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

Explanation -

Let a_n = n sin(2 π en!) we have 

⇒ 

Where r is positive integer. so we have

= 

Further, observe that 

By squeeze principle, we have 

 and 

So using the result that  we get 

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  

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

Therefore     ∀ n  where M is bound on f(x).

Thus the sequence {b_n} is bounded.

integration by parts, we get 

b_n  = 

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) -

 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  then the series  ∑ an is convergent.

Explanation -

We have the series  

Now for Absolutely convergent -

Now using Ratio Test -

Hence the series  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)   then 

Explanation -

We have the sequence 

Now as n →  ∞ ,

a_n = 0 + (-1)ⁿ cos³(0) + (-1)ⁿ

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  = 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  = -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

T = { x2 - 2x

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.

Solution:

Given function is:

F(x, y) = x3 + y3 – 3xy – 4 = 0

And x = 2 and g(2) = 2

Now,

F(2, 2) = (2)3 + (2)3 – 3(2)(2) – 4

= 8 + 8 – 12 – 4

= 0

So, F(2, 2) = 0

∂F/∂x = ∂/∂x (x3 + y3 – 3xy – 4) = 3x2 – 3y

∂F/∂y = ∂/∂y (x3 + y3 – 3xy – 4) = 3y2 – 3x

Let us calculate the value of ∂F/∂y at (2, 2).

That means, ∂F(2, 2)/∂y = 3(2)2 – 3(2) = 12 – 6 = 6 ≠ 0.

Thus, ∂F/∂y is continuous everywhere.

Hence, by the implicit function theorem, we can say that there exists a unique function g defined in the neighborhood of x = 2 by g(x) = y, where F(x, y) = 0 such that g(2) = 2.

Also, we know that ∂F/∂x is continuous.

Now, by implicit function theorem, we get;

g’(x) = -[∂F(x, y)/∂x]/ [∂F(x, y)/ ∂y]

= -(3x2 – 3y)/(3y2 – 3x)

= -3(x2 – y)/ 3(y2 – x)

= -(x2 – y)/(y2 – x)

Hence, option 3 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 

Explanation:

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

 where x

⇒ f(x) - f(y) = (x - y)f'(c)

⇒ |f(x) - f(y)| = |x - y||f'(c)|

For a given ε > 0 ∃,  0\) such that

 |f(x) - f(y)|

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] → , let a

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 =  and RHD = 

(ii) 

(iii)   

Explanation -

For option (1) -

We have f(x) = sin( |x|x )

Now use the definition of differentiability - 

0 \\ 0 & x = 0 \end{cases}\)

⇒  as 

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

For option (2) -

We have 

Now use the definition of differentiability - 

⇒  as 

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

For option (3) -

We have 

Now use the definition of differentiability - 

0 \\ 0 & otherwise \end{cases}\)

⇒  and  as 

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

For option (4) -

We have f(x) = [x] sin2(πx) = 

Now use the definition of differentiability - 

⇒  and  

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    

Explanation -

We have |sin2 x - sin2y| ≤ K |x - y|

⇒ 

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   

Hence A is not dence in R.

Hence option(2) is true.