Solve the following for x, where |x| is modulus function, [x] is the g

1.	Solve the following for x, where \|x\| is modulus function, [x] is the greatest integer function, {x} is a fractional part function. \|x2 – 9\| + \|x2 – 4\| = 5
Answer» \|x² – 9\| + \|x² – 4\| = 5 ∴ \|(x – 3) (x + 3)\| + \|(x – 2) ( x + 2)\| = 5 ………(i) Case I: x < -3 Also, x < -2, x < 2, x < 3 ∴ (x – 3) (x + 3) > 0 and (x – 2) (x + 2) > 0 Equation (i) reduces to x² – 9 + x² – 4 = 5 ∴ 2x² = 18 ∴ x = -3 or 3 (both rejected as x < -3) Case II: -3 ≤ x < -2 As x < -2, x < 3 ∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0 Equation (i) reduces to -(x² – 9) + x² – 4 = 5 ∴ 5 = 5 (true) -3 ≤ x < -2 is a solution ….(ii) Case III: -2 ≤ x < 2 As x > -3, x < 3 ∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) < 0 Equation (i) reduces to 9 – x² + 4 – x2 = 5 ∴ 2x² = 13 – 5 ∴ x² = 4 ∴ x = -2 is a solution …..(iii) Case IV: 2 ≤ x < 3 As x > -3, x > -2 ∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0 Equation (i) reduces to 9 – x² + x² – 4 = 5 ∴ 5 = 5 (true) ∴ 2 ≤ x < 3 is a solution ……(iv) Case V: 3 ≤ x As x > -3, x > -2, x > 2 ∴ (x + 3) (x – 3) > 0, (x – 2) (x + 2) > 0 Equation (i) reduces to x² – 9 + x² – 4 = 5 ∴ 2x = 18 ∴ x = 9 ∴ x = 3 …..(v) (x = -3 rejected as x ≥ 3) From (ii), (iii), (iv), (v), we get ∴ Solution set = [-3, -2] ∪ [2, 3]

Solve the following for x, where |x| is modulus function, [x] is the greatest integer function, {x} is a fractional part function. |x2 – 9| + |x2 – 4| = 5

Answer»

|x² – 9| + |x² – 4| = 5

∴ |(x – 3) (x + 3)| + |(x – 2) ( x + 2)| = 5 ………(i)

Case I: x < -3

Also, x < -2, x < 2, x < 3

∴ (x – 3) (x + 3) > 0 and (x – 2) (x + 2) > 0

Equation (i) reduces to x² – 9 + x² – 4 = 5

∴ 2x² = 18

∴ x = -3 or 3 (both rejected as x < -3)

Case II: -3 ≤ x < -2

As x < -2, x < 3

∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0

Equation (i) reduces to -(x² – 9) + x² – 4 = 5

∴ 5 = 5 (true)

-3 ≤ x < -2 is a solution ….(ii)

Case III: -2 ≤ x < 2 As x > -3, x < 3

∴ (x – 3) (x + 3) < 0,

(x – 2) (x + 2) < 0

Equation (i) reduces to 9 – x² + 4 – x2 = 5

∴ 2x² = 13 – 5

∴ x² = 4

∴ x = -2 is a solution …..(iii)

Case IV: 2 ≤ x < 3 As x > -3, x > -2

∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0

Equation (i) reduces to 9 – x² + x² – 4 = 5

∴ 5 = 5 (true)

∴ 2 ≤ x < 3 is a solution ……(iv)

Case V: 3 ≤ x As x > -3, x > -2, x > 2

∴ (x + 3) (x – 3) > 0, (x – 2) (x + 2) > 0

Equation (i) reduces to x² – 9 + x² – 4 = 5

∴ 2x = 18

∴ x = 9

∴ x = 3 …..(v)

(x = -3 rejected as x ≥ 3) From (ii), (iii), (iv), (v), we get

∴ Solution set = [-3, -2] ∪ [2, 3]