} θ$
Then, $- 1 \leq θ \geq 1$
So,
$- 1 \leq \frac{3 x^{2} + x - 1}{(x - 1)^{2}} \leq 1$
$- (x - 1)^{2} \leq 3 x^{2} + x - 1 \geq (x - 1)^{2}$
Now,
$- (x - 1)^{2} \leq 3 x^{2} + x - 1$
$- (x^{2} + 1 - 2 x) \leq 3 x^{2} + x - 1$
$- x^{2} - 1 + 2 x \leq 3 x^{2} + x - 1$
$4 x^{2} - x \geq 0$
$x (4 x - 1) \geq 0$
$4 x - 1 \geq 0, x \geq 0$
$x \geq 0, x \geq \frac{1}{4}$
original image
$\begin{aligned} x \in (- \infty, 0] \cup [\frac{1}{4}, \infty) \\ And, 3 x^{2} + x - 1 \leq (x - 1)^{2} \\ 3 x^{2} + x - 1 \leq x^{2} - 2 x + 1 \\ 2 x^{2} + 3 x - 2 \leq 0 \\ 2 x^{2} + 4 x - x - 2 \leq 0 \\ 2 x (x + 2) - 1 (x + 2) \leq 0 \\ (x + 2) (2 x - 1) \leq 0 \\ x \leq \frac{1}{2}, x \leq - 2 \end{aligned}$
original image
$x \in [- 2, \frac{1}{2}]$
Now, $- 1 \leq \frac{x - 1}{x + 1} \leq 1$
$\begin{aligned} - (x + 1) \leq x - 1 \leq x + 1 \\ - (x + 1) \leq x - 1 \\ - x - 1 \leq x - 1 \\ 2 x \geq 0 \\ x \geq 0 \\ x \in [0, \infty) \end{aligned}$
True for all values
Hence, $x \in [- 2, 0] \cup [\frac{1}{4}, \frac{1}{2}]$
So, $x \in {0} \cup [\frac{1}{4}, \frac{1}{2}]$

JEE Main 31.08.2021 Shift-II

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Question Bank Series

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