Let $$f(x) = \begin{cases} e^{x-1}, & x < 0 \\ x^2 - 5x + 6, & x \geq 0 \end{cases}$$ and $$g(x) = f(|x|) + |f(x)|$$. If the number of points where $$g$$ is not continuous and is not differentiable are $$\alpha$$ and $$\beta$$ respectively, then $$\alpha + \beta$$ is equal to :

$$f(x) = \begin{cases} e^x - 1, & x < 0 \\ x^2 - 5x + 6, & x \ge 0 \end{cases}$$

$$|f(x)| = \begin{cases} 1 - e^x, & x < 0 \\ x^2 - 5x + 6, & 0 \le x \le 2 \\ -(x^2 - 5x + 6), & 2 < x < 3 \\ x^2 - 5x + 6, & x \ge 3 \end{cases}$$

$$f(|x|) = |x|^2 - 5|x| + 6 = x^2 - 5|x| + 6$$

$$f(|x|) = \begin{cases} x^2 + 5x + 6, & x < 0 \\ x^2 - 5x + 6, & x \ge 0 \end{cases}$$

$$g(x) = f(|x|) + |f(x)|$$:

For $$x < 0$$: $$g(x) = (x^2 + 5x + 6) + (1 - e^x) = x^2 + 5x + 7 - e^x$$

For $$0 \le x \le 2$$: $$g(x) = (x^2 - 5x + 6) + (x^2 - 5x + 6) = 2x^2 - 10x + 12$$

For $$2 < x < 3$$: $$g(x) = (x^2 - 5x + 6) - (x^2 - 5x + 6) = 0$$

For $$x \ge 3$$: $$g(x) = (x^2 - 5x + 6) + (x^2 - 5x + 6) = 2x^2 - 10x + 12$$

$$\lim_{x \to 0^-} g(x) \neq \lim_{x \to 0^+} g(x)$$, $$g(x)$$ is discontinuous at $$x = 0$$.

Number of points of discontinuity, $$\alpha = 1$$ (at $$x = 0$$)

$$g'(x) = \begin{cases} 2x + 5 - e^x, & x < 0 \\ 4x - 10, & 0 < x < 2 \\ 0, & 2 < x < 3 \\ 4x - 10, & x > 3 \end{cases}$$

At $$x = 2$$: $$LHD = 4(2) - 10 = -2$$, $$RHD = 0$$

At $$x = 3$$: $$LHD = 0$$, $$RHD = 4(3) - 10 = 2$$

Number of points of non-differentiability, $$\beta = 3$$ (at $$x = 0, 2, 3$$)

$$\alpha + \beta = 1 + 3 = 4$$