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

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 = \(2n^{\frac{1}{n}}\) 

and   \(\lim _{n \rightarrow \infty} x_n = 2\) 

yn = \(e^{1-x_n}\) 

\(y_n = e^{1- 2n ^{1/n}} \)  

\(\lim _{n \rightarrow \infty} y_n = \lim _{n \rightarrow \infty} e^{1 -2 n^{1/n}} \)   = \(e^{1 - 2} = e^{-1} = 0.36\)   

Hence 0.36 is the Answer.

Explanation:

Since Under continuous map image of compact set is compact 

⇒ f: [a , b] → (a, b) cam not be onto 

But By plotting the Graph , 

Option(1) , Option(3) and Option(4) are possible 

Explanation: 

By Plotting the Graph, We can observe that only Option(1) , Option(3) and Option(4) satisfies the conditions 

Concept:

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

continuous on all of \(\mathbb{R}\) , especially on \( R \setminus S \), the complement of \( S\)  in \(\mathbb{R} \) .

Explanation:

Option 1: Continuity on a dense subset \(S\) does not imply continuity on the whole set \(R\).

A function can be continuous on a dense subset but exhibit discontinuities on \(R \setminus S \).

Therefore, this option is incorrect.

Option 2: \(g\) is defined only on \(S\), so even if \(g\) is continuous on \(S\), it says nothing about \( f \)'s

continuity on the rest of \(​​R\). Continuity of \(g\) does not guarantee the continuity of \( f \) everywhere.

Hence, this option is incorrect.

Option 3:  If \(g(x) = f(x) = 0\) for all \(x \in S \) (which is dense in \(R\)), and \( f \) is continuous on \( R \setminus S \),

by the density of \(S\), \( f \) must be 0 everywhere on \(R\), 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: \(g\) being identically 0 on \(S \) and \( f \) being continuous on \( S \)) does not imply \( f \) is identically 0

on \( R \setminus S \). Continuity on \(S\) 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\(\rm\frac{1}{x}\)

\(\lim_{x\to0}\sin\frac1x\) does not exist so f(x) = sin\(\rm\frac{1}{x}\) is not uniformly continuous on (0, 1)

Option (1) is false

(3): f(x) = ex cos\(\rm\frac{1}{x}\)

\(\lim_{x\to0}\)ex cos\(\rm\frac{1}{x}\) does not exist so f(x) = ex cos\(\rm\frac{1}{x}\) is not uniformly continuous on (0, 1)

Option (3) is false

(4): f(x) = cos x cos\(\rm\frac{\pi}{x}\)

\(\lim_{x\to0}\)cos x cos\(\rm\frac{\pi}{x}\) does not exist as \(\lim_{x\to0}\)cos\(\rm\frac{\pi}{x}\) does not exist so f(x) = cos x cos\(\rm\frac{\pi}{x}\)  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 = \(\rm\sqrt{2 n \pi + \frac{\pi}{2}}\) and bn = \(\rm\sqrt{2 n \pi − \frac{\pi}{2}}\)

then \(\rm\displaystyle\lim_{n → ∞}\) an − bn = \(\rm\displaystyle\lim_{n → ∞}\) \(\rm\sqrt{2 n \pi + \frac{\pi}{2}} − \sqrt{2 n \pi − \frac{\pi}{2}}\) = 0

But sin(2nπ + \(\frac{\pi}{2}\)) − sin(2nπ − \(\frac{\pi}{2}\)) = 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)| < ε, whenever |x − y| < δ, x, y ∈ x.

Suppose h(x) is uniformly continuous.

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

Take x ∈ ℝ and y = x + δ/2 then |x − y| = |x − x − δ/2| < δ and |f(x) − f(y)| = |x2 sin x − (x + δ/2)2 sin(x + δ/2)|

≤ |x2 − (x + \(\frac{\delta}{2}\))2| = |x2 − x2 − \(\frac{\delta^2}{2}\) − δx| < 1

⇒ \(\rm\left| − \delta x − \frac{\delta^2}{4}\right|\) < 1 ⇒ δx + \(\frac{\delta^2}{4}\) < 1.

⇒ x < \(\frac{1 − \frac{\delta^2}{4}}{\delta}\) 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| < 1 ⇒ h(x) is uniformly continuous.

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 \(\lim_{x\to c}f(x)\) = \(\lim_{x\to c}g(x)\) = 0 or ± ∞ and g'(x) ≠ 0 for all x in I with x ≠ c and \(\lim_{x\to c}\frac{f'(x)}{g'(x)}\) exist then \(\lim_{x\to c}\frac{f(x)}{g(x)}\) = \(\lim_{x\to c}\frac{f'(x)}{g'(x)}\)

Explanation:

\(\lim_{x\to0}\frac{x(1-\cos x)-ax\sin x}{x^4}\) (0/0 form so using L'hospital rule)

= \(\lim_{x\to0}\frac{x sin x + 1 - cosx - ax cos x - asinx }{4x^3}\) 

= \(\lim_{x\to0}\frac{1 + (x-a) sin x - (ax + 1) cos x}{4x^3}\)

Again using L'hospital rule

= \(\lim_{x\to0}\frac{(x-a) cos x + sin x + (ax + 1) sin x - acos x}{12x^2}\)

= \(\lim_{x\to0}\frac{(x-2a) cos x + (ax + 2) sin x }{12x^2}\)

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 \(\lim _{x \rightarrow 0} g(x) = 1\)

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.

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 -

(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.

Explanation:

f(x) = \(\begin{cases}7 x^2+5 x+3 & ; x<0 \\ 7 x^2+5 x & ; x \geq 0\end{cases}\)

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) = \(\lim_{x\to0}\frac{f(x) -f(0)}{x-0}\) = \(\lim_{x\to0}\frac{x|x| -0}{x-0}\) = \(\lim_{x\to0}|x|\) = 0

f is differentiable 

g(x) = x | cos x |

|cos x| = \(\begin{cases}cos x, x< 0\\ cos x, x\geq 0\end{cases}\)

Now x | cos x | = \(\begin{cases}xcos x, x< 0\\ xcos x, x\geq 0\end{cases}\)

So, LHD = \(\lim_{x\to0}\frac{x\cos x -0}{x-0}\) = \(\lim_{x\to0}(\cos x) = 1\) 

RHD = \(\lim_{x\to0}\frac{x\cos x -0}{x-0}\) = \(\lim_{x\to0}(\cos x)\) = 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 ∈ \(\mathbb{R}\)

So There does not exists x0 ∈ \(\mathbb{R}\) 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.