CBSE Class 11 Mathematics Relations Functions Worksheet Set 06

CBSE Class 11 Mathematics Worksheet - Relations Functions (4). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Draw the graph of the function \( f(x) = [x - 1] \).
Answer:
We have, \( f(x) = [x - 1] \)
For example:
When \( x = 1 \), then \( f(x) = [0] = 0 \)
When \( x = 2 \), then \( f(x) = [1] = 1 \)

Domain = \( R \)
Range = \( Z \)

Question. Let \( R \) be a relation on \( N \) defined by \( R = \{(a, b): a, b \in N \text{ and } a = b^2\} \). Are the following true?
(i) \( (a, a) \in R \) for all \( a \in N \)
(ii) \( (a, b) \in R \Rightarrow (b, a) \in R \)
(iii) \( (a, b) \in R, (b, c) \in R \Rightarrow (a, c) \in R \).
Answer:
We have, \( R = \{(a, b): a = b^2\} \)

(i) \( 2 \in N \)
But \( (2, 2) \notin R \) because \( 2 \neq 2^2 \).
\( \therefore \) False

(ii) \( (4, 2) \in R \) as \( 4 = 2^2 \)
But \( (2, 4) \notin R \) because \( 2 \neq 4^2 \).
\( \dots \) False

(iii) \( (16, 4) \in R \) and \( (4, 2) \in R \) as \( 16 = 4^2 \) and \( 4 = 2^2 \)
But \( (16, 2) \notin R \) because \( 16 \neq 2^2 \).
\( \dots \) False

Question. Let \( R \) be a relation on \( Q \) defined by \( R = \{(a, b): a, b \in Q \text{ and } a - b \in Z \text{ i.e., } a - b \text{ is an integer}\} \). Show that:
(i) \( (a, a) \in R \) for all \( a \in Q \)
(ii) \( (a, b) \in R \Rightarrow (b, a) \in R \)
(iii) \( (a, b) \in R \) and \( (b, c) \in R \Rightarrow (a, c) \in R \).
Answer:
We have, \( R = \{(a, b): a - b \text{ is an integer}\} \) where \( Q \rightarrow \) set of rational numbers.

(i) For any \( a \in Q \), \( a - a = 0 \), which is an integer.
\( \therefore (a, a) \in R \)

(ii) Let \( (a, b) \in R \)
\( \Rightarrow a - b \) is an integer.
\( \Rightarrow a - b = m \) where \( m \in Z \).
\( \Rightarrow b - a = -m \), which is also an integer.
\( \Rightarrow (b, a) \in R \)

(iii) Let \( (a, b) \in R \) and \( (b, c) \in R \)
\( \Rightarrow a - b = m \) and \( b - c = n \) where \( m, n \in Z \).
Now, \( a - c = (a - b) + (b - c) \)
\( = m + n \qquad \{ \because \text{sum of two integers is also an integer} \} \)
\( = \text{integer} \)
\( \therefore (a, c) \in R \)

Question. If \( A = \{2, 3\} \) and \( B = \{1, 2, 3\} \). Find the number of relations.
Answer:
Given, \( A = \{2, 3\} \) and \( B = \{1, 2, 3\} \).
Here, number of elements in \( A \), \( m = 2 \) and number of elements in \( B \), \( n = 3 \).

\( \text{Number of relations} = 2^{mn} \)
\( = 2^{2 \times 3} \)
\( = 2^6 = 64 \) ans.

Question. If \( A \times A \) has 9 elements, among which two are \( (-1, 0) \) and \( (0, 1) \). Find Set \( A \) & remaining elements of \( A \times A \).
Answer:
Given, \( n(A \times A) = 9 \Rightarrow n(A) = 3 \).

Since \( (-1, 0) \in A \times A \) and \( (0, 1) \in A \times A \):
\( \Rightarrow -1, 0 \in A \) and \( 0, 1 \in A \)
\( \Rightarrow -1, 0, 1 \in A \)

Since \( n(A) = 3 \):
\( \therefore A = \{-1, 0, 1\} \)

The complete set \( A \times A \) is:
\( A \times A = \{(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)\} \)

The remaining elements of \( A \times A \) are:
\( \{(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)\} \)

Question. Find the domain of \( f(x) = \frac{1}{\sqrt{x-|x|}} \).
Answer:
We have, \( f(x) = \frac{1}{\sqrt{x-|x|}} \)

By definition of absolute value:
\( |x| = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases} \)
\( \Rightarrow -|x| = \begin{cases} -x, & x \geq 0 \\ x, & x < 0 \end{cases} \)
\( \Rightarrow x - |x| = \begin{cases} x - x, & x \geq 0 \\ x + x, & x < 0 \end{cases} \)
\( \Rightarrow x - |x| = \begin{cases} 0, & x \geq 0 \\ 2x, & x < 0 \end{cases} \) ............ (1)

Since \( f(x) \) requires \( x - |x| > 0 \):
From (1), we see that \( x - |x| \) is either \( 0 \) (for \( x \geq 0 \)) or negative (for \( x < 0 \)).
\( \therefore \frac{1}{\sqrt{x-|x|}} \) does not take real values for any \( x \in R \).
\( \therefore \text{Domain} = \emptyset \)

Question. Find the domain of \( f(x) = \frac{1}{\sqrt{x-[x]}} \).
Answer:
\( f(x) \) is defined when \( x - [x] > 0 \).

We know that for all \( x \in R \):
\( 0 \leq x - [x] < 1 \)
But \( x - [x] = 0 \) when \( x \in Z \) (set of integers).

\( \therefore x - [x] > 0 \) when \( x \in R - Z \).
\( \therefore \text{Domain} = R - Z \) ans.

Question. Find the domain for which the functions \( f(x) = 2x^2 - 1 \) and \( g(x) = 1 - 3x \) are equal.
Answer:
We have, \( f(x) = g(x) \)
\( \Rightarrow 2x^2 - 1 = 1 - 3x \)
\( \Rightarrow 2x^2 + 3x - 2 = 0 \)
\( \Rightarrow 2x^2 + 4x - x - 2 = 0 \)
\( \Rightarrow 2x(x + 2) - 1(x + 2) = 0 \)
\( \Rightarrow (2x - 1)(x + 2) = 0 \)
\( \Rightarrow x = \frac{1}{2} \text{ or } x = -2 \)

The domain consists of all values of \( x \) that satisfy this condition:
\( \therefore \text{Domain} = \left\{\frac{1}{2}, -2\right\} \) ans.

Question. Redefine the function \( f(x) = |x - 2| + |2 + x| \) where \( -3 \leq x \le 3 \).
Answer:
The critical point of \( |x - 2| \) is \( 2 \), and the critical point of \( |2 + x| \) is \( -2 \).

First, we arrange the terms in ascending order:
\( f(x) = |2 + x| + |x - 2| \)

There are 3 cases:
(i) When \( -3 \leq x < -2 \):
\( f(x) = -(2 + x) - (x - 2) = -2x \)

(ii) When \( -2 \leq x < 2 \):
\( f(x) = (2 + x) - (x - 2) = 4 \)

(iii) When \( 2 \leq x \leq 3 \):
\( f(x) = (2 + x) + (x - 2) = 2x \)

Thus, the redefined function is:
\( f(x) = \begin{cases} -2x, & -3 \leq x < -2 \\ 4, & -2 \leq x < 2 \\ 2x, & 2 \leq x \leq 3 \end{cases} \) ans.

Question. Find the Range of \( f(x) = \frac{1}{1-2\cos x} \).
Answer:
We know that:
\( -1 \leq \cos x \leq 1 \)
\( \Rightarrow -2 \leq 2\cos x \leq 2 \)
\( \Rightarrow 2 \geq -2\cos x \geq -2 \)
\( \Rightarrow 3 \geq 1 - 2\cos x \geq -1 \)

Taking the reciprocal of the term \( 1 - 2\cos x \):
\( \Rightarrow f(x) \leq -1 \text{ or } f(x) \geq \frac{1}{3} \)

\( \therefore \text{Range} = (-\infty, -1] \cup \left[\frac{1}{3}, \infty\right) \) ans.