# 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)

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