JEE Main 2026 — Mathematics Trigonometry

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

$$-1 \le \dfrac{x+[x]}{3} \le 1$$−1≤3x+[x]​≤1

$$-3 \le x + [x] \le 3$$−3≤x+[x]≤3

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

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

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

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

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

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

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

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

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

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

Therefore, $\alpha^2 + \beta^2 = (-1)^2 + (2)^2 = 1 + 4 = 5$α2+β2=(−1)2+(2)2=1+4=5.

Answer: $5$5