Continuity & Differentiability MCQ Quiz - Objective Question with Answer for Continuity & Differentiability - Download Free PDF

Concept:

Limit of Riemann Sum vs Definite Integral:

• The given expression compares a Riemann sum to the corresponding definite integral.
• Let , which is continuous on .
• We are given:
• Note that is a right Riemann sum approximation of .
• Hence, the given expression simplifies to:
• Using the Euler–Maclaurin expansion: 

Calculation:

Given,

⇒ Apply Euler–Maclaurin expansion:

⇒

⇒

∴ The correct answer is , i.e., Option 3.

Concept:

Multivariable Integral as Expectation:

• The n-fold integral over is the expected value of the integrand when are i.i.d. random variables.
• Set and . The integrand equals .
• Law of Large Numbers (LLN): and in probability as .
• Thus in probability.
• Since , the Dominated Convergence Theorem allows passing the limit inside the expectation.

Calculation:

Given,

 for 

⇒

⇒

⇒ Need

⇒ By LLN, ,

⇒ Hence in probability

⇒ Integrand bounded by 1

⇒ dominated

⇒

∴ Value of the limit = .

Concept:

Greatest Integer Function:

• The greatest integer function, denoted by [x], returns the largest integer less than or equal to x.
• The function is also known as the floor function. Mathematically, [x] is defined as the greatest integer less than or equal to x.
• The greatest integer function is continuous everywhere except at integer points, where it is not differentiable.
• For differentiability, the function must have no "sharp corners" at the points of discontinuity.

Calculation:

Let's analyze each function in the options to match with the correct descriptions.

• (B)  x − |x|: This is a combination of absolute value functions. These are continuous and differentiable everywhere except at the points where the absolute values change, which are x = 1 and x = 2. Therefore, this function is differentiable everywhere except at x = 0.
• (A) |x – 1| + |x – 2| : This function involves the absolute value function. The greatest integer function has a discontinuity at integer points, and this function involves absolute values, which means it is continuous everywhere but not differentiable at x = 0. Hence, it is continuous everywhere.
• (C) x − [x]: This function involves the greatest integer function (floor function), which is continuous but not differentiable at integer points. Therefore, this function is not differentiable at x = 1 because there is a discontinuity at integer points.
• (D) |x|: This function is continuous and differentiable at all points, including x = 0. Therefore, it is differentiable at x = 1.

∴ Correct Matching: (A) - (II), (B) - (I), (C) - (III), (D) - (IV)

Explanation:  

Let xn =  

and    

yn =  

   =    

Hence 0.36 is the Answer.

Concept:

If a function is continuous on a dense set , it doesn't necessarily imply that the function is

continuous on all of  , especially on , the complement of   in  .

Explanation:

Option 1: Continuity on a dense subset  does not imply continuity on the whole set .

A function can be continuous on a dense subset but exhibit discontinuities on .

Therefore, this option is incorrect.

Option 2:  is defined only on , so even if  is continuous on , it says nothing about 's

continuity on the rest of . Continuity of  does not guarantee the continuity of  everywhere.

Hence, this option is incorrect.

Option 3:  If  for all  (which is dense in ), and  is continuous on ,

by the density of ,  must be 0 everywhere on , because a continuous function on a dense set that is

0 must be 0 on the entire set. Therefore, this option is correct.

Option 4:  being identically 0 on  and  being continuous on ) does not imply  is identically 0

on . Continuity on  does not extend to the whole set without further conditions.

Therefore, this option is incorrect.

The correct answer is Option 3).

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:

Recall: x2 is not uniformly continuous function on ℝ. but sin x is uniformly continuous on ℝ.

(1) h(x) = g(f(x)) = sin(x2)

Recall: For f ∶ ℝ → ℝ, of ∃ sequence {an} & {bn} s.t. |an − bn| → 0 but |f(an) − f(bn)| → 0 then f is not uniformly continuous.

Take an =  and bn = 

then  an − bn =   = 0

But sin(2nπ + ) − sin(2nπ − ) = 2 → as n → ∞

∴ h(x) is not uniformly continuous.

option (1) is false.

(2) h(x) = g(x) f(x) = sin x ⋅ x2

Recall: f is uniformly continuous on a set x if for given ε > 0, ∃ δ > 0 s.t. |f(x) − f(y)|

Suppose h(x) is uniformly continuous.

Then, for ε = 1 > 0 ∃ δ > 0 s.t. |h(x) − h(y)|

Take x ∈ ℝ and y = x + δ/2 then |x − y| = |x − x − δ/2| 2 sin x − (x + δ/2)2 sin(x + δ/2)|

≤ |x2 − (x + )2| = |x2 − x2 −  − δx|

⇒   

⇒ x  but we have taken x ∈ ℝ.

which is a contradiction.

∴ h(x) is not uniformly continuous. option (2) is false.

(3) h(x) = (sin x)2

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

here, h'(x) = 2 sin x cos x = sin 2x

then |sin 2x|

option (3) is true

(4) h(x) = x2 + sin x

Recall: If f + g is U.C. then f(x) ± g(x) is also U.C.

Now, let us supose h(x) is U.C. on ℝ.

Take, f1(x) = sin x and sin x is U.C. on ℝ.

then h(x) ± f1(x) is U.C. on ℝ.

⇒ x2 + sin x − sin x = x2 is U.C. on ℝ.

which is a contradiction as x2 is not U.C. on ℝ.

option (4) is false.

Concept:

L’Hospital’s Rule: If  =  = 0 or ± ∞ and g'(x) ≠ 0 for all x in I with x ≠ c and  exist then  = 

Explanation:

 (0/0 form so using L'hospital rule)

=  

= 

Again using L'hospital rule

= 

= 

It will be 0/0 form if

x - 2a = 0

⇒ a = 0

Option (1) 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.

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.

Explanation:

f(x) = 

f'(x) = 14x + 5, ∀ x ∈ ℝ / {0}

So, f''(x) = 14, ∀ x ∈ ℝ / {0}

Hence f'''(x) = 0, ∀ x ∈ ℝ / {0}

(2) is correct

Explanation:

f(x) = x|x| 

Using definition of differentiability,

f'(0) =  =  =  = 0

f is differentiable 

g(x) = x | cos x |

|cos x| = 

Now x | cos x | = 

So, LHD =  =  

RHD =  =  = 1

As LHD = RHD  at x = 0 so g(x) is differentiable at x = 0.

(3) correct

Explanation:

Option (1): Let k = 2, f(x) = x² and g(x) = - 1

as  x2  ≠ -1 ∀ x ∈ 

So There does not exists x0 ∈  such that f(x0) = g(x0) always.

Option (1) is false.

Option (2): Let k = 1 so f(x) = x and g(x) = 1

as  f(x)  = g(x) for x = 1

So there is x0 = 1 such that f(x0) = g(x0)

Option (2) is false.

Hence option (4) is true.