[4 marks sum]

Question 14 Marks

Find the solution set of the following inequalities and draw the graph of their solutions sets : $\frac{3}{|x-2|} > 5$.

Answer

We have $\frac{3}{|x-2|}>5$
$3 > 5 | x - 2 |$
$5 | x - 2 | < 3$
|$x-2 \left\lvert\,<\frac{3}{5}\right.$
Using property $| x | < a =- a < x < a$
$\therefore-\frac{3}{5}$
$\Rightarrow-\frac{3}{5}+2$
$\Rightarrow \frac{7}{5}\text { So } \frac{3}{|x-2|} > 5$
$\Rightarrow {x}: \frac{7}{5}$

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Question 24 Marks

Find the solution set of the following inequalities and draw the graph of their solutions sets :
| 3 - 2x | ≥ 2

Answer

We have
| 3 - 2x | ≥ 2
Using prop. | x | ≥ a ⇔ x ≥ a or x ≤ - a
| 3 - 2x | ≥ 2
or
3 - 2x ≤ -2 or (3 - 2x) ≥ 2
⇒ -2x ≤ -5 or -2x ≥ -1
⇒ 2x ≤ 1 or 2x ≥ 5
$\Rightarrow x \leq \frac{1}{2} \quad$ or $\quad x \geq \frac{5}{2}$
or
| 3 - 2x | ≥ 2
$\Rightarrow\left\{x: x \leq \frac{1}{2}\right.$ or $\left.\geq \frac{5}{2}\right\}$
The graph of this set is

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Question 34 Marks

Solve the following inequalities in the given universal set :
2x - 5 ≤ 5x + 4 < 11, where x ∈ I.

Answer

2x - 5 ≤ 5x + 4 < 11, x ∈ I.
2x - 5 ≤ 5x + 4
2x - 5x ≤ 4 + 5
-3x ≤ 9
3x ≥ -3
or
-3 ≤ x
5x + 4 < 11
5x < 11 - 4
5x < 7
$x<\frac{7}{5}$
$x<1 \frac{2}{5}$
From (1) and (2) $-3 \leq x<1 \frac{2}{5}, x \in I$
∴ Solution set = {-3, -2, -1, 0, 1}

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Question 44 Marks

Find the values of x, which satisfy the inequation
$-2 \frac{5}{6}<\frac{1}{2}-\frac{2 x}{3} \leq 2, x \in W$. Graph the solution set on the number line.

Answer

$-2 \frac{5}{6}<\frac{1}{2}-\frac{2 x}{3} \leq 2$
Taking, $-2 \frac{5}{6}<\frac{1}{2}-\frac{2 x}{3}$
$-\frac{17}{6}<\frac{1}{2}-\frac{2 x}{3}$
$-\frac{17}{6}-\frac{1}{2}<-\frac{2 x}{3}$
$\frac{-17-3}{6}<-\frac{2 x}{3}$
$-\frac{20}{6}<-\frac{2 x}{3}$
$\Rightarrow \frac{10}{3}>\frac{2 x}{3}$
$5 > x ...(1)$
Now taking, $\frac{1}{2}-\frac{2 x}{3} \leq 2$
$-\frac{2 x}{3} \leq 2-\frac{1}{2}$
$-\frac{2 x}{3} \leq \frac{3}{2}$
$-x \leq \frac{9}{4}$
$\Rightarrow x \geq-\frac{9}{4} ...(2)$
From $(1)$ and $(2),$ we get
$-\frac{9}{4} \leq x<5$
$\Rightarrow-2 \frac{1}{4} \leq x<5$
Required number line,

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Question 54 Marks

Solve the following inequation, write the solution set and represent it on the number line:
$-\frac{x}{3} \leq \frac{x}{2}-1 \frac{1}{3}<\frac{1}{6}, x \in R$

Answer

$-\frac{x}{3} \leq \frac{x}{2}-1 \frac{1}{3}<\frac{1}{6}, x \in R $
$ -\frac{x}{3} \leq \frac{x}{2}-1 \frac{1}{3}$
$ -\frac{x}{3} \leq \frac{x}{2}-\frac{4}{3}$
$\frac{4}{3} \leq \frac{x}{2}+\frac{x}{3}$
$ \frac{4}{3} \leq \frac{5 x}{6} $
$ \frac{6}{5} \times \frac{4}{3} \leq x$
$\frac{8}{5} \leq x $
$ \frac{x}{2}-1 \frac{1}{3}<\frac{1}{6} $
$\frac{x}{2}<\frac{1}{6}+\frac{4}{3} $
$ \frac{x}{2}<\frac{1+8}{6}$
$x<\frac{9 \times 2}{6}$
$x < 3$
From $(1)$ and $(2)$
$\frac{8}{5} \leq x<3$
or $1·6 \leq x < 3$
$\therefore$ Solution set $\{x : 1·6 \leq x < 3, x \in R\}$

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Question 64 Marks

Solve the following inequalities and graph their solution set
A = {x : 11x -5 ≥ 7x + 3, x ∈ R} and
B = {x : 18x - 9 ≥ 15 + 12x, x ∈ R}

Answer

A = {x : 11x -5 ≥ 7x + 3, x ∈ R}
= {x : 11x - 7x ≥ 3 + 5, x ∈ R}
= {x : 4x ≥ 8, x ∈ R}
= {x : x ≥ 2, x ∈ R} ...(i)
Also B = {x : 18x - 9 ≥ 15 + 12x, x ∈ R}
= {x : 18x - 12x ≥ 15 + 9, x ∈ R}
= (x : 6x ≥ 24, x ∈ R}
= {x : x ≥ 4, x ∈ R} ...(ii)
∴ on number line,

∴ A ∩ B : {x : x ≥ 4, x ∈ R}
i.e. A ∩ B :

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Mathematics [4 marks sum]

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