JEE Main 2026 — Inverse Trigonometric Functions Question with Solution

For the function $f(x)={\sin }^{-1}\left(\frac{x+[x]}{3}\right)$ to be defined, the argument of the inverse sine function must lie in the interval $[-1,1]$.

$-1\le \frac{x+[x]}{3}\le 1$

$-3\le x+[x]\le 3$

Using the fractional part function $\{x\}$, we can write $x=[x]+\{x\}$, where $0\le \{x\}<1$. Substituting this into the inequality gives:

$-3\le 2[x]+\{x\}\le 3$

We analyze the possible integer values for $[x]$:

If $[x]\le -2$, then $2[x]+\{x\}\le -4+\{x\}<-3$, which does not satisfy the inequality.

If $[x]=-1$, then $2[x]+\{x\}=-2+\{x\}$. Since $0\le \{x\}<1$, we have $-2\le -2+\{x\}<-1$, which satisfies the inequality. This gives $x\in [-1,0)$.

If $[x]=0$, then $2[x]+\{x\}=\{x\}$. Since $0\le \{x\}<1$, this satisfies the inequality. This gives $x\in [0,1)$.

If $[x]=1$, then $2[x]+\{x\}=2+\{x\}$. Since $0\le \{x\}<1$, we have $2\le 2+\{x\}<3$, which satisfies the inequality. This gives $x\in [1,2)$.

If $[x]\ge 2$, then $2[x]+\{x\}\ge 4+\{x\}\ge 4>3$, which does not satisfy the inequality.

Taking the union of the valid intervals, the domain of $f(x)$ is $[-1,0)\cup [0,1)\cup [1,2)=[-1,2)$.

Comparing this with the given domain $[\alpha ,\beta )$, we get $\alpha =-1$ and $\beta =2$.

Therefore, ${\alpha }^2+{\beta }^2=(-1{)}^2+(2{)}^2=1+4=5$.

Answer: $5$