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 $a_n$ on natural numbers.
• Convergence: A sequence $a_n$ converges if it approaches a unique real number as $n\to \mathrm{\infty }$.
• 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: "$lim sup{a}_n=lim{b}_n$"

→ False.

$lim sup{a}_n=1$

$b_n=max\{a_1,a_2,...,a_n\}$

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: "$lim inf{a}_n=lim{c}_n$"

→ False.

$lim inf{a}_n=-1$

$a_1=-1-e^{-1}$ 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:

• Pointwise Convergence: A sequence of functions $f_n$ converges pointwise to $f$ on an interval if for each $x$ in the interval, $\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)=f(x)$.
• Uniform Convergence: A sequence $f_n$ converges uniformly to $f$ if the convergence is uniform across the entire domain, i.e., $\underset{x\in [0,1]}{sup}|f_n(x)-f(x)|\to 0$.
• Dirac-like Behavior: Functions like $f_n(x)=nx(1-x{)}^n$ 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,

$f_n(x)=nx(1-x{)}^n$ on $[0,1]$

Step 1: Evaluate pointwise limit

⇒ Fix $x\in [0,1)$

⇒ $(1-x{)}^n\to 0\Rightarrow f_n(x)\to 0$

⇒ At $x=1\Rightarrow f_n(1)=0$

⇒ At $x=0\Rightarrow f_n(0)=0$

⇒ So $f_n(x)\to 0$ for all $x\in [0,1]$

⇒ Pointwise limit is $f(x)=0$ (a continuous function)

Step 2: Uniform convergence?

⇒ $f_n(x)$ peaks near $x=\frac{1}{n+1}$

⇒ Maximum value: $f_n\left(\frac{1}{n+1}\right)=\frac{n}{n+1}{\left(\frac{n}{n+1}\right)}^n\to 1$

⇒ So sup norm $\Vert f_n-0{\Vert }_{\mathrm{\infty }}\nrightarrow 0$

⇒ Not uniformly convergent

Step 3: Summary of options

⇒ Statement 1: FALSE (pointwise convergence occurs)

⇒ Statement 2: TRUE (converges to $f(x)=0$)

⇒ 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 $f(r)=\frac{2^{2^{24}r}}{e}$ is strictly increasing and continuous.
• Since $\mathbb{Q}$ is dense in $\mathbb{R}$, the image $f(\mathbb{Q})$ is dense in the range $(0,\mathrm{\infty })$.
• Hence, for any $x<y\in \mathbb{R}$, there exists a rational number $r$ such that $x<\frac{2^{2^{24}r}}{e}<y$.
• Statement 1 is true.

2. Limit Superior and Growth Rate:

• Given $lim sup\frac{a_n}{\log n}=L$, this implies that $a_n$ grows asymptotically like $\log n$.
• However, since $\log n\to \mathrm{\infty }$, the sequence $a_n$ can still diverge.
• So, $lim sup{a}_n$ can be ∞ even though the ratio with $\log n$ is finite.
• Statement 2 is false.

3. Countability of Finite Subsets:

• The set $\mathbb{Q}$ is countable.
• A countable union of finite subsets of a countable set is countable.
• There are infinitely many finite subsets of $\mathbb{Q}$.
• Statement 3 is true.

4. Continuous Functions from $\mathbb{R}$ to $\{0,1\}$:

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

Calculation:

Given,

Let $x,y\in \mathbb{R}$ with $x<y$

Function: $f(r)=\frac{2^{2^{24}r}}{e}$

⇒ f is strictly increasing and continuous

⇒ Image of $\mathbb{Q}$ under f is dense in $(0,\mathrm{\infty })$

⇒ ∃ rational $r$ such that $x<f(r)<y$

⇒ Statement 1 is true

Given: $lim sup\frac{a_n}{\log n}=L$

⇒ a_n ≈ L × log n

⇒ But $\log n\to \mathrm{\infty }$

⇒ So $lim sup{a}_n=\mathrm{\infty }$ is possible

⇒ Statement 2 is false

Set of finite subsets of $\mathbb{Q}$ is:

⇒ Countable union of finite sets

⇒ countable

⇒ Infinite in number

⇒ Statement 3 is true

Continuous functions $f:\mathbb{R}\to \{0,1\}$

⇒ 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 $\mathbf{1}$ and $\mathbf{3}$ are true.

Concept:

• Monotonic Function: A function $f:[0,1]\to \mathbb{R}$ 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,

$f:[0,1]\to \mathbb{R}$ 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)

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

Note:

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

Calculation:

Given:

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

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

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

f(a,b) = f(0,0) ⇒ 0 (given)

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

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

fy(0, 0) = $\underset{\left(0,k\right)\to \left(0,0\right)}{lim}${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 ${\displaystyle \sum _{\mathrm{n}=1}^{\mathrm{\infty }}}$(-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) $\underset{n\to \mathrm{\infty }}{lim}{b}_n=0$

Explanation:

an = (−1)n+1$(\sqrt{\mathrm{n}+1}-\sqrt{\mathrm{n}})$

    = (−1)n+1  $\frac{(\sqrt{\mathrm{n}+1}-\sqrt{\mathrm{n}})(\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}})}{(\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}})}$ 

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

So series is ${\displaystyle \sum _{\mathrm{n}=1}^{\mathrm{\infty }}}$$\frac{(-1{)}^{\mathrm{n}+1}}{(\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}})}$ 

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

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

Also $\underset{n\to \mathrm{\infty }}{lim}{b}_n$ = $\underset{n\to \mathrm{\infty }}{lim}$ $\frac{1}{(\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}})}$ = 0

Hence by Leibnitz's test ${\displaystyle \sum _{\mathrm{n}=1}^{\mathrm{\infty }}}$ an is convergent.

Now the series is ${\displaystyle \sum _{\mathrm{n}=1}^{\mathrm{\infty }}}$$|\frac{(-1{)}^{\mathrm{n}+1}}{(\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}})}|$ =  ${\displaystyle \sum _{\mathrm{n}=1}^{\mathrm{\infty }}}$$\frac{1}{(\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}})}$ = ${\displaystyle \sum _{\mathrm{n}=1}^{\mathrm{\infty }}}$ $\frac{1}{\sqrt{\mathrm{n}}(\sqrt{1+\frac{1}{\mathrm{n}}}+1)}$

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 

$\underset{n}{lim sup}{E}_n=\bigcap _{k=1}^{\mathrm{\infty }}\bigcup _{n=k}^{\mathrm{\infty }}{E}_n$  and 

$\underset{n}{lim inf}{E}_n=\bigcup _{k=1}^{\mathrm{\infty }}\bigcap _{n=k}^{\mathrm{\infty }}{E}_n$

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

option 1 is incorrect

x $\in$ $\cap$ A_i imply x ∈ A_i 

x $\in$$\cap$($\cup$E_n )

x $\in$$\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, $\left[\begin{array}{c}2\\ 1\\ 0\\ 3\end{array}\right]\mathrm{a}\mathrm{n}\mathrm{d}\left[\begin{array}{c}1\\ 0\\ 1\\ 2\end{array}\right]$

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

Hence, $\mathrm{X}-\mathrm{Y}=\left[\begin{array}{c}1\\ 1\\ -1\\ 1\end{array}\right]$

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

$\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)=f(x).$

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

$\underset{n\to \mathrm{\infty }}{lim}\underset{x\in D}{sup}|f_n(x)-f(x)|=0.$

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 $n\to \mathrm{\infty }$, denoted by $f(x)$ . We are tasked with determining which statement about f(x) is true.

We are asked to take the limit $n\to \mathrm{\infty }$ of the function:

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

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

 $li{m}_{n\to \mathrm{\infty }}{f}_n(x)=\frac{x^2}{\sqrt{x^2}}=\frac{x^2}{|x|}$
 

Case 1: $x\ne 0$

For $x\ne 0$ 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 $n\to \mathrm{\infty }$, we get f(0) = 0 .

The function f(x) , $n\to \mathrm{\infty }$ , is given by

   
  $f(x)=\{\begin{array}{ll}|x|,& \text{if }x\ne 0\\ 0,& \text{if }x=0\end{array}$

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. $\rho (T)=1and\eta (T)=3$

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 = $\frac{1}{2^{\mathrm{n}}}$ < 1 ∀ n ∈ ℕ 

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

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