Find the domain of f x = 1 x 2 - x - 6 , where · denotes the greatest integer function.

Now, f x = 1 x 2 - x - 6 x 2 - x - 6 > 0 ⇒ x 2 - 3 x + 2 x - 6 > 0 ⇒ x x - 3 + 2 x - 3 > 0 ⇒ x - 3 x + 2 > 0 ⇒ x ∈ - ∞ , - 2 ∪ 3 , ∞ So, domain of x ∈ - ∞ , - 2 ∪ [ 4 , ∞ )

Find the domain of f x = 1 x 2 - x - 6 , where · denotes the greatest integer function.

Now, f x = 1 x 2 - x - 6 x 2 - x - 6 > 0 ⇒ x 2 - 3 x + 2 x - 6 > 0 ⇒ x x - 3 + 2 x - 3 > 0 ⇒ x - 3 x + 2 > 0 ⇒ x ∈ - ∞ , - 2 ∪ 3 , ∞ So, domain of x ∈ - ∞ , - 2 ∪ [ 4 , ∞ )

Find the domain of f x = log 10 x - 4 + 6 - x .

f x = log 10 x - 4 + 6 - x The logarithmic function is defined as, x - 4 + 6 - x > 0 Case I : x - 4 is defined when x - 4 ≥ 0 ⇒ x ≥ 4 Case II : 6 - x is defined when 6 - x ≥ 0 ⇒ x ≤ 6 x ∈ 4 , 6

Find the domain of f x = sin - 1 3 - 2 x 5 + 3 - x .

Given that, f x = sin - 1 3 - 2 x 5 + 3 - x The inverse of sine is defined when, - 1 ≤ 3 - 2 x 5 ≤ 1 - 5 ≤ 3 - 2 x ≤ 5 - 8 ≤ - 2 x ≤ 2 - 2 ≤ 2 x ≤ 8 - 1 ≤ x ≤ 4 x ∈ - 1 , 4 and 3 - x is defined when, 3 - x ≥ 0 x - 3 ≤ 0 ⇒ x ∈ ( - ∞ , 3 ] So the domain of f x is x ∈ - 1 , 3 .

Find the domain of f x = 2 - x + 1 + x .

Given that, 2 - x is defined when 2 - x ≥ 0 x - 2 ≤ 0 x ≤ 2 x ∈ - 2 , 2 1 + x is defined when 1 + x ≥ 0 which is true ∀ x ∈ R So, the domain of f x is x ∈ - 2 , 2

Find the domain of f x = x 2 - x - 2 .

Given that f x = x 2 - x - 2 x 2 - x - 2 is defined when x 2 - x - 2 ≥ 0 Case I: When x ≥ 0 x 2 - x - 2 ≥ 0 x 2 - 2 x + x - 2 ≥ 0 x x - 2 + 1 x - 2 ≥ 0 x + 1 x - 2 ≥ 0 ⇒ x ∈ ( - ∞ , - 1 ] ∪ [ 2 , ∞ ) But x ≥ 0 So x ∈ [ 2 , ∞ ) Case II : When x < 0 x 2 + x - 2 ≥ 0 x + 2 x - 1 ≥ 0 x ∈ ( - ∞ , - 2 ] ∪ [ 1 , ∞ ) But x < 0 So x ∈ ( - ∞ , - 2 ] So the domain of f x is x ∈ ( - ∞ , - 2 ] ∪ [ 2 , ∞ ) .

Find the domain of f x = 1 - 1 - 1 - x 2

Given that f x = 1 - 1 - 1 - x 2 1 - 1 - 1 - x 2 is defined when 1 - 1 - 1 - x 2 ≥ 0 ⇒ 1 - 1 - x 2 ≤ 1 Squaring both sides, ⇒ 1 - 1 - x 2 ≤ 1 ⇒ - 1 - x 2 ≤ 0 ⇒ 1 - x 2 ≥ 0 ⇒ x + 1 x - 1 ≤ 0 ⇒ x ∈ - 1 , 1 Now 1 - 1 - x 2 is defined when 1 - 1 - x 2 ≥ 0 ⇒ 1 - x 2 ≤ 1 Squaring both sides, 1 - x 2 ≤ 1 ⇒ x 2 ≥ 0 ⇒ x ∈ R 1 - x 2 is defined when 1 - x 2 ≥ 0 ⇒ 1 - x 1 + x ≥ 0 ⇒ x ∈ - 1 , 1 So the domain of f x is - 1 , 1 .

Find the domain of f x = log 0 . 3 3 x - x 2 x - 1 .

f x is defined when log 0 . 3 3 x - x 2 x - 1 ≥ 0 Using the function of change of base of logarithm, log e 3 x - x 2 x - 1 log e 0 . 3 ≥ 0 Here log e 0 . 3 < 0 So to make log e 3 x - x 2 x - 1 log e 0 . 3 ≥ 0 log e 3 x - x 2 x - 1 ≤ 0 Since, log e x ≤ 0 when x ∈ ( 0 , 1 ] Therefore, 0 < 3 x - x 2 x - 1 ≤ 1 Case I: 3 x - x 2 x - 1 > 0 x 2 - 3 x x - 1 < 0 On solving, x ∈ - ∞ , 0 ∪ 1 , 3 . . . i Case II: 3 x - x 2 x - 1 ≤ 1 x 2 - 2 x - 1 x - 1 ≥ 0 On solving x ∈ [ 1 - 2 , 1 ) ∪ [ 1 + 2 , ∞ ) . . . ii From equation i and ii , x ∈ [ 1 - 2 , 0 ) ∪ [ 1 + 2 , 3 )