Analysis MCQ Quiz - Objective Question with Answer for Analysis - Download Free PDF

Last updated on Jul 7, 2025

Latest Analysis MCQ Objective Questions

Analysis Question 1:

Let (an)n≥1, (bn)n≥1 and (Cn)n≥1 be sequences given by

an = (-1)n (1 + e-n)

bn = max{a1.....an}, and

Cn = min{a1....an},

Which of the following statements are true? 

  1. (an)n≥1 does not converge.
  2. limnsupan=limnbn
  3. limninf an=limncn
  4. limnbn=limncn

Answer (Detailed Solution Below)

Option :

Analysis Question 1 Detailed Solution

Concept:

Understanding Sequences and Their Limits:

  • Sequence: An ordered list of numbers defined by a function an on natural numbers.
  • Convergence: A sequence an converges if it approaches a unique real number as n.
  • lim sup: The greatest accumulation point (limit of suprema of tails).
  • lim inf: The smallest accumulation point (limit of infima of tails).
  • Max sequence: bn=max{a1,a2,,an} is non-decreasing.
  • Min sequence: cn=min{a1,a2,,an} is non-increasing.

 

Calculation:

Given,

an=(1)n(1+en)

⇒ For even n: an1+en1

⇒ For odd n: an1en1

⇒ So the sequence oscillates between numbers close to +1 and -1, hence:

liman does not exist 

Option 1: "an does not converge"

⇒ Correct.

Since the sequence oscillates between two values without settling on one, it diverges.

Option 2: "lim supan=limbn"

→ False.

lim supan=1

bn=max{a1,a2,...,an}

Since a2=1+e2>1, and this is the max forever, bn1+e2

⇒ not equal to 1

Option 3: "lim infan=limcn"

→ False.

lim infan=1

a1=1e1 which is minimum forever ⇒ cn=1e1 for all large n

So limcn=1e11

Option 4: "limbn=limcn"

→ False.

As shown, limbn=1+e2

limcn=1e1

Hence, not equal

∴ Only correct statement is Option 1.

Analysis Question 2:

Let F : ℝ → ℝ be a continuous function such that f(x) = 0 for all x ≤ 0 and for all x ≥ 1, Define

F(x)=Σn=f(x+n),xR

Which of the following statements are true? 

  1. F is bounded. 
  2. F is continuous on ℝ.  
  3. F is uniformly continuous on ℝ. 
  4. F is not uniformly continuous on ℝ. 

Answer (Detailed Solution Below)

Option :

Analysis Question 2 Detailed Solution

We will update the solution soon.

Analysis Question 3:

For each positive integer n, define fn : [0, 1] → ℝ by fn(x) = nx(1 - x)n.

Which of the following statements are true? 

  1. (fn)n≥1 does not converge pointwise on [0, 1]
  2. (fn)n≥1 converges pointwise to a continuous function on [ [0, 1]
  3. (fn)n≥1 converges pointwise to a discontinuous function on [0, 1]
  4. (fn)n≥1 does not converge uniformly on [0, 1]

Answer (Detailed Solution Below)

Option :

Analysis Question 3 Detailed Solution

Concept:

  • Pointwise Convergence: A sequence of functions fn converges pointwise to f on an interval if for each x in the interval, limnfn(x)=f(x).
  • Uniform Convergence: A sequence fn converges uniformly to f if the convergence is uniform across the entire domain, i.e., supx[0,1]|fn(x)f(x)|0.
  • Dirac-like Behavior: Functions like fn(x)=nx(1x)n become sharply peaked near a point and converge to zero everywhere except potentially at a specific location.
  • Continuity of Limit Function: If pointwise limit is not continuous, then the convergence cannot be uniform.

 

Calculation:

Given,

fn(x)=nx(1x)n on [0,1]

Step 1: Evaluate pointwise limit

⇒ Fix x[0,1)

(1x)n0fn(x)0

⇒ At x=1fn(1)=0

⇒ At x=0fn(0)=0

⇒ So fn(x)0 for all x[0,1]

⇒ Pointwise limit is f(x)=0 (a continuous function)

Step 2: Uniform convergence?

fn(x) peaks near x=1n+1

⇒ Maximum value: fn(1n+1)=nn+1(nn+1)n1

⇒ So sup norm fn00

⇒ Not uniformly convergent

Step 3: Summary of options

⇒ Statement 1: FALSE (pointwise convergence occurs)

⇒ Statement 2: TRUE (converges to f(x)=0)

⇒ Statement 3: FALSE (limit function is continuous)

⇒ Statement 4: TRUE (not uniformly convergent)

∴ Final Answer: Statements 2 and 4 are TRUE.

Analysis Question 4:

Which of the following statements are true? 

  1. Let x, y ∈ ℝ with x < y. Then there exists r ∈ ℚ such that \(\rm x<\frac{2^{2024}r}{e}
  2. Let (an)n≥2 be a sequence of positive real numbers. If there exists a positive real number L such that limnanlogn=L, then, limnan<
  3. The set of all finite subsets of ℚ is countably infinite. 
  4. The set of continuous functions from ℝ to the set {0, 1} is infinite.

Answer (Detailed Solution Below)

Option :

Analysis Question 4 Detailed Solution

Concept:

1. Density and Continuous Functions:

  • A function of the form f(r)=2224re is strictly increasing and continuous.
  • Since Q is dense in R, the image f(Q) is dense in the range (0,).
  • Hence, for any x<yR, there exists a rational number r such that x<2224re<y.
  • Statement 1 is true.

2. Limit Superior and Growth Rate:

  • Given lim supanlogn=L, this implies that an grows asymptotically like logn.
  • However, since logn, the sequence an can still diverge.
  • So, lim supan can be ∞ even though the ratio with logn is finite.
  • Statement 2 is false.

3. Countability of Finite Subsets:

  • The set Q is countable.
  • A countable union of finite subsets of a countable set is countable.
  • There are infinitely many finite subsets of Q.
  • Statement 3 is true.

4. Continuous Functions from R to {0,1}:

  • The set {0,1} is discrete and totally disconnected.
  • The image of a connected space under a continuous map must be connected.
  • So the only continuous functions from R to {0,1} are constant functions.
  • Only two such functions exist: one always 0 and one always 1.
  • Statement 4 is false.

 

Calculation:

Given,

Let x,yR with x<y

Function: f(r)=2224re

⇒ f is strictly increasing and continuous

⇒ Image of Q under f is dense in (0,)

⇒ ∃ rational r such that x<f(r)<y

⇒ Statement 1 is true

Given: lim supanlogn=L

⇒ an ≈ L × log n

⇒ But logn

⇒ So lim supan= is possible

⇒ Statement 2 is false

Set of finite subsets of Q is:

⇒ Countable union of finite sets

⇒ countable

⇒ Infinite in number

⇒ Statement 3 is true

Continuous functions f:R{0,1}

⇒ ℝ is connected

⇒ {0,1} is disconnected

⇒ Image must be a singleton

⇒ only two such functions exist

⇒ Statement 4 is false

∴ The correct answer is: Only Statements 1 and 3 are true.

Analysis Question 5:

Let f : [0, 1] → ℝ be a monotonic function. Which of the following statements are true? 

  1. f is Riemann integrable on [0, 1]. 
  2. The set of discontinuities of f cannot contain a non-empty open set. 
  3. f is a Lebesgue measurable function. 
  4. f is a Borel measurable function. 

Answer (Detailed Solution Below)

Option :

Analysis Question 5 Detailed Solution

Concept:

  • Monotonic Function: A function f:[0,1]R is monotonic if it is either non-increasing or non-decreasing throughout the interval.
  • Riemann Integrability: A function is Riemann integrable on a closed interval if the set of its discontinuities has measure zero.
  • Discontinuities of Monotonic Functions: A monotonic function can only have jump discontinuities, and the set of such points is at most countable. Therefore, it cannot contain any non-empty open interval.
  • Lebesgue Measurability: A function is Lebesgue measurable if the preimage of any Borel set is a Lebesgue measurable set. Monotonic functions are always Lebesgue measurable.
  • Borel Measurability: A function is Borel measurable if the preimage of any Borel set lies in the Borel sigma-algebra. All monotonic functions are Borel measurable.

 

Calculation:

Given,

f:[0,1]R is a monotonic function

Step 1: Check Riemann integrability

⇒ Monotonic functions have at most countably many discontinuities

⇒ Such sets have measure zero

⇒ Riemann integrable

Step 2: Discontinuities and open sets

⇒ Discontinuity set of monotonic function is at most countable

⇒ Countable set cannot contain non-empty open set

⇒ Statement 2 is TRUE

Step 3: Lebesgue measurability

⇒ Every monotonic function is Lebesgue measurable

⇒ Statement 3 is TRUE

Step 4: Borel measurability

⇒ Monotonic functions are Borel measurable

⇒Statement 4 is TRUE

∴ All four statements are TRUE.

Top Analysis MCQ Objective Questions

Let f(x,y)={xyx2+y2x2+y200x=y=0, Then Which of the following is not Correct ?

  1. f(x, y) is not differentiable at the origin
  2. f(x, y) is continuous at the origin
  3. fx (0,0) = f(0,0)
  4. fy (0,0) = f(0,0)

Answer (Detailed Solution Below)

Option 1 : f(x, y) is not differentiable at the origin

Analysis Question 6 Detailed Solution

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Concept:

A function f(x,y) is defined for (x,y) = (a,b) is said to be continuous at (x,y) = (a,b) if:

i) f(a,b) = value of f(x,y) at (x,y) = (a,b) is finite.

ii) The limit of the function f(x,y) as (x,y) → (a,b) exists and equal to the value of f(x,y) at (x,y) = (a,b)

lim(x,y)(0,0)f(x,y)=f(a,b)

Note:

For a function to be differentiable at a point, it should be continuous at that point too.

Calculation:

Given:

f(x,y)={xyx2+y2x2+y200x=y=0

For function f(x,y) to be continuous:

lim(x,y)(0,0)f(x,y)=f(a,b) and finite.

f(a,b) = f(0,0) ⇒ 0 (given)

lim(r,θ)(0,0)f(r,θ)=r2cosθrsinθr = 0 

fx(0, 0) = lim(h,0)(0,0){f(h, 0) - f(0, 0)} / h = 0 

fy(0, 0) = lim(0,k)(0,0){f(0, k) - f(0, 0)} / k = 0 

 

∵ the limit value is defined and function value is 0 at (x,y) = (0,0), ∴ the function f(x,y) is continuous.

Hence, Option 2, 3 & 4 all are correct 

Hence, Option 1 is not correct 

Hence, The Correct Answer is option 1.

Consider the series n=1 an, where an = (−1)n+1(n+1n). Which of the following statements is true?

  1. The series is divergent.
  2. The series is convergent.
  3. The series is conditionally convergent.
  4. The series is absolutely convergent.

Answer (Detailed Solution Below)

Option 3 : The series is conditionally convergent.

Analysis Question 7 Detailed Solution

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Concept:

Leibniz's test: A series of the form n=1(-1)nbn, where either all bn are positive or all bn are negative is convergent if

(i) |bn| decreases monotonically i.e., |bn+1| ≤ |bn|

(ii) limnbn=0

Explanation:

an = (−1)n+1(n+1n)

    = (−1)n+1  (n+1n)(n+1+n)(n+1+n) 

   = (−1)n+11(n+1+n)

So series is n=1(1)n+1(n+1+n) 

So here bn1(n+1+n)bn+1 = 1(n+2+n+1)

bn+1bn<1 so bn+1 < bn  

Also limnbn = limn 1(n+1+n) = 0

Hence by Leibnitz's test n=1 an is convergent.

Now the series is n=1|(1)n+1(n+1+n)| =  n=11(n+1+n) = n=1 1n(1+1n+1)

Hence by Limit comparison Test, it is divergent series by P - Test.

Hence the given series is conditionally convergent.

Option (3) is correct.

In Official answer key - Options (2) & (3) both are correct.

Let {En} be a sequence of subsets of R.
Define
lim supnEn=k=1n=kEn

lim infnEn=k=1n=kEn

Which of the following statements is true?

  1. limsupn En = liminfn En 
  2. limsupn En = {x ∶ x ∈ En for some n}
  3. liminfn En = {x ∶ x ∈ Efor all but finitely many n}
  4. liminfn En = {x ∶ x ∈ E for infinitely many n}

Answer (Detailed Solution Below)

Option 3 : liminfn En = {x ∶ x ∈ Efor all but finitely many n}

Analysis Question 8 Detailed Solution

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Concept -

(i) If the sequence xconvergent  then limsupn En = liminfn En 

Calculation:

Let {En} be a sequence of subsets of R 

lim supnEn=k=1n=kEn  and 

lim infnEn=k=1n=kEn

for option 1, if convergent  then limsupn En = liminfn En 

option 1 is incorrect

   Ai imply x ∈ Ai 

 ( En )

  En ( finite )

Hence option (2) & (4) are incorrect 

Hence option (3) is correct 

How many real roots does the polynomial x3 + 3x − 2023 have?

  1. 0
  2. 1
  3. 2
  4. 3

Answer (Detailed Solution Below)

Option 2 : 1

Analysis Question 9 Detailed Solution

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Concept: 

Every odd degree polynomial p(x) ∈ R(x) has at least a real root

Explanation:

p(x) = x3 + 3x − 2023

p'(x) = 3x2 + 3

Since x2 ≥ 0 for all x so

3x2 + 3 > 0 ⇒ p'(x) > 0

Therefore p'(x) has no real roots

We know that between two distinct real roots of p(x) there exist a real root of p'(x).

Since here p'(x) no real roots, so p(x) can't have more than one real root

Option (2) correct

Two vectors [2 1 0 3]𝑇 and [1 0 1 2]𝑇 belong to the null space of a 4 × 4 matrix of rank 2. Which one of the following vectors also belongs to the null space?

  1. [1 1 −1 1]T
  2. [2 0 1 2]T
  3. [0 −2 1 −1]T
  4. [3 1 1 2]T

Answer (Detailed Solution Below)

Option 1 : [1 1 −1 1]T

Analysis Question 10 Detailed Solution

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ρ(A4 × 4) = 2

N (A) = Number of column – Rank

          = 4 – 2 = 2    

i.e. Null space of A will consist only two linearly independent vectors which is given as x and y.

Eigen vectors of matrix A, [2 1 0 3]and[1 0 1 2]

As these are linearly independent eigen vectors so remaining eigen vectors of null space must be linearly dependent.

Hence, XY=[1 1 1 1]

Consider ℝ2 with the usual Euclidean metric. Let 

𝑋 = {(𝑥, 𝑥 sin 1x) ∈ ℝ2 ∶ 𝑥 ∈ (0,1]} ⋃ {(0, 𝑦) ∈ ℝ2 : −∞ < 𝑦 < ∞} and

𝑌 = {(𝑥, sin  1x) ∈ ℝ2 : 𝑥 ∈ (0,1]} ⋃ {(0, 𝑦) ∈ ℝ2 : −∞ < 𝑦 < ∞}.

Consider the following statements:

𝑃: 𝑋 is a connected subset of ℝ2 .

𝑄: 𝑌 is a connected subset of ℝ2 .

Then 

  1. both 𝑃 and 𝑄 are TRUE
  2. 𝑃 is FALSE and 𝑄 is TRUE
  3. 𝑃 is TRUE and 𝑄 is FALSE
  4. both 𝑃 and 𝑄 are FALSE

Answer (Detailed Solution Below)

Option 1 : both 𝑃 and 𝑄 are TRUE

Analysis Question 11 Detailed Solution

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Explanation -

𝑋 = {(𝑥, 𝑥 sin 1x) ∈ ℝ2 ∶ 𝑥 ∈ (0,1]} ⋃ {(0, 𝑦) ∈ ℝ2 : −∞ < 𝑦 < ∞} 

 {(0, 𝑦) ∈ ℝ2 : −∞ < 𝑦 < ∞}  represents the whole y - axis and 𝑥 sin 1x is oscillates between 0 and 1. Hence it is connected.

So Clearly X is also connected subset of ℝ2 .

Hence Statement P is correct.

𝑌 = {(𝑥, sin  1x) ∈ ℝ2 : 𝑥 ∈ (0,1]} ⋃ {(0, 𝑦) ∈ ℝ2 : −∞ < 𝑦 < ∞}.

{(0, 𝑦) ∈ ℝ2 : −∞ < 𝑦 < ∞}  represents the whole y - axis and  sin 1x is oscillates between 0 and 1. Hence it is connected.

So Clearly Y is also connected subset of ℝ2 .

Hence Statement Q is correct.

Hence option (1) is correct.

For each n ≥ 1 define fn : ℝ → ℝ by fn(x)=x2x2+1n, x ∈ ℝ

where √ denotes the non-negative square root. Wherever limnfn(x) exists, denote it by f(x). Which of the following statements is true? 

  1. There exists x ∈ ℝ such that f(x) is not defined 
  2. f(x) = 0 for all x ∈ ℝ 
  3. f(x) = x for all x ∈ ℝ 
  4. f(x) = |x| for all x ∈ ℝ

Answer (Detailed Solution Below)

Option 4 : f(x) = |x| for all x ∈ ℝ

Analysis Question 12 Detailed Solution

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Concept:

Limit of a Sequence of Functions:

1. Let {fn} be a sequence of functions defined on a set D . We say that fn converges pointwise to a function

 f  on D if, for every x  D

limnfn(x)=f(x).

2. A stronger form of convergence is uniform convergence. The sequence {fn} converges uniformly to a function  f  on D if

limnsupxD|fn(x)f(x)|=0.

Explanation: The problem gives a sequence of functionsfn:RR defined by

fn(x)=x2x2+1n

and asks about the limit of fn(x) as n, denoted by f(x) . We are tasked with determining which statement about f(x) is true.

We are asked to take the limit n of the function:

 fn(x)=x2x2+1n   

As n, the term 1n0. So, for large n , the function fn(x) approaches

 limnfn(x)=x2x2=x2|x|
 

Case 1: x0

For x0 we have, f(x)=x2|x|=|x|
 

Case 2: x=0
When x=0 , the function becomes fn(0)=0202+1n=0

Therefore, as n, we get f(0) = 0 .

The function f(x) , n , is given by

   
  f(x)={|x|,if x00,if x=0

This function is equal to |x| for all xR

Therefore, The correct option is 4).

Let C be the collection of all sets S such that the power set of S is countably infinite. Which of the following statements is true? 

  1. There exists a non-empty finite set in C
  2. There exists a countably infinite set in C 
  3. There exists an uncountable set in C 
  4. C is empty 

Answer (Detailed Solution Below)

Option 4 : C is empty 

Analysis Question 13 Detailed Solution

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Concept:
 

1. Power Set: The power set of a set S, denoted P(S), is the set of all subsets of  S.

If S has n  elements, then P(S) contains 2n elements. 
 

2. Countably Infinite Set: A set is countably infinite if its elements can be put into a one-to-one

correspondence with the natural numbers (i.e., it has the same cardinality as N ).
 

3. Uncountably Infinite Set: A set is uncountable if it is not countably infinite (for example, the real numbers R ).

4. Power Set and Cardinality: If the power set P(S) is countably infinite, then S cannot be finite.

This is because for any finite set S, its power set P(S) has 2n elements, where n is the number of

elements in S, and 2n is always finite. Additionally, if S is uncountably infinite, then its power

set P(S) will be uncountably infinite. 

Explanation:

 

Option 1: This cannot be true because if S is finite, its power set will also be finite, not countably infinite.
  
Option 2: This is incorrect option because if S is any countably infinite set then its power set must be uncountable.

Option 3: This cannot be true because if S were uncountable, its power set would be uncountable as well.

Option 4:  This is true, as there is no countably infinite set whose power set is countably infinite, so C is empty.

The correct option is 4).

Let 𝑇 ∶ ℝ4 → ℝ4 be a linear transformation and the null space of 𝑇 be the subspace of ℝ4 given by

{(𝑥1, 𝑥2, 𝑥3, 𝑥4) ∈ ℝ4 ∶ 4𝑥1 + 3𝑥2 + 2𝑥3 + 𝑥4 = 0}.

If 𝑅𝑎𝑛𝑘(𝑇 − 3𝐼) = 3, where 𝐼 is the identity map on ℝ4 , then the minimal polynomial of 𝑇 is 

  1. 𝑥(𝑥 − 3) 
  2. 𝑥(𝑥 − 3)3
  3. 𝑥3 (𝑥 − 3) 
  4. 𝑥2 (𝑥 − 3)2

Answer (Detailed Solution Below)

Option 1 : 𝑥(𝑥 − 3) 

Analysis Question 14 Detailed Solution

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Given -

Let 𝑇 ∶ ℝ4 → ℝ4 be a linear transformation and the null space of 𝑇 be the subspace of ℝ4 given by

{(𝑥1, 𝑥2, 𝑥3, 𝑥4) ∈ ℝ4 ∶ 4𝑥1 + 3𝑥2 + 2𝑥3 + 𝑥4 = 0}.

If 𝑅𝑎𝑛𝑘(𝑇 − 3𝐼) = 3, where 𝐼 is the identity map on ℝ4

Concept -

(i) The dimension of subspace = dim (V) - number of restriction

(ii) Rank - Nullity theorem -

 η (T) + ρ (T) = n  where n is the dimension of the vector space or the order of the matrix.

(iii) The formula for Geometric multiplicity (GM) is = η(T - λ I)

(iv) AM ≥ GM

(v) If the rank of A is less than n this implies that |A| = 0

Explanation -

we have null space {(𝑥1, 𝑥2, 𝑥3, 𝑥4) ∈ ℝ4 ∶ 4𝑥1 + 3𝑥2 + 2𝑥3 + 𝑥4 = 0}

Now the dimension of null space = dim (V) - number of restriction = 4 - 1 = 3

Hence the nullity of T is 3 so this implies rank of T is 1.   [by rank - Nullity theorem]

i.e. ρ(T)=1  and  η(T)=3

Now the formula for Geometric multiplicity (GM) is = η(T - λ I)

if we take λ = 0 then GM = 3 for λ = 0 and we know that AM ≥ GM then AM = 3, 4 only because it is not greater than the dimension of vector space.

But we have the another condition  𝑅𝑎𝑛𝑘(𝑇 − 3𝐼) = 3 < dim (ℝ4) then |𝑇 − 3𝐼| = 0

hence λ = 3 is another eigen value of the transformation. and we have 𝑅𝑎𝑛𝑘(𝑇 − 3𝐼) = 3 ⇒ η(𝑇 − 3𝐼) = 1

Hence the eigen values of T is 0,0,0 and 3.

Now the characteristic polynomial of T is 𝑥3 (𝑥 − 3)  and the minimal polynomial of T is x(x - 3).

Hence the option(1) is correct.

Let x, y ∈ [0, 1] be such that x ≠ y. Which of the following statements is true for every ϵ > 0?

  1. There exists a positive integer N such that |x − y| < 2n ϵ for every integer n ≥ N.
  2. There exists a positive integer N such that 2n ϵ < |x − y| for every integer n ≥ N.
  3. There exists a positive integer N such that |x − y| < 2−n ϵ for every integer n ≥ N.
  4. For every positive integer N, |x − y| < 2−n ϵ for some integer n ≥ N.

Answer (Detailed Solution Below)

Option 1 : There exists a positive integer N such that |x − y| < 2n ϵ for every integer n ≥ N.

Analysis Question 15 Detailed Solution

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Concept -

Archimedian Property of real:

Let a, b ∈ ℝ and a > 0 then ∃ N ∈ ℕ such that na > b, ∀ n ≥ N (fix natural number)

Explanation -

Let ε  = a and b = |x - y|

⇒ ∃ N ∈ ℕ such that nε > b = |x - y| ∀ n > N

→ 2n ε > nε > |x - y| ∀ n ≥ N

⇒ 2n ε > |x - y| ∀ n ≥ N

So, option (1) is true

For option (2):

Let x = 0, y = 1 and ε = 12

⇒ |x - y| = 1

If possible let 2n ε < |x - y|

i.e. 2n 12 < 1, a contradiction

So, option (2) is false.

For option (3) and (4):

Let ε = 1, x = 0 and y = 1

⇒ |x - y| = 1 but 2-n ε = 12n < 1 ∀ n ∈ ℕ 

So, |x - y| < 2-n ε is not true for any n ∈ ℕ.

So, Option (3) and (4) are false.

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