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JEE Main20264 April 2026Evening ShiftMathematicsContinuity and DifferentiabilityActual

Let f(x)= cases e^ x-1 , & x<0 x^2-5x+6, & x 0 cases and g(x)=f(|x|)+|f(x)| . If the number of points where g is not continuous and is not differentiable are and respectively, then + is equal to ______

Correct answer

0

Step-by-step solution

We are given the function: f(x) = cases e^ x-1 , & x We need to analyze the continuity and differentiability of g(x) = f(|x|) + |f(x)| . For x 0 . Thus, f(|x|) = (-x)^2 - 5(-x) + 6 = x^2 + 5x + 6 . Also, for x 0 , so |f(x)| = e^ x-1 . Therefore, for x For x 0 , |x| = x . Thus, f(|x|) = f(x) = x^2 - 5x + 6 . Therefore, for x 0 , g(x) = x^2 - 5x + 6 + |x^2 - 5x + 6| . Let us check the continuity of g(x) at x = 0 : _ x 0^- g(x) = _ x 0^- (x^2 + 5x + 6 + e^ x-1 ) = 6 + 1 e _ x 0^+ g(x) = _ x 0^+ (x^2 - 5x + 6 + |x^2 -

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