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 , where is the greatest integer function, and g(x) = |x| , the absolute value function.

We are tasked with finding:

For , we know that:

For , we know that:

For , , so f(x) = sin(0) = 0 .

For , , so , which is a nonzero constant.

Evaluating the limit:

For ,

For , , which becomes undefined as 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, whereis the greatest integer function, and g(x) = |x| , the absolute value function.

We are tasked with finding:

For , we know that:

For , we know that:

For ), , so f(x) = sin(0) = 0 .

For , , so , which is a nonzero constant.

Evaluating the limit:

For,

For, , which approaches 0 as .

∴ The value of is 0.

The correct answer is Option (2).

Calculation: 

Since,

⇒ –2 ≤ k ≤ 2 …(i) 

⇒ 

Given,

0 ≤ f(x) ≤ 2

⇒ 

⇒ 

⇒ 

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)

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.

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

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

Formula of Logarithms:

If  then x = ab (Here a ≠ 1 and a > 0)

Calculation:

Given: 

        (∵ )

               (∵ ) 

Concept:

, where  and a > 0 and x be any number.

Calculation:

Given: 92^1/5 = 4.

As we know that, 

Comparing 92^1/5 = 4 with  we have,

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

So, the logarithmic form of 92^1/5 = 4 is .

Concept:

Logarithm properties

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

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

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

4. Change of base rule

If m = n;
⇒

Calculation:

Here, we have to find the value of

= 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

= 1 – 3 log₇ 2

Concept:

Logarithm properties:  

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

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

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

Formula of Logarithms:

If  then x = ab (Here a ≠ 1 and a > 0)

Calculation:

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

        (∵ )

⇒ (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.

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

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

4. Change of base rule

If m = n;

⇒ 

5. 

Calculation:

Here, we have to find the value of 

Now,

 = 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 M + log_a N = log_a (MN)

Factorial:

• n! = 1 × 2 × 3 × ⋯ × (n – 1) × n

Calculation:

Using 

= log_N (2 × 3 × ⋯ × 100)

= log_N (100!)

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 _x (x) = 1

CALCULATION:

Given that 

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.