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Let f(x) = cases |x^2 - 2x|, & x 5 cases Let a be the number of points where f is not continuous, and b be the number of points where f is not differentiable. Then the value of a + b is

Correct answer

4

Step-by-step solution

We analyze the function f(x) in its respective intervals and at the junction points. For x f(x) = |x(x - 2)| . The roots of x^2 - 2x = 0 are x = 0 and x = 2 . Since the domain is x At x = 0 , f(x) is continuous but not differentiable (LHD = -2 , RHD = 2 ). For 2 x 5 : f(x) = (3x - 6, 6 - x) . The two lines intersect when 3x - 6 = 6 - x 4x = 12 x = 3 . For 2 x 3 , 6 - x 3x - 6 , so f(x) = 6 - x . For 3 6 - x , so f(x) = 3x - 6 . At x = 3 , f(x) is continuous but not differentiable (LHD = -1 , RHD = 3 ). Now we check

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