MATHS M.C.Q (1 Marks)

3. $\text{x}\in(-\infty,-4)\cup(6,\infty)$

Solution:

|x−1| > 5

⇒ x − 1 > 5 or x − 1 < −5

⇒ x > 5 + 1 or x < −5 + 1

⇒ x > 6 or x < −4

$\Rightarrow\text{x}\in(-\infty,-4)\cup(6,\infty)$

1. $\text{x}\in(10,\infty)$

Solution:

− 3x + 17 < −13

Subtracting 17 on both sides, we get

⇒ −3x + 17 − 17 < −13 − 17

⇒ −3x < − 30

Dividing −3 on both sides, we get

$\Rightarrow\frac{-3\text{x}}{-3}>\frac{-30}{-3}$

$\Rightarrow\text{x}>10$

$\Rightarrow\text{x}\in(10,\infty)$

2. [-7, 3]

Solution:

$|\text{x}+2|\leq5$

$\Rightarrow-5\leq\text{x}+2\leq5$

$\Rightarrow-5-2\leq\text{x}+2-2\leq5-2$

$\Rightarrow-7\leq\text{x}\leq3$

$\Rightarrow\text{x}\in[-7,3]$

2. $-5<\text{x}<5$

Solution:

If x is a real number.

|x| < 5

⇒ -5 < x < 5

3. $\text{x}\in[-11,7]$

Solution:

$|\text{x}+2|\leq9$

$\Rightarrow-9\leq\text{x}+2\leq9$

$\Rightarrow-9-2\leq\text{x}+2-2\leq9-2$

$\Rightarrow-11\leq\text{x}\leq7$

$\Rightarrow\text{x}\in[-11,7]$

4. $\text{x}\in(-\infty,-13]\cup[7,\infty)$

Solution:

$|\text{x}+3|\geq10$

$\Rightarrow\text{x}+3\geq10\text{ or }\text{x}+3;\leq-10$

$\Rightarrow\text{x}\geq10-3\text{ or }\text{x}\leq-10-3$

$\Rightarrow\text{x}\geq7\ \text{or}\ \text{x}\leq-13$

$\Rightarrow\text{x}\in(-\infty,-13)\cup[7,\infty)$

3. $-\text{x}>-7$

Solution:

x < 7

subtracting x on both sides, we get

⇒ x − x < 7 − x

⇒ 0 < 7 − x

subtracting 7 on both sides, we get

⇒ 0 − 7 < 7 − x − 7

⇒ −7 < − x

⇒ − x > −7

4. $\text{x}\in(-\infty,-\text{a})\cup(\text{a},\infty)$

Solution:

If x and a are real numbers such that a > 0.

|x| > a

⇒ x > a or x < −a

$\Rightarrow\text{x}\in(-\infty,-\text{a})\cup(\text{a},\infty)$

3. $\text{x}\in[-11,7]$

Solution:

As according to the graph,

x lies between $(-\infty,-5]$ and $[5,\infty)$

$\Rightarrow\text{x}\geq5$ or $\text{x}\leq-5$

$\Rightarrow|\text{x}|\geq5$

1. $\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$

Solution:

Given that x, y and b are real numbers and x < y, b > 0.

Both sides of an inequality can be multiplied or divided by the same positive number.

$\therefore\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$

2. $\text{|x|}\leq3$

Solution:

As according to the graph,

x lies between −3 and 3

$\Rightarrow-3\leq\text{x}\leq3$

$\Rightarrow|\text{x}|\leq3$

2. [-7, 3]

Solution:

$|\text{x}+2|\leq5$

$\Rightarrow-5\leq\text{x}+2\leq5$

$\Rightarrow-5-2\leq\text{x}+2-2\leq5-2$

$\Rightarrow-7\leq\text{x}\leq3$

$\Rightarrow\text{x}\in[-7,3]$