Let f ( x ) = ∏ k = 1 ∞ 3 1 + 2 c o s ( 2 x / 3 k ) . How many points in ( 0 , 3 π ) make ⌊ x f ( x )⌋ + ∣ x f ( x ) ∣ + ( x − 1 ) x 2 − 3 x + 2 nondifferentiable?

Question

Let f ( x ) = ∏ k = 1 ∞ ​ 3 1 + 2 c o s ( 2 x / 3 k ) ​ . How many points in ( 0 , 3 π ) make ⌊ x f ( x )⌋ + ∣ x f ( x ) ∣ + ( x − 1 ) ​ x 2 − 3 x + 2 ​ nondifferentiable?

TestHub · Accepted Answer

Using sin 3 u = sin u ( 1 + 2 cos 2 u ) , the product telescopes to f ( x ) = sin x / x . Thus x f ( x ) = sin x . Checking jumps of ⌊ sin x ⌋ , cusps of ∣ sin x ∣ , and zeros of the last absolute-value factor gives 5 distinct nondifferentiability points.

Mathematics - Continuity - Differentiability Question with Solution | TestHub

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