**Question:**

**Find the domain of each of the following functions given by**

**(i) $f(x)=\frac{1}{\sqrt{1-\cos x}}$**

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

(iii) $f(x)=x|x|$

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

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

**Solution:**

**(i)**

$f(x)=\frac{1}{\sqrt{1-\cos x}}$

According to the question,

We know the value of cos x lies between –1, 1,

–1 ≤ cos x ≤ 1

Multiplying by negative sign, we get

Or 1 ≥ – cos x ≥ –1

Adding 1, we get

2 ≥ 1– cos x ≥ 0 …(i)

Now,

$f(x)=\frac{1}{\sqrt{1-\cos x}}$

1– cos x ≠ 0

⇒ cos x ≠ 1

Or, x ≠ 2nπ ∀ n ∈ Z

Therefore, the domain of f = R–{2nπ:n∈Z}

(ii)

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

According to the question,

For real value of f,

x + |x| > 0

When x > 0,

x + |x| > 0⇒ x + x > 0 ⇒ 2x > 0⇒ x > 0

When x < 0,

x + |x| > 0⇒ x – x > 0 ⇒ 2x > 0⇒ x > 0

So, x > 0, to satisfy the given equation.

Therefore, the domain of f = R+

(iii)

f(x) = x|x|

According to the question,

We know x and |x| are defined for all real values.

Therefore, the domain of f = R

(iv)

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

According to the question,

For real value of

x2–1≠0

⇒ (x–1)(x + 1)≠0

⇒ x–1≠0 or x + 1≠0

⇒ x≠1 or x≠–1

Therefore, the domain of f = R–{–1, 1}

(v)

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

According to the question,

For real value of

28 – x ≠0

⇒ x≠ 28

Therefore, the domain of f = R–{28}