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}  

Answer:

\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} }  

Answer:

\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}

Answer:

\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)