Analysis MCQ Quiz in मल्याळम - Objective Question with Answer for Analysis - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given:

f(x, y) = \(\frac{siny}{x}\) (x, y) → (0, 0)

Concept Used:

Putting y = mx in the function and checking whether the function is free from m then limit will exist if not then limit will not exist.

Solution:

We have,

f(x, y) = \(\frac{siny}{x}\) (x, y) → (0, 0)

Put y = mx

So, 

lim (x, y) → (0, 0) \(\frac{siny}{x}\)

⇒ lim x → 0 \(\frac{sin mx}{x}\)
 

We cannot eliminate m from the above function.

Hence limit does not exist.

\(\therefore\) Option 4 is correct.

Explanation:

Here, it is given that

(i) f(x + y) = f(x).f(y) and

(ii) f(x) = 1 + x g(x), where \(\lim _{x \rightarrow 0} g(x) = 1\)

Now, writing for y in the given condition. We have

f(x + h) = f(x).f(h)

Then, f(x + h) - f(x) = f(x)f(h) - f(x)

Or \(\frac{f(x+h)-f(x)}{h}= \frac{f(x)[f(h) - 1]}{h}\)

                      = \( \frac{f(x)h. g(h)}{h}=f(x). g(h)\) (using (ii))

Hence, \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} f(x) \cdot g(h)=f(x) \cdot 1\)

Since, by hypothesis \(\lim_{h \rightarrow 0} g(h) = 1\)

It follows that f'(x) = f(x)

Since, f(x) exists, f'(x) also exists

and f'(x) = f(x) 

⇒ \(\frac{d}{dx} f(x) = f(x)\)

(2) is true.

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 use:

Cauchy's first theorem on limits: If an is a sequence of positive terms such that 

\(\rm\displaystyle\lim_{n \rightarrow \infty} a_n = l\) then \(\rm\displaystyle\lim_{n \rightarrow \infty}\left(\frac{a_1+a_2+\ldots+a_n}{n}\right) = l\)

Explanation:

Let an = \(\rm \frac{n}{(n+n)^3}\) = \(\frac{n}{8n^3}\) = \(\frac{1}{8n^2}\)

⇒ \(\rm\displaystyle\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty}\frac{1}{8n^2}\) = 0

Hence, by Cauchy's first theorem on limits, we have

\(\rm\displaystyle\lim_{n \rightarrow \infty}\left(\frac{a_1+a_2+\ldots+a_n}{n}\right)\) = 0

⇒ \(\rm\displaystyle\lim_{n \rightarrow \infty} \frac{1}{n}\left[\frac{n}{(n+1)^3}+\frac{n}{(n+2)^3}+\ldots+\frac{n}{(2 n)^3}\right]\) = 0

⇒ \(\rm\displaystyle\lim_{n \rightarrow \infty}\left[\frac{1}{(n+1)^3}+\frac{1}{(n+2)^3}+\ldots+\frac{1}{(2 n)^3}\right]\) = 0

Now, \(\rm\displaystyle\lim_{n \rightarrow \infty}\left[\frac{1}{n^3}+\frac{1}{(n+1)^3}+\frac{1}{(n+2)^3}+\ldots+\frac{1}{(2 n)^3}\right]\) = 0

Sequence converges to 0

Option (1) is correct

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.