CBSE Class 11 Mathematics Relations Functions Worksheet Set 05

CBSE Class 11 Mathematics Worksheet - Relations Functions (3). 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. \( f(x) = |x - 3| \). Find the Domain and Range.
Answer:
We have, \( f(x) = |x - 3| \)

Domain: \( f(x) \) is real for all values of \( x \) such that \( x \in R \).
\( \therefore \text{Domain} = R \)

Range: Let \( y = f(x) \)
\( \Rightarrow y = |x - 3| \)
\( \Rightarrow y = \begin{cases} x - 3, & x \geq 3 \\ -(x - 3), & x < 3 \end{cases} \)

For \( x \geq 3 \):
\( y = x - 3 \Rightarrow x = y + 3 \)
We have, \( x \geq 3 \Rightarrow y + 3 \geq 3 \Rightarrow y \geq 0 \Rightarrow y \in [0, \infty) \)

For \( x < 3 \):
\( y = -(x - 3) \Rightarrow y = -x + 3 \Rightarrow x = 3 - y \)
We have, \( x < 3 \Rightarrow 3 - y < 3 \Rightarrow -y < 0 \Rightarrow y > 0 \Rightarrow y \in (0, \infty) \)

Since \( y \in [0, \infty) \) and \( y \in (0, \infty) \):
\( \therefore \text{Range} = [0, \infty) \text{ ans. } \{ \text{since when } x = 3 \text{ then } y = 0 \} \)

Question. Find Domain and Range of \( f(x) = 1 - |x - 2| \).
Answer:
We have, \( f(x) = 1 - |x - 2| \)

Domain: \( f(x) \) is real for all values of \( x \) such that \( x \in R \).
\( \dots \text{Domain} = R \)

Range: Let \( y = f(x) \)
\( \Rightarrow y = 1 - |x - 2| \)
\( \Rightarrow y = \begin{cases} 1 - (x - 2), & x \geq 2 \\ 1 + (x - 2), & x < 2 \end{cases} \)
\( \Rightarrow y = \begin{cases} -x + 3, & x \geq 2 \\ x - 1, & x < 2 \end{cases} \)

For \( x \geq 2 \):
\( y = -x + 3 \Rightarrow x = 3 - y \)
We have, \( x \geq 2 \Rightarrow 3 - y \geq 2 \Rightarrow -y \geq -1 \Rightarrow y \leq 1 \Rightarrow y \in (-\infty, 1] \)

For \( x < 2 \):
\( y = x - 1 \Rightarrow x = y + 1 \)
We have, \( x < 2 \Rightarrow y + 1 < 2 \Rightarrow y < 1 \Rightarrow y \in (-\infty, 1) \)

\( \therefore \text{Range} = (-\infty, 1] \text{ ans. } \{ \text{when } x = 2 \text{ then } y = 1 \therefore y = 1 \text{ is included in Range} \} \)

Question. Find the Domain & Range of \( f(x) = \frac{1}{2-\sin(3x)} \).
Answer:
We have, \( f(x) = \frac{1}{2 - \sin(3x)} \)

Domain: We have, \( -1 \leq \sin(3x) \leq 1 \)
\( \Rightarrow 2 - \sin(3x) \neq 0 \)
\( \therefore \text{Domain} = R \)

Range: We have, \( -1 \leq \sin(3x) \leq 1 \)
\( \Rightarrow 1 \geq -\sin(3x) \geq -1 \qquad \{ \text{multiplying by } (-1) \} \)
\( \Rightarrow -1 \leq -\sin(3x) \leq 1 \)
\( \Rightarrow 1 \leq 2 - \sin(3x) \leq 3 \qquad \{ \text{adding } 2 \} \)
\( \Rightarrow \frac{1}{3} \leq \frac{1}{2-\sin(3x)} \leq 1 \qquad \left\{ \text{if } a < x < b \text{ then } \frac{1}{b} < \frac{1}{x} < \frac{1}{a} \text{ for positive } a, b \right\} \)
\( \therefore \text{Range} = \left[\frac{1}{3}, 1\right] \text{ ans.} \)

Question. Find the Domain of \( f(x) = \sqrt{\frac{1-|x|}{2-|x|}} \).
Answer:
We have, \( f(x) = \sqrt{\frac{1 - |x|}{2 - |x|}} \)

Domain: \( f(x) \) is real for all values of \( x \) such that:
\( \frac{1-|x|}{2-|x|} \geq 0 \text{ and } 2 - |x| \neq 0 \)
\( \Rightarrow \frac{-(|x|-1)}{-(|x|-2)} \geq 0 \text{ and } |x| \neq 2 \text{ and } x \neq \pm 2 \)
\( \Rightarrow \frac{(|x|-1)(|x|-2)}{(|x|-2)^2} \geq 0 \qquad \{ \text{multiply \& divide by } |x| - 2 \} \)
\( \Rightarrow (|x| - 1)(|x| - 2) \geq 0 \)

Case 1: \( |x| \leq 1 \) and \( |x| > 2 \) (No common solution)
Case 2: \( |x| \leq 1 \) and \( |x| \leq 2 \Rightarrow |x| \leq 1 \Rightarrow -1 \leq x \leq 1 \)
Case 3: \( |x| \geq 1 \) and \( |x| > 2 \Rightarrow |x| > 2 \Rightarrow x > 2 \text{ or } x < -2  />
Clearly, the common solution is:
\( x \in (-\infty, -2) \cup [-1, 1] \cup (2, \infty) \text{ ans.} \)

Question. If \( f(x) = \frac{x+1}{x-1} \), find \( f(f(x)) \).
Answer:
We have, \( f(x) = \frac{x+1}{x-1} \)
Now, \( f(f(x)) = f\left(\frac{x+1}{x-1}\right) \)
\( = \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} \)
\( = \frac{\frac{x+1+x-1}{x-1}}{\frac{x+1-(x-1)}{x-1}} \)
\( = \frac{2x}{2} \)
\( = x \text{ ans.} \)

Question. Let \( f = \{(1,1), (2,3), (0, -1), (-1, -3)\} \) be a linear function. Find \( f(x) \).
Answer:
Let \( f(x) = ax + b \)
\( (1,1) \in f \Rightarrow x = 1 \text{ and } f(x) = 1 \)
\( \therefore 1 = a + b \) ............ (1)
\( (2,3) \in f \Rightarrow x = 2 \text{ and } f(x) = 3 \)
\( \dots 3 = 2a + b \) ............ (2)

Solving (1) and (2):
Subtracting (1) from (2) gives:
\( a = 2 \)
Substituting \( a = 2 \) into (1):
\( b = -1 \)
\( \therefore f(x) = 2x - 1 \text{ ans.} \)

Question. \( f(x) = \begin{cases} x^2, & 0 \leq x \leq 3 \\ 3x, & 3 \leq x \leq 10 \end{cases} \) and \( g(x) = \begin{cases} x^2, & 0 \leq x \leq 2 \\ 3x, & 2 \leq x \leq 10 \end{cases} \). Show that \( f \) is a function but \( g \) is not a function.
Answer:
We have, \( f(x) = \begin{cases} x^2, & 0 \leq x \leq 3 \\ 3x, & 3 \leq x \leq 10 \end{cases} \)
\( \therefore f = \{(0,0), (1,1), (2,4), (3,9), (4,12), (5,15), (6,18), (7,21), (8,24), (9,27), (10,30)\} \)
Clearly, the first element of each ordered pair is unique (or each element in the domain has a unique image).
\( \therefore f \) is a function.

Now, \( g(x) = \begin{cases} x^2, & 0 \leq x \leq 2 \\ 3x, & 2 \leq x \leq 10 \end{cases} \)
\( \therefore g = \{(0,0), (1,1), (2,4), (2,6), (3,9), (4,12), (5,15), (6,18), (7,21), (8,24), (9,27), (10,30)\} \)
Clearly, the element 2 has two different images, 4 and 6.
\( \therefore g \) is not a function.

Question. Let \( A = \{9, 10, 11, 12, 13\} \); let \( f: A \rightarrow N \) be defined by \( f(n) = \text{highest prime factor of } n \). Find the range of \( f \).
Answer:
\( A = \{9, 10, 11, 12, 13\} \)
\( f(n) = \text{highest prime factor of } n \)

\( f(9) = \text{highest prime factor of } 9 = 3 \)
\( f(10) = \text{highest prime factor of } 10 = 5 \)
\( f(11) = \text{highest prime factor of } 11 = 11 \)
\( f(12) = \text{highest prime factor of } 12 = 3 \)
\( f(13) = \text{highest prime factor of } 13 = 13 \)

\( \therefore \text{Range} = \{3, 5, 11, 13\} \text{ ans.} \)

Question. Let \( f \) be a subset of \( Z \times Z \) defined by \( f = \{(ab, a+b): a, b \in Z\} \). Is \( f \) a function? Justify your answer.
Answer:
We have, \( f = \{(ab, a+b): a, b \in Z\} \)
Let \( a = 2 \) and \( b = 3 \):
\( \therefore ab = 6 \text{ and } (a+b) = 5 \Rightarrow (6,5) \in f \)

Now, let \( a = 6 \) and \( b = 1 \):
\( \therefore ab = 6 \text{ and } (a+b) = 7 \Rightarrow (6,7) \in f \)

Since both \( (6,5) \in f \) and \( (6,7) \in f \), the element 6 has two different images, 5 and 7.
\( \therefore f \) is not a function ans.