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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\}: 

Option: 1

contains exactly two elements 


Option: 2

contains exactly one element
 


Option: 3

is an empty set


Option: 4

is an infinite set


Answers (1)

best_answer

equation of parabola which have focus \left(-\frac{1}{2}, 0\right)  and directrix  y=\frac{1}{2}   is

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

y = f(x) = (x2 + x) 

\begin{aligned} & \because S=\left\{\mathrm{x} \in \mathrm{R}: \tan ^{-1}\left((\sqrt{f(x)})+\sin ^{-1}(\sqrt{f(x)+1})=\frac{\pi}{2}\right)\right\} \\ & \tan ^{-1}(\sqrt{f(x)})+\sin ^{-1}(\sqrt{f(x)+1})=\frac{\pi}{2} \end{aligned}

f(x) \geq 0 \& \sqrt{f(x)+1}     can not greater then 1, so f(x) must be 0

\begin{aligned} & \text { i.e. } f(x)=0 \\ & \Rightarrow x^2+x=0 \\ & x(x+1)=0 \\ & x=0, x=-1 \end{aligned}

S contain 2 element.

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Gaurav

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