If the local maximum value of the function f(x)=(√(3 e)/(2 sin x))^(sin ² x), x ∈(0, π/2) , is k/e, then (k/e)⁸+k⁸/e⁵+k⁸ is equal to

Solution

Maths sir · 5 steps

Put t = sin² x, take logs, and the power-of-a-function becomes an easy one-variable maximum.

1. Take logs. Let t = sin² x, so t ∈ (0, 1) and 2 sin x = 2√t. Taking ln brings the power down.

   g(t) = ln f = t/2(ln 3e/4 - ln t)

2. Differentiate. g'(t) = 1/2(ln 3e/4 - ln t) - 1/2. Setting it to zero gives ln t = ln 3e/4 - 1 = ln 3/4, so t = 3/4. Since g''(t) = -1/2t < 0, this is a maximum.
3. Maximum value. t = 3/4 means sin x = √3/2, which is x = π/3. The base is √3e/√3 = √e, so f = (√e)^(3/4) = e^(3/8).
4. Find k. k/e = e^(3/8) gives k = e^(11/8), so k⁸ = e¹¹. Then (k/e)⁸ = e³ and k⁸/e⁵ = e⁶.
5. So. The sum is e³ + e⁶ + e¹¹, which is option (c).

The idea to keep. When a function sits in both the base and the exponent, take ln first. Substitute for the repeated trig expression so the maximum becomes a one-variable calculus problem.

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