Answer : |x| > 4 Square ⇒ x2 > 16 ⇒ x2 – 16 > 0 ⇒ x2 – 42 > 0 ⇒ (x + 4)(x – 4) > 0 Observe that when x is greater than 4, (x + 4)(x – 4) is positive And for each root the sign changes...

Answer : |x| < 4 Square ⇒ x2 < 16 ⇒ x2 – 16 < 0 ⇒ x2 – 42 < 0 ⇒ (x + 4)(x – 4) < 0 Observe that when x is greater than 4, (x + 4)(x – 4) is positive And for each root the sign changes...

Answer :Observe that       is zero at x = -5 and not defined at x = 5 Hence plotting these two points on number line Now for x > 5  is positive For every root and not defined value of the sign...

Answer :Observe that zero at x = 5 and not defined at x = 2 Hence plotting these two points on number line Now for x > 5,  is negative For every root and not defined value of the sign will change...

Answer : x + 5 > 4x – 10 ⇒ 5 + 10 > 4x – x ⇒ 15 > 3x Divide by 3 ⇒ 5 > x ⇒ x < 5 x < 5 means x is from -∞ to 5 that is x ∈ (-∞, 5) Hence solution of x + 5 > 4x – 10 is x ∈ (-∞,...

Answer : We have to find integer values of x for which -4x > 16 (why only integer values because it is given that x ∈ Z that is set of integers) -4x > 16 ⇒ -x > 4 ⇒ x < -4 The integers...

Answer : We have to find values of x for which both the equations hold true x – 2 ≥ 0 and 2x – 5 ≤ 3 We will solve both the equations separately and then their intersection set will be solution of...

Answer :                means we have to find values of x for which         is negative Observe that the numerator |x – 2| is always positive because of mod, hence for negative quantity the...

Answer : |x – 1| < 2 Square both sides ⇒ (x – 1)2 < 4 ⇒ x2 – 2x + 1 < 4 ⇒ x2 – 2x – 3 < 0 ⇒ x2 – 3x + x – 3 < 0 ⇒ x(x – 3) + 1(x – 3) < 0 ⇒ (x + 1)(x – 3) < 0 Observe that when...

Answer : We have to find values of x for which is less than zero that is negative Now for negative x – 2 should be negative that is x – 2 < 0 ⇒ x – 2 < 0 ⇒ x < 2 Hence x should be less than...