JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Continuity and Differentiability

Let $g(x)=x\sin x$. Then $g^{\prime }(x)=\sin x+x\cos x$.

For $x\in [0,\pi /2)$: $g^{\prime }(x)\ge 0$, so $g$ is increasing.
For $x\in (-\pi /2,0]$: $g^{\prime }(x)\le 0$, so $g$ is decreasing.

The function $[x\sin x]$ has jump discontinuities at points $x_1,-x_1$ where $x_1\sin x_1$ is an integer in $(0,\pi /2)$. There are exactly 2 such points.

The function $(|x|+|x-1|)\sin x$ is continuous everywhere but not differentiable at $x=0$ and $x=1$. However, at $x=0$, both terms are differentiable, so $f$ is differentiable there. At $x=1$, the first term is not differentiable but the second is, so $f$ is not differentiable.

At $x_1,-x_1$: $[x\sin x]$ is discontinuous, so $f$ is discontinuous. Also not differentiable.

So $\alpha =2$ (discontinuities at $x_1,-x_1$)
$\beta =3$ (non-differentiable at $x_1,-x_1,1$)

$\alpha +\beta =5$.