Examples on Domains and Ranges of Functions Set 4

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Example- 13

(a) \(f\left( x \right) = {x^2} + x + 2\) (b) \(f\left( x \right) = 6{x^2} + 5x + 1\) (c) \(f\left( x \right) = {x^2} + 7x + 1\)
(d) \(f\left( x \right) = {x^2} + 2x + 3\) (e) \(f\left( x \right) = \sqrt {{x^2} + 2x + 3} \)  (f) \(f\left( x \right) = \begin{align}\frac{1}{{{x^2} + 2x + 3}}\end{align}\)
(g) \(f\left( x \right) = \begin{align}\frac{1}{{\sqrt {{x^2} + 2x + 3} }}\end{align}\)  (h) \(f\left( x \right) = \begin{align}\frac{1}{{{x^2} - 2x + 1}}\end{align}\)   (i) \(f\left( x \right) = \sqrt {{x^2} - 7x + 6} \)
(j) \(f\left( x \right) = \begin{align}\frac{1}{{1 + 2x + 3{x^2}}}\end{align}\)    

Solution:

(a) \(f(x) = {x^2} + x + 2 = {\left( {x + \frac{1}{2}} \right)^2} + \frac{7}{4} \ge\frac{7}{4}\,\,\, \Rightarrow \,\,\,\mathbb{R} = \left[ {\frac{7}{4},\infty } \right)\)


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

\(\begin{align}&= 6\left( {{x^2} + \frac{{5x}}{6} + \frac{{25}}{{144}} - \frac{{25}}{{144}} + \frac{1}{6}} \right) \hfill\qquad\qquad\qquad \\&= 6{\left( {x + \frac{5}{{12}}} \right)^2} - \frac{1}{{24}} \ge - \frac{1}{{24}}\,\, \Rightarrow \,\,\mathbb{R} = \left[ { - \frac{1}{{24}},\infty } \right) \hfill \\\end{align} \)

 

(c) \(f(x) = {x^2} + 7x + 1 = {x^2} + 7x + {\left( {\frac{7}{2}} \right)^2} + 1 - {\left( {\frac{7}{2}} \right)^2}\)

\( = {\left( {x + \frac{7}{2}} \right)^2} - \frac{{45}}{4} \ge - \frac{{45}}{4}\,\,\, \Rightarrow \,\mathbb{R} = \left[ { - \frac{{45}}{4},\infty } \right)\)

 

(d) \(f(x) = {x^2} + 2x + 3 = {(x + 1)^2} + 2 \ge 2\,\, \Rightarrow \,\,R = [2,\infty ).\)

 

(e) \(f(x) = \sqrt {{x^2} + 2x + 3} \)

Since \({x^2} + 2x + 3 \ge 2,\,\,\sqrt {{x^2} + 2x + 3} \ge \sqrt 2 \,\, \Rightarrow \,\,R = [\sqrt 2 ,\infty )\)

 

(f) \(f(x) =\begin{align} \frac{1}{{{x^2} + 2x + 3}}\end{align}\)

\({x^2} + 2x + 3 \ge 2\,\, \Rightarrow \,\,0 < \begin{align}\frac{1}{{{x^2} + 2x + 3}} \end{align}\le\frac{1}{2}\,\, \Rightarrow \,\,R = \left( {0,\frac{1}{2}} \right]\)

Note again that 0 is not inthe range, for the same reason as in Q.15.

 

(g) \(f(x) = \begin{align}\frac{1}{{\sqrt {{x^2} + 2x + 3} }}\end{align}\,\, \Rightarrow \,\,0 < f(x) \le \frac{1}{{\sqrt 2 }}\,\) (by extending the result of Q.19)

\( \Rightarrow R = \left( {0,\frac{1}{{\sqrt 2 }}} \right]\)

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

The domain for \(f(x)\)is \(\mathbb{R}\backslash \{ 1\} \)

For this domain, \({(x - 1)^2} > 0\,\, \Rightarrow \,\,\begin{align}\frac{1}{{{{(x - 1)}^2}}}\end{align} > 0\,\, \Rightarrow \,\,R = (0,\infty )\)

(i) \(f(x) = \sqrt {{x^2} - 7x + 6} \)

For the domain, \({x^2} - 7x + 6 \ge 0\,\, \Rightarrow \,\,x \le 1\)or \(x \ge 6\)

Now, since \({x^2} - 7x + 6 \ge 0\,\, \Rightarrow \,\,\sqrt {{x^2} - 7x + 6} \ge 0\,\, \Rightarrow \,\,R = [0,\infty )\)

(Reread this solution carefully till you understand it properly).

(j)

\(\begin{align}f\left( x \right) =\frac{1}{{1 + 2x + 3{x^2}}} =\frac{1}{{3\left( {{x^2} + \frac{2}{3}x + \frac{1}{3}} \right)}} = \frac{1}{{3{{\left( {x + \frac{1}{3}} \right)}^2} + \frac{2}{3}}} \le \frac{1}{{2/3}} = \frac{3}{2}\end{align}\)

\( \Rightarrow R = \left[ {\frac{3}{2},\,\infty } \right)\)

 

Example- 14

(a) \(f\left( x \right) =\begin{align} \frac{1}{{\left\{ x \right\}}}\end{align}\)  (b) \(f\left( x \right) = \begin{align}\frac{1}{{1 + {x^2}}}\end{align}\) (c) \(f\left( x \right) = \begin{align}\frac{1}{{1 + {{\sin }^2}x}}\end{align}\)
(d) \(f\left( x \right) = \begin{align}\frac{1}{{\left[ x \right]}}\end{align}\)  (e) \(f\left( x \right) = \begin{align}\frac{1}{{{{\left[ x \right]}^2}}}\end{align}\) (f) \(f\left( x \right) = \begin{align}\frac{1}{{1 + {{\left\{ x \right\}}^2}}}\end{align}\)
(g) \(f\left( x \right) = \begin{align}\frac{1}{{1 + \left| {\cos x} \right|}}\end{align}\) (h) \(f\left( x \right) =\begin{align} \frac{1}{{1 - \left\{ x \right\}}}\end{align}\)  

Solution:

(a) \(f(x) = \begin{align}\frac{1}{{\{ x\} }}\end{align}\)

The domain for \(f(x)\) is \(\mathbb{R}\backslash \mathbb{Z}\) ; for this domain \(0 < \{ x\} < 1\,\, \Rightarrow \,\,1 < \frac{1}{{\{ x\} }}\,\, \Rightarrow \,\,R = (1,\infty )\)


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

\({x^2} \ge 0\,\, \Rightarrow \,\,1 + {x^2} \ge 1\,\, \Rightarrow \,\,0 < \frac{1}{{1 + {x^2}}} \le 1\,\, \Rightarrow \,\,R = (0,1]\)

Note that 0 is not included in the range because no value of x can make \(f(x)\) equal to 0.

(c) \(f(x) = \begin{align}\frac{1}{{1 + {{\sin }^2}x}}\end{align}\)

Since \(\sin x \in [ - 1,1],\)\(\begin{align}1 \le 1 + {\sin ^2}x \le 2\,\, \Rightarrow \,\,\frac{1}{2} \le \frac{1}{{1 + {{\sin }^2}x}} \le 1\,\, \Rightarrow \,\,R = \left[ {\frac{1}{2},1} \right]\end{align}\)

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

Since \([x] \in \mathbb{Z}\,\, \Rightarrow \,\,R = \left\{ {\frac{1}{n};\,\,n \in \mathbb{Z}\backslash \{ 0\} } \right\}\)

(e) \(f(x) = \begin{align}\frac{1}{{{{[x]}^2}}}\end{align}\)

Again, since \([x] \in \mathbb{Z}\,\, \Rightarrow \,\,R = \left\{ {\frac{1}{{{n^2}}};\,\,n \in \mathbb{Z}\backslash \{ 0\} } \right\}\)

(f) \(f(x) = \begin{align}\frac{1}{{1 + {{\{ x\} }^2}}}\end{align}\)

\(0 \le \{ x\} < 1\,\, \Rightarrow \,\,1 \le 1 + {\{ x\} ^2} < 2\,\, \Rightarrow \,\,\frac{1}{2} < \frac{1}{{1 + {{\{ x\} }^2}}} \le 1\,\, \Rightarrow \,\,R = \left( {\frac{1}{2},1} \right]\)

(g) \(f(x) = \begin{align}\frac{1}{{1 + |\cos x|}}\end{align}\)

\( - 1 \le \cos x \le 1 \Rightarrow 1 \le 1 + |\cos x| \le 2 \Rightarrow \frac{1}{2} \le \frac{1}{{1 + |\cos x|}} \le 1 \Rightarrow R = \left[ {\frac{1}{2},1} \right]\)

(h) \(f(x) = \begin{align}\frac{1}{{1 - \{ x\} }}\end{align}\)

We have, \(0 < 1 - \{ x\} \le 1\,\, \Rightarrow \,\,1 \le \frac{1}{{1 - \{ x\} }}\, \Rightarrow \,R = [1,\infty )\)

 

Example- 15

Find the domain of \(\begin{align}f(x) = \frac{1}{{\sqrt {{{[x]}^2} - 3[x] + 2} }}\end{align}\)

Solution: For \(f(x)\) to be defined, we require \({[x]^2} - 3[x] + 2 > 0\) (note that it cannot be equal to 0)

\[\begin{align} \Rightarrow \quad ([x] - 1)([x] - 2) > 0 \quad\Rightarrow &\quad [x] < 1{\text{ }}or{\text{ }}[x] > 2\\ \Rightarrow & \quad x < 1{\text{ }}or{\text{ }}x > 3 \end{align}\]

(Note carefully where to place the equality sign also, and where there will be a pure inequality)

\[ \Rightarrow D = ( - \infty ,1) \cup [3,\infty )\]