Discontinuity MCQ Quiz - Objective Question with Answer for Discontinuity - Download Free PDF

Calculation:

Given: f(x) = \(\left\{\begin{array}{cc} x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0 \end{array}\right.\)

As we know function is said to be continuous at a point if limiting value of the function at that point is equal to the functional value of the function at that point.

Now, \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x^2 \sin \left(\frac{1}{x}\right)=0=f(0)\)

∴  f(x) is continuous at x = 0.

As we know function is said to be differentiable at a point if LHD = RHD at that point.

Now, L.H.D. at x = 0, f '(0^–) = \(\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h}\)

= \(\lim _{h \rightarrow 0} \frac{-h^2 \sin \left(\frac{1}{h}\right)-0}{-h}=\lim _{h \rightarrow 0} h \sin \left(\frac{1}{h}\right)=0\)

Now, R.H.D. at x = 0, f′(0⁺) = \(\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\)

= \(\lim _{h \rightarrow 0} h \sin \left(\frac{1}{h}\right)=0\)

∵ L.H.D. = R.H.D. at x = 0

∴ f(x) is differentiable at x = 0

Now, f′(x) = \(\begin{cases}2 x \sin \left(\frac{1}{x}\right)+x^2 \cos \left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right) & , x \neq 0 \\ 0 & , x=0\end{cases}\)

f′(x) = \(\begin{cases}2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0\end{cases}\)

Since, limit of f′(x) = 0 oscillates

∴ f′(x) is not continuous at x = 0

∴ f is continuous but f′ is not continuous.

The correct answer is Option 4.

Concept: 

 The greatest integer function ⌊x⌋ \lfloor x \rfloor" id="MathJax-Element-196-Frame" role="presentation" style="position: relative;" tabindex="0">⌊x⌋$\lfloor x \rfloor$ $\lfloor x \rfloor$  is known to have discontinuities at integer points because it jumps from one integer value to the next. Specifically, at any integer k k" id="MathJax-Element-197-Frame" role="presentation" style="position: relative;" tabindex="0">k$k$ $k$ , we have:

This jump discontinuity occurs at every integer.

Calculation:

Since f(x) f(x)" id="MathJax-Element-199-Frame" role="presentation" style="position: relative;" tabindex="0">f(x)$f(x)$ $f(x)$  is defined as:

At integer points k k" id="MathJax-Element-201-Frame" role="presentation" style="position: relative;" tabindex="0">k$k$ $k$ , we analyze the left-hand and right-hand limits:

For n≥2" id="MathJax-Element-203-Frame" role="presentation" style="position: relative;" tabindex="0">n≥2n≥2 n≥2

n \geq 2The two limits are not equal:

Thus, f(x) f(x)" id="MathJax-Element-205-Frame" role="presentation" style="position: relative;" tabindex="0">f(x)$f(x)$ $f(x)$  is discontinuous at all integer points.

Hence option 4 is true

Calculation:

Given: f(x) = \(\left\{\begin{array}{cc} x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0 \end{array}\right.\)

As we know function is said to be continuous at a point if limiting value of the function at that point is equal to the functional value of the function at that point.

Now, \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x^2 \sin \left(\frac{1}{x}\right)=0=f(0)\)

∴  f(x) is continuous at x = 0.

As we know function is said to be differentiable at a point if LHD = RHD at that point.

Now, L.H.D. at x = 0, f '(0^–) = \(\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h}\)

= \(\lim _{h \rightarrow 0} \frac{-h^2 \sin \left(\frac{1}{h}\right)-0}{-h}=\lim _{h \rightarrow 0} h \sin \left(\frac{1}{h}\right)=0\)

Now, R.H.D. at x = 0, f′(0⁺) = \(\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\)

= \(\lim _{h \rightarrow 0} h \sin \left(\frac{1}{h}\right)=0\)

∵ L.H.D. = R.H.D. at x = 0

∴ f(x) is differentiable at x = 0

Now, f′(x) = \(\begin{cases}2 x \sin \left(\frac{1}{x}\right)+x^2 \cos \left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right) & , x \neq 0 \\ 0 & , x=0\end{cases}\)

f′(x) = \(\begin{cases}2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0\end{cases}\)

Since, limit of f′(x) = 0 oscillates

∴ f′(x) is not continuous at x = 0

∴ f is continuous but f′ is not continuous.

The correct answer is Option 4.

Concept:

f(x) is continuous at x = a, if LHL = RHL = f(a)

 \(\rm\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=lim _{x \rightarrow a} f(x)\)

 f(x) is differentiable if LHD = RHD

\(\begin{array}{l} \rm L H D=\lim _{h \rightarrow 0^{-}} \frac{f(a-h)-f(a)}{-h} \\ \rm R H D=\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h} \end{array}\)

Discontinuity of the First Kind: A function f(x) is said to have a discontinuity of the first kind from the right at x = a if the right hand of the function exists but not equal to f(a).

Discontinuity of the Second Kind: A function f(x) is said to have discontinuity of the second kind at x = a, if neither left-hand limit of f(x) at x = a nor right-hand limit of f(x) at x = a exists.

Removable Discontinuity: A function f(x) is said to have a removable discontinuity at x = a if the left-hand limit at x tends to point ‘a’ is equal to the right-hand limit at x tends to point ‘a’ but their common value is not equal to f(a). 

Calculation:

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\left| x \right|}}{x}}\\ {0,} \end{array}\begin{array}{*{20}{c}} {,x ≠ 0}\\ {x = 0} \end{array}} \right.\)

For x ≠ 0,

f(x) = -x/x = -1,   if x < 0

f(x) = x/x = 1. if x > 0

Now,

LHL = \(\mathop {\lim }\limits_{x \to 0^- }f(x)=\mathop {\lim }\limits_{x \to 0^- }-x/x=-1\)

RHL = \(\mathop {\lim }\limits_{x \to 0^+ }f(x)=\mathop {\lim }\limits_{x \to 0^+}x/x=1\)

\(\mathop {\lim }\limits_{x \to 0 }f(x) = 0\)

Since,

LHL ≠ RHL we can say that the function is not continuous at x = 0

Only x = 0, is the point of discontinuity. 

Based on the definition, the function has discontinuity of the first kind. 

Explanation -

We have the function f(x) = [x] where [.] denotes the greatest integer function.

So the limit of this function at every integers is not exist.

e.g. - let x = 1 then 

\(lim_{ x \to 1^-} f(x) = 0\) and \(lim_{ x \to 1^+} f(x) = 1\)

Hence limit is not equal so not exist.

let x = -1 then ​​​

\(lim_{ x \to -1^-} f(x) = -2\)​ and \(lim_{ x \to 1^+} f(x) = -1\)

Hence limit is not equal so not exist.

Therefore, Points of discontinuity of the greatest integer function is all the integers.

Hence Option(3) is correct.

Concept:

If a function is continuous at a point a, then

\(\rm \displaystyle \lim_{x\rightarrow a} f(x) = f(a)\)

sin(∞) = a, Where -1≤ a ≤ 1

Calculation:

Given:

f(0) = 1

f(x) = x sin (1/x)

Checking continuity at x = 0

\(\rm \displaystyle \lim_{x\rightarrow 0} f(x) = f(0)\)

L.H.L

= \(\rm \displaystyle \lim_{x\rightarrow 0} f(x) = \rm \displaystyle \lim_{x\rightarrow 0} \ x\ \sin \left ( \frac 1 x\right ) = 0× \sin \left ( \frac {1}{ 0} \right )\)

= 0 × sin(∞)

= 0 

Similarly R.H.L = 0

L.H.L = R.H.L ≠ f(0)

Hence, function is discontinuous at x = 0.

Explanation:

To find the point of discontinuity of the function f(x) = [x], draw the graph of the function f(x) = [x].

Greatest Integer Function: (Floor function)

The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.

The domain of [x] is R and range is I, where R is the set of real numbers and I is the set of integers.

From the graph, we can say that the function is discontinuous at every integer.

Hence, the function f(x) = [x] discontinuous is at infinite points.

Concept:

f(x) is continuous at x = a, if LHL = RHL = f(a)

 \(\rm\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=lim _{x \rightarrow a} f(x)\)

 f(x) is differentiable if LHD = RHD

\(\begin{array}{l} \rm L H D=\lim _{h \rightarrow 0^{-}} \frac{f(a-h)-f(a)}{-h} \\ \rm R H D=\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h} \end{array}\)

Discontinuity of the First Kind: A function f(x) is said to have a discontinuity of the first kind from the right at x = a if the right hand of the function exists but not equal to f(a).

Discontinuity of the Second Kind: A function f(x) is said to have discontinuity of the second kind at x = a, if neither left-hand limit of f(x) at x = a nor right-hand limit of f(x) at x = a exists.

Removable Discontinuity: A function f(x) is said to have a removable discontinuity at x = a if the left-hand limit at x tends to point ‘a’ is equal to the right-hand limit at x tends to point ‘a’ but their common value is not equal to f(a). 

Calculation:

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\left| x \right|}}{x}}\\ {0,} \end{array}\begin{array}{*{20}{c}} {,x ≠ 0}\\ {x = 0} \end{array}} \right.\)

For x ≠ 0,

f(x) = -x/x = -1,   if x < 0

f(x) = x/x = 1. if x > 0

Now,

LHL = \(\mathop {\lim }\limits_{x \to 0^- }f(x)=\mathop {\lim }\limits_{x \to 0^- }-x/x=-1\)

RHL = \(\mathop {\lim }\limits_{x \to 0^+ }f(x)=\mathop {\lim }\limits_{x \to 0^+}x/x=1\)

\(\mathop {\lim }\limits_{x \to 0 }f(x) = 0\)

Since,

LHL ≠ RHL we can say that the function is not continuous at x = 0

Only x = 0, is the point of discontinuity. 

Based on the definition, the function has discontinuity of the first kind. 

Concept:

Let f(x) = \(\rm \frac {p(x)}{q(x)}\)

There are three conditions that need to be met by a function f(x) in order to be continuous at a number a. These are:

• f(a) is defined [you can’t have a hole in the function]
• \(\rm \lim_{x \to a} f(x)\) exists
• \(\rm \lim_{x \to a} f(x) = f(a)\)

Note:

if any of the three conditions of continuity is violated, the function is said to be discontinuous.

If sin x = 0 then x = nπ, n ∈ Z

Calculation:

Given: f(x) = cosec x

\(\rm cosec \;x = \frac{1}{\sin x}\)

Check where denominator becomes zero

sin x = 0

x = x = nπ, n ∈ Z

∴ Given function is discontinuous at x = nπ, n ∈ Z

Hence, option (3) is correct.

Important Points

• When dealing with a rational expression in which both the numerator and denominator are continuous.
• The only points in which the rational expression will be discontinuous where denominator becomes zero.

Alternate Method

To find the discontinuity of the function f(x) = cosec x, draw the graph.

From the graph, the function f(x) = cosec x is discontinuous at the point x = 0, π , 2π ...

Hence, the function f(x) = cosec x is discontinuous on the set {x = nπ, n ∈ Z}.

Concept:

f(x) is continuous at x = a, if LHL = RHL = f(a)

 \(\rm\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=lim _{x \rightarrow a} f(x)\)

 f(x) is differentiable if LHD = RHD

\(\begin{array}{l} \rm L H D=\lim _{h \rightarrow 0^{-}} \frac{f(a-h)-f(a)}{-h} \\ \rm R H D=\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h} \end{array}\)

Discontinuity of the First Kind: A function f(x) is said to have a discontinuity of the first kind from the right at x = a if the right hand of the function exists but not equal to f(a).

Discontinuity of the Second Kind: A function f(x) is said to have discontinuity of the second kind at x = a, if neither left-hand limit of f(x) at x = a nor right-hand limit of f(x) at x = a exists.

Removable Discontinuity: A function f(x) is said to have a removable discontinuity at x = a if the left-hand limit at x tends to point ‘a’ is equal to the right-hand limit at x tends to point ‘a’ but their common value is not equal to f(a). 

Calculation:

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\left| x \right|}}{x}}\\ {0,} \end{array}\begin{array}{*{20}{c}} {,x ≠ 0}\\ {x = 0} \end{array}} \right.\)

For x ≠ 0,

f(x) = -x/x = -1,   if x < 0

f(x) = x/x = 1. if x > 0

Now,

LHL = \(\mathop {\lim }\limits_{x \to 0^- }f(x)=\mathop {\lim }\limits_{x \to 0^- }-x/x=-1\)

RHL = \(\mathop {\lim }\limits_{x \to 0^+ }f(x)=\mathop {\lim }\limits_{x \to 0^+}x/x=1\)

\(\mathop {\lim }\limits_{x \to 0 }f(x) = 0\)

Since,

LHL ≠ RHL we can say that the function is not continuous at x = 0

Only x = 0, is the point of discontinuity. 

Based on the definition, the function has discontinuity of the first kind. 

Concept:

Continuity:- A function f(x) is said to be continuous at the point x = a if right hand limit and left hand limit of f(x) as x → a are equal and finite and both limits coincide with the value of function f(x) at x = a

f(x) is said to be continuous at x = a if

\(\lim\limits_{h \to 0} f(a + h) = \lim\limits_{h \to 0} f(a - h) = f(a)\)

The above definition can be stated in short form as: 

\(\lim\limits_{x \to a} f(x) = f(a)\)

Discontinuous Function:- If a function f(x) is not continuous at the point x = a, then f(x) is said to be discontinuous at x = a

Formula Used:

\(\lim\limits_{x \to 0}\frac{sinx}{x} = 1 \)

Calculation:

We have function,

f(x) = \(\frac{{{\rm{sin}}\,{\rm{x}}}}{{\left| {\rm{x}} \right|}}\)

Right hand limit

⇒ f(0 + 0) = \(\lim\limits_{h \to 0} \frac{sin \ h}{|h|}\)

⇒ f(0 + 0) = \(\lim\limits_{h \to 0} \frac{sinh}{h} = 1\)

Left hand limit

⇒ f(0 - 0) = \(\lim\limits_{h \to 0} \frac{sin \ (-h)}{|-h|}\)

⇒ f(0 - 0) = \(\lim\limits_{h \to 0} \frac{-sin \ h}{h}\) = - 1

⇒ f(0 + 0) ≠ f(0 - 0)

Hence, limit does not exists

So, the function is discontinuous.

∴ The function is discontinuous.

Concept:

Jump Discontinuity: 

A jump Discontinuity occurs when the right-hand and left-hand limits exist but are not equal.

eg. \(f\left( x \right) = \;\left\{ {\begin{array}{*{20}{c}} {x + 1}&{x > 0}\\ { - x}&{x \le 0} \end{array}} \right.\)

Calculation:

(a) g(x) = ln|x| ⇒

 (b) \(g\left( x \right) = \frac{1}{{{x^2}}} \Rightarrow \)

(c) \(g\left( x \right) = \;\left\{ {\begin{array}{*{20}{c}} {\frac{{\sin \left( x \right)}}{x}}&{x \ne 0}\\ 0&{x = 0} \end{array}} \right.\)

∴ only option (3) satisfy condition for jump discontinuity.

Concept:

If a function is continuous at a point a, then

\(\rm \displaystyle \lim_{x\rightarrow a} f(x) = f(a)\)

sin(∞) = a, Where -1≤ a ≤ 1

Calculation:

Given:

f(0) = 1

f(x) = x sin (1/x)

Checking continuity at x = 0

\(\rm \displaystyle \lim_{x\rightarrow 0} f(x) = f(0)\)

L.H.L

= \(\rm \displaystyle \lim_{x\rightarrow 0} f(x) = \rm \displaystyle \lim_{x\rightarrow 0} \ x\ \sin \left ( \frac 1 x\right ) = 0× \sin \left ( \frac {1}{ 0} \right )\)

= 0 × sin(∞)

= 0 

Similarly R.H.L = 0

L.H.L = R.H.L ≠ f(0)

Hence, function is discontinuous at x = 0.

Explanation:

To find the point of discontinuity of the function f(x) = [x], draw the graph of the function f(x) = [x].

Greatest Integer Function: (Floor function)

The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.

The domain of [x] is R and range is I, where R is the set of real numbers and I is the set of integers.

From the graph, we can say that the function is discontinuous at every integer.

Hence, the function f(x) = [x] discontinuous is at infinite points.

Concept:

f(x) is continuous at x = a, if LHL = RHL = f(a)

 \(\rm\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=lim _{x \rightarrow a} f(x)\)

 f(x) is differentiable if LHD = RHD

\(\begin{array}{l} \rm L H D=\lim _{h \rightarrow 0^{-}} \frac{f(a-h)-f(a)}{-h} \\ \rm R H D=\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h} \end{array}\)

Discontinuity of the First Kind: A function f(x) is said to have a discontinuity of the first kind from the right at x = a if the right hand of the function exists but not equal to f(a).

Discontinuity of the Second Kind: A function f(x) is said to have discontinuity of the second kind at x = a, if neither left-hand limit of f(x) at x = a nor right-hand limit of f(x) at x = a exists.

Removable Discontinuity: A function f(x) is said to have a removable discontinuity at x = a if the left-hand limit at x tends to point ‘a’ is equal to the right-hand limit at x tends to point ‘a’ but their common value is not equal to f(a). 

Calculation:

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\left| x \right|}}{x}}\\ {0,} \end{array}\begin{array}{*{20}{c}} {,x ≠ 0}\\ {x = 0} \end{array}} \right.\)

For x ≠ 0,

f(x) = -x/x = -1,   if x < 0

f(x) = x/x = 1. if x > 0

Now,

LHL = \(\mathop {\lim }\limits_{x \to 0^- }f(x)=\mathop {\lim }\limits_{x \to 0^- }-x/x=-1\)

RHL = \(\mathop {\lim }\limits_{x \to 0^+ }f(x)=\mathop {\lim }\limits_{x \to 0^+}x/x=1\)

\(\mathop {\lim }\limits_{x \to 0 }f(x) = 0\)

Since,

LHL ≠ RHL we can say that the function is not continuous at x = 0

Only x = 0, is the point of discontinuity. 

Based on the definition, the function has discontinuity of the first kind.