Domain of a Function MCQ Quiz - Objective Question with Answer for Domain of a Function - Download Free PDF
Last updated on Apr 7, 2025
Latest Domain of a Function MCQ Objective Questions
Domain of a Function Question 1:
\(\mathrm{f}(x)=\cos x-1+\frac{x^{2}}{2!}, x \in \mathbb{R}\) Then f(x) is
Answer (Detailed Solution Below)
Domain of a Function Question 1 Detailed Solution
Calculation:
Give, f(x) = \(\cos x-1+\frac{x^{2}}{2!}, x ∈ \mathbb{R}\)
⇒ f '(x) = - sin x + x
Now, ∀ x ∈ ℝ, x > sin x
⇒ x - sin x > 0
⇒ f '(x) > 0
⇒ f(x) is an increasing function.
∴ f(x) is an increasing function.
The correct answer is Option 2.
Domain of a Function Question 2:
If the domain of the function \(f(x)=\cos ^{-1}\left(\frac{2-|x|}{4}\right)+\left(\log _e(3-x)\right)^{-1}\) is [-α, β)-{y}, then α + β + γ is equal to:
Answer (Detailed Solution Below)
Domain of a Function Question 2 Detailed Solution
Calculation
Given
\(f(x)=\cos ^{-1}\left(\frac{2-|x|}{4}\right)+\left(\log _e(3-x)\right)^{-1}\)
⇒ \(-1 \leq\left|\frac{2-|x|}{4}\right| \leq 1\)
\( \Rightarrow\left|\frac{2-|x|}{4}\right| \leq 1\)
⇒ –4 < 2 – |x| < 4
⇒ –6 < – |x| < 2
⇒ –2 < |x| < 6
⇒ |x| < 6
⇒ x ∈ [–6, 6] …(1)
Now, 3 – x ≠ 1
And x ≠ 2 …(2)
And 3 – x > 0
⇒ x < 3 …(3)
From (1), (2) and (3)
⇒ x ∈ [–6, 3) – {2}
⇒ α = 6
⇒ β = 3
⇒ γ = 2
⇒ α + β + γ = 11
Hence option(3) is correct
Domain of a Function Question 3:
If the domain of the function log5(18x - x2 - 77) is (α, β) and the domain of the function \(\rm \log _{(x-1)}\left(\frac{2 x^{2}+3 x-2}{x^{2}-3 x-4}\right)\) is (γ, δ), then α2 + β2 + γ2 is equal to :
Answer (Detailed Solution Below)
Domain of a Function Question 3 Detailed Solution
f1(x) = log5(18x – x2 – 77)
∴ 18x – x2 – 77 > 0
x2 – 18x + 77 < 0
x ∈ (7, 11) α = 7, β = 11
x > 1 , x ≠ 2 ,
∴ x ∈ (4, ∞)
∴ γ = 4
∴ α2 + β2 + γ2 = 49 + 121 + 16
= 186
Domain of a Function Question 4:
Let [x] denote the greatest integer less than or equal to x. Then domain of f(x) = sec–1 (2[x] + 1) is :
Answer (Detailed Solution Below)
Domain of a Function Question 4 Detailed Solution
Calculation
f(x) = sec–1 (2[x] + 1)
2[x] + 1 ≤ –1 or 2 [x] + 1 ≥ 1
⇒ [x] ≤ -1 ∪ [x] ≥ 0
⇒ x ∈ (-∞, 0) ∪ x ∈ [0, ∞)
⇒ x ∈ (-∞, ∞)
Hence option 2 is correct
Domain of a Function Question 5:
Let f(x) = loge x and g(x) = \(\frac{x^{4}-2 x^{3}+3 x^{2}-2 x+2}{2 x^{2}-2 x+1}\) Then the domain of fog is
Answer (Detailed Solution Below)
Domain of a Function Question 5 Detailed Solution
Calculation
f(x) = lnx
\(g(x)=\frac{x^{4}-2 x^{3}+3 x^{2}-2 x+2}{2 x^{2}-2 x+1}\)
Dg ∈ R
Df ∈ (0, ∞)
For Dfog ⇒ g(x) > 0
\(\frac{x^{4}-2 x^{3}+3 x^{2}-2 x+2}{2 x^{2}-2 x+1}>0\)
⇒ x4 – 2x3 + 3x2 – 2x + 2 > 0
Clearly x < 0 satisfies which are included in option (1) only.
Hence option 1 is correct
Top Domain of a Function MCQ Objective Questions
What is the domain of the function f(x) = sin-1 (x + 1) ?
Answer (Detailed Solution Below)
Domain of a Function Question 6 Detailed Solution
Download Solution PDFConcept:
Domin of sin-1 x is [-1, 1]
Adding or subtracting the same quantity from both sides of an inequality leaves the inequality symbol unchanged.
Calculation:
Given: f(x) = sin-1 (x + 1)
As we know, domin of sin1 x is [-1, 1]
Therefore, -1 ≤ (x + 1) ≤ 1
subtracting 1 in above inequality,
⇒ -1 - 1 ≤ x + 1 - 1 ≤ 1 - 1
⇒ -2 ≤ x ≤ 0
∴ Domin of sin-1 (x + 1) is [-2, 0]
Mistake Points[-2, 0] is different from [-2, 0). '[' and ']' indicates that the end number (2 and 0) is also included. '(' and ')' indicates that 2 and 0 are not taken into consideration.
Find domain of the function \({\rm{f}}\left( {\rm{x}} \right) = {\rm{\;}}\frac{4}{{\sqrt {{\rm{x}} - 2} }}\).
Answer (Detailed Solution Below)
Domain of a Function Question 7 Detailed Solution
Download Solution PDFConcept:
1. Domain of a functions:
- The domain of a function is the set of all possible values of the independent variable. That is all the possible inputs for a function.
Calculation:
Observe that the given function is in the form of numerator and denominator. The function will be well defined for all non zero values of the denominator.
Therefore, \({\rm{x}} - 2{\rm{\;}} \ne 0\) that implies that \({\rm{x\;}} \ne 2\).
Similarly square root function is well defined for all non-negative values.
Therefore, \({\rm{x}} - 2 > 0\) that implies \({\rm{x}} > 2.\)
Thus, domain of the given function is \(\left( {2,{\rm{\;}}\infty } \right).\)
What is the domain and range of the function (x) = \(\sqrt {(16 - x^2)}\)?
Answer (Detailed Solution Below)
Domain of a Function Question 8 Detailed Solution
Download Solution PDFConcept:
We know that, the domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes.
Calculations:
Given function is f(x) = \(\sqrt {(16 - x^2)}\)
The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes.
For domain, \(\rm f(x) \ge 0\)
⇒\( \rm16 - x^2\ge 0 \)
⇒\( \rm16 \ge x^2 \)
⇒\( \rm x^2\le 16 \)
⇒ -4 ≤ x ≤ 4
Hence, domain of f(x) = [-4, 4]
For Range,
f(x) is maximum at x = 0 i.e. f(0) = 4
f(x) is minimum at x = 4 i.e. f(4) = 0
Hence, Range of f(x) = [0, 4]
Hence, the domain and range of the function (x) = of the function f(x) = \(\sqrt {(16 - x^2)}\) are [-4, 4], [0, 4].
The domain of cos-1 (2x + 1) is:
Answer (Detailed Solution Below)
Domain of a Function Question 9 Detailed Solution
Download Solution PDFConcept:
- The domain of a function f(x) is the set of values of x for which the function is defined.
- The value of cos θ always lies in the interval [-1, 1].
- cos-1 (cos θ) = θ.
- cos (cos-1 x) = x.
Calculation:
Let's say that cos-1 (2x + 1) = θ
⇒ cos (cos-1 (2x + 1)) = cos θ
⇒ cos θ = 2x + 1
Since, -1 ≤ cos θ ≤ 1
⇒ -1 ≤ 2x + 1 ≤ 1
⇒ -1 - 1 ≤ 2x + 1 - 1 ≤ 1 - 1
⇒ -2 ≤ 2x ≤ 0
⇒ \(\rm -\dfrac{2}{2}≤ x ≤ \dfrac{0}{2}\)
⇒ -1 ≤ x ≤ 0
⇒ x ∈ [-1, 0]
∴ The domain of the function is the closed interval [-1, 0].
The domain of the function f : R → R defined by \(\rm \sqrt {x^2 \ - \ x - 110}\) is.
Answer (Detailed Solution Below)
Domain of a Function Question 10 Detailed Solution
Download Solution PDFConcept:
The domain of a function is the complete set of possible values of the independent variable.
To find domain of √f(x), set f(x) ≥ o
Calculations:
Given: the function f : R → R defined by \(\rm \sqrt {x^2 \ - \ x - 110}\)
We know that the domain of a function is the complete set of possible values of the independent variable.
To find the domain
= x2 - x - 110 ≥ 0
= x2 - 11x + 10x - 110 ≥ 0
= x(x - 11) + 10(x - 11) ≥ 0
= (x + 10)(x - 11) ≥ 0
= x ≤ - 10 or x ≥ 11
= x ∈ (- ∞, - 10] ∪ [11, ∞)
Hence, the domain of the function f : R → R defined by f(x) = \(\rm \sqrt {x^2 \ - \ x - 110}\) is (- ∞, - 10] ∪ [11, ∞)
What is the domain of the function f(x) = 3x?
Answer (Detailed Solution Below)
Domain of a Function Question 11 Detailed Solution
Download Solution PDFConcept:
The domain is the set of all possible value of x which have a finite value of f(x).
Calculation:
Given function f(x) = 3x
The function will have a finite value for all x ∈ (-∞, ∞)
Mistake PointsThe range of the given function will be from (0,∞). 0, when x = -∞, and ∞ when x = ∞.
The domain of the function f(x) = \(\rm \frac{1}{\sqrt{|x|-x}}\) is :
Answer (Detailed Solution Below)
Domain of a Function Question 12 Detailed Solution
Download Solution PDFConcept:
Domain of a function:
- The domain of a function is the set of all input values (x-values) for which the function is defined.
- For a function containing a square root, the expression inside the root must be ≥ 0.
- If the root is in the denominator, then the expression must be > 0 (since division by zero is undefined).
- Here, the function is f(x) = 1 / √(|x| - x).
- So, we must ensure that |x| - x > 0 for f(x) to be defined.
Calculation:
Given,
f(x) = 1 / √(|x| - x)
We need: |x| - x > 0
⇒ Consider two cases for x:
⇒ Case 1: x ≥ 0 ⇒ |x| = x ⇒ |x| - x = x - x = 0 (Not allowed)
⇒ Case 2: x < 0 ⇒ |x| = -x ⇒ |x| - x = -x - x = -2x > 0
⇒ This is true for all x < 0
∴ Domain of the function is (-∞, 0)
What is the period of the function f(x) = sin x?
Answer (Detailed Solution Below)
Domain of a Function Question 13 Detailed Solution
Download Solution PDFConcept:
Period of a Function:
- If a function repeats over at a constant period we say that is a periodic function.
- It is represented like f(x) = f(x + T), T is the real number and this is the period of the function.
Calculation:
We have to find the period of the function f(x) = sin x
Now,
f(x + 2π) = sin (x + 2π) = sin x
⇒ f(x + 2π) = f(x)
∴ Period of sin x is 2πThe domain of the function \({\rm{f}}\left( {\rm{x}} \right) = \frac{1}{{\sqrt {\left| {\rm{x}} \right| - {\rm{x}}} }}\) is
Answer (Detailed Solution Below)
Domain of a Function Question 14 Detailed Solution
Download Solution PDFConcept:
Domain: Domain of function f(x) is define as values of x for which function f(x) exist.
\({\rm{f}}\left( {\rm{x}} \right) = \left| {\rm{x}} \right| = {\rm{\;}}\left\{ {\begin{array}{*{20}{c}} { - x,\;\;x < 0}\\ {x,\;\;x \ge 0} \end{array}} \right.\)
Calculation:
We have to find the domain of the function \({\rm{f}}\left( {\rm{x}} \right) = \frac{1}{{\sqrt {\left| {\rm{x}} \right| - {\rm{x}}} }}\)
We know that square root is always positive.
Therefore, |x| - x > 0 (|x| - x ≠ 0)
⇒ |x| > x
As we can see |x| is greater than x in (-∞, 0)
What is the period of the function f(x) = sin x?
Answer (Detailed Solution Below)
Domain of a Function Question 15 Detailed Solution
Download Solution PDFConcept:
Period of a Function:
- If a function repeats over at a constant period we say that is a periodic function.
- It is represented like f(x) = f(x + T), T is the real number and this is the period of the function.
Calculation:
We have to find the period of the function f(x) = sin x
Now,
f(x + 2π) = sin (x + 2π) = sin x
⇒ f(x + 2π) = f(x)
∴ Period of sin x is 2π