Tips For Finding The Domain of a Function –
1. Algebraic Functions –
(i) Denominator should be non-zero.
(ii) Expression under the even root should be non-negative.
2. Trigonometric Functions -
(i) sin x and cos x are defined for all real values of x except x = (2n + 1)
π
------, where n ∈ I
2
(ii) cot x and cosec x are defined for all real values of x except x = nπ, where n ∈ I.
3. Logarithmic Functions –
logₑ a is defined when a > 0, e > 0 and e ≠ 1.
4. Exponential Functions –
a˟ is defined for all real values of x, where a > 0.
Example.1) If f is a real function, find the domain of
(i) f(x) = √(a² - x²)
Ans.) (i) Since f is real, a²- x² ≥ 0
=> (a + x) (a – x) ≥ 0
=> - a ≤ x ≤ a
So, Domain of f(x) = √(a²- x²) where {x ǀ - a ≤ x ≤ a} (Ans.)
1
(ii) f(x) = ----------
(3x + 2)
Ans.) Since f is real, x can take all real values except the value for 3x + 2 = 0
x ≠ - 2/3
1 2
Domain of f(x) = -------- is R – {- ------} (Ans.)
(3x + 2) 3
1
(iii) f(x) = -----------
Log ǀ x ǀ
Ans.) Since f is real, x can take all real values except 0, 1, - 1.
1
So, domain of f(x) = ------- is R – {0, 1, - 1}
log ǀ x ǀ
(iv) f(x) = 10⁻˟
1
Ans.) f(x) = -------- , so Domain of f is R (Ans.)
10˟
Example.2) Find the domain of the following functions.
(i) f(x) = log₃₊ₓ (x²- 1)
Ans.)
f(x) = log₃₊ₓ (x²- 1)
f(x) is defined when x²- 1 > 0
=> x² > 1 and 3 + x > 0
=> x < - 1 or x > 1 and x > - 3 and x ≠ - 2, since base 3 + x ≠ 1.
So, Domain is (-3, - 2) ∪ (- 2, - 1) ∪ (1, ∞).
[Recall that in the definition of a logarithm, logₐ n = x, the a, n are positive real numbers, x is any real number and a ≠ 1.] (Ans.)
sin⁻¹ (3 – x)
(ii) f(x) = -----------------
ln (ǀ x ǀ - 2)
Ans.) Let g (x) = sin⁻¹ (3 – x)
=> - 1 ≤ 3 – x ≤ 1
So, Domain of g(x) is [2, 4].
Let h(x) = log [ ǀ x ǀ - 2]
=> ǀ x ǀ - 2 > 0
=> ǀ x ǀ > 2
=> x < - 2 or x > 2
=> Domain h(x) = ( - ∞, - 2) ∪ (2, ∞)
f f(x)
Now, we know that (-----) x = ------ ⩝ x ∈ D₁ ∩ D₂ - {x ∈ R : g(x) = 0}
g g(x)
so, Domain of f(x) = (2, 4] – {3} = (3, 4] (Ans.)
x
(iii) f(x) = sin⁻¹ log₂ ------
3
x
Ans.) Let f(x) = sin⁻¹ z where, z = log₂ ------, since sin⁻¹ z is defined
3
only for – 1 ≤ z ≤ 1, therefore, the domain is given by
x x x
– 1 ≤ log₂ ----- ≤ 1 and ------ > 0 => 2⁻¹ ≤ ------ ≤ 2¹
3 3 3
1 x 3
=> ------ ≤ ------ ≤ 2 and x > 0 = > ------ ≤ x ≤ 6 and 6 > 0
2 3 2
3
=> Domain is [-----, 6] (since logₐ x = k => x = aᵏ) (Ans.)
2
log₂ (x + 3)
(iv) f(x) = ---------------
(x²+ 3x + 2)
Ans.) For the function to be defined x + 3 > 0 and x²+ 3x + 2 ≠ 0
=> x > - 3 and (x + 1) (x + 2) ≠ 0, i.e., x ≠ - 1, - 2
Domain is (- 3, ∞) – {- 1, - 2} (Ans.)
log₂ (x + 3)
(v) f(x) = --------------
√(9 - x²)
Ans.) For f(x) to be defined –
(9 - x²) > 0
=> -3 < x < 3 …………..(1)
- 1 ≤ x – 3 ≤ 1 [sin θ is defined for - 1 ≤ θ ≤ 1]
=> 2 ≤ x ≤ 4 ………………..(2)
From (1) & (2), 2 ≤ x < 3, i.e., the domain is [2, 3) (Ans.)
2 + x
(vi) f(x) = logₑ ---------
2 – x
(2 + x)
Ans.) f(x) will be defined when ---------- > 0
(2 – x)
i.e., when 2 + x > 0 and 2 – x > 0 or 2 + x < 0 and 2 – x < 0
i.e., when x > - 2 and x < 2 or x < - 2 and x > 2
i.e., when – 2 < x < 2 or x < - 2 and x > 2.
As the second option is absurd, the domain is – 2 < x < 2. (Ans.)