Question

The domain of the function \[f(x) = \dfrac{1}{{\sqrt {\left\{ {\sin x} \right\} + \left\{ {\sin (\pi + x)} \right\}} }}\] where \[\left\{ . \right\}\] denotes the fractional part is,
(A) \[[0,\pi ]\]
(B) \[(2n + 1)\pi /2,n \in Z\]
(C) \[(0,\pi )\]
(D) None of these.

Answer 1

Hint: The domain of the function means the possible input values for the given function, that is, all values of \[x\] for which \[f(x)\] stays defined or gives a valid answer. For finding the domain of a function one should remember that the value of the denominator is not equal to \[0\] and if a square root function is used then the denominator should be greater than zero. Let's find the domain and solve \[f(x)\] step by step.

Complete answer:
We are given the function,
 \[f(x) = \dfrac{1}{{\sqrt {\left\{ {\sin x} \right\} + \left\{ {\sin (\pi + x)} \right\}} }}\]
Here the numerator is one so we have nothing to do with, where as the denominator have square function with \[\sin \] which is a fractional part.
In denominator we have \[\left\{ {\sin (\pi + x)} \right\}\], which is in the form \[\sin (a + b)\] so let apply the formula: \[\sin (a + b) = \sin a\cos b + \cos a\sin b\]
\[\sin (\pi + x) = \sin \pi \cos x + \cos \pi \sin x\]
The value of \[\sin \pi = 0\] and \[\cos \pi = - 1\], substituting this,
\[ = 0\cos x + ( - 1)\sin x\]
Any number multiplying by \[0\] is \[0\].
\[ = - \sin x\].
Now, \[f(x) = \dfrac{1}{{\sqrt {\left\{ {\sin x} \right\} + \left\{ { - \sin x} \right\}} }}\]
We know that the denominator \[ \ne 0\] and \[\sqrt {({\text{expression}})} > 0\],
\[f(x) = \{ \sin x\} + \{ - \sin x\} \ne 0\]
\[\{ - x\} = 1 - \{ x\} ,x \notin {\text{I}}\] and \[\{ x\} = [0,1)\]
By applying this in \[f(x)\],
\[f(x)\left[
  \{ \sin x\} + 1 - \{ \sin x\} {\text{ if }}\sin x{\text{ is not an integer}} \\
  0{\text{ }}\sin x{\text{ is integer}} \\
  \right.\]
For \[f(x)\]to be defined \[\{ \sin x\} + 1 - \{ \sin x\} \ne 0\],
That is, \[\sin x \ne 0\] and \[1 - \sin x \ne 0 \Rightarrow \sin x \ne \pm 1\]
Now by applying general solution for the trigonometric equation,
\[\sin \theta = 0\] then \[\theta = n\pi \] and \[\sin \theta = \pm 1\] then \[\theta = \dfrac{\pi }{2}\].
Now by using this, \[\sin x \ne 0 \Rightarrow x = n\pi \] and \[\sin x \ne \pm 1 \Rightarrow x = \dfrac{\pi }{2}\]
We know that the value of \[\sin \pi = 0\] similarly the value of \[\sin 2\pi ,\sin 3\pi ,... = 0\], simply we can say that \[\sin n\pi = 0\]
\[\{ \sin x\} + 1 - \{ \sin x\} \ne 0\]
\[\sin x \ne \] Integer
\[\sin x \ne \pm 1,0\]
We already found that \[\sin n\pi = 0\], thus excluding it,
\[x \ne n\dfrac{\pi }{2},n \in {\text{I}}\], here the value of \[n\] is an integer where the value can be both positive and negative.
The \[x \ne n\dfrac{\pi }{2}\] thus subtract \[\dfrac{{n\pi }}{2}\] from the range to get the domain of the function.
Hence, the domain is \[R - \left\{ {\dfrac{{n\pi }}{2},n \in {\text{I}}} \right\}\], where I is the integer, that is, \[x \in R - \left\{ {\dfrac{{n\pi }}{2},n \in {\text{I}}} \right\}\].
Therefore the provided options do not match with the domain we got.
Hence option (D) None of these, is the correct answer.

Note:
 If the square root function is in the numerator then the numerator should be greater than or equal to zero, if \[\sqrt {({\text{expression}})} \] is in numerator then, \[\sqrt {({\text{expression}})} \geqslant 0\].
To find the domain in a denominator the student can simply omit the functions which gives \[0\] because the denominator should not be equal to \[0\]. In domain values the brackets play an important rule, the square bracket \[\left[ {\text{ }} \right]\] is used to indicate the endpoint included in the interval. For example \[\left[ {2,4} \right]\] means the value start from \[2\] and ends in \[4\], and if the round bracket \[\left( {\text{ }} \right)\] indicates that the point is not included. For example, \[{\text{(2,4)}}\] means it includes a number greater than \[2\] and less than \[4\] but does not include \[2\] and 4.