Test Index

Functions Part 2

Section: Mathematics

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Question : 64 of 73

Marks: +1, -0

Let the sets A and B denote the domain and range respectively of the function

f(x)=\frac{1}{\sqrt{[x]-x}}

[x]

denotes the smallest integer greater than or equal to

x

. Then among the statements :

(S1): A \cap B = (1, \infty) - N

A \cup B = (1, \infty)

[6-Apr-2023 shift 2]

only (S1) is true
neither (S1) nor (S2) is true
only (S2) is true
both (S1) and (S2) are true

Solution:

f(x)=\frac{1}{\sqrt{[x]-x}}

If

x \in I [x] = [x]

(greatest integer function)

If

x \notin I [x] = [x] + 1

\Rightarrow f(x) = \begin{cases} \frac{1}{\sqrt{[x]-x}}, & x \in I \\ \frac{1}{\sqrt{[x]+1-x}}, & x \notin I \end{cases}

\Rightarrow f(x) = \begin{cases} \frac{1}{\sqrt{-\{x\}}}, & x \in I,\ (\text{ does not exist }) \\ \frac{1}{\sqrt{1-\{x\}}}, & x \notin I \end{cases}.

\Rightarrow \text{ domain of } f(x) = \mathbb{R} - I

\text{ Now, } f(x) = \frac{1}{\sqrt{1-\{x\}}},\ x \notin I

\Rightarrow x < \{x\} < 1

\Rightarrow 0 < 1\sqrt{1-\{x\}} < 1

\Rightarrow \frac{1}{\sqrt{1-\{x\}}} > 1

\Rightarrow \text{Range}(1, \infty)

\Rightarrow A = \mathbb{R} - I

\Rightarrow B = (1, \infty)

\text{ So, } A \cap B = (1, \infty) - N

A \cup B \neq (1, \infty)

\Rightarrow S1 \text{ is only correct.}

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