# CLASS-11RELATION & FUNCTION - TIPS FOR FINDING DOMAIN OF A FUNCTION

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

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.)