Solved examples on Domain-II

Solved examples on Domain - I (click here )

5. $f(x) = \sqrt {{x^2} - |x| - 2}$

therefore, ${{x^2} - |x| - 2 \ge 0}$

or, $|x{|^2} - 2|x| + |x| - 2 \ge 0$ ( since ${x^2} = |x{|^2}$ )

or, $(|x| - 2)(|x| + 1) \ge 0$

Therefore, by wavy cury method ,

$|x| \le - 1$ or $|x| \ge 2$

When, $|x| \le - 1$

Since |x| is always positive. Therefore, it will never less than or equal to -1.

Therefore, $x \in \phi$ i.e. no solution is possible

When, $|x| \ge 2$

To know how to solve modulus inequality (click here)

Clearly from graph $|x| \ge 2$ , when

$x \ge 2$ or $x \le - 2$

Therefore, $x \in ( - \infty , - 2]\bigcup {[2,\infty )}$

$Domain = ( - \infty , - 2]\bigcup {[2,\infty )} = R - ( - 2,2)$

5. $f(x) = \frac{1}{{\sqrt {1 + 2\sin x} }}$

Since, ${\rm{denominator}} \ne {\rm{0}}$

And we cannot find the root of negative number.

Therefore, 1 + 2sin x > 0

sin x > -1/2

First we have to find the interval or values of 'x' for which sin x > -1/2 in a period of sin x i.e. in the interval of [0, 2pi] , then we generalise the result.

see the graph of sin x,

In a period of 2pi or in interval [0, 2pi] ,

sin x > -1/2 when $\sin x \notin \left[ {\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right]$ ( from the above graph you can see that between $\left[ {\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right]$ ,the graph is below the line y = -1/2, i.e. value of sin x is less than -1/2 between $\left[ {\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right]$ )

Generalize the result for entire number line,

To find the values of 'x' for which sin x > -1/2, we have to eliminate those value of 'x' for which $\sin x \le - 1/2$

Therefore, $Domain = R - \left[ {2k\pi + \frac{{7\pi }}{6},2k\pi + \frac{{11\pi }}{6}} \right]$

6. $f(x) = {\sin ^{ - 1}}[2{x^2} - 3]$ , where [.] denotes the greatest integer function.

To know about the Greatest integer function and its properties 1. (click here) and 2. (click here)

Since for sin inverse function is only defined for the interval [-1,1]

Therefore, $- 1 \le [2{x^2} - 3] \le 1$

or, $- 1 \le [2{x^2}] - 3 \le 1$ ( Since, [x - I] = [x] - I )

or, $2 \le [2{x^2}] \le 4$

To solve this type of inequality i.e. inequality of greatest integer function [ f(x) ] , find the points where f(x) will get the integer values.

Therefore we have to find the values where $2{x^2}$ will get the integer values

at x = 0, $2{x^2} = 0$

at $x = \pm \frac{1}{{\sqrt 2 }}$ , $2{x^2} = 1$

at, $x = \pm 1$ , $2{x^2} = 2$ ..............................................(1)

at, $x = \pm \frac{{\sqrt 3 }}{{\sqrt 2 }}$ , $2{x^2} = 3$

at $x = \pm \sqrt 2$ , $2{x^2} = 4$

at $x = \pm \frac{{\sqrt 5 }}{{\sqrt 2 }}$ , $2{x^2} = 5$ ............................................(2)

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* There is no need to find all the integer values, here we have to solve $2 \le [2{x^2}] \le 4$ , we are only concerned about where $[2{x^2}]$ get the value less than 5 (i.e. 4 because [.] only have integer values) and more than 1 i.e. 2

see the graph (BLUE GRAPH)

You can see from the above graph(BLUE GRAPH) that at x = root(5/2), value of [2x^2] = 5 and if x < root(5/2) slightly then it dropped to 4 and same for x = -root(5/2) i.e. if x > -root(5/2) slightly then it dropped to 4.

Also you can observe at x = +/- 1 , value of [2x^2] = 2 and for any x >/= 1 or any x </= -1, [2x^2] >/= 2.

Therefore, $Domain = ( - \frac{{\sqrt 5 }}{{\sqrt 2 }}, - 1]\bigcup {[1,} \frac{{\sqrt 5 }}{{\sqrt 2 }})$

OR

When   $x \in [\frac{1}{{\sqrt 2 }},1)$ or $x \in ( - 1, - \frac{1}{{\sqrt 2 }}]$ , $[2{x^2}] = 1$

and when x = 1 or x = -1 , $[2{x^2}] = 2$

Therefore, $[2{x^2}] \ge 2$ , when $x \ge 1$ or $x \le - 1$ ............... (1)

When $x = \frac{{\sqrt 5 }}{{\sqrt 2 }}$ or $x = - \frac{{\sqrt 5 }}{{\sqrt 2 }}$ , $[2{x^2}] = 5$

Therefore, $[2{x^2}] \le 4$ , when $x < \frac{{\sqrt 5 }}{{\sqrt 2 }}$ or $x > - \frac{{\sqrt 5 }}{{\sqrt 2 }}$

Because if x is slightly less than $\frac{{\sqrt 5 }}{{\sqrt 2 }}$ or slightly more than $- \frac{{\sqrt 5 }}{{\sqrt 2 }}$ the value of $[2{x^2}]$ is equal to 4.

Therefore, $- \frac{{\sqrt 5 }}{{\sqrt 2 }} < x \le - 1$ or $1 < x \le \frac{{\sqrt 5 }}{{\sqrt 2 }}$ then $2 \le [2{x^2}] \le 4$

Therefore, $Domain = ( - \frac{{\sqrt 5 }}{{\sqrt 2 }}, - 1]\bigcup {[1,} \frac{{\sqrt 5 }}{{\sqrt 2 }})$

REFERENCE:

Domain: The Possible point of  for which we can get finite value of

To know more about domain and its properties (click here)

1. If $y = \frac{{f(x)}}{{g(x)}}$ then, $g(x) \ne 0$ , i.e. ${\rm{Denominator}} \ne {\rm{0}}$

2. If $y = \sqrt {f(x)}$   then, $f(x) \ge 0$

3. If $y = {\log _{f(x)}}g(x)$ then $f(x) > 0,f(x) \ne 1$ and

4. If $y = {[f(x)]^{g(x)}}$ then $f(x) > 0,f(x) \ne 1$ and g(x) must be defined.

5. If $y = {\sin ^{ - 1}}f(x)$ or $y = {\cos ^{ - 1}}f(x)$ then $- 1 \le f(x) \le 1$

if a function  is defined in some interval or points( by definition domain), and another function  is defined in some other interval and points (by definition domain) .... then sum , subtraction , multiplication and division of the functions are defined on the interval or points which are common to both.

1) If domain of  and  are ${D_1}$ and ${D_2}$ respectively then domain of

a) $y = f(x) \pm g(x)$ , $Domain = {D_1}\bigcap {{D_2}}$

b) , $Domain = {D_1}\bigcap {{D_2}}$

c) $y = \frac{{f(x)}}{{g(x)}}$ , Since here, $g(x) \ne 0$ , therefore $Domain = {D_1}\bigcap {{D_2}} - \{ g(x) = 0\}$ , i.e. we have to eliminate those points from ${D_1}\bigcap {{D_2}}$ for which , because $denominator \ne 0$

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Solved examples on Domain-II

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