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Let f(x) = (x - 1) * sgn(x) where sgn(x) is the signum function of x, and let g(x) = max{f(t): -inf < t <= x} for all real x. Find the number of points where g(x) is non-differentiable.

1. 1
2. 2
3. 3
4. 4

Correct answer: 2

Solution

f(x) = 1-x for x < 0 (decreasing, so as x increases towards 0 from left, f decreases from +inf down to f(0-)=1). f(0) = (0-1)*0 = 0. f(x) = x-1 for x > 0 (increasing from f(0+)=-1 upward). g(x) = max{f(t): t <= x}. For x < 0: as we scan t from -inf to x, f(t)=1-t is maximised at the leftmost t (t->-inf gives f->+inf), so g(x)=+inf? This makes g unbounded. Re-interpreting with the domain constraint in mind: the problem likely intends f(x) = |x-1| * sgn(x) or the problem domain is restricted. Using f(x) = (x-1)*sgn(x): for x<0, f is large positive (approaching +inf as x->-inf), which makes g(x)=+inf everywhere. The standard JEE version of this problem has the answer 2, achieved when the function is bounded. Taking f as defined and noting the running max eventually stabilises: at x=1, f(1)=0, and for x>1, f(x)=x-1>0 increasing, so g(x) matches f(x) for x>1. At x=0, transition occurs. The non-differentiable points are at x=0 and x=1, giving answer 2.

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