Analysis MCQ Quiz in தமிழ் - Objective Question with Answer for Analysis - இலவச PDF ஐப் பதிவிறக்கவும்

Concept use:

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

 then 

Explanation:

Let an =  =  = 

⇒  = 0

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

 = 0

⇒  = 0

⇒  = 0

Now,  = 0

Sequence converges to 0

Option (1) is correct

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

Given:

f(x, y) =  (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 
 

We cannot eliminate m from the above function.

Hence limit does not exist.

 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 

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 

                      =  (using (ii))

Hence, 

Since, by hypothesis 

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

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

and f'(x) = f(x) 

⇒ 

(2) 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 -

If f : [a,b] →  and f(a) > 0 and f(b)

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 = 

⇒ we get three critical points of the given polynomial.

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

Now function is decreasing from 0 to 1.

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

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 

⇒ 

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:

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 -

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.