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

Last updated on Jul 3, 2025

Latest Special Functions MCQ Objective Questions

Special Functions Question 1:

For α,β,γR, if limx0x2sinαx+(γ1)ex2sin2xβx=3, thenβ+γα is equal to:

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

Answer (Detailed Solution Below)

Option 1 : 7

Special Functions Question 1 Detailed Solution

Concept:

Limits and Expansion of Elementary Functions:

  • To solve limits involving indeterminate forms like 0/0, use series expansion (Maclaurin expansion) of functions such as sinx and ex.
  • Standard expansions:
    • sin x = x − x3/3! + x5/5! − ...
    • ex = 1 + x + x2/2! + x3/3! + ...
  • Compare powers of x from numerator and denominator and simplify accordingly.

 

Calculation:

Given,

limx→0 [ x2·sin(αx) + (γ−1)·ex2 ] / [ sin(2x) − βx ] = 3

Numerator:

x2·sin(αx) = x2·(αx − (αx)3/6 + ... ) = αx3 − α3x5/6 + ...

(γ − 1)·ex2 = (γ − 1)·(1 + x2 + x4/2! + ...) = (γ − 1) + (γ − 1)x2 + ...

So, numerator ≈ (γ − 1) + (γ − 1)x2 + αx3 + ...

Denominator:

sin(2x) = 2x − (2x)3/6 + ... = 2x − 8x3/6 + ...

So, denominator = 2x − βx − (8x3/6) + ... = (2 − β)x − (4/3)x3 + ...

Now rewrite limit as:

limx→0 [ (γ−1) + (γ−1)x2 + αx3 + ... ] / [ (2−β)x − (4/3)x3 + ... ] = 3

Now, use expansion up to x0 term after dividing:

⇒ limx→0 (γ−1) / (2−β)x = 0 unless (γ−1) = 0

So, γ − 1 = 0 ⇒ γ = 1

Then numerator becomes αx3 + ...

Denominator: (2 − β)x − (4/3)x3 + ...

⇒ lim = αx3 / [ (2 − β)x − (4/3)x3 ]

Now multiply numerator and denominator by x−3:

⇒ α / [ (2 − β)x−2 − (4/3) ]

Now take limit x → 0, x−2 → ∞

For limit to be finite, coefficient of x−2 in denominator must be 0

⇒ (2 − β) = 0

⇒ β = 2

Now denominator = −(4/3),

numerator = α ⇒ α / (−4/3) = 3

⇒ α = −4

Given γ = 1, β = 2, α = −4

β + γ − α = 2 + 1 − (−4) = 7

∴ β + γ − α = 7

Special Functions Question 2:

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

  1. (2π)410
  2. (2π)9/210
  3. (π)9/25
  4. (2π)9/25

Answer (Detailed Solution Below)

Option 2 : (2π)9/210

Special Functions Question 2 Detailed Solution

Explanation:

We know that

Γ(1/n)Γ(2/n)....Γ(1-1/n) = (2π)n12n

Putting n = 10 we get

Γ(0.1) Γ(0.2) Γ(0.3)....Γ(0.9) = (2π)9/210

Option (2) is true.

Special Functions Question 3:

If f(x)=[x]|x|, 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 3 Detailed Solution

Concept:

Modullus funtion: f(x) = |x| 

f(x) = {xx>0xx<0

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:

limxaf(x)=limxa+f(x)=limxaf(x)

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

LHD = limxaf(x)=limh0f(ah)f(a)h

RHD = limxa+f(x)=limh0+f(a+h)f(a)h

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

Calculation:

f(x)=[x]|x|

f(x) = {1x=1x1<x<00x=00<x<11x=1x1<x<2

At x = 1

RHL = limx0+=limx0+1x=1

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

Special Functions Question 4:

The value of  limx0{x}sin{x} is equal to?

Where {x} denote the fractional part of x.

  1. 1sin1
  2. -1
  3. 1
  4. Limit doesn't exist

Answer (Detailed Solution Below)

Option 4 : Limit doesn't exist

Special Functions Question 4 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 < 1

0

1 ≤ x < 2

1

2 ≤ x < 3

2

 

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

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

 

Existence of Limit:

limxaf(x) is exists if limxaf(x) and limxa+f(x) exist and limxaf(x)=limxa+f(x)

 

Calculation:

To Find: Value of limx0{x}sin{x}

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

limx0{x}sin{x}=limx0x[x]sin(x[x])

 

RHL = limx0+x[x]sin(x[x])

If x → 0+ then [x] = 0

RHL = limx0+xsin(x)        [Form (0/0)]

Apply L-Hospital Rule,

RHL=limx0+1cos(x)=1

RHL = 1

 

LHL = limx0x[x]sin(x[x])

If x → 0- then [x] = -1

LHL=limx0x(1)sin(x(1))limx0x+1sin(x+1)1sin1

RHL ≠ LHL

So, the limit doesn't exist.

Special Functions Question 5:

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 5 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.

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 nxn+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]

0x1

0

1x2

1

 

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

f(x) is Continuous at x = a ⇔ limxa+f(x)=limxaf(x)=limxaf(x)

Calculation:

For f(x) = [x]:

LHL = limx0f(x)=limx0[x]=[0h]=1

RHL=limx0+f(x)=limx0+[x]=[0+h]=0

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

For g(x) = sin x

LHL=limx0g(x)=limx0sinx=0

RHL=limx0+g(x)=limx0+sinx=0

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  limx0{x}sin{x} is equal to?

Where {x} denote the fractional part of x.

  1. 1sin1
  2. -1
  3. 1
  4. 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 < 1

0

1 ≤ x < 2

1

2 ≤ x < 3

2

 

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

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

 

Existence of Limit:

limxaf(x) is exists if limxaf(x) and limxa+f(x) exist and limxaf(x)=limxa+f(x)

 

Calculation:

To Find: Value of limx0{x}sin{x}

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

limx0{x}sin{x}=limx0x[x]sin(x[x])

 

RHL = limx0+x[x]sin(x[x])

If x → 0+ then [x] = 0

RHL = limx0+xsin(x)        [Form (0/0)]

Apply L-Hospital Rule,

RHL=limx0+1cos(x)=1

RHL = 1

 

LHL = limx0x[x]sin(x[x])

If x → 0- then [x] = -1

LHL=limx0x(1)sin(x(1))limx0x+1sin(x+1)1sin1

RHL ≠ LHL

So, the limit doesn't exist.

If f(x)=[x]|x|, 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) = {xx>0xx<0

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:

limxaf(x)=limxa+f(x)=limxaf(x)

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

LHD = limxaf(x)=limh0f(ah)f(a)h

RHD = limxa+f(x)=limh0+f(a+h)f(a)h

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

Calculation:

f(x)=[x]|x|

f(x) = {1x=1x1<x<00x=00<x<11x=1x1<x<2

At x = 1

RHL = limx0+=limx0+1x=1

∴ 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 nxn+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]

0x1

0

1x2

1

 

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

f(x) is Continuous at x = a ⇔ limxa+f(x)=limxaf(x)=limxaf(x)

Calculation:

For f(x) = [x]:

LHL = limx0f(x)=limx0[x]=[0h]=1

RHL=limx0+f(x)=limx0+[x]=[0+h]=0

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

For g(x) = sin x

LHL=limx0g(x)=limx0sinx=0

RHL=limx0+g(x)=limx0+sinx=0

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  limx0{x}sin{x} is equal to?

Where {x} denote the fractional part of x.

  1. 1sin1
  2. -1
  3. 1
  4. 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 < 1

0

1 ≤ x < 2

1

2 ≤ x < 3

2

 

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

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

 

Existence of Limit:

limxaf(x) is exists if limxaf(x) and limxa+f(x) exist and limxaf(x)=limxa+f(x)

 

Calculation:

To Find: Value of limx0{x}sin{x}

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

limx0{x}sin{x}=limx0x[x]sin(x[x])

 

RHL = limx0+x[x]sin(x[x])

If x → 0+ then [x] = 0

RHL = limx0+xsin(x)        [Form (0/0)]

Apply L-Hospital Rule,

RHL=limx0+1cos(x)=1

RHL = 1

 

LHL = limx0x[x]sin(x[x])

If x → 0- then [x] = -1

LHL=limx0x(1)sin(x(1))limx0x+1sin(x+1)1sin1

RHL ≠ LHL

So, the limit doesn't exist.

Special Functions Question 12:

If f(x)=[x]|x|, 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) = {xx>0xx<0

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:

limxaf(x)=limxa+f(x)=limxaf(x)

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

LHD = limxaf(x)=limh0f(ah)f(a)h

RHD = limxa+f(x)=limh0+f(a+h)f(a)h

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

Calculation:

f(x)=[x]|x|

f(x) = {1x=1x1<x<00x=00<x<11x=1x1<x<2

At x = 1

RHL = limx0+=limx0+1x=1

∴ 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:

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

  1. (2π)410
  2. (2π)9/210
  3. (π)9/25
  4. (2π)9/25

Answer (Detailed Solution Below)

Option 2 : (2π)9/210

Special Functions Question 14 Detailed Solution

Explanation:

We know that

Γ(1/n)Γ(2/n)....Γ(1-1/n) = (2π)n12n

Putting n = 10 we get

Γ(0.1) Γ(0.2) Γ(0.3)....Γ(0.9) = (2π)9/210

Option (2) is true.

Special Functions Question 15:

Let a1, a2, a3. ...an be n positive consecutive terms of an arithmetic progression. If d > 0 is its common difference, then limn1a1+a2+1a2+a3++1an1+ann

  1. 1
  2. d
  3. 1d
  4. 0

Answer (Detailed Solution Below)

Option 1 : 1

Special Functions Question 15 Detailed Solution

Calculation: 

limndn(1a1+a2+1a2+a3++1an1+an)n

On rationalising each term

limndnana1dn

limndn((n1)d(an+a1)d)n=1

Hence, the correct answer is Option 1. 

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