Question 1: Find the domain of each of the following real valued functions of real variable:

$\displaystyle \text{(i) } f(x) = \frac{1}{x} \hspace{1.0cm} \text{(ii) } f(x) = \frac{1}{x-7} \hspace{1.0cm} \text{(iii) } f(x) = \frac{3x-2}{x-1} \hspace{1.0cm} \\ \\ \\ \text{(iv) } f(x) = \frac{2x+1}{x^2-9} \hspace{1.0cm} \text{(v) } f(x) = \frac{x^2 + 2x +1}{x^2 - 8x + 12}$

$\displaystyle \text{(i) } \text{Given } f(x) = \frac{1}{x}$

$\displaystyle \therefore f(x) \text{ assumes real values for all } x \text{ except } \text{For } x = 0$

$\displaystyle \text{Hence Domain } (f) = R - \{ 0 \}$

$\displaystyle \text{(ii) } \text{Given } f(x) = \frac{1}{x-7}$

$\displaystyle \therefore f(x) \text{ assumes real values for all } x \text{ except } \text{For } x = 7$

$\displaystyle \text{Hence Domain } (f) = R - \{ 7 \}$

$\displaystyle \text{(iii) } \text{Given } f(x) = \frac{3x-2}{x-1}$

$\displaystyle \therefore f(x) \text{ assumes real values for all } x \text{ except } \text{For } x = 1$

$\displaystyle \text{Hence Domain } (f) = R - \{ 1 \}$

$\displaystyle \text{(iv) } \text{Given } f(x) = \frac{2x+1}{x^2-9} = \frac{2x+1}{(x-3)(x+3)}$

$\displaystyle \therefore f(x) \text{ assumes real values for all } x \text{ except } \text{For } x = -3, 3$

$\displaystyle \text{Hence Domain } (f) = R - \{ -3, 3 \}$

$\displaystyle \text{(v) } \text{Given } f(x) = \frac{x^2 + 2x +1}{x^2 - 8x + 12} = \frac{x^2 + 2x +1}{(x-6)(x-2)}$

$\displaystyle \therefore f(x) \text{ assumes real values for all } x \text{ except } \text{For } x = 6, 2$

$\displaystyle \text{Hence Domain } (f) = R - \{ 6, 2 \}$

$\displaystyle \\$

Question 2: Find the domain of each of the following real valued functions of real variable:

$\displaystyle \text{(i) } f(x) = \sqrt{x-2} \hspace{1.0cm} \text{(ii) } f(x) = \frac{1}{ \sqrt{x^2 - 1} } \hspace{1.0cm} \text{(iii) } f(x) = \sqrt{9-x^2} \hspace{1.0cm} \\ \\ \text{(iv) } f(x) = \sqrt{ \frac{x-2}{3-x} }$

$\displaystyle \text{(i) } \text{Given } f(x) = \sqrt{x-2}$

$\displaystyle \therefore f(x) \text{ assumes real values for all } x \geq 2$

$\displaystyle \therefore x \in [ 2, \infty )$

$\displaystyle \text{Hence Domain } (f) = [ 2, \infty )$

$\displaystyle \text{(ii) } \text{Given } f(x) = \frac{1}{ \sqrt{x^2 - 1} }$

$\displaystyle f(x)$ assumes real values $\displaystyle \text{For } \sqrt{x^2-1} > 0$

$\displaystyle \Rightarrow (x-1)(x+1) > 0$

$\displaystyle \Rightarrow x > 1 or x < -1$

$\displaystyle \Rightarrow x \in (- \infty , -1) \cup (1, \infty)$

Hence the Domain $\displaystyle (f) = (- \infty , -1) \cup (1, \infty)$

$\displaystyle \text{(iii) } \text{Given } f(x) = \sqrt{9-x^2}$

$\displaystyle f(x) \text{ assumes real value } \text{For } 9-x^2 \geq 0$

$\displaystyle \Rightarrow (3-x)(3+x) \geq 0$

$\displaystyle \Rightarrow x \geq -3 \text{ and } x \leq 3$

$\displaystyle \therefore -3 \leq x \leq 3$

$\displaystyle \text{Hence Domain } (f) = [ -3 , 3 ]$

$\displaystyle \text{(iv) } \text{Given } f(x) = \sqrt{ \frac{x-2}{3-x} }$

$\displaystyle f(x)$ assumes real value if

$\displaystyle x-2 \geq 2 \text{ and } 3-x > 0$

$\displaystyle \Rightarrow x \geq 2 \text{ and } x < 3$

$\displaystyle \therefore x \in [2, 3)$

Hence the Domain $\displaystyle (f) = [2, 3)$

$\displaystyle \\$

Question 3: Find the domain of each of the following real valued functions of real variable:

$\displaystyle \text{(i) } f(x) = \frac{ax+b}{bx-a} \hspace{1.0cm} \text{(ii) } f(x) = \frac{ax-b}{cx-d} \hspace{1.0cm} \text{(iii) } f(x) = \sqrt{x-1}$

$\displaystyle \text{(iv) } f(x) = \sqrt{x-3} \hspace{1.0cm} \text{(v) } f(x) = \frac{x-2}{2-x} \hspace{1.0cm} \text{(vi) } f(x) = | x-1 |$

$\displaystyle \text{(vii) } f(x) = - |x|$ ($\displaystyle \text{(viii) } f(x) = \sqrt{9-x^2} \hspace{1.0cm} \text{(ix) } f(x) = \frac{1}{\sqrt{16-x^2}}$

$\displaystyle \text{(x) } f(x) = \sqrt{x^2- 16}$

$\displaystyle \text{(i) } \text{Given } f(x) = \frac{ax+b}{bx-a}$

$\displaystyle f(x)$ will assume real values for all $\displaystyle x$ except for the value of $\displaystyle x$ for

$\displaystyle bx - a = 0 \Rightarrow x = \frac{a}{b}$

$\displaystyle \text{Therefore Domain } (f) = R - \{ \frac{a}{b} \}$

$\displaystyle \text{Let } f(x) = y$

$\displaystyle y = \frac{ax+b}{bx-a}$

$\displaystyle \Rightarrow ax + b = bxy - ay$

$\displaystyle \Rightarrow b + ay = x(by - a)$

$\displaystyle \Rightarrow x = \frac{b + ay}{by-a}$

Clearly $\displaystyle x$ will take real value for all $\displaystyle x \in R$ except for

$\displaystyle by-a = 0 \Rightarrow y = \frac{a}{b}$

$\displaystyle \therefore \text{Range } (f) = R - \{ \frac{a}{b} \}$

$\displaystyle \text{(ii) } \text{Given } f(x) = \frac{ax-b}{cx-d}$

$\displaystyle f(x)$ will assume real values for all $\displaystyle x$ except for the value of $\displaystyle x$ for

$\displaystyle cx - d = 0 \Rightarrow x = \frac{d}{c}$

$\displaystyle \text{Therefore Domain } (f) = R - \{ \frac{d}{c} \}$

$\displaystyle \text{Let } f(x) = y$

$\displaystyle y = \frac{ax-b}{cx-d}$

$\displaystyle \Rightarrow ax - b = cxy - dy$

$\displaystyle \Rightarrow dy-b = cxy-ax$

$\displaystyle \Rightarrow x = \frac{dy-b}{cy-a}$

Clearly $\displaystyle x$ will take real value for all $\displaystyle y \in R$ except for

$\displaystyle cy-a = 0 \Rightarrow y = \frac{a}{c}$

$\displaystyle \therefore \text{Range } (f) = R - \{ \frac{a}{c} \}$

$\displaystyle \text{(iii) } \text{Given } f(x) = \sqrt{x-1}$

$\displaystyle \therefore f(x)$ assumes real value if $\displaystyle x- 1 \geq 0 \Rightarrow x \geq 1 \Rightarrow x \in [1, \infty)$

$\displaystyle \text{Hence Domain } (f) = [1, \infty)$

$\displaystyle \text{For } x \geq 1$, we have $\displaystyle x - 1 \geq 0 \Rightarrow \sqrt{x-1} \geq 0 \Rightarrow f(x) \geq 0$

Thus $\displaystyle f(x)$ takes all real values greater than equal to $\displaystyle 0$

$\displaystyle \text{Hence Range } (f) = [0, \infty)$

$\displaystyle \text{(iv) } \text{Given } f(x) = \sqrt{x-3}$

$\displaystyle \therefore f(x)$ assumes real value if $\displaystyle x- 3 \geq 0 \Rightarrow x \geq 3 \Rightarrow x \in [3, \infty)$

$\displaystyle \text{Hence Domain } (f) = [3, \infty)$

$\displaystyle \text{For } x \geq 3$, we have $\displaystyle x - 3 \geq 0 \Rightarrow \sqrt{x-3} \geq 0 \Rightarrow f(x) \geq 0$

Thus $\displaystyle f(x)$ takes all real values greater than equal to $\displaystyle 0$

$\displaystyle \text{Hence Range } (f) = [0, \infty)$

$\displaystyle \text{(v) } \text{Given } f(x) = \frac{x-2}{2-x}$

$\displaystyle \therefore f(x)$ is defined for all $\displaystyle x \in R$ except for which $\displaystyle 2-x = 0 \Rightarrow x = 2$

$\displaystyle \text{Hence Domain } (f) = R - \{ 2 \}$

Left $\displaystyle f(x) = y$

$\displaystyle \Rightarrow \frac{x-2}{2-x} = y$

$\displaystyle \Rightarrow x-2 = 2y - xy$

$\displaystyle \Rightarrow y = \frac{x-2}{2-x} = - 1$

$\displaystyle \text{Hence Range } (f) = \{ -1 \}$

$\displaystyle \text{(vi) } \text{Given } f(x) = | x- 1|$

$\displaystyle \therefore f(x)$ is defined for all $\displaystyle x \in R$

$\displaystyle \text{Hence Domain } (f) = R$

$\displaystyle \text{Let } f(x) = y$

$\displaystyle \Rightarrow |x-1| = y$

$\displaystyle \Rightarrow f(x) \geq 0$ for all $\displaystyle x \in R$

$\displaystyle \text{Hence Range } (f) = [0, \infty)$

$\displaystyle \text{(vii) } \text{Given } f(x) = -| x|$

$\displaystyle \therefore f(x)$ is defined for all $\displaystyle x \in R$

$\displaystyle \text{Hence Domain } (f) = R$

$\displaystyle \text{Let } f(x) = y$

$\displaystyle \Rightarrow -|x| = y$

It can be observed that the range of $\displaystyle f(x) = -|x|$ is all real except positive real numbers.

$\displaystyle \text{Hence Range } (f) = (- \infty, 0)$

$\displaystyle \text{(viii) } \text{Given } f(x) = \sqrt{9-x^2}$

$\displaystyle \therefore f(x)$ will assume real values for all $\displaystyle x$ except when

$\displaystyle 9-x^2 < 0 \Rightarrow (3-x)(3+x) < 0$

$\displaystyle \Rightarrow x > 3 \text{ and } x <-3$

$\displaystyle \text{Therefore } f(x)$ will be real $\displaystyle \text{For } -3 \leq x \leq 3$

$\displaystyle \text{Hence Domain } (f) = \{ -3, 3 \}$

$\displaystyle \text{Let } f(x) = y$

$\displaystyle \Rightarrow \sqrt{9-x^2} = y$

$\displaystyle y$ will assume real values for all

$\displaystyle 9-x^2 \geq 0 \Rightarrow (3-x)(3+x) \geq 0 \Rightarrow -3 \leq x \leq 3$

For any value such that $\displaystyle -3 \leq x \leq 3$, the value of $\displaystyle y$ will lie between $\displaystyle 0 \text{ and } 3$

$\displaystyle \text{Hence Range } (f) = [0, 3]$

$\displaystyle \text{(ix) } \text{Given } f(x) = \frac{1}{\sqrt{16-x^2}}$

$\displaystyle \therefore f(x)$ will assume real values when

$\displaystyle 16-x^2 > 0 \Rightarrow (4-x)(4+x) > 0 \Rightarrow -4 \leq x \leq 4$

$\displaystyle \Rightarrow x \in (-4, 4)$

$\displaystyle \text{Hence Domain } (f) = (-4, 4)$

$\displaystyle \text{Let } f(x) = y$

$\displaystyle \Rightarrow \frac{1}{\sqrt{16-x^2}} = y$

$\displaystyle \text{Since } -4 \leq x \leq 4$ mean $\displaystyle \frac{1}{4} < y < \infty$

$\displaystyle \text{Hence Range } (f) = [ \frac{1}{4} , \infty )$

$\displaystyle \text{(x) } \text{Given } f(x) = \sqrt{x^2 - 16}$

$\displaystyle \therefore f(x)$ will assume real values for all $\displaystyle x \geq 4 \text{or } x \leq -4$

$\displaystyle \Rightarrow x \in [4, \infty) or x \in (-\infty, -4]$

$\displaystyle \text{Hence Domain } (f) = (-\infty, -4] \cup [4, \infty)$

$\displaystyle \text{For } x \geq 4, f(x) \geq 0$

$\displaystyle \text{For } x \leq -4 , f(x) \geq 0$

$\displaystyle \therefore f(x)$ will take real values greater that equal to $\displaystyle 0$.

$\displaystyle \text{Hence Range } (f) = [0, \infty)$