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

Concept:

• log A + log B = log AB
• log A - log B = log(A/B)
• log a^k = k log a
• \(\sum_{k=1}^n k = {n(n+1)\over 2}\)
• \(\sum_{k=1}^n k^2 = {n(n+1)(2n+1)\over 6}\)

Calculation:

Given,

\(\displaystyle \frac{logx+logx^4+logx^9+....+logx^{n^2}}{logx+logx^2+logx^3+....+logx^n}\)

⇒ \(\displaystyle \frac{log[x(x^4)(x^9)....(x^{n^2})]}{log[x(x^2)(x^3)...(x^n)]}\)

⇒ \(\displaystyle \frac{log[x^{1 +4 + 9+ ...n^2}]}{log[x^{1+2+3+...+n}]}\)

⇒ \(\displaystyle \frac{log[x^{\sum_{k=1}^n k^2} ]}{log[x^{\sum_{k=1}^n k}]}\)

⇒ \(\displaystyle \frac{{\sum_{k=1}^n k^2} \ log[x]}{{\sum_{k=1}^n k}\ log[x]}\)

⇒ \(\displaystyle \frac{{\sum_{k=1}^n k^2} }{{\sum_{k=1}^n k}\ }\)

⇒ \(\displaystyle \frac{{n(n+1)(2n+1)\over 6}}{{n(n+1)\over 2}}\)

⇒ \({2n+1\over 3}\)

∴ The correct answer is option (1).

Explanation:

Since, {x}+[x] = x

⇒x – [x] = {x}

⇒ 0≤ x – [x] < 1 (0 ≤ {x} < 1)

⇒ –1 ≤ [x] –x ≤ 0

But x is positive and non-integer; then

⇒ –1 < [x]–x < 0

⇒ –1 < y < 0

⇒ [y] = –1 

∴ Option (a) is correct.

Given:

8k = 2

Formula used:

log_b a = c ^{if and only if} b^c = a

Calculation:

Given: 8k = 2

⇒ log₈ 2 = x

∴ The correct answer is option (3).

Given:

log₁₀ 10000 = ?

Formula used:

log_b a = c ⟺ b^c = a

Calculation:

log₁₀ 10000 = ?

10000 = 10⁴

⇒ log₁₀ 10000 = 4

∴ The correct answer is option (4).

Explanation -

We have,

\(4^x+2^{2 x-1}=3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}}\)

⇒ 2 × 2^2x-1 + 2^2x-1 = \(3^{x-\frac{1}{2}} \) × 3 + \(3^{x-\frac{1}{2}}\)

⇒ 2^2x-1 (2 + 1) = \(3^{x-\frac{1}{2}}\) (3 + 1)

⇒ 22x-1 × 3 = \(3^{x-\frac{1}{2}}\) × 4

⇒ 22x-3 = \(3^{x-\frac{3}{2}}\)

⇒ \(\left(2^2\right)^{x-\frac{3}{2}}\) = \(3^{x-\frac{3}{2}} \) 

⇒ \(4^{x-\frac{3}{2}}\) = \(3^{x-\frac{3}{2}}\) 

⇒ \(x-\frac{3}{2}\) = 0 

⇒ \(x=\frac{3}{2}\)

​Hence Option (2) is correct.

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.