Solve the following inequality : ( x + 2 ) x 2 - 2 x + 1 - 4 + 3 x - x 2 ≥ 0

( x + 2 ) ( x 2 - 2 x + 1 ) - 4 + 3 x - x 2 ≥ 0 ⇒ ( x + 2 ) ( x - 1 ) 2 - 4 + 3 x - x 2 ≥ 0 ∵ a - b 2 = a 2 + b 2 - 2 a b ∵ Quadratic equation - x 2 + 3 x - 4 = 0 does not have real roots because its discriminant D = 9 - 4 - 1 - 4 = - 7 < 0 As we know, if a < 0 , D < 0 of the quadratic expression a x 2 + b x + c = 0 , then quadratic expression gives a negative value for all values of x . So, - x 2 + 3 x - 4 < 0 So, in given inequality, denominator is always negative. So, ( x + 2 ) ( x - 1 ) 2 ( - ve ) ≥ 0 ⇒ ( x + 2 ) ( x - 1 ) 2 ≤ 0 {sign of inequality will be reversed if we multiply by negative number} From wavy curve method Taking negative region So, x ∈ ( - ∞ , - 2 ]

Solve the following inequality : ( x + 2 ) x 2 - 2 x + 1 - 4 + 3 x - x 2 ≥ 0

( x + 2 ) ( x 2 - 2 x + 1 ) - 4 + 3 x - x 2 ≥ 0 ⇒ ( x + 2 ) ( x - 1 ) 2 - 4 + 3 x - x 2 ≥ 0 ∵ a - b 2 = a 2 + b 2 - 2 a b ∵ Quadratic equation - x 2 + 3 x - 4 = 0 does not have real roots because its discriminant D = 9 - 4 - 1 - 4 = - 7 < 0 As we know, if a < 0 , D < 0 of the quadratic expression a x 2 + b x + c = 0 , then quadratic expression gives a negative value for all values of x . So, - x 2 + 3 x - 4 < 0 So, in given inequality, denominator is always negative. So, ( x + 2 ) ( x - 1 ) 2 ( - ve ) ≥ 0 ⇒ ( x + 2 ) ( x - 1 ) 2 ≤ 0 {sign of inequality will be reversed if we multiply by negative number} From wavy curve method Taking negative region So, x ∈ ( - ∞ , - 2 ]

Solve the following inequality: 3 x x 2 - 4 ≤ 1

Given, 3 x x 2 - 4 ≤ 1 As we know from the property of modulus function, if f x ≤ a ⇒ - a ≤ f x ≤ a So, - 1 ≤ 3 x x 2 - 4 ≤ 1 Case i : 3 x x 2 - 4 ≤ 1 ⇒ 3 x x 2 - 4 - 1 ≤ 0 ⇒ - x 2 + 3 x + 4 x 2 - 4 ≤ 0 ⇒ x 2 - 3 x - 4 x 2 - 4 ≥ 0 { ∵ multiplying by - 1 } ⇒ ( x + 1 ) ( x - 4 ) ( x + 4 ) ( x - 4 ) ≥ 0 {using factorization and a 2 - b 2 = a - b a + b } ⇒ ( x + 1 ) ( x + 4 ) ≥ 0 and x ≠ 4 From wavy curve method Taking positive region, we get, x ∈ ( - ∞ , - 4 ) ∪ [ - 1 , ∞ ) - 4 Case ii : 3 x x 2 - 4 ≥ - 1 ⇒ 3 x x 2 - 4 + 1 ≥ 0 ⇒ x 2 + 3 x - 4 x 2 - 4 ≥ 0 ⇒ ( x - 1 ) ( x + 4 ) ( x - 4 ) ( x + 4 ) ≥ 0 {using factorization and a 2 - b 2 = a - b a + b } ⇒ ( x - 1 ) ( x - 4 ) ≥ 0 and x ≠ - 4 From wavy curve method, Taking positive region, we get x ∈ ( - ∞ , 1 ] ∪ ( 4 , ∞ ) - - 4 . Taking the intersection of values of x from case i and case ii , we get x ∈ ( - ∞ , - 4 ) ∪ [ - 1 , 1 ] ∪ ( 4 , ∞ )

Solve the following inequality: | x + 3 | + x x + 2 > 1

| x + 3 | + x x + 2 > 1 In inequalities involving modulus function, we have to make case as per modulus function Case i x < - 3 ⇒ x ∈ - ∞ , - 3 . . . i ⇒ - x + 3 + x x + 2 - 1 > 0 ∵ x = - x ; x < 0 ⇒ - x - 3 + x x + 2 - 1 > 0 ⇒ - 3 - x - 2 x + 2 > 0 ⇒ - x - 5 x + 2 > 0 ⇒ x + 5 x + 2 < 0 From the wavy curve method Taking negative region So, x ∈ ( - 5 , - 2 ) . . . ii Taking the intersection of equation i and equation ii , we get x ∈ ( - 5 , - 3 ) Case ii x ≥ - 3 ⇒ x ∈ [ - 3 , ∞ ) . . . iii ⇒ x + 3 + x x + 2 - 1 > 0 ⇒ 2 x + 3 - x - 2 x + 2 > 0 ⇒ x + 1 x + 2 > 0 From the wavy curve method Taking positive region So, x ∈ ( - ∞ , - 2 ) ∪ ( - 1 , ∞ ) . . . iv Taking the intersection of the equation iii and equation iv , we get x ∈ [ - 3 , - 2 ) ∪ - 1 , ∞ Now, taking the union of values of x from case i and case ii x ∈ ( - 5 , - 2 ) ∪ ( - 1 , ∞ )

Solve the following inequality: x 2 + 3 x + x 2 - 2 ≥ 0

| x 2 + 3 x | + x 2 - 2 ≥ 0 ⇒ | x ( x + 3 ) | + x 2 - 2 ≥ 0 Case-I: If x < - 3 , then x x + 3 > 0 By the property of modulus function, x x + 3 can be written as x x + 3 x 2 + 3 x + x 2 - 2 ≥ 0 ⇒ 2 x 2 + 3 x - 2 ≥ 0 By using the factorisation method, 2 x 2 + 4 x - x - 2 ≥ 0 ⇒ 2 x ( x + 2 ) - 1 ( x + 2 ) ≥ 0 ⇒ ( 2 x - 1 ) ( x + 2 ) ≥ 0 Taking positive region, x ∈ - ∞ , - 2 ∪ 1 2 , ∞ After taking intersection with the case i.e., x < - 3 , we get x ∈ ( - ∞ , - 3 ) Case-II: If - 3 ≤ x < 0 x x + 3 ≤ 0 By the property of modulus function, x x + 3 can be written as - x x + 3 ⇒ - x 2 - 3 x + x 2 - 2 ≥ 0 ⇒ - 3 x - 2 ≥ 0 If we multiply by the negative sign, the sign of inequality will be reversed. 3 x + 2 ≤ 0 ∴ x ∈ - ∞ , - 2 3 After taking intersection with the case i.e., - 3 ≤ x < 0 x ∈ - 3 , - 2 3 Case-III: If x ≥ 0 x x + 3 ≥ 0 By the property of modulus function, x x + 3 can be written as x x + 3 . x 2 + 3 x + x 2 - 2 ≥ 0 ⇒ 2 x 2 + 3 x - 2 ≥ 0 By using the factorization method, ( 2 x + 1 ) ( x + 2 ) ≥ 0 Taking positive region, x ∈ - ∞ , - 2 ∪ 1 2 , ∞ After taking intersection with the case of domain, we get x ∈ 1 2 , ∞ After taking the union of the Case I, Case II and Case III, we get x ∈ - ∞ , - 2 3 ∪ 1 2 , ∞

Solve the inequality : x 2 + x + 1 ≥ x

Given, x 2 + x + 1 ≥ x Case I : When x < 0 , then x 2 + x + 1 ≥ x is always true as L.H.S. is always positive. Case II: When x ≥ 0 x 2 + x + 1 ≥ x On both the sides, the positive expression is present, so, we can square both the sides. x 2 + x + 1 ≥ x 2 x + 1 ≥ 0 As x ≥ 0 Then, x + 1 ≥ 0 is true for x ≥ 0 . From Case I and Case II, x ∈ - ∞ , ∞ .