Which of the following statements is TRUE ?

First write the two auxiliary functions in a convenient form.
$$\psi_1(x)=e^{-x}+x,\; x\ge 0$$
$$\psi_2(x)=x^{2}-2x-2e^{-x}+2,\; x\ge 0$$

Checking Option A
Put $$x=\sqrt{\ln 3}$$. For $$x\gt 0$$ we have $$f(x)=\int_{-x}^{x}(|t|-t^{2})e^{-t^{2}}\;dt =2\int_{0}^{x}(t-t^{2})e^{-t^{2}}\;dt$$ (because the integrand is even). Separate the two integrals:

$$I_1=\int_{0}^{x}t\,e^{-t^{2}}\;dt, \qquad I_2=\int_{0}^{x}t^{2}e^{-t^{2}}\;dt$$

1. For $$I_1$$, substitute $$u=t^{2}$$, $$du=2t\,dt$$: $$I_1=\tfrac12\int_{0}^{x^{2}}e^{-u}\;du=\tfrac12\left(1-e^{-x^{2}}\right).$$

2. To evaluate $$I_2$$ use integration by parts. Note that $$\dfrac{d}{dt}\!\left(te^{-t^{2}}\right)=e^{-t^{2}}-2t^{2}e^{-t^{2}}.$$ Hence $$t^{2}e^{-t^{2}}=\tfrac12e^{-t^{2}}-\tfrac12\dfrac{d}{dt}\! \left(te^{-t^{2}}\right).$$ Therefore $$2I_2=\int_{0}^{x}e^{-t^{2}}\;dt-xe^{-x^{2}}.$$ (All calculations are on $$[0,x]$$, so the lower limit contributes zero.)

Now $$f(x)=2(I_1-I_2)= \bigl(1-e^{-x^{2}}\bigr)-\Bigl(\int_{0}^{x}e^{-t^{2}}\;dt-xe^{-x^{2}}\Bigr) =1-e^{-x^{2}}-\int_{0}^{x}e^{-t^{2}}\;dt+xe^{-x^{2}}.$$

For $$g(x)=\int_{0}^{x^{2}}\!\sqrt{t}\,e^{-t}\;dt,$$ set $$t=u^{2}$$, $$dt=2u\,du$$, to get $$g(x)=\int_{0}^{x}2u^{2}e^{-u^{2}}\;du=2I_2 =\int_{0}^{x}e^{-t^{2}}\;dt-xe^{-x^{2}}.$$

Adding the two results, all integral terms cancel: $$f(x)+g(x)=1-e^{-x^{2}}.$$ Insert $$x=\sqrt{\ln 3}$$: $$f\!\bigl(\sqrt{\ln 3}\bigr)+g\!\bigl(\sqrt{\ln 3}\bigr) =1-e^{-\ln 3}=1-\dfrac13=\dfrac23\neq\dfrac13.$$ Hence Option A is false.

Checking Option B
For $$x\gt 1$$ define $$\alpha=\dfrac{\psi_1(x)-1}{x} =\dfrac{x+e^{-x}-1}{x}=1-\dfrac{1-e^{-x}}{x}.$$ Because $$1-e^{-x}\gt 0$$ for $$x\gt 0$$, we have $$\alpha\lt 1.$$ Thus $$\alpha$$ cannot belong to the interval $$(1,x)$$ when $$x\gt 1$$. Option B is false.

Checking Option C
First observe that $$\psi_1(t)-1=e^{-t}+t-1,$$ so it is continuous on every closed interval contained in $$[0,\infty).$$ Integrate from $$0$$ to $$x\gt 0$$:

$$\int_{0}^{x}2\bigl(\psi_1(t)-1\bigr)\;dt =2\int_{0}^{x}\!\bigl(e^{-t}+t-1\bigr)\;dt =\bigl[x^{2}-2x-2e^{-t}+2\bigr]_{0}^{x} =x^{2}-2x-2e^{-x}+2 =\psi_2(x).$$

All the hypotheses of the Mean Value Theorem for Integrals are satisfied; therefore there exists some $$\beta\in(0,x)$$ such that

$$\psi_2(x)=2x\bigl(\psi_1(\beta)-1\bigr).$$

This is exactly the statement in Option C, so Option C is true.

Checking Option D
Differentiate the earlier expression $$f(x)=2\int_{0}^{x}(t-t^{2})e^{-t^{2}}\;dt.$$ By the First Fundamental Theorem of Calculus, $$f'(x)=2(x-x^{2})e^{-x^{2}}=2x(1-x)e^{-x^{2}}.$$ On $$[0,1)$$ the derivative is positive; on $$(1,\infty)$$ it is negative. Hence $$f(x)$$ is increasing on $$[0,1]$$ and decreasing afterwards, so it is **not** increasing on the full interval $$\left[0,\frac32\right].$$ Option D is false.

Only Option C is correct.

Final answer: Option C which is: For every $$x\gt 0$$, there exists a $$\beta\in(0,x)$$ such that $$\psi_2(x)=2x\bigl(\psi_1(\beta)-1\bigr).$$