JEE Main 2023 (Online) 31st January Morning ShiftQuestion 1
Let $y=f(x)$ represent a parabola with focus $\left(-\frac12,0\right)$ and directrix $y=-\frac12$. Then $S=\left\{x\in\mathbb R:\tan^{-1}\sqrt{f(x)}+\sin^{-1}\sqrt{f(x)+1}=\frac\pi2\right\}$
Inverse Trigonometric Functions
Solution
$(x+\frac12)^2+y^2=(y+\frac12)^2$ gives $f(x)=x^2+x$. Need $f\ge0$ and $f+1\le1$, so $f(x)=0$, which satisfies $0+\frac\pi2=\frac\pi2$. $x=0,-1$: exactly two elements.