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

Last updated on Apr 22, 2025

Latest Special Functions MCQ Objective Questions

Special Functions Question 1:

\(\frac{logx+logx^4+logx^9+....+logx^{n^2}}{logx+logx^2+logx^3+....+logx^n}\) is equal to

  1. \(\frac{2n+1}{3}\)
  2. \(\frac{2n-1}{3}\)
  3. \(\frac{3(n+2)}{2}\)
  4. None of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : \(\frac{2n+1}{3}\)

Special Functions Question 1 Detailed Solution

Concept:

  • log A + log B = log AB
  • log A - log B = log(A/B)
  • log ak = 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).

Special Functions Question 2:

Let z = [y] and y = [x] - x, where [.] is the greatest integer function. If x is not an integer but positive, then what is the value of z? 

  1. -1
  2. 0
  3. 1
  4. 2

Answer (Detailed Solution Below)

Option 1 : -1

Special Functions Question 2 Detailed Solution

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.

Special Functions Question 3:

8k = 2 then select the correct option from below? 

  1. logx 8 = 2
  2. logx 2 = 8
  3. log8 2 = x
  4. log2 x = 8

Answer (Detailed Solution Below)

Option 3 : log8 2 = x

Special Functions Question 3 Detailed Solution

Given:

8k = 2

Formula used:

logb a = c if and only if bc = a

Calculation:

Given: 8k = 2

⇒ log8 2 = x

∴ The correct answer is option (3).

Special Functions Question 4:

log10 10000 = _______?

  1. 104
  2. 10
  3. 100
  4. 4

Answer (Detailed Solution Below)

Option 4 : 4

Special Functions Question 4 Detailed Solution

Given:

log10 10000 = ?

Formula used:

logb a = c ⟺ bc = a

Calculation:

log10 10000 = ?

10000 = 104

⇒ log10 10000 = 4

∴ The correct answer is option (4).

Special Functions Question 5:

If 4x + 22x-1 = 3x+\(\frac{1}{2}\) + 3x-\(\frac{1}{2}\) then x =

  1. \(\frac{1}{2}\)
  2. \(\frac{3}{2}\)
  3. \(\frac{5}{2}\)
  4. 1
  5. \(\frac{1}{8}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{3}{2}\)

Special Functions Question 5 Detailed Solution

Explanation -

We have,

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

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

⇒ 22x-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.

Top Special Functions MCQ Objective Questions

If \(\rm \log_{3}{(x^{4} - x^3)} - \log_{3} (x - 1) = 3\) then x is equal to ?

  1. 1
  2. 6
  3. 3
  4. 9

Answer (Detailed Solution Below)

Option 3 : 3

Special Functions Question 6 Detailed Solution

Download Solution PDF

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

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

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

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

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

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

Write the logarithmic form of 921/5 = 4.

  1. \(lo{g_{92}}4 = \frac{1}{5}\)
  2. \(lo{g_{\frac{1}{5}}}4 = 92\)
  3. \(lo{g_{92}}\left( {\frac{1}{5}} \right) = 3\)
  4. None of these

Answer (Detailed Solution Below)

Option 1 : \(lo{g_{92}}4 = \frac{1}{5}\)

Special Functions Question 7 Detailed Solution

Download Solution PDF

Concept:

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

Calculation:

Given: 921/5 = 4.

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

Comparing 921/5 = 4 with \({a^b} = x\) we have,

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

So, the logarithmic form of 921/5 = 4 is \(lo{g_{92}}4 = \frac{1}{5}\).

What is the value of \({\log _7}{\rm{\;}}{\log _7}\sqrt {7\sqrt {7\sqrt 7 } } \) equal to?

  1. 3 log2 7
  2. 1 – 3 log2 7
  3. 1 – 3 log7 2
  4. \(\frac{7}{8}\)

Answer (Detailed Solution Below)

Option 3 : 1 – 3 log7 2

Special Functions Question 8 Detailed Solution

Download Solution PDF

Concept:

Logarithm properties

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

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

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

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

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

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

  1. Change of base rule

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

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

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

 

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 (71/2 × 71/4 × 71/8)

= log7 log7 (7(1/2 + 1/4 + 1/8))

= log7 log7 (7(4 + 2 + 1)/8)

= log7 log7 (77/8)

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

If \(\rm \log_{4}{(x^{2} - 1)} - \log_{4} (x + 1) = 1\) then x is equal to ?

  1. 1
  2. 2
  3. 4
  4. 5

Answer (Detailed Solution Below)

Option 4 : 5

Special Functions Question 9 Detailed Solution

Download Solution PDF

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

If log10 2 = 0.3010, then log10 80 = ?

  1. 1.240
  2. 0.9030
  3. 3.010
  4. 1.9030

Answer (Detailed Solution Below)

Option 4 : 1.9030

Special Functions Question 10 Detailed Solution

Download Solution PDF

Concept:

Logarithms:

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

If 5x-1 = (2.5)log105, then what is the value of x ?

  1. 1
  2. log102
  3. log10​5
  4. 2log10​5

Answer (Detailed Solution Below)

Option 4 : 2log10​5

Special Functions Question 11 Detailed Solution

Download Solution PDF

Given:

5x-1 = (2.5)log105

Formula Used:

If ax = n then x = logan

logab = logeb/logea

Calculation:

We have 5x-1 = (2.5)log105

⇒ (2.5)log10= 5x-1 

⇒ log105  = log2.55x-1 

⇒ log105  = (x - 1) log2.55

⇒ (x - 1) = (log105)/(log2.55)

⇒ (x - 1) = log102.5

⇒ 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.

What is the value of \({\log _3}{\log _3}\sqrt {3\sqrt 3 }\) equal to?

  1. 3 log2 (3)
  2. 1 – 3 log2 (2)
  3. 1 – 2 log3 (2)
  4. \(\frac{3}{8}\)

Answer (Detailed Solution Below)

Option 3 : 1 – 2 log3 (2)

Special Functions Question 12 Detailed Solution

Download Solution PDF

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 (31/2 × 31/4)

= log3 log3 (3(1/2 + 1/4))

= log3 log3 (3(2 + 1)/4)

= log3 log3 (33/4)

From power rule;

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

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

= log3 (3/4) = log3 3 – log3 4

= 1 – log3 4 = 1 – log3 22

= 1 – 2 log3 2

What is \(\frac{1}{{{{\log }_2}N}} + \frac{1}{{{{\log }_3}N}} + \frac{1}{{{{\log }_4}N}} + \ldots + \frac{1}{{{{\log }_{100}}N\;}}\;\) equal to (N ≠ 1)?

  1. \(\frac{1}{{{{\log }_{100!}}N}}\)
  2. \(\frac{1}{{{{\log }_{99!}}N}}\)
  3. \(\frac{{99}}{{{{\log }_{100!}}N}}\)
  4. \(\frac{{99}}{{{{\log }_{99!}}N}}\)

Answer (Detailed Solution Below)

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

Special Functions Question 13 Detailed Solution

Download Solution PDF

Concept:

Formula used:

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


Factorial:

  • n! = 1 × 2 × 3 × ⋯ × (n – 1) × n

 

Calculation:

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

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

= logN (2 × 3 × ⋯ × 100)

= logN (100!)

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

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

  1. log y
  2. log (y/2)
  3. log (2y)
  4. 2 log (y)

Answer (Detailed Solution Below)

Option 4 : 2 log (y)

Special Functions Question 14 Detailed Solution

Download Solution PDF

Concept:

Logarithm Rule

log m= 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

The value of \(\left\{ {\frac{1}{{{{\log }_9}60}} + \frac{1}{{{{\log }_{16}}60}} + \frac{1}{{{{\log }_{25}}60}}} \right\}\) is - 

  1. 0
  2. 1
  3. 2
  4. 3

Answer (Detailed Solution Below)

Option 3 : 2

Special Functions Question 15 Detailed Solution

Download Solution PDF

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.

Get Free Access Now
Hot Links: teen patti star login teen patti bliss teen patti neta teen patti noble teen patti app