Let f(x)= cases x^3+8, & x<0 x^2-4, & x 0 cases and g(x)= cases (x-8)^ 1/3 , & x<0 (x+4)^ 1/2 , & x 0 cases . Then the number of points, where the function g∘f is discontinuous, is _______.

JEE Main 2026 Mathematics question

Let and . Then the number of points, where the function g∘f is discontinuous, is .

Answer: 3

For x<−2, f<0 so g(f)=x; for −2≤x<0, f≥0 so g(f)=√(x³+12); for 0≤x<2, f<0 so g(f)=(x²−; for x≥2, g(f)=x. At x=−2 the

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Question ID VK-015584