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

${\left(x+\frac{1}{2}\right)}^2=\left(y+\frac{1}{4}\right)$
$y=\left(x^2+x\right)$
${\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+1}=\pi /2$
$0\le x^2+x+1\le 1$
$x^2+x\le 0$....(1)
$\text{Also}{x}^2+x\ge 0$....(2)
$\therefore x^2+x=0\Rightarrow x=0,-1$
$S$ contains 2 element.

Answer 2

Step 1: The equation of a parabola with focus \( \left( -\frac{1}{2}, 0 \right) \) and directrix \( y = -\frac{1}{2} \) is given by: \[ y = x^2 + x \] Step 2: We are also given the equation for \( S \): \[ \tan^{-1} \left( \sqrt{f(x)} + \sin \left( \sqrt{f(x) + 1} \right) \right) = \frac{\pi}{2} \] This implies: \[ 0 \leq x^2 + x + 1 \quad \text{and} \quad x^2 + x \leq 0 \] Step 3: Solving these inequalities, we get: \[ x^2 + x \leq 0 \quad \Rightarrow \quad x \leq 0 \] Hence, the set \( S \) contains exactly two elements.