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

Explanation:

Each  fn(x) = xe−nx, is a continuous function on [0, 1] because it is the product of two continuous functions: x and \(e^{-nx}\) .

A family of functions S is equi-continuous if, for every \(\epsilon > 0 \), there exists a \(\delta > 0\) such that for

all \(f_n \in S\) and for all x, y \(\in\) [0, 1] with \(|x - y| < \delta\) , we have \(|f_n(x) - f_n(y)| < \epsilon\) .

For fn(x) = xe−nx :

1. As n increases, the term \(e^{-nx}\) decays very rapidly to zero for x > 0 , so the functions \(f_n(x)\) become

concentrated near x = 0 .

2. For any two points x, y close to each other, the difference \(|f_n(x) - f_n(y)|\) can be made small by choosing a

sufficiently small \(\delta\) that works uniformly for all n , because the factor e−nx does not affect small variations near x and y significantly for fixed x, y .

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 \(d(f, g) = \sup_{x \in [0, 1]} |f(x) - g(x)| .\)

As \(n \to \infty\) ,fn(x) = xe−nx  converges pointwise to the zero function f(x) = 0 on [0, 1] because \(e^{-nx} \to 0\) for x > 0 .

However, f(x) = 0 is not in S since \( f_n \)(x) for any finite n is not identically zero.

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 \(\|f_n\| = \sup_{x \in [0, 1]} |f_n(x)|\) for each \( f_n \).

For fn(x) = xe−nx  , the maximum value occurs near \(x = \frac{1}{n} \).

Substituting \(x = \frac{1}{n} \) :
   
  \( f_n\left(\frac{1}{n}\right) = \frac{1}{n} e^{-1} = \frac{e^{-1}}{n}.\)
   
Since \(\frac{e^{-1}}{n} \to 0 \) as \(n \to \infty\) , the functions \( f_n \)(x) are uniformly bounded by a constant M = \(e^{-1}\) .

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.

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, ..., x_n}

Then there exist an open ball containing x_i such that B(xi) ⊆ {xi}

Then each subset of X is open.

(1) is false

\(\left\{x_{i_1}, x_{i_2}, ..., x_{i_k}\right\}\) = \(∪_{j=1}^k\{x_{i_j}\}\) is open

So, every open cover of X has a finite subcover

Hence X is compact.

(2) is correct

X can be written as X = {x_i} ∪ {x_i}^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

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 ∈ \(\mathbb{R}\) and for all x, y.

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 \(30 \nleq 10\) so p(x1+x2,y1+y2) \(\nleq\) p(x1+x2,y1+y2)

Option (4) is not correct.

Concept:

Understanding Sequences and Their Limits:

• Sequence: An ordered list of numbers defined by a function \( a_n \) on natural numbers.
• Convergence: A sequence \( a_n \) converges if it approaches a unique real number as \( n \to \infty \).
• lim sup: The greatest accumulation point (limit of suprema of tails).
• lim inf: The smallest accumulation point (limit of infima of tails).
• Max sequence: \( b_n = \max\{a_1, a_2, \dots, a_n\} \) is non-decreasing.
• Min sequence: \( c_n = \min\{a_1, a_2, \dots, a_n\} \) is non-increasing.

Calculation:

Given,

\( a_n = (-1)^n (1 + e^{-n}) \)

⇒ For even \( n \): \( a_n \approx 1 + e^{-n} \to 1 \)

⇒ For odd \( n \): \( a_n \approx -1 - e^{-n} \to -1 \)

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

\( \lim a_n \) does not exist 

Option 1: "\( a_n \) does not converge"

⇒ Correct.

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

Option 2: "\( \limsup a_n = \lim b_n \)"

→ False.

\( \limsup a_n = 1 \)

\( b_n = \max\{a_1, a_2, ..., a_n\} \)

Since \( a_2 = 1 + e^{-2} > 1 \), and this is the max forever, \( b_n \to 1 + e^{-2} \)

⇒ not equal to 1

Option 3: "\( \liminf a_n = \lim c_n \)"

→ False.

\( \liminf a_n = -1 \)

\( a_1 = -1 - e^{-1} \) which is minimum forever ⇒ \( c_n = -1 - e^{-1} \) for all large \( n \)

So \( \lim c_n = -1 - e^{-1} \ne -1 \)

Option 4: "\( \lim b_n = \lim c_n \)"

→ False.

As shown, \( \lim b_n = 1 + e^{-2} \)

\( \lim c_n = -1 - e^{-1} \)

Hence, not equal

∴ Only correct statement is Option 1.

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)

\(\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right) = f\left( {a,b} \right)\)

Note:

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

Calculation:

Given:

\(f\left( {x,y} \right) = \;\left\{ {\begin{array}{*{20}{c}} {\frac{{xy}}{{\sqrt {{x^2} + {y^2}} }}\;\;\;{x^2} + {y^2} \ne 0}\\ {0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x = y = 0} \end{array}} \right.\)

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

\(\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right) = f\left( {a,b} \right)\) and finite.

f(a,b) = f(0,0) ⇒ 0 (given)

\(\mathop {\lim }\limits_{\left( {r,\theta } \right) \to \left( {0,0} \right)} f\left( {r,\theta} \right) =\frac{ r^2cos\theta rsin\theta }{r} \) = 0 

f_x(0, 0) = \(\mathop {\lim }\limits_{\left( {h,0 } \right) \to \left( {0,0} \right)}\){f(h, 0) - f(0, 0)} / h = 0 

fy(0, 0) = \(\mathop {\lim }\limits_{\left( {0,k } \right) \to \left( {0,0} \right)}\){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 \(\rm\displaystyle\sum_{n=1}^{\infty}\)(-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) \(\lim_{n\to\infty}b_n=0\)

Explanation:

an = (−1)n+1\(\rm (\sqrt{n+1}−\sqrt{n})\)

    = (−1)n+1  \(\rm \frac{(\sqrt{n+1}−\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})}\) 

   = (−1)n+1\(\rm \frac{1}{(\sqrt{n+1}+\sqrt{n})}\)

So series is \(\rm\displaystyle\sum_{n=1}^{\infty}\)\(\rm \frac{(-1)^{n+1}}{(\sqrt{n+1}+\sqrt{n})}\) 

So here b_n = \(\rm \frac{1}{(\sqrt{n+1}+\sqrt{n})}\), bn+1 = \(\rm \frac{1}{(\sqrt{n+2}+\sqrt{n+1})}\)

\(\frac{b_{n+1}}{b_n}<1\) so bn+1 < bn  

Also \(\lim_{n\to\infty}b_n\) = \(\lim_{n\to\infty}\) \(\rm \frac{1}{(\sqrt{n+1}+\sqrt{n})}\) = 0

Hence by Leibnitz's test \(\rm\displaystyle\sum_{n=1}^{\infty}\) an is convergent.

Now the series is \(\rm\displaystyle\sum_{n=1}^{\infty}\)\(\rm |\frac{(-1)^{n+1}}{(\sqrt{n+1}+\sqrt{n})}|\) =  \(\rm\displaystyle\sum_{n=1}^{\infty}\)\(\rm \frac{1}{(\sqrt{n+1}+\sqrt{n})}\) = \(\rm\displaystyle\sum_{n=1}^{\infty}\) \(\rm \frac{1}{\sqrt n(\sqrt{1+\frac{1}{n}}+1)}\)

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.

Concept -

(i) If the sequence x_n convergent  then limsupn En = liminfn En 

Calculation:

Let {En} be a sequence of subsets of R 

\(\limsup _n E_n=\bigcap_{k=1}^{\infty} \bigcup_{n=k}^{\infty} E_n\)  and 

\(\liminf _n E_n=\bigcup_{k=1}^{\infty} \bigcap_{n=k}^{\infty} E_n\)

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

option 1 is incorrect

x \(∈\) \(\ {\cap}\) A_i imply x ∈ A_i 

x \(∈\)\(\ {\cap}\)(\(\ {\cup}\)E_n )

x \(∈\)\(\ {\cup}\) 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, \(\begin{bmatrix}2\\\ 1\\\ 0 \\\ 3 \end{bmatrix}\rm and \begin{bmatrix}1\\\ 0 \\\ 1 \\\ 2 \end{bmatrix}\)

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

Hence, \(\rm X - Y=\begin{bmatrix}1\\\ 1\\\ -1 \\\ 1 \end{bmatrix}\)

Explanation -

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

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

{(0, 𝑦) ∈ ℝ2 : −∞ < 𝑦 < ∞}  represents the whole y - axis and  sin \(\frac{1}{x}\) 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 \(\{f_n\}\) be a sequence of functions defined on a set D . We say that \(f_n\) converges pointwise to a function

 f  on D if, for every x \(\in\) D

\(\lim_{n \to \infty} f_n(x) = f(x).\)

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

\(\lim_{n \to \infty} \sup_{x \in D} |f_n(x) - f(x)| = 0.\)

Explanation: The problem gives a sequence of functions\( f_n: \mathbb{R} \to \mathbb{R}\) defined by

\(f_n(x) = \frac{x^2}{\sqrt{x^2 + \frac{1}{n}}}\)

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

We are asked to take the limit \(n \to \infty \) of the function:

 \(f_n(x) = \frac{x^2}{\sqrt{x^2 + \frac{1}{n}}} \)   

As \(n \to \infty \), the term \(\frac{1}{n} \to 0 \). So, for large n , the function \(f_n(x) \) approaches

 \(lim_{n \to \infty} f_n(x) = \frac{x^2}{\sqrt{x^2}} = \frac{x^2}{|x|}\)
 

Case 1: \(x \neq 0\)

For \(x \neq 0\) we have, \( f(x) = \frac{x^2}{|x|} = |x|\)
 

Case 2: \(x =0\)
When \(x =0\) , the function becomes \(f_n(0) = \frac{0^2}{\sqrt{0^2 + \frac{1}{n}}} = 0\)

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

The function f(x) , \(n \to \infty \) , is given by

   
  \( f(x) = \begin{cases} |x|, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} \)

This function is equal to |x| for all \(x \in \mathbb{R} \), 

Therefore, The correct option is 4).

Concept:
 

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

If S has \(n\)  elements, then \( \mathcal{P}(S) \) contains \( 2^n \) 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 \(\mathbb{N} \) ).
 

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

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

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

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

set \( \mathcal{P}(S) \) 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).

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.

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 = \(\frac{1}{2}\)

⇒ |x - y| = 1

If possible let 2n ε < |x - y|

i.e. 2n \(\frac{1}{2}\) < 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 = \(\rm \frac{1}{2^n}\) < 1 ∀ n ∈ ℕ 

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

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