1. Key Concepts and Formulas

To solve this problem, we'll need to utilize the fundamental definitions and properties of parabolas and inverse trigonometric functions.

• Definition of a Parabola: A parabola is defined as the locus of all points $P(x, y)$ that are equidistant from a fixed point, called the focus $F$, and a fixed line, called the directrix $L$. Mathematically, this means $PF = PL$, where $PF$ is the distance from point $P$ to the focus $F$, and $PL$ is the perpendicular distance from point $P$ to the directrix $L$.

• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

• Perpendicular Distance from a Point to a Line: The perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

• Domain and Range of Inverse Trigonometric Functions (Principal Values):

  • For $\tan^{-1}(A)$ to be defined, $A$ can be any real number ($A \in \mathbb{R}$). Its principal value range is $(-\frac{\pi}{2}, \frac{\pi}{2})$.
  • For $\sin^{-1}(B)$ to be defined, $B$ must be in the interval $[-1, 1]$. Its principal value range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

2. Step-by-Step Derivation of the Parabola Equation $y=f(x)$

The problem provides the focus $F = \left(-\frac{1}{2}, 0\right)$ and the directrix $L: y = -\frac{1}{2}$. Let $P(x, y)$ be an arbitrary point on the parabola.

• Step 2.1: Calculate the distance from $P(x, y)$ to the focus $F\left(-\frac{1}{2}, 0\right)$. We use the distance formula for two points: $PF = \sqrt{\left(x - \left(-\frac{1}{2}\right)\right)^2 + (y - 0)^2}$ $PF = \sqrt{\left(x + \frac{1}{2}\right)^2 + y^2}$ Explanation: This is a direct application of the distance formula to find the length of the segment connecting the point on the parabola to the focus.

• Step 2.2: Calculate the perpendicular distance from $P(x, y)$ to the directrix $L: y = -\frac{1}{2}$. First, rewrite the directrix equation in the general form $Ax + By + C = 0$: $0x + 1y + \frac{1}{2} = 0$. Now, apply the formula for the perpendicular distance from a point $(x, y)$ to this line: $PL = \frac{\left|0 \cdot x + 1 \cdot y + \frac{1}{2}\right|}{\sqrt{0^2 + 1^2}}$ $PL = \frac{\left|y + \frac{1}{2}\right|}{\sqrt{1}}$ $PL = \left|y + \frac{1}{2}\right|$ Explanation: The perpendicular distance from a point to a horizontal line $y=k$ is simply $|y-k|$. Here, $k = -\frac{1}{2}$, so the distance is $|y - (-\frac{1}{2})| = |y + \frac{1}{2}|$. The absolute value is crucial because distance must always be non-negative.

• Step 2.3: Equate the distances and simplify to find $y=f(x)$. According to the definition of a parabola, $PF = PL$. To eliminate the square root and the absolute value, we square both sides of the equation. This is a valid operation because both $PF$ and $PL$ represent distances, which are always non-negative. $PF^2 = PL^2$ $\left(\sqrt{\left(x + \frac{1}{2}\right)^2 + y^2}\right)^2 = \left(\left|y + \frac{1}{2}\right|\right)^2$ $\left(x + \frac{1}{2}\right)^2 + y^2 = \left(y + \frac{1}{2}\right)^2$ Now, expand both sides using the formula $(a+b)^2 = a^2 + 2ab + b^2$: $x^2 + 2 \cdot x \cdot \frac{1}{2} + \left(\frac{1}{2}\right)^2 + y^2 = y^2 + 2 \cdot y \cdot \frac{1}{2} + \left(\frac{1}{2}\right)^2$ $x^2 + x + \frac{1}{4} + y^2 = y^2 + y + \frac{1}{4}$ We can cancel the $y^2$ term and the $\frac{1}{4}$ term from both sides: $x^2 + x = y$ Thus, the equation of the parabola is $y = f(x) = x^2 + x$. Explanation: Squaring both sides is a standard algebraic technique to simplify equations involving square roots or absolute values. The subsequent expansion and cancellation of common terms help to isolate $y$ and express it as a function of $x$.

3. Determining the Domain of the Inverse Trigonometric Equation

The given equation is $\tan ^{-1}(\sqrt{f(x)})+\sin ^{-1}(\sqrt{f(x)+1})=\frac{\pi}{2}$. For this equation to be defined, the arguments of both inverse trigonometric functions must satisfy their respective domain requirements.

• Step 3.1: Domain constraints for $\tan^{-1}(\sqrt{f(x)})$. For $\tan^{-1}(A)$ to be defined, $A$ can be any real number. However, our argument is $\sqrt{f(x)}$. For $\sqrt{f(x)}$ to be a real number, the expression inside the square root must be non-negative. Therefore, we must have: $f(x) \ge 0 \quad \text{(Condition 1)}$ Explanation: The square root function is only defined for non-negative real numbers in the real number system. If $f(x)$ were negative, $\sqrt{f(x)}$ would be an imaginary number, and $\tan^{-1}$ of an imaginary number is not typically considered in standard JEE calculus context.

• Step 3.2: Domain constraints for $\sin^{-1}(\sqrt{f(x)+1})$. For $\sin^{-1}(B)$ to be defined, $B$ must be in the interval $[-1, 1]$. Our argument is $B = \sqrt{f(x)+1}$. First, for $\sqrt{f(x)+1}$ to be a real number, the expression inside the square root must be non-negative: $f(x)+1 \ge 0 \implies f(x) \ge -1 \quad \text{(Condition 2)}$ Second, the value of $\sqrt{f(x)+1}$ must lie within $[-1, 1]$. So, we need: $-1 \le \sqrt{f(x)+1} \le 1$ Since the square root symbol $\sqrt{\cdot}$ by definition always yields a non-negative real value (when its argument is non-negative), the condition $\sqrt{f(x)+1} \ge -1$ is automatically satisfied whenever $f(x)+1 \ge 0$. Therefore, we only need to satisfy the upper bound: $\sqrt{f(x)+1} \le 1$ Since both sides of this inequality are non-negative ( $\sqrt{f(x)+1} \ge 0$ and $1 \ge 0$), we can safely square both sides without changing the direction of the inequality: $(\sqrt{f(x)+1})^2 \le 1^2$ $f(x)+1 \le 1$ $f(x) \le 0 \quad \text{(Condition 3)}$ Explanation: This is a crucial step. We first ensure the square root is defined. Then, we apply the domain constraint for $\sin^{-1}$. The non-negativity of the square root simplifies the lower bound, leaving only the upper bound to be actively solved. Squaring both sides of an inequality is permissible only when both sides are non-negative.

• Step 3.3: Combine all domain constraints. For the entire equation to be defined, all three conditions must be met simultaneously:

  1. $f(x) \ge 0$
  2. $f(x) \ge -1$
  3. $f(x) \le 0$

  Let's analyze these:

  • Conditions 1 ($f(x) \ge 0$) and 3 ($f(x) \le 0$) together imply that the only possible value for $f(x)$ is $0$.
  • If $f(x) = 0$, then Condition 2 ($f(x) \ge -1$) is also satisfied ($0 \ge -1$).

  Therefore, the given inverse trigonometric equation is defined only if $f(x) = 0$.

  Common Mistake Alert: Forgetting to consider the domain restrictions of inverse trigonometric functions and square roots is a common error that can lead to incorrect solutions. Always start by identifying the domain for which the expression is meaningful.

4. Solving for $x$ and Verifying the Solution

We have established that the equation is defined if and only if $f(x) = 0$. Substitute the expression for $f(x)$ we found in Step 2.3: $x^2 + x = 0$ Factor out $x$: $x(x+1) = 0$ This gives us two possible values for $x$: $x = 0 \quad \text{or} \quad x = -1$

Now, we must verify if these values of $x$ (which make $f(x)=0$) actually satisfy the original inverse trigonometric equation. If $f(x) = 0$, the equation becomes: $\tan ^{-1}(\sqrt{0})+\sin ^{-1}(\sqrt{0+1})=\frac{\pi}{2}$ $\tan ^{-1}(0)+\sin ^{-1}(\sqrt{1})=\frac{\pi}{2}$ $\tan ^{-1}(0)+\sin ^{-1}(1)=\frac{\pi}{2}$ We know that $\tan^{-1}(0) = 0$ (since $\tan(0)=0$) and $\sin^{-1}(1) = \frac{\pi}{2}$ (since $\sin(\frac{\pi}{2})=1$). $0 + \frac{\pi}{2} = \frac{\pi}{2}$ $\frac{\pi}{2} = \frac{\pi}{2}$ This is a true statement. Therefore, the values of $x$ for which $f(x)=0$ are indeed the solutions to the given equation.

5. Conclusion

The set $S$ consists of all $x \in \mathbb{R}$ that satisfy the given equation. We found that the equation is defined and satisfied precisely when $f(x) = 0$, which occurs for $x=0$ and $x=-1$. So, the set $S = \{-1, 0\}$.

The set $S$ contains exactly two elements.

The final answer is $\boxed{\text{C}}$.

Key Takeaway: When dealing with equations involving inverse trigonometric functions, always begin by carefully determining the domain for which all functions are defined. This often provides significant constraints on the possible values of the variables, sometimes narrowing down the solution space considerably, as seen in this problem where it forced $f(x)$ to be exactly zero. After finding potential solutions, it's good practice to substitute them back into the original equation to ensure they are valid.