Inverse Trigonometric Functions — JEE Main practice

1. For $\cos^{-1}$ to be defined, its argument must lie in $[-1,1]$, so set up two inequalities for the rational expression. Given $\frac{2x-5}{11-3x}\leq1$ and $\frac{2x-5}{11-3x}\geq-1$, solve each separately.

2. Solve the first inequality by combining into a single fraction. Given

$$\frac{5x-16}{11-3x}\leq0 \quad\Rightarrow\quad x\in\left(-\infty,\frac{16}{5}\right]\cup\left(\frac{11}{3},\infty\right)$$

3. Solve the second inequality similarly. Given

$$\frac{x-6}{3x-11}\geq0 \quad\Rightarrow\quad x\in\left(-\infty,\frac{11}{3}\right)\cup[6,\infty)$$

4. For $\sin^{-1}$ to be defined, its argument must also lie in $[-1,1]$, so set up two more inequalities for the quadratic expression. Given $2x^2-3x+1\leq1$,

$$x(2x-3)\leq0 \quad\Rightarrow\quad x\in\left[0,\frac32\right]$$

Given $2x^2-3x+1\geq-1$, $2x^2-3x+2\geq0 \quad\Rightarrow\quad x\in\mathbb{R} \text{ (always true, since discriminant is negative)}$

5. Take the intersection of all four conditions to find the overall domain. Since the quadratic conditions restrict $x\in[0,3/2]$, and this range is well within the ranges satisfying the rational expression conditions, the final domain is

$$x\in\left[0,\frac32\right]$$

6. Identify $\alpha=0$ and $\beta=\frac32$, then compute $\alpha+2\beta$. Hence,

$$\alpha+2\beta = 0+2\times\frac32 = 3$$

Hence, the answer is option A.

1. Combine the two inverse tangent terms using the tangent addition formula, since the given interval ensures $24x^2<1$ so the formula applies directly without adjustment. Given

$$\tan^{-1}\left(\frac{10x}{1-24x^2}\right) = \frac{\pi}{6}$$

2. Take tangent of both sides, since $\tan(\pi/6)=\frac{1}{\sqrt3}$, and cross-multiply to form a quadratic in $x$. This leads to

$$24x^2+10\sqrt3x-1=0$$

3. Solve this quadratic using the quadratic formula. Given

$$x = \frac{-5\sqrt3\pm3\sqrt{14}}{24}$$

4. Check which of these two roots actually lies within the given interval $-\frac{1}{2\sqrt6}<x<\frac{1}{2\sqrt6}$, since only solutions within this range are valid (the interval restricts $24x^2<1$, needed for the addition formula to hold without a $\pi$ correction). Only one of the two roots satisfies this constraint.

Hence, there is exactly 1 valid solution, so the answer is option B.

1. Since $\sin^{-1}\theta$ requires $-1\leq\theta\leq1$, set up two separate inequalities for the given rational function and find their intersection. Given $f(x)$ is defined when $-1\leq\frac{1}{x^2-2x-2}\leq1$, split this into $\frac{1}{x^2-2x-2}\geq-1$ and $\frac{1}{x^2-2x-2}\leq1$.

2. Solve the first inequality by combining into a single fraction and finding the roots of numerator and denominator. Given

$$\frac{x^2-2x-1}{x^2-2x-2}\geq0$$

The roots of $x^2-2x-1$ are $1\pm\sqrt2$, and the roots of $x^2-2x-2$ are $1\pm\sqrt3$. Using sign analysis, this gives $x\in(-\infty,1-\sqrt3)\cup[1-\sqrt2,1+\sqrt2]\cup(1+\sqrt3,\infty) \quad (S_1)$

3. Solve the second inequality similarly. Given

$$\frac{x^2-2x-3}{x^2-2x-2}\geq0$$

The roots of $x^2-2x-3$ are $3,-1$, and the roots of $x^2-2x-2$ are again $1\pm\sqrt3$. This gives

$$x\in(-\infty,-1]\cup(1-\sqrt3,1+\sqrt3)\cup[3,\infty) \quad (S_2)$$

4. Take the intersection of $S_1$ and $S_2$, since both conditions must hold simultaneously. Hence,

$$x\in(-\infty,-1]\cup[1-\sqrt2,1+\sqrt2]\cup[3,\infty)$$

5. Compare this with the given form to identify $\alpha,\beta,\gamma,\delta$. Hence, $\alpha=-1$, $\beta=1-\sqrt2$, $\gamma=1+\sqrt2$, $\delta=3$. Therefore,

$$\alpha+\beta+\gamma+\delta = -1+(1-\sqrt2)+(1+\sqrt2)+3 = 4$$

Hence, the answer is option A.