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 ∈ - ∞ , ∞ .

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 ∈ - ∞ , ∞ .

The complete solution set of the inequality x 4 - 3 x 3 + 2 x 2 x 2 - x - 30 ≥ 0 is

( - ∞ , - 5 ) ∪ [ 1 , 2 ] ∪ ( 6 , ∞ ) ∪ { 0 }

( - ∞ , - 5 ) ∪ ( 1 , 2 ) ∪ ( 6 , ∞ ) ∪ { 0 }

( - ∞ , - 5 ] ∪ [ 1 , 2 ] ∪ [ 6 , ∞ ) ∪ { 0 }

Complete set of value(s) of x satisfying the inequality x - 3 < x 2 + 4 x - 5 is(are)

( - ∞ , - 5 ] ∪ [ 1 , ∞ )

Find the range of the function f x = x 1 + x 2 .

Let, y = x 1 + x 2 ⇒ y 1 + x 2 = x ⇒ y x 2 - x + y = 0 , which is a quadratic in x . y ≠ 0 for this equation to be quadratic but y = 0 ⇒ x = 0 which is a real solution in x and hence y = 0 will be included in Range. Now, since x ∈ R , D ≥ 0 i.e., b 2 - 4 a c ≥ 0 , where a = y , b = - 1 , c = y ⇒ - 1 2 - 4 y y ≥ 0 ⇒ 1 - 4 y 2 ≥ 0 ⇒ 4 y 2 - 1 ≤ 0 ⇒ 2 y - 1 2 y + 1 ≤ 0 (solve by wavy-curve method) y ∈ - 1 2 , 1 2 Therefore, the range of the function f x is - 1 2 , 1 2 .