Analysis MCQ Quiz - Objective Question with Answer for Analysis - Download Free PDF
Last updated on Jul 16, 2025
Latest Analysis MCQ Objective Questions
Analysis Question 1:
For an integer n, let fn(x) = xe−nx, where x ∈ [0, 1]. Let S := {fn : n ≥ 1}. Consider the metric space (C([0, 1]), d), where
Which of the following statement(s) is/are false?
Answer (Detailed Solution Below)
Analysis Question 1 Detailed Solution
Explanation:
Each fn(x) = xe−nx, is a continuous function on [0, 1] because it is the product of two continuous functions: x and
A family of functions S is equi-continuous if, for every
all
For fn(x) = xe−nx :
1. As n increases, the term
concentrated near x = 0 .
2. For any two points x, y close to each other, the difference
sufficiently small
Thus, S is an equi-continuous family of functions. So, Option 1 is correct.
To determine if S is closed, we need to check if it contains all its limit points in the metric defined by
As
However, f(x) = 0 is not in S since
Therefore, S is not closed in C([0, 1]) because it does not contain its pointwise limit f(x) = 0 . So, Option 2 is incorrect.
To determine if S is bounded in C([0, 1]) , we compute
For fn(x) = xe−nx , the maximum value occurs near
Substituting
Since
Thus, S is bounded in C([0, 1]) . So, Option 3 is correct.
A subset of C([0, 1]) is compact if it is closed, bounded, and equi-continuous (by the Arzelà-Ascoli theorem). We have shown that:
S is bounded , S is equi-continuous
However, S is not closed.
Since S is not closed, it cannot be compact. Therefore, Option 4 is incorrect.
Analysis Question 2:
Let (X, d) be a finite non‐singleton metric space. Which of the following statement/ statements is /are false?
Answer (Detailed Solution Below)
Analysis Question 2 Detailed Solution
Concept:
(i) All metric spaces are Hausdorff spaces.
(ii) Finite union of open set is open
(iii) Compact set: A metric space X is said to be compact if every open covering has a finite sub-covering.
(iv) Connected set: A metric space (M, d) is said connected metric space if and only if M cannot be written as a disjoint union N = X ∪ Y where X and Y are both non-empty open subsets of M.
(v) A function f : X → Y is continuous if and only if G is open in Y implies f-1(G) is open in X
Explanation:
(X, d) be a finite non‐singleton metric space.
Let X = {x1, x2, ..., xn}
Then there exist an open ball containing xi such that B(xi) ⊆ {xi}
Then each subset of X is open.
(1) is false
So, every open cover of X has a finite subcover
Hence X is compact.
(2) is correct
X can be written as X = {xi} ∪ {xi}c
So X is not connected
(3) is correct
X is a discrete metric space so every function f : X → ℝ must be continuous.
(4) is fasle
Analysis Question 3:
Let p ∶
Which of the following statements is/are false?
Answer (Detailed Solution Below)
Analysis Question 3 Detailed Solution
Explanation:
p(x,y) = 0 ⇒ |x| = 0 ⇒ x = 0, when x ≠ 0
and |y| = 0 ⇒ y=0 when x=0
∴ p(x,y) = 0 if and only if x = y = 0.
Option (1) is correct.
as |x| ≥ 0, |y| ≥ 0 for all x, y so p(x, y) ≥ 0 for all x, y.
Option (2) is correct.
for x ≠ 0, p(αx, αy) = |α x| = |α||x| = |α|p(x,y)
for x = 0, p(αx, αy) = |α y| = |α||y| = |α|p(x,y)
Hence p(αx, αy) = |α| p(x, y) for all α ∈
Option (3) is correct.
A counterexample for option (4):
Let (x1, y1) = (5,10), (x2, y2) = (-5,20) then (x1+x2,y1+y2) = (0,30)
So p(x1+x2,y1+y2) = 30 as x1+x2 = 0
nand p(x1, y1) + p(x2, y2) = |5| + |-5| = 5 + 5 = 10
as
Option (4) is not correct.
Analysis Question 4:
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 4 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 5:
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 5 Detailed Solution
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 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?
Answer (Detailed Solution Below)
Analysis Question 13 Detailed Solution
Download Solution PDFConcept:
1. Power Set: The power set of a set S, denoted
If S has
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
3. Uncountably Infinite Set: A set is uncountable if it is not countably infinite (for example, the real numbers
4. Power Set and Cardinality: If the power set
This is because for any finite set S, its power set
elements in
set
Explanation:
Option 1: This cannot be true because if
Option 2: This is incorrect option because if
Option 3: This cannot be true because if
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
Answer (Detailed Solution Below)
Analysis Question 14 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 15 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.