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Home » But what about \( m = 2 \): no - ECD Germany

But what about \( m = 2 \): no - ECD Germany

📅 May 4, 2026 👤 scraface

May 04, 2026

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Now, since \(0 < x < \frac{\pi}{2}\), \(u = \sin x + \cos x = \sqrt{2}...
Let \(v = u^2\), so \(v \in (1, 2]\). Then:
f(x) = \frac{4v}{(v - 1)^2}
Let \(g(v) = \frac{4v}{(v - 1)^2}\). Find maximum on \(v \in (1, 2]\).
Take derivative:
g'(v) = \frac{4(v - 1)^2 - 4v \cdot 2(v - 1)}{(v - 1)^4} = \frac{4(v -...
For \(v > 1\), \(g'(v) < 0\), so \(g(v)\) is decreasing on \((1, 2]\)....
Wait — since derivative is negative, function decreases from \(v = 1^+...
But observe: as \(v \to 1^+\), g(v) → ∞? Let’s test:
At \(v = 1.1\), \(g = 4(1.1)/(0.1)^2 = 4.4 / 0.01 = 440\)

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