Let g(x)=f(x)+f(1-x) and f^(′ ′)(x) > 0, x ∈(0,1). If g is decreasing in the interval (0, a) and increasing in the interval (α, 1), then tan ⁻¹(2 α)+tan ⁻¹(1/α)+tan ⁻¹((α+1)/α) is equal to :

Solution

Maths sir · 4 steps

Since f' is increasing, g' changes sign exactly where x = 1-x, so α = 1/2.

1. Differentiate g. g'(x) = f'(x) - f'(1-x). Because f''(x) > 0 on (0,1), f' is an increasing function there.
2. Sign of g′. For x < 1/2 we have x < 1-x, so f'(x) < f'(1-x) and g'(x) < 0. For x > 1/2 the inequality flips and g'(x) > 0. So g decreases on (0, 1/2) and increases on (1/2, 1), which gives α = 1/2.
3. Substitute alpha. With α = 1/2: 2α = 1, 1/α = 2 and (α+1)/α = 3. The expression becomes tan⁻¹ 1 + tan⁻¹ 2 + tan⁻¹ 3. Here 2 × 3 = 6 > 1 with both numbers positive, so the sum formula needs an extra π.

   tan⁻¹ 2 + tan⁻¹ 3 = π + tan⁻¹((2+3)/(1-2 × 3)) = π + tan⁻¹(-1) = 3π/4

4. So. The total is π/4 + 3π/4 = π, which is option (b).

The idea to keep. For f(x)+f(1-x) with f''>0, symmetry puts the minimum at x=1/2. When adding tan⁻¹ a+tan⁻¹ b with a,b>0 and ab>1, remember to add π.

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