Examples on Domains and Ranges of Functions Set 1

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Example- 8

Find the domains of the following functions, assuming the functions to be real.

(a) \(f(x) = \sqrt {{x^2} - 1} \) 

(b) \(f(x) = \begin{align}\frac{1}{{\sqrt {{x^2} - 1} }}\end{align}\)

(c) \(f(x) = \begin{align}\frac{1}{{1 + \sin x}}\end{align}\)

(d) \(f(x) = \sqrt {2 - x} + \sqrt {3 + x} \)

(e) \(f(x) = \sqrt {\sin x - 1} \) 

 (f) \(f(x) = \begin{align}\frac{1}{{\left[ x \right]}}\end{align}\)

(g) \(f(x) = k\) 

(h) \(f(x) = \sin \sqrt x \)

Solution:

(a) A square root function is defined when its argument is non-negative. Therefore, we require that

\[\begin{align}{x^2} - 1 \ge 0 \quad \Rightarrow \quad x \ge 1\,\,\,{\rm{or}}\,\,\,x \le - 1\\D =\quad( – \infty, –1] \cup [1, \infty)\end{align}\]

(b) Here again, we require the argument of the square root function to be non-negative. In addition, since the term\(\sqrt {{x^2} - 1} \) is in the denominator, we require it to be non-zero. Hence

\[\begin{align}{x^2} - 1 > 0 \quad &\Rightarrow \quad x > 1\; {\rm{or}}\,\,\,\,x < - 1\\D & = (– \infty , –1 \cup  (1, \infty)\end{align}\]

(c) For this reciprocal function, we require the denominator to be non-zero, that is, the domain should exclude all those points where

\[\begin{align}\text{sin }x = {\rm{ - }}1{\rm{ }} \quad &\Rightarrow \quad {\rm{ }}x = 2n\pi {\rm{ }} + 3{\rm{ }}\pi /2\\D &=\mathbb{R}  –\left\{ {2n\pi + \frac{{3\pi }}{2}} \right\} n  \in \mathbb{Z}\end{align}\]

(d) Here we require both the square root functions to be defined separately.

\[\begin{align}2 - x \ge 0\,\,\,\,{\rm{and}}\,\,\,\,3 + x \ge 0 \quad &\Rightarrow \quad  x \le 2\,\,\,\,\,\,{\rm{and}}\,\,\,\,\,x \ge - 3\\D &= [–3, 2]\end{align}\]

(e) This is the square- root function again. We require \(sin\,x \ge {\rm{ }}1.\) Now, sin x can vary in [–1, 1], (the range of sin x).

Hence the given function will assume real values at only those points where \({\rm{sin}}\,x = {\rm{ }}1.\)

(At all other values of \(x,{\rm{ }}sin\,x-{\rm{ }}1{\rm{ }} < {\rm{ }}0\))

\[\begin{align}\sin \;x = 1 &\Rightarrow x = 2n\pi + \frac{\pi }{2} \hfill \\ D &= \{ 2n\pi + \frac{\pi }{2}\} \;\;n \in \mathbb{Z} \hfill \\\end{align} \]

(f) We require the denominator to be non-zero.

\[\begin{align}[x] \ne 0 \quad & \Rightarrow \quad \text{x does not lie between 0 and 1 (where [x] is 0)}\\&\Rightarrow \quad x < 0 \; \text{and}\   x > 1\\D &= – \infty, 0 \cup [1, \infty )\end{align}\]

(g) This is a constant function, whose output is k regardless of the input.

D =\(\mathbb{R} \)

(h) There is no constraint on the argument of ‘sin’ function (its domain is \(\mathbb{R} \)). All we require here is that \(x \ge 0\) (for the square root function to be defined)

D = [0, \(\infty \) )

 

Example- 9

Find the domains of the following functions:

(a) \(f\left( x \right) = \sqrt {\left[ x \right] - 1} \)

(b) \(f\left( x \right) =\begin{align} \frac{1}{{\sqrt {\left[ x \right] - 1} }}\end{align}
\)

(c) \(f\left( x \right) = \sqrt {\tan x} \)

(d) \(f\left( x \right) =\begin{align} \frac{1}{{\sin x}} + \frac{1}{{\cos x}}\end{align}\) (e) \(f\left( x \right) = \begin{align}\frac{1}{{{{\sin }^2}x - 1}}\end{align}\) (f) \(f\left( x \right) = \sqrt {6x - 1} + \sqrt {1 - x} \)

(g) \(f\left( x \right) = \begin{align}\frac{1}{{\sqrt {{{\left[ x \right]}^2} - 4} }}\end{align}\)

(h) \(f\left( x \right) = \sqrt {\left\{ x \right\} - 1} \)  

Solution:

(a) \(f\left( x \right) = \sqrt {\left[ x \right] - 1} \,\, \Rightarrow \left[ x \right] - 1 \ge 0 \Rightarrow \left[ x \right] \ge 1 \Rightarrow D = \left[ {1,\infty } \right)\)

(b) \(f\left( x \right) =\begin{align} \frac{1}{{\sqrt {\left[ x \right] - 1} }}\end{align}\,\, \Rightarrow \left[ x \right] - 1 > 0 \Rightarrow x \ge 2 \Rightarrow D = \left[ {2,\infty } \right)\)

(c)

\(\begin{align} f\left( x \right) = \sqrt {\tan x} \,  \Rightarrow \quad & \tan x \ge 0 \Rightarrow x \in \left[ {n\pi ,\left( {n + \frac{1}{2}} \right)\pi } \right)\\  \Rightarrow \quad & D = \left\{ {\left[ {n\pi ,\left( {n + \frac{1}{2}} \right)\pi } \right)} \right\},\,\,n \in \mathbb{Z}\end{align}\)

 

(d)

\(\begin{align}f\left( x \right) = \frac{1}{{\sin x}} + \frac{1}{{\cos x}} \Rightarrow \quad & \sin x \ne 0\,\,{\rm{and}}\,\,\cos x \ne 0\\ \Rightarrow \quad & D = \mathbb{R}\backslash \left\{ {n\pi } \right\}\backslash \left\{ {\left( {2n \pm \frac{1}{2}} \right)\pi } \right\},\;n \in \mathbb{Z}\end{align}\)

 

(e)

\(\begin{align}f\left( x \right) = \frac{1}{{{{\sin }^2}\,x - 1}} \Rightarrow \quad & \sin x \ne - 1,1 \Rightarrow \quad x \ne \left( {2n + \frac{1}{2}} \right)\pi ,\left( {2n - \frac{1}{2}} \right)\pi ; \;n \in \,\mathbb{Z}\\  \Rightarrow \quad &  D = \mathbb{R}\backslash \left\{ {\left( {2n + \frac{1}{2}} \right)\pi } \right\}\backslash \left\{ {\left( {2n - \frac{1}{2}} \right)\pi } \right\}; \;n \in \mathbb{Z}\end{align}\)

 

(f)

\(\begin{align}f\left( x \right) = \sqrt {6x - 1} + \sqrt {1 - x} \Rightarrow \quad & 6x - 1 \ge 0\,\,{\rm{and}}\,\,1 - x \ge 0\\   \Rightarrow \quad & x \ge \frac{1}{6}\,\,{\rm{and}}\,\,x \le 1 \Rightarrow \quad  D = \left[ {\frac{1}{6},1} \right]\end{align}\)

 

(g)

\(\begin{align} f\left( x \right) = \frac{1}{{\sqrt {{{\left[ x \right]}^2} - 4} }} \Rightarrow \quad & {\left[ x \right]^2} - 4 > 0 \Rightarrow \quad \left[ x \right] > 2\,\,{\rm{or}}\,\,\left[ x \right] < - 2\\ \Rightarrow \quad &  x \ge 3\,\,{\rm{or}}\,\,x < - 2 \Rightarrow \quad D = \left( { - \infty , - 2} \right) \cup \left[ {3,\infty } \right) \end{align}\)

 

(h) \(f\left( x \right) = \sqrt {\left\{ x \right\} - 1} \)

      Notice that \(\left\{ x \right\}\)is always less than 1 and hence\(f\left( x \right)\)can never be defined for any real value of \(x \Rightarrow D = \phi \)