Let f : (0, 2) → R be defined as f(x) = log₂( 1 + tan ( (π x)/4 ) ). Then, lim _(n → ∞) 2/n( f( 1/n ) + f( 2/n ) + ... + f(1) ) is equal to ___________.

Solution

Maths sir · 4 steps

Turn the sum into 2∫₀¹ f(x) dx, then use x → 1-x to make the integral solve itself.

1. Sum to integral. The terms are f(r/n) for r = 1 to n, so r/n runs over (0, 1] in steps of 1/n. Write 2/n = 2 × 1/n. Then the limit is 2∫₀¹ f(x) dx.
2. Use the reflection. Let I = ∫₀¹ log₂(1 + tan(π x)/4) dx. Since ∫₀¹ g(x) dx = ∫₀¹ g(1-x) dx, replace x by 1 - x. With t = tan(π x)/4, the new tangent is tan(π/4 - (π x)/4) = (1-t)/(1+t).

   1 + (1-t)/(1+t) = 2/(1+t) ⇒ log₂ 2/(1+t) = 1 - log₂(1+t)

3. Solve for I. So I = ∫₀¹ (1 - log₂(1 + tan(π x)/4)) dx = 1 - I. This gives 2I = 1, so I = 1/2.
4. So. The limit is 2I = 2 × 1/2 = 1, so the answer is 1.

The common slip: 0.5. Dropping the factor 2 from 2/n leaves only ∫₀¹ f(x) dx, which gives 0.5, not 1.

The idea to keep. When you see 1 + tan θ with angles adding to π/4, use x → a - x: the product (1+tan A)(1+tan B) = 2 makes the log integral equal to its own complement.

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