The number of points, where the function f: R → R,
f(x)=|x-1| cos |x-2| sin |x-1|+(x-3)|x²-5 x+4|, is NOT differentiable, is :

Solution

Maths sir · 4 steps

Tidy away the harmless modulus signs first, then check whether the cofactor of each surviving kink is zero.

1. Cosine ignores sign. Cosine is an even function, so cos|x-2| = cos(x-2). This is smooth everywhere, so x = 2 gives no trouble.

   |t| sin|t| = t sin t for all t

2. First term is smooth. Put t = x-1. For t < 0, |t| = -t and sin|t| = -sin t, so the two minus signs cancel. The first term is (x-1)sin(x-1)cos(x-2), which is differentiable for every x.
3. Second term. Factor x² - 5x + 4 = (x-1)(x-4), so the term is (x-3)|x-1||x-4|. A product |x-a| · g(x), with g differentiable, fails to be differentiable at a only when g(a) ≠ 0.

   Point	Cofactor of the modulus	Value there	Differentiable?
   x = 1	(x-3)|x-4|	(-2)(3) = -6 ≠ 0	No
   x = 2	from cos|x-2|	cos|x-2| = cos(x-2)	Yes
   x = 3	no modulus here	(x-3) is a polynomial	Yes
   x = 4	(x-3)|x-1|	(1)(3) = 3 ≠ 0	No

4. So. The function is not differentiable only at x = 1 and x = 4, which is 2 points.

The idea to keep. A modulus causes a kink only if it survives simplification. |x-a| · g(x) is non-differentiable at a exactly when g(a) ≠ 0, and cos|u| or |u|sin|u| are always smooth.

Answer from the NTA final answer key. Solution by PrepWiser.