Question

What is the Solution set of the inequality $\tt \frac{1}{ |x| - 3 } < \frac{1}{2}$ ?

• $\displaystyle \sf x \in ( - \infty , - 3) \cup \: (3 , \infty )$

• $\sf \: x \in( - \infty , - 5) \cup( - 3,3) \cup(5, \infty )$

• $\sf \: x \in( - \infty , - 5) \cup(3 , \infty )$

• $\sf \: x \in \: ( - \infty , - 3) \cup \: (5, \infty )$​

Answer 1

EXPLANATION.

$\sf \implies \dfrac{1}{|x| - 3} < \dfrac{1}{2}$

As we know that,

⇒ |x| = ± x.

$\sf \implies \dfrac{1}{|x| - 3} \ - \dfrac{1}{2} < 0$

Two cases exists, we get.

⇒ (a) = x < 0.

⇒ (b) = x ≥ 0.

$\sf \implies \dfrac{2 - [ |x| - 3]}{2[|x| - 3]} < 0.$

$\sf \implies \dfrac{2 - |x| + 3}{2[|x| - 3]} < 0.$

$\sf \implies \dfrac{5 - |x|}{2[|x| - 3]} < 0.$

⇒ (a) = x < 0.

$\sf \implies \dfrac{5 - |x|}{2[|x| - 3]} < 0.$

$\sf \implies \dfrac{5 + x}{2[-x - 3 ]} < 0.$

$\sf \implies \dfrac{5 + x}{2[x + 3]} > 0.$

Find zeroes of the equation, we get.

⇒ 5 + x = 0.

⇒ x = -5.

⇒ x + 3 = 0.

⇒ x = -3.

Put this point on wavy curve method, we get.

⇒ x ∈ (-∞,-5) ∪ (-3,∞). - - - - - (1).

⇒ (b) = x ≥ 0.

$\sf \implies \dfrac{5 - x }{2(x - 3)} < 0$

$\sf \implies \dfrac{x - 5}{2(x - 3)} > 0.$

Find zeroes of the equation, we get.

⇒ x - 5 = 0.

⇒ x = 5.

⇒ x - 3 = 0.

⇒ x = 3.

⇒ x = 0.

Put this point on wavy curve method, we get.

⇒ x ∈ (0,3) ∪ (5,∞). - - - - - (2).

Taking equation (1) ∪ (2), we get.

⇒ x ∈ (-∞,-5) ∪ (-3,∞) ∪ (0,3) ∪ (5,∞).

⇒ x ∈ (-∞,-5) ∪ (-3,3) ∪ (5,∞).

Option [B] is correct answer.