$\begin{aligned} & \text { (A) } \psi_1(x)=e^{-x}+x, x \geq 0 \\ & \psi_1^{\prime}(x)=1-e^{-x}>0 \quad \Rightarrow \psi_1(x) \text { is ↑ } \\ & \psi_1(x) \geq \psi_1(0) \quad \forall x \geq 0 \quad \Rightarrow \psi_1(x) \geq 1 \\ & \text { (B) } \psi_2(x)=x^2-2 x+2-2 e^{-x} \quad x \geq 0\end{aligned}$ $\begin{aligned} & \psi_2^{\prime}(x)=2 x-2+2 e^{-x}=2 \psi_2(x)-2 \geq 0 \quad \forall x \geq 0 \\ & \Rightarrow \psi_2(x) \text { is } \uparrow \Rightarrow \psi_2(x) \geq \psi_2(0) \quad \Rightarrow \psi_2(x) \geq 0 \\ & \text { (C) } f(x)=2 \int_0^x(t-t)^2 e^{-t^2} d t \quad \& \quad x \in\left(0, \frac{1}{2}\right) \\ & =\int_0^x 2 t e^{-t^2} d t-\int_0^x 2 t^2 e^{-t^2} d l \\ & =-\left.e^{-x^2}\right|_0 ^x- \\ & \text { Let } H(x)=f(x)-1+e^{-x^2}+\frac{2}{3} x^3-\frac{2}{5} x^5, x \in\left(0, \frac{1}{2}\right) \\ & H(0)=0 \\ & H^{\prime}(x)=2\left(x-x^2\right) e^{-x^2}-2 x e^{-x^2}+2 x^2-2 x^4 \\ & =-2 x^2 e^{-x^2}+2 x^2-2 x^4 \\ & =2 x^2\left(1-x^2-e^{-x^2}\right) \\ & \because e^{-x} \geq 1-x \quad \forall x \geq 0 \\ & \Rightarrow H^{\prime}(x) \leq 0 \\ & \Rightarrow H(x) \text { is } \downarrow \Rightarrow 1-1(x)<0 \quad \forall x \in\left(0, \frac{1}{2}\right) \\ & \text { Let } P(x)=f(x)-\frac{2}{3} x^3+\frac{2}{5} x^5-\frac{1}{7} x^7 x \in\left(0, \frac{1}{2}\right)\end{aligned}$ $ \begin{aligned} & \text { Let } P(x)=f(x)-\frac{2}{3} x^3+\frac{2}{5} x^5-\frac{1}{7} x^7 x \in\left(0, \frac{1}{2}\right) \\ & \begin{aligned} P^{\prime}(x)-2 x^2 e^{-x^2}-2 x^2+2 x^4-x^6 & \\ & =2 x^2\left(1-\frac{x^2}{1}+\frac{x^4}{2}-\frac{x^6}{3}+\ldots\right)-2 x^2+2 x^4-x^6 \\ & =-\frac{x^8}{3}+\frac{x^{10}}{12} \ldots \ldots \\ \Rightarrow P^{\prime}(x) & \leq 0 \\ \Rightarrow P(x) \text { is } & \\ \Rightarrow P(x) & \leq 0 \end{aligned} \end{aligned} $ option (D) is correct