CBSE Class 11 Mathematics Relations Functions Worksheet Set 07

CBSE Class 11 Mathematics Worksheet - Relations Functions (5). 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.

FUNCTIONS

Question. Find the domain of function, \( f(x) = \frac{x^2+3x+5}{x^2-5x+4} \).
Answer: We have, \( f(x) = \frac{x^2+3x+5}{x^2-5x+4} \)
\( f(x) \) is real for all values of \( x \) such that:
\( x^2 - 5x + 4 \neq 0 \)
\( \Rightarrow (x - 4)(x - 1) \neq 0 \)
\( \therefore \text{Domain} = R - \{1,4\} \)

Question. Find the domain of the function, \( f(x) = \frac{2x-3}{x^2-3x+2} \).
Answer: \( R - \{1,2\} \)

Question. Find the domain of the function, \( f(x) = \sqrt{4-x} + \frac{1}{\sqrt{x^2-1}} \).
Answer: \( f(x) \) is real for all values of \( x \) such that:
\( 4 - x \geq 0 \) and \( x^2 - 1 > 0 \)
\( \Rightarrow x - 4 \leq 0 \) and \( (x + 1)(x - 1) > 0 \)
\( \Rightarrow x \leq 4 \) and \( x \in (-\infty, -1) \cup (1, \infty) \)

The common solution is:
\( \therefore \text{Domain is } (-\infty, -1) \cup (1, 4] \text{ ans.} \)

Question. Find the domain and range of function \( f(x) = \frac{x-2}{3-x} \).
Answer: We have, \( f(x) = \frac{x-2}{3-x} \)

Domain: \( f(x) \) is real for all values of \( x \) such that:
\( 3 - x \neq 0 \Rightarrow x \neq 3 \)
\( \therefore \text{Domain} = R - \{3\} \)

Range: Let \( y = f(x) \)
\( \Rightarrow y = \frac{x-2}{3-x} \)
\( \Rightarrow 3y - xy = x - 2 \)
\( \Rightarrow x + xy = 3y + 2 \)
\( \Rightarrow x(1 + y) = 3y + 2 \)
\( \Rightarrow x = \frac{3y+2}{y+1} \)

\( x \) is real for all values of \( y \) such that:
\( y + 1 \neq 0 \Rightarrow y \neq -1 \)
\( \therefore \text{Range} = R - \{-1\} \text{ ans.} \)

Question. Find the domain & range of the function \( f(x) = \sqrt{16-x^2} \).
Answer: We have, \( f(x) = \sqrt{16-x^2} \)

Domain: \( f(x) \) is real for all values of \( x \) such that:
\( 16 - x^2 \geq 0 \)
\( \Rightarrow x^2 - 16 \leq 0 \)
\( \Rightarrow (x + 4)(x - 4) \leq 0 \)
\( \therefore \text{Domain: } x \in [-4, 4] \)

Range: Let \( y = f(x) \)
\( \Rightarrow y = \sqrt{16-x^2} \) ............. (1)
Squaring both sides:
\( y^2 = 16 - x^2 \)
\( \Rightarrow x^2 = 16 - y^2 \)

\( x \) is real for all values of \( y \) such that:
\( 16 - y^2 \geq 0 \)
\( \Rightarrow y^2 - 16 \leq 0 \)
\( \Rightarrow (y + 4)(y - 4) \leq 0 \)
\( \Rightarrow y \in [-4, 4] \)

But \( y \geq 0 \) from equation (1):
\( \therefore \text{Range} = [0, 4] \text{ ans.} \)

Question. Find the Domain & Range of \( f(x) = \frac{1}{\sqrt{x-5}} \).
Answer: We have, \( f(x) = \frac{1}{\sqrt{x-5}} \)

Domain: \( f(x) \) is real for all values of \( x \) such that:
\( x - 5 > 0 \Rightarrow x > 5 \)
\( \Rightarrow x \in (5, \infty) \)
\( \dots \text{Domain: } (5, \infty) \)

Range: Let \( y = f(x) \)
\( \Rightarrow y = \frac{1}{\sqrt{x-5}} \) ............ (1)
Squaring both sides:
\( y^2 = \frac{1}{x-5} \)
\( \Rightarrow x - 5 = \frac{1}{y^2} \)
\( \Rightarrow x = \frac{1}{y^2} + 5 \)
\( \Rightarrow x = \frac{1+5y^2}{y^2} \)

\( x \) is real for all values of \( y \) such that:
\( y^2 \neq 0 \Rightarrow y \neq 0 \)
\( \therefore y \in R - \{0\} \)
But \( y > 0 \) from equation (1):
\( \therefore \text{Range: } (0, \infty) \text{ ans.} \)

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

Domain: \( f(x) \) is real for all values of \( x \) such that:
\( 2 - x^2 \neq 0 \)
\( \Rightarrow x^2 - 2 \neq 0 \)
\( \Rightarrow (x + \sqrt{2})(x - \sqrt{2}) \neq 0 \)
\( \Rightarrow x \neq -\sqrt{2} \text{ and } x \neq \sqrt{2} \)
\( \dots \text{Domain: } R - \{-\sqrt{2}, \sqrt{2}\} \)

Range: Let \( y = f(x) \)
\( \Rightarrow y = \frac{3}{2-x^2} \)
\( \Rightarrow 2y - x^2y = 3 \)
\( \Rightarrow x^2y = 2y - 3 \)
\( \Rightarrow x^2 = \frac{2y-3}{y} \)
\( \Rightarrow x = \sqrt{\frac{2y-3}{y}} \)

\( x \) is real for all values of \( y \) such that:
\( \frac{2y-3}{y} \geq 0 \text{ and } y \neq 0 \)
\( \Rightarrow \frac{y(2y-3)}{y^2} \geq 0 \text{ and } y \neq 0 \qquad \{ \text{multiply \& divide by } y \} \)
\( \Rightarrow y(2y - 3) \geq 0 \text{ and } y \neq 0 \)
\( \Rightarrow y \in (-\infty, 0) \cup [\frac{3}{2}, \infty) \)
\( \therefore \text{Range: } (-\infty, 0) \cup [\frac{3}{2}, \infty) \text{ ans.} \)

Question. Let \( f = \left\{\left(x, \frac{x^2}{1+x^2}\right) : x \in R\right\} \) be a function from R to R. Determine Domain & Range.
Answer: We have, \( f(x) = \frac{x^2}{1+x^2} \)

Domain: \( f(x) \) is real for all values of \( x \) such that:
\( x \in R \qquad \{ \because 1 + x^2 \neq 0 \text{ for any } x \in R \} \)
\( \therefore \text{Domain} = R \)

Range: Let \( y = f(x) \)
\( \Rightarrow y = \frac{x^2}{1+x^2} \)
\( \Rightarrow y + x^2y = x^2 \)
\( \Rightarrow x^2y - x^2 = -y \)
\( \Rightarrow x^2(y - 1) = -y \)
\( \Rightarrow x^2 = \frac{-y}{y-1} \)
\( \Rightarrow x = \sqrt{\frac{-y}{y-1}} \)

\( x \) is real for all values of \( y \) such that:
\( \frac{-y}{y-1} \geq 0 \text{ and } y - 1 \neq 0 \)
\( \Rightarrow \frac{y}{y-1} \leq 0 \text{ and } y \neq 1 \)
\( \Rightarrow \frac{y(y-1)}{(y-1)^2} \leq 0 \text{ and } y \neq 1 \qquad \{ \text{multiply \& divide by } (y-1) \} \)
\( \Rightarrow y(y - 1) \leq 0 \)
\( \Rightarrow y \in [0, 1] \text{ but } y \neq 1 \)
\( \therefore \text{Range: } [0, 1) \text{ ans.} \)

Question. Find Domain & Range of \( f(x) = \frac{x}{1+x^2} \).
Answer: We have, \( f(x) = \frac{x}{1+x^2} \)

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

Range: Let \( y = f(x) \)
\( \Rightarrow y = \frac{x}{1+x^2} \)
\( \Rightarrow y + x^2y = x \)
\( \Rightarrow x^2y - x + y = 0 \)
Here, \( a = y, b = -1, c = y \)
By quadratic formula:
\( \Rightarrow x = \frac{1 \pm \sqrt{1-4y^2}}{2y} \)

\( x \) is real for all values of \( y \) such that:
\( 1 - 4y^2 \geq 0 \text{ and } 2y \neq 0 \)
\( \Rightarrow 4y^2 - 1 \leq 0 \text{ and } y \neq 0 \)
\( \Rightarrow (2y + 1)(2y - 1) \leq 0 \)
\( \Rightarrow y \in [-\frac{1}{2}, \frac{1}{2}] \text{ and } y \neq 0 \)

Additionally, checking \( y = 0 \):
If \( y = 0 \), then \( \frac{x}{1+x^2} = 0 \Rightarrow x = 0 \), which is real. Thus, \( y = 0 \) is included in the range.
\( \therefore \text{Range} = [-\frac{1}{2}, \frac{1}{2}] \text{ ans.} \)

Question. Find the Domain and Range of \( f(x) = \frac{x^2-9}{x-3} \).
Answer: We have, \( f(x) = \frac{x^2-9}{x-3} \)

Domain: \( f(x) \) is real for all values of \( x \) such that:
\( x - 3 \neq 0 \Rightarrow x \neq 3 \)
\( \therefore \text{Domain} = R - \{3\} \br />
Range: Let \( y = f(x) \)
\( \Rightarrow y = \frac{x^2-9}{x-3} \)
\( \Rightarrow y = \frac{(x+3)(x-3)}{(x-3)} \)
\( \Rightarrow y = x + 3 \)
\( \Rightarrow x = y - 3 \)

Clearly, \( x \) is real for all values of \( y \) such that \( y \in R \).
However, \( x \neq 3 \) from the domain:
If \( x = 3 \Rightarrow y = 3 + 3 = 6 \)
Therefore, \( y \neq 6 \).
\( \therefore \text{Range} = R - \{6\} \text{ ans.} \)