Consider the function

$f:\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\to \left(-\infty ,\infty \right)$

defined by $f\left(x\right)=\left(\left|\mathrm{(x}\right|+\left|x-\mathrm{1)}\right|\right)\sin x+\left[\mathrm{[x}\sin \mathrm{x] }\right]$where $\left[x\sin x\right]$ denotes the greatest integer less than or equal to $\mathrm{x }\sin x$.Let $\alpha$ be the total number of points in the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ at which $f$ is NOT continuous, and let $\beta$ be the total number of points in the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ at which $f$ is NOT differentiable.Then the value of$\alpha +\beta$is ________.

The correct answer is 3.

Step 1: Breakdown of the Function

The given function is defined as:

$f\left(x\right)=\left(\left|x\right|+\left|x-1\right|\right)\sin x+\left[x\sin x\right]$

We can separate $f\left(x\right)$ into two parts:

$f\left(x\right)=g\left(x\right)+h\left(x\right)$

where:

$g\left(x\right)=\left(\left|x\right|+\left|x-1\right|\right)\sin x$

$h\left(x\right)=\left[x\sin x\right]$

Step 2: Analyzing $g\left(x\right)$

Let us rewrite $g\left(x\right)$ by simplifying the absolute value expressions in different sub-intervals:

1. For $x\in \left[0,1\right]$:

$\left|x\right|=x$ and $\left|x-1\right|=1-x$

$\left|x\right|+\left|x-1\right|=x+1-x=1$

$g\left(x\right)=\sin x$

2. For $x>1$:

$\left|x\right|=x$ and $\left|x-1\right|=x-1$

$\left|x\right|+\left|x-1\right|=2x-1$

$g\left(x\right)=(2x-1)\sin x$

Now, let's check continuity and differentiability of $g\left(x\right)$ at $x=1$:

- Continuity at $x=1$:

$\underset{x\to 1^{-}}{\lim }g\left(x\right)=\sin 1$

$\underset{x\to 1^{+}}{\lim }g\left(x\right)=\left(2\right(1)-1)\sin 1=\sin 1$

Thus, $g\left(x\right)$ is continuous at $x=1$.

- Differentiability at $x=1$:

Left hand derivative (LHD): $g^{\prime }\left(1^{-}\right)=\cos 1$

Right hand derivative (RHD): $g^{\prime }\left(1^{+}\right)=2\sin 1+\cos 1$

Since LHD ≠ RHD, $g\left(x\right)$ is NOT differentiable at $x=1$.

Step 3: Analyzing $h\left(x\right)=\left[x\sin x\right]$

Consider the function $u\left(x\right)=x\sin x$:

At $x=0$, $u\left(0\right)=0$.

At $x=\frac{\pi }{2}$, $u\left(\frac{\pi }{2}\right)=\frac{\pi }{2}\approx 1.57$.

Since $u\left(x\right)$ is continuous and strictly increasing on $\left[0,\frac{\pi }{2}\right]$, there exists a unique point $x_0\in \left(0,\frac{\pi }{2}\right)$ where:

$x_0\sin x_0=1$

For $x<x_0$, $0\le x\sin x<1\Rightarrow \left[x\sin x\right]=0$.

For $x\ge x_0$, $1\le x\sin x<2\Rightarrow \left[x\sin x\right]=1$.

Thus, $h\left(x\right)$ has a step jump at $x=x_0$, making it NOT continuous (and consequently NOT differentiable) at $x=x_0$.

Step 4: Finding $\alpha$ and $\beta$

1. Points of Non-continuity ($\alpha$):

$f\left(x\right)$ is not continuous only at the jump point of the greatest integer function:

$x=x_0$

Therefore, $\alpha =1$.

2. Points of Non-differentiability ($\beta$):

$f\left(x\right)$ is not differentiable at:

• $x=1$ (due to the corner point in $g\left(x\right)$)

• $x=x_0$ (due to the discontinuity in $h\left(x\right)$)

Therefore, $\beta =2$.

Step 5: Final Calculation

$\alpha +\beta =1+2=3$