Analysis MCQ Quiz in বাংলা - Objective Question with Answer for Analysis - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept -

If f : [a,b] → \(\mathbb{R}\) and f(a) > 0 and f(b) < 0 then there exist c ∈ (a,b) such that f(c) = 0

Explanation -

We have the polynomial f(x) = x4 - 3x3 - x2 + 4

Now f'(x) = 4x³ - 9x² - 2x = x( 4x² - 9x - 2) 

Now for the critical points 

f'(x) = 0

⇒  x( 4x2 - 9x - 2) = 0

⇒ x = 0 or 4x2 - 9x - 2 = 0

Now for 4x2 - 9x - 2 = 0 ⇒ x = \(\frac{9\pm\sqrt{81+ 32}}{8}= \frac{9\pm\sqrt{113}}{8}\)

⇒ we get three critical points of the given polynomial.

Now f(0) = 4 and f(1/2) = 1/16 - 3/8 -1/4 + 4 < 4 and f(1) = 1 - 3 - 1 + 4= 1

Now function is decreasing from 0 to 1.

Now f(2) = 16 - 24 - 4 + 4 = -8 < 0

Hence we get a one real roots in between 1 & 2.

Now f(3) > 0 and f(4) > f(3) 

Hence we get a one real roots in between 2 & 3.

Therefore we get two real roots in between  [1,4].

Hence option(3) is correct. 

Explanation -

Let a_n = n sin(2 π en!) we have 

\(e = \sum_{k=0}^n \frac{1}{k!} + \sum_{k=n+1}^{\infty} \frac{1}{k!} \)

⇒ \(2 \pi e n! = 2\pi r + 2\pi n!( \sum_{k=n+1}^{\infty} \frac{1}{k!} )\)

Where r is positive integer. so we have

\(lim_{ n \to \infty } n sin( 2\pi r + 2\pi n!( \sum_{k=n+1}^{\infty} \frac{1}{k!} ))\)

= \(lim_{ n \to \infty } n \ sin( 2\pi n!( \sum_{k=n+1}^{\infty} \frac{1}{k!} ))\)

Further, observe that 

\(\frac{1}{n+1} < ​​n! ( \sum_{k=n+1}^{\infty}\frac{1}{k!}) = \frac{1}{n+1}+ \frac{1}{(n+1)(n+2)}+..... < \frac{1}{n}+ \frac{1}{n^2} + ... = \frac{1}{n-1}\)

By squeeze principle, we have 

\(lim_{n \to \infty }​​n! ( \sum_{k=n+1}^{\infty}\frac{1}{k!}) = lim_{n \to \infty } b_n = 0\) and \(lim_{n \to \infty }n b_n = 1\)

So using the result that \(lim_{x \to 0 } \frac{sinx}{x} = 1\) we get 

\(lim_{n \to \infty } a_n = ​​lim_{n \to \infty } n sin(2 \pi b_n) = ​​lim_{n \to \infty }2 \pi b_n n \frac{sin(2 \pi b_n)}{2 \pi b_n} = 2\pi\)

Hence Option(3) is correct.

Concept -

Reimann Lebesgue Lemma -

If the Lebesgue Integral of |f| is finite then the fourier transform of |f| vanishes as its argument does to infinity.

Explanation -

We have the sequence  \(b_n = \int_{-\pi}^{\pi} f(t) sin(nt) dt\)

Note that f(x) being continuous on a compact set is bounded and |sin t | ≤ 1

Therefore  \(|a_n| \le \int_{-\pi}^{\pi} |f(t)|dt \le 2 \pi M\)   ∀ n  where M is bound on f(x).

Thus the sequence {b_n} is bounded.

\(b_n = \int_{-\pi}^{\pi} f(t) sin(nt) dt\)

integration by parts, we get 

b_n  = \([\frac{f(t) cos (nt)}{n}]_{-\pi}^{\pi} - \frac{1}{n} \int_{-\pi}^{\pi} f'(t) cos(nt) dt = - \frac{1}{n} \int_{-\pi}^{\pi} f'(t) cos(nt) dt \)

Since f'(t) is continuous then by Reimann Lebesgue Lemma, which is " If the Lebesgue Integral of |f| is finite then the fourier transform of |f| vanishes as its argument does to infinity. "

Thus in particular, b_n and n b_n → 0 as n → ∞ 

Hence option (1) and (2) is correct.

For option(iii) -

\(\sum _{n=1}^{\infty} n^3 b_n^3\) is also absolutely convergent because b_n being bounded and and cgs to 0.

Hence the option (3) is correct. 

Hence option(4) is the correct option.

Concept -

(i) ∑ |a_n | is convergent then ∑ an is absolutely convergent.

(ii) Ratio Test - 

If \(lim_{n \to \infty } \frac{a_{n+1}}{a_n}= p < 1\) then the series  ∑ an is convergent.

Explanation -

We have the series \(∑{3^n sin(\frac{1}{5^n x})}\) 

Now for Absolutely convergent -

\(∑ |{3^n sin(\frac{1}{5^n x})}| \le ∑ \frac{1}{x} \times (\frac{3}{5})^n\)

Now using Ratio Test -

\(lim_{n \to \infty } \frac{a_{n+1}}{a_n}= lim_{n \to \infty } (\frac{3}{5})= \frac{3}{5}< 1\)

Hence the series \(∑ \frac{1}{x} \times (\frac{3}{5})^n\) is convergent and the given series is absolutely convergent.

Hence Option (i) is true.

Concept -

(i)  If n is even then (-1)n = 1 

(ii)  If n is odd then (-1)n = -1

(iii)  \(\frac{19}{e^3}< 1\) then \((\frac{19}{e^3})^n \to 0 \ \ as \ \ n \to \infty\)

Explanation -

We have the sequence \(a_n = e^{-n} + (-1)^n cos^3(\frac{19}{e^3})^n+ (-1)^n (sin(\frac{1}{n^2}+ \frac{(-1)^n \pi }{2}))\)

Now as n →  ∞ ,

a_n = 0 + (-1)ⁿ cos³(0) + (-1)ⁿ\(sin(\frac{(-1)^n\pi}{2})\)

Now we make the cases -

Case - I - If n is even then put (-1)ⁿ = 1 in the above equation we get

an = 0 + 1 x cos3(0) + 1 x \(sin(\frac{\pi}{2})\) = 1 + 1 = 2

Case - II - If n is odd then put (-1)n = -1 in the above equation, we get

an = 0 - 1 x cos3(0) - 1 x \(sin(\frac{-\pi}{2})\) = -1 + 1 = 0

Hence largest and smallest limit points are 2 & 0.

So Options (i) & (iv) are wrong.

And we know that limit of the sequence is different in both the cases so not convergent.

Hence option (iii) is correct and (ii) is wrong.

Concept use:

Bounded set : A set S is bounded if it has both upper and lower bounds. 

Closed set: If a set contain each of its limit point in the set 

Calculations:

S = {x5 - x4<=100 where x ∈ R} is unbounded and Closed 

T = { x2 - 2x <67 where x ∈ (0, ∞)} is Open and bounded

Hence the Intersection of the Closed set and Open Set need not be closed set, but it is bounded also.

So, The Correct option is 2.

Concept:

Lagrange's mean value theorem:  Let f(x) be a continuous function in [a, b] and differentiable in (a, b) then there exist a point c ∈ (a, b) such that 

 \(f'(c)= \frac{f(b)-f(a)}{b-a}\)

Explanation:

For x, y ∈ I, by Lagrange's mean value theorem

\(\frac{f(x) - f(y)}{x - y} = f'(c)\) where x < c < y

⇒ f(x) - f(y) = (x - y)f'(c)

⇒ |f(x) - f(y)| = |x - y||f'(c)|

For a given ε > 0 ∃, \(δ = \frac{ε}{\alpha} > 0\) such that

 |f(x) - f(y)| < ε, ∀ |x - y| < δ, x, y ∈ I

Hence, f(x) is uniformly continuous on I.

We know that every uniformly continuous function is also continuous.

Given f(x) is differentiable.

Hence option (4) is true

Concept -

(i) Differentiability -

Let f(x) be a real-valued function defined on an interval [a,b], i.e. f : [a,b] → \(\mathbb{R}\), let a < c < b

If left-hand derivative of f(x) at c is equal to right-hand derivative of f(x) at c then f(x) is differentiable at c. where LHD = \(lim_{x → c^-} \frac{f(x) -f(c)}{x-c} \) and RHD = \(lim_{x → c^+} \frac{f(x) -f(c)}{x-c} \)

(ii) \(sin(x) = x - \frac{x^3}{3!}+ \frac{x^5}{5!}+....\)

(iii) \( lim_{x → 0} \frac{sin x^2}{x} = lim_{x → 0} \frac{x^2-\frac{x^6}{3!}+...}{x}=0\)  

Explanation -

For option (1) -

We have f(x) = sin( |x|x )

Now use the definition of differentiability - 

\(lim_{x → 0} \frac{f(x) -f(0)}{x-0} =\begin{cases} lim_{x → 0} \frac{-sin x^2}{x} & x< 0 \\ lim_{x → 0} \frac{sin x^2}{x} &x>0 \\ 0 & x = 0 \end{cases}\)

⇒ \(lim_{x → 0} \frac{f(x) -f(0)}{x-0} = 0\) as \( lim_{x → 0} \frac{sin x^2}{x} = lim_{x → 0} \frac{x^2-\frac{x^6}{3!}+...}{x}=0\)

Hence the function is differentiable at x = 0. So option (1) is true.

For option (2) -

We have \(f(x) =\begin{cases} sin(x^2) & if \ \ x\in \mathbb{Q} \\ 0 & otherwise\\ \end{cases}\)

Now use the definition of differentiability - 

\(lim_{x → 0} \frac{f(x) -f(0)}{x-0} =\begin{cases} lim_{x → 0} \frac{sin x^2}{x} & x \in \mathbb{Q} \\ 0 & otherwise \end{cases}\)

⇒ \(lim_{x → 0} \frac{f(x) -f(0)}{x-0} = 0\) as \( lim_{x → 0} \frac{sin x^2}{x} = lim_{x → 0} \frac{x^2-\frac{x^6}{3!}+...}{x}=0\)

Hence the function is differentiable at x = 0. So option (2) is true.

For option (3) -

We have \(f(x) =\begin{cases} sin(|x|) & if \ \ x\in \mathbb{Q} \\ 0 & otherwise\\ \end{cases}\)

Now use the definition of differentiability - 

\(lim_{x → 0} \frac{f(x) -f(0)}{x-0} =\begin{cases} lim_{x → 0} \frac{-sin x}{x} & x \in \mathbb{Q}, x< 0 \\ lim_{x → 0} \frac{sin x}{x} & x \in \mathbb{Q},x>0 \\ 0 & otherwise \end{cases}\)

⇒ \(lim_{x → 0^-} \frac{f(x) -f(0)}{x-0} = -1 \) and \(lim_{x → 0^+} \frac{f(x) -f(0)}{x-0} = 1 \) as \( lim_{x → 0} \frac{sin x}{x} = lim_{x → 0} \frac{x-\frac{x^3}{3!}+...}{x}=1\)

Hence the function is not differentiable at x = 0. So option (3) is false.

For option (4) -

We have f(x) = [x] sin2(πx) = \(\begin{cases} -sin^2 \pi x & 1 \le x < 0 \\ 0 & 0\leq x\leq 1 \\ \end{cases}\)

Now use the definition of differentiability - 

⇒ \(LHD =lim_{x → 0^-} \frac{f(x) -f(0)}{x-0} = lim_{x → 0^-} \frac{-sin^2 \pi x}{x} = 0\) and \(RHD =lim_{x → 0^+} \frac{f(x) -f(0)}{x-0} = lim_{x → 0^-} \frac{0}{x} = 0\) 

Hence the function is differentiable at x = 0. So option (4) is true.

Therefore option(3) is correct option.

Concept -

Mean Value Theorem -

If f(x) is differentiable then \(\frac{|f(x) -f(y)|}{|x-y|} \le sup|f'(t)|\)   

Explanation -

We have |sin2 x - sin2y| ≤ K |x - y|

⇒ \(\frac{|sin^2x-sin^2y|}{|x-y|} \le K\)

Now use Mean Value Theorem, we get -

⇒  K = sup |f'(t)| where f(t) = sin²(t)

⇒  f'(t) = 2 sin(t) cos(t) = sin(2t)

⇒  K = sup |sin(2t)| = 1 

Hence option(3) is correct.

Concept -

(i) Image of an interval under continuity is an interval.

(ii)  Image of compact set under continuity is compact.

(iii) Every interval in connected set.

Explanation -

We have B is closed interval in (o, ∞) and f(t) = sin(t) 

So clearly f(t) is a continuous function.

And B is a closed interval in (o, ∞) implies B is compact set because of boundedness.

We know that image of compact set under continuity is compact.

Hence A is compact set ⇒ closed set.

So option(1) and (4) are false.

we know that image of an interval under continuity is interval. Hence A is connected.

So option (3) is false.

Now the closure of \(\bar{A} = A \neq R\)  

Hence A is not dence in R.

Hence option(2) is true.