Let g(x)=3 f(x/3)+f(3-x) and f^(′ ′)(x)>0 for all x ∈(0,3). If g is decreasing in (0, α) and increasing in (α, 3), then 8 α is :

Solution

Maths sir · 4 steps

Since f''>0, f' is increasing, so the sign of g' depends only on which input is bigger.

1. Differentiate g. Use the chain rule on both terms. The 3 outside cancels the 1/3 from x/3, and f(3-x) brings out a factor of -1.

   g'(x) = f'(x/3) - f'(3-x)

2. Use f'' > 0. f'' > 0 means f' is strictly increasing. So g'(x) < 0 exactly when x/3 < 3 - x, and g'(x) > 0 exactly when x/3 > 3 - x.
3. Find the switch. x/3 < 3 - x gives x < 9 - 3x, so 4x < 9, that is x < 9/4. So g decreases on (0, 9/4) and increases on (9/4, 3).
4. So. α = 9/4, and 8α = 18, which is option (c).

The idea to keep. When f''>0, f' is increasing, so comparing f'(u) with f'(v) is the same as comparing u with v. Turn the sign of g' into a simple inequality on the inputs.

Answer from the NTA final answer key. Solution by PrepWiser.