Let S be the set of positive integral values of a for which a x 2 + 2 a + 1 x + 9 a + 4 x 2 - 8 x + 32 < 0 , ∀ x ∈ ℝ . Then, the number of elements in S is:

Step-by-step solution

Given: a x 2 + 2 ( a + 1 ) x + 9 a + 4 x 2 - 8 x + 32 < 0 ∀ x ∈ R For quadratic x 2 - 8 x + 32 = 0 , D 1 = - 8 2 - 4 32 = - 64 Since the discriminant is less than zero and the leading coefficient is positive, this quadratic will always be positive. Now, solving a x 2 + 2 ( a + 1 ) x + 9 a + 4 < 0 We know that, for a quadratic to be always negative, the coefficient of x 2 < 0 , D < 0 . ⇒ a < 0 But we want positive values. So, no positive integral value exist.