1) ∣((3+2x)/(3x))∣ ≤1 2) 1≤ ∣ ((x−3)/(1−2x))∣≤ 2 3) ((x^2 +2x−35)/(x+2))


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Question Number 201595 by mokys last updated on 09/Dec/23

$1) ∣ \frac{3 + 2 x}{3 x} ∣ \leq 1$ $2) 1 \leq ∣ \frac{x - 3}{1 - 2 x} ∣\leq 2$ $3) \frac{x^{2} + 2 x - 35}{x + 2} > 0$ $4) - 1 \leq \frac{x + 1}{x - 2} \leq 2$
	Answered by deleteduser1 last updated on 09/Dec/23

	$4) x > 2 \Rightarrow - x + 2 ⩽ x + 1 ⩽ 2 x - 4 \Rightarrow x ⩾ 5$ $x < 2 \Rightarrow - x + 2 ⩾ x + 1 ⩾ 2 x - 4$ $\Rightarrow x ⩽ \frac{1}{2} \land x ⩽ 5 \Rightarrow x ⩽ \frac{1}{2} \Rightarrow x \in (- \infty, \frac{1}{2}) \cup [5, + \infty)$
	Answered by deleteduser1 last updated on 09/Dec/23

	$3) x < - 2 \Rightarrow x^{2} + 2 x - 35 < 0 \Rightarrow (x + 7) (x - 5) < 0$ $\Rightarrow - 7 < x < 5 \Rightarrow x ([- 7, 2)$ $x > - 2 \Rightarrow x > 5 \Rightarrow x \in (- 7, 2) \cup (5, + \infty)$
	Answered by shunmisaki007 last updated on 09/Dec/23

	$1 . ∣ \frac{3 + 2 x}{3 x} ∣\leq 1$ $\Rightarrow {(\frac{3 + 2 x}{3 x})}^{2} \leq 1$ $\Rightarrow 9 + 12 x + 4 x^{2} \leq 9 x^{2} ∣ x \neq 0$ $\Rightarrow 5 x^{2} - 12 x - 9 \geq 0$ $\Rightarrow (5 x + 3) (x - 3) \geq 0$ $\Rightarrow x \leq - \frac{3}{5} and x \geq 3$ $∴ x \in (- \infty, - \frac{3}{5}] \cup [3, \infty) ★$ $2 . 1 \leq∣ \frac{x - 3}{1 - 2 x} ∣\leq 2$ $\Rightarrow 1 \leq {(\frac{x - 3}{1 - 2 x})}^{2} \leq 4$ $\Rightarrow 1 - 4 x + 4 x^{2} \leq x^{2} - 6 x + 9 \leq 4 - 16 x + 16 x^{2} ∣ x \neq \frac{1}{2}$ $\Rightarrow 1 - 4 x + 4 x^{2} \leq x^{2} - 6 x + 9 and x^{2} - 6 x + 9 \leq 4 - 16 x + 16 x^{2} ∣ x \neq \frac{1}{2}$ $\Rightarrow 3 x^{2} + 2 x - 8 \leq 0 and 15 x^{2} - 10 x - 5 \geq 0$ $\Rightarrow (3 x - 4) (x + 2) \leq 0 and 5 (3 x + 1) (x - 1) \geq 0$ $\Rightarrow x \leq \frac{4}{3}, x \geq - 2, x \leq - \frac{1}{3} and x \geq 1$ $∴ x \in [- 2, - \frac{1}{3}] \cup [1, \frac{4}{3}] ★$ $3 . \frac{x^{2} + 2 x - 35}{x + 2} > 0$ $\Rightarrow (x^{2} + 2 x - 35) (x + 2) > 0 ∣ x \neq - 2$ $\Rightarrow (x - 5) (x + 7) (x + 2) > 0$ $\Rightarrow x > 5, x < - 2 and x > - 7$ $∴ x \in (- 7, - 2) \cup (5, \infty) ★$ $4 . - 1 \leq \frac{x + 1}{x - 2} \leq 2$ $\Rightarrow - 1 {(x - 2)}^{2} \leq (x + 1) (x - 2) \leq 2 {(x - 2)}^{2} ∣ x \neq 2$ $\Rightarrow - x^{2} + 4 x - 4 \leq x^{2} - x - 2 and x^{2} - x - 2 \leq 2 x^{2} - 8 x + 8$ $\Rightarrow 2 x^{2} - 5 x + 2 \geq 0 and x^{2} - 7 x + 10 \geq 0$ $\Rightarrow (2 x - 1) (x - 2) \geq 0 and (x - 2) (x - 5) \geq 0$ $\Rightarrow x \leq \frac{1}{2}, x \geq 2, x \leq 2 and x \geq 5$ $but x \neq 2$ $∴ x \in (- \infty, \frac{1}{2}] \cup [5, \infty) ★$
	Commented by deleteduser1 last updated on 09/Dec/23

	$3 i s w r o n g, t r y x < - 7 (e . g . - 8)$
	Commented by shunmisaki007 last updated on 09/Dec/23

	$I^{'} ve corrected it . Thanks .$
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