CUET UG Mathematics Booster Test 2 - Inverse Sin and Inverse Cosine Functions -

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CUET UG Mathematics Booster Test 2 - Inverse Sin and Inverse Cosine Functions

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Mathematics Booster Test 2 - Inverse Sin and Inverse Cosine Functions

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QUESTION 1 OF 20

If a custom function f(x) = sin x is strictly defined only on the interval [π/2, 3π/2], does an inverse function exist for it, and if so, what is its resulting range?

Answer Review

1 If a custom function f(x) = sin x is strictly defined only on the interval [π/2, 3π/2], does an inverse function exist for it, and if so, what is its resulting range?

A No, the inverse does not exist here.

B Yes, the inverse exists with range [-1, 1].

C Yes, the inverse exists with range [π/2, 3π/2].

D Yes, the inverse exists with range [-π/2, π/2].

💡 Short Explanation

Sine is one-one on [π/2,3π/2]. Inverse exists on bijective intervals. Inverse range equals restricted domain.

✅ Correct Answer Explanation

On the interval [π/2,3π/2], the sine function becomes one-one and covers the range [-1,1]. Therefore an inverse exists. For inverse functions, the range of the inverse equals the restricted domain of the original function. Hence the inverse range is [π/2,3π/2], making option C correct.

❓ Why Other Options Are Incorrect

Option A → Inverse exists because sine becomes bijective on this interval.
Option B → [-1,1] is the domain of the inverse, not its range.
Option D → [-π/2,π/2] is the principal branch, not this custom branch.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply inverse-function rule: original restricted domain becomes inverse range.

Final Logic:

Inverse range always equals the original restricted domain.

🧠 Memory Trick

"Inverse swaps domain and range."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Alternate Branches of sin⁻¹x

2 Accurately match the following principal inverse trigonometric evaluations to their correct radian outcomes.

List 1	List 2
1. sin⁻¹(1/2)	a. π/6
2. sin⁻¹(−1/2)	b. −π/6
3. cos⁻¹(1/2)	c. π/3
4. cos⁻¹(−1/2)	d. 2π/3

A 1-b, 2-a, 3-c, 4-d

B 1-d, 2-c, 3-b, 4-a

C 1-a, 2-b, 3-c, 4-d

D 1-c, 2-a, 3-d, 4-b

💡 Short Explanation

sin⁻¹(1/2)=π/6 cos⁻¹(1/2)=π/3 Negative inputs follow principal branches.

✅ Correct Answer Explanation

Using standard principal values: sin⁻¹(1/2)=π/6, sin⁻¹(−1/2)=−π/6, cos⁻¹(1/2)=π/3, cos⁻¹(−1/2)=2π/3. Thus the matching becomes: 1-a, 2-b, 3-c, 4-d. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Swaps positive and negative sine inverse values.
Option B → Completely mismatches standard principal values.
Option D → Incorrectly exchanges cosine inverse values.

🎯 Strategy

Used: Substitution

Application:

Recall exact trigonometric standard-angle values.

Final Logic:

Principal values directly determine the matching.

🧠 Memory Trick

"Half gives 30°, negative half gives −30°."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Values

3 Which distinct pair of intervals below BOTH represent mathematically valid, bijective branches capable of defining an inverse sine function?

A [-3π/2, -π/2] and [π/2, 3π/2]

B [0, π] and [π, 2π]

C [-π, 0] and [0, π]

D [-π/4, π/4] and [π/4, 3π/4]

💡 Short Explanation

Valid branch requires one-one behaviour. Sine is injective on option A intervals. Other intervals repeat values.

✅ Correct Answer Explanation

For inverse sine, the sine function must remain one-one on the chosen interval. On both intervals in option A, sine changes monotonically without repeating values. Intervals in other options contain repeated sine outputs or incomplete ranges. Therefore only option A provides valid bijective branches.

❓ Why Other Options Are Incorrect

Option B → Sine repeats values on both intervals.
Option C → Intervals contain symmetry causing repeated outputs.
Option D → First interval does not cover full range [-1,1].

🎯 Strategy

Used: Elimination

Application:

Reject intervals where sine repeats outputs.

Final Logic:

Bijective intervals must avoid repeated sine values.

🧠 Memory Trick

"No repetition means valid inverse."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Valid Branches of Sine

4 Which of the following analytical statements is mathematically INCORRECT concerning the principal branch of sin⁻¹?

A It contains radian values where the standard cosine function evaluates to strictly negative outputs.

B It directly corresponds to angle mappings inside the first and fourth quadrants of the unit circle.

C It continuously spans a total interval length exactly equal to π radians.

D It acts as a strictly increasing and continuous mathematical function.

💡 Short Explanation

Principal branch is [-π/2,π/2]. Cosine stays nonnegative there. Hence option A is false.

✅ Correct Answer Explanation

The principal branch of sin⁻¹x is: [-π/2,π/2]. In this interval, cosine values are always nonnegative. Therefore option A is incorrect. Statements B, C, and D are correct because the branch covers quadrants I and IV, spans length π, and remains increasing and continuous.

❓ Why Other Options Are Incorrect

Option B → Principal branch indeed lies in quadrants I and IV.
Option C → Interval length equals π radians exactly.
Option D → sin⁻¹x is continuous and strictly increasing.

🎯 Strategy

Used: Elimination

Application:

Check cosine sign behaviour in the principal branch.

Final Logic:

Cosine never becomes negative on [-π/2,π/2].

🧠 Memory Trick

"Cos positive in principal sine branch."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Branch Properties

5 An angle y is analytically mapped from the expression sin⁻¹(√3/2). What is the exact fractional numerical value of this mapping interval's principal output in radians?

A π/6

B π/3

C π/4

D π/2

💡 Short Explanation

sin(π/3)=√3/2 π/3 lies in principal branch. Hence inverse value equals π/3.

✅ Correct Answer Explanation

The principal value of inverse sine lies in: [-π/2,π/2]. Since: sin(π/3)=√3/2, the principal output becomes: sin⁻¹(√3/2)=π/3. Hence option B is correct. Other angles produce different sine values.

❓ Why Other Options Are Incorrect

Option A → sin(π/6)=1/2.
Option C → sin(π/4)=1/√2.
Option D → sin(π/2)=1.

🎯 Strategy

Used: Substitution

Application:

Recall the exact angle corresponding to √3/2.

Final Logic:

Principal angle with sine √3/2 is π/3.

🧠 Memory Trick

"√3/2 belongs to 60°."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Standard Principal Values

6 In the standard (x, y) Cartesian plane, the entire area strictly enclosed by the curve y = sin⁻¹x, the vertical y-axis, and the horizontal lines y = -π/2 and y = π/2 is bounded horizontally on its extremes precisely at:

A x = -π/2 and π/2

B x = -1 and 1

C x = 0 and 1

D y = -1 and 1

💡 Short Explanation

Domain of sin⁻¹x is [-1,1]. Horizontal bounds come from domain. Hence extremes are x=−1 and 1.

✅ Correct Answer Explanation

The graph of y=sin⁻¹x exists only for: x∈[-1,1]. Thus the horizontal extent of the enclosed region is bounded by: x=−1 and x=1. Hence option B is correct. Other options confuse x-bounds with y-values or angle limits.

❓ Why Other Options Are Incorrect

Option A → These are angle values, not x-coordinates.
Option C → Excludes half the graph region.
Option D → Represents vertical limits instead of horizontal bounds.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Identify whether the question asks about horizontal or vertical limits.

Final Logic:

Horizontal graph bounds come from the function domain.

🧠 Memory Trick

"Domain controls left-right spread."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Domain of sin⁻¹x

7 A processor computes y = sin(sin⁻¹x) for all inputs from -1 to 1. Because this relationship evaluates to y = x, what is the exact moving average of any two symmetrically opposed domain values (x and -x)?

A x

B -x

C 1

D 0

💡 Short Explanation

sin(sin⁻¹x)=x Average of x and −x is zero. Symmetry cancels values exactly.

✅ Correct Answer Explanation

Using the identity: sin(sin⁻¹x)=x, the outputs become x and −x for symmetric inputs. Their average is: (x+(−x))/2=0. Hence option D is correct. Symmetric opposite values always cancel perfectly around the origin.

❓ Why Other Options Are Incorrect

Option A → Represents only one value, not the average.
Option B → Same mistake with negative value only.
Option C → Opposite values never average to 1 generally.

🎯 Strategy

Used: Substitution

Application:

Simplify the inverse identity before averaging.

Final Logic:

Opposite symmetric values always average to zero.

🧠 Memory Trick

"Opposites cancel instantly."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Property sin(sin⁻¹x)=x

8 A continuous variable x is chosen completely uniformly from the interval [0, 2π]. What is the exact mathematical probability that the property sin⁻¹(sin x) = x evaluates as true?

A 1

B 1/2

C 0

D 1/4

💡 Short Explanation

Identity works only in principal branch. Valid interval inside [0,2π] is [0,π/2]. Probability becomes 1/4.

✅ Correct Answer Explanation

The identity: sin⁻¹(sin x)=x holds only for: x∈[-π/2,π/2]. Inside [0,2π], this reduces to [0,π/2], whose length is π/2. Total interval length is 2π. Thus probability: (π/2)/(2π)=1/4. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Identity fails outside principal branch.
Option B → Overestimates valid interval length.
Option C → Identity is true on part of the interval.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Use interval length ratio for continuous probability.

Final Logic:

Valid branch length divided by total length equals 1/4.

🧠 Memory Trick

"Only first quadrant survives."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Property of sin⁻¹(sin x)

9 A spatial vector transformation T strictly swaps the components of any vector (x,y) to (y,x). If T is applied to every coordinate vector making up the locus of y = sin x strictly where x ∈ [-π/2, π/2], the resulting new vector locus defines:

A y = sin⁻¹x limited to its principal branch

B y = cos⁻¹x

C y = (sin x)⁻¹

D x = sin⁻¹y

💡 Short Explanation

Inverse graphs swap coordinates. Reflection occurs along y=x. Resulting graph becomes inverse sine.

✅ Correct Answer Explanation

Swapping coordinates transforms the graph of a function into the graph of its inverse. Restricting sine to: [-π/2,π/2] ensures invertibility. Therefore the transformed locus becomes: y=sin⁻¹x. Hence option A is correct. Other options confuse reciprocal and inverse notation.

❓ Why Other Options Are Incorrect

Option B → Coordinate swapping does not create cosine inverse.
Option C → Reciprocal sine differs from inverse sine.
Option D → This is equivalent relation form but not standard graph equation.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use graph reflection property of inverse functions.

Final Logic:

Coordinate interchange generates inverse-function graph.

🧠 Memory Trick

"Swap coordinates to get inverse."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Reflection Along y=x

10 The region bounded by y = sin x (for x ∈ [0, π/2]), the y-axis, and y = 1 has an exact geometric area of (π/2 - 1). By utilizing the reflection along y = x, the integral ∫sin⁻¹x dx from 0 to 1 directly corresponds to this area. What is its numerical value?

A 1

B π/2

C π/2 - 1

D 1 - π/2

💡 Short Explanation

Reflection preserves enclosed area. Inverse graph area equals original reflected area. Integral value becomes π/2−1.

✅ Correct Answer Explanation

Using reflection symmetry between y=sin x and y=sin⁻¹x about y=x, the enclosed areas remain equal. The known area bounded by y=sin x on [0,π/2] equals: π/2−1. Therefore: ∫₀¹ sin⁻¹x dx = π/2−1. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores geometric area relation entirely.
Option B → Represents interval length only, not enclosed area.
Option D → Gives negative area, impossible geometrically.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use inverse-function reflection area property.

Final Logic:

Reflected regions preserve identical area.

🧠 Memory Trick

"Mirror areas stay equal."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Graph Reflection and Area Interpretation

11 Because y = sin⁻¹x behaves as a strictly increasing function on the sub-interval, what must identically be the mathematical sign of the definite integral of its derivative over this exact interval?

A Negative

B Positive

C Zero

D Undefined

💡 Short Explanation

Increasing functions have positive derivatives. Integral of positive quantity stays positive. Hence result is positive.

✅ Correct Answer Explanation

The function y = sin⁻¹x is strictly increasing on its domain (-1,1). Therefore its derivative: 1/√(1−x²) remains positive throughout the interval. The definite integral of a positive function over any nonzero interval is positive. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Negative sign would imply decreasing behaviour.
Option C → Positive derivative cannot integrate to zero on nonzero interval.
Option D → Derivative exists and is integrable inside the interval.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Relate increasing behaviour directly to derivative sign.

Final Logic:

Increasing functions have positive derivative integrals.

🧠 Memory Trick

"Increasing means positive slope."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Increasing Nature of sin⁻¹x

12 Assertion (A): The mathematical evaluation of sin⁻¹(2) strictly equals π/2.
Reason (R): The defined domain limits of the sin⁻¹x function strictly exclude any numerical values greater than 1.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

sin⁻¹x exists only on [-1,1]. 2 lies outside the domain. Therefore assertion is false.

✅ Correct Answer Explanation

The inverse sine function is defined only for: x∈[-1,1]. Since 2 lies outside this domain, sin⁻¹(2) has no real value. Therefore the Assertion is false. The Reason correctly states the domain restriction, making it true. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Reason is mathematically correct.
Option B → Assertion is false, not true.
Option C → Assertion itself is invalid, so explanation fails.

🎯 Strategy

Used: Elimination

Application:

Verify the allowed domain before evaluating inverse sine.

Final Logic:

Inputs greater than 1 are invalid for sin⁻¹x.

🧠 Memory Trick

"Beyond ±1, inverse sine fails."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Domain of sin⁻¹x –

13 Arrange the following inverse cosine evaluations strictly in descending order of their numerical radian results:
1. cos⁻¹(0)
2. cos⁻¹(-1)
3. cos⁻¹(1)
4. cos⁻¹(-1/2)

A 2, 4, 1, 3

B 1, 2, 3, 4

C 4, 2, 1, 3

D 3, 1, 4, 2

💡 Short Explanation

cos⁻¹(−1)=π cos⁻¹(−1/2)=2π/3 cos⁻¹(0)=π/2, cos⁻¹(1)=0

✅ Correct Answer Explanation

Using principal values: cos⁻¹(−1)=π, cos⁻¹(−1/2)=2π/3, cos⁻¹(0)=π/2, cos⁻¹(1)=0. Descending order becomes: π > 2π/3 > π/2 > 0. Thus the correct arrangement is: 2,4,1,3. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Places π/2 above π incorrectly.
Option C → Reverses first two largest values.
Option D → Starts with smallest value instead of largest.

🎯 Strategy

Used: Substitution

Application:

Evaluate each inverse cosine value individually.

Final Logic:

Arrange evaluated angles from greatest to least.

🧠 Memory Trick

"−1 gives biggest cosine inverse."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Values of cos⁻¹x

14 The exact mathematical solution to the equation cos⁻¹x = 2π/3 falls validly within the principal range [0, π]. What is the specific real value of x?

A 1/2

B √3/2

C -1/2

D -√3/2

💡 Short Explanation

Apply cosine to both sides. cos(2π/3)=−1/2. Therefore x=−1/2.

✅ Correct Answer Explanation

Given: cos⁻¹x = 2π/3, apply cosine to both sides: x = cos(2π/3). Since: cos(2π/3)=−1/2, the required value becomes: x=−1/2. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → cos(π/3)=1/2, not 2π/3.
Option B → Corresponds to cosine π/6.
Option D → Gives cosine value for 5π/6.

🎯 Strategy

Used: Substitution

Application:

Use direct inverse-function cancellation.

Final Logic:

Applying cosine gives x=cos(2π/3).

🧠 Memory Trick

"120° gives −1/2."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Evaluation of cos⁻¹x

15 If the expression cos⁻¹x is analyzed strictly in an alternative, valid branch such as [π, 2π], the corresponding function behavior on this domain is:

A Strictly increasing

B Bijective

C Mathematically undefined

D Perfectly constant

💡 Short Explanation

Valid inverse branch requires bijection. Restricted cosine becomes one-one. Therefore inverse exists uniquely.

✅ Correct Answer Explanation

An inverse function exists only when the original function becomes bijective on the chosen interval. Restricting cosine to a valid branch such as [π,2π] allows unique inverse mapping over an appropriate range. Therefore the function behaves bijectively, making option B correct.

❓ Why Other Options Are Incorrect

Option A → Cosine is not strictly increasing throughout this interval.
Option C → Valid restricted branches remain mathematically defined.
Option D → Cosine clearly changes values over the interval.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Focus on the condition required for inverse existence.

Final Logic:

Inverse branches must always be bijective.

🧠 Memory Trick

"Valid branch means bijection."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Restricted Branches of Cosine

16 By examining the analytical graph of y = cos⁻¹x, the specific planar coordinate point where the curve intersects the horizontal x-axis is:

A (0, π/2)

B (-1, π)

C (0, 0)

D (1, 0)

💡 Short Explanation

x-axis means y=0. cos⁻¹x=0 implies x=1. Intersection point becomes (1,0).

✅ Correct Answer Explanation

The graph intersects the x-axis where: y=0. So: cos⁻¹x=0. Applying cosine: x=cos0=1. Therefore the intersection point is: (1,0). Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Lies on graph but not x-axis.
Option B → Gives y-value π instead of 0.
Option C → cos⁻¹(0)=π/2, not 0.

🎯 Strategy

Used: Substitution

Application:

Set y=0 and solve inverse equation directly.

Final Logic:

Only x=1 makes cos⁻¹x equal zero.

🧠 Memory Trick

"Cos inverse touches x-axis at 1."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Graph of cos⁻¹x

17 When analytically comparing sin⁻¹x and cos⁻¹x, evaluating the expression (sin⁻¹x + cos⁻¹x) for any value of x strictly confined within [-1, 1] yields a universal constant. This constant is:

A π/2

B π

C 0

D -π/2

💡 Short Explanation

Standard inverse identity applies. Sum remains constant always. Constant equals π/2.

✅ Correct Answer Explanation

A standard inverse trigonometric identity states: sin⁻¹x + cos⁻¹x = π/2 for every: x∈[-1,1]. Therefore the expression always evaluates to the constant π/2. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Too large for the standard identity.
Option C → Sum never becomes zero generally.
Option D → Incorrect negative constant value.

🎯 Strategy

Used: Substitution

Application:

Recall the standard NCERT inverse identity.

Final Logic:

Inverse sine and cosine always complement to π/2.

🧠 Memory Trick

"Sine inverse plus cosine inverse equals 90°."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Relation Between sin⁻¹x and cos⁻¹x

18 The identical domain similarity [-1, 1] necessary for defining both the inverse sine and inverse cosine mathematically originates from the fundamental fact that:

A Both original functions are perfectly odd mathematically.

B The unrestricted output range of both the original sine and cosine functions is exactly [-1, 1].

C They are strict mathematical reciprocals.

D Their inverse graphs mutually intersect at the coordinate origin.

💡 Short Explanation

Inverse domain equals original range. Sine and cosine ranges are [-1,1]. Therefore inverse domains match.

✅ Correct Answer Explanation

For inverse functions, the domain equals the range of the original function. Since both sine and cosine have range: [-1,1], their inverses sin⁻¹x and cos⁻¹x also possess domain: [-1,1]. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Oddness does not determine inverse domains.
Option C → Sine and cosine are not reciprocals.
Option D → Graph intersections do not define domains.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use inverse-function relationship between domain and range.

Final Logic:

Original function range becomes inverse domain.

🧠 Memory Trick

"Inverse domain comes from original range."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Domains of Inverse Functions

📄 Passage

Whenever no branch of an inverse trigonometric function is mentioned, we mean the principal value branch of that function. The value of an inverse trigonometric function which lies in the range of the principal branch is called the principal value of that inverse trigonometric function. These concepts guarantee unique solutions when evaluating algebraic expressions involving these inverses.

Based on the passage's demand for unique principal values, the innate origin symmetry of sin⁻¹x enables the analytical conclusion that the expression sin⁻¹(-x) is perfectly equivalent to:

A π - sin⁻¹x

B sin⁻¹x

C -sin⁻¹x

D cos⁻¹x

💡 Short Explanation

sin⁻¹x is an odd function. Odd functions satisfy f(−x)=−f(x). Hence required relation follows.

✅ Correct Answer Explanation

The inverse sine function is odd because its graph is symmetric about the origin. Therefore: sin⁻¹(−x)=−sin⁻¹x. This property holds for all x in the domain [-1,1]. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Relation belongs to cosine-type identities.
Option B → Would imply even symmetry, which is false.
Option D → Inverse cosine is a different function entirely.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Identify symmetry type of inverse sine graph.

Final Logic:

Odd symmetry gives f(−x)=−f(x).

🧠 Memory Trick

"Inverse sine is odd."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Properties of sin⁻¹x

📄 Passage

Utilizing the graph reflection rule conceptually, if the coordinate point (x, y) constitutes the principal value on the graph of an inverse function, which corresponding analytical relationship must strictly hold true regarding the original function 'f'?

A f(x) = y where x remains completely unrestricted.

B f(-x) = -y where y is restricted to positive values only.

C f(x) = y where y dictates the restricted domain.

D f(y) = x where y must strictly reside within the defined restricted domain.

💡 Short Explanation

Inverse swaps x and y coordinates. Original function reverses the relation. Restricted domain ensures uniqueness.

✅ Correct Answer Explanation

If: y=f⁻¹(x), then by definition: f(y)=x. For inverse functions to exist uniquely, y must belong to the restricted domain where the original function is one-one. Therefore option D correctly expresses the reflection and inverse relationship.

❓ Why Other Options Are Incorrect

Option A → Original function domain cannot remain unrestricted.
Option B → Restriction to positive y-values is unnecessary.
Option C → Relation incorrectly keeps original coordinate order.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the defining property of inverse functions carefully.

Final Logic:

Inverse relation always satisfies f(y)=x.

🧠 Memory Trick

"Inverse swaps input and output."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Reflection Rule and Inverse Mapping

CUET UG Mathematics Booster Test 2 - Inverse Sin and Inverse Cosine Functions

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