CUET UG Mathematics Booster Test 1 - Inverse Sin and Inverse Cosine Functions -

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CUET UG Mathematics Booster Test 1 - 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 1 - Inverse Sin and Inverse Cosine Functions

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QUESTION 1 OF 20

Why is it mathematically necessary to restrict the natural domain of the sine function to define its inverse?

Answer Review

1 Why is it mathematically necessary to restrict the natural domain of the sine function to define its inverse?

A Because the sine function is naturally unbounded.

B To ensure the sine function becomes bijective (one-one and onto).

C To artificially expand its range.

D Because inverse functions only apply to straight lines.

💡 Short Explanation

Inverse exists only for bijections. Sine repeats values on R. Restriction makes sine one-one.

✅ Correct Answer Explanation

A function must be bijective to possess an inverse. The sine function on all real numbers is many-one because values repeat periodically. Restricting the domain to [-π/2, π/2] makes sine one-one and onto [-1,1]. Hence option B is correct. Other options misrepresent sine function properties.

❓ Why Other Options Are Incorrect

Option A → Sine is bounded between −1 and 1, not unbounded.
Option C → Domain restriction does not expand the range artificially.
Option D → Inverse functions apply to many nonlinear functions as well.

🎯 Strategy

Used: Elimination

Application:

Check the mathematical requirement for inverse functions.

Final Logic:

Only bijective functions can possess inverses.

🧠 Memory Trick

"No bijection, no inverse."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Need for Domain Restriction

2 Match the restricted domain branches of sin x to their inverse function classifications.

List 1	List 2
1. [-π/2, π/2]	a. Principal value branch
2. [π/2, 3π/2]	b. Other valid branch
3. [0, π]	c. Invalid for sin⁻¹
4. [-π/2, 0]	d. Partial branch

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

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

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

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

💡 Short Explanation

Principal branch is [-π/2,π/2]. [π/2,3π/2] is another valid branch. [0,π] repeats sine values.

✅ Correct Answer Explanation

The interval [-π/2,π/2] is the principal value branch. Interval [π/2,3π/2] also makes sine one-one and is another valid branch. Interval [0,π] is invalid because sine repeats values. Interval [-π/2,0] is only partial since it does not cover full range [-1,1]. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrectly swaps principal and secondary branches.
Option C → [0,π] cannot serve as principal branch.
Option D → Principal branch classification becomes incorrect.

🎯 Strategy

Used: Option Grouping

Application:

Classify intervals according to injectivity and completeness of range.

Final Logic:

Only option A matches all branch properties correctly.

🧠 Memory Trick

"Principal sine inverse: ±90° branch."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal and Alternate Branches

3 Which of the following intervals is strictly NOT a valid branch for the inverse sine function? (Choose the option identifying the incorrect branch)
I. [-3π/2, -π/2]
II. [π/2, 3π/2]
III. [0, π]

A I only

B II only

C I and II only

D III only

💡 Short Explanation

Valid branch requires one-one behaviour. Sine repeats values on [0,π]. Therefore III is invalid.

✅ Correct Answer Explanation

A valid branch for inverse sine requires sine to be one-one. On intervals [-3π/2,-π/2] and [π/2,3π/2], sine remains injective. However, on [0,π], sine repeats values symmetrically around π/2. Thus interval III is not a valid branch, making option D correct.

❓ Why Other Options Are Incorrect

Option A → Interval I is actually a valid branch.
Option B → Interval II is also injective for sine.
Option C → Both I and II are mathematically acceptable branches.

🎯 Strategy

Used: Elimination

Application:

Check whether sine repeats outputs within the interval.

Final Logic:

Only [0,π] fails the one-one condition.

🧠 Memory Trick

"No repeated sine values allowed."

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

4 Identify the incorrect statement regarding the standard notation of inverse branches.

A The inverse of sine is denoted by sin⁻¹.

B The principal branch for sin⁻¹ is [-π/2, π/2].

C The notation sin⁻¹x is identical to the algebraic reciprocal (sin x)⁻¹.

D The chosen interval dictates the branch of the inverse function.

💡 Short Explanation

sin⁻¹x means inverse sine. Reciprocal means cosec x. Both notations differ completely.

✅ Correct Answer Explanation

The notation sin⁻¹x denotes inverse sine, not reciprocal sine. Reciprocal sine is written as cosec x or 1/sin x. Statements A, B, and D are standard NCERT facts about inverse trigonometric branches. Therefore option C is the incorrect statement.

❓ Why Other Options Are Incorrect

Option A → Standard notation for inverse sine is correct.
Option B → Principal branch is indeed [-π/2,π/2].
Option D → Choice of interval determines inverse branch.

🎯 Strategy

Used: Odd One Out

Application:

Differentiate between inverse notation and reciprocal notation.

Final Logic:

Inverse and reciprocal are fundamentally different concepts.

🧠 Memory Trick

"sin⁻¹ ≠ 1/sin."

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

5 In a structural engineering calculation, an angle is mapped using f(x) = sin⁻¹x. If the input is x = 1/2, what is the exact numerical output of the mapped angle in radians?

A π/3

B π/6

C π/4

D π/2

💡 Short Explanation

sin(π/6)=1/2 π/6 lies in principal branch. Hence inverse sine gives π/6.

✅ Correct Answer Explanation

The inverse sine function returns the principal angle in [-π/2,π/2]. Since: sin(π/6)=1/2, the principal value becomes: sin⁻¹(1/2)=π/6. Thus option B is correct. Other angles correspond to different sine values.

❓ Why Other Options Are Incorrect

Option A → sin(π/3)=√3/2, not 1/2.
Option C → sin(π/4)=1/√2.
Option D → sin(π/2)=1, not 1/2.

🎯 Strategy

Used: Substitution

Application:

Recall the standard angle whose sine equals 1/2.

Final Logic:

Principal angle for sine value 1/2 is π/6.

🧠 Memory Trick

"Half belongs to 30°."

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

6 When plotting the region corresponding to the principal range of sin⁻¹x on the y-axis, what is the total continuous vertical height (span) of this interval?

A π

B 2π

C π/2

D 1

💡 Short Explanation

Principal range is [-π/2,π/2]. Vertical span equals upper minus lower. Total height becomes π.

✅ Correct Answer Explanation

The principal range of sin⁻¹x is: [-π/2,π/2]. The vertical span equals: π/2 − (−π/2)=π. Hence option A is correct. Other options either double the interval or confuse angle measures with unit lengths.

❓ Why Other Options Are Incorrect

Option B → Doubles the actual vertical span.
Option C → Represents only half the interval length.
Option D → Not measured in angular units properly.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Compute total interval width using upper minus lower limit.

Final Logic:

Vertical height of the principal range equals π.

🧠 Memory Trick

"From −90° to 90° totals 180°."

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

7 An electronic sensor records data points using the formula y = sin(sin⁻¹x) for inputs x = -0.5, 0, and 0.5. What is the moving average of these three respective data points?

A 0.5

B -0.5

C 1.0

D 0

💡 Short Explanation

sin(sin⁻¹x)=x Outputs are −0.5, 0, 0.5 Their average equals zero.

✅ Correct Answer Explanation

Using the identity: sin(sin⁻¹x)=x, outputs become: −0.5, 0, 0.5. Average: (−0.5+0+0.5)/3 = 0. Hence option D is correct. Other options incorrectly use individual values instead of the arithmetic mean.

❓ Why Other Options Are Incorrect

Option A → Represents only the positive data value.
Option B → Represents only the negative data value.
Option C → Gives incorrect total average calculation.

🎯 Strategy

Used: Substitution

Application:

Simplify the identity before computing the average.

Final Logic:

Symmetric values cancel, leaving zero average.

🧠 Memory Trick

"Opposites cancel perfectly."

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

8 A number x is picked entirely at random from a continuous uniform distribution between -π/2 and π/2. What is the probability that the equation sin⁻¹(sin x) = x holds exactly true?

A 0

B 0.5

C 1.0

D 0.25

💡 Short Explanation

Entire interval is principal branch. Identity holds throughout the interval. Probability equals one.

✅ Correct Answer Explanation

The identity: sin⁻¹(sin x)=x holds for every x in the principal interval: [-π/2,π/2]. Since the random selection occurs entirely inside this interval, every possible value satisfies the equation. Hence the probability is 1.0, making option C correct.

❓ Why Other Options Are Incorrect

Option A → The identity certainly holds in this interval.
Option B → Underestimates valid outcomes.
Option D → No restriction removes valid points from interval.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Check whether the chosen interval equals the principal branch.

Final Logic:

All points satisfy the identity, giving probability 1.

🧠 Memory Trick

"Inside principal branch, identity always works."

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

9 Let a position vector p = 0i + 1j rest on the graph of y = sin x. After the prescribed axis interchange to formulate the inverse graph, the newly generated position vector is:

A 0i + 0j

B 1i + 0j

C 1i + 1j

D -1i + 0j

💡 Short Explanation

Inverse reflection swaps coordinates. Original point is (0,1). New point becomes (1,0).

✅ Correct Answer Explanation

The graph of an inverse function is obtained by interchanging x and y coordinates. The vector: 0i + 1j represents the point (0,1). After reflection across y=x: (0,1) → (1,0). Hence the new vector becomes: 1i + 0j, making option B correct.

❓ Why Other Options Are Incorrect

Option A → Reflection does not move point to origin.
Option C → Coordinates are not preserved unchanged.
Option D → Reflection introduces no negative sign.

🎯 Strategy

Used: Substitution

Application:

Apply coordinate interchange rule directly.

Final Logic:

Swapping (0,1) gives (1,0).

🧠 Memory Trick

"Inverse graph swaps coordinates."

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

10 A triangle formed by points (0,0), (1,0), and (0,1) lies in the first quadrant. Reflecting this entire specific geometric area strictly across the line y = x yields a new total area of:

A 0.5 sq units

B 1.0 sq units

C 2.0 sq units

D 1.5 sq units

💡 Short Explanation

Reflection preserves area. Original triangle area is 1/2. Reflected triangle has same area.

✅ Correct Answer Explanation

The triangle with vertices (0,0), (1,0), and (0,1) is a right triangle with base=1 and height=1. Its area: (1/2)×1×1 = 0.5. Reflection across y=x preserves geometric area. Therefore the reflected triangle also has area 0.5 square units. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Doubles the actual triangular area.
Option C → Reflection never changes area magnitude.
Option D → Does not follow triangle area formula.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Compute original area and use reflection invariance.

Final Logic:

Reflection changes orientation, not area.

🧠 Memory Trick

"Mirror keeps area unchanged."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Reflection Property of Inverse Graphs –

11 Evaluate the definite integral of a constant function f(x) = 1, taking the lower domain limit of sin⁻¹x as the lower bound and the upper domain limit as the upper bound (∫ dx from -1 to 1).

A 0

B 1

C -2

D 2

💡 Short Explanation

Domain limits are −1 and 1. Integral of 1 gives interval length. Result equals 2.

✅ Correct Answer Explanation

The domain of sin⁻¹x is [-1,1]. Evaluating: ∫₋₁¹ dx gives: [x]₋₁¹ = 1−(−1)=2. Hence option D is correct. Other options either ignore interval width or incorrectly evaluate the definite integral.

❓ Why Other Options Are Incorrect

Option A → Integral over a nonzero interval cannot vanish.
Option B → Gives only half the interval length.
Option C → Interval width cannot be negative here.

🎯 Strategy

Used: Substitution

Application:

Apply definite integration directly over the domain interval.

Final Logic:

Upper bound minus lower bound gives 2.

🧠 Memory Trick

"From −1 to 1 spans 2 units."

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

12 Assertion (A): It is impossible to evaluate the real function sin⁻¹(2).
Reason (R): The strict domain of the sin⁻¹x function is limited to [-1, 1].

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

sin⁻¹x accepts inputs only in [-1,1]. 2 lies outside the domain. Hence real evaluation is impossible.

✅ Correct Answer Explanation

The inverse sine function is defined only for inputs in the interval [-1,1]. Since 2 lies outside this domain, sin⁻¹(2) has no real value. Therefore both the Assertion and Reason are true, and the Reason correctly explains the Assertion. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Both statements are mathematically true.
Option B → The reason is correct, not false.
Option D → Assertion is true because input 2 is invalid.

🎯 Strategy

Used: Elimination

Application:

Check the allowed domain of inverse sine carefully.

Final Logic:

Inputs outside [-1,1] produce no real inverse sine value.

🧠 Memory Trick

"sin⁻¹ works only between −1 and 1."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Domain Restrictions –

13 Order the following sequential valid branches of cos⁻¹x starting from the negative real axis moving sequentially toward the positive:
1. [0, π]
2. [-π, 0]
3. [π, 2π]
4. [2π, 3π]

A 1, 2, 3, 4

B 2, 1, 3, 4

C 3, 2, 1, 4

D 4, 3, 2, 1

💡 Short Explanation

Start from negative interval first. Move progressively toward positive direction. Ordering becomes 2,1,3,4.

✅ Correct Answer Explanation

The intervals arranged from the negative real axis toward the positive side are: [-π,0], [0,π], [π,2π], [2π,3π]. Thus the correct chronological ordering is: 2,1,3,4. Hence option B is correct. Other options disrupt the natural increasing sequence of intervals.

❓ Why Other Options Are Incorrect

Option A → Starts from nonnegative interval instead of negative axis.
Option C → Incorrectly places positive interval before negative interval.
Option D → Entire sequence is reversed incorrectly.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Arrange intervals according to increasing real-number direction.

Final Logic:

Negative interval precedes all positive intervals.

🧠 Memory Trick

"Negative first, then move rightward."

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

14 For the function y = cos⁻¹x located within its principal branch, what is the absolute maximum value y can achieve?

A π

B π/2

C 0

D 2π

💡 Short Explanation

Principal range is [0,π]. Maximum occurs at upper endpoint. Largest possible value is π.

✅ Correct Answer Explanation

The principal range of cos⁻¹x is [0,π]. Therefore the maximum possible output value is π, achieved when x=−1 since: cos⁻¹(−1)=π. Hence option A is correct. Other values either lie inside the interval or exceed the principal range.

❓ Why Other Options Are Incorrect

Option B → Represents midpoint of principal range only.
Option C → Minimum value, not maximum.
Option D → Outside the defined principal branch entirely.

🎯 Strategy

Used: Extreme Word Filter

Application:

Focus on the phrase "absolute maximum value."

Final Logic:

Upper endpoint of [0,π] is π.

🧠 Memory Trick

"Cos inverse tops at 180°."

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

15 The principal value of cos⁻¹(-1) lies strictly at the upper boundary of its defined branch. What is this exact radian value?

A 0

B -π

C π/2

D π

💡 Short Explanation

cosπ = −1 π lies in principal branch [0,π]. Therefore cos⁻¹(−1)=π.

✅ Correct Answer Explanation

The inverse cosine function returns principal values from [0,π]. Since: cos(π)=−1, the principal value becomes: cos⁻¹(−1)=π. Thus option D is correct. Options A and π/2 produce cosine values 1 and 0 respectively, while −π is outside the principal branch.

❓ Why Other Options Are Incorrect

Option A → cos0=1, not −1.
Option B → −π is not inside the principal range.
Option C → cos(π/2)=0, not −1.

🎯 Strategy

Used: Substitution

Application:

Recall the standard angle whose cosine equals −1.

Final Logic:

The principal angle for cosine value −1 is π.

🧠 Memory Trick

"Cosine becomes −1 at 180°."

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

16 As the input variable x sweeps sequentially from -1 to 1, the corresponding graph of the principal branch of y = cos⁻¹x:

A Remains completely constant

B Increases strictly throughout

C Decreases strictly throughout

D Oscillates between peaks

💡 Short Explanation

cos⁻¹x is monotonic decreasing. Output falls from π to 0. Graph decreases continuously.

✅ Correct Answer Explanation

For x increasing from −1 to 1: cos⁻¹(−1)=π, cos⁻¹(1)=0. Thus the graph continuously decreases throughout its domain. Hence option C is correct. The function neither remains constant nor oscillates periodically like cosine itself.

❓ Why Other Options Are Incorrect

Option A → Function values clearly change with x.
Option B → cos⁻¹x decreases instead of increasing.
Option D → Inverse cosine is not periodic or oscillatory.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Observe endpoint behaviour of inverse cosine graph.

Final Logic:

Outputs move downward from π to 0.

🧠 Memory Trick

"More x means smaller cos inverse."

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

17 If you sum the maximum attainable principal value of sin⁻¹x and the maximum attainable principal value of cos⁻¹x, the combined result is:

A π

B 3π/2

C π/2

D 2π

💡 Short Explanation

Max of sin⁻¹x is π/2. Max of cos⁻¹x is π. Sum equals 3π/2.

✅ Correct Answer Explanation

The principal range of sin⁻¹x is: [-π/2,π/2], so its maximum value is π/2. The principal range of cos⁻¹x is: [0,π], so its maximum value is π. Adding: π/2 + π = 3π/2. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores one of the maximum contributions.
Option C → Represents only maximum of sin⁻¹x.
Option D → Exceeds the actual combined value.

🎯 Strategy

Used: Substitution

Application:

Use principal ranges to identify maximum outputs.

Final Logic:

π/2 plus π gives 3π/2.

🧠 Memory Trick

"90° plus 180° equals 270°."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Comparison of Principal Ranges

18 Because both sin⁻¹x and cos⁻¹x identically share the restricted domain [-1, 1], their respective graphs are strictly confined horizontally between the lines:

A x = -1 and x = 1

B y = -1 and y = 1

C x = -π and x = π

D y = -π and y = π

💡 Short Explanation

Domain controls horizontal spread. Both inverse functions use domain [-1,1]. Graphs stay between x=−1 and x=1.

✅ Correct Answer Explanation

The domain of both sin⁻¹x and cos⁻¹x is [-1,1]. Domain determines the horizontal extent of a graph. Therefore both graphs remain confined between the vertical lines: x=−1 and x=1. Hence option A is correct. Other options refer to ranges instead of domains.

❓ Why Other Options Are Incorrect

Option B → Represents y-values, not horizontal confinement.
Option C → Domain does not extend to ±π.
Option D → These are angular ranges, not graph boundaries.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Differentiate between horizontal domain and vertical range.

Final Logic:

Domain boundaries create vertical limiting lines.

🧠 Memory Trick

"Domain controls x-direction."

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

📄 Passage

Like sine function, the cosine function is a function whose domain is the set of all real numbers and range is the set [-1, 1]. If we restrict the domain of cosine function to [0, π], then it becomes one-one and onto with range [-1, 1]. We can, therefore, define the inverse of cosine function in each of these intervals. The branch with range [0, π] is called the principal value branch.

Based directly on the passage and graphical mapping principles, the specified restriction to [0, π] guarantees what property for mapping the inverse?

A The inverse is symmetrically centered on the y-axis.

B The inverse perfectly reflects the x-axis.

C The inverse covers all possible real numbers.

D The inverse maps perfectly one-to-one to the domain [-1, 1].

💡 Short Explanation

Restriction makes cosine bijective. One-one mapping becomes possible. Inverse is then well-defined.

✅ Correct Answer Explanation

Restricting cosine to [0,π] makes it one-one and onto the interval [-1,1]. Therefore every value in [-1,1] corresponds to exactly one angle in [0,π]. This guarantees a valid inverse mapping. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Passage discusses bijection, not y-axis symmetry.
Option B → Reflection across x-axis is unrelated here.
Option C → Inverse cosine does not cover all real numbers.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Focus on the phrase "one-one and onto" in the passage.

Final Logic:

Bijection guarantees existence of inverse mapping.

🧠 Memory Trick

"One-one plus onto gives inverse."

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

📄 Passage

According to the standard reflection rule implied by the passage, the restricted domain [0, π] of the original cosine function becomes what specific parameter for the inverse function?

A Its independent domain

B Its bounding asymptote

C Its principal range

D Its exclusive x-intercept

💡 Short Explanation

Domain and range interchange in inverses. Restricted cosine domain becomes inverse range. Therefore it becomes principal range.

✅ Correct Answer Explanation

For inverse functions, domain and range interchange. Since cosine is restricted to [0,π] before inversion, this interval becomes the range of cos⁻¹x. NCERT calls this the principal value range or principal branch. Therefore option C is correct.

❓ Why Other Options Are Incorrect

Option A → Domain of cos⁻¹x is actually [-1,1].
Option B → Asymptotes are unrelated to cosine inverse here.
Option D → x-intercepts are not determined by branch restriction.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the inverse-function rule that swaps domain and range.

Final Logic:

Restricted original domain becomes inverse range.

🧠 Memory Trick

"Original domain becomes inverse range."

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

CUET UG Mathematics Booster Test 1 - Inverse Sin and Inverse Cosine Functions

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