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 5 of 5 pages

l o g_{5} a . l o g_{a} x = 2

, then x is equal to

125
a2
25
None of these

Discuss Question

Answer: Option C. -> 25
:
C

l o g_{5} a . l o g_{a} x = 2 \Rightarrow l o g_{5} x = 2 \Rightarrow x = 5^{2} = 25

Question 42.

- 8 > 2 x + 10 + 6 x > - 20

, Which is the possible range of values of 4x + 5 ?

Any value greater than -4 or less than -10
Any value greater than -10 & less than -4
Any value greater than −94 or less than −154
Any value greater than −154 or less than −94

Discuss Question

Answer: Option B. -> Any value greater than -10 & less than -4
:
B

- 8 > 2 x + 10 + 6 x > - 20

- 8 > 8 x + 10 > - 20

Dividing by 2 on each side

- \frac{8}{2} > \frac{8 x + 10}{2} > - \frac{20}{2}

- 4 > 4 x + 5 > - 10

Question 43. 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]

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 44. If

A = l o g_{2} l o g_{2} l o g_{4} 256 + 2 l o g_{\sqrt{2}} 2

, then A is equal to

Discuss Question

Answer: Option C. -> 5
:
C

A = l o g_{2} l o g_{2} l o g_{4} 256 + 2 l o g_{2_{\frac{1}{2}}} 2

= l o g_{2} l o g_{2} l o g_{4} 4^{4} + 2 \times \frac{1}{(\frac{1}{2})} l o g_{2} 2

= l o g_{2} l o g_{2} 4 + 4 = l o g_{2} l o g_{2} 2^{2} + 4

= l o g_{2} 2 + 4 = 1 + 4 = 5

Question 45. Find the value of

{log}_{e}

144.

2loge12
4loge2 + 2loge3
3loge2 + loge18
All of these

Discuss Question

Answer: Option D. -> All of these
:
D

{log}_{e}

144 =

{log}_{e} 12^{2}

2 {log}_{e} 12

(option A)
Factors of 144 =

2^{4}

\times

3^{2}

{log}_{e}

144 =

{log}_{e}

(

2^{4}

\times

3^{2}

)
=

{log}_{e} 2^{4}

{log}_{e} 3^{2}

4 {log}_{e} 2

2 {log}_{e} 3

(option B)

{log}_{e}

144 =

{log}_{e}

(

2^{3}

\times

18) =

{log}_{e} (2^{3})

log 18

3 {log}_{e} 2

log 18

(option C)

Question 46. If 2 - 4x < -6 , what are the possible values for 2x -1?

All values less than 3
All values greater than 3
All values less than -3
All values greater than -3

Discuss Question

Answer: Option B. -> All values greater than 3
:
B

2 - 4 x < - 6

\Rightarrow 2 (1 - 2 x) < - 6

\Rightarrow (1 - 2 x) < - 3

(on division by 2)

\Rightarrow (2 x - 1) > 3

(on multiplication with -1), multiplying with a negative number inverts the inequality.
Hence, B is the correct choice

Question 47. A woman has m

$ 20

notes and n

$ 50

notes. The total amount of money in her possession cannot be less than

$ 800

. Represent this as an inequality.

20m+50n
20m+50n≤800
20m+50n>800
20m+50n≥800

Discuss Question

Answer: Option D. -> 20m+50n≥800
:
D
Total value of m

$ 20

notes and n

$ 50

notes = 20m + 50 n
The least amount of money the woman can have in her possession is

$ 800

.
Therefore,

20 m + 50 n \geq 800

The correct choice is D.

Question 48. Find the interval(s) in which the expression (4-x)(x+2) is positive.

(−∞,−2)
(−2,4)
(4,∞)
The expression is always less than zero

Discuss Question

Answer: Option B. -> (−2,4)
:
B

L e t P = (4 - x) (x + 2)

Find the points on the number line where P=0

P = 0 \Rightarrow x ϵ {- 2, 4}

Split the number line into 3 using these points.

(- \infty, - 2), (- 2, 4) & (4, \infty) (- \infty, - 2) : (4 - x) > 0 & (x + 2) < 0 \Rightarrow P < 0 (- 2, 4) : (4 - x) > 0 & (x + 2) > 0 \Rightarrow P > 0 (4, \infty) : (4 - x) < 0 & (x + 2) > 0 \Rightarrow P < 0

∴ P > 0, w h e n x ϵ (- 2, 4)

Question 49. Which of the following inequalities represent the statement - ' k cannot exceed 5' ?

k>5
k
k≥5
k≤5

Discuss Question

Answer: Option D. -> k≤5
:
D
k cannot exceed 5 means that k can be less than or equal to 5.
Hence, D is the correct choice.

Question 50.

6 x - 9 y > 12

Which of the following inequalities is equivalent to the inequality above?

x−y>2
2x−3y>4
3x−2y>4
3y−2x>2

Discuss Question

Answer: Option B. -> 2x−3y>4
:
B
We know that an inequality remains unchanged if the entire inequality is multiplied by a positive constant. If an inequality is multiplied by a positive constant, the ratios of the coefficients of x, coefficients of y and the constant terms will be equal.
It can be observed that in option B,
Ratio of coefficients of