Let g:(0,∞)→R be a differentiable function such that ∫(ex+1x(cosx−sinx)+(ex+1)2g(x)(ex+1−xex))dx=ex+1xg(x)+C, for all x>0, where C is an arbitrary constant. Then
Held on 25 Jun 2022 · Verified 6 Jul 2026.
g is decreasing in (0,4π)
g−g′ is increasing in (0,2π)
g′ is increasing in (0,4π)
g+g′ is increasing in (0,2π)
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