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

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.

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.

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:

⇒ a_n ≈ 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.

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.

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 

f_x(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.

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 b_n = , bn+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.

Concept -

(i) If the sequence x_n convergent  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

x   A_i imply x ∈ A_i 

x (E_n )

x  E_n ( finite )

Hence option (2) & (4) are incorrect 

Hence option (3) is correct 

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

ρ(A_{4 × 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, 

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.

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).

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.

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.

Explanation:

Let f: ℝ² → ℝ 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