JEE Main 2026 — Functions Question with Solution

For the domain of $f(x)={\cos }^{-1}\left(\frac{4x+2[x]}{3}\right)$, the argument must satisfy:

$-1\le \frac{4x+2[x]}{3}\le 1$

$-3\le 4x+2[x]\le 3$

Let $x=I+f$, where $I=[x]\in \mathbb{Z}$ and $f=\{x\}\in [0,1)$. Substituting $x=I+f$ into the inequality:

$-3\le 4(I+f)+2I\le 3$

$-3\le 6I+4f\le 3$

Since $0\le f<1$, we have $0\le 4f<4$. We can check the possible integer values of $I$:
If $I\ge 1$, $6I+4f\ge 6(1)+0=6>3$, which gives no solution.
If $I\le -2$, $6I+4f<6(-2)+4=-8<-3$, which gives no solution.

Thus, the only possible values for $I$ are $0$ and $-1$.

Case 1: $I=0$
$-3\le 6(0)+4f\le 3\Rightarrow -3\le 4f\le 3\Rightarrow -\frac{3}{4}\le f\le \frac{3}{4}$
Since $f\in [0,1)$, we get $0\le f\le \frac{3}{4}$.
Therefore, $x=I+f\in \left[0,\frac{3}{4}\right]$.

Case 2: $I=-1$
$-3\le 6(-1)+4f\le 3\Rightarrow 3\le 4f\le 9\Rightarrow \frac{3}{4}\le f\le \frac{9}{4}$
Since $f\in [0,1)$, we get $\frac{3}{4}\le f<1$.
Therefore, $x=I+f\in \left[-1+\frac{3}{4},-1+1\right)=\left[-\frac{1}{4},0\right)$.

Taking the union of the intervals from both cases, the domain is $x\in \left[-\frac{1}{4},\frac{3}{4}\right]$.

Comparing this with $[\alpha ,\beta ]$, we get $\alpha =-\frac{1}{4}$ and $\beta =\frac{3}{4}$.

Finally, evaluating the required expression:

$12(\alpha +\beta )=12\left(-\frac{1}{4}+\frac{3}{4}\right)=12\left(\frac{2}{4}\right)=12\left(\frac{1}{2}\right)=6$

Answer: $6$