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  f(x) = f(x) = f(a).

Calculation:

Given: f(x) = \frac{\pi}{2}\end{array}\right.\)

f() = m ×  + 1

Left-hand limit = 

Applying the limits:

Left- hand limit = m ×  + 1

Right-hand limit = 

Applying the limits:

Right-hand limit = 1 + n

For the function to be continuous at x = ,

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

⇒ m×  + 1 = 1 + n

⇒ n = 

The correct answer is n =  .

Concept:

If a function is continuous at a point a, then

sin(∞) = a, Where -1≤ a ≤ 1

Calculation:

Given:

f(0) = 1

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

Checking continuity at x = 0

L.H.L

= 

= 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  

Right hand limit (R.H.L) of f(x) at x = a is 

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

So function is continuous for x > 0 and x

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

So function is continuous for x > 0 and x

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  exists or or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ .

Formulae:

Calculation: 

Since f(x) is given to be continuous at x = 0, .

Also,  because f(x) is same for x > 0 and x

.

Concept:

Definition:

• A function f(x) is said to be continuous at a point x = a in its domain, if  exists or or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ .

Calculation:

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

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

= 

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

⇒ K = .

Concept:

For a function say f,  exists

, where l is a finite value.

Any function say f is said to be continuous at point say a if and only if , where l is a finite value.

Calculation:

 Given:  is not continuous at x = 3.

So, if any function is not continuous at x = a then 

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) = 

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]
•  exists

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

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  = f(a)

 = 1

Explanation:

LHL = f(0^-) =  =  = 2 = 2

RHL = f(0⁺) = 

Since f(x) is continuous at x = 0

So, LHL = RHL = f(0)

i.e., 2 =  = α

So, α = 2 and β = 2√2

∴ 

Option (2) is true.

Concept:

a² - b² = (a - b) (a + b)

Calculation:

Given that,

 [∵ a² - b² = (a-b) (a+b)]

⇒

⇒

Given f(x) is continuous at x = 3

Concept:

For the function to be continuous:

LHL = RHL = f(x)

Where LHL = f(x - α) and RHL = f(x + α)

Calculation:

Given that f(x) is continuous function

LHL = f(x) = RHL

 f(1 - α) = f(1)

 [a + b(1 - α)] = 5

 [a + b - bα] = 5

a + b = 5

The correct answer is option 4.

Given: 

Calculation:

⇒ 

For the value of x = 2

The function f(2) =  = 0

For the value of x = 0; f(0) =  = Impossible value

For the value of x = -2; f(-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

A function f(x) is continuous at x = a, if 

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

Calculation:

Here, f(x) =  e-|x| 

So, the function is continuous at x = 0

f(x) =  e-|x| 

f'(x) = ex  for x -x  for x > 0

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.