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?
Answer (Detailed Solution Below)
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:
Option 1: "
⇒ Correct.
Since the sequence oscillates between two values without settling on one, it diverges.
Option 2: "
→ False.
Since
⇒ not equal to 1
Option 3: "
→ False.
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?
Answer (Detailed Solution Below)
Analysis Question 2 Detailed Solution
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?
Answer (Detailed Solution Below)
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,
Step 1: Evaluate pointwise limit
⇒ Fix
⇒
⇒ At
⇒ At
⇒ So
⇒ Pointwise limit is
Step 2: Uniform convergence?
⇒
⇒ 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?
Answer (Detailed Solution Below)
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
- 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
Function:
⇒ f is strictly increasing and continuous
⇒ Image of
⇒ ∃ rational
⇒ Statement 1 is true
Given:
⇒ an ≈ L × log n
⇒ But
⇒ So
⇒ Statement 2 is false
Set of finite subsets of
⇒ 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
Analysis Question 5:
Let f : [0, 1] → ℝ be a monotonic function. Which of the following statements are true?
Answer (Detailed Solution Below)
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,
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
Answer (Detailed Solution Below)
Analysis Question 6 Detailed Solution
Download Solution PDFConcept:
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:
f(a,b) = f(0,0) ⇒ 0 (given)
fx(0, 0) =
fy(0, 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
Answer (Detailed Solution Below)
Analysis Question 7 Detailed Solution
Download Solution PDFConcept:
Leibniz's test: A series of the form
(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 bn =
Also
Hence by Leibnitz's test
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?
Answer (Detailed Solution Below)
Analysis Question 8 Detailed Solution
Download Solution PDFConcept -
(i) If the sequence xn convergent then limsupn En = liminfn En
Calculation:
Let {En} be a sequence of subsets of R
for option 1, if convergent then limsupn En = liminfn En
option 1 is incorrect
x
x
x
Hence option (2) & (4) are incorrect
Hence option (3) is correct
How many real roots does the polynomial x3 + 3x − 2023 have?
Answer (Detailed Solution Below)
Analysis Question 9 Detailed Solution
Download Solution PDFConcept:
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?
Answer (Detailed Solution Below)
Analysis Question 10 Detailed Solution
Download Solution PDFρ(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
𝑌 = {(𝑥, sin
Consider the following statements:
𝑃: 𝑋 is a connected subset of ℝ2 .
𝑄: 𝑌 is a connected subset of ℝ2 .
Then
Answer (Detailed Solution Below)
Analysis Question 11 Detailed Solution
Download Solution PDFExplanation -
𝑋 = {(𝑥, 𝑥 sin
{(0, 𝑦) ∈ ℝ2 : −∞ represents the whole y - axis and 𝑥 sin
So Clearly X is also connected subset of ℝ2 .
Hence Statement P is correct.
𝑌 = {(𝑥, sin
{(0, 𝑦) ∈ ℝ2 : −∞ represents the whole y - axis and sin
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
where √ denotes the non-negative square root. Wherever
Answer (Detailed Solution Below)
Analysis Question 12 Detailed Solution
Download Solution PDFConcept:
Limit of a Sequence of Functions:
1. Let
f on D if, for every x
2. A stronger form of convergence is uniform convergence. The sequence
Explanation: The problem gives a sequence of functions
and asks about the limit of
We are asked to take the limit
As
Case 1:
For
Case 2:
When
Therefore, as
The function f(x) ,
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
Answer (Detailed Solution Below)
Analysis Question 13 Detailed Solution
Download Solution PDFGiven -
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?
Answer (Detailed Solution Below)
Analysis Question 14 Detailed Solution
Download Solution PDFConcept -
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?
Answer (Detailed Solution Below)
Analysis Question 15 Detailed Solution
Download Solution PDFExplanation:
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