JEE Main 2026 — Mathematics Algebra

For the domain of $f(x) = \cos^{-1}\left(\dfrac{4x+2[x]}{3}\right)$f(x)=cos−1(34x+2[x]​), the argument must satisfy:

$$-1 \le \dfrac{4x+2[x]}{3} \le 1$$−1≤34x+2[x]​≤1

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

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

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

$$-3 \le 6I + 4f \le 3$$−3≤6I+4f≤3

Since $0 \le f < 1$0≤f<1, we have $0 \le 4f < 4$0≤4f<4. We can check the possible integer values of $I$I:

If $I \ge 1$I≥1, $6I + 4f \ge 6(1) + 0 = 6 > 3$6I+4f≥6(1)+0=6>3, which gives no solution.

If $I \le -2$I≤−2, $6I + 4f < 6(-2) + 4 = -8 < -3$6I+4f<6(−2)+4=−8<−3, which gives no solution.

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

Case 1: $I = 0$I=0

$$-3 \le 6(0) + 4f \le 3 \Rightarrow -3 \le 4f \le 3 \Rightarrow -\dfrac{3}{4} \le f \le \dfrac{3}{4}$$−3≤6(0)+4f≤3⇒−3≤4f≤3⇒−43​≤f≤43​

Since $f \in [0, 1)$f∈[0,1), we get $0 \le f \le \dfrac{3}{4}$0≤f≤43​.

Therefore, $x = I + f \in \left[0, \dfrac{3}{4}\right]$x=I+f∈[0,43​].

Case 2: $I = -1$I=−1

$$-3 \le 6(-1) + 4f \le 3 \Rightarrow 3 \le 4f \le 9 \Rightarrow \dfrac{3}{4} \le f \le \dfrac{9}{4}$$−3≤6(−1)+4f≤3⇒3≤4f≤9⇒43​≤f≤49​

Since $f \in [0, 1)$f∈[0,1), we get $\dfrac{3}{4} \le f < 1$43​≤f<1.

Therefore, $x = I + f \in \left[-1 + \dfrac{3}{4}, -1 + 1\right) = \left[-\dfrac{1}{4}, 0\right)$x=I+f∈[−1+43​,−1+1)=[−41​,0).

Taking the union of the intervals from both cases, the domain is $x \in \left[-\dfrac{1}{4}, \dfrac{3}{4}\right]$x∈[−41​,43​].

Comparing this with $[\alpha, \beta]$[α,β], we get $\alpha = -\dfrac{1}{4}$α=−41​ and $\beta = \dfrac{3}{4}$β=43​.

Finally, evaluating the required expression:

$$12(\alpha + \beta) = 12\left(-\dfrac{1}{4} + \dfrac{3}{4}\right) = 12\left(\dfrac{2}{4}\right) = 12\left(\dfrac{1}{2}\right) = 6$$12(α+β)=12(−41​+43​)=12(42​)=12(21​)=6

Answer: $6$6