JEE Main 2022 (Online) 26th June Morning ShiftQuestion 1
Let $f,g:\mathbf{R}\to\mathbf{R}$ be two real valued functions defined as $f(x)=\begin{cases}-|x+3|, & x<0\\ e^{x}, & x\ge0\end{cases}$ and $g(x)=\begin{cases}x^{2}+k_1x, & x<0\\ 4x+k_2, & x\ge0\end{cases}$, where $k_1$ and $k_2$ are real constants. If $(g\circ f)$ is differentiable at $x=0$, then $(g\circ f)(-4)+(g\circ f)(4)$ is equal to :
Functions
Solution
Near $0^-$, $(g\circ f)(x)=(x+3)^2-k_1(x+3)$; near $0^+$, $4e^x+k_2$. Equal values and slopes at $0$: $9-3k_1=4+k_2$, $6-k_1=4$, so $k_1=2$, $k_2=-1$. $(g\circ f)(-4)=g(-1)=1-2=-1$, $(g\circ f)(4)=4e^4-1$. Sum $=4e^4-2=2(2e^4-1)$.