Let `y=f(x)` represent a parabola with focus `(-(1)/(2),0)` and directrix `y=-(1)/(2)`.Then `S={x in R:tan^(-1)(sqrt(f(x)))+sin^(-1)(sqrt(f(x)+1))=(pi)/(2)}`.

Options

A. is an empty set
B. contains exactly one element
C. contains exactly two elements
D. is an infinite set

Answer

To solve the problem, we need to find the set \( S \) defined by the equation:

\[ \tan^{-1}(\sqrt{f(x)}) + \sin^{-1}(\sqrt{f(x) + 1}) = \frac{\pi}{2} \]

where \( y = f(x) \) represents a parabola with focus at \((-1/2, 0)\) and directrix \( y = -1/2 \).

### Step 1: Find the equation of the parabola

The standard form of a parabola that opens upwards with a focus at \((h, k + p)\) and directrix \(y = k - p\) is given by:

\[ (x - h)^2 = 4p(y - k) \]

In our case, the focus is \((-1/2, 0)\) and the directrix is \(y = -1/2\). The distance \(p\) from the focus to the directrix is:

\[ p = 0 - (-1/2) = 1/2 \]

The vertex of the parabola is halfway between the focus and the directrix, which is at:

\[ \left(-\frac{1}{2}, -\frac{1}{4}\right) \]

Thus, the equation of the parabola is:

\[ \left(x + \frac{1}{2}\right)^2 = 2\left(y + \frac{1}{4}\right) \]

Expanding this, we get:

\[ \left(x + \frac{1}{2}\right)^2 = 2y + \frac{1}{2} \]

Rearranging gives:

\[ y = \frac{1}{2}\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} \]

### Step 2: Simplify the equation for \(f(x)\)

From the above equation, we can express \(f(x)\):

\[ f(x) = \frac{1}{2}\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} \]

### Step 3: Substitute \(f(x)\) into the given equation

We need to substitute \(f(x)\) into the equation:

\[ \tan^{-1}(\sqrt{f(x)}) + \sin^{-1}(\sqrt{f(x) + 1}) = \frac{\pi}{2} \]

Using the identity \( \sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2} \), we can rewrite the equation as:

\[ \tan^{-1}(\sqrt{f(x)}) + \cos^{-1}(\sqrt{f(x) + 1}) = \frac{\pi}{2} \]

This implies:

\[ \sqrt{f(x) + 1} = \frac{1}{\sqrt{f(x)}} \]

### Step 4: Solve for \(f(x)\)

Squaring both sides gives:

\[ f(x) + 1 = \frac{1}{f(x)} \]

Multiplying through by \(f(x)\) leads to:

\[ f(x)^2 + f(x) - 1 = 0 \]

Using the quadratic formula:

\[ f(x) = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 5: Find \(S\)

Since \(f(x)\) must be non-negative (as it is under a square root), we take:

\[ f(x) = \frac{-1 + \sqrt{5}}{2} \]

Now we need to find the values of \(x\) such that:

\[ \frac{1}{2}\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} = \frac{-1 + \sqrt{5}}{2} \]

Solving this will give us the values of \(x\) that belong to the set \(S\).

### Final Solution

After solving, we find that the set \(S\) contains the elements \(0\) and \(1\).

Thus, the final answer is:

\[ S = \{0, 1\} \]