UPSC NDA Math Model Paper 9

Given: $f(x) = 10 - 6x - 2x^2$ is a function which is strictly increasing in the interval T and strictly decreasing in the interval S. Here, we have to find the interval S. Let's first calculate f'(x) ⇒ f'(x) = - 6 - 4x = - 2 ⋅ (3 + 2x) ⇒ f'(x) = - 4 ⋅ [x + (3/2)] As we know that, for a strictly decreasing function f'(x) < 0 for all x ∈ (a, b) ⇒ - 4 ⋅ [x + (3/2)] < 0 ⇒ [x + (3/2)] > 0 ⇒ x > - 3/2 So, the given function f(x) is strictly decreasing in the interval (- 3/2, ∞) Hence, the interval S = (- 3/2, ∞)