CUET UG Mathematics Booster Test 1 - Properties of Inverse Trigonometric Functions -

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CUET UG Mathematics Booster Test 1 - Properties of Inverse Trigonometric Functions

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 1 - Properties of Inverse Trigonometric Functions

1 / 20

QUESTION 1 OF 20

The domain of y = cosec⁻¹ x is mathematically given as R - (-1, 1). Which of the following inequalities correctly bounds this domain?

Answer Review

1 The domain of y = cosec⁻¹ x is mathematically given as R - (-1, 1). Which of the following inequalities correctly bounds this domain?

A |x| < 1

B |x| ≤ 1

C |x| ≥ 1

D x > 1

💡 Short Explanation

cosec⁻¹x excludes values between −1 and 1. Domain includes endpoints ±1. Hence modulus inequality is |x| ≥ 1.

✅ Correct Answer Explanation

The domain of cosec⁻¹x is: R−(−1,1), which means all real numbers except values strictly between −1 and 1. This can be written compactly as: |x| ≥ 1. Thus option C correctly represents the domain condition.

❓ Why Other Options Are Incorrect

Option A → Represents excluded interval only.
Option B → Incorrectly includes interior values between −1 and 1.
Option D → Ignores negative valid values like −2.

🎯 Strategy

Used: Substitution

Application:

Convert interval notation into modulus inequality form.

Final Logic:

R−(−1,1) directly becomes |x| ≥ 1.

🧠 Memory Trick

"Outside one survives."

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

2 Match the inverse function to its appropriate restricted principal value branch range.

List 1	List 2
1. sin⁻¹	a. [−π/2, π/2] − {0}
2. cosec⁻¹	b. [−π/2, π/2]
3. tan⁻¹	c. (−π/2, π/2)
4. sec⁻¹	d. [0, π] − {π/2}

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

sin⁻¹ → closed symmetric range. tan⁻¹ → open symmetric range. sec⁻¹ and cosec⁻¹ exclude undefined points.

✅ Correct Answer Explanation

Principal ranges are: sin⁻¹x → [-π/2, π/2], cosec⁻¹x → [-π/2, π/2]−{0}, tan⁻¹x → (−π/2, π/2), sec⁻¹x → [0,π]−{π/2}. Thus the correct matching becomes: 1-b, 2-a, 3-c, 4-d.

❓ Why Other Options Are Incorrect

Option A → Swaps sin⁻¹ and cosec⁻¹ ranges.
Option C → Gives incorrect open interval for sin⁻¹.
Option D → sec⁻¹ does not use symmetric interval around zero.

🎯 Strategy

Used: Option Grouping

Application:

Recall standard principal branches directly.

Final Logic:

Only option B matches all standard inverse ranges correctly.

🧠 Memory Trick

"sin closed, tan open, sec skips π/2."

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

3 Which of the following evaluations yield a value that strictly lies inside the principal branch of cosec⁻¹?
(i) cosec⁻¹(2)
(ii) cosec⁻¹(-2)

A Both (i) and (ii)

B Only (i)

C Only (ii)

D Neither

💡 Short Explanation

cosec⁻¹(2)=π/6. cosec⁻¹(−2)=−π/6. Both belong to principal branch.

✅ Correct Answer Explanation

The principal branch of cosec⁻¹x is: [-π/2,π/2]−{0}. Now: cosec⁻¹(2)=π/6, cosec⁻¹(−2)=−π/6. Both values lie within the principal branch interval. Therefore both evaluations are valid and option A is correct.

❓ Why Other Options Are Incorrect

Option B → Ignores valid negative branch value.
Option C → Positive branch value is also valid.
Option D → Both expressions are properly defined.

🎯 Strategy

Used: Substitution

Application:

Evaluate inverse cosecant values directly.

Final Logic:

Both π/6 and −π/6 lie inside principal range.

🧠 Memory Trick

"±2 gives ±π/6."

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

4 Regarding the properties and the exclusion of zero for cosec⁻¹, identify the incorrect statement:

A The value y = 0 is included in the principal range of cosec⁻¹.

B The principal branch is [-π/2, π/2] - {0}.

C x = 0 is not in the domain of cosec⁻¹ x.

D The range excludes 0 because the original function is undefined at integral multiples of π.

💡 Short Explanation

0 is excluded from principal range. cosec 0 is undefined. Hence statement A is incorrect.

✅ Correct Answer Explanation

The principal branch of cosec⁻¹x is: [-π/2,π/2]−{0}. Since: cosec0 is undefined, 0 cannot belong to the range. Therefore statement A is mathematically incorrect, while the remaining statements are correct.

❓ Why Other Options Are Incorrect

Option B → Correct standard principal branch.
Option C → Domain excludes values between −1 and 1 including 0.
Option D → Undefined nature at multiples of π explains exclusion.

🎯 Strategy

Used: Elimination

Application:

Check whether zero belongs to principal branch.

Final Logic:

0 must be excluded because cosec0 is undefined.

🧠 Memory Trick

"No zero in cosec inverse."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Exclusion of Zero in cosec⁻¹x

5 A case requires evaluating sec⁻¹(x) precisely when x = -2. Find the numerical principal value.

A π/3

B -π/3

C 5π/6

D 2π/3

💡 Short Explanation

sec y = −2. Hence cos y = −1/2. Principal angle becomes 2π/3.

✅ Correct Answer Explanation

Given: sec⁻¹(−2)=y, then: sec y=−2, cos y=−1/2. In the principal range: [0,π]−{π/2}, the angle satisfying cos y=−1/2 is: 2π/3. Thus option D is correct.

❓ Why Other Options Are Incorrect

Option A → sec(π/3)=2, positive.
Option B → Negative angles not in principal range.
Option C → cos(5π/6)=−√3/2, not −1/2.

🎯 Strategy

Used: Substitution

Application:

Convert secant equation into cosine form.

Final Logic:

cos y=−1/2 corresponds to 2π/3.

🧠 Memory Trick

"Negative half cosine → 120°."

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

6 Assertion (A): sec⁻¹ x is a bijective mapping from R - (-1,1) to [0, π] - {π/2}.
Reason (R): The secant function is restricted to [0, π] - {π/2} to make it one-one and onto, guaranteeing the existence of its inverse.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

sec x restricted to principal branch. Restriction makes function bijective. Hence inverse exists properly.

✅ Correct Answer Explanation

The inverse secant function exists because sec x is restricted to: [0,π]−{π/2}, where it becomes one-one and onto for the range: R−(−1,1). Thus sec⁻¹x becomes a bijection from: R−(−1,1) → [0,π]−{π/2}. Both Assertion and Reason are true.

❓ Why Other Options Are Incorrect

Option A → Both statements are mathematically correct.
Option B → Reason correctly explains bijection.
Option D → Assertion is valid under principal branch restriction.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Connect inverse existence with bijective restriction.

Final Logic:

Restriction ensures one-one correspondence and inverse existence.

🧠 Memory Trick

"Restrict first, invert later."

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

7 When analyzing the visual graph of y = sec⁻¹x, where does the valid curve exist on the coordinate plane?

A In the region where strictly -1 < x < 1.

B In the region where x ≤ -1 or x ≥ 1, with y bounded between 0 and π (excluding π/2).

C Bounded exclusively in the positive x quadrants.

D Crossing exactly through the origin (0,0).

💡 Short Explanation

Domain excludes interval (−1,1). Principal range excludes π/2. Graph exists only outside ±1.

✅ Correct Answer Explanation

The domain of sec⁻¹x is: x≤−1 or x≥1. Its principal range is: [0,π]−{π/2}. Hence the graph exists only in those x-regions while y remains within the stated principal branch. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → Interior interval excluded from domain.
Option C → Negative x-values are also valid.
Option D → sec⁻¹(0) is undefined.

🎯 Strategy

Used: Elimination

Application:

Use standard domain and range restrictions.

Final Logic:

sec⁻¹x exists only outside the interval (−1,1).

🧠 Memory Trick

"sec inverse lives outside one."

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

8 Arrange the following outputs in ascending numerical order based on the principal branch of sec⁻¹x:
1. sec⁻¹(-1)
2. The point of discontinuity
3. sec⁻¹(1)

A 1, 2, 3

B 3, 2, 1

C 2, 3, 1

D 1, 3, 2

💡 Short Explanation

sec⁻¹(1)=0. Discontinuity occurs at π/2. sec⁻¹(−1)=π.

✅ Correct Answer Explanation

Principal values: sec⁻¹(1)=0, discontinuity point=π/2, sec⁻¹(−1)=π. Ascending order becomes: 0 < π/2 < π, which corresponds to: 3, 2, 1. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Reverses ascending order.
Option C → Places discontinuity before zero incorrectly.
Option D → π cannot come before π/2 numerically.

🎯 Strategy

Used: Substitution

Application:

Evaluate each inverse secant value numerically.

Final Logic:

0, π/2, π is the correct ascending sequence.

🧠 Memory Trick

"1 gives 0, −1 gives π."

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

9 A continuous random variable X is normally distributed from -∞ to ∞. If y = tan⁻¹(X) is computed for any valid draw, what is the theoretical probability that y successfully evaluates to a real number?

A 0

B 0.5

C -1

D 1

💡 Short Explanation

tan⁻¹x accepts all real numbers. Every draw is valid. Probability equals one.

✅ Correct Answer Explanation

The domain of tan⁻¹x is: R, meaning every real number input produces a valid real output. Since a normal distribution only produces real values, every draw from X is acceptable. Therefore the probability of successful evaluation is: 1.

❓ Why Other Options Are Incorrect

Option A → No restriction excludes inputs.
Option B → Half probability has no basis here.
Option C → Probability cannot be negative.

🎯 Strategy

Used: Extreme Word Filter

Application:

Check whether tan⁻¹ has any domain restriction.

Final Logic:

All real inputs are valid for tan⁻¹x.

🧠 Memory Trick

"tan inverse accepts all reals."

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

10 If a moving average filter tracks a signal defined by y = tan⁻¹x as the input x approaches positive infinity, what exact asymptotic value does the data converge to?

A 0

B π

C π/2

D -π/2

💡 Short Explanation

tan⁻¹x has horizontal asymptote. As x→∞, output approaches π/2. Never actually equals π/2.

✅ Correct Answer Explanation

For the inverse tangent function: y=tan⁻¹x, the principal range is: (−π/2,π/2). As: x→∞, the graph approaches the horizontal asymptote: y=π/2. Therefore the limiting asymptotic value is π/2.

❓ Why Other Options Are Incorrect

Option A → tan⁻¹x equals 0 only at x=0.
Option B → π lies outside principal range.
Option D → Negative asymptote occurs as x→−∞.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recall asymptotic behavior of inverse tangent graph.

Final Logic:

Positive infinity maps toward π/2.

🧠 Memory Trick

"Big tan inverse → π/2."

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

11 A vector creates an angle θ with the x-axis defined by θ = tan⁻¹(-√3). What represents the principal value of this directional angle?

A -π/3

B 2π/3

C π/3

D -π/6

💡 Short Explanation

tan(−π/3)=−√3. Principal range of tan⁻¹ is open symmetric interval. Therefore θ=−π/3.

✅ Correct Answer Explanation

The principal range of tan⁻¹x is: (−π/2, π/2). We need an angle in this interval whose tangent equals: −√3. Since: tan(−π/3)=−√3, the principal value becomes: −π/3. Thus option A is correct.

❓ Why Other Options Are Incorrect

Option B → Lies outside principal range of tan⁻¹.
Option C → Gives positive √3 instead of negative.
Option D → tan(−π/6)=−1/√3, not −√3.

🎯 Strategy

Used: Substitution

Application:

Check standard tangent values directly.

Final Logic:

Only −π/3 satisfies tan θ=−√3 within principal range.

🧠 Memory Trick

"√3 belongs to π/3 family."

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

12 A periodic waveform uses multiple valid interval branches of tan⁻¹ x. If you calculate the area under y = tan⁻¹ x for x ∈ using the principal branch, and compare it to the branch (π/2, 3π/2), which concept describes how the values structurally shift?

A Division by π

B Squaring the original domain

C The vertical range is completely unchanged

D Translation of the range values by multiples of π

💡 Short Explanation

Tangent is periodic with period π. Different branches differ by π shifts. Vertical outputs translate accordingly.

✅ Correct Answer Explanation

Different inverse tangent branches arise from restricting tan x to intervals differing by π. Hence inverse outputs differ by integer multiples of π. The structure remains similar but translated vertically. Therefore the correct interpretation is translation of range values by multiples of π.

❓ Why Other Options Are Incorrect

Option A → No division operation occurs.
Option B → Domains are shifted, not squared.
Option C → Different branches produce shifted ranges.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use periodicity property of tangent.

Final Logic:

tan(x+π)=tan x causes branch shifts by π.

🧠 Memory Trick

"Tan repeats every π."

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

13 Evaluating an integral of y = cot⁻¹ x from -∞ to ∞ requires understanding its domain, which is all real numbers R. If a boundary point requires evaluating the exact center at x = 0, what is cot⁻¹(0)?

A 0

B π

C -π/2

D π/2

💡 Short Explanation

cot(π/2)=0. Principal range of cot⁻¹ is (0,π). Hence cot⁻¹(0)=π/2.

✅ Correct Answer Explanation

The principal range of cot⁻¹x is: (0,π). We seek an angle in this interval whose cotangent equals zero. Since: cot(π/2)=0, the principal value becomes: π/2. Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → cot0 is undefined.
Option B → cotπ is undefined.
Option C → Negative values are outside principal range.

🎯 Strategy

Used: Substitution

Application:

Use standard cotangent identity.

Final Logic:

cot(π/2)=0 gives required principal value.

🧠 Memory Trick

"cot zero-cross at π/2."

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

14 The value of cot⁻¹(1/√3) must fall inside its strict principal range. What is this value?

A π/6

B π/2

C π/3

D 5π/6

💡 Short Explanation

cot(π/3)=1/√3. π/3 lies inside principal branch. Therefore cot⁻¹(1/√3)=π/3.

✅ Correct Answer Explanation

We require an angle θ in: (0,π), such that: cotθ=1/√3. Since: cot(π/3)=1/√3, the principal value is: π/3. Thus option C is correct.

❓ Why Other Options Are Incorrect

Option A → cot(π/6)=√3.
Option B → cot(π/2)=0.
Option D → cot(5π/6)=−√3.

🎯 Strategy

Used: Substitution

Application:

Recall standard cotangent values.

Final Logic:

π/3 uniquely gives cot value 1/√3.

🧠 Memory Trick

"cot and tan swap √3 positions."

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

15 Identify the incorrect statement about the cot⁻¹ branch:

A The principal branch includes the solid boundaries 0 and π.

B The range of the principal branch is strictly (0, π).

C cot⁻¹ is continuously defined for all real x.

D It acts as the inverse of the restricted cotangent function.

💡 Short Explanation

Principal range is open interval. Endpoints 0 and π excluded. Hence statement A is incorrect.

✅ Correct Answer Explanation

The principal range of cot⁻¹x is: (0,π), which excludes both endpoints. Therefore statement A is false. The remaining statements correctly describe cot⁻¹x as the inverse of restricted cotangent defined continuously over all real inputs.

❓ Why Other Options Are Incorrect

Option B → Correct standard principal range.
Option C → cot⁻¹ exists for every real number.
Option D → Inverse arises after restricting cotangent domain.

🎯 Strategy

Used: Extreme Word Filter

Application:

Check whether boundaries are included or excluded.

Final Logic:

Open interval means endpoints are excluded.

🧠 Memory Trick

"cot inverse keeps brackets open."

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

16 Apart from the principal branch (0, π), which other periodic intervals can strictly restrict the cotangent function to be bijective?
(i) (-π, 0)
(ii) (π, 2π)

A Only (i)

B Both (i) and (ii)

C Only (ii)

D Neither

💡 Short Explanation

cotangent repeats every π. Any interval of length π works. Both listed intervals are valid branches.

✅ Correct Answer Explanation

The cotangent function has period π and remains one-one over any open interval of length π avoiding discontinuities. Both: (−π,0) and (π,2π) satisfy this condition. Hence both can define valid inverse branches.

❓ Why Other Options Are Incorrect

Option A → Ignores second valid interval.
Option C → Ignores first valid interval.
Option D → Both intervals are legitimate restrictions.

🎯 Strategy

Used: Option Grouping

Application:

Use periodicity and interval length properties.

Final Logic:

Any π-length interval avoiding asymptotes works.

🧠 Memory Trick

"cot repeats every π."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Branches of cotangent Function

17 Assertion (A): The curve of y = tan⁻¹ x never crosses the origin (0,0) in its principal branch.
Reason (R): tan(0) = 1, hence tan⁻¹(1) = 0.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

tan⁻¹0=0, so graph crosses origin. tan0=0, not 1. Therefore both statements are false.

✅ Correct Answer Explanation

The graph of: y=tan⁻¹x passes through: (0,0), because: tan⁻¹0=0. Thus Assertion is false. Also: tan0=0, not 1. Hence the Reason is also false. Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → Assertion itself is false.
Option C → Both statements are not true.
Option D → Reason incorrectly states tan0=1.

🎯 Strategy

Used: Substitution

Application:

Directly evaluate tan0 and tan⁻¹0.

Final Logic:

Origin belongs to tan inverse graph.

🧠 Memory Trick

"tan zero is zero."

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

18 Match the geometrical features to the graph of y = cot⁻¹ x based on its properties.

List 1	List 2
1. x-axis bounds	a. π/2
2. y-intercept (at x = 0)	b. (−∞, ∞)
3. y-axis bounds	c. (0, π)

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

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

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

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

💡 Short Explanation

Domain of cot⁻¹ is all reals. cot⁻¹(0)=π/2. Principal range is (0,π).

✅ Correct Answer Explanation

For y=cot⁻¹x: domain → (−∞,∞), y-intercept → cot⁻¹0=π/2, range → (0,π). Thus matching becomes: 1-b, 2-a, 3-c. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → x-axis bounds are not π/2.
Option C → Swaps domain and range.
Option D → y-intercept incorrectly matched with range interval.

🎯 Strategy

Used: Option Grouping

Application:

Recall graph features systematically.

Final Logic:

Domain, intercept, and range uniquely determine option B.

🧠 Memory Trick

"cot inverse: all x, π/2 center."

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

📄 Passage

"The inverse trigonometric functions play an important role in calculus for they serve to define many integrals. For suitable values of domain... sin(sin⁻¹x) = x, x ∈ [-1, 1] and sin⁻¹(sin x) = x, x ∈ [-π/2, π/2]. Whenever no branch of an inverse trigonometric function is mentioned, we mean the principal value branch."

Based on the passage, if one evaluates sin(sin⁻¹(0.5)), the output is exactly 0.5. This property mathematically holds because 0.5 safely lies within which established domain?

A [-π/2, π/2]

B R

C [-1, 1]

D (0, π)

💡 Short Explanation

sin⁻¹x defined on [-1,1]. 0.5 belongs to this interval. Hence composition returns original value.

✅ Correct Answer Explanation

The identity: sin(sin⁻¹x)=x holds for: x∈[-1,1]. Since: 0.5∈[-1,1], the expression evaluates exactly to: 0.5. Therefore option C correctly identifies the required domain condition.

❓ Why Other Options Are Incorrect

Option A → Interval belongs to inverse range, not input domain.
Option B → sin⁻¹ not defined for all reals.
Option D → Incorrect interval for composition identity.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Identify correct domain from standard identity.

Final Logic:

Composition identity valid only on [-1,1].

🧠 Memory Trick

"sin inverse inputs stay between ±1."

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

📄 Passage

According to the passage, what specific term or condition is implied when an inverse trigonometric function is written without mentioning a specific branch?

A It is considered undefined.

B Any valid branch can be chosen.

C The domain strictly restricts it to negative values.

D It universally refers to the principal value branch.

💡 Short Explanation

Default inverse notation uses principal branch. Ensures uniqueness of values. Mentioned explicitly in passage.

✅ Correct Answer Explanation

NCERT defines that whenever no branch is specified for an inverse trigonometric function, the principal value branch is automatically assumed. This convention guarantees uniqueness and consistency in inverse evaluations. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Function remains well-defined using principal branch.
Option B → Arbitrary branch choice destroys uniqueness.
Option C → No restriction only to negative values exists.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use direct statement from passage.

Final Logic:

Unspecified inverse trig functions always imply principal branch.

🧠 Memory Trick

"No branch mentioned = principal branch."

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

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