Question

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

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

Hint:

For problems involving parabolas and their conditions, always check the inequalities and solve for the values that satisfy the system.

Answer 1

The Correct Option is C

To solve this problem, we need to determine the set \( S \) for which the condition \(\tan^{-1}(\sqrt{f(x)}) + \sin^{-1}(\sqrt{f(x)+1}) = \frac{\pi}{2}\) is satisfied, where \( y = f(x) \) is a parabola.

First, let’s determine the equation of the parabola with the given focus and directrix.

1. The parabola has a focus at \(\left( -\frac{1}{2}, 0 \right)\) and a directrix at \(y = -\frac{1}{2}\).
2. The general equation of a parabola with vertical axis is given by \((x - h)^2 = 4p(y - k)\), where \((h, k)\) is the vertex, and \( p \) is the distance from the vertex to the focus (also equal to the distance from the vertex to the directrix).
3. The vertex of the parabola is midway between the focus and the directrix. Thus, the vertex is at \(\left( -\frac{1}{2}, -\frac{1}{4} \right)\).
4. The distance \( p \) is \(\frac{1}{4}\).
5. The equation of the parabola is: \((x + \frac{1}{2})^2 = \frac{1}{2}(y + \frac{1}{4})\).

Now, considering the condition \(\tan^{-1}(\sqrt{f(x)}) + \sin^{-1}(\sqrt{f(x)+1}) = \frac{\pi}{2}\), we can apply the identity:

• \(\sin^{-1}(A) + \cos^{-1}(A) = \frac{\pi}{2}\) implies \(A = \cos^{-1}(B)\).
• Given that \(\tan^{-1}(\sqrt{f(x)}) + \sin^{-1}(\sqrt{f(x)+1}) = \frac{\pi}{2}\), we can deduce:
• \(\tan^{-1}(\sqrt{f(x)}) = \cos^{-1}(\sqrt{f(x)+1})\).
• Therefore, \(\sqrt{f(x)} = \sin(\cos^{-1}(\sqrt{f(x)+1}))\).

Simplifying further:

• \(\sqrt{f(x)} = \sqrt{1-(f(x)+1)}\)
• This implies \(\sqrt{f(x)} = \sqrt{-f(x)}\).

We now transform and solve for the valid \( f(x) \):

• This holds true for non-negative values, thus: \(f(x) = -1\).

Now, solve the parabola's equation for \( f(x) = -1 \):

• Substitute \( y = -1 \) into the parabola equation \((x + \frac{1}{2})^2 = \frac{1}{2}(-1 + \frac{1}{4})\), and solve for \( x \).
• The solutions yield exactly two values satisfying the given condition.

Therefore, the set \( S \) contains exactly two elements.

Correct Answer: contains exactly two elements