The function
f(x) = | x² - 2x - 3 | . e^(| 9x² - 12x + 4 |) is not differentiable at exactly :

Solution

Maths sir · 4 steps

Factor both quadratics first: one is a perfect square, so its modulus does nothing at all.

1. Factor the exponent. 9x² - 12x + 4 = (3x-2)², which is never negative. So |9x²-12x+4| = (3x-2)² everywhere, and e^((3x-2)²) is a smooth function with no corners.
2. Factor the outside. x² - 2x - 3 = (x-3)(x+1), which changes sign at x = 3 and x = -1. A modulus of a function that crosses zero like this makes a sharp corner at each crossing.

   f(x) = |(x-3)(x+1)| · e^((3x-2)²)

3. Check the corners survive. Near a simple root a of g(x) = x²-2x-3, the left and right derivatives of f are ∓ g'(a) e^((3a-2)²). Here g'(3) = 4 and g'(-1) = -4 are non-zero, and the exponential is never zero, so the two one-sided derivatives differ at both points.
4. So. f is not differentiable only at x = -1 and x = 3, which is exactly two points.

The idea to keep. A modulus creates a corner only where the inside changes sign at a simple root. If the inside is a perfect square, it never goes negative, so the modulus can be dropped.

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