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

Concept:

The given function is a power series:

$$f(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$

This is the Taylor (Maclaurin) series expansion of $\sin (x)$.

Calculation:

Differentiate both sides with respect to x:

$$\frac{d}{dx}(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots )=1-\frac{3x^2}{3!}+\frac{5x^4}{5!}-\frac{7x^6}{7!}+\cdots$$

This new series is the Maclaurin expansion of $\cos (x)$.

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   $\underset{n\to \mathrm{\infty }}{lim}{x}_n=2$ 

yn = $e^{1-x_n}$ 

$y_n=e^{1-2n^{1/n}}$  

$\underset{n\to \mathrm{\infty }}{lim}{y}_n=\underset{n\to \mathrm{\infty }}{lim}{e}^{1-2n^{1/n}}$   = $e^{1-2}=e^{-1}=0.36$   

Hence 0.36 is the Answer.

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).

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.

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