Domain or Range MCQ Quiz - Objective Question with Answer for Domain or Range - Download Free PDF

Concept:

• The domain of a function is the set of values that we are allowed to plug into our function.
•  The range is the set of all possible values that the function will give when we give in the domain as input.

Calculation:

The above graph is of the function cosecx

Since \(cosec x = \frac{1}{sinx}\), the domain of the cosec function is the set {x : x ∈ R and x ≠ nπ, n ∈ Z} or R - {x : x = nπ, n ∈ Z} and 

the range set is the set {y : y ∈ R and y ≥ 1 or y ≤ -1} 

i.e., the set R - (-1,1).

It means that y = cosec x assumes all real values except -1 < y < 1 and is not defined for the integral multiple of π.

Hence, the correct answer is option 2).

Concept: 

Where the function is defined is called the range of the inverse trigonometric function and the values we get are called the domain of the inverse trigonometric function.

Solutions  -

So the final answer is  [-1,1] hence option 2 is correct.

Alternate Method

Let \(sin^{-1}x=A \Rightarrow x = sinA\) 

and we all know that the Sin function is defined on a whole real line so the range of the given function is \(\mathbb{R}\). and the values which we get of sin function for the whole real line is in between [-1,1] because sin function is periodic and oscillates between 1 and -1 on the whole real line.

So the final answer is  [-1,1] hence option 2 is correct.

Concept:

The range of a function is the set of all its outputs.

Example:

Let us consider the function f: A→ B,

where f(x) = 2x and A and B = {set of natural numbers}.

Here we say A is the domain and B is the co-domain.

The set of ƒ images of all the elements of A is known as the range of ƒ.  

Calculation:

f(x) = 1 - sinx 

Here the domain of f(x) is (-∞, ∞)

At this domain, the range of sinx is [-1, 1]

Range of f(x) = 1 - sin2x = 1 - [-1, 1] 

⇒ f(x) = [0 , 2]

∴ Range of f(x) is [0 , 2].

Concept Used:

The range of cos(x) is [-1, 1]

Calculation

Given:

f(x) = 7 cos(10x + 4π)

The range of cos(10x + 4π) is [-1, 1]

Multiply by 7:

⇒ 7 × [-1, 1] = [-7, 7]

∴ The range of f(x) = 7 cos(10x + 4π) is [-7, 7]

Hence option 4 is correct

Concept:

Invertibility of a Function:

• A function is invertible if it is one-one (injective) and onto (surjective).
• For composition g(f(x)) to be invertible, f(x) must be one-one in its domain, and g(x) must be one-one in the range of f(x).
• f(x) = sin 2x + cos 2x is a periodic trigonometric function. We need to find the interval where it is one-one.
• g(x) = x² − 1 is not one-one over ℝ since it is a parabola, but it is one-one over intervals where x is either increasing or decreasing (monotonic intervals).
• So, we need to choose an interval where:
  • f(x) is one-one
  • g(f(x)) is one-one, which means f(x) should not take both +a and −a, as g(x) = x² − 1 gives same output for ±a

Calculation:

Given,

f(x) = sin 2x + cos 2x

⇒ f(x) = √2 sin(2x + π/4)

This is a sine function with amplitude √2 and period π

We want to restrict domain such that f(x) is one-one and doesn't take same absolute values with opposite sign.

The function sin(θ) is one-one in interval [−π/2, π/2]

So, 2x + π/4 ∈ [−π/2, π/2]

⇒ −π/2 ≤ 2x + π/4 ≤ π/2

⇒ Subtract π/4: −3π/4 ≤ 2x ≤ π/4

⇒ Divide by 2: −3π/8 ≤ x ≤ π/8

But we need symmetric interval around 0 that ensures f(x) is one-one and g(f(x)) is also one-one.

Try x ∈ [−π/8, π/8]

Then 2x ∈ [−π/4, π/4] ⇒ sin(2x + π/4) lies in a monotonic interval

⇒ f(x) = √2 sin(2x + π/4) becomes one-one

Also, f(x) will not take both a and −a in this range, so g(f(x)) = (f(x))² − 1 will be one-one too

∴ The domain in which g(f(x)) is invertible is x ∈ [−π/8, π/8]

Concept:

The range of sine function is [-1,1] i.e \(\rm \Rightarrow -1\leq sin(x) \leq 1\)

Calculation:

We can write, \(\rm sin(A - \frac{\pi}{3}) \in [-1,1]\)

\(\rm \Rightarrow -1\leq sin(A - \frac{\pi}{3}) \leq 1\)

\(\rm \Rightarrow -2\leq 2sin(A - \frac{\pi}{3}) \leq 2\)

So, from the relation, we can conclude that the minimum value of 2sin(A - \(\rm \frac{\pi}{3}\)) is -2.

Concept:

Calculation: 

Let f(x) = sin-1 3x

As we know domain of sin^-1 x is x ∈ [-1, 1]

So, -1 ≤ 3x ≤ 1

\(\Rightarrow \rm \frac{-1}{3} \leq x \leq \frac{1}{3}\)

Hence domain of sin-1 3x is \(\left[\frac{-1}{3} , \frac{1}{3} \right ]\)

Concept:

If cos^-1 x = y then x = cos y

Calculation:

As we know that, if cos-1 x = y then x = cos y

So,y = cos-1 (x2 - 4) --------(Given)

⇒ x² - 4 = cos y

As we know that, -1 ≤ cos y ≤ 1, where y ∈ R

⇒ - 1 ≤ x2 - 4 ≤ 1

Adding 4 on all the sides of the inequation we get

⇒ 3 ≤ x² ≤ 5

We can re-write the above inequation as

⇒ √3 ≤ |x| ≤ √5

⇒ x ∈ [-√5, -√3] ∪ [√3, √5]

Hence, the domain of the given function is [-√5, -√3] ∪ [√3, √5]

Concept:

-1 ≤ sin x ≤ 1

0 ≤ sin2 x ≤ 1

Calculation:

Given:

f(x) = 3 sin2 x + 4 cos2 x

= 3 sin2 x + 3 cos2 x +  cos2 x

= cos2 x + 3 (sin2 x + cos2 x)

= cos2 x + 3                                    (∵sin2 x + cos2 x = 1)

As we know that, 0 ≤ cos2 x ≤ 1

⇒ 3 + 0 ≤ 3 + cos2 x ≤ 3 + 1

⇒ 3 ≤ f(x) ≤ 4

∴ Range of the f(x) is [3, 4]

Concept:

The range of a function is the set of all its outputs.

Example:

Let us consider the function f: A→ B,

where f(x) = 2x and A and B = {set of natural numbers}.

Here we say A is the domain and B is the co-domain.

The set of ƒ images of all the elements of A is known as the range of ƒ.  

Calculation:

f(x) = 1 - sinx 

Here the domain of f(x) is (-∞, ∞)

At this domain, the range of sinx is [-1, 1]

Range of f(x) = 1 - sin2x = 1 - [-1, 1] 

⇒ f(x) = [0 , 2]

∴ Range of f(x) is [0 , 2].

Concept:

If cos-1 x = y then x = cos y

Calculation:

As we know that, if cos-1 x = y then x = cos y

So,y = cos-1 (x2 - 9) --------(Given)

⇒ x2 - 9 = cos y

As we know that, -1 ≤ cos y ≤ 1, where y ∈ R

⇒ - 1 ≤ x2 - 9 ≤ 1

Adding 9 on all the sides of the inequation we get

⇒ 8 ≤ x2 ≤ 10

We can re-write the above inequation as

⇒ 2√2 ≤ |x| ≤ √10

⇒ x ∈ [-√10, -2√2] ∪ [2√2, √10]

Hence, the domain of the given function is [-√10, -2√2] ∪ [2√2, √10]

Concept:

-1 ≤ sin x ≤ 1

0 ≤ sin² x ≤ 1

Calculation:

Given:

f(x) = 4 sin² x + 3 cos² x

= sin² x + 3 sin² x + 3 cos² x

= sin² x + 3 (sin² x + cos² x)

= sin² x + 3             (∵sin² x + cos² x = 1)

As we know that, 0 ≤ sin² x ≤ 1

⇒ 3 + 0 ≤ 3 + sin² x ≤ 3 + 1

⇒ 3 ≤ f(x) ≤ 4

∴ Range of the f(x) is [3, 4]

Concept:

The graphs of Sin θ and Cos θ are given by:

From above graph,

-1 ≤ sin θ ≤ 1

-1 ≤ cos θ ≤ 1

Since, both sin θ and cos θ oscillates between -1 and 1. So option 1 is correct.

Concept:

• The domain of a function f(x) is the set of values of x for which the function is defined.
• The value of sin θ always lies in the interval [-1, 1].
• sin-1 (sin θ) = θ.
• sin (sin-1 x) = x.

Calculation: 

Let f(x) = cos-1 2x

As we know domain of cos-1 x is x ∈ [-1, 1]

So, -1 ≤ 2x ≤ 1

⇒ \(\rm \frac{-1}{2}\leq x\leq \frac{1}{2}\)

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Concept:

Leibnitz integral rule,

\({d\over dx} [\int_{a(x)}^{b(x)} f(t,x)dt] = \int_a^b {\partial f \over \partial x} dt + b'f(b(x),x) - a'f(a(x),x)\)

Calculation:

Given, g(x) = \(\int^x_0\sqrt{1-t^2}\ dt\), 

By Leibnitz rule,

g'(x) = \(\int^x_0 0\ dt + (x)'\sqrt{1-x^2} - (0)'\sqrt{1-0^2} \)

⇒ g'(x) = \(0 + 1.\sqrt{1-x^2} + 0\)

⇒ g'(x) = \(\sqrt{1-x^2}\)

If g'(x) is defined, then 1 - x² ≥ 0

⇒ x ∈ [-1, 1] 

∴ The correct option is (3).