Let S = t ∈ R : f x = x - π ⋅ e x - 1 sin x is not differentiable at t . Then, the set S is equal to:

Step-by-step solution

f x = x - π . e x - 1 . sin x In the neighbourhood of x = π , f x = x - π . e x - 1 . sin x R f ′ π = lim h → 0 ⁡ f π + h - f π h = lim h → 0 h sin ⁡ π + h e π + h - 1 h = 0 L f ′ π = lim h → 0 f π - f π - h h = - lim h → 0 - h sin ⁡ π - h e π - h - 1 h = 0 Differentiable at x = π R f ′ 0 = lim h → 0 f h - f 0 h = lim h → 0 h - π × sin h e h - 1 h = 0 L f ′ 0