Continuity and Differentiability of Function with Absolute Values

To determine the value of $\alpha + \beta$, we analyze the continuity and differentiability of the function $f(x) = (|x| + |x - 1|)\sin x + [x \sin x]$ in the domain $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

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Step 1: Simplify $g(x) = |x| + |x - 1|$

The critical points for the absolute value terms are $x = 0$ and $x = 1$. Both points lie within the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

• For $x \in \left(-\frac{\pi}{2}, 0\right]$: $|x| = -x, \quad |x - 1| = 1 - x \implies g(x) = 1 - 2x$

• For $x \in (0, 1]$: $|x| = x, \quad |x - 1| = 1 - x \implies g(x) = 1$

• For $x \in \left(1, \frac{\pi}{2}\right)$: $|x| = x, \quad |x - 1| = x - 1 \implies g(x) = 2x - 1$

Thus, the first term $g(x) \sin x$ is given by:

$$g(x) \sin x = \begin{cases} (1 - 2x)\sin x, & x \in \left(-\frac{\pi}{2}, 0\right] \\ \sin x, & x \in (0, 1] \\ (2x - 1)\sin x, & x \in \left(1, \frac{\pi}{2}\right) \end{cases}$$

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Step 2: Differentiability of $g(x) \sin x$

The function $g(x) \sin x$ is continuous for all $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. We test its differentiability at the critical points $x = 0$ and $x = 1$:

1. At $x = 0$: $\text{LHD} = \lim_{h \to 0^-} \frac{(1 - 2h)\sin h - 0}{h} = 1$ $\text{RHD} = \lim_{h \to 0^+} \frac{\sin h - 0}{h} = 1$ Since $\text{LHD} = \text{RHD}$, $g(x)\sin x$ is differentiable at $x = 0$.

2. At $x = 1$: $\text{LHD} = \left.\frac{d}{dx}(\sin x)\right|_{x=1^-} = \cos(1)$ $\text{RHD} = \left.\frac{d}{dx}((2x - 1)\sin x)\right|_{x=1^+} = 2\sin(1) + \cos(1)$ Since $2\sin(1) \neq 0$, $\text{LHD} \neq \text{RHD}$, so $g(x)\sin x$ is NOT differentiable at $x = 1$.

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Step 3: Analyze the term $[x \sin x]$

Let $k(x) = x \sin x$.

• $k(x)$ is an even function, $k(0) = 0$, and as $x \to \pm \frac{\pi}{2}$, $k(x) \to \frac{\pi}{2} \approx 1.5708$.
• $k(x)$ is strictly increasing on $\left[0, \frac{\pi}{2}\right)$.
• At $x = 1$, $k(1) = \sin(1) \approx 0.8415 < 1$.

By the Intermediate Value Theorem, there exists a unique point $x_0 \in \left(1, \frac{\pi}{2}\right)$ such that: $x_0 \sin x_0 = 1$

Due to symmetry, $k(-x_0) = 1$ as well. Therefore:

$$[x \sin x] = \begin{cases} 1, & x \in \left(-\frac{\pi}{2}, -x_0\right] \cup \left[x_0, \frac{\pi}{2}\right) \\ 0, & x \in (-x_0, x_0) \end{cases}$$

The step function $[x \sin x]$ has jump discontinuities at $x = -x_0$ and $x = x_0$.

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Step 4: Count Points of Non-Continuity ($\alpha$)

Since $g(x)\sin x$ is continuous throughout $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, the points of discontinuity of $f(x)$ are determined solely by $[x \sin x]$.

• Discontinuous points: $x = -x_0$ and $x = x_0$.

$\alpha = 2$

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Step 5: Count Points of Non-Differentiability ($\beta$)

1. A function cannot be differentiable where it is not continuous. Thus, $f(x)$ is not differentiable at $x = -x_0$ and $x = x_0$.
2. In the interval $(-x_0, x_0)$, $[x \sin x] = 0$, so $f(x) = g(x) \sin x$. As established in Step 2, $g(x) \sin x$ is not differentiable at $x = 1$.

Therefore, $f(x)$ is not differentiable at $x = -x_0, x_0, 1$.

$\beta = 3$

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Conclusion

$\alpha + \beta = 2 + 3 = 5$