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 \( \alpha, \beta, \gamma \in \mathbf{R}, \text{ if } \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1) e^{x^2}}{\sin 2x - \beta x} = 3, \\ \) \({then } \beta + \gamma - \alpha \text{ is equal to:}\)

  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. \(\rm \frac{(2\pi)^4}{\sqrt{10}}\)
  2. \(\rm \frac{(2\pi)^{9/2}}{\sqrt{10}}\)
  3. \(\rm \frac{(\pi)^{9/2}}{\sqrt{5}}\)
  4. \(\rm \frac{(2\pi)^{9/2}}{\sqrt{5}}\)

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{(2\pi)^{9/2}}{\sqrt{10}}\)

Special Functions Question 2 Detailed Solution

Explanation:

We know that

Γ(1/n)Γ(2/n)....Γ(1-1/n) = \((2\pi)^{\frac{n-1}2}\over\sqrt n\)

Putting n = 10 we get

Γ(0.1) Γ(0.2) Γ(0.3)....Γ(0.9) = \(\rm \frac{(2\pi)^{9/2}}{\sqrt{10}}\)

Option (2) is true.

Special Functions Question 3:

If \(f(x) = \frac{[x]}{|x|}\), 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) = \(\rm \left\{\begin{matrix} x & x > 0\\ -x & x < 0 \end{matrix}\right.\)

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:

\(\rm \lim_{x\rightarrow a^{-}}f(x) = \lim_{x\rightarrow a^{+}} f(x) = \lim_{x\rightarrow a}f(x) \)

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

LHD = \(\rm \lim_{x\rightarrow a^{-}}f'(x) = \rm \lim_{h\rightarrow 0^{-}}\frac{f(a-h) - f(a)}{-h}\)

RHD = \(\rm \lim_{x\rightarrow a^{+}}f'(x) = \rm \lim_{h\rightarrow 0^{+}}\frac{f(a+h) - f(a)}{h}\)

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

Calculation:

\(f(x) = \frac{[x]}{|x|}\)

f(x) = \(\rm \left\{\begin{matrix} \frac{-1}{-x} = \frac{1}{x} & -1 < x < 0 & \\ \frac{0}{x} = 0 & 0 < x < 1 & \\ \frac{1}{x} = \frac{1}{x} & 1 < x < 2 & \end{matrix}\right.\)

At x = 1

RHL = \(\lim_{x \to 0^+} =\lim_{x \to 0^+} \frac{1}{x} = 1\)

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

Special Functions Question 4:

The value of  \(\rm \displaystyle \lim_{x\rightarrow 0}\frac{\{x\}}{\sin\{x\}}\) is equal to?

Where {x} denote the fractional part of x.

  1. \(\rm \frac{1}{\sin 1}\)
  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:

\( \rm \displaystyle \lim_{x \to a}f(x)\) is exists if \(\rm \displaystyle \lim_{x \to a^{-}}f(x)\) and \(\rm \displaystyle \lim_{x \to a^{+}}f(x)\) exist and \( \rm \displaystyle \lim_{x \to a^{-}}f(x) = \rm \displaystyle \lim_{x \to a^{+}}f(x)\)

 

Calculation:

To Find: Value of \(\rm \displaystyle \lim_{x→ 0}\frac{\{x\}}{\sin\{x\}}\)

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

\(\rm \displaystyle \lim_{x→ 0}\frac{\{x\}}{\sin\{x\}}= \rm \displaystyle \lim_{x→ 0}\frac{x- [x]}{\sin (x- [x])}\)

 

RHL = \(\rm \displaystyle \lim_{x→ 0^+}\frac{x- [x]}{\sin (x- [x])}\)

If x → 0+ then [x] = 0

RHL = \(\rm \displaystyle \lim_{x\rightarrow 0^+}\frac{x}{\sin (x)} \)        [Form (0/0)]

Apply L-Hospital Rule,

\(\rm RHL=\rm \displaystyle \lim_{x\rightarrow 0^+}\frac{1}{\cos (x)} = 1\)

RHL = 1

 

LHL = \(\rm \displaystyle \lim_{x→ 0^-}\frac{x- [x]}{\sin (x- [x])}\)

If x → 0- then [x] = -1

\(\rm LHL = \rm \displaystyle \lim_{x \to 0^{-}}\frac{x- (-1)}{\sin (x- (-1))}\\\Rightarrow \rm \displaystyle \lim_{x \to 0^{-}}\frac{x+1}{\sin (x+1)}\\ \Rightarrow \frac{1}{\sin 1}\)

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 \(n \le x\; \le n + 1\) 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]

\(0 \le x\; \le 1\)

0

\(1 \le x\; \le 2\)

1

 

2. A function f(x) is said to be continuous at a point x = a, in its domain if \(\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right) = f\left( a \right)\) exists or its graph is a single unbroken curve.

f(x) is Continuous at x = a ⇔ \({\rm{\;}}\mathop {\lim }\limits_{x \to {{\rm{a}}^ + }} f\left( x \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{x \to {{\rm{a}}^ - }} f\left( x \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right)\)

Calculation:

For f(x) = [x]:

LHL = \(\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ - }} \left[ {\rm{x}} \right] = \left[ {0 - {\rm{h}}} \right] = - 1\)

\({\rm{RHL}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \left[ {\rm{x}} \right] = \left[ {0 + {\rm{h}}} \right] = 0\)

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

For g(x) = sin x

\({\rm{LHL}} = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ - }} {\rm{g}}\left( {\rm{x}} \right) = {\rm{\;}}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ - }} {\rm{sinx}} = 0\)

\({\rm{RHL}} = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ + }} {\rm{g}}\left( {\rm{x}} \right) = {\rm{\;}}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ + }} {\rm{sinx}} = 0\)

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  \(\rm \displaystyle \lim_{x\rightarrow 0}\frac{\{x\}}{\sin\{x\}}\) is equal to?

Where {x} denote the fractional part of x.

  1. \(\rm \frac{1}{\sin 1}\)
  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:

\( \rm \displaystyle \lim_{x \to a}f(x)\) is exists if \(\rm \displaystyle \lim_{x \to a^{-}}f(x)\) and \(\rm \displaystyle \lim_{x \to a^{+}}f(x)\) exist and \( \rm \displaystyle \lim_{x \to a^{-}}f(x) = \rm \displaystyle \lim_{x \to a^{+}}f(x)\)

 

Calculation:

To Find: Value of \(\rm \displaystyle \lim_{x→ 0}\frac{\{x\}}{\sin\{x\}}\)

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

\(\rm \displaystyle \lim_{x→ 0}\frac{\{x\}}{\sin\{x\}}= \rm \displaystyle \lim_{x→ 0}\frac{x- [x]}{\sin (x- [x])}\)

 

RHL = \(\rm \displaystyle \lim_{x→ 0^+}\frac{x- [x]}{\sin (x- [x])}\)

If x → 0+ then [x] = 0

RHL = \(\rm \displaystyle \lim_{x\rightarrow 0^+}\frac{x}{\sin (x)} \)        [Form (0/0)]

Apply L-Hospital Rule,

\(\rm RHL=\rm \displaystyle \lim_{x\rightarrow 0^+}\frac{1}{\cos (x)} = 1\)

RHL = 1

 

LHL = \(\rm \displaystyle \lim_{x→ 0^-}\frac{x- [x]}{\sin (x- [x])}\)

If x → 0- then [x] = -1

\(\rm LHL = \rm \displaystyle \lim_{x \to 0^{-}}\frac{x- (-1)}{\sin (x- (-1))}\\\Rightarrow \rm \displaystyle \lim_{x \to 0^{-}}\frac{x+1}{\sin (x+1)}\\ \Rightarrow \frac{1}{\sin 1}\)

RHL ≠ LHL

So, the limit doesn't exist.

If \(f(x) = \frac{[x]}{|x|}\), 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) = \(\rm \left\{\begin{matrix} x & x > 0\\ -x & x < 0 \end{matrix}\right.\)

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:

\(\rm \lim_{x\rightarrow a^{-}}f(x) = \lim_{x\rightarrow a^{+}} f(x) = \lim_{x\rightarrow a}f(x) \)

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

LHD = \(\rm \lim_{x\rightarrow a^{-}}f'(x) = \rm \lim_{h\rightarrow 0^{-}}\frac{f(a-h) - f(a)}{-h}\)

RHD = \(\rm \lim_{x\rightarrow a^{+}}f'(x) = \rm \lim_{h\rightarrow 0^{+}}\frac{f(a+h) - f(a)}{h}\)

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

Calculation:

\(f(x) = \frac{[x]}{|x|}\)

f(x) = \(\rm \left\{\begin{matrix} \frac{-1}{-x} = \frac{1}{x} & -1 < x < 0 & \\ \frac{0}{x} = 0 & 0 < x < 1 & \\ \frac{1}{x} = \frac{1}{x} & 1 < x < 2 & \end{matrix}\right.\)

At x = 1

RHL = \(\lim_{x \to 0^+} =\lim_{x \to 0^+} \frac{1}{x} = 1\)

∴ 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 \(n \le x\; \le n + 1\) 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]

\(0 \le x\; \le 1\)

0

\(1 \le x\; \le 2\)

1

 

2. A function f(x) is said to be continuous at a point x = a, in its domain if \(\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right) = f\left( a \right)\) exists or its graph is a single unbroken curve.

f(x) is Continuous at x = a ⇔ \({\rm{\;}}\mathop {\lim }\limits_{x \to {{\rm{a}}^ + }} f\left( x \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{x \to {{\rm{a}}^ - }} f\left( x \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right)\)

Calculation:

For f(x) = [x]:

LHL = \(\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ - }} \left[ {\rm{x}} \right] = \left[ {0 - {\rm{h}}} \right] = - 1\)

\({\rm{RHL}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \left[ {\rm{x}} \right] = \left[ {0 + {\rm{h}}} \right] = 0\)

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

For g(x) = sin x

\({\rm{LHL}} = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ - }} {\rm{g}}\left( {\rm{x}} \right) = {\rm{\;}}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ - }} {\rm{sinx}} = 0\)

\({\rm{RHL}} = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ + }} {\rm{g}}\left( {\rm{x}} \right) = {\rm{\;}}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ + }} {\rm{sinx}} = 0\)

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  \(\rm \displaystyle \lim_{x\rightarrow 0}\frac{\{x\}}{\sin\{x\}}\) is equal to?

Where {x} denote the fractional part of x.

  1. \(\rm \frac{1}{\sin 1}\)
  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:

\( \rm \displaystyle \lim_{x \to a}f(x)\) is exists if \(\rm \displaystyle \lim_{x \to a^{-}}f(x)\) and \(\rm \displaystyle \lim_{x \to a^{+}}f(x)\) exist and \( \rm \displaystyle \lim_{x \to a^{-}}f(x) = \rm \displaystyle \lim_{x \to a^{+}}f(x)\)

 

Calculation:

To Find: Value of \(\rm \displaystyle \lim_{x→ 0}\frac{\{x\}}{\sin\{x\}}\)

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

\(\rm \displaystyle \lim_{x→ 0}\frac{\{x\}}{\sin\{x\}}= \rm \displaystyle \lim_{x→ 0}\frac{x- [x]}{\sin (x- [x])}\)

 

RHL = \(\rm \displaystyle \lim_{x→ 0^+}\frac{x- [x]}{\sin (x- [x])}\)

If x → 0+ then [x] = 0

RHL = \(\rm \displaystyle \lim_{x\rightarrow 0^+}\frac{x}{\sin (x)} \)        [Form (0/0)]

Apply L-Hospital Rule,

\(\rm RHL=\rm \displaystyle \lim_{x\rightarrow 0^+}\frac{1}{\cos (x)} = 1\)

RHL = 1

 

LHL = \(\rm \displaystyle \lim_{x→ 0^-}\frac{x- [x]}{\sin (x- [x])}\)

If x → 0- then [x] = -1

\(\rm LHL = \rm \displaystyle \lim_{x \to 0^{-}}\frac{x- (-1)}{\sin (x- (-1))}\\\Rightarrow \rm \displaystyle \lim_{x \to 0^{-}}\frac{x+1}{\sin (x+1)}\\ \Rightarrow \frac{1}{\sin 1}\)

RHL ≠ LHL

So, the limit doesn't exist.

Special Functions Question 12:

If \(f(x) = \frac{[x]}{|x|}\), 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) = \(\rm \left\{\begin{matrix} x & x > 0\\ -x & x < 0 \end{matrix}\right.\)

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:

\(\rm \lim_{x\rightarrow a^{-}}f(x) = \lim_{x\rightarrow a^{+}} f(x) = \lim_{x\rightarrow a}f(x) \)

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

LHD = \(\rm \lim_{x\rightarrow a^{-}}f'(x) = \rm \lim_{h\rightarrow 0^{-}}\frac{f(a-h) - f(a)}{-h}\)

RHD = \(\rm \lim_{x\rightarrow a^{+}}f'(x) = \rm \lim_{h\rightarrow 0^{+}}\frac{f(a+h) - f(a)}{h}\)

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

Calculation:

\(f(x) = \frac{[x]}{|x|}\)

f(x) = \(\rm \left\{\begin{matrix} \frac{-1}{-x} = \frac{1}{x} & -1 < x < 0 & \\ \frac{0}{x} = 0 & 0 < x < 1 & \\ \frac{1}{x} = \frac{1}{x} & 1 < x < 2 & \end{matrix}\right.\)

At x = 1

RHL = \(\lim_{x \to 0^+} =\lim_{x \to 0^+} \frac{1}{x} = 1\)

∴ 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. \(\rm \frac{(2\pi)^4}{\sqrt{10}}\)
  2. \(\rm \frac{(2\pi)^{9/2}}{\sqrt{10}}\)
  3. \(\rm \frac{(\pi)^{9/2}}{\sqrt{5}}\)
  4. \(\rm \frac{(2\pi)^{9/2}}{\sqrt{5}}\)

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{(2\pi)^{9/2}}{\sqrt{10}}\)

Special Functions Question 14 Detailed Solution

Explanation:

We know that

Γ(1/n)Γ(2/n)....Γ(1-1/n) = \((2\pi)^{\frac{n-1}2}\over\sqrt n\)

Putting n = 10 we get

Γ(0.1) Γ(0.2) Γ(0.3)....Γ(0.9) = \(\rm \frac{(2\pi)^{9/2}}{\sqrt{10}}\)

Option (2) is true.

Special Functions Question 15:

\(\text{If} \quad \lim_{x \to 1^+} \frac{(x-1)(6 + \lambda \cos (x-1)) + \mu \sin (1 - x)}{(x-1)^3} = -1, \\ \text{where } \lambda, \mu \in \mathbb{R}, \text{ then } \lambda + \mu \text{ is equal to} \)

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

Answer (Detailed Solution Below)

Option 1 : 18

Special Functions Question 15 Detailed Solution

Calculation:

\(\text{Put } x = 1 + h\)

\(\lim_{h \to 0} \frac{h(6 + \lambda \cosh h) - \mu \sinh h}{h^3} = -1 \)

\(\implies \lim_{h \to 0} \frac{h \left( 6 + \lambda \left( 1 - \frac{h^2}{2!} + \cdots \right) \right) - \mu \left( h - \frac{h^3}{3!} + \cdots \right)}{h^3} = -1 \)

\(\implies \lim_{h \to 0} \frac{6h + \lambda h - \frac{\lambda h^3}{2} - \mu h + \frac{\mu h^3}{6} + \cdots}{h^3} = -1\)

\(\implies \lim_{h \to 0} \frac{h (6 + \lambda - \mu) - \frac{\lambda h^3}{2} + \frac{\mu h^3}{6} + \cdots}{h^3} = -1 \)

\(\implies \lim_{h \to 0} \left( \frac{6 + \lambda - \mu}{h^2} - \frac{\lambda}{2} + \frac{\mu}{6} + \cdots \right) = -1 \)

\(\text{For the limit to be finite, the coefficient of } \frac{1}{h^2} \text{ must be zero:}\)

\(\quad 6 + \lambda - \mu = 0\)

And the remaining terms give

\(\quad -\frac{\lambda}{2} + \frac{\mu}{6} = -1 \)

\(\text{Solving these simultaneously:} \quad \lambda + \mu = 18 \)

Hence, the correct answer is Option 1. 

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