(i) f (x) = | x – 3 | is defined for all x ∈ R, so domain of f (x) = R. 

 Now | x – 3 | ≥ 0 for all x ∈ R ⇒ 0 ≤ | x – 3 | < ∞ for all x ∈ R 

⇒ f (x) ∈ [0, ∞) for all x ∈ R ⇒ Range of f (x) = | x – 3 | is [0, ∞)

(ii) Let y = f (x) = \(\sqrt{x-5}\)

f (x) is defined for x – 5 ≥ 0 ⇒ x ≥ 5 ⇒ x ∈ [5, ∞)

Now y = \(\sqrt{x-5}\)  

⇒ x – 5 = y² ⇒ x = y² + 5 

⇒ x > 0 for all value of y 

For x to be real and x ∈ [5, ∞), y ∈ [0, ∞). 

∴ Range of f is [0, ∞ ) 

(iii) Let y = f (x) = \(\sqrt{3x^2-4x+5}\) 

f (x) is defined if 3x² – 4x + 5 ≥ 0 

⇒ 3\(\big(x^2-\frac{4}{3}x+\frac{5}{3}\big)\)≥ 0

⇒ \(3\big(x^2-\frac{4}{3}x+\frac{4}{9}-\frac{4}{9}+\frac{5}{3}\big) \geq 0\)

⇒ 3 \(\big(\big(x-\frac{2}{3}\big)^2 +\frac{11}{9}\big) \geq 0\) , which is true for all real numbers, 

i.e., Domain of f is (– ∞, ∞)

y = \(\sqrt{3x^2-4x+5}\)  

⇒ 3x² - 4x +5 =y² 

⇒ 3x² - 4x +(5-y²) = 0

⇒ x = \(\frac{4 \,\pm \sqrt{(-4)^2 -4 \times 3 \times (5-y^2)}}{2 \times 3}\) 

For x to be real, (– 4)² – 4 × 3 × (5 – y²) ≥ 0 

⇒ 16 – 60 + 12y² ≥ 0 

⇒ – 44 + 12y² ≥ 0 

⇒ 12y² ≥ 44 

⇒ \(y^2 \geq \frac{11}{3} \implies y \geq \sqrt{\frac{11}{3}}\) 

∴ Range of y = [ \(\sqrt{\frac{11}{3}}\) ,  ∞ )

(iv) Let y = f(x) = \(\frac{x}{1+x^2}\) 

The given function f(x) = \(\frac{x}{1+x^2}\) is defined for all real numbers, so domain of f is R.

y = \(\frac{x}{1+x^2}\) 

⇒ x²y -x+y = 0  ⇒ x = \(\frac{-(-1)\, \pm \sqrt{1-4y^2}}{2y}\)

Now, \(\frac{1\, \pm \sqrt{1-4y^2}}{2y}\) will be a real number if and only if 

1 – 4y² ≥ 0 and y ≠ 0 

⇒ 4y² – 1 ≤ 0 and y ≠ 0 

⇒ \(\big( y^2 -\frac{1}{4}\big) \leq 0\) and y ≠ 0

⇒ \(\big(y-\frac{1}{2}\big)\big(y+\frac{1}{2}\big) \leq 0\) and y ≠ 0

⇒ \(-\frac{1}{2} \) \(\leq\) \(\frac{1}{2}\)  and  y ≠ 0

⇒ y ∈ [ \(-\frac{1}{2} \), 0 ) \(\cup\) ( 0, \(\frac{1}{2}\) ]

For x = 0, y = 0 

∴ range is [ \(-\frac{1}{2} \), \(\frac{1}{2}\) ]