JEE Main 2023 (Online) 13th April Morning ShiftQuestion 1
For $x\in\mathbb{R}$, two real valued functions $f(x)$ and $g(x)$ are such that, $g(x)=\sqrt{x}+1$ and $f\circ g(x)=x+3-\sqrt{x}$. Then $f(0)$ is equal to
Functions
Solution
Put $t=\sqrt x+1$, so $\sqrt x=t-1$ and $x=(t-1)^2$. $f(t)=(t-1)^2+3-(t-1)=t^2-3t+5$. So $f(0)=5$.