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

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 $f(x)$,

\\{ \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.