NDA 2024 · PYQ · Functional Equations / Integration · hard
Let 3f(x) + f(1/x) = 1/x + 1. What is 8∫_1^2 f(x) dx equal to?
- A.ln(8√e)✓ Correct
- B.ln(4√e)
- C.ln 2
- D.ln 2 - 1
From f(x) = 3/(8x) - x/8 + 1/4, we get 8f(x) = 3/x - x + 2. So 8∫_1^2 f(x) dx = ∫_1^2 (3/x - x + 2) dx = [3 ln x - x²/2 + 2x]_1^2 = (3 ln 2 - 2 + 4) - (0 - 1/2 + 2) = (3 ln 2 + 2) - 3/2 = 3 ln 2 + 1/2 = ln 8 + ln √e = ln(8√e).
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