# Examples on Domains and Ranges of Functions Set 7

Go back to  'Functions'

Example- 23

Find the range for

(a) f(x) = \begin{align}\frac{1}{{2 + \sin 3x + \cos 3x}}\end{align}                                                                     (b) $$f(x) = [{x^2}] - {[x]^2}$$

(c) $$f(x) = \sqrt {a - x} \,\,\, + \,\,\,\sqrt {x - b}$$  $$\quad a > b > 0$$                                   (d) $$f(x) = {x^3} + 3{x^2} + 4x + 5$$

Solution: In case of linear, quadratic and other simple functions, we can express x in terms of f(x) and find the values of f(x) for which x is defined (These values form the range)

We did so in the last two questions. However, this expression is not always easily possible. So we have to find other ways that could yield the answer more easily:

(a) f (x) =  \begin{align} \frac{1}{{2 + \sin 3x + \cos 3x}} = \frac{1}{{2 + \sqrt 2 \sin (3x + \pi /4)}}\end{align}

The denominator can vary from $$2 - \sqrt 2$$ to $$2 + \sqrt 2$$ because $$-{\rm{ }}1 \le sin\theta \le 1$$

(The denominator is never 0 and hence D = $$\mathbb{R}$$ )

Therefore,

$\frac{1}{{2 + \sqrt 2 }} \le f(x){\rm{ }} \le \frac{1}{{2 - \sqrt 2 }}$

(b) Let x be expressed as I + f where I is the integral part and f the fractional part of x.

f (x) = $$[{(I + f)^2}] - {I^2} = [{I^2} + {f^2} + 2If] - {I^2} = [{f^2} + 2If]$$

A little thinking will show that the right side can take on any integral value, whether positive, zero or negative.

(Try out some examples: for x = 0.5, f(x) = 0; for x = –0.5, f(x) = –1

for x = 100.9 f(x) =180; for x = –100.9 f(x) = –21)

Also assume any value for f(x) and see whether you can find a corresponding value for x)

{you could also note that the difference between $${x^2}$$ and $${\left[ x \right]^2}$$ can be increased arbitrarily in magnitude due to the term 2If (which contains I). Try visualizing this in the form of a graph}

Hence, f(x) can take on any integer value.

i.e. R = $$\mathbb{Z}$$(the set of integers).

(c) For the domain, we require a x > 0 and xb > 0

$$\Rightarrow D = \left[ {b,\,\,a} \right]$$

For the range, let y = f(x) = $$\sqrt {a - x} + \sqrt {x - b}$$

We see that y > 0.

Observe the expression for f(x) carefully. The expression is ‘symmetric w.r.t ‘a’ and ‘b’. This gives us a hint that f(x) should attain an extremum at $$x = \frac{{a + b}}{2}$$. (We can of course, prove this)

At $$x = a$$, f(x) = $$\sqrt {a - b}$$

$$x = b$$, f(x) = $$\sqrt {a - b}$$

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

Hence, the observation of some symmetry in the expression directly allows us to write the range as

R = $$\left[ {\sqrt {a - b} ,\sqrt {2(a - b)} } \right]$$

Still not convinced about the symmetry part? Consider

$g(x) = {\rm{ }}{(x - {a_1})^2} + {(x - {a_2})^2} + ......... + {(x - {a_n})^2}$

Obviously, the maximum for $$g\left( x \right)$$ is unbounded. What is the minimum, and for what value of x is it attained? Symmetry in the expression hints that {x_{min}} =\begin{align} {\rm{ }}\frac{{{a_1} + {a_2} + ...... + {a_n}}}{n}.\end{align} Lets verify this.

$$g(x) = {\rm{ }}n{x^2} - 2({a_1} + {a_2} + .... + {a_n})x + a_1^2 + a_2^2 + ...... + a_n^2$$

$$p{x^2} + qx + r$$ = \begin{align}q = - 2({a_2} + {a_2} + ... + {a_n})\\r = a_1^2 + a_2^2 + .... + a_n^2\end{align} where

From quadratic expressions, we know that this expression is minimum for x = $$\frac{{ - q}}{{2p}}$$, (verify this), which for this case, becomes $$\frac{{{a_1} + {a_2} + ..... + {a_n}}}{n}!$$

Symmetry directly tells us the answer.

In our original question, suppose you want to solve it analytically

y = $$\sqrt {a - x} + \sqrt {x - b}$$

Squaring and rearranging gives

$${y^2}$$ = $$(a - b) + \,\,\underbrace {2\sqrt {(a - x)(x - b)} }_{{\rm{always positive}}}$$ > ab

$$\Rightarrow$$  y > $$\sqrt {a - b}$$ ... (i)

Now $${y^2} - (a - b)$$ = $$2\sqrt {(a - x)(x - b)}$$

Squaring and arranging in the form of a quadratic in x gives

$$4{x^2} - 4(a + b)x + {y^4} - 2(a - b){y^2} + {(a + b)^2}$$ = 0

Since x is real, the discriminant for this equation should satisfy D > 0 (This gives a constraint on y, or the range, as in the earlier cases)

$$\Rightarrow$$  $${(a + b)^2}$$> $${y^4} - 2(a - b){y^2} + {(a + b)^2}$$

$$\Rightarrow {\rm{}}{y^{2{\rm{ }}}} \le 2(a - b)$$

$$\Rightarrow$$  y $$\le \sqrt {2(a - b)}$$ .....(ii)

Combining (i) and (ii) gives R = $$\left[ {\sqrt {a - b} ,\sqrt {2(a - b)} } \right]$$

(d) Although the expression for f(x) look a bit complicated, we can at once determine the range as follows.

As x increases (or as x $$\rightarrow \infty$$ ), f (x) will keep on increasing in an unbounded fashion $$\left( {f\left( x \right) \to \infty } \right)$$. Similarly, as x decreases (or as x $$\rightarrow -\infty$$), f (x) will keep on decreasing in an unbounded fashion $$\left( {{\rm{or}}\,f\left( x \right) \to - \infty } \right)$$

Also, since f (x) is a polynomial function, it is continuous (and hence will vary continuously).

Hence, f(x) will vary between – $$\infty$$ and + $$\infty$$

R = $$\mathbb{R}$$

Example- 24

Find the range for

 (i) f\left( x \right) =\begin{align} \frac{1}{{\sin x + 2\cos x + 3}}\end{align} (ii) f\left( x \right) =\begin{align} \frac{1}{{{x^4} + 2{x^2} + 2}}\end{align} (iii) $$f\left( x \right) = \sqrt {3{x^2} - 4x + 5}$$ (iv) f\left( x \right) = \begin{align}\frac{1}{{1 + 3{{\left\{ x \right\}}^2}}}\end{align}

Solution:

(i) f\left( x \right) = \begin{align} \frac{1}{{\sin x + 2\cos x + 3}}\end{align}

To evaluate the range, our approach should be to somehow determine the range of the variable term {sin x + 2 cos x} in the denominator; this can be determined by somehow reducing this term to a simpler form; we are describing the general approach to reduce A sin x + B cos x :

$A\sin x + B\cos x = \sqrt {{A^2} + {B^2}} \left( {\frac{A}{{\sqrt {{A^2} + {B^2}} }}\sin x + \frac{B}{{\sqrt {{A^2} + {B^2}} }}\cos x} \right)$

Put $$\frac{A}{{\sqrt {{A^2} + {B^2}} }} = \cos \phi \,{\rm{and }}\frac{B}{{\sqrt {{A^2} + {B^2}} }} = \sin \phi \,\,{\rm{where }}\tan \phi = \frac{B}{A}$$

(verify that this substitution is valid)

Therefore, $$A\sin x + B\cos x = \sqrt {{A^2} + {B^2}} \sin \left( {x + \phi } \right)$$

For this particular question,

$$\sin x + 2\cos x = \sqrt 5 \sin \left( {x + \phi } \right)\,\,\,\,\,\,\,\,\,{\rm{where }}\tan \phi = 2$$

\begin{align}&\Rightarrow - \sqrt 5 \le \sqrt 5 \sin \left( {x + \phi } \right) \le \sqrt 5 \\&\Rightarrow - \sqrt 5 \le \sin x + 2\cos x \le \sqrt 5 \Rightarrow 3 - \sqrt 5 \le \sin x + 2\cos x + 3 \le 3 + \sqrt 5 \\&\Rightarrow \frac{1}{{3 + \sqrt 5 }} \le \frac{1}{{\sin x + 2\cos x + 3}} \le \frac{1}{{3 - \sqrt 5 }}\,\,\,\,\,\,\,\,\,\,\, \Rightarrow R = \left[ {\frac{1}{{3 + \sqrt 5 }},\frac{1}{{3 - \sqrt 5 }}} \right]\end{align}

(ii) f\left( x \right) = \begin{align} \frac{1}{{{x^4} + 2{x^2} + 2}} = \frac{1}{{{{\left( {{x^2} + 1} \right)}^2} + 1}}\end{align}

Now, $${x^2} + 1 \ge 1$$

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

(iii) $$f\left( x \right) = \sqrt {3{x^2} - 4x + 5}$$

The expression inside the square root function is

$$3{x^2} - 4x + 5 = 3\left( {{x^2} - \frac{4}{3}x + \frac{5}{3}} \right) = 3{\left( {x - \frac{2}{3}} \right)^2} + \frac{{11}}{3} \ge \frac{{11}}{3}$$

Therefore, $$\sqrt {3{x^2} - 4x + 5} \ge \sqrt {\frac{{11}}{3}} \Rightarrow R = \left[ {\sqrt {\frac{{11}}{3}} ,\infty } \right)$$

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

\begin{align}&0 \le \left\{ x \right\} < {\kern 1pt} \,1\,\,\,\, \Rightarrow 0 \le 3{\left\{ x \right\}^2} < 3\,\,\,\,\,\, \Rightarrow 1 \le 1 + 3{\left\{ x \right\}^2} < 4\\& \Rightarrow \frac{1}{4} < \frac{1}{{1 + 3{{\left\{ x \right\}}^2}}} \le 1\qquad\qquad \Rightarrow R = \left( {\frac{1}{4},1} \right]\end{align}

## TRY YOURSELF - II

Q. 1 Find the domains of the following functions

 (a) $$f\left( x \right) = \sqrt {2x + 1}$$ (b) f\left( x \right) = \begin{align}\frac{1}{{\sqrt {\sin x - 1} }}\end{align} (c) f\left( x \right) =\begin{align} \frac{1}{{\sqrt {1 - \sin x} }}\end{align} (d) f\left( x \right) =\begin{align} \frac{1}{{1 - {{\cos }^2}x}}\end{align} (e) f\left( x \right) =\begin{align} \frac{1}{{\sqrt {\left\{ x \right\}} }}\end{align} (f) $$f\left( x \right) = \sqrt {{x^2} - 3x + 2}$$ (g) f\left( x \right) =\begin{align} \frac{1}{{{x^2} + 2x + 4}}\end{align} (h) f\left( x \right) =\begin{align} \frac{1}{{2{x^2} + 5x + 2}}\end{align} (i) f\left( x \right) =\begin{align} \frac{1}{{{x^2} + 7x + 1}}\end{align} (j) f\left( x \right) = \begin{align}\frac{1}{{1 + 2\left[ x \right] + {{\left[ x \right]}^2}}}\end{align} (k) f\left( x \right) =\begin{align} \frac{1}{{1 + {x^2}}} + \sqrt {\left[ x \right]} \end{align} (l) f\left( x \right) =\begin{align} \frac{1}{{\left[ {{x^2}} \right]}}\end{align} (m) f\left( x \right) = \begin{align}\frac{1}{{\left[ {\left| x \right|} \right]}}\end{align} (n) f\left( x \right) =\begin{align} \frac{1}{{\left| {\left[ x \right]} \right|}}\end{align} (o) f\left( x \right) =\begin{align} \cdot \frac{1}{{\sqrt {\left[ x \right]} }}\end{align}

Q. 2 Find the range of the following functions

 (a) $$f\left( x \right) = \left[ x \right] + 1$$ (b) $$f\left( x \right) = 1 + \sin x$$ (c) $$f\left( x \right) = \left| {\left[ x \right]} \right|$$ (d) f\left( x \right) = \begin{align}\frac{1}{{{x^2}}}\end{align} (e) $$f\left( x \right) = \left| {\tan x} \right|$$ (f) $$f\left( x \right) = \left| {{x^2} - 3x + 2} \right|$$ (g) $$f\left( x \right) = \left[ {{x^2}} \right]$$ (h) $$f\left( x \right) = {\left[ x \right]^2}$$ (i) f\left( x \right) =\begin{align} 1 - x - {x^2}\end{align} (j) $$f\left( x \right) = 4{x^2} + 3x + 2$$ (k) $$f\left( x \right) = - 1 - 2x - 3{x^2}$$ (l) f\left( x \right) = \begin{align}\frac{1}{{\left| x \right|}}\end{align} (m) $$f\left( x \right) = 3\sin x + 4\cos x\,$$ (n) $$f\left( x \right) = \left[ {{x^2} - 2} \right]$$ (o) $$f\left( x \right) = \left[ {\sin \left| x \right|} \right]$$
Functions
grade 11 | Questions Set 1
Functions
Functions
grade 11 | Questions Set 2
Functions