Let f:(0,π)→ℝ be a twice differentiable function such that limt→x (f(x)sint − f(t)sinx)/(t − x) = sin2x for all x ∈ (0,π). If f(π/6) = −π/12, then which of the following statement(s) is (are) TRUE?

limt→x (f(x)sint − f(t)sinx)/(t − x) = sin2x
By using L'Hospital's Rule
limt→x (f(x)cost − f'(t)sinx)/1 = sin2x ⇒ f(x)cosx − f'(x)sinx = sin2x
⇒ −(f'(x)sinx − f(x)cosx)/sin2x = 1 ⇒ −d(f(x)sinx) = 1 ⇒ f(x)sinx = x + c
Put x = π/6 and f(π/6) = −π/12 ∴ c = 0 ⇒ f(x) = −x/sinx
(A) f(π/4) = −π/(4√2)
(B) f(x) = −x/sinx
As sinx > x − x³/6, −x/sinx < −x/(x − x³/6) ∴ f(x) < −x/(x − x³/6) which is not −x²/4 + x/6 for all x ∈ (0,π)
(C) f'(x) = −sinx − xcosx
f'(x) = 0 ⇒ tanx = −x ⇒ there exist α ∈ (0,π) for which f'(α) = 0
(D) f''(x) = −cosx + xsinx − cosx
f''(π/2) = π/2, f(π/2) = −π/2 ∴ f''(π/2) + f(π/2) = 0.