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Find the domain of each of the following functions given by (i) f(x)=\frac{1}{\sqrt{1-\cos x}} (ii) f(x)=\frac{1}{\sqrt{x+|x|}}
CBSE | Class 11 | Maths | NCERT Exemplar | Relations and Functions
Find the domain of each of the following functions given by (i) f(x)=\frac{1}{\sqrt{1-\cos x}} (ii) f(x)=\frac{1}{\sqrt{x+|x|}}

Solution:

(i)

f(x)=11–cosx
\mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{1-\cos \mathrm{x}}}

As per the question,

It is known that the value of \cos x lies between -1,1

-1 \leq \cos x \leq 1

Multiply by negative sign, we have

Or 1 \geq-\cos x \geq-1

When we add 1, we have

2 \geq 1-\cos x \geq 0 \ldots \dots \dots (i)

So now,

\begin{array}{l} \mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{1-\cos \mathrm{x}}} \\ 1-\cos \mathrm{x} \neq 0 \\ \Rightarrow \cos \mathrm{x} \neq 1 \end{array}

Or, x \neq 2 n \pi \forall n \in Z

As a result, the domain of \mathrm{f}=\mathrm{R}-\{2 \mathrm{n} \pi: \mathrm{n} \in \mathrm{Z}\}

(ii)

f(x)=\frac{1}{\sqrt{x+|x|}}

As per the question,

For real value of f,

x+|x|>0

When x>0

x+|x|>0 \Rightarrow x+x>0 \Rightarrow 2 x>0 \Rightarrow x>0

When x<0

x+|x|>0 \Rightarrow x-x>0 \Rightarrow 2 x>0 \Rightarrow x>0

Therefore, x>0, to satisfy the provided equation.

As a result, the domain of \mathrm{f}=\mathrm{R}^{+}

i

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