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.

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 on natural numbers.
  • Convergence: A sequence converges if it approaches a unique real number as .
  • lim sup: The greatest accumulation point (limit of suprema of tails).
  • lim inf: The smallest accumulation point (limit of infima of tails).
  • Max sequence: is non-decreasing.
  • Min sequence: is non-increasing.

 

Calculation:

Given,

⇒ For even :

⇒ For odd :

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

does not exist 

Option 1: " does not converge"

⇒ Correct.

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

Option 2: ""

→ False.

Since 1 \), and this is the max forever,

⇒ not equal to 1

Option 3: ""

→ False.

which is minimum forever ⇒ for all large

So

Option 4: ""

→ False.

As shown,

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

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

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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 converges pointwise to on an interval if for each in the interval, .
  • Uniform Convergence: A sequence converges uniformly to if the convergence is uniform across the entire domain, i.e., .
  • Dirac-like Behavior: Functions like 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,

on

Step 1: Evaluate pointwise limit

⇒ Fix

⇒ At

⇒ At

⇒ So for all

⇒ Pointwise limit is (a continuous function)

Step 2: Uniform convergence?

peaks near

⇒ Maximum value:

⇒ So sup norm

⇒ Not uniformly convergent

Step 3: Summary of options

⇒ Statement 1: FALSE (pointwise convergence occurs)

⇒ Statement 2: TRUE (converges to )

⇒ 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 
  2. Let (an)n≥2 be a sequence of positive real numbers. If there exists a positive real number L such that 
  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 is strictly increasing and continuous.
  • Since is dense in , the image is dense in the range .
  • Hence, for any , there exists a rational number such that .
  • Statement 1 is true.

2. Limit Superior and Growth Rate:

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

3. Countability of Finite Subsets:

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

4. Continuous Functions from to :

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

 

Calculation:

Given,

Let with

Function:

⇒ f is strictly increasing and continuous

⇒ Image of under f is dense in

⇒ ∃ rational such that

⇒ Statement 1 is true

Given:

⇒ an ≈ L × log n

⇒ But

⇒ So is possible

⇒ Statement 2 is false

Set of finite subsets of is:

⇒ Countable union of finite sets

⇒ countable

⇒ Infinite in number

⇒ Statement 3 is true

Continuous functions

⇒ 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 and 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 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,

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 , 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)

Note:

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

Calculation:

Given:

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

 and finite.

f(a,b) = f(0,0) ⇒ 0 (given)

 = 0 

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

fy(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  an, where an = (−1)n+1. 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 (-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) 

Explanation:

an = (−1)n+1

    = (−1)n+1   

   = (−1)n+1

So series is  

So here bnbn+1 = 

 so bn+1 n  

Also  =   = 0

Hence by Leibnitz's test  an is convergent.

Now the series is  =   =  

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 .
Define

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 

  and 

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, 

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

Hence, 

Consider ℝ2 with the usual Euclidean metric. Let 

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

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

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

 {(0, 𝑦) ∈ ℝ2 : −∞  represents the whole y - axis and 𝑥 sin  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  ) ∈ ℝ2 : 𝑥 ∈ (0,1]} ⋃ {(0, 𝑦) ∈ ℝ2 : −∞

{(0, 𝑦) ∈ ℝ2 : −∞  represents the whole y - axis and  sin  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  x ∈ ℝ

where √ denotes the non-negative square root. Wherever  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  be a sequence of functions defined on a set D . We say that  converges pointwise to a function

 f  on D if, for every x  D

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

Explanation: The problem gives a sequence of functions defined by

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

We are asked to take the limit  of the function:

    

As , the term . So, for large n , the function  approaches

 
 

Case 1: 

For  we have, 
 

Case 2: 
When  , the function becomes 

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

The function f(x) ,  , is given by

   
  

This function is equal to |x| for all

Therefore, 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 13 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. 

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 4) 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 14 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 = 

⇒ |x - y| = 1

If possible let 2n ε

i.e. 2n  

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 =  

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

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

Which of the following statements is true?

  1. There are at most countably many continuous maps from to .
  2. There are at most finitely many continuous surjective maps from  to .
  3. There are infinitely many continuous injective maps from to .
  4. There are no continuous bijective maps from to .

Answer (Detailed Solution Below)

Option 4 : There are no continuous bijective maps from to .

Analysis Question 15 Detailed Solution

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Explanation:

Let f: ℝ2 →  is defined by

f(x, y) = cx, c ∈ ℝ\{0} then f is continuous function.

(1) and (2) are false.

If possible let there are infinitely many continuous injective maps f: ℝ2 → ℝ.

Then it will map a connected set to a connected set.

If we consider f such that f(0) = c, c ∈ ℝ\{0} then

f(ℝ\{0}) = (-∞, c) ∪ (c, ∞), which is not connected. So we are getting a contradiction.

(3) is false.

Hence option (4) is correct

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