Which of the following statements is TRUE ?

For each option we test the stated inequality on $$x \in (0,\tfrac12)$$ (for Options C, D) or on $$x\gt 0$$ (for Options A, B).

$$\psi_1(x)=e^{-x}+x,\qquad x\ge 0$$ Differentiate: $$\psi_1'(x)=-e^{-x}+1.$$ For every $$x\gt 0$$ we have $$e^{-x}\lt 1\;\Longrightarrow\;\psi_1'(x)&gt0,$$ so $$\psi_1$$ is strictly increasing on $$[0,\infty).$$ Hence $$\psi_1(x)&gt\psi_1(0)=1$$ for all $$x&gt0.$$ Therefore $$\psi_1(x)\le1$$ is FALSE ⇒ Option A is wrong.

$$\psi_2(x)=e^{2}-2x-2e^{-x}+2,\qquad x\ge0$$ At $$x=0: \psi_2(0)=e^{2}&gt0.$$ Thus the claimed inequality $$\psi_2(x)\le0$$ already fails at $$x=0^+$$, so Option B is wrong.

The integral defining $$f(x)$$ is even because the integrand is even: $$f(x)=\int_{-x}^{x}(|t|-t^{2})e^{-t^{2}}\mathrm dt =2\int_{0}^{x}(t-t^{2})e^{-t^{2}}\mathrm dt.$$ Split the two terms:

First part $$2\int_{0}^{x}t\,e^{-t^{2}}\mathrm dt =\Bigl[-e^{-t^{2}}\Bigr]_{0}^{x}=1-e^{-x^{2}}.$$

Second part $$2\int_{0}^{x}t^{2}e^{-t^{2}}\mathrm dt.$$ Hence $$f(x)=1-e^{-x^{2}}-2\int_{0}^{x}t^{2}e^{-t^{2}}\mathrm dt.$$

The proposed lower bound would require $$2\int_{0}^{x}t^{2}e^{-t^{2}}\mathrm dt\;\le\;\frac23x^{3}-\frac25x^{5}\qquad (0\lt x&lt\tfrac12).$$ For $$0\le t\le x&lt\tfrac12$$ we have $$e^{-t^{2}}\ge1-t^{2}$$ (because $$e^{-y}\ge1-y$$ for all real $$y$$). Thus $$t^{2}e^{-t^{2}}\ge t^{2}-t^{4},$$ and integrating gives $$\int_{0}^{x}t^{2}e^{-t^{2}}\mathrm dt\;\ge\;\int_{0}^{x}(t^{2}-t^{4})\mathrm dt =\frac13x^{3}-\frac15x^{5}.$$ Therefore $$2\int_{0}^{x}t^{2}e^{-t^{2}}\mathrm dt\;\ge\;\frac23x^{3}-\frac25x^{5},$$ the opposite of what we need. So Option C is wrong.

Write $$g(x)=\int_{0}^{x^{2}}t^{1/2}e^{-t}\mathrm dt,\qquad 0\lt x&lt\tfrac12.$$ For $$0\le t\le x^{2}\le\left(\tfrac12\right)^{2}=0.25$$ the Maclaurin expansion with alternating remainders gives the bound $$e^{-t}\le1-t+\frac{t^{2}}{2}.$$ Hence $$g(x)\le\int_{0}^{x^{2}}t^{1/2}\Bigl(1-t+\frac{t^{2}}{2}\Bigr)\mathrm dt =\int_{0}^{x^{2}}\Bigl(t^{1/2}-t^{3/2}+ \tfrac12 t^{5/2}\Bigr)\mathrm dt.$$

Integrating term-by-term:

$$\int_{0}^{x^{2}}t^{1/2}\mathrm dt =\frac23t^{3/2}\Big|_{0}^{x^{2}}=\frac23x^{3},$$ $$\int_{0}^{x^{2}}t^{3/2}\mathrm dt =\frac25t^{5/2}\Big|_{0}^{x^{2}}=\frac25x^{5},$$ $$\frac12\int_{0}^{x^{2}}t^{5/2}\mathrm dt=\frac12\cdot\frac27t^{7/2}\Big|_{0}^{x^{2}}=\frac17x^{7}.$$

Combining:

$$g(x)\le\frac23x^{3}-\frac25x^{5}+\frac17x^{7},\qquad 0\lt x&lt\tfrac12.$$

This is exactly the inequality stated in Option D, so Option D is TRUE.

Therefore, the only correct statement is:
Option D which is: $$g(x)\le\frac23x^{3}-\frac25x^{5}+\frac17x^{7}\;,\; \forall\,x\in\left(0,\tfrac12\right).$$