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

iii) $f(x) =x\left | x \right |$

iv) $f(x) =\frac{x^3-x+3}{x^2-1}$

v) $f(x) =\frac{3x}{2x-8}$

Answers (1)

i) Given data: $f(x) = \frac{1}{\sqrt{1-\cos x}}$

Now that we know that,

$-1 \leq cosx \leq 1$

$1 \geq -cosx \geq -1$

$1 + 1 \geq 1 - cos x \geq -1+1$

$2 \geq 1 - cos x \geq 0$

$0 \leq 1 - cos x \leq 2$

Now, for the real value of the domain,

$1 - cos x\neq0,cosx \neq 1$

But, $x \neq 0\: \: 2n\pi \: \: \forall n \in Z$

Thus, domain of $f = R - \left \{2n\pi, n \in Z\right \}$

ii)Given data: $f(x) = \frac{1}{\sqrt{x+\left | x \right |}}$

Now, $x + \left |x \right | = x+x = 2x ...... if \: \: x \geq0$

& $x + \left |x \right | = x - x = 0 ...... if \: \: x < 0$

Now, $x<0$ is not defined so far, hence,

The domain = $R^+$

iii)Given data: $f(x) =x\left | x \right |$

For all x $\in$ R, f(x) is defined

Thus, the domain of f = R.

iv)Given data: $f(x) =\frac{x^3-x+3}{x^2-1}$

Here, only if $x^2 - 1 \neq 0$ , f(x) is defined,

$(x-1)(x+1) \neq 0,$

$Thus, x \neq 1$

$and\: \: x \neq -1$

Therefore, domain of $f = R - \left \{-1,1\right \}$

v)Given data: $f(x) =\frac{3x}{2x-8}$

F(x) is only defined at $2x - 8 \neq 0, x \neq 4$ ,

Thus, the domain = $R - \left \{4\right \}$ .

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infoexpert21

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