Why do we not need to use the formula to solve the equation?

Because the domain restrictions already force a unique value for : forces , while requires . The only real number satisfying both is , making algebraic manipulation unnecessary.

JEE MainMathematics

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Conic Sections: JEE Main Mathematics Question with Solution

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 \mathbb{R}: \tan ^{-1}(\sqrt{f(x)})+\sin ^{-1}(\sqrt{f(x)+1})=\frac{\pi}{2}\right\}

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Acontains exactly two elementsBcontains exactly one elementCis an empty setDis an infinite set

Step-by-step solutionView

Correct answer

The inverse trigonometric equation requires $f(x)=0$ , which yields exactly two real solutions from the parabola's equation.

Option analysis

Why each option works or fails

A · contains exactly two elements

None. This option is correct. Because $\sin^{-1}(\sqrt{f(x)+1}) \le \frac{\pi}{2}$ requires $\sqrt{f(x)+1} \le 1$ , we must have $f(x) \le 0$ . Combined with the square root condition $f(x) \ge 0$ , this forces $f(x) = 0$ . Solving $f(x) = 0$ yields two distinct real values for $x$ .

B · contains exactly one element

Assuming the vertex of the parabola is the only point satisfying $f(x) = 0$ , or erroneously treating the quadratic equation $f(x)=0$ as having a double root. Find the equation of the parabola explicitly using focus-directrix distance: $(x + 1/2)^2 + y^2 = (y + 1/2)^2$ . Setting $y = 0$ gives $(x + 1/2)^2 = 1/4$ , which gives two distinct roots: $x = 0$ and $x = -1$ .

C · is an empty set

Believing that $\sqrt{f(x)+1} > 1$ always, making the argument of $\sin^{-1}$ fall outside $[-1, 1]$ and yielding no solutions. Check the boundary of the domain: $\sin^{-1}(u)$ is defined for $u \le 1$ . Here $u = \sqrt{f(x)+1}$ , so $f(x)+1 \le 1 \implies f(x) \le 0$ . Since $f(x) \ge 0$ is also required for $\sqrt{f(x)}$ , $f(x) = 0$ is valid and yields solutions.

D · is an infinite set

Treating $\tan^{-1}(u) + \sin^{-1}(v) = \pi/2$ as an identity that holds over an entire continuous interval of $x$ . The argument $\sqrt{f(x)+1}$ is restricted to values $\le 1$ , meaning $f(x) \le 0$ . Together with $f(x) \ge 0$ , only the discrete condition $f(x)=0$ can hold, not an interval.

Reviewed route

Solution

StepWorking

01given

Focus of the parabola is $S = \left(-\frac{1}{2}, 0\right)$ , directrix is the horizontal line $y = -\frac{1}{2}$ , and the set is defined by $S = \left\{x \in \mathbb{R}: \tan^{-1}\left(\sqrt{f(x)}\right) + \sin^{-1}\left(\sqrt{f(x)+1}\right) = \frac{\pi}{2}\right\}$ .

02goal

Find the number of real elements in the set $S$ .

03approach

First, find $y = f(x)$ using the locus definition of a parabola: distance from $(x, y)$ to the focus equals the perpendicular distance to the directrix. Second, analyze the domain of the inverse trigonometric functions $\tan^{-1}(\sqrt{f(x)})$ and $\sin^{-1}(\sqrt{f(x)+1})$ to constrain the possible values of $f(x)$ without solving complicated inverse trig equations.

04execute

By the focus-directrix property, $\left(x + \frac{1}{2}\right)^2 + (y - 0)^2 = \left(y - \left(-\frac{1}{2}\right)\right)^2 = \left(y + \frac{1}{2}\right)^2$ . Expanding this: $\left(x + \frac{1}{2}\right)^2 + y^2 = y^2 + y + \frac{1}{4} \implies x^2 + x + \frac{1}{4} = y + \frac{1}{4} \implies y = f(x) = x^2 + x$ .

05execute

Now analyze the domain for the equation $\tan^{-1}\left(\sqrt{f(x)}\right) + \sin^{-1}\left(\sqrt{f(x)+1}\right) = \frac{\pi}{2}$ : 1) For the square root $\sqrt{f(x)}$ to be defined in real numbers, $f(x) \ge 0$ . 2) For $\sin^{-1}\left(\sqrt{f(x)+1}\right)$ to be defined, the argument must satisfy $0 \le \sqrt{f(x)+1} \le 1 \implies 0 \le f(x) + 1 \le 1 \implies -1 \le f(x) \le 0$ . Intersecting both conditions: $f(x) \ge 0$ and $f(x) \le 0 \implies f(x) = 0$ .

06execute

Substitute $f(x) = 0$ into the original equation to check validity: $\tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$ , which is identically satisfied. Thus, we solve $f(x) = 0 \implies x^2 + x = 0 \implies x(x + 1) = 0 \implies x = 0$ or $x = -1$ .

✓verify

For $x = 0$ , $f(0) = 0$ , giving $\tan^{-1}(0) + \sin^{-1}(1) = \frac{\pi}{2}$ . For $x = -1$ , $f(-1) = 0$ , giving the same. Both solutions are valid, distinct real numbers. Hence, $S = \{0, -1\}$ , which has exactly 2 elements.

Hints that build this answer step by step

What restrictions do the domains of $\tan^{-1}(\sqrt{f(x)})$ and $\sin^{-1}(\sqrt{f(x)+1})$ impose on $f(x)$ ?
$f(x) = 0$
What is the Cartesian equation representing the parabola with focus $\left(-\frac{1}{2}, 0\right)$ and directrix $y = -\frac{1}{2}$ ?
$\left(x + \frac{1}{2}\right)^2 + y^2 = \left(y + \frac{1}{2}\right)^2$
Setting $y = f(x) = 0$ in the parabola's equation, how many real values of $x$ satisfy the equation?
Two distinct values: $x = 0$ and $x = -1$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we not need to use the formula

\tan^{-1}(A) + \tan^{-1}(B)

to solve the equation?

Because the domain restrictions already force a unique value for $f(x)$ : $\sqrt{f(x)} \ge 0$ forces $f(x) \ge 0$ , while $\sin^{-1}(\sqrt{f(x)+1})$ requires $\sqrt{f(x)+1} \le 1 \implies f(x) \le 0$ . The only real number satisfying both is $f(x) = 0$ , making algebraic manipulation unnecessary.

Answer

The inverse trigonometric equation requires $f(x)=0$ , which yields exactly two real solutions from the parabola's equation.

Why each option works or fails

A · correct: contains exactly two elements - None. This option is correct. Because $\sin^{-1}(\sqrt{f(x)+1}) \le \frac{\pi}{2}$ requires $\sqrt{f(x)+1} \le 1$ , we must have $y=-\frac{1}{2}$ . Combined with the square root condition $f(x) \ge 0$ , this forces $f(x) = 0$ . Solving $f(x) = 0$ yields two distinct real values for $x$ .
B: contains exactly one element - Assuming the vertex of the parabola is the only point satisfying $f(x) = 0$ , or erroneously treating the quadratic equation $f(x)=0$ as having a double root. Find the equation of the parabola explicitly using focus-directrix distance: $(x + 1/2)^2 + y^2 = (y + 1/2)^2$ . Setting $y = 0$ gives $(x + 1/2)^2 = 1/4$ , which gives two distinct roots: $x = 0$ and $x = -1$ .
C: is an empty set - Believing that $\sqrt{f(x)+1} > 1$ always, making the argument of $\sin^{-1}$ fall outside $[-1, 1]$ and yielding no solutions. Check the boundary of the domain: $\sin^{-1}(u)$ is defined for $u \le 1$ . Here $u = \sqrt{f(x)+1}$ , so $f(x)+1 \le 1 \implies f(x) \le 0$ . Since $f(x) \ge 0$ is also required for $\sqrt{f(x)}$ , $f(x) = 0$ is valid and yields solutions.
D: is an infinite set - Treating $\tan^{-1}(u) + \sin^{-1}(v) = \pi/2$ as an identity that holds over an entire continuous interval of $x$ . The argument $\sqrt{f(x)+1}$ is restricted to values $\le 1$ , meaning $f(x) \le 0$ . Together with $f(x) \ge 0$ , only the discrete condition $y = 0$ can hold, not an interval.

Step-by-step solution

given: Focus of the parabola is $S = \left(-\frac{1}{2}, 0\right)$ , directrix is the horizontal line $y = -\frac{1}{2}$ , and the set is defined by $S = \left\{x \in \mathbb{R}: \tan^{-1}\left(\sqrt{f(x)}\right) + \sin^{-1}\left(\sqrt{f(x)+1}\right) = \frac{\pi}{2}\right\}$ .
goal: Find the number of real elements in the set $S$ .
approach: First, find $y = f(x)$ using the locus definition of a parabola: distance from $(x, y)$ to the focus equals the perpendicular distance to the directrix. Second, analyze the domain of the inverse trigonometric functions $\tan^{-1}(\sqrt{f(x)})$ and $\sin^{-1}(\sqrt{f(x)+1})$ to constrain the possible values of $f(x)$ without solving complicated inverse trig equations.
execute: By the focus-directrix property, $\left(x + \frac{1}{2}\right)^2 + (y - 0)^2 = \left(y - \left(-\frac{1}{2}\right)\right)^2 = \left(y + \frac{1}{2}\right)^2$ . Expanding this: $\left(x + \frac{1}{2}\right)^2 + y^2 = y^2 + y + \frac{1}{4} \implies x^2 + x + \frac{1}{4} = y + \frac{1}{4} \implies y = f(x) = x^2 + x$ .
execute: Now analyze the domain for the equation $\tan^{-1}\left(\sqrt{f(x)}\right) + \sin^{-1}\left(\sqrt{f(x)+1}\right) = \frac{\pi}{2}$ : 1) For the square root $\sqrt{f(x)}$ to be defined in real numbers, $f(x) \ge 0$ . 2) For $\sin^{-1}\left(\sqrt{f(x)+1}\right)$ to be defined, the argument must satisfy $0 \le \sqrt{f(x)+1} \le 1 \implies 0 \le f(x) + 1 \le 1 \implies -1 \le f(x) \le 0$ . Intersecting both conditions: $f(x) \ge 0$ and $f(x) \le 0 \implies f(x) = 0$ .
execute: Substitute $f(x) = 0$ into the original equation to check validity: $\tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$ , which is identically satisfied. Thus, we solve $f(x) = 0 \implies x^2 + x = 0 \implies x(x + 1) = 0 \implies x = 0$ or $x = -1$ .
verify: For $x = 0$ , $f(0) = 0$ , giving $\tan^{-1}(0) + \sin^{-1}(1) = \frac{\pi}{2}$ . For $x = -1$ , $f(-1) = 0$ , giving the same. Both solutions are valid, distinct real numbers. Hence, $S = \{0, -1\}$ , which has exactly 2 elements.

Conic Sections: JEE Main Mathematics Question with Solution

What feels right?

The inverse trigonometric equation requires $f(x)=0$ , which yields exactly two real solutions from the parabola's equation.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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Conic Sections: JEE Main Mathematics Question with Solution

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