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

Concept:

Exponential Generating Function & Series Coefficients:

• The expression uses combinations ⁿC_r and factorials, hinting at exponential and binomial expansion identities.
• Function f(x) = 1 + (1 + x)/1! + (1 + x)²/2! + (1 + x)³/3! + ... is considered to evaluate the coefficient of x².
• This function can be transformed using the identity: e^1+x / (1 + x)
• The coefficient of x² in this expansion corresponds to the RHS of 'a' series.
• The value of b is derived using the identity: 1 + 2/1! + 2²/2! + 2³/3! + ... = e²

Calculation:

Let f(x) = 1 + (1 + x)/1! + (1 + x)²/2! + (1 + x)³/3! + ...

⇒ f(x) = e^(1+x) / (1 + x)

Expand RHS:

= (1 + x + x²/2! + ...) / (1 + x)

⇒ (1 + x + (1 + x)²/2! + (1 + x)³/3! + (1 + x)⁴/4! + ...)

So, coefficient of x² in RHS is:

²C₂/3! + ³C₂/4! + ⁴C₂/5! + ... = a - 1

coefficient of x2 in RHS:

e × (1 + x+ x2/2!) × (1 -x+ x2/2!)

is e- e+ e/2! =a 

Now, expand LHS expression:

e × (1 + x+ x²/2!) × (1 -x+ x²/2!)

⇒ e × (1 - (x⁴/4!)) = e 

So, coefficient of x² = e × e = e²

Thus, b = 1 + 2/1! + 2²/2! + 2³/3! + ... = e²

a = e\2!

⇒ 8b / a² = 2 × e² / (e/2!)² = 32

∴ 8b / a² = 32

Calculation:

Given,

The function is \( f(x) = \sin(\lfloor x \rfloor) \), where \(\lfloor x \rfloor \) is the greatest integer function, and g(x) = |x| , the absolute value function.

We are tasked with finding:

\( \lim_{x \to 0} \frac{f(x)}{g(x)} \)

For \( g(x) = |x| \), we know that:

\( \lim_{x \to 0} g(x) = 0 \)

For \( f(x) = \sin(\lfloor x \rfloor) \), we know that:

For \( x \to 0^+ \), \( \lfloor x \rfloor = 0 \), so f(x) = sin(0) = 0 .

For \( x \to 0^- \), \( \lfloor x \rfloor = -1 \), so \( f(x) = \sin(-1) \), which is a nonzero constant.

Evaluating the limit:

For \( x \to 0^+ \), \( \frac{f(x)}{g(x)} = \frac{0}{x} = 0 \)

For \( x \to 0^- \), \( \frac{f(x)}{g(x)} = \frac{\sin(-1)}{-x} \), which becomes undefined as \( x \to 0^- \)because the denominator approaches 0, but the numerator remains a nonzero constant.

∴ Since the left-hand and right-hand limits do not match, the limit does not exist.

The correct answer is Option (4):

Calculation:

Given,

The function is\(f(x) = \sin(\lfloor x \rfloor) \), where\(\lfloor x \rfloor \)is the greatest integer function, and g(x) = |x| , the absolute value function.

We are tasked with finding:

\( \lim_{x \to 0} f(x) g(x) \)

For \(g(x) = |x| \), we know that:

\( \lim_{x \to 0} g(x) = 0 \)

For \(f(x) = \sin(\lfloor x \rfloor) \), we know that:

For \(x \to 0^+ \)), \( \lfloor x \rfloor = 0 \), so f(x) = sin(0) = 0 .

For \( x \to 0^- \), \( \lfloor x \rfloor = -1 \), so \(f(x) = \sin(-1) \), which is a nonzero constant.

Evaluating the limit:

For\(x \to 0^+ \),\(f(x)g(x) = 0 \times x = 0 \)

For\(x \to 0^- \), \(f(x)g(x) = \sin(-1) \times (-x) \), which approaches 0 as \(x \to 0^- .\).

∴ The value of \(\lim_{x \to 0} f(x) g(x) \)is 0.

The correct answer is Option (2).

Calculation: 

Since,

\(-\sqrt{2} \leq \sin \mathrm{x}-\cos \mathrm{x} \leq \sqrt{2} \)

\(\therefore -2 \leq \sqrt{2}(\sin \mathrm{x}-\cos \mathrm{x}) \leq 2 \)

\((\text { Assume } \sqrt{2}(\sin \mathrm{x}-\cos \mathrm{x})=\mathrm{k})\)

⇒ –2 ≤ k ≤ 2 …(i) 

⇒ \(f(\mathrm{x})=\log _{\sqrt{\mathrm{m}}}(\mathrm{k}+\mathrm{m}-2)\)

Given,

0 ≤ f(x) ≤ 2

⇒ \(0 \leq \log _{\sqrt{\mathrm{m}}}(\mathrm{k}+\mathrm{m}-2) \leq 2 \)

⇒ \(1 \leq \mathrm{k}+\mathrm{m}-2 \leq \mathrm{m} \)

⇒ \(-\mathrm{m}+3 \leq \mathrm{k} \leq 2 \ldots \text { (ii) }\)

From eq. (i) & (ii), we get –m + 3 = –2

⇒ m = 5

Hence, the correct answer is Option 1.

Concept:

Greatest Integer Function and Definite Integral:

• The greatest integer function, denoted by [x], gives the largest integer less than or equal to x.
• To integrate a greatest integer function, divide the integral into intervals where the function is constant.
• The function inside the integral is f(x) = [1 / e^x−1] = [e^1−x].
• We need to evaluate ∫₀³ [e^1−x] dx = α − logₑ2.

Calculation:

f(x) = [e^1−x] is a decreasing function

f(0) = [e¹] = [2.718] = 2

f(1−ln2) = e^{1−(1−ln2)} = e^ln2 = 2

⇒ boundary point

f(x) = 2 for x ∈ [0, 1−ln2)

f(1) = [e⁰] = [1] = 1

f(x) = 1 for x ∈ [1−ln2, 1)

f(x) < 1 for x ≥ 1 ⇒ [f(x)] = 0

Now break the integral accordingly:

∫₀³ [e^1−x] dx = ∫₀^1−ln2 2 dx + ∫_1−ln2¹ 1 dx + ∫₁³ 0 dx

⇒ 2(1 − ln2) + (1 − (1 − ln2)) + 0

⇒ 2 − 2ln2 + ln2 = 2 − ln2

Given: ∫₀³ [e^1−x] dx = α − ln2

Comparing both sides:

α − ln2 = 2 − ln2 ⇒ α = 2

Now, α³ = 2³ = 8

∴ The value of α³ is 8.

Concept:

Logarithm properties:  

Product rule: The log of a product equals the sum of two logs.

\(\rm {\log _a}\left( {mn} \right) = \;{\log _a}m + \;{\log _a}n\)

Quotient rule: The log of a quotient equals the difference of two logs.

\(\rm {\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\)

Power rule: In the log of power the exponent becomes a coefficient.

\(\rm {\log _a}{m^n} = n{\log _a}m\)

Formula of Logarithms:

If \(\rm lo{g_a}x = b \) then x = ab (Here a ≠ 1 and a > 0)

Calculation:

Given: \(\rm \log_{3}{(x^{4} - x^3)} - \log_{3} (x - 1) = 3\)

\(\rm \Rightarrow \log_{3} \left[{\frac{(x^{4} - x^3)}{(x - 1)}} \right ] = 3\)        (∵ \(\rm {\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\))

\(\rm \Rightarrow \log_{3} \left[{\frac{x^3(x-1)}{(x - 1)}} \right ] = 3\)

\(\rm \Rightarrow \log_{3} x^3 = 3\)

\(\Rightarrow \rm 3\log_3 x = 3\)               (∵ \(\rm {\log _a}{m^n} = n{\log _a}m\)) 

\(\Rightarrow \rm \log_3 x = 1 \\\therefore x=3\)

Concept:

\({a^b} = x \Leftrightarrow lo{g_a}x = b\), where \(a \ne 1\) and a > 0 and x be any number.

Calculation:

Given: 92^1/5 = 4.

As we know that, \({a^b} = x \Leftrightarrow lo{g_a}x = b.\)

Comparing 92^1/5 = 4 with \({a^b} = x\) we have,

Here, a = 92, b = 1 / 5 and x = 4.

Concept:

Logarithm properties

1. Product rule: The log of a product equals the sum of two logs.

\({\log _a}\left( {mn} \right) = \;{\log _a}m + \;{\log _a}n\)

2. Quotient rule: The log of a quotient equals the difference of two logs.

\({\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\)

3. Power rule: In the log of a power the exponent becomes a coefficient.

\({\log _a}{m^n} = n{\log _a}m\)

4. Change of base rule

\({\log _m}n = \frac{{{{\log }_a}n}}{{{{\log }_a}m}}\)

If m = n;
⇒ \({\log _m}m = \frac{{{{\log }_a}m}}{{{{\log }_a}m}} = 1\)

5. \({\log _m}n = \frac{1}{{{{\log }_n}m}}\)

Calculation:

Here, we have to find the value of \({\log _7}{\rm{\;}}{\log _7}\sqrt {7\sqrt {7\sqrt 7 } } \)

\({\log _7}{\rm{\;}}{\log _7}\sqrt {7\sqrt {7\sqrt 7 } } \)

= log₇ log₇ (7^1/2 × 7^1/4 × 7^1/8)

= log₇ log₇ (7(^{1/2 + 1/4 + 1/8)})

= log₇ log₇ (7^{(4 + 2 + 1)/8})

= log₇ log₇ (7^7/8)

From power rule;

= log₇ (7/8) log₇7

= log₇ (7/8) × 1 = log₇ (7/8) = log₇ 7 – log₇ 8

= 1 – log₇ 8 = 1 – log₇ 2³

Concept:

Logarithm properties:  

Product rule: The log of a product equals the sum of two logs.

\(\rm {\log _a}\left( {mn} \right) = \;{\log _a}m + \;{\log _a}n\)

Quotient rule: The log of a quotient equals the difference of two logs.

\(\rm {\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\)

Power rule: In the log of power the exponent becomes a coefficient.

\(\rm {\log _a}{m^n} = n{\log _a}m\)

Formula of Logarithms:

If \(\rm lo{g_a}x = b \) then x = ab (Here a ≠ 1 and a > 0)

Calculation:

Given: \(\rm \log_{4}{(x^{2} - 1)} - \log_{4} (x + 1) = 1\)

\(\rm ⇒ \log_{4} \left[{\frac{(x^{2} - 1)}{(x + 1)}} \right ] = 1\)        (∵ \(\rm {\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\))

\(\rm ⇒ \log_{4} \left[{\frac{(x - 1)(x+1)}{(x + 1)}} \right ] = 1\)

\(\rm ⇒ \log_{4} (x -1) = 1\)

⇒ (x - 1) = 4

∴ x = 5

Concept:

Logarithms:

• If ab = x, then we say that loga x = b.
• log_a a = 1.
• log_a (xy) = log_a x + log_a y.

Calculation:

We know that 80 = 23 × 10.

Changing the given logarithms to log 2, we get:

log₁₀ 80

= log₁₀ (23 × 10)

= log10 2³ + log10 10

= 3 (log10 2) + 1

= 3(0.3010) + 1

= 1.9030.

Given:

5x-1 = (2.5)log105

Formula Used:

If a^x = n then x = log_aⁿ

log_a^b = log_e^b/log_e^a

Calculation:

We have 5x-1 = (2.5)log105

⇒ (2.5)log105 = 5x-1 

⇒ log₁₀⁵  = log_2.5^5x-1 

⇒ log105  = (x - 1) log2.55

⇒ (x - 1) = (log105)/(log2.55)

⇒ (x - 1) = log₁₀^2.5

⇒ x = log102.5 _{+ 1}

⇒ x = log102.5 ₊ log₁₀¹⁰

⇒ x = log1010 × 2.5

⇒ x = log1025

⇒ x = log105²

⇒ x = 2log10​5

∴ The value of x is 2log10​5.

Concept:

Logarithm properties

1. Product rule: The log of a product equals the sum of two logs.

\({\log _a}\left( {mn} \right) = \;{\log _a}m + \;{\log _a}n\)

2. Quotient rule: The log of a quotient equals the difference of two logs.

\({\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\)

3. Power rule: In the log of a power the exponent becomes a coefficient.

\({\log _a}{m^n} = n{\log _a}m\)

4. Change of base rule

\({\log _m}n = \frac{{{{\log }_a}n}}{{{{\log }_a}m}}\)

If m = n;

⇒ \({\log _m}m = \frac{{{{\log }_a}m}}{{{{\log }_a}m}} = 1\)

5. \({\log _m}n = \;\frac{1}{{{{\log }_n}m}}\)

Calculation:

Here, we have to find the value of \({\log _3}{\rm{\;}}{\log _3}\sqrt {3\sqrt 3 }\)

Now,

\({\log _3}{\rm{\;}}{\log _3}\sqrt {3\sqrt 3 }\) = log₃ log₃ (3^1/2 × 3^1/4)

= log₃ log₃ (3^{(1/2 + 1/4)})

= log₃ log₃ (3^{(2 + 1)/4})

= log₃ log₃ (3^3/4)

From power rule;

= log₃ [(3/4)× log₃3]                [∵ log_a (m) ⁿ = n × log_a (m)]

= log₃ (3/4)                  (∵ log_m m = 1)

= log₃ (3/4) = log₃ 3 – log₃ 4

= 1 – log₃ 4 = 1 – log₃ 2²

= 1 – 2 log₃ 2

Concept:

Formula used:

• \({\log _a}b = \frac{1}{{{{\log }_b}a}}\)
• Log_a M + log_a N = log_a (MN)

Factorial:

• n! = 1 × 2 × 3 × ⋯ × (n – 1) × n

Calculation:

Using \({\log _a}b = \frac{1}{{{{\log }_b}a}}\)

\(\frac{1}{{{{\log }_2}{\rm{N}}}} + \frac{1}{{{{\log }_3}{\rm{N}}}} + \ldots + \frac{1}{{{{\log }_{100}}{\rm{N\;}}}} = {\log _{\rm{N}}}2 + {\log _{\rm{N}}}3 + \ldots + {\log _{\rm{N}}}100\)

= log_N (2 × 3 × ⋯ × 100)

= log_N (100!)

\(= \frac{1}{{{{\log }_{100!}}N}}\)

Concept:

Logarithm Rule

log mⁿ = n log m

Calculations:

Let x, y, z are three consecutive positive integers.

⇒ y = x + 1 and z = y + 1

⇒ z = x + 2

Consider, log (1 + xz)

= log [1 + x(x+2)]

= log [1 + x² + 2x]

= log (1 + x)²

= 2 log (1 + x)

= 2 log y

Hence, If x, y, z are three consecutive positive integers, then log (1 + xz) is 2 log y

CONCEPT:

• By base changing theorem we know that \({\log _b}a = \frac{{{{\log }_x}a}}{{{{\log }_x}b}} = \frac{1}{{{{\log }_a}b}}\).
• log _x (x) = 1

CALCULATION:

Given that \(\left\{ {\frac{1}{{{{\log }_9}60}} + \frac{1}{{{{\log }_{16}}60}} + \frac{1}{{{{\log }_{25}}60}}} \right\}\)

By base changing we can write it as - 

⇒ log₆₀9 + log6016 +log6025

By product rule of logarithms we can write it again as - 

⇒ log₆₀(9 x 16 x 25) = log₆₀(3600)

⇒ log₆₀ (60)² = 2 log₆₀(60) = 2

Therefore, option (3) is the correct answer.