If a = lim _(n → ∞) Σₖ₌₁ⁿ 2n/(n² + k²) and f(x) = √((1 - cos x)/(1 + cos x)), x ∈ (0,1), then :

Solution

Maths sir · 4 steps

Pull 1/n out of the sum, read it as ∫₀¹, then turn f into tan x/2.

1. Turn into integral. Divide the top and bottom by n²: 2n/(n² + k²) = 1/n · 2/(1 + (k/n)²). As n → ∞, k/n becomes x and 1/n becomes dx. Since k runs from 1 to n, x runs from 0 to 1.

   a = ∫₀¹ 2/(1 + x²) dx = 2[tan⁻¹ x]₀¹ = 2 × π/4 = π/2

2. Simplify f. Use 1 - cos x = 2 sin² x/2 and 1 + cos x = 2 cos² x/2. Then f(x) = tan x/2, which is positive for x ∈ (0, 1), and f'(x) = 1/2 sec² x/2.
3. Evaluate at a/2. Here a/2 = π/4, so f(π/4) = tan π/8 = √2 - 1. Then f'(π/4) = 1/2(1 + tan² π/8) = 1/2(4 - 2√2) = 2 - √2.
4. So. f'(a/2) = 2 - √2 = √2(√2 - 1) = √2 f(a/2). This is option (c).

The idea to keep. For a limit of a sum, take out 1/n and write the rest in k/n. That gives ∫₀¹ g(x) dx. Also, √((1-cos x)/(1+cos x)) is |tan x/2|.

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