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

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.

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

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.

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:

\( \rm \displaystyle \lim_{x \to a}f(x)\) 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.

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

Example:

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.

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:

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.

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:

\( \rm \displaystyle \lim_{x \to a}f(x)\) 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.

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

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.

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

Example:

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.

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:

\( \rm \displaystyle \lim_{x \to a}f(x)\) 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.

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

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.

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.

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. 

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,

\[ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2}}{\sin 2x - \beta x} = 3 \]

Using expansions near zero:

\[ \sin \alpha x \approx \alpha x, \quad e^{x^2} \approx 1 + x^2, \quad \sin 2x \approx 2x - \frac{(2x)^3}{3!} \]

Simplify the numerator and denominator:

\[ \text{Numerator} \approx x^2 (\alpha x) + (\gamma - 1)(1 + x^2) = \alpha x^3 + (\gamma - 1) + (\gamma - 1)x^2 \]

\[ \text{Denominator} \approx 2x - \frac{8x^3}{6} - \beta x = (2 - \beta) x - \frac{4}{3} x^3 \]

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

\[ \gamma - 1 = 0 \Rightarrow \gamma = 1 \]

\[ 2 - \beta = 0 \Rightarrow \beta = 2 \]

Next, equate coefficients of x³ terms:

\[ \lim_{x \to 0} \frac{\alpha x^3}{ - \frac{4}{3} x^3 } = 3 \Rightarrow \frac{\alpha}{-\frac{4}{3}} = 3 \Rightarrow \alpha = -4 \]

\[ \beta + \gamma - \alpha = 2 + 1 - (-4) = 7 \]

∴ The correct answer is Option 1