Question 10

Let $g(x) = (|x| + |x-1|)\sin x$ and $h(x) = [x \sin x]$.

1. For $g(x)$:

• At $x=0$: $g(x)$ is continuous. LHD at $x=0$ is $1$, RHD at $x=0$ is $1$. So $g(x)$ is differentiable at $x=0$.
• At $x=1$: $g(x)$ is continuous but not differentiable (LHD = $\cos 1$, RHD = $2\sin 1 + \cos 1$).

2. For $h(x) = [x \sin x]$:

• On $(-\pi/2, \pi/2)$, $x \sin x$ is an even function ranging from $0$ to $\pi/2 \approx 1.57$.
• $[x \sin x]$ is discontinuous when $x \sin x = 1$. This occurs at $x = x_0$ and $x = -x_0$, where $x_0 \in (1, \pi/2)$.
• At $x=0$, $x \sin x = 0$ and $\lim_{x \to 0} [x \sin x] = 0$, so it is continuous at $x=0$.

3. Summary:

• Points of non-continuity ($\alpha$): $x_0, -x_0$. So $\alpha = 2$.
• Points of non-differentiability ($\beta$): $x_0, -x_0$ (due to discontinuity) and $x=1$ (due to $|x-1|$ term in $g(x)$). So $\beta = 3$. Total $\alpha + \beta = 2 + 3 = 5$.