Let y = f(x) represent a parabola with focus (-1/2, 0) and directrix y = -1/2. Then S = {x ∈ R : tan

Q: Wait, why can't I just use the identity $\tan^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$?

That is a very common temptation! However, the identity $\tan^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ only applies when the second term is $\cot^{-1}$. Here, we have $\sin^{-1}$. You cannot swap them blindly. Always check the specific functions given in the problem before applying identities.

Q: Why do we have to check the domain? Isn't the algebra enough?

In JEE Advanced problems, the domain is often the key to the entire solution. If you ignore the domain, you might find algebraic solutions that are actually impossible in the real number system. By checking the domain, we often find that the equation is only defined for specific values, which makes the problem much easier to solve.

Q: How did you know $f(x)$ had to be zero?

Great question! We found two constraints: $f(x) \geq 0$ (from the $\tan^{-1}$ term) and $f(x) \leq 0$ (from the $\sin^{-1}$ term). If a real number must be both greater than or equal to zero AND less than or equal to zero, the only logical conclusion is that the number must be exactly zero. This is a powerful technique to simplify complex equations.

Question

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

Options:

(A) contains exactly two elements

(B) contains exactly one element

(C) is an infinite set

(D) is an empty set

Sigma Percentile · Accepted Answer

### The Geometric Foundation: Defining the Parabola

Imagine you are standing on the Cartesian plane. You are given a focus $S = (-\frac{1}{2}, 0)$ and a directrix $L: y = -\frac{1}{2}$. These two elements are the DNA of a parabola.

By definition, any point $P(x, y)$ on the parabola must satisfy the condition that its distance to the focus $S$ is equal to its perpendicular distance to the directrix $L$. Let $M$ be the foot of the perpendicular from $P$ to the directrix. The condition is $PS = PM$.

Using the distance formula, the distance $PS$ is $\sqrt{(x + \frac{1}{2})^2 + y^2}$. The perpendicular distance $PM$ to the horizontal line $y = -\frac{1}{2}$ is simply $|y + \frac{1}{2}|$.

Equating these, we get:
$$ \sqrt{(x + \frac{1}{2})^2 + y^2} = |y + \frac{1}{2}| $$

Squaring both sides, we have:
$$ (x + \frac{1}{2})^2 + y^2 = (y + \frac{1}{2})^2 $$

Expanding this, we get $x^2 + x + \frac{1}{4} + y^2 = y^2 + y + \frac{1}{4}$. The $y^2$ and $\frac{1}{4}$ terms cancel out beautifully, leaving us with $y = x^2 + x$. Thus, our function is $f(x) = x^2 + x$.

### The Inverse Trig Trap

Now, we face the equation:
$$ 	an^{-1}(\sqrt{f(x)}) + \sin^{-1}(\sqrt{f(x)+1}) = \frac{\pi}{2} $$

Substituting $f(x) = x^2 + x$, we get:
$$ 	an^{-1}(\sqrt{x^2+x}) + \sin^{-1}(\sqrt{x^2+x+1}) = \frac{\pi}{2} $$

In the world of JEE, when you see inverse trigonometric functions, your first instinct should be to check the domain. For $	an^{-1}(\sqrt{x^2+x})$ to be defined, we need $x^2+x \geq 0$.

For $\sin^{-1}(\sqrt{x^2+x+1})$ to be defined, the argument must be in the interval $[-1, 1]$. Since it is a square root, it is already non-negative, so we just need:
$$ \sqrt{x^2+x+1} \leq 1 $$

Squaring this gives $x^2+x+1 \leq 1$, which simplifies to $x^2+x \leq 0$.

### The Domain Squeeze

We now have two conflicting conditions: $x^2+x \geq 0$ and $x^2+x \leq 0$. The only way both can be true is if $x^2+x = 0$.

This is the **'Aha!' moment**. The entire equation collapses into this simple quadratic.

Solving $x^2+x = 0$ gives $x(x+1) = 0$, so $x = 0$ or $x = -1$.

Finally, we verify: if $x^2+x = 0$, the equation becomes:
$$ 	an^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2} $$

It works perfectly! The set $S = \{0, -1\}$ contains exactly **two elements**. Always remember: when in doubt, check the domain constraints first. It is often the shortest path to the solution.

Animated Solution for Mathematics - Conic Sections: Let $y = f (x)$ represent a parabola with focus $(- \frac{1}{2}, 0)$ and directrix $y = - \frac{1}{2}$ . Then $S = {x \in R : tan^{- 1} (f (x)) + sin^{- 1} (f (x) + 1) = \frac{π}{2}}$ :

Select Answer:

Visualized Solution

Visualizing the Parabola Setup

The Parabola Definition

Applying the Distance Formula

Expanding the Equation

Simplifying to find $f (x)$

The Inverse Trig Equation

Domain Constraint: The Square Root

Domain Constraint: Sine Inverse

Finding the Intersection

Solving for x

Final Verification

Conclusion and Key Takeaway

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Geometric Foundation

The Inverse Trig Trap

The Domain Squeeze

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