JEE Advanced 2021 — Paper 2 — Question 49

Define $\psi_1(x)=e^{-x}+x$ and $\psi_2(x)=x^2-2x-2e^{-x}+2$. Compute $\psi_2'(x)=2x-2+2e^{-x}=2(x+e^{-x}-1)$. Note $\psi_1(x)-1 = e^{-x}+x-1$, so $\psi_2'(x)=2(\psi_1(x)-1)$. We have $\psi_2(0)=0$. For any $x>0$, $\psi_2$ is continuous on $[0,x]$ and differentiable on $(0,x)$. By Lagrange’s Mean Value Theorem, there exists $\beta\in(0,x)$ such that $\frac{\psi_2(x)-\psi_2(0)}{x-0}=\psi_2'(\beta)$. Substituting: $\frac{\psi_2(x)}{x}=2(\psi_1(\beta)-1)$. Multiply by $x$: $\psi_2(x)=2x(\psi_1(\beta)-1)$. Hence option C holds. The other options can be verified to be false, confirming that C is the correct statement.