CUET UG Applied Mathematics Booster Test 3 - Advanced Arithmetic Functions -

🚀

CUET UG Applied Mathematics Booster Test 3 - Advanced Arithmetic Functions

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 3 - Advanced Arithmetic Functions

1 / 20

QUESTION 1 OF 20

Determine the fundamental reason why phi(1) = 1 according to the definition of Euler's totient function.

Answer Review

1 Determine the fundamental reason why phi(1) = 1 according to the definition of Euler's totient function.

A Because 1 is a prime number.

B Because 1 is the only strictly positive integer less than or equal to 1 that is coprime to 1 (gcd(1, 1) = 1).

C Because it is an exception to the rule.

D Because 1 has zero prime factors.

💡 Short Explanation

�� Euler's totient function counts positive integers that are coprime to n. �� For n = 1, the only positive integer less than or equal to 1 is 1. �� Since gcd(1, 1) = 1, phi(1) = 1.

✅ Correct Answer Explanation

Euler's totient function is defined as the number of positive integers x such that 1 ≤ x ≤ n and gcd(x, n) = 1. For n = 1, the only positive integer satisfying 1 ≤ x ≤ 1 is x = 1. Since gcd(1, 1) = 1, the number 1 is coprime to itself. Therefore, phi(1) = 1. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because 1 is neither a prime number nor a composite number.
�� Option C → Incorrect because phi(1) = 1 follows directly from the definition of Euler's totient function and is not an exception.
�� Option D → Incorrect because the number of prime factors does not determine the value of the totient function.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply the definition of Euler's totient function directly to n = 1.

Final Logic:

Only one positive integer satisfies the coprime condition for n = 1.

🧠 Memory Trick

Only 1 counts for phi(1).

📚 Class XI – Mathematics – Chapter: Number Theory – Topic: Euler's Totient Function

2 Match the value of n = 12 with its corresponding arithmetic function values.

List 1	List 2
1. phi(12)	a. 0
2. tau(12)	b. 28
3. sigma(12)	c. 4
4. mu(12)	d. 6

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

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

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

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

💡 Short Explanation

�� phi(12) = 4. �� tau(12) = 6 and sigma(12) = 28. �� mu(12) = 0 because 12 contains a repeated prime factor.

✅ Correct Answer Explanation

First express 12 = 2^2 × 3. Now evaluate each arithmetic function. • phi(12) phi(12) = 12 × (1 - 1/2) × (1 - 1/3) = 12 × 1/2 × 2/3 = 4. Therefore, 1 → c. • tau(12) tau(12) = (2 + 1)(1 + 1) = 3 × 2 = 6. Therefore, 2 → d. • sigma(12) The positive divisors of 12 are 1, 2, 3, 4, 6 and 12. Their sum is 1 + 2 + 3 + 4 + 6 + 12 = 28. Therefore, 3 → b. • mu(12) Since 12 = 2^2 × 3 contains the repeated prime factor 2^2, mu(12) = 0. Therefore, 4 → a. Thus, the correct matching is 1 → c 2 → d 3 → b 4 → a Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect because all four arithmetic function values are mismatched.
�� Option C → Incorrect because phi(12) equals 4, not 6.
�� Option D → Incorrect because sigma(12) equals 28, not 6.

🎯 Strategy

Used: Option Grouping

Application:

Evaluate each arithmetic function separately and then compare the complete mapping.

Final Logic:

Only Option A correctly matches all four arithmetic function values.

🧠 Memory Trick

For 12: 4 - 6 - 28 - 0.

📚 Class XI – Mathematics – Chapter: Number Theory – Topic: Arithmetic Functions

3 If p = 11, which of the following expressions correctly reflect the properties of simple arithmetic functions applied to this prime?
1. phi(11) = 10
2. tau(11) = 2
3. sigma(11) = 12
4. mu(11) = -1

A 1, 2

B 1, 2, 3

C 2, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

�� Standard arithmetic function formulas apply directly to prime numbers. �� Every prime has exactly two divisors. �� The Möbius function of a prime number is -1.

✅ Correct Answer Explanation

For the prime number p = 11, evaluate each statement. • Statement 1: phi(11) = 11 - 1 = 10. Hence, Statement 1 is correct. • Statement 2: The divisors of 11 are 1 and 11. Therefore, tau(11) = 2. Hence, Statement 2 is correct. • Statement 3: sigma(11) = 1 + 11 = 12. Hence, Statement 3 is correct. • Statement 4: Since 11 has exactly one distinct prime factor, mu(11) = (-1)^1 = -1. Hence, Statement 4 is correct. Therefore, all four statements are correct. Hence, Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because Statements 3 and 4 are also correct.
�� Option B → Incorrect because it omits Statement 4.
�� Option C → Incorrect because it omits Statement 1.

🎯 Strategy

Used: Substitution

Application:

Evaluate each arithmetic function directly using the standard formulas for a prime number.

Final Logic:

All four statements are true for p = 11, so Option D is correct.

🧠 Memory Trick

Prime p ⇒ phi = p - 1, tau = 2, sigma = p + 1, mu = -1.

📚 Class XI – Mathematics – Chapter: Number Theory – Topic: Arithmetic Functions of Prime Numbers

4 Which of the following is an INCORRECT interpretation of the bounds and nature of φ(n)?

A For any n > 1, 1 ≤ φ(n) < n.

B φ(p) = p - 1 for any prime p.

C φ(n) represents the sum of positive divisors of n.

D φ(n) is considered a multiplicative function.

💡 Short Explanation

�� φ(n) counts coprimes �� σ(n) sums divisors �� φ(n) is multiplicative

✅ Correct Answer Explanation

Euler's totient function counts positive integers coprime to n. It does not sum divisors. The divisor-sum function is: Thus Option C is incorrect. Other statements are valid: and for primes: Also, φ(n) is multiplicative for coprime arguments.

❓ Why Other Options Are Incorrect

�� Option A → Correct standard bound for φ(n).
�� Option B → True property for primes.
�� Option D → Euler totient is multiplicative.

🎯 Strategy

Used: Odd One Out

Application: Identify which statement mismatches the definition of φ(n).

Final Logic: Summing divisors belongs to σ(n), not φ(n)

🧠 Memory Trick

"Phi counts, Sigma sums."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Euler Totient Properties

5 Calculate the mixture of arithmetic operations required to find the value of: φ(90) + τ(42) + σ(72) + μ(70). First, what is φ(90)?

A 24

B 12

C 48

D 90

💡 Short Explanation

�� Factorize 90 �� Apply Euler product formula �� Final value is 24

✅ Correct Answer Explanation

Prime factorization: Apply Euler's formula: Simplifying: Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incomplete multiplication.
�� Option C → Incorrect simplification.
�� Option D → φ(n) is less than n for n>1.

🎯 Strategy

Used: Substitution

Application: Substitute prime factors into Euler's formula.

Final Logic: Product simplifies to 24.

🧠 Memory Trick

"90 → ×1/2 ×2/3 ×4/5."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Euler Product Formula

6 The constraint formula φ(n) = n(1 - 1/x)(1 - 1/y) requires x and y to be distinct prime factors of n. If n = 100, what are x and y?

A 2 and 50

B 4 and 25

C 2 and 5

D 10 and 10

💡 Short Explanation

�� Prime factorization of 100 is needed �� Distinct prime factors are used �� 100=2²×5²

✅ Correct Answer Explanation

Prime factorization: Euler's totient formula uses distinct prime factors only: Thus the distinct primes are: Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → 50 is not prime.
�� Option B → 4 and 25 are composite numbers.
�� Option D → 10 is composite and not a prime factor.

🎯 Strategy

Used: Elimination

Application: Keep only distinct prime factors.

Final Logic: Distinct prime divisors of 100 are 2 and 5.

🧠 Memory Trick

"Totient uses distinct primes only."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Euler Totient Formula

7 To exactly calculate τ(42), we factorize 42 = 2 × 3 × 7. What is τ(42)?

A 3

B 6

C 8

D 12

💡 Short Explanation

�� Use divisor-count formula �� Exponents are all 1 �� Multiply (1+1)(1+1)(1+1)

✅ Correct Answer Explanation

Prime factorization: Number-of-divisors formula: Thus: Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Too few divisors counted.
�� Option B → Misses combinations of factors.
�� Option D → Overcounts divisor combinations.

🎯 Strategy

Used: Substitution

Application: Insert exponents into divisor-count formula.

Final Logic: 2×2×2=8.

🧠 Memory Trick

"Three primes → 2×2×2."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Divisor Counting Function

8 Moving to the prime relation identity, if p is a prime number, what is the value of the expression τ(p) + σ(p) mathematically equivalent to?

A p + 3

B p + 2

C p + 1

D p

💡 Short Explanation

�� τ(p)=2 for primes �� σ(p)=p+1 �� Sum becomes p+3

✅ Correct Answer Explanation

For prime number p: and: Therefore: Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Misses one extra unit.
�� Option C → Represents only σ(p).
�� Option D → Ignores divisor contributions.

🎯 Strategy

Used: Substitution

Application: Replace τ(p) and σ(p) using prime identities.

Final Logic: 2+(p+1)=p+3.

🧠 Memory Trick

"Tau gives 2; sigma gives p+1."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Prime Arithmetic Identities

9 When computing σ(72) where 72 = 2^3 × 3^2, what is the actual sum of all its positive divisors?

A 72

B 195

C 144

D 100

💡 Short Explanation

�� Use sigma product formula �� Separate powers of 2 and 3 �� Multiply divisor sums

✅ Correct Answer Explanation

Prime factorization: Sigma formula: Thus: So: Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Equals n only, not divisor sum.
�� Option C → Incorrect multiplication.
�� Option D → Underestimates divisor total.

🎯 Strategy

Used: Substitution

Application: Apply sigma formula directly to prime powers.

Final Logic: 15×13=195.

🧠 Memory Trick

"72 → 15×13."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Divisor Sum Function

10 Based on the prime identity φ(p) + τ(p) = σ(p), evaluate the vector components for p = 13.

A 12 + 2 = 14

B 13 + 1 = 14

C 12 + 1 = 13

D 13 + 2 = 15

💡 Short Explanation

�� φ(13)=12 �� τ(13)=2 �� σ(13)=14

✅ Correct Answer Explanation

For prime number: Euler totient: Divisor count: Sigma value: Hence: Thus Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Uses incorrect φ(13).
�� Option C → τ(13) is not 1.
�� Option D → Incorrect sigma relation.

🎯 Strategy

Used: Substitution

Application: Evaluate φ(p), τ(p), and σ(p) for p=13.

Final Logic: Prime identities give 12+2=14.

🧠 Memory Trick

"For prime 13: 12, 2, 14."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Prime Arithmetic Function Identities

11 Which area of number theory relies heavily on the fact that prime numbers p exclusively possess exactly two distinct positive divisors, locking τ(p) = 2?

A Geometry bounds

B Coprime limits

C Simple Arithmetic Functions

D Integrals

💡 Short Explanation

�� Prime numbers have exactly two divisors �� τ(p) = 2 is a divisor-function property �� This belongs to arithmetic function theory

✅ Correct Answer Explanation

The divisor-counting function is defined as: For a prime number p, the only divisors are: Hence: This property is fundamentally studied under simple arithmetic functions in number theory, where functions like φ(n), τ(n), σ(n), and μ(n) are analyzed. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Geometry bounds are unrelated to divisor-counting functions.
�� Option B → Coprime limits mainly concern gcd and totient concepts, not τ(p)=2 specifically.
�� Option D → Integrals belong to calculus, not arithmetic function theory.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match the divisor property τ(p)=2 with the branch of number theory discussing arithmetic functions.

Final Logic: τ(p)=2 is a standard arithmetic-function property.

🧠 Memory Trick

"Tau belongs to arithmetic functions."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Arithmetic Functions

12 If the integral definition of σ(n) maps n to the sum of its divisors, what uniquely defines σ(n) for a composite number compared to a prime?

A σ(n) = n + 1

B σ(n) > n + 1

C σ(n) < n

D σ(n) = 0

💡 Short Explanation

�� Prime numbers have only two divisors �� Composite numbers have additional divisors �� Extra divisors increase σ(n)

✅ Correct Answer Explanation

For a prime number p: because its only divisors are 1 and p. However, a composite number has extra divisors besides 1 and itself. Therefore: For example: Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → σ(n)=n+1 is true only for primes.
�� Option C → Sum of divisors is generally greater than n for composite numbers.
�� Option D → σ(n) can never be zero since 1 and n are always divisors.

🎯 Strategy

Used: Option Grouping

Application: Compare divisor structures of primes and composite numbers.

Final Logic: Composite numbers have additional divisors, making σ(n) exceed n+1.

🧠 Memory Trick

"Composite → extra divisors → bigger sigma."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Divisor Sum Function

13 If an integer n has 5 distinct prime factors and no repeated prime factors, what is μ(n)?

A 1

B -1

C 0

D 5

💡 Short Explanation

�� Möbius function uses distinct prime count �� μ(n)=(-1)^k for square-free n �� Here k=5

✅ Correct Answer Explanation

For a square-free integer with k distinct prime factors: Since n has 5 distinct prime factors: Therefore: Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Would occur when k is even.
�� Option C → μ(n)=0 only when repeated prime factors exist.
�� Option D → Möbius values are restricted to −1, 0, or 1.

🎯 Strategy

Used: Substitution

Application: Substitute k=5 into μ(n)=(-1)^k.

Final Logic: Odd number of distinct primes gives −1.

🧠 Memory Trick

"Odd distinct primes → negative Möbius."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Möbius Function

14 Why does the Repeated Prime Factor Rule force μ(75) to equal 0?

A Because 75 is odd.

B Because 75 = 3 × 5^2, containing the repeated prime factor 5^2.

C Because 75 is a multiple of 15.

D Because 75 = 5 × 15, and 15 is not prime.

💡 Short Explanation

�� Möbius function checks repeated prime factors �� 75 contains 5² �� Repeated prime factor gives zero

✅ Correct Answer Explanation

Prime factorization: The Möbius rule states: • If n contains any repeated prime factor, then: Since 5² appears, 75 is not square-free. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Oddness has no role in determining μ(n).
�� Option C → Being a multiple of 15 does not imply μ(n)=0.
�� Option D → Factorization is incomplete and conceptually incorrect.

🎯 Strategy

Used: Elimination

Application: Identify whether repeated prime factors exist.

Final Logic: Presence of 5² directly forces μ(75)=0.

🧠 Memory Trick

"Square factor ⇒ Möbius zero."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Möbius Function Rules

15 Calculate the sum of μ(75) + μ(85).

A 0

B -1

C 1

D 2

💡 Short Explanation

�� μ(75)=0 due to repeated factor �� 85=5×17 has two distinct primes �� μ(85)=1

✅ Correct Answer Explanation

First evaluate: Since 5² repeats: Now: This has 2 distinct primes and no repetition. Thus: Therefore: Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Ignores μ(85)=1.
�� Option B → Wrong sign evaluation.
�� Option D → Sum cannot become 2 because μ(75)=0.

🎯 Strategy

Used: Substitution

Application: Evaluate each Möbius value separately before adding.

Final Logic: 0+1=1.

🧠 Memory Trick

"Repeated square gives 0; two primes give +1."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Möbius Function Evaluation

16 For the product of three distinct prime numbers, what is the defined output of the Mobius function?

A 1

B -1

C 0

D 3

💡 Short Explanation

�� Three distinct primes mean k=3 �� Use μ(n)=(-1)^k �� Odd exponent gives −1

✅ Correct Answer Explanation

For a square-free integer with k distinct primes: Here: Therefore: Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Applies when k is even.
�� Option C → Would require repeated prime factors.
�� Option D → Möbius outputs are never 3.

🎯 Strategy

Used: Substitution

Application: Apply the Möbius formula directly with k=3.

Final Logic: Odd number of distinct primes gives −1.

🧠 Memory Trick

"Three primes → negative sign."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Möbius Function

17 Additive functions require f(xy) = f(x) + f(y) specifically when gcd(x, y) = 1. If gcd(x, y) ≠ 1, can the function still be universally categorized as a strictly additive function under these rules?

A Yes, gcd does not matter.

B No, the definition strict constraints require x and y to be coprime.

C Only for prime outputs.

D Only if x = y.

💡 Short Explanation

�� Additive functions require coprime inputs �� gcd(x,y)=1 is essential �� Without coprimality, definition fails

✅ Correct Answer Explanation

An additive arithmetic function satisfies: The coprime condition is part of the formal definition. If: the additive rule cannot be universally guaranteed. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → gcd condition is fundamental to the definition.
�� Option C → Output type is irrelevant.
�� Option D → Equality of x and y is unrelated.

🎯 Strategy

Used: Extreme Word Filter

Application: Watch for absolute claims like "gcd does not matter."

Final Logic: Coprimality is mandatory in additive-function definitions.

🧠 Memory Trick

"Additive needs coprime partners."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Additive Arithmetic Functions

18 Show that τ(n) is a multiplicative function for n = 24. Since 24 = 3 × 8, and gcd(3, 8) = 1, calculate τ(24) using τ(3) × τ(8).

A 2 × 4 = 8

B 3 × 8 = 24

C 1 × 4 = 4

D 4 × 4 = 16

💡 Short Explanation

�� τ is multiplicative for coprime numbers �� τ(3)=2 and τ(8)=4 �� Product gives τ(24)=8

✅ Correct Answer Explanation

Since: and: the multiplicative property applies: Now: • τ(3)=2 because divisors are 1 and 3 • τ(8)=4 because divisors are 1,2,4,8 Thus: Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect interpretation of τ-values.
�� Option C → τ(3) is not 1.
�� Option D → τ(8) is not 4 multiplied by itself.

🎯 Strategy

Used: Substitution

Application: Compute τ-values individually and multiply.

Final Logic: τ(24)=2×4=8.

🧠 Memory Trick

"Tau multiplies for coprimes."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Multiplicative Arithmetic Functions

📄 Passage

We observed the unique interrelations of simple arithmetic functions. Specifically, let us examine the relation n + φ(n) + τ(n) = σ(n). This formula is a special identity that holds true when n = 2p, and p is an odd prime number less than 14. For instance, if we select the prime p = 5, the derived value of n becomes 10. We can then individually evaluate the Euler's totient function φ(10), the number of divisors τ(10), and the sum of divisors σ(10) to verify if the equation naturally balances out.

Question:
Given n = 10 (from p = 5), what are the individual values of φ(10) and τ(10)?

A φ(10) = 4, τ(10) = 4

B φ(10) = 5, τ(10) = 2

C φ(10) = 4, τ(10) = 6

D φ(10) = 6, τ(10) = 4

💡 Short Explanation

�� Coprimes to 10 are 1,3,7,9 �� Total coprimes = 4 �� Divisors of 10 are 1,2,5,10

✅ Correct Answer Explanation

For Euler's totient: because the integers coprime to 10 are: For divisor counting: because divisors are: Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → τ(10) is not 2.
�� Option C → 10 has only four divisors.
�� Option D → φ(10) is not 6.

🎯 Strategy

Used: Substitution

Application: Directly compute coprimes and divisors.

Final Logic: Both values equal 4.

🧠 Memory Trick

"10 → four coprimes, four divisors."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Euler Totient and Divisor Functions

📄 Passage

Question:
For n = 10, determine σ(10) and verify if the identity n + φ(n) + τ(n) = σ(10) is mathematically true.

A σ(10) = 10; False

B σ(10) = 18; True (10 + 4 + 4 = 18)

C σ(10) = 14; False

D σ(10) = 24; True

💡 Short Explanation

�� Divisors of 10 are 1,2,5,10 �� Their sum is 18 �� Identity balances correctly

✅ Correct Answer Explanation

The positive divisors of 10 are: Thus: Also: and: Therefore: Hence the identity is true and Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → σ(10) is not 10.
�� Option C → Sum of divisors exceeds 14.
�� Option D → σ(10) does not equal 24.

🎯 Strategy

Used: Substitution

Application: Evaluate σ(10), φ(10), and τ(10) individually.

Final Logic: Both sides simplify to 18.

🧠 Memory Trick

"10 + 4 + 4 = 18."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Relations Among Arithmetic Functions

CUET UG Applied Mathematics Booster Test 3 - Advanced Arithmetic Functions

Test Complete!

Answer Review