Continuity of a function MCQ Quiz - Objective Question with Answer for Continuity of a function - Download Free PDF

Concept:

​A function f(x) is continuous at x = a, if $\underset{x\to a^{-}}{lim}$ f(x) = $\underset{x\to a^{+}}{lim}$f(x) = f(a).

Calculation:

Given: f(x) = $\{\begin{array}{l}\mathrm{m}\mathrm{x}+1,\text{ if }\mathrm{x}\le \frac{\pi }{2}\\ \sin \mathrm{x}+\mathrm{n},\text{ if }\mathrm{x}>\frac{\pi }{2}\end{array}$

f($\frac{\pi }{2}$) = m × $\frac{\pi }{2}$ + 1

Left-hand limit = $\underset{h\to 0}{lim}f(\frac{\pi }{2}-h)$

$=\underset{h\to 0}{lim}m\times (\frac{\pi }{2}-h)+1$

Applying the limits:

Left- hand limit = m × $\frac{\pi }{2}$ + 1

Right-hand limit = $\underset{h\to 0}{lim}f(\frac{\pi }{2}+h)$

$=\underset{h\to 0}{lim}\sin (\frac{\pi }{2}+h)+n$

Applying the limits:

 $=\sin \frac{\pi }{2}+n$

Right-hand limit = 1 + n

For the function to be continuous at x = $\frac{\pi }{2}$,

Left-hand limit = Right-hand limit = f(π/2)

⇒ m× $\frac{\pi }{2}$ + 1 = 1 + n

⇒ n = $\frac{m\pi }{2}$

The correct answer is n = $\frac{m\pi }{2}$ .

Concept:

If a function is continuous at a point a, then

${\displaystyle \underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})}$

sin(∞) = a, Where -1≤ a ≤ 1

Calculation:

Given:

f(0) = 1

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

Checking continuity at x = 0

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(0)}$

L.H.L

= ${\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})={\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{x}\sin \left(\frac{1}{\mathrm{x}}\right)=0\times \sin \left(\frac{1}{0}\right)}}$

= 0 × sin(∞)

= 0 

R.H.L

= f(0) = 1

L.H.L ≠ R.H.L

Hence, function is discontinuous at x = 0.

Concept:

If a function is continuous at x = a, then L.H.L = R.H.L = f(a).

Left hand limit (L.H.L) of f(x) at x = a is $\underset{h\to 0}{lim}f(a-h)$ 

Right hand limit (R.H.L) of f(x) at x = a is $\underset{h\to 0}{lim}f(a+h)$

Calculation:

Concept:

The function f(x) is continuous at x = a if

 f(a-) = f(a) = f(a+)

Calculation:

Given, f(x) = |x| 

For x ≥ 0, f(x) = x

and for x < 0, f(x) = - x 

So function is continuous for x > 0 and x < 0

At x = 0, 

f(0-) = f(0) = f(0+) = 0

⇒ f(x) is continuous at x = 0

∴ The correct answer is option (1).

Concept:

The function f(x) is continuous at x = a if

 f(a-) = f(a) = f(a+)

Calculation:

Given, f(x) = |x| 

For x ≥ 0, f(x) = x

and for x < 0, f(x) = - x 

So function is continuous for x > 0 and x < 0

At x = 0, 

f(0-) = f(0) = f(0+) = 0

⇒ f(x) is continuous at x = 0

∴ The correct answer is option (1).

Concept:

Definition:

• A function f(x) is said to be continuous at a point x = a in its domain, if ${\displaystyle \underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}(\mathrm{x})}$ exists or or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ ${\displaystyle \underset{\mathrm{x}\to {\mathrm{a}}^{+}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})}$.

Formulae:

• ${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{\sin \mathrm{x}}{\mathrm{x}}}=1}$
• ${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}}=1}$

Calculation: 

Since f(x) is given to be continuous at x = 0, ${\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(0)}$.

Also, ${\displaystyle \underset{\mathrm{x}\to {\mathrm{a}}^{+}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{lim}\mathrm{f}(\mathrm{x})}$ because f(x) is same for x > 0 and x < 0.

 $\therefore {\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(0)}$

$\Rightarrow {\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{\sin 3\mathrm{x}}{{\mathrm{e}}^{2\mathrm{x}}-1}}=\mathrm{k}-2}$

$\Rightarrow {\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{\sin 3\mathrm{x}}{3\mathrm{x}}}\times 3\mathrm{x}}{{\displaystyle \frac{{\mathrm{e}}^{2\mathrm{x}}-1}{2\mathrm{x}}}\times 2\mathrm{x}}}=\mathrm{k}-2}$

$\Rightarrow {\displaystyle \frac{3}{2}}=\mathrm{k}-2$

$\Rightarrow \mathrm{k}={\displaystyle \frac{7}{2}}$.

Concept:

Definition:

• A function f(x) is said to be continuous at a point x = a in its domain, if ${\displaystyle \underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}(\mathrm{x})}$ exists or or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ ${\displaystyle \underset{\mathrm{x}\to {\mathrm{a}}^{+}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})}$.

Calculation:

For x ≠ 0, the given function can be re-written as:

$\mathrm{f}(\mathrm{x})=\{\begin{array}{cc}\frac{2\mathrm{x}-{\sin }^{-1}\mathrm{x}}{2\mathrm{x}+{\tan }^{-1}\mathrm{x}};& \mathrm{x}\ne 0\\ \mathrm{K};& \mathrm{x}=0\end{array}$

Since the equation of the function is same for x < 0 and x > 0, we have:

${\displaystyle \underset{\mathrm{x}\to 0^{+}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to 0^{-}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to 0}{lim}\frac{2\mathrm{x}-{\sin }^{-1}\mathrm{x}}{2\mathrm{x}+{\tan }^{-1}\mathrm{x}}}$

= ${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{2-\frac{{\sin }^{-1}\mathrm{x}}{\mathrm{x}}}{2+\frac{{\tan }^{-1}\mathrm{x}}{\mathrm{x}}}=\frac{2-1}{2+1}=\frac{1}{3}}$

For the function to be continuous at x = 0, we must have:

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(0)}$

⇒ K = ${\displaystyle \frac{1}{3}}$.

Concept:

For a function say f, $\underset{x\to a}{lim}f(x)$ exists

$\Rightarrow \underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=l=\underset{x\to a}{lim}f(x)$, where l is a finite value.

Any function say f is said to be continuous at point say a if and only if $\underset{x\to a}{lim}f(x)=l=f\left(a\right)$, where l is a finite value.

Calculation:

 Given: $\mathrm{f}(\mathrm{x})=\frac{{\mathrm{x}}^2+\mathrm{x}-6}{{\mathrm{x}}^2+\mathrm{k}\mathrm{x}-3}$ is not continuous at x = 3.

So, if any function is not continuous at x = a then $\underset{x\to a}{lim}f(x)=l\ne f\left(a\right)$

So, for the function f(x) if denominator is 0 at x = 3 then we can say that f(3) is infinite and limit cannot exist.

Let's find the value of k for which the denominator of f(x) is 0 for x = 3.

So, substitute x = 3 in x2 + kx - 3 = 0

⇒ 3² + 3k - 3 = 0.

⇒ 6 + 3k = 0.

⇒ k = - 2.

Hence, option 1 is correct.

Concept:

Let f(x) = $\frac{\mathrm{p}(\mathrm{x})}{\mathrm{q}(\mathrm{x})}$

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]
• $\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}(\mathrm{x})$ exists
• $\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})$

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) = cot x

$\cot \mathrm{x}=\frac{\cos \mathrm{x}}{\sin \mathrm{x}}$

Check where denominator becomes zero

sin x = 0

x = nπ, n ∈ Z

∴ Given function is discontinuous at x = nπ

Hence, option (1) 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.

Concept:

A function y = f(x) is said to be continuous at a point x = a if $\underset{x\to a^{-}}{lim}f(x)=\underset{x\to a^{+}}{lim}f(x)$ = f(a)

$\underset{x\to a}{lim}\frac{\sin x}{x}$ = 1

Explanation:

LHL = f(0^-) = $\underset{x\to 0}{lim}\frac{1-\cos 2x}{x^2}$ = ${\displaystyle \underset{x\to 0}{lim}\frac{2{\sin }^{2}x}{x^2}}$ = 2${\displaystyle \underset{x\to 0}{lim}(\frac{\sin x}{x}{)}^2}$ = 2

RHL = f(0⁺) = $\underset{x\to 0}{lim}\frac{\beta \surd 1-\cos x}{x}$${\displaystyle \underset{x\to 0^{+}}{lim}\beta \surd 2\frac{\sin \frac{x}{2}}{2\frac{x}{2}}=\frac{\beta }{\surd 2}}$

Since f(x) is continuous at x = 0

So, LHL = RHL = f(0)

i.e., 2 = $\frac{\beta }{\surd 2}$ = α

So, α = 2 and β = 2√2

∴ ${\alpha }^2+{\beta }^2=4+8=12$

Option (2) is true.

Concept:

a² - b² = (a - b) (a + b)

Calculation:

Given that,

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2-9}{{\mathrm{x}}^2-2\mathrm{x}-3}$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{x}-3\right)\left(\mathrm{x}+3\right)}{{\mathrm{x}}^2-3\mathrm{x}+\mathrm{x}-3}$ [∵ a² - b² = (a-b) (a+b)]

⇒ $\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{x}-3\right)\left(\mathrm{x}+3\right)}{\mathrm{x}\left(\mathrm{x}-3\right)+1\left(\mathrm{x}-3\right)}$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{x}-3\right)\left(\mathrm{x}+3\right)}{\left(\mathrm{x}-3\right)\left(\mathrm{x}+1\right)}$

⇒ $\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{x}+3\right)}{\mathrm{x}+1}$

Given f(x) is continuous at x = 3

Concept:

For the function to be continuous:

LHL = RHL = f(x)

Where LHL = $\underset{\alpha \to 0}{lim}$f(x - α) and RHL = $\underset{\alpha \to 0}{lim}$f(x + α)

Calculation:

Given that f(x) is continuous function

LHL = f(x) = RHL

$\underset{\alpha \to 0}{lim}$ f(1 - α) = f(1)

$\underset{\alpha \to 0}{lim}$ [a + b(1 - α)] = 5

$\underset{\alpha \to 0}{lim}$ [a + b - bα] = 5

a + b = 5

The correct answer is option 4.

Given: $f(x)={\displaystyle \frac{x-|x|}{x}}$

Calculation:

⇒ $f(x)={\displaystyle \frac{x-|x|}{x}}$

For the value of x = 2

The function f(2) = ${\displaystyle \frac{2-|2|}{2}}$ = 0

For the value of x = 0; f(0) = ${\displaystyle \frac{0-|0|}{0}}$ = Impossible value

For the value of x = -2; f(-2) = ${\displaystyle \frac{-2-|-2|}{-2}}$ = 2

So, the function has some definite solution for all the values of x except x = 0.

Hence, the function is a continuous function for all the values of x except x = 0. 

Sin x

|sinx|

The graph of f(x) = 1 + |sin x| is as shown in the figure:

From the graph, it is clear that function is continuous everywhere but not differentiable at integral multiplies of π (∴ at these points curve has sharp turnings).

Concept:

f(x) = |x| ⇒ f(x) = x if x > 0,  and f(x) =  -x, x < 0

A function f(x) is continuous at x = a, if $\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{lim}\mathrm{f}(\mathrm{x})=\mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{x}\to \mathrm{a}}\mathrm{f}(\mathrm{x})$

A function f(x) is differentiable at x = a, if  LHD = RHD

$\begin{array}{l}\mathrm{L}\mathrm{H}\mathrm{D}=\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})=\underset{\mathrm{h}\to 0^{-}}{lim}\frac{\mathrm{f}(\mathrm{a}-\mathrm{h})-\mathrm{f}(\mathrm{a})}{-\mathrm{h}}\\ \mathrm{R}\mathrm{H}\mathrm{D}=\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})=\underset{\mathrm{h}\to 0^{+}}{lim}\frac{\mathrm{f}(\mathrm{a}+\mathrm{h})-\mathrm{f}(\mathrm{a})}{\mathrm{h}}\end{array}$

Calculation:

Here, f(x) =  e-|x| 

 $$\begin{gathered} \text{LHL =}\underset{\mathrm{x}\to 0^{-}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to 0^{-}}{lim}{\mathrm{e}}^{-|\mathrm{x}|}={\mathrm{e}}^0=1 \\ \text{RHL =}\underset{\mathrm{x}\to 0^{+}}{lim}\mathrm{f}(\mathrm{x})=\mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{x}\to 0^{+}}{\mathrm{e}}^{-|\mathrm{x}|}={\mathrm{e}}^0=1 \\ \mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{x}\to 0}\mathrm{f}(\mathrm{x})=\mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{x}\to 0}{\mathrm{e}}^{-|\mathrm{x}|}={\mathrm{e}}^0=1 \end{gathered}$$

So, the function is continuous at x = 0

f(x) =  e-|x| 

f'(x) = ex  for x < 0 and f'(x) =  -e-x  for x > 0

 $$\begin{gathered} \mathrm{L}\mathrm{H}\mathrm{D}=\underset{\mathrm{x}\to 0^{-}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})=\underset{\mathrm{x}\to 0^{-}}{lim}{\mathrm{e}}^{\mathrm{x}}={\mathrm{e}}^0=1 \\ \mathrm{R}\mathrm{H}\mathrm{D}=\underset{\mathrm{x}\to 0^{+}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})=\underset{\mathrm{x}\to 0^{+}}{lim}-{\mathrm{e}}^{-\mathrm{x}}=-{\mathrm{e}}^0=-1 \end{gathered}$$

Here, LHD ≠ RHD so f(x) is not differentiable at x = 0

Hence, option (1) is correct.

Alternate MethodReferring to the graph for the function,

 f(x) = e-|x| 

 f(x) = ex for x > 0 

 f(x) = e-x for x > 0

 f(x) = 1 for x = 0

• The graph can be as,

• This will be an even function as it is symmetric about y-axis.
• We can see that the function is continuous at x = 0 as, there is no discontinuity at x = 0.
• You can see there is a sharp corner at x = 0  for the graph so this not differentiable at x = 0

• Hence, option (1) is correct.