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

Concept:

Greatest Integer Function:

• The greatest integer function, denoted by [x], returns the largest integer less than or equal to x.
• The function is also known as the floor function. Mathematically, [x] is defined as the greatest integer less than or equal to x.
• The greatest integer function is continuous everywhere except at integer points, where it is not differentiable.
• For differentiability, the function must have no "sharp corners" at the points of discontinuity.

Calculation:

Let's analyze each function in the options to match with the correct descriptions.

• (B)  x − |x|: This is a combination of absolute value functions. These are continuous and differentiable everywhere except at the points where the absolute values change, which are x = 1 and x = 2. Therefore, this function is differentiable everywhere except at x = 0.
• (A) |x – 1| + |x – 2| : This function involves the absolute value function. The greatest integer function has a discontinuity at integer points, and this function involves absolute values, which means it is continuous everywhere but not differentiable at x = 0. Hence, it is continuous everywhere.
• (C) x − [x]: This function involves the greatest integer function (floor function), which is continuous but not differentiable at integer points. Therefore, this function is not differentiable at x = 1 because there is a discontinuity at integer points.
• (D) |x|: This function is continuous and differentiable at all points, including x = 0. Therefore, it is differentiable at x = 1.

∴ Correct Matching: (A) - (II), (B) - (I), (C) - (III), (D) - (IV)

Explanation:

f(x) = 3/((1-x)(1+2x)) = A/(1-x) + B/(1+2x)

Solving for A and B, we get:

A = 1, B = 2

Therefore,  f(x) = 1/(1-x) + 2/(1+2x)

Now, we can use the geometric series expansion:

1/(1-x) = 1 + x + x² + x³ + x⁴ + ⋯ 

and 1/(1+2x) = 1 - 2x + 4x² - 8x³ + 16x⁴ + ⋯ 

Multiplying the second series by 2:

2/(1+2x) = 2 - 4x + 8x² - 16x³ + 32x⁴ + ⋯  

Now, we can add the two series to get the expansion of f(x):

f(x) = (1 + x + x² + x³ + x⁴ + ⋯ ) + (2 - 4x + 8x² - 16x³ + 32x⁴ + ⋯ )

To find the coefficient of x²:

x2 from the first series +x2 from the second series = +8x2

Therefore, the coefficient of x2 in the expansion of f(x) is 9

Hence 9 is the correct answer.

Explanation:  

Let xn = \(2n^{\frac{1}{n}}\) 

and   \(\lim _{n \rightarrow \infty} x_n = 2\) 

yn = \(e^{1-x_n}\) 

\(y_n = e^{1- 2n ^{1/n}} \)  

\(\lim _{n \rightarrow \infty} y_n = \lim _{n \rightarrow \infty} e^{1 -2 n^{1/n}} \)   = \(e^{1 - 2} = e^{-1} = 0.36\)   

Hence 0.36 is the Answer.

Explanation:

If   \(a_n = |\frac{f^n(0)}{n!}|^{\frac{1}{n}} \leq \frac{k^{\frac{1}{n}}}{(n!)^{1/n}} \) 

Now \(k^{\frac{1}{n}} \rightarrow 1 \)  and  \((n!)^{1/n} \rightarrow \infty as n \rightarrow \infty \) 

\( \frac{k^{\frac{1}{n}}}{(n!)^{1/n}} \rightarrow 0 \quad as \quad n \rightarrow \infty \)     

So, (1) is true and (2) is false

Consider the function:

\(f(x) = \begin{cases} x; & x \in (-\infty, 1) \\ x + 1; & x \in (1, \infty) \end{cases} \)   

Then  \(|f^n(0)| \leq 1 \forall n \in \mathbb{N} \)  but f'(x) does not exist at x = 1, so (3) is false. 

Consider \(\sum_{n=1}^{\infty} \frac{f^n(0)}{(n-1)!}\)   

Now  \( |\frac{f^n(0)}{(n-1)!}| \leq |\frac{k}{(n-1)!}| \) 

But  \(\sum_{n=1}^{\infty} \frac{k}{(n-1)!} \)  Converges, so  \(\sum_{n=1}^{\infty} |\frac{f^n(0)}{(n-1)!}| \)  Converges by comparison test

So,  \(\sum_{n=1}^{\infty} \frac{f^n(0)}{(n-1)!} \)  Converges absolutely\(\sum_{n=0}^{\infty} \frac{f^n(0)x^n}{n!} \)   converges to f(x) for each \(x \in (-1, 1) \)  

\(f(x) = \sum_{n=0}^{\infty} \frac{f^n(0)x^n}{n!} \)
 
f(0) = 0

Also, f(x) = 0,  \(\forall n \in \mathbb{N} \) 

Hence Option(1) and Option(4) are correct 

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).

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.

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