CUET UG Mathematics Booster Test 2 - Advanced Applications of Inverse Trigonometric Functions -

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CUET UG Mathematics Booster Test 2 - Advanced Applications 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 2 - Advanced Applications of Inverse Trigonometric Functions

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QUESTION 1 OF 20

Evaluate:
sec⁻¹(2)

Answer Review

1 Evaluate:
sec⁻¹(2)

A π/6

B π/3

C π/2

D 2π/3

💡 Short Explanation

sec θ = 2 implies cos θ = 1/2. Principal range of sec⁻¹x is [0, π] excluding π/2. cos(π/3)=1/2.

✅ Correct Answer Explanation

To evaluate sec⁻¹(2), rewrite as: sec θ = 2 ⇒ cos θ = 1/2. The principal value range of sec⁻¹x is [0, π], excluding π/2. In this interval, the angle satisfying cos θ = 1/2 is π/3. Therefore sec⁻¹(2)=π/3. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → sec(π/6)=2/√3, not 2.
Option C → sec(π/2) is undefined.
Option D → cos(2π/3)=-1/2, giving sec=-2.

🎯 Strategy

Used: Substitution

Application:

Convert secant into cosine and use standard values.

Final Logic:

cos θ = 1/2 gives θ = π/3.

🧠 Memory Trick

"sec 2 → cos 1/2 → π/3."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: sec⁻¹ Function

2 Which of the following are equivalent to cot⁻¹x for x>0?
1. tan⁻¹(1/x)
2. π/2 − tan⁻¹x
3. π − tan⁻¹x

A 1 only

B 2 only

C 1 and 2 only

D 1, 2, and 3

💡 Short Explanation

cot⁻¹x = tan⁻¹(1/x) for x>0. Also equals π/2 − tan⁻¹x. π − tan⁻¹x exceeds principal range.

✅ Correct Answer Explanation

For positive x: cot⁻¹x = tan⁻¹(1/x) and cot⁻¹x = π/2 − tan⁻¹x. However, π − tan⁻¹x generally lies outside the principal branch (0, π) for standard equivalence conditions. Hence only statements 1 and 2 are valid. Therefore option C is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores second correct identity.
Option B → Ignores tan⁻¹(1/x) equivalence.
Option D → Third expression is not generally equivalent.

🎯 Strategy

Used: Option Grouping

Application:

Use standard inverse trigonometric identities together.

Final Logic:

Only statements 1 and 2 satisfy cotangent identities.

🧠 Memory Trick

"cot = reciprocal tan."

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

3 Simplify:
sin(cos⁻¹x), where -1≤x≤1

A √(1−x²)

B 1−x²

C x

D −√(1−x²)

💡 Short Explanation

Let θ = cos⁻¹x. Then cos θ = x. Using Pythagoras, sin θ = √(1−x²).

✅ Correct Answer Explanation

If θ = cos⁻¹x, then θ lies in [0, π], where sine is nonnegative. Since cos θ = x, using: sin²θ + cos²θ = 1, we get: sin θ = √(1−x²). Thus: sin(cos⁻¹x)=√(1−x²). Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Missing square root.
Option C → Represents cosine value, not sine.
Option D → Negative sign invalid in principal interval [0,π].

🎯 Strategy

Used: Substitution

Application:

Replace inverse expression with an angle variable.

Final Logic:

Right-triangle identity gives positive square root.

🧠 Memory Trick

"sin(arccos x) → triangle root."

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

4 Arrange the correct steps to simplify:
sin(tan⁻¹x)
1. Let θ=tan⁻¹x ⇒ tanθ=x
2. Construct a right triangle with opposite = x, adjacent = 1
3. Hypotenuse becomes √(1+x²)
4. Then sinθ=x/√(1+x²)

A 1, 2, 3, 4

B 4, 3, 2, 1

C 2, 1, 4, 3

D 2, 4, 1, 3

💡 Short Explanation

Start with substitution. Build right triangle from tangent ratio. Use Pythagoras to find hypotenuse.

✅ Correct Answer Explanation

Correct simplification sequence: First let θ=tan⁻¹x, so tanθ=x. Then construct a triangle with opposite=x and adjacent=1. Using Pythagoras, hypotenuse=√(1+x²). Finally: sinθ = opposite/hypotenuse = x/√(1+x²). Thus the correct order is 1,2,3,4. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Reverses logical derivation order.
Option C → Triangle construction before defining θ is incomplete.
Option D → Uses final result before derivation.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Follow natural mathematical derivation sequence.

Final Logic:

Substitution must precede geometric construction.

🧠 Memory Trick

"Define → draw → compute → conclude."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Simplification Techniques

5 Assertion (A): sin(sin⁻¹x)=x
Reason (R): This identity holds for all x∈[-1,1]

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

arcsin domain is [-1,1]. Composition restores original input. Reason correctly explains assertion.

✅ Correct Answer Explanation

The identity: sin(sin⁻¹x)=x is valid because sin⁻¹x is defined only for x ∈ [-1,1]. Over this domain, arcsine produces the angle whose sine equals x. Thus both 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 actually correct.
Option B → Reason is not false.
Option D → Assertion is a standard identity.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Check domain restriction behind inverse identities.

Final Logic:

Reason directly justifies the identity.

🧠 Memory Trick

"arcsin returns back x."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Basic Identities

6 A point (a,b) lies on the graph of y=sin⁻¹x. What is the corresponding point on the graph of y=sinx representing this inverse relation?

A (a,b)

B (b,a)

C (-a,b)

D (b,-a)

💡 Short Explanation

Inverse functions interchange coordinates. Reflection occurs across y=x. x and y coordinates swap positions.

✅ Correct Answer Explanation

Graphs of inverse functions are reflections across the line y=x. Therefore, if (a,b) lies on y=sin⁻¹x, then the corresponding point on y=sinx is obtained by interchanging coordinates: (b,a). Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Same point does not represent inverse mapping.
Option C → Only changes sign of x-coordinate.
Option D → Incorrect sign reversal.

🎯 Strategy

Used: Odd One Out

Application:

Identify inverse-function coordinate swapping property.

Final Logic:

Inverse graphs interchange x and y.

🧠 Memory Trick

"Inverse means flip coordinates."

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

7 For the function y=tan⁻¹x, which of the following horizontal lines is NOT included in its range?

A y=0

B y=π/4

C y=−π/4

D y=π/2

💡 Short Explanation

arctan range is (-π/2, π/2). Endpoints excluded from range. π/2 is asymptotic only.

✅ Correct Answer Explanation

The principal range of tan⁻¹x is: (-π/2, π/2). Therefore values like 0, π/4, and −π/4 are included, but π/2 is excluded because it is only a horizontal asymptote. Thus option D is correct.

❓ Why Other Options Are Incorrect

Option A → Zero lies inside principal range.
Option B → π/4 is attainable at x=1.
Option C → −π/4 occurs at x=−1.

🎯 Strategy

Used: Elimination

Application:

Recall open interval endpoints for arctangent.

Final Logic:

π/2 is excluded from arctan range.

🧠 Memory Trick

"arctan never reaches ±π/2."

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

8 To define integrals involving tan⁻¹x, what is its principal value range?

A (-π/2,π/2)

B [-π/2,π/2]

C (0,π)

D (-π,π)

💡 Short Explanation

tan⁻¹x uses open interval endpoints. Tangent undefined at ±π/2. Principal branch excludes boundaries.

✅ Correct Answer Explanation

The principal value branch of tan⁻¹x is: (-π/2, π/2). Endpoints are excluded because tangent is undefined at ±π/2. This interval ensures tangent remains one-one and invertible. Therefore option A correctly gives the principal value range.

❓ Why Other Options Are Incorrect

Option B → Endpoints incorrectly included.
Option C → This is cotangent-type range.
Option D → Tangent is not one-one there.

🎯 Strategy

Used: Elimination

Application:

Recall standard principal branch definitions.

Final Logic:

tan⁻¹ range is open between ±π/2.

🧠 Memory Trick

"tan opens, never touches ±π/2."

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

9 Solve for x∈[0,2π]:
sinx=cosx

A π/4,5π/4

B π/2,3π/2

C 0,π

D π/3,4π/3

💡 Short Explanation

Divide by cosx when possible. tanx=1. Solutions repeat every π.

✅ Correct Answer Explanation

Given: sinx = cosx. Dividing by cosx: tanx = 1. General solution: x = π/4 + nπ. In [0,2π], solutions are: π/4 and 5π/4. Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → Gives sinx ≠ cosx.
Option C → Produces tanx=0.
Option D → Corresponds to tanx=√3.

🎯 Strategy

Used: Substitution

Application:

Convert equation into tangent form.

Final Logic:

tanx=1 gives π/4+nπ.

🧠 Memory Trick

"sin = cos at 45°."

📚 Class XI – NCERT Mathematics – Chapter: Trigonometric Functions – Topic: Trigonometric Equations

10 Solve:
tan⁻¹x=π/4

A x=0

B x=1

C x=−1

D x=√3

💡 Short Explanation

Apply tangent on both sides. tan(π/4)=1. Hence x=1.

✅ Correct Answer Explanation

Given: tan⁻¹x = π/4. Applying tangent to both sides: x = tan(π/4). Since tan(π/4)=1, we get: x=1. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → tan⁻¹0=0.
Option C → tan⁻¹(−1)=−π/4.
Option D → tan⁻¹(√3)=π/3.

🎯 Strategy

Used: Substitution

Application:

Apply direct inverse-function property.

Final Logic:

arctan value π/4 corresponds to x=1.

🧠 Memory Trick

"tan 45° = 1."

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

11 Which of the following is INCORRECT (considering principal value branches)?

A sin(sin⁻¹x)=x, x∈[-1,1]

B cos(cos⁻¹x)=x, x∈[-1,1]

C tan(tan⁻¹x)=x, x∈R

D sin⁻¹(sinx)=x, x∈R

💡 Short Explanation

Inverse identities require restricted domains. sin⁻¹(sinx)=x only on [-π/2, π/2]. Statement D ignores restriction.

✅ Correct Answer Explanation

The identity sin⁻¹(sinx)=x is not valid for all real x because sine is periodic and not one-one everywhere. It holds only for x ∈ [-π/2, π/2]. Statements A, B, and C are standard inverse identities valid on their proper domains. Hence option D is incorrect.

❓ Why Other Options Are Incorrect

Option A → Correct standard identity on [-1,1].
Option B → Valid because cos⁻¹x maps correctly over [-1,1].
Option C → tan and tan⁻¹ are inverses over all real numbers.

🎯 Strategy

Used: Elimination

Application:

Check each inverse identity with its valid restricted domain.

Final Logic:

Only option D ignores principal branch restrictions.

🧠 Memory Trick

"sin⁻¹(sinx) works only in principal zone."

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

12 Match each expression to its principal value:

List I	List II
1. sin⁻¹(sin 2π/3)	a. π/3
2. cos⁻¹(cos 5π/3)	b. 2π/3
3. tan⁻¹(tan 3π/4)	c. -π/4
4. sin⁻¹(sin 7π/6)	d. -π/6

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

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

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

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

💡 Short Explanation

Use principal branch ranges. Convert angles into allowed intervals. Match equivalent principal values carefully.

✅ Correct Answer Explanation

sin⁻¹(sin 2π/3)=π/3 since arcsin range is [-π/2,π/2]. cos⁻¹(cos 5π/3)=2π/3 since arccos range is [0,π]. tan⁻¹(tan 3π/4)=-π/4 because arctan range is (-π/2,π/2). sin⁻¹(sin 7π/6)=-π/6. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrectly maps first and fourth values.
Option B → Principal values mismatched entirely.
Option D → Does not respect standard inverse ranges.

🎯 Strategy

Used: Substitution

Application:

Reduce each angle into the principal value branch.

Final Logic:

Correct branch adjustment gives option C.

🧠 Memory Trick

"Bring angles back to principal interval."

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

13 Find the set of lower bounds of principal value ranges for:
sin⁻¹x, cos⁻¹x, tan⁻¹x

A {-π/2,0,-π/2}

B {-π/2,0,-π/2}

C {0,-π/2,-π/2}

D {-π,0,-π}

💡 Short Explanation

arcsin range starts at -π/2. arccos starts at 0. arctan starts at -π/2.

✅ Correct Answer Explanation

Principal ranges are: sin⁻¹x → [-π/2, π/2], cos⁻¹x → [0, π], tan⁻¹x → (-π/2, π/2). Their lower bounds are therefore: {-π/2, 0, -π/2}. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Duplicate of correct answer formatting variation.
Option C → Incorrectly swaps cosine lower bound.
Option D → Uses ranges not associated with principal branches.

🎯 Strategy

Used: Option Grouping

Application:

Recall standard principal value intervals together.

Final Logic:

Only option B lists correct lower limits.

🧠 Memory Trick

"sin and tan start at −π/2; cos starts at 0."

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

14 Which inverse trigonometric function has domain R?

A tan⁻¹x

B sin⁻¹x

C cos⁻¹x

D sec⁻¹x

💡 Short Explanation

tan⁻¹x accepts all real numbers. sin⁻¹x and cos⁻¹x need [-1,1]. sec⁻¹x excludes (-1,1).

✅ Correct Answer Explanation

The domain of tan⁻¹x is all real numbers R because tangent covers all real outputs on its principal branch. In contrast, sin⁻¹x and cos⁻¹x are restricted to [-1,1], while sec⁻¹x excludes values between -1 and 1. Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → Restricted to [-1,1].
Option C → Also restricted to [-1,1].
Option D → Domain excludes (-1,1).

🎯 Strategy

Used: Elimination

Application:

Compare standard domains of inverse trigonometric functions.

Final Logic:

Only tan⁻¹x has unrestricted real domain.

🧠 Memory Trick

"tan⁻¹ takes every real input."

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

15 What happens if inverse trigonometric functions are not restricted to principal branches?

A They become linear

B They become constant

C They are no longer one–one (not invertible)

D They map only to negative infinity

💡 Short Explanation

Inverse functions need bijections. Trigonometric functions repeat values periodically. Restrictions create invertibility.

✅ Correct Answer Explanation

Without restricting trigonometric functions to principal branches, they remain periodic and repeat outputs. Hence they fail the one-one condition necessary for inverses. Restricting domains makes them bijective and allows inverse trigonometric functions to exist uniquely. Therefore option C is correct.

❓ Why Other Options Are Incorrect

Option A → No linear behavior results from restrictions.
Option B → Functions do not become constant.
Option D → No connection with negative infinity exists.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the inverse-function requirement of injectivity.

Final Logic:

Periodic repetition destroys invertibility.

🧠 Memory Trick

"No restriction → no inverse."

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

16 Evaluate:
sin⁻¹(1/2)+cos⁻¹(1/2)

A π/6

B π/3

C π

D π/2

💡 Short Explanation

sin⁻¹(1/2)=π/6. cos⁻¹(1/2)=π/3. Sum equals π/2.

✅ Correct Answer Explanation

Using principal values: sin⁻¹(1/2)=π/6 and cos⁻¹(1/2)=π/3. Adding: π/6 + π/3 = π/6 + 2π/6 = 3π/6 = π/2. This also follows from the identity: sin⁻¹x + cos⁻¹x = π/2. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Only equals arcsine value.
Option B → Only equals arccosine value.
Option C → Sum is much smaller than π.

🎯 Strategy

Used: Substitution

Application:

Use standard special-angle inverse values.

Final Logic:

Adding π/6 and π/3 gives π/2.

🧠 Memory Trick

"arcsin + arccos = 90°."

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

17 The work Yuktibhasa is associated with which tradition?

A Aryabhata

B Brahmagupta

C A 16th-century Malayalam author tradition (Kerala School)

D Bhaskara I

💡 Short Explanation

Yuktibhasa belongs to Kerala School. Written in Malayalam tradition. Known for infinite series proofs.

✅ Correct Answer Explanation

Yuktibhasa is a famous mathematical work from the Kerala School of Mathematics written in Malayalam during the 16th century. It contains important derivations of infinite series expansions and advanced trigonometric ideas. Therefore option C correctly identifies its tradition.

❓ Why Other Options Are Incorrect

Option A → Aryabhata predates Yuktibhasa by centuries.
Option B → Brahmagupta was unrelated to Kerala School texts.
Option D → Bhaskara I also belonged to an earlier era.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recall historical associations of Kerala School mathematics.

Final Logic:

Yuktibhasa is uniquely linked to Malayalam mathematical tradition.

🧠 Memory Trick

"Yuktibhasa → Kerala School."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Historical Notes

18 Before the Indian approach was adopted in Europe, knowledge was primarily transmitted through:

A China

B Egypt

C Greece

D Arabia

💡 Short Explanation

Indian mathematics spread westward. Arabia acted as transmission center. Europe later adopted these methods.

✅ Correct Answer Explanation

Historical accounts mention that mathematical knowledge from India first traveled to Arabia and then reached Europe. Arabian scholars preserved and transmitted Indian trigonometric developments. Therefore option D correctly identifies the transmission route before European adoption.

❓ Why Other Options Are Incorrect

Option A → China is not the historical transmission route here.
Option B → Egypt played no stated intermediary role.
Option C → Greece developed separate earlier methods.

🎯 Strategy

Used: Elimination

Application:

Use the historical flow India → Arabia → Europe.

Final Logic:

Arabia served as the transmission bridge.

🧠 Memory Trick

"India → Arabia → Europe."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Historical Development

19 Aryabhata's foundational work in trigonometry is dated around:

A 1114 A.D.

B 476 A.D.

C 598 A.D.

D 600 A.D.

💡 Short Explanation

Aryabhata was earliest among listed scholars. 476 A.D. is historically associated with him. Important Indian trigonometric contributor.

✅ Correct Answer Explanation

Aryabhata's contributions to mathematics and trigonometry are historically dated around 476 A.D. His works strongly influenced later Indian and global mathematical developments. Hence option B correctly identifies the associated year.

❓ Why Other Options Are Incorrect

Option A → Refers to Bhaskara II.
Option C → Associated with Brahmagupta.
Option D → Associated with Bhaskara I.

🎯 Strategy

Used: Elimination

Application:

Match mathematicians with their known dates.

Final Logic:

Only 476 A.D. corresponds to Aryabhata.

🧠 Memory Trick

"Aryabhata → 476."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Historical Notes

20 Bhaskara II provided exact trigonometric values such as:

A sin30°=1/2

B sin45°=1/√3

C cos60°=1/√2

D sin90°=0

💡 Short Explanation

sin30° equals 1/2 exactly. Other listed trigonometric values are incorrect. Standard special-angle result.

✅ Correct Answer Explanation

The exact trigonometric value: sin30°=1/2 is correct. However: sin45°=1/√2, not 1/√3; cos60°=1/2, not 1/√2; sin90°=1, not 0. Therefore option A is the only correct statement.

❓ Why Other Options Are Incorrect

Option B → Incorrect value for sin45°.
Option C → cos60° equals 1/2, not 1/√2.
Option D → sin90° equals 1.

🎯 Strategy

Used: Elimination

Application:

Recall standard trigonometric ratios for special angles.

Final Logic:

Only sin30° correctly equals 1/2.

🧠 Memory Trick

"30–60–90 starts with 1/2."

📚 Class XI – NCERT Mathematics – Chapter: Trigonometric Functions – Topic: Standard Trigonometric Values

CUET UG Mathematics Booster Test 2 - Advanced Applications of Inverse Trigonometric Functions

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