UPSC NDA Math Model Paper 9

Given: f(x)=10−6x−2x2 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, ∞)