Continuity & Differentiability MCQ Quiz in తెలుగు - Objective Question with Answer for Continuity & Differentiability - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

A function y = f(x) is uniformly continuous at an open interval (a, b) if f(x) is continuous on (a, b) and limit exist at end pints a, b.

Explanation:

(1): f(x) = sin

 does not exist so f(x) = sin is not uniformly continuous on (0, 1)

Option (1) is false

(3): f(x) = ex cos

ex cos does not exist so f(x) = ex cos is not uniformly continuous on (0, 1)

Option (3) is false

(4): f(x) = cos x cos

cos x cos does not exist as cos does not exist so f(x) = cos x cos  is not uniformly continuous on (0, 1)

Option (4) is false

(2): f(x) = e−1/x2

(Here f(x) is continuous in (0, 1) and limit exist at x = 0 and x = 1

so f(x) = e−1/x2 is uniformly continuous on (0, 1)

Option (2) is correct

Explanation:

f : ℝ → ℝ be defined by 

 and f' be its derivative

S = {c ∈ ℝ : f'(x) ≤ cf(x) for all x ∈ ℝ}.

i.e., S = {c ∈ ℝ : f'(x) - cf(x) ≤ 0 for all x ∈ ℝ}.

Let g(x) = e^-cx f(x)

Then g'(x) = e-cx f'(x) - ce-cx f(x) = e-cx {f'(x) - cf(x)}

Since e-cx > 0 for all x ∈ ℝ and f'(x) - cf(x) ≤ 0

So, g'(x) ≤ 0

then g(x) is decreasing function

So, for all x > 0, g(x) ≤ g(0) ⇒ g(x) ≤ 0 (as g(0) = 1.f(0) = 0)

i.e., for x ∈ (0, 1), g(x) = e-cx f(x) ≤ 0 ⇒ f(x) ≤ 0 , which is a contradiction as for x ∈ (0, 1), f(x) = (1-x)²sin(x²) ≥ 0 

Therefore we will not get any such c which satisfies f'(x) ≤ cf(x) for all x ∈ ℝ.   

Hence S = Ø

(1) is correct

Concept:

Let D = 

W√3 = √3 not contained in f(D) because f(D) = f() = 

Concept: 

If function is continuous and limit exist at the end points of the interval a and b then the function is uniformly continuous at (a, b)

Solution:

Here, a = 0 and b = 1 

Here  

At x = 0,  

so limit does not exist at end points 0

so,   is not uniformly continuous at (0, 1)

Therefore, Correct option is Option 2).

Concept:

Let  D be a nonempty subset of  R. A function  f: D→R is called uniformly continuous on  D if for any ε > 0, there exists δ > 0 such that if u, v ∈ D and  |u − v|

Explanation:

Let δ = ϵ then

||||  δ

⇒ |x_i - y_i|

⇒ |Max{|xi|} - Max{|yi|}| 

So f is uniformly continuous.

Option (4) is correct

Note: Here option (1) is also correct but we have to choose best option

Explanation:

Given 

f(0) = 0 and f(1)(0) = 0. So x = 0 is a repeated root.

Hence x² is factor of given polynomial.

Also f(-1) = 0 so (x + 1) is a factor of f.

let f(x) = x²(x  + 1)(αx + b)

Given f(1) = 1

So 1 = 2(α + b) ⇒ α + b = 1/2...(i)

Hence

f(x) = x2(x  + 1)(αx + b)

     = (x³ + x²)(αx + b)

    = αx⁴ + (α + b)x³ +bx²

   =  αx4 + (1/2)x3 + bx2 (by using (i))

∴ f'(x) = 4αx3 + (3/2)x2 + 2bx

⇒ f''(x) = 12αx² + 3x + 2b

⇒ f⁽³⁾(x) = 24αx + 3

So, f(3)(0) = 3

Option (4) is correct and option (3) is false

For α = -1 f(3)(1/2) = - 12 + 3 = - 9

Option (2) false.

As f(3)(0) = 3, there exists a ∈ (-1, 1) such that f(3)(a) ≥ 3

Option (1) is correct

Explanation:

 ≠ 1

So, f(x) is not continuous at x = 0

Option (2) and (4) false.

For, any irrational number f(x) = 0

so, continuous at for irrational number.

Option (1) is true.

Explanation -

For option(1) -

We have  f be continuously differentiable 2 π periodic real valued function,

= 

which shows that f' is also a 2π periodic.

Hence Option(1) is true and Option(2) is false.

For option(3) -

Since f is continuous and 2 π periodic means it is bounded, being uniformly continuous. Let M  be bound on f, then

Observe that   depends on n which means that no required C exists.

Hence Option(3) is false.

For Option(4) -

By Riemann Lebesgue lemma, If f be continuously differentiable 2 π periodic real valued function on the real line and 

then a_n → 0 as n → ∞ 

Hence Option(4) is false.

Concept:

Removal discontinuity: A real-valued univariate function f(x) is said to have a removable discontinuity at a point x₀ in its domain if both f(x₀) and f(x) = L 0) ≠ L.

Uniformly continuous: A function f(x) is said to be uniformly continuous at in an interval [a, ∞) if it is continuous on that interval and f(x) exists.

Explanation:

f(x) = 1 - |2x - 1| for x ∈ [0, 1].

f(x) = 

i.e., f(x) = 

LHL = f(x) = 2x = 1

RHL = f(x) = (2 - 2x) = 2 - 1 = 1

f(1/2) = 1

So, LHL = RHL = f(1/2)

hence f(x) is continuous in [0, ∞).

(1) is correct

f(x) is uniformly continuous in [0, ∞) since ax + b is unifromly continuous in [0, ∞).

(2) is correct

g(x) = f(x²) = 1 - |2x² - 1| for x ∈ [0, 1].

      =  

LHL = g(x) = 2x² = 1

RHL = f(x) = (2 - 2x²) = 2 - 1 = 1

f() = 1

So, LHL = RHL = f(1/2)

hence g(x) is continuous in [0, ∞).

(3) is correct

g(x) is not uniformly continuous in [0, ∞) as x² is not uniformly continuous in [0, ∞)

(4) is false

Concept:

(i) A function f(x) is uniformly continuous then it is continuous

(ii) If |f'(x)| ≤ K then f is uniformly continuous

Explanation:

(1): f(x) = sin does not exist at x = 0 so not continuous at x = 0 so not uniformly continuous

Option (1) is false

(2): f(x) = (sin x)2

f'(x) = 2 sin x cos x = sin 2x

Here |f'(x)| = |sin 2x|

⇒ f(x) =  (sin x)2 is uniformly continuous.

option (2) is true

(3) f(x) = x2 is uniformly continuous only on [-M, M] for fixed M > 0

Option (3) is false