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

Concept:

1. The graph for the function 1 - x³ will be as follows:

We can see that the value of y = 1 - x³ decreases as x increases in the interval [0,1]

2. The graph for the function x - 1 will be as follows:

The value of y = x - 1 is increasing in the interval [0,1]

3. The graph for the function 1/x² is as follows:

The value of y = 1/x² decreases as x increases in the interval [0,1].

From the above options, only the Graph of "x-1" is increasing in the interval [0,1]

and the Graph of   and  decreases as x increases.

Hence Option(4) is the correct answer.

Concept:

To determine differentiability, we need to check if the limit



exists and is finite. 

Explanation:

Checking Continuity at (0, 0)

To check if f(x, y) is continuous at (0, 0) , we examine the limit of f(x, y) as .

1. If y = 0 , f(x, 0) = 0 , so f(0, 0) = 0 .

2. If y  0 , then we consider:
   

   
The term  is bounded by [0, 2] , so we can bound f(x, y) as follows:
   

   
 As  , which implies f(x, y) \to 0 . Thus, f(x, y) is continuous at (0, 0) , and Option 1 is correct.

In this case:



As (  , the term  oscillates, and hence the limit does not exist. Therefore, f(x, y) is not

differentiable at (0, 0) , and Option 2 is incorrect.

The directional derivative of f in the direction of a unit vector  at (0, 0) is given by:



For small t , with ta and tb approaching zero, f(ta, tb) tends to zero. Therefore, all directional derivatives of f at (0, 0) doed not exist. Thus, Option 3 is incorrect.

The partial derivatives of f at (0, 0) are:

Partial derivative with respect to x :
   

   
 Partial derivative with respect to y :
   
 
   
Both partial derivatives exist and are equal to zero. Therefore, Option 4 is correct.

Hence option 1) and 4) are correct.

Concept Used:-

When the left-hand limit and right-hand limits value of a function at any given point, x=a, are equal, then the function is called the continuous at that point.

L.H.L = R.H.L = f(a) = 0.

When the left-hand derivative and right-hand derivative value of a function at any given point, x=a, are equal, then the function is called the differentiable at that point.

L.H.D= R.H.D = f(a) = 0.

Explanation:-

Given function is,

Rewrite this function as,

Here, the value of Left-hand limit at x=0,

⇒L.H.L= 0(1)=0

Which is equal to the right-hand limit at x=0.

⇒R.H.L= f(0)

So, the function is continuous at x= 0.

On differentiating the above function with respect to x we get,

Here, the value of the Left-hand derivative at x=0,

⇒L.H.D = 0 - 0

⇒L.H.D = 0

Which is equal to the right-hand derivative at x=0.

⇒R.H.D = f'(0)

So, the function is differentiable at x= 0.

Therefore, the given function is continuous and differentiable at x = 0.

The correct option is 1.

Concept: 

If f is continuous at given interval \([a,b]\) and at end points 

f is defined then f is Uniformly continuous at given interval \([a,b] \) 

Explanation: 

Here given, \(f(x) = \frac{x}{x+1}\) and it is continuous except at x = -1 

so, f is continuous at \([0,2]\) 

At x = 0, f(0) is 0 

similarly at x = 2, f (2) = \(\frac{2}{3}\) 

So given function is defined at endpoints 

Hence it is Uniformly Continuous at \([0,2] \) 

Therefore, Correct Option is Option 4).

Concept: 

If f(x) is defined over [a,b] and it is continuous, and at endpoints 

of the interval f is defined then f is uniformly continuous. 

Explanation: 

Here f(x) =  and it is continuous in [0, 2]

Now,  f(0) = 0 and f(2) =  

So f(x) is defined on end-points.

Hence f is uniformly continuous on [0, 2]

Therefore, Correct option is Option 4).

Concept:

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

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Analysis:

This function is not defined for:

x² + 3x – 4 = 0

⇒ (x + 4)(x - 1) = 0 i.e.

At x = 1 and x = - 4

∴ This function f(x) is not continuous at x = 1, -4

Concept:

A function is said to be continuous if 

Calculation:

Given:

f(x) = (x + 1)cotx 

Taking log on both sides

log (f(x)) = cot x log (x + 1)

For checking continuity at x = 0

 = 0/0 form

Using L-Hospital rule

⇒ log (f(0)) = 1

⇒ f(0) = e¹ = e

Concept:

A function is written in the form of the ratio of two polynomial functions is called a rational function.

Rational functions are continuous at all the points except for the points where the denominator becomes zero.

where P(x) and Q(x) are polynomials and Q(x) ≠ 0.

f(x) will be discontinuous at points where Q(x) = 0.

Calculation:

Given:

This is a rational function, so it will be discontinuous at points where the denominator becomes zero.

4x - x³ = 0

x(4 - x²) = 0

x(2² - x2) = 0

x(2 + x)(2 - x) = 0

x = 0, x = - 2 and x = 2

Hence the function  will be discontinuous at exactly three points 0, - 2 and 2.

Mistake Points

There may be a doubt that some factors are eliminating each other so first, we have to simplify this. 

Note that, 

If (4 - x2) = 0 then the f(x) will come indeterminant or 0/0 form.

So, the function will not have any value for x  = ± 2. So, these will also be the points of discontinuity.

Also, if  (4 - x2) ≠ 0

⇒ 

Here, x = 0 is also the point of discontinuity.

There will be  exactly three points 0, - 2 and 2.

Given that,

A function f(x) is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

i.e.,

Analysis:

f(x) = x², x ≥ 0

= -x², x ≤ 0

f^'(x) = 2x, x ≥ 0

= -2x, x

f^'(x) = 2|x|

f^’(x) is continuous but not differentiable at x = 0.

Hence f(x) is differentiable but its first derivative is not differentiable at x = 0.

Option(4) is the correct answer.

Concept:

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

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Calculation:

Given:

f(x) = |x|

|x| = x for x ≥ 0

|x|= -x for x

At x = 0

Left limit = 0, Right limit = 0, f(0) = 0

As

Left limit = Right limit = Function value = 0

∴ |X| is continuous at x = 0.

Now

Left derivative (at x = 0) = -1

Right derivative (at x = 0) = 1

Left derivative ≠ Right derivative

∴ |x| is not differentiable at x = 0

Concept:

A function f(x) is said to be continuous at a point x = a, in its domain if,

  exists or its graph is a single unbroken curve.

In other words,

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

Left limit = Right limit = Function value = Real and finite

Analysis:

0} \end{array}} \right.\)

For x ≤ 0, f(x) = 1 – x

g (x) = -x

(fog)(x) = f(g(x)) = f(-x) = 1 + x

y = 1 + x is a continuous function.

Hence, the number of discontinuities = 0

We know that, (Intermediate value theorem)

If  then f(x) has at least one root in (a, b)

Since there is no root in [a, b] this implies f(a) and f(b) are of same sign 

i.e. either they both are positive or they both are negative

In both cases  0\)

Concept:

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

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Calculation:

Given:

f(x) = |2 − 3x|

|2 − 3x| = 2 - 3x for x

|2 − 3x| = -2 + 3x for x ≥ 2/3

At x = 2/3,

Left limit = 0, Right limit = 0, f(0) = 0

As

Left limit = Right limit = Function value = 0

∴ |X| is continuous at x = 2/3.

Now

Left derivative (at x = 2/3) = -3

Right derivative (at x = 2/3) = 3

Left derivative ≠ Right derivative

∴ |x| is not differentiable at x = 2/3

Concept:

Given:

A function will be discontinuous at the point where the denominator becomes zero.

X² – 1 will become zero at x = +1 and x = -1

In the numerator, ln(x) is there and it is not defined for negative values.

∴ Function f(x) will be continuous in the interval  (0, 1),(1, ∞)

We know that

(a + b + c)² ≥ 0

a² + b² + c² + 2(ab + bc + ac) ≥ 0

1 + 2(ab + bc + ac) ≥ 0

(ab + bc + ac) ≥ -1/2

(a - b)² + (b - c)² + (c - a)² ≥ 0

2(a2 + b2 + c2) - 2(ab + bc + ac) ≥ 0

(a2 + b2 + c2) - (ab + bc + ac) ≥ 0

(1) - (ab + bc + ac) ≥ 0

(ab + bc + ac) ≤ 1

⇒ -1/2 ≤ (ab + bc + ac) ≤ 1

(ab + bc + ac) ∈ [-1/2, 1]