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

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 e^x.
• Standard expansions:
  • sin x = x − x³/3! + x⁵/5! − ...
  • e^x = 1 + x + x²/2! + x³/3! + ...
• Compare powers of x from numerator and denominator and simplify accordingly.

Calculation:

Given,

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

Numerator:

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

(γ − 1)·e^x2 = (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:

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

Now, use expansion up to x⁰ term after dividing:

⇒ lim_x→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

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.

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 = 

Calculation:

f(x) = 

At x = 1

RHL = 

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

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:

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.

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.

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:

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.

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:

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.

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 = 

Calculation:

f(x) = 

At x = 1

RHL = 

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

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.

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:

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.

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:

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.

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 = 

Calculation:

f(x) = 

At x = 1

RHL = 

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

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.

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.

Calculation: 

0 \)

= g(–1) = 1

0\)

0\)

= 1

Hence, the correct answer is Option 1.