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$x\ge 3$, $x-3=3$, 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] :

${\int }_0^3(3-x)dx={[3x-\frac{x^2}{2}]}_0^3=9-4.5=4.5$

- For x in [3, 6] :

${\int }_3^6(x-3)dx={[\frac{x^2}{2}-3x]}_3^6=4.5$

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 $(-\mathrm{\infty },\mathrm{\infty })$

Hence, the correct answer is Option 3.

Calculation:

Give, f(x) = $\cos x-1+\frac{x^2}{2!},x\in \mathbb{R}$

⇒ 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

$f(x)={\cos }^{-1}\left(\frac{2-|x|}{4}\right)+{({\log }_e(3-x))}^{-1}$

⇒ $-1\le \left|\frac{2-|x|}{4}\right|\le 1$

$\Rightarrow \left|\frac{2-|x|}{4}\right|\le 1$

⇒ –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, $\mathrm{x}-2\ne 0$ that implies that $\mathrm{x}\ne 2$.

Similarly square root function is well defined for all non-negative values.

Therefore, $\mathrm{x}-2>0$ that implies $\mathrm{x}>2.$

Thus, domain of the given function is $\left(2,\mathrm{\infty }\right).$

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, $\mathrm{f}(\mathrm{x})\ge 0$

⇒$16-{\mathrm{x}}^2\ge 0$

⇒$16\ge {\mathrm{x}}^2$

⇒${\mathrm{x}}^2\le 16$

⇒ -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

⇒ $-{\displaystyle \frac{2}{2}}\le \mathrm{x}\le {\displaystyle \frac{0}{2}}$

⇒ -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 $\sqrt{{\mathrm{x}}^2-\mathrm{x}-110}$

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) = $\sqrt{{\mathrm{x}}^2-\mathrm{x}-110}$ 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.

$\mathrm{f}\left(\mathrm{x}\right)=\left|\mathrm{x}\right|=\{\begin{array}{c}-x,x<0\\ x,x\ge 0\end{array}$

Calculation:

We have to find the domain of the function $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\sqrt{\left|\mathrm{x}\right|-\mathrm{x}}}$

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)