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

(A) contains exactly two elements

(B) contains exactly one element

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Parabola

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}}

(A)contains exactly two elements✓ Correct
(B)contains exactly one element
(C)is an infinite set
(D)is an empty set

Answer: (A)

Solution

The parabola with focus

(- \frac{1}{2}, 0)

and directrix

y = - \frac{1}{2}

satisfies

(x + \frac{1}{2})^{2} = y + \frac{1}{4}

, that is

y = x^{2} + x

, so

f (x) = x^{2} + x

. In

tan^{- 1} f (x) + sin^{- 1} f (x) + 1 = \frac{π}{2}

, the square roots require

f (x) \geq 0

and

f (x) + 1 \leq 1

, i.e.

f (x) \leq 0

; together these force

f (x) = 0

. Solving

x^{2} + x = 0

gives

x = 0

and

x = - 1

, both of which satisfy the equation. Hence the set contains exactly two elements.

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option

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