Let y = f ( x ) represent a parabola with focus ( - 1 2 , 0 ) and directrix y = - 1 2 . Then S = { x : tan - 1 ( f ( x ) ) + sin - 1 ( f ( x ) + 1 ) = 2 } : [2023]

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Solution

Q.
Let $y = f (x)$ represent a parabola with focus $(- \frac{1}{2}, 0)$ and directrix $y = - \frac{1}{2}$ . Then $S = {x \in ℝ : \tan^{- 1} (\sqrt{f (x)}) + \sin^{- 1} (\sqrt{f (x) + 1}) = \frac{π}{2}} :$ [2023]

1 is an empty set

2 contains exactly one element

3 contains exactly two elements

4 is an infinite set

Ans.
(3)

Equations of parabola, $\sqrt{{(x + \frac{1}{2})}^{2} + {(y - 0)}^{2}} =$ $| y + \frac{1}{2} |$ $\Rightarrow x^{2} + \frac{1}{4} + x + y^{2} = y^{2} + \frac{1}{4} + y$

$\Rightarrow x^{2} + x = y = f (x) (let)$

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

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

We know, $\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2},$

$\Rightarrow \frac{1}{\sqrt{1 + f (x)}} = \sqrt{f (x) + 1} \Rightarrow f (x) + 1 = 1 \Rightarrow f (x) = 0$

$x^{2} + x = 0 \Rightarrow x (x + 1) = 0 \Rightarrow x = 0, - 1$ $\Rightarrow S = {- 1, 0}$

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