# Examples on Domains and Ranges of Functions Set 5

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Example 16

Find the domain of \begin{align}\ f\left( x \right) = \frac{1}{{{{\left\{ x \right\}}^2} - \left\{ x \right\} + 3/16}}\end{align}

Solution: f\left( x \right) =\begin{align} \frac{1}{{{{\left\{ x \right\}}^2} - \left\{ x \right\} + \frac{3}{{16}}}}\end{align} \Rightarrow {\left\{ x \right\}^2} - \left\{ x \right\} + \frac{3}{{16}} \ne 0 \Rightarrow \left( {\left\{ x \right\} - \frac{1}{4}} \right)\left( {\left\{ x \right\} - \frac{3}{4}} \right) \ne 0

$$\begin{gathered}\Rightarrow \left\{ x \right\} \ne \frac{1}{4}\,\,{\text{or}}\,\,\frac{3}{4} \hfill \\\qquad\qquad\quad\Rightarrow x \ne n + \frac{1}{4},\,\,n + \,\frac{3}{4};n \in \mathbb{Z} \hfill \\\qquad\qquad\qquad\qquad\Rightarrow D = \mathbb{R}\backslash \left\{ {n + \frac{1}{4},n + \frac{3}{4}} \right\};n \in \mathbb{Z} \hfill \\\end{gathered}$$

Example- 17

Find the domains of the following functions.

(i) $$f\left( x \right) =\sqrt {1 - \left| {{x^2} + 3x + 2} \right|}$$           (ii) f\left( x \right) = \begin{align}\frac{1}{{\sqrt {{x^2} + 3x + 3} - 1}}\end{align}

Solution:

(i) $$f\left( x \right) = \sqrt {1 - \left| {{x^2} + 3x + 2} \right|} \Rightarrow 1 - \left| {{x^2} + 3x + 2} \right| \ge 0$$

\begin{align}& \Rightarrow \left| {{x^2} + 3x + 2} \right| \le 1\\ &\Rightarrow \underbrace { - 1 \le {x^2}}_A + 3x + \overbrace {2 \le 1}^B\end{align}

Inequality - A: $${x^2} + 3x + 2 \ge - 1 \Rightarrow {x^2} + 3x + 3 \ge 0$$

This is satisfied for all real values of x.

Inequality - B: $${x^2} + 3x + 2 \le 1 \Rightarrow {x^2} + 3x + 1 \le 0$$

The roots of LHS are \begin{align}\frac{{ - 3 \pm \sqrt 5 }}{2} \Rightarrow \frac{{ - 3 - \sqrt 5 }}{2} \le x \le \frac{{ - 3 + \sqrt 5 }}{2}\end{align}

$$\Rightarrow D = \left[ {\frac{{ - 3 - \sqrt 5 }}{2},\frac{{ - 3 + \sqrt 5 }}{2}} \right]$$

(ii)  f\left( x \right) =\begin{align}\frac{1}{{\sqrt {{x^2} + 3x + 3} - 1}} \Rightarrow \sqrt {{x^2} + 3x + 3} - 1 \ne 0\end{align}

$$\begin{gathered}\Rightarrow {x^2} + 3x + 3 \ne 1 \hfill \\\Rightarrow {x^2} + 3x + 2 \ne 0 \hfill \\\Rightarrow x \ne - 1, - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow D = \mathbb{R}\backslash \left\{ { - 1, - 2} \right\} \hfill \\\end{gathered}$$

Example 18

Find the domains of the following functions.

 (i) $$f\left( x \right) = \sqrt {\sin x} + \sqrt {\cos x}$$ \begin{align} (ii) f\left( x \right) =\frac{1}{{{{\sin }^2}x - {{\cos }^2}x}}\end{align} \begin{align}(iii) f\left( x \right) = \sin \left( {\sqrt {{x^2} - 2} } \right) + \cos \sqrt {2 - {x^2}} + \frac{1}{{x - \sqrt 2 }}\end{align} \begin{align} (iv) f\left( x \right) = \frac{1}{{\sin x + \cos x}}\end{align}

Solution:

(i) $$f\left( x \right) = \sqrt {\sin x} + \sqrt {\cos x} \Rightarrow \sin x \ge 0\,{\rm{and }}\cos x \ge 0$$

$$\Rightarrow x \in \left[ {2n\pi ,\left( {2n + 1} \right)\pi } \right]{\rm{and }}x \in \left[ {\left( {2n - \frac{1}{2}} \right)\pi ,\left( {2n + \frac{1}{2}} \right)\pi } \right]$$

$$\Rightarrow x \in \left[ {2n\pi ,\left( {2n + \frac{1}{2}} \right)\pi } \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow D = \left\{ {\left[ {2n\pi ,\left( {2n + \frac{1}{2}} \right)\pi } \right]} \right\};n \in \mathbb{Z}$$

(ii) f\left( x \right) = \begin{align}\frac{1}{{{{\sin }^2}x - {{\cos }^2}x}} \Rightarrow {\sin ^2}x - {\cos ^2}x \ne 0 \Rightarrow \left( {\sin x + \cos x} \right)\left( {\sin x - \cos x} \right) \ne 0\end{align}

$$\begin{gathered}\qquad\qquad\qquad\qquad\Rightarrow \sin x \ne - \cos x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin x \ne \cos x \hfill \\\Rightarrow \tan x \ne 1\,{\text{or }} - 1 \hfill \\\qquad\qquad\qquad\qquad\Rightarrow x \ne n\pi \pm \frac{\pi }{4} \Rightarrow D = \mathbb{R}\backslash \left\{ {n\pi \pm \frac{\pi }{4}} \right\};n \in \mathbb{Z} \hfill \\\end{gathered}$$

(you could also have evaluated the domain by observing that the denominator is – cos 2x; so we require $$\cos 2x \ne 0$$$$\Rightarrow 2x \ne \left( {2n \pm \frac{1}{2}} \right)\pi$$$$\Rightarrow x \ne \left( {n \pm \frac{1}{4}} \right)\pi ;n \in \mathbb{Z}$$

(iii) $$f\left( x \right) = \sin \left( {\sqrt {{x^2} - 2} } \right) + \cos \left( {\sqrt {2 - {x^2}} } \right) + \frac{1}{{x - \sqrt 2 }}$$

$$\Rightarrow {x^2} - 2 \ge 0,\,\,\,\,\,\,2 - {x^2} \ge 0,\,\,\,\,\,\,\,\,\,\,\,x \ne \sqrt 2$$

Notice that only $$x = - \sqrt 2$$ satisfies all the three constraints simultaneously $$\Rightarrow D = \left\{ { - \sqrt 2 } \right\}$$

(iv) f\left( x \right) = \begin{align}\frac{1}{{\sin x + \cos x}} \Rightarrow \sin x + \cos x \ne 0\end{align}

$$\begin{gathered}\Rightarrow \tan x \ne - 1 \hfill \\\qquad\qquad\qquad\Rightarrow x \ne n\pi - \frac{\pi }{4}\, \Rightarrow D = \mathbb{R}\backslash \left\{ {n\pi - \frac{\pi }{4}} \right\};n \in \mathbb{Z} \hfill \\\end{gathered}$$

Example- 19

Find the domains of the following functions

(i) $$f\left( x \right) = \frac{1}{{\left| {{x^2} + 3x + 2} \right|}} + \frac{1}{{\left| {{x^2} - 1} \right|}} + \frac{1}{{\left| {6{x^2} + 5x + 1} \right|}}$$             (ii) $$f\left( x \right) = \frac{1}{{\sqrt {\left| {{x^2} - 1} \right|} - 1}}$$

Solution:

(i) f\left( x \right) = \begin{align}\frac{1}{{\left| {{x^2} + 3x + 2} \right|}} + \frac{1}{{\left| {{x^2} - 1} \right|}} + \frac{1}{{\left| {6{x^2} + 5x + 1} \right|}}\end{align}

$$\Rightarrow {x^2} + 3x + 2 \ne 0,\qquad\quad{x^2} - 1 \ne 0,\qquad\quad 6{x^2} + 5x + 1 \ne 0$$

\begin{align}\Rightarrow x \ne - 1, - 2\qquad\qquad x \ne - 1,1\qquad\qquad x \ne \frac{{ - 1}}{2},\frac{{ - 1}}{3} \hfill \\\Rightarrow D = \mathbb{R}\backslash \left\{ { - 2, - 1,\frac{{ - 1}}{2},\frac{{ - 1}}{3},1} \right\}. \hfill \\\end{align}

(ii) f\left( x \right) =\begin{align} \frac{1}{{\sqrt {\left| {{x^2} - 1} \right|} - 1}} \Rightarrow \sqrt {\left| {{x^2} - 1} \right|} \ne 1\end{align}

$$\Rightarrow \left| {{x^2} - 1} \right| \ne 1$$

$$\Rightarrow {x^2} - 1 \ne - 1,1$$

$$\Rightarrow {x^2} \ne 0,2\qquad \Rightarrow D = \mathbb{R}\backslash \left\{ {0, - \sqrt 2 ,\sqrt 2 } \right\}$$