Domain of a Function MCQ Quiz - Objective Question with Answer for Domain of a Function - Download Free PDF

Calculation:

Given,

The function is f(x) = |x - 3| , and we need to find the area bounded by the curve and the line y = 3.

To find the points of intersection, we set the function equal to 3:

\( |x - 3| = 3 \)

Solving for x :

- For\( x \geq 3 \), \(x - 3 = 3 \), which gives x = 6 .
- For ( x < 3 ), 3 - x = 3 , which gives x = 0 .

Therefore, the points of intersection are x = 0  and x = 6 .

The area can be calculated by integrating the difference between the curve and the line from x = 0 to x = 6. The integral is split into two parts due to the absolute value function:

\( A = \int_{0}^{3} (3 - x) \, dx + \int_{3}^{6} (x - 3) \, dx \)

For x in [0, 3] , ( f(x) = 3 - x ), and for ( x in [3, 6] ), ( f(x) = x - 3 ).

Compute both integrals:

- For x in [0, 3] :

\( \int_{0}^{3} (3 - x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{0}^{3} = 9 - 4.5 = 4.5 \)

- For x in [3, 6] :

\( \int_{3}^{6} (x - 3) \, dx = \left[ \frac{x^2}{2} - 3x \right]_{3}^{6} = 4.5 \)

Step 4: The total area is the sum of the two areas:

\( A = 4.5 + 4.5 = 9 \, \text{square units} \)

∴ The total area bounded by the curve and the line is 9 square units.

The correct answer is Option (4):

Calculation:

Given,

The function is \(f(x) = |x - 3| \), which is an absolute value function.

The domain of an absolute value function \(f(x) = |x - a| \) is all real numbers, because the absolute value function is defined for all values of x. The function handles both positive and negative inputs for x .

Since there are no restrictions or undefined points for the absolute value function, the domain is all real numbers.

∴ The domain of the function is \((-\infty, \infty) \)

Hence, the correct answer is Option 3.

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.

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

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 

Concept:

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.

Concept:

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 \(\left( {2,{\rm{\;}}\infty } \right).\)

Concept:

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].

Concept:

• 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].

Concept:

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, ∞)

Concept:

The domain is the set of all possible value of x which have a finite value of f(x).

Calculation:

Given function f(x) = 3^x

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 = ∞. 

Concept:

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)

Concept:

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)

Concept:

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)

Concept:

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)