Find the domain of the following functions: 9 - x 2

To find :- Domain 9 - x 2 Let f ( x ) = 9 - x 2 As, we know that square root function i.e. p ( x ) is defined when p ( x ) ≥ 0 ∴ f ( x ) is defined when 9 - x 2 ≥ 0 ∴ x 2 - 9 ≤ 0 ⇒ ( x - 3 ) ( x + 3 ) ≤ 0 ∴ By sign-scheme method. ∴ x ∈ [ - 3 , 3 ] . ∴ Domain of f ( x ) is [ - 3 , 3 ] .

Find the domain of the following functions: 9 - x 2

To find :- Domain 9 - x 2 Let f ( x ) = 9 - x 2 As, we know that square root function i.e. p ( x ) is defined when p ( x ) ≥ 0 ∴ f ( x ) is defined when 9 - x 2 ≥ 0 ∴ x 2 - 9 ≤ 0 ⇒ ( x - 3 ) ( x + 3 ) ≤ 0 ∴ By sign-scheme method. ∴ x ∈ [ - 3 , 3 ] . ∴ Domain of f ( x ) is [ - 3 , 3 ] .

Find the domain of the following functions: x 1 + x 2

To find :- Domain of x 1 + x 2 Let f ( x ) = x 1 + x 2 f ( x ) is rational function i.e. of the form p ( x ) q ( x ) and Rational function is defined for all real values except when q ( x ) = 0 So, f ( x ) is defined for all values of x except 1 + x 2 = 0 But as, x 2 ≥ 0 ∴ 1 + x 2 ≥ 1 ∴ f ( x ) is defined for all real values of x . ∴ Domain of f ( x ) = R

Find the domain of the following functions: 1 x 2 - 1

To find:- Domain of = 1 x 2 - 1 Let f x = 1 x 2 - 1 As, Domain of a function f x is defined at the all values of x for which f x is defined. Here, f x is defined when x 2 - 1 ≠ 0 ⇒ x - 1 x + 1 ≠ 0 i.e. x ≠ - 1 , 1 ∴ Domain of f x = R - - 1 , 1

Find the domain of the following functions: sin x 1 + tan x

To find:- Domain of sin x 1 + tan x Let f x = sin x 1 + tan x As, Domain of sin x = R Domain of tan x = R - 2 x + 1 π 2 ; n ∈ I Also, f x is defined when 1 + tan x ≠ 0 i.e. tan x ≠ - 1 ∴ x ≠ m π + 3 π 4 ; m ∈ I ∴ Domain of f x is R - 2 n + 1 π 2 - m π + 3 π 4 Where n , m ∈ I

Find the domain of the following functions: 1 x + x

To find:- Domain of 1 x + x As, x = x ; x ≥ 0 = - x ; x < 0 So, 1 x + x is defined when x + x ≠ 0 So, x + x = 0 when x < 0 & x = 0 ∴ Domains of 1 x + x = R - - ∞ , 0 or R +

Find the domain of the following functions: log 12 x 2 - x

Let f x log 10 12 x 2 - x So, f x is defined when log 10 12 x 2 - x ≥ 0 , 12 x 2 - x > 0 & x 2 - x ≠ 0 12 x 2 - x ≥ 1 , 12 x x - 1 > 0 & x x - 1 ≠ 0 Now 12 x 2 - x - 1 ≥ 0 ⇒ 12 - x 2 + x x x - 1 ≥ 0 ⇒ x 2 - x - 12 x x - 1 ≤ 0 ⇒ x 2 - 4 x + 3 x - 12 x x - 1 ≤ 0 ⇒ x x - 4 + 3 x - 4 x x - 1 ≤ 0 ⇒ x + 3 x - 4 x x - 1 ≤ 0 By sign-scheme Method ∴ x ∈ [ - 3 , 0 ) ∪ ( 1 , 4 ] . . . . . . 1 And 12 x x - 1 > 0 By sign-scheme Method ∴ x ∈ - ∞ , 0 ∪ 1 , ∞ . . . . . . 2 And x x - 1 ≠ 0 ∴ x ≠ 0 , 1 . . . . . . 3 From 1 , 2 , 3 Domain of f x = [ - 3 , 0 ) ∪ ( 1 , 4 ]