Inequalities Modulus And Logarithms(11th Grade > Mathematics ) Questions and answers for exam Preparation

11th Grade > Mathematics

INEQUALITIES MODULUS AND LOGARITHMS MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

If $x ϵ R$ and $m = \frac{x^{2}}{(x^{4} - 2 x^{2} + 4)}$ , then m lies in the interval

$[0, \frac{1}{4}]$
$[0, \frac{1}{3}]$
$[0, \frac{1}{2}]$
$[0, \frac{1}{5}]$

Discuss Question

Answer: Option C. ->

[0, \frac{1}{2}]

:
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 2.

If $l o g_{4}$ 5 = a and $l o g_{5}$ 6 = b, then $l o g_{3}$ 2 is equal to

$\frac{1}{2 a + 1}$
$\frac{1}{2 b + 1}$
$2 a b + 1$
$\frac{1}{2 a b - 1}$

Discuss Question

Answer: Option D. ->

\frac{1}{2 a b - 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 3.

To pass a bill in the Lower house, the Indian government requires at least $\frac{2}{3}$ majority in the house. There are 545 seats in the lower house. 248 members are in favor of passing a particular bill. Which of the following inequality represents the number of additional lower house members, p, who would support the bill, to achieve the required majority ? Round off your answer to nearest tenth digit.

$p \geq 363.3$
$p \geq 445.5$
$p \geq 198$
$p \geq 115.3$

Discuss Question

Answer: Option D. ->

p \geq 115.3

:
D

Step 1 : Underline the key words.
To pass a bill in the Lower house, the Indian government requires at least $\frac{2}{3}$ majority in the house. There are 545 seats in the lower house. 248 members are in favor of passing a particular bill. Which of the following inequality represents the number of additional lower house members, p, who would support the bill, to achieve the required majority.

Step 2 : Recognize the variable
Additional members required to support the bill

Step 3 : Decipher the relation
Given, there are 248 members already supporting the bill and we require 'p' members more.
Thus, 248 + p is required to pass the bill

Step 4 : decide the inequality
Look out for cues :at least $\frac{2}{3}$ majority. We also, know there are 545 seats.
Therefore, $248 + p \geq (\frac{2}{3})$ 545.
Since, the answer options are simplified. Simplify the given inequation :
$p \geq 115.333$
Hence, option D is correct.

Question 4.

$l o g_{5} a . l o g_{a} x = 2$ , then x is equal to

125
$a^{2}$
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 5.

A worker uses a forklift to move boxes that weigh either 40 pounds or 65 pounds each. Let x be the number of 40-pound boxes and y be the number of 65-pound boxes. The forklift can carry up to either 45 boxes or a weight of 2,400 pounds. Which of the following systems of inequalities represents this relationship?

$40 x + 65 \leq 2400; x + y \leq 45$
$\frac{x}{40} + \frac{y}{65} \leq 2400; x + y \leq 45$
$40 x + 65 \leq 45; x + y \leq 2400$
$40 x + 65 \leq 2400; x + y \leq 2400$

Discuss Question

Answer: Option A. ->

40 x + 65 \leq 2400; x + y \leq 45

:
A

The first constraint is on the weight of the boxes. The total weight of the boxes cannot exceed 2400. Since there are x 40 pound boxes and y 65 pound boxes, total weight is (40x + 65 y). So,
$40 x + 65 y \leq 2400$
The total number of boxes cannot exceed 45. Total number of boxes is (x + y). So,
$x + y \leq 45$
Hence, A is the correct choice.

Question 6.

Solve for $- 2 x + 5 \leq 10$

$x \geq - \frac{5}{2}$
$x \geq \frac{5}{2}$
$x \leq \frac{5}{2}$
None of the above

Discuss Question

Answer: Option A. ->

x \geq - \frac{5}{2}

:
A

$- 2 x + 5 \leq 10$
Subtract 5 on both side
$- 2 x + 5 - 5 \leq 10 - 5$
$- 2 x \leq 5$
Divide by -2 on both side
$- \frac{2 x}{- 2} \geq \frac{5}{- 2}$
$x \geq - \frac{5}{2}$
The correct answer is option A

Question 7.

If $3 < p < 7$ and $- 9 < q < - 4$ , then what is the range of p - q?

(-16, -7)
(16, 7)
(7, 16)
(-7, -16)

Discuss Question

Answer: Option C. -> (7, 16)
:
C

3 < p < 7

and

- 9 < q < - 4

Multiply the second inequality with - 1. This causes the sign to flip.
So,

3 < p < 7

and

4 < - q < 9

Adding these two inequalities, we get

7 < p - q < 16

Question 8.

$- 8 > 2 x + 10 + 6 x > - 20$ , 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 $- \frac{9}{4}$ or less than $- \frac{15}{4}$
Any value greater than $- \frac{15}{4}$ or less than $- \frac{9}{4}$

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 9.

Select the solution set of $- 1 < 2 x + 3 \leq 10$ from given options

{1,2,3}
{1,2,3,4}
{2,3,4}
{2,3,4,5}

Discuss Question

Answer: Option A. -> {1,2,3}
:
A

$- 1 < 2 x + 3 \leq 10$
Subtract 3 from each sides
$- 1 - 3 < 2 x + 3 - 3 \leq 10 - 3$
$- 4 < 2 x \leq 7$
Divide by 2 each side
$- \frac{4}{2} < \frac{2 x}{2} \leq \frac{7}{2}$
$- 2 < x \leq 3.5$
From here we can say that x will that all the value between -2 to 3.5 [including 3.5 because $\leq$ sign].

Now lets look at options

A) {1,2,3}
B) {1,2,3,4} $\to$ not possible because 4 is not in the range
C) {2,3,4} $\to$ same reason
D) {2,3,4,5} $\to$ Same reason

Question 10.

Find the intervals in which $(x - 3) (x^{2} - 1) > 0$

$(- \infty, - 1)$
$(- 1, 1)$
$(1, 3)$
$(3, \infty)$

Discuss Question

Answer: Option B. ->

(- 1, 1)

:
B and D

$(x - 3) (x^{2} - 1) = (x - 3) (x - 1) (x + 1) L e t P = (x - 3) (x - 1) (x + 1)$
Find the points on the number line where P=0
$P = 0 \Rightarrow x ϵ {- 1, 1, 3}$
Split the number line into 4 using these points.
$(- \infty, - 1), (- 1, 1), (1, 3) & (3, \infty) (- \infty, - 1) : (x - 3) < 0, (x - 1) < 0 & (x + 1) < 0 \Rightarrow P < 0 (- 1, 1) : (x - 3) < 0, (x - 1) < 0 & (x + 1) > 0 \Rightarrow P > 0 (1, 3) : (x - 3) < 0, (x - 1) > 0 & (x + 1) > 0 \Rightarrow P < 0 (3, \infty) : (x - 3) > 0, (x - 1) > 0 & (x + 1) > 0 \Rightarrow P > 0$
$∴ P > 0, w h e n x ϵ (- 1, 1) \cup (3, \infty)$