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 = 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 )

Find the domain of f x = x 12 - x 9 + x 4 - x + 1 .

Given that f x = x 12 - x 9 + x 4 - x + 1 Case I: x ∈ [ 0 , 1 ] We know that: 1 - x ≥ 0 , x 4 - x 9 ≥ 0 , and x 12 ≥ 0 So, x 12 - x 9 - x 4 - x + 1 ≥ 0 for x ∈ 0 , 1 Case II: x ∈ [ 1 , ∞ ) We know that: x 12 > x 9 ⇒ x 12 - x 9 ≥ 0 x 4 > x ⇒ x 4 - x ≥ 0 ∴ x 12 - x 9 + x 4 - x + 1 ≥ 0 ∀ x ∈ [ 1 , ∞ ) Case III: x ∈ - ∞ , 0 x 12 > 0 ∀ x ∈ - ∞ , 0 - x 9 > 0 ∀ x ∈ - ∞ , 0 x 4 > 0 ∀ x ∈ - ∞ , 0 - x > 0 ∀ x ∈ - ∞ , 0 1 > 0 Combining the above cases, x 12 - x 9 + x 4 - x + 1 > 0 ∀ x ∈ - ∞ , 0 So the domain of f x is x ∈ - ∞ , ∞

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

We know that f x = log 2 x 3 cos - 1 2 x - 1 Case I: Base of log is always positive but not equal to 1 . So, 2 x > 0 and 2 x ≠ 1 x > 0 and x ≠ 1 2 . . . i Case II: cos - 1 2 x - 1 ≠ 0 (denominator term) 2 x - 1 ≠ 1 As per the definition of inverse of cos function, - 1 ≤ 2 x - 1 < 1 0 ≤ 2 x < 2 , and 0 ≤ x < 1 So, domain of f x is x ∈ 0 , 1 - 1 2 .

Find the domain of f x = log 0 . 3 x - 1 x 2 - 2 x - 8 .

We know that f x = log 0 . 3 x - 1 x 2 - 2 x - 8 log 0 . 3 x - 1 x 2 - 2 x - 8 ≥ 0 (only defined when values under root are non-negative) Applying the change of base of logarithm, log e x - 1 log e 0 . 3 x 2 - 2 x - 8 ≥ 0 Here log e 0 . 3 < 0 log e x - 1 x 2 - 2 x - 8 ≤ 0 Case I: log e x - 1 ≥ 0 and x 2 - 2 x - 8 < 0 x - 1 ≥ 1 ⇒ x ≥ 2 … i x 2 - 2 x - 8 < 0 On solving x ∈ - 2 , 4 … ii From equations i and ii x ∈ [ 2 , 4 ) … i i i Case II: log e x - 1 ≤ 0 and x 2 - 2 x - 8 > 0 For definition of logarithm, x - 1 > 0 x > 1 … iv log e x - 1 ≤ 0 1 < x ≤ 2 … v x 2 - 2 x - 8 > 0 On solving x ∈ ( - ∞ , - 2 ) ∪ 4 , ∞ … vi From equations ( v ) and ( vi ) , x ∈ ϕ … vii On combining equations i i i and vii , x ∈ [ 2 , 4 )

Find the domain of f x = log 10 log 2 log 2 π tan - 1 x - 1 .

Given, f x = log 10 log 2 log 2 π tan - 1 x - 1 f x is defined only when, log 2 log 2 π tan - 1 x - 1 > 0 ⇒ log 2 π tan - 1 x - 1 > 2 0 ⇒ log 2 π 1 tan - 1 x > 1 ⇒ 1 tan - 1 x < 2 π ∴ tan - 1 x > π 2 . . . 1 1 tan - 1 x > 0 ( log a x is only defined for values x > 0 ) ⇒ tan - 1 x < ∞ (reciprocal of terms) . . . 2 From equations 1 and 2 we get, π 2 < tan - 1 x < ∞ From the above graph we can say, tan - 1 x ∈ - π 2 , π 2 Hence, tan - 1 x ≯ π 2 So x ∈ ϕ (has a empty domain)

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

Given, f x = cos - 1 log x x x Let y = cos - 1 log x x x Now, log x x x is defined when x > 0 and x ≠ 1 x ≥ 2 And x x > 0 ⇒ x > 0 ⇒ x ∈ [ 2 , ∞ ) Now, cos - 1 log x x x is defined when - 1 ≤ log x x x ≤ 1 x x for x ∈ [ 2 , ∞ ) is always 1 So, for x ∈ [ 2 , ∞ ) log x x x = log x 1 = 0 x ∈ [ 2 , ∞ ) So, the domain of f x is [ 2 , ∞ ) .

Find the domain of f x = C 2 x 2 + 3 x 2 + 4 x .

Here, f x = C 2 x 2 + 3 x 2 + 4 x So, C 2 x 2 + 3 x 2 + 4 x ≥ 0 Now, to define C 2 x 2 + 3 x 2 + 4 x x ∈ natural numbers . . . i Case I: x 2 + 4 x > 0 x ( x + 4 ) > 0 On solving x ∈ - ∞ , - 4 ∪ 0 , ∞ … ii Case II: 2 x 2 + 3 ≥ 0 On solving x ∈ R … iii Case III: x 2 + 4 x ≥ 2 x 2 + 3 x 2 - 4 x + 3 ≤ 0 x - 1 x - 3 ≤ 0 On solving x ∈ 1 , 3 … iv From equation i , ii , iii & iv x ∈ 1 , 2 , 3