JEE Mathematics 2026 Question with Solution

Given: $f(x) = \cos^{-1}\left(\frac{2x-5}{11-3x}\right) + \sin^{-1}(2x^2-3x+1)$

Find: The value of $\alpha + 2\beta$ where the domain is $[\alpha, \beta]$.

For the sum to be defined, both inverse trigonometric expressions must be defined. Hence we need:

$$-1 \le \frac{2x-5}{11-3x} \le 1$$

and

$$-1 \le 2x^2-3x+1 \le 1$$

First solve

$$-1 \le \frac{2x-5}{11-3x} \le 1$$

This gives two inequalities.

$$\frac{2x-5}{11-3x} \le 1$$

So,

$$\frac{2x-5}{11-3x} - 1 \le 0$$ $$\frac{2x-5-(11-3x)}{11-3x} \le 0$$ $$\frac{5x-16}{11-3x} \le 0$$

Using the critical points $x = \frac{16}{5}$ and $x = \frac{11}{3}$, this holds for

$$x \in (-\infty, 16/5] \cup (11/3, \infty)$$

$$\frac{2x-5}{11-3x} \ge -1$$

So,

$$\frac{2x-5}{11-3x} + 1 \ge 0$$ $$\frac{2x-5+11-3x}{11-3x} \ge 0$$ $$\frac{-x+6}{11-3x} \ge 0$$ $$\frac{x-6}{3x-11} \ge 0$$

Using the critical points $x = 6$ and $x = \frac{11}{3}$, this holds for

$$x \in (-\infty, 11/3) \cup [6, \infty)$$

Hence,

$$D_1 = (-\infty, 16/5] \cup [6, \infty)$$

Now solve

$$-1 \le 2x^2-3x+1 \le 1$$

Again split into two parts.

$$2x^2-3x+1 \ge -1$$ $$2x^2-3x+2 \ge 0$$

Its discriminant is

$$\Delta = (-3)^2 - 4(2)(2) = 9-16 = -7 < 0$$

Since the coefficient of $x^2$ is positive, this inequality is true for all real $x$.

$$2x^2-3x+1 \le 1$$ $$2x^2-3x \le 0$$ $$x(2x-3) \le 0$$

So,

$$0 \le x \le \frac{3}{2}$$

Thus,

$$D_2 = [0, 3/2]$$

Now take the intersection:

$$D_1 \cap D_2 = \left[0, \frac{3}{2}\right]$$

Therefore,

$$[\alpha, \beta] = \left[0, \frac{3}{2}\right]$$

So,

$$\alpha = 0, \quad \beta = \frac{3}{2}$$

Then,

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

Therefore, the correct option is A.](streamdown:incomplete-link)