Inverse Trigonometric Functions – JEE Main 21 Jan 2026 Shift 1

Solution:

For the domain of $f(x)$, both inverse trigonometric functions must be defined.

Condition for $\sin^{-1}(2x^2 – 3x + 1)$:
$$-1 \le 2x^2 – 3x + 1 \le 1$$

Taking the right part of the inequality:
$$2x^2 – 3x + 1 \le 1 \Rightarrow 2x^2 – 3x \le 0$$
$$\Rightarrow x(2x – 3) \le 0$$
$$\Rightarrow x \in \left[0, \frac{3}{2}\right]$$

Taking the left part of the inequality:
$$2x^2 – 3x + 1 \ge -1 \Rightarrow 2x^2 – 3x + 2 \ge 0$$
Discriminant $D = (-3)^2 – 4(2)(2) = 9 – 16 = -7 < 0$. Since $a > 0$ and $D < 0$, the expression is always positive. Thus, this holds for all $x \in R$. So, from the first term, the valid range is $x \in [0, \frac{3}{2}]$.
Condition for $\cos^{-1}\left(\frac{2x-5}{11-3x}\right)$:
$$-1 \le \frac{2x-5}{11-3x} \le 1$$
For $x \in [0, \frac{3}{2}]$, the denominator $11 – 3x$ is positive (ranging from $6.5$ to $11$). Thus, we can cross-multiply without changing the inequality sign.

$$-(11 – 3x) \le 2x – 5 \le 11 – 3x$$

Solving left part:
$$-11 + 3x \le 2x – 5 \Rightarrow x \le 6$$

Solving right part:
$$2x – 5 \le 11 – 3x \Rightarrow 5x \le 16 \Rightarrow x \le \frac{16}{5} = 3.2$$

Intersection of all conditions:
$$x \in \left[0, \frac{3}{2}\right] \cap (-\infty, 6] \cap \left(-\infty, \frac{16}{5}\right]$$
$$x \in \left[0, \frac{3}{2}\right]$$

Thus, the domain is $[\alpha, \beta] = [0, \frac{3}{2}]$.
Here, $\alpha = 0$ and $\beta = \frac{3}{2}$.

Value of $\alpha + 2\beta$:
$$0 + 2\left(\frac{3}{2}\right) = 3$$

Ans. (2)