Special Functions MCQ Quiz - Objective Question with Answer for Special Functions - Download Free PDF

Last updated on May 17, 2025

Latest Special Functions MCQ Objective Questions

Special Functions Question 1:

Γ(0.1) Γ(0.2) Γ(0.3)....Γ(0.9) is equal to

Answer (Detailed Solution Below)

Option 2 :

Special Functions Question 1 Detailed Solution

Explanation:

We know that

Γ(1/n)Γ(2/n)....Γ(1-1/n) = 

Putting n = 10 we get

Γ(0.1) Γ(0.2) Γ(0.3)....Γ(0.9) = 

Option (2) is true.

Special Functions Question 2:

If , x ≠ 0, where [⋅] denotes the greatest integer function, then what is the right-hand limit of f(x) at x = 1?

  1. -1
  2. 0
  3. 1
  4. Right-hand limit of f(x) at x = 1 does not exist.

Answer (Detailed Solution Below)

Option 3 : 1

Special Functions Question 2 Detailed Solution

Concept:

Modullus funtion: f(x) = |x| 

f(x) =  0\\ -x & x

Greatest Integer Function:

Greatest Integer Function [x] indicates an integral part of the real number x which is the nearest and smaller integer to x. It is also known as the floor of x

  • In general, If n≤ x ≤ n+1 Then [x] = n     (n ∈ Integer)
  • This means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

Continuity of function at any point:

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

LHD = 

RHD = 

\lim_{x\rightarrow a^{-1}}xa \lim_{x\rightarrow a^{+}xa+

Calculation:

f(x) = 

At x = 1

RHL = 

∴ The right-hand limit of f(x) at x = 1 is 1.

Special Functions Question 3:

The value of   is equal to?

Where {x} denote the fractional part of x.

  1. -1
  2. 1
  3. Limit doesn't exist

Answer (Detailed Solution Below)

Option 4 : Limit doesn't exist

Special Functions Question 3 Detailed Solution

Concept:

Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is a nearest and smaller integer to x. It is also known as floor of x

  • In general, If n ≤ x ≤ n+1 Then [x] = n   (n ∈ Integer)
  • Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

 

Example:

x

[x]

0 ≤ x

0

1 ≤ x

1

2 ≤ x

2

 

Fractional part of x: fractional part will always be non-negative.

  • It is denoted by {x}
  • {x} = x - [x]

 

Existence of Limit:

 is exists if  and  exist and 

 

Calculation:

To Find: Value of 

As we know {x} = x - [x]

 

RHL = 

If x → 0+ then [x] = 0

RHL =         [Form (0/0)]

Apply L-Hospital Rule,

RHL = 1

 

LHL = 

If x → 0- then [x] = -1

RHL ≠ LHL

So, the limit doesn't exist.

Special Functions Question 4:

Which of the following statements is/are true:

1. sin x is a periodic function with period 2π

2. cos x is a periodic function with period 2π

  1. Only 1
  2. Only 2
  3. Both 1 and 2
  4. Neither 1 nor 2

Answer (Detailed Solution Below)

Option 3 : Both 1 and 2

Special Functions Question 4 Detailed Solution

Concept:

Periodic function:

A function f is said to be periodic function with period p if f(x + p) = f(x) ∀ x.

The period of sin x and cos x is 2π 

Calculation:

Statement 1: sin x is a periodic function with period 2π

As we know that, sin x and cos x are periodic function with period 2π.

So, statement 1 is true.

Statement 2: cos x is a periodic function with period 2π

As we know that, sin x and cos x are periodic function with period 2π.

So, statement 2 is true.

Hence, option C is the right answer.

Special Functions Question 5:

Let f(x) = [x], where [.] is the greatest integer function and g(x) = sin x be two real valued functions over R.

Which of the following statements is correct?

  1. Both f(x) and g(x) are continuous at x = 0.
  2. f(x) is continuous at x = 0, but g(x) is not continuous at x = 0.
  3. g(x) is continuous at x = 0, but f(x) is not continuous at x = 0
  4. both f(x) and g(x) are discontinuous at x = 0.

Answer (Detailed Solution Below)

Option 3 : g(x) is continuous at x = 0, but f(x) is not continuous at x = 0

Special Functions Question 5 Detailed Solution

Concept:

1. Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is nearest and smaller integer to x. It is also known as floor of x

  • In general, If  Then [x] = n     (n ∈ Integer)
  • Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.


Example:

x

[x]

0

1

 

2. 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.

f(x) is Continuous at x = a ⇔ 

Calculation:

For f(x) = [x]:

LHL = 

LHL ≠ RHL, so f(x) is discontinuous at x = 0

For g(x) = sin x

g (0) = sin (0) = 0

LHL = RHL = g (0), so g(x) is continuous at x = 0

Hence, option (3) is correct.

Top Special Functions MCQ Objective Questions

Let f(x) = [x], where [.] is the greatest integer function and g(x) = sin x be two real valued functions over R.

Which of the following statements is correct?

  1. Both f(x) and g(x) are continuous at x = 0.
  2. f(x) is continuous at x = 0, but g(x) is not continuous at x = 0.
  3. g(x) is continuous at x = 0, but f(x) is not continuous at x = 0
  4. both f(x) and g(x) are discontinuous at x = 0.

Answer (Detailed Solution Below)

Option 3 : g(x) is continuous at x = 0, but f(x) is not continuous at x = 0

Special Functions Question 6 Detailed Solution

Download Solution PDF

Concept:

1. Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is nearest and smaller integer to x. It is also known as floor of x

  • In general, If  Then [x] = n     (n ∈ Integer)
  • Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.


Example:

x

[x]

0

1

 

2. 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.

f(x) is Continuous at x = a ⇔ 

Calculation:

For f(x) = [x]:

LHL = 

LHL ≠ RHL, so f(x) is discontinuous at x = 0

For g(x) = sin x

g (0) = sin (0) = 0

LHL = RHL = g (0), so g(x) is continuous at x = 0

Hence, option (3) is correct.

The value of   is equal to?

Where {x} denote the fractional part of x.

  1. -1
  2. 1
  3. Limit doesn't exist

Answer (Detailed Solution Below)

Option 4 : Limit doesn't exist

Special Functions Question 7 Detailed Solution

Download Solution PDF

Concept:

Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is a nearest and smaller integer to x. It is also known as floor of x

  • In general, If n ≤ x ≤ n+1 Then [x] = n   (n ∈ Integer)
  • Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

 

Example:

x

[x]

0 ≤ x

0

1 ≤ x

1

2 ≤ x

2

 

Fractional part of x: fractional part will always be non-negative.

  • It is denoted by {x}
  • {x} = x - [x]

 

Existence of Limit:

 is exists if  and  exist and 

 

Calculation:

To Find: Value of 

As we know {x} = x - [x]

 

RHL = 

If x → 0+ then [x] = 0

RHL =         [Form (0/0)]

Apply L-Hospital Rule,

RHL = 1

 

LHL = 

If x → 0- then [x] = -1

RHL ≠ LHL

So, the limit doesn't exist.

If , x ≠ 0, where [⋅] denotes the greatest integer function, then what is the right-hand limit of f(x) at x = 1?

  1. -1
  2. 0
  3. 1
  4. Right-hand limit of f(x) at x = 1 does not exist.

Answer (Detailed Solution Below)

Option 3 : 1

Special Functions Question 8 Detailed Solution

Download Solution PDF

Concept:

Modullus funtion: f(x) = |x| 

f(x) =  0\\ -x & x

Greatest Integer Function:

Greatest Integer Function [x] indicates an integral part of the real number x which is the nearest and smaller integer to x. It is also known as the floor of x

  • In general, If n≤ x ≤ n+1 Then [x] = n     (n ∈ Integer)
  • This means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

Continuity of function at any point:

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

LHD = 

RHD = 

\lim_{x\rightarrow a^{-1}}xa \lim_{x\rightarrow a^{+}xa+

Calculation:

f(x) = 

At x = 1

RHL = 

∴ The right-hand limit of f(x) at x = 1 is 1.

Which of the following statements is/are true:

1. sin x is a periodic function with period 2π

2. cos x is a periodic function with period 2π

  1. Only 1
  2. Only 2
  3. Both 1 and 2
  4. Neither 1 nor 2

Answer (Detailed Solution Below)

Option 3 : Both 1 and 2

Special Functions Question 9 Detailed Solution

Download Solution PDF

Concept:

Periodic function:

A function f is said to be periodic function with period p if f(x + p) = f(x) ∀ x.

The period of sin x and cos x is 2π 

Calculation:

Statement 1: sin x is a periodic function with period 2π

As we know that, sin x and cos x are periodic function with period 2π.

So, statement 1 is true.

Statement 2: cos x is a periodic function with period 2π

As we know that, sin x and cos x are periodic function with period 2π.

So, statement 2 is true.

Hence, option C is the right answer.

Special Functions Question 10:

Let f(x) = [x], where [.] is the greatest integer function and g(x) = sin x be two real valued functions over R.

Which of the following statements is correct?

  1. Both f(x) and g(x) are continuous at x = 0.
  2. f(x) is continuous at x = 0, but g(x) is not continuous at x = 0.
  3. g(x) is continuous at x = 0, but f(x) is not continuous at x = 0
  4. both f(x) and g(x) are discontinuous at x = 0.

Answer (Detailed Solution Below)

Option 3 : g(x) is continuous at x = 0, but f(x) is not continuous at x = 0

Special Functions Question 10 Detailed Solution

Concept:

1. Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is nearest and smaller integer to x. It is also known as floor of x

  • In general, If  Then [x] = n     (n ∈ Integer)
  • Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.


Example:

x

[x]

0

1

 

2. 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.

f(x) is Continuous at x = a ⇔ 

Calculation:

For f(x) = [x]:

LHL = 

LHL ≠ RHL, so f(x) is discontinuous at x = 0

For g(x) = sin x

g (0) = sin (0) = 0

LHL = RHL = g (0), so g(x) is continuous at x = 0

Hence, option (3) is correct.

Special Functions Question 11:

The value of   is equal to?

Where {x} denote the fractional part of x.

  1. -1
  2. 1
  3. Limit doesn't exist

Answer (Detailed Solution Below)

Option 4 : Limit doesn't exist

Special Functions Question 11 Detailed Solution

Concept:

Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is a nearest and smaller integer to x. It is also known as floor of x

  • In general, If n ≤ x ≤ n+1 Then [x] = n   (n ∈ Integer)
  • Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

 

Example:

x

[x]

0 ≤ x

0

1 ≤ x

1

2 ≤ x

2

 

Fractional part of x: fractional part will always be non-negative.

  • It is denoted by {x}
  • {x} = x - [x]

 

Existence of Limit:

 is exists if  and  exist and 

 

Calculation:

To Find: Value of 

As we know {x} = x - [x]

 

RHL = 

If x → 0+ then [x] = 0

RHL =         [Form (0/0)]

Apply L-Hospital Rule,

RHL = 1

 

LHL = 

If x → 0- then [x] = -1

RHL ≠ LHL

So, the limit doesn't exist.

Special Functions Question 12:

If , x ≠ 0, where [⋅] denotes the greatest integer function, then what is the right-hand limit of f(x) at x = 1?

  1. -1
  2. 0
  3. 1
  4. Right-hand limit of f(x) at x = 1 does not exist.

Answer (Detailed Solution Below)

Option 3 : 1

Special Functions Question 12 Detailed Solution

Concept:

Modullus funtion: f(x) = |x| 

f(x) =  0\\ -x & x

Greatest Integer Function:

Greatest Integer Function [x] indicates an integral part of the real number x which is the nearest and smaller integer to x. It is also known as the floor of x

  • In general, If n≤ x ≤ n+1 Then [x] = n     (n ∈ Integer)
  • This means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

Continuity of function at any point:

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

LHD = 

RHD = 

\lim_{x\rightarrow a^{-1}}xa \lim_{x\rightarrow a^{+}xa+

Calculation:

f(x) = 

At x = 1

RHL = 

∴ The right-hand limit of f(x) at x = 1 is 1.

Special Functions Question 13:

Which of the following statements is/are true:

1. sin x is a periodic function with period 2π

2. cos x is a periodic function with period 2π

  1. Only 1
  2. Only 2
  3. Both 1 and 2
  4. Neither 1 nor 2

Answer (Detailed Solution Below)

Option 3 : Both 1 and 2

Special Functions Question 13 Detailed Solution

Concept:

Periodic function:

A function f is said to be periodic function with period p if f(x + p) = f(x) ∀ x.

The period of sin x and cos x is 2π 

Calculation:

Statement 1: sin x is a periodic function with period 2π

As we know that, sin x and cos x are periodic function with period 2π.

So, statement 1 is true.

Statement 2: cos x is a periodic function with period 2π

As we know that, sin x and cos x are periodic function with period 2π.

So, statement 2 is true.

Hence, option C is the right answer.

Special Functions Question 14:

  1. 18
  2. 20
  3. 19
  4. 17

Answer (Detailed Solution Below)

Option 1 : 18

Special Functions Question 14 Detailed Solution

Calculation:

And the remaining terms give

Hence, the correct answer is Option 1. 

Special Functions Question 15:

For  

  1. 7
  2. 4
  3. 6
  4. -1

Answer (Detailed Solution Below)

Option 1 : 7

Special Functions Question 15 Detailed Solution

Concept:

Limit Evaluation Using Series Expansion:

  • We use expansions of sine and exponential functions near zero to simplify limits.
  • Substitute the series expansions into the expression and simplify the numerator and denominator.
  • Compare powers of x in numerator and denominator to evaluate the limit as \(x \to 0\).

 

Calculation:

Given,

Using expansions near zero:

Simplify the numerator and denominator:

For limit to be finite and non-zero, coefficient of lowest power terms must satisfy:

Next, equate coefficients of x3 terms:

∴ The correct answer is Option 1 

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