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 \(\lim_{x\to a^{-}}\) f(x) = \(\lim_{x\to a^{+}}\)f(x) = f(a).

Calculation:

Given: f(x) = \(\left\{\begin{array}{l}\rm mx+1, \text { if } x \leq \frac{\pi}{2} \\ \sin \rm x+n, \text { if } x>\frac{\pi}{2}\end{array}\right.\)

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

Left-hand limit = \(\lim_{h\to0}f(\frac{π}{2}-h)\,\)

\(=\,\lim_{h\to0}\,m\times \left(\frac{π}{2}-h\right) + 1 \)

Applying the limits:

Left- hand limit = m × \(\frac{π}{2}\) + 1

Right-hand limit = \(\lim_{h\to0}f(\frac{π}{2}+h)\,\)

\(=\,\lim_{h\to0}\,\sin\left(\frac{π}{2}+h\right) +n\)

Applying the limits:

 \(=\sin \frac{π}{2}+n\)

Right-hand limit = 1 + n

For the function to be continuous at x = \(\frac{π}{2}\),

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

⇒ m× \(\frac{π}{2}\) + 1 = 1 + n

⇒ n = \(\frac{mπ}{2}\)

The correct answer is n = \(\frac{mπ}{2}\) .

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 

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 \(\lim_{h\to0}\,f(a-h)\) 

Right hand limit (R.H.L) of f(x) at x = a is \(\lim_{h\to0}\,f(a+h)\)

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 \(\rm \displaystyle \lim_{x\to a}f(x)\) exists or or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ \(\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a}f(x)=f(a)\).

Formulae:

• \(\rm \displaystyle \lim_{x\to 0}\dfrac{\sin x}{x}=1\)
• \(\rm \displaystyle \lim_{x\to 0}\dfrac{e^x-1}{x}=1\)

Calculation: 

Since f(x) is given to be continuous at x = 0, \(\rm \displaystyle \lim_{x\to 0}f(x)=f(0)\).

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

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

\(\rm \Rightarrow \displaystyle \lim_{x\to 0}\dfrac{\sin 3x}{e^{2x}-1}=k-2\)

\(\rm \Rightarrow \displaystyle \lim_{x\to 0}\dfrac{\dfrac{\sin 3x}{3x}\times3x}{\dfrac{e^{2x}-1}{2x}\times2x}=k-2\)

\(\rm \Rightarrow \dfrac{3}{2}=k-2\)

\(\rm \Rightarrow k=\dfrac{7}{2}\).

Concept:

Definition:

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

Calculation:

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

\(\rm f(x) = \left\{ \begin{matrix} \rm \frac{2x-\sin^{-1} x}{2x + \tan^{-1} x}; & \rm x \ne 0 \\ \rm K; & \rm x = 0 \end{matrix}\right.\)

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

\(\rm \displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to 0^-}f(x)=\lim_{x\to 0}\frac{2x-\sin^{-1} x}{2x + \tan^{-1} x}\)

= \(\rm \displaystyle \lim_{x\to 0}\frac{2- \frac {\sin^{-1} x} x}{2 +\frac {\tan^{-1} x} x} = \frac {2-1}{2+1} = \frac 1 3\)

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

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

⇒ K = \(\dfrac{1}{3}\).

Concept:

For a function say f, \(\mathop {\lim }\limits_{x \to a} f(x)\) exists

\( ⇒ \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = l = \;\mathop {\lim }\limits_{x \to a} f(x)\), where l is a finite value.

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

Calculation:

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

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

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) = \(\rm \frac {p(x)}{q(x)}\)

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) = cot x

\(\rm \cot x = \frac{\cos x}{\sin x}\)

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 \(\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)\) = f(a)

\(\lim_{x\to a}{\sin x\over x}\) = 1

Explanation:

LHL = f(0^-) = \(\lim_{x\to0}{1-\cos2x\over x^2}\) = \(\displaystyle\lim_{x\rightarrow 0} \frac{2 \sin^2 x}{x^2}\) = 2\(\displaystyle\lim_{x\rightarrow 0} (\frac{\sin x}{x})^2\) = 2

RHL = f(0⁺) = \(\lim_{x\to0}\frac{β √{1-\cos x}}{x} \)\(\displaystyle\lim_{x\rightarrow 0^+} β √{2} \frac{\sin \frac{x}{2}}{2\frac{x}{2}} = \frac{β}{√{2}}\)

Since f(x) is continuous at x = 0

So, LHL = RHL = f(0)

i.e., 2 = \({β\over √2}\) = α

So, α = 2 and β = 2√2

∴ \(α^2 + β^2 = 4 + 8 = 12\)

Option (2) is true.

Concept:

a² - b² = (a - b) (a + b)

Calculation:

Given that,

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

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

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

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

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

Given f(x) is continuous at x = 3

Concept:

For the function to be continuous:

LHL = RHL = f(x)

Where LHL = \(\rm\mathop {\lim }\limits_{α \to 0}\)f(x - α) and RHL = \(\rm\mathop {\lim }\limits_{α \to 0}\)f(x + α)

Calculation:

Given that f(x) is continuous function

LHL = f(x) = RHL

\(\rm\mathop {\lim }\limits_{α \to 0}\) f(1 - α) = f(1)

\(\rm\mathop {\lim }\limits_{α \to 0}\) [a + b(1 - α)] = 5

\(\rm\mathop {\lim }\limits_{α \to 0}\) [a + b - bα] = 5

a + b = 5

The correct answer is option 4.

Given: \(f(x)=\dfrac{x-|x|}{x}\)

Calculation:

⇒ \(f(x)=\dfrac{x-|x|}{x}\)

For the value of x = 2

The function f(2) = \(\dfrac{2-|2|}{2}\) = 0

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

For the value of x = -2; f(-2) = \(\dfrac{-2-|-2|}{-2}\) = 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 \(\rm\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=lim _{x \rightarrow a} f(x)\)

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

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

Calculation:

Here, f(x) =  e-|x| 

 \(\text {LHL =}\rm\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}e^{-|x|}=e^0=1\\ \text {RHL =}\lim _{x \rightarrow 0^{+}} f(x)=lim _{x \rightarrow 0^+} e^{-|x|}=e^0=1\\ lim _{x \rightarrow 0}f(x)=lim _{x \rightarrow 0}e^{-|x|}=e^0=1\)

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

 \(\rm LHD=\rm \lim _{x \rightarrow 0^{-}}f'(x)=\lim _{x \rightarrow 0^{-}}e^x=e^0=1\\ \rm RHD=\rm \lim _{x \rightarrow 0^{+}}f'(x)=\lim _{x \rightarrow 0^{+}}-e^{-x}=-e^0=-1\)

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.