Logarithms(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

LOGARITHMS MCQs

Inequalities Modulus And Logarithms (11th And 12th Grade)

Total Questions : 50 | Page 1 of 5 pages

l o g_{10} x = y t h e n l o g_{1000} x^{2}

is equal to

Discuss Question

Answer: Option D. -> 2y3
:
D

l o g_{1000} x^{2}

= l o g_{10^{3}} x^{2}

= 2 l o g_{10^{3}} x

= \frac{2}{3} l o g_{10} x

= \frac{2}{3} y

Question 2. If

x ϵ R

and

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

, then m lies in the interval

[0,14]
[0,13]
[0,12]
[0,15]
b2+c2b+c+c2+a2c+a+a2+b2a+b>a+b+c

Discuss Question

Answer: Option C. -> [0,12]
:
C

m = \frac{x^{2}}{(x^{2} - 1)^{2} + 3} = + i v e i . e . \geq 0

Again

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

m ϵ [0, \frac{1}{2}]

Question 3. The value of

81^{(\frac{1}{l o g_{5} 3})} + 27^{l o g_{9} 36} + 3^{\frac{4}{l o g_{7} 9}}

is equal to

Discuss Question

Answer: Option D. -> 890
:
D

81^{(\frac{1}{l o g_{5} 3})} + 27^{l o g_{9} 36} + 3^{\frac{4}{l o g_{7} 9}}

= 3^{l o g_{3} 5^{4}} + 3^{l o g_{3} 36^{\frac{3}{2}}} + 3^{l o g_{3} 7^{\frac{4}{2}}}

= 5^{4} + 36^{\frac{3}{2}} + 7^{2} = 890

Question 4. The value of

l o g_{3} 4 l o g_{4} 5 l o g_{5} 6 l o g_{6} 7 l o g_{7} 8 l o g_{8} 9

Discuss Question

Answer: Option B. -> 2
:
B

l o g_{3} 4. l o g_{4} 5. l o g_{5} 6. l o g_{6} 7. l o g_{7} 8. l o g_{8} 9

= \frac{l o g 4}{l o g 3} . \frac{l o g 5}{l o g 4} . \frac{l o g 6}{l o g 5} . \frac{l o g 7}{l o g 6} . \frac{l o g 8}{l o g 7} . \frac{l o g 9}{l o g 8} = \frac{l o g 9}{l o g 3}

= l o g_{3} 9 = l o g_{3} 3^{2} = 2

Question 5. If

y = 3^{x - 1} + 3^{- x - 1}

(x real), then the least value of y is

2
6
23
None of these

Discuss Question

Answer: Option C. -> 23
:
C
We have

3^{x - 1} + 3^{- x - 1} = \frac{1}{3} (3^{x} + 3^{- x}) \geq \frac{1}{3} . 2 \sqrt{3^{x} . 3^{- x}}

[∵ A . M . \geq G . M .]

3^{x - 1} + 3^{- x - 1} \geq \frac{2}{3}

Hence the least value of

3^{x - 1} + 3^{- x - 1}

\frac{2}{3}

Question 6. If a and b are two positive quantities whose sum is

λ

, then the minimum value of

\sqrt{(1 + \frac{1}{a}) (1 + \frac{1}{b})}

λ−1λ
λ−2λ
1+1λ
1+2λ

Discuss Question

Answer: Option D. -> 1+2λ
:
D

E^{2} = 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{a b} = 1 + \frac{a + b + 1}{a b} = 1 + \frac{λ + 1}{a b}

Above will be minimum when ab is maximum.
Now we know that if sum of two quantities is constant then their product is maximum when the quantities are equal.

∴ a + b = λ \Rightarrow a = b = \frac{λ}{2}

∴ E^{2} = \frac{λ^{2} + 4 λ + 4}{λ^{2}} = {(\frac{λ + 2}{λ})}^{2}

∴ E = \frac{λ + 2}{λ} = 1 + \frac{2}{λ} \Rightarrow (d)

Question 7. If

x = l o g_{5} 3 + l o g_{7} 5 + l o g_{9} 7

, then x

\geq

32
1213
3213
None

Discuss Question

Answer: Option C. -> 3213
:
C
Apply

A . M . \geq G . M .

∴ x \geq 3. (l o g_{5} 3. l o g_{7} 5. l o g_{9} 7)^{\frac{1}{3}}

= 3 (l o g_{9} 3)^{\frac{1}{3}} = 3 (l o g_{3^{2}} 3)^{\frac{1}{3}} = 3 {(\frac{1}{2})}^{\frac{1}{3}}

Question 8. If a, b, c be three real numbers of the same sign then the value of

\frac{a}{b} + \frac{b}{c} + \frac{c}{a}

lies in the interval

[3,∞)
(3,∞)
[2,∞)
(−∞,3)

Discuss Question

Answer: Option A. -> [3,∞)
:
A
Since a, b, c are of same sign,

\frac{a}{b}, \frac{b}{c}, \frac{c}{a}

are all +ive.
Apply A.M.

\geq

G.M.

∴ E \geq 3 {[\frac{a}{b} . \frac{b}{c} . \frac{c}{a}]}^{\frac{1}{3}} = 3

therefore E lies in the interval

[3, \infty)

Question 9. If

l o g_{4}

5 = a and

l o g_{5}

6 = b, then

l o g_{3}

2 is equal to

12a+1
12b+1
2ab+1
12ab−1

Discuss Question

Answer: Option D. -> 12ab−1
:
D

a b = l o g_{4} 5. l o g_{5} 6 = l o g_{4} 6 = \frac{1}{2} l o g_{2} 6

a b = \frac{1}{2} (1 + l o g_{2} 3) \Rightarrow 2 a b - 1 = l o g_{2} 3

∴ l o g_{3} 2 = \frac{1}{2 a b - 1}

Question 10. If the entry fee to a fair is

$ 5

and each game at the fair costs

$ 2

to play, write an inequality for the total cost when a person plays n games and the maximum amount the person can spend is

$ 25

2n+5≤25
2n−5≤25
5n+2≤25
5n−2≤25

Discuss Question

Answer: Option A. -> 2n+5≤25
:
A
Cost of entry (dollars) = 5
Cost of playing n games (dollars) = 2n
Total cost (dollars) = 2n + 5
Total cost can be at most