Domain or Range MCQ Quiz in తెలుగు - Objective Question with Answer for Domain or Range - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Use simply trigonometric identities.

\(sin^{-1}\left ( sinx \right )= x\)

Solutions: 

We know that \(sin30^\circ =\frac{1}{2}\) and this implies 

\(sin^{-1}\left ( sin30^\circ \right )=tan^{-1}x\)

Using the rule \(sin^{-1}\left ( sinx \right )= x\) then 

\(30^\circ = tan^{-1}x\)

Now taking tan of both side 

\(tan30^\circ = x\)

we again know that the value of \(tan30^\circ = \frac{1}{\sqrt{3}}\) then 

\(x= \frac{1}{\sqrt{3}}\)

The final answer to this question is  \(x= \frac{1}{\sqrt{3}}\), and option 2 is correct.

Concept:

Range of sin θ is [-1, 1]

Calculation:

Given that

f(x) = π + sin2 x.

The given function is a combination of a constant term (π) and the

square of the sine function sin2 x.

We know that 

-1 ≤ sin x ≤ 1 

⇒ 0 ≤ sin²x ≤ 1 

⇒ π + 0  ≤ π + sin2x ≤ π + 1 

Therefore, the range of π + sin2 x is [π, π + 1].

Concept:

We know that, \(\rm -1 \leq sin \;\theta \leq 1\)

Calculations:

We know that, \(\rm -1 \leq sin \;\theta \leq 1\)

Adding - 2 on both side, we get

⇒\(\rm -2 -1 \leq -2 + sin \;\theta \leq -2+ 1\)

⇒\(\rm -3 \leq -2 + sin \;\theta \leq -1\)

If we take mod inequality will be changed.

⇒\(\rm |-3| \leq |-2 + sin \;\theta| \leq |-1|\)

∴  \(\rm 1 \leq |-2 + sin \;\theta| \leq 3\)

Hence, the maximum value of |-2 + sin θ| is 3.

Concept:

The minimum and maximum value of a.sin x ± b.cos x is given by:

\(\rm - \sqrt {{a^2 + b^2}} ≤ a\sin x \pm b\cos x ≤ \sqrt{a^2 + b^2}\)

Calculation:

Given: sin x - √8cos x - 2 

Compare sin x - √8cox x is a.sinx - b.cosx 

Here a = 1 and b = √8

⇒  \(\rm - \sqrt {{(1^2 + √{8}^2}}) ≤ \sin x - \sqrt 8 \cos x ≤ \sqrt {{(1^2 + √{8}^2}}\)

⇒ \(\rm - 3 ≤ sin x - √ 8.cosx ≤ 3\)

Subtract 2 in the above equation

⇒ -3 - 2 ≤ sin x - √8.cos x - 2 ≤ 3 - 2

⇒ -5 ≤ sin x - √8.cos x - 2 ≤ 1

Range [- 5, 1]

Concept: 

The period of the function f(x) = π + sin2 x is determined by the

periodicity of the sine function, as the sin2 x.

Calculation:

Given that 

f(x) = π + sin2 x    ----(1)

The cos x function has a period of 2π.

The cos 2x function has a period of π.

We know that 

sin²x = (1 = cos 2x)/2

As, adding or subtracting a constant number does not affect the period

of the function.

Hence, The period of the function π + sin2 x is π.

Concept:

• If the given equation is in quadratic form ax² + bx + c=0, then its solution is found as \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\).
• -1 ≤ sin x ≤ 1
• cos 2x = 1 - 2sin2x

Calculation:

k sin x + cos 2x = 2k - 7

⇒ k sin x + 1 - 2sin²x = 2k - 7

⇒ -2sin2x + k sin x + 1 - 2k + 7 = 0 

⇒ -2sin²x + k sin x - 2k + 8 = 0

⇒ 2sin2x - k sinx + 2k - 8 = 0  ...(Multiply by -)

After comparing with Quadratic equation, a = 2, b = -k, c = (2k - 8)

⇒ \(sin~x=\frac{k\pm\sqrt{k^2-8(2k-8)}}{4}\)

\(\Rightarrow sin~x=\frac{k\pm\sqrt{k^2-16k+64}}{4}\)

\(\Rightarrow sin~x=\frac{k\pm\sqrt{(k-8)^2}}{4}\)

\(\Rightarrow sin~x=\frac{k\pm{(k-8)}}{4}\)

\(\Rightarrow sin~x=\frac{k+{(k-8)}}{4},\frac{k-{(k-8)}}{4}\)

\(\Rightarrow sin~x=\frac{2k{-8}}{4},\frac{{8}}{4}\)

⇒ sin x = \(\frac{k-4}{2}\) and sin x = 2,

Since, sin x cannot be greater than 1,

Thus, sin x ≤ 1

\(\frac{k-4}{2}≤1\)

k - 4 ≤ 2

k ≤ 6

Thus, the maximum value of k is 6.

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