Find the range of values of k , such that f x = k x 2 + 2 k + 1 x + 9 k + 4 x 2 - 8 x + 17 is always negative.

Given expression, f x = k x 2 + 2 k + 1 x + 9 k + 4 x 2 - 8 x + 17 . . . i For f x to be always negative first we check denominator is always positive or negative, x 2 - 8 x + 17 For the quadratic equation a x 2 + b x + c = 0 , D = b 2 - 4 a c a = coefficient of x 2 b = coefficient of x c = constant term Here a > 0 , D = b 2 - 4 a c = - 4 < 0 D = discriminant So, the denominator is always positive. Now, numerator should be negative for the whole expression to be negative. So, condition for that a < 0 , D < 0 k x 2 + 2 k + 1 x + 9 k + 4 Here, a = k , b = 2 k + 1 , c = 9 k + 4 k < 0 . . . ii D = 2 k + 1 2 - 4 · k · 9 k + 4 < 0 ⇒ - 8 k 2 - 2 k + 1 < 0 = 4 k - 1 2 k + 1 > 0 k < - 1 2 or k > 1 4 , Now from ii , k < 0 , Range for k ∈ - ∞ , - 1 2 .

Find the condition on a for roots of the equation x 2 + ( a - b ) x + ( 1 - a - b ) = 0 , a , b ∈ R are imaginary ∀ b ∈ R .

x 2 + ( a - b ) x + ( 1 - a - b ) = 0 , a , b ∈ R As we know for a quadratic equation p x 2 + q x + r = 0 , the discriminant D = q 2 - 4 p r . After comparing we can write ⇒ D = a - b 2 - 4 1 1 - a - b ⇒ D = a 2 + b 2 - 2 a b + 4 a + 4 b - 4 ⇒ D = b 2 + 4 - 2 a b + a 2 + 4 a - 4 Also we know if roots are imaginary ⇒ D < 0 ⇒ b 2 + 4 - 2 a b + a 2 + 4 a - 4 < 0 this is true ∀ b ∈ R . ⇒ Coefficient of b 2 < 0 & D < 0 Coefficient of b 2 = 1 which is greater than zero means it is not true for any value of a . a ∈ ϕ .