Analysis of the Options

We will analyze each statement to determine which one is true.

A: $f(\sqrt {\ln 3} ) + g(\sqrt {\ln 3} ) = {1 \over 3}$

1. Simplify f(x): The integrand in f(x) is $h(t) = (|t| - {t^2}){e^{ - {t^2}}}$. Let's check if h(t) is an even or odd function: $h( - t) = (| - t| - {( - t)^2}){e^{ - {{( - t)}^2}}} = (|t| - {t^2}){e^{ - {t^2}}} = h(t)$ Since h(t) is an even function, the integral over a symmetric interval becomes: $f(x) = \int_{ - x}^x h(t)dt = 2\int_0^x h(t)dt$ For $t \ge 0$, we have $|t| = t$. So, for $x > 0$: $f(x) = 2\int_0^x {(t - {t^2}){e^{ - {t^2}}}dt}$

2. Simplify g(x): $g(x) = \int_0^{{x^2}} {\sqrt t {e^{ - t}}dt}$ Let's use the substitution $t = u^2$. Then $dt = 2u\,du$. The limits of integration change as follows: When $t = 0$, $u = 0$. When $t = x^2$, $u = x$ (since $x > 0$). $g(x) = \int_0^x {\sqrt {{u^2}} {e^{ - {u^2}}}(2u\,du)} = \int_0^x u \cdot {e^{ - {u^2}}} \cdot 2u\,du = 2\int_0^x {{u^2}{e^{ - {u^2}}}du}$ Renaming the integration variable back to t, we get: $g(x) = 2\int_0^x {{t^2}{e^{ - {t^2}}}dt}$

3. Calculate f(x) + g(x): $f(x) + g(x) = 2\int_0^x {(t - {t^2}){e^{ - {t^2}}}dt} + 2\int_0^x {{t^2}{e^{ - {t^2}}}dt}$ $f(x) + g(x) = 2\int_0^x {((t - {t^2}) + {t^2}){e^{ - {t^2}}}dt} = 2\int_0^x {t{e^{ - {t^2}}}dt}$

4. Evaluate the integral: Let $u = t^2$, so $du = 2t\,dt$. $f(x) + g(x) = \int_0^{x^2} {{e^{ - u}}du} = \left[ { - {e^{ - u}}} \right]_0^{{x^2}} = ( - {e^{ - {x^2}}}) - ( - {e^0}) = 1 - {e^{ - {x^2}}}$

5. Substitute $x = \sqrt{\ln 3}$: $f(\sqrt {\ln 3} ) + g(\sqrt {\ln 3} ) = 1 - {e^{ - {{(\sqrt {\ln 3} )}^2}}} = 1 - {e^{ - \ln 3}} = 1 - {e^{\ln ({3^{ - 1}})}} = 1 - {1 \over 3} = {2 \over 3}$ The statement claims the value is $1/3$. Therefore, statement A is FALSE.

B: For every x > 1, there exists an $\alpha\in$(1, x) such that ${\psi _1}(x) = 1 + \alpha x$

1. Given ${\psi _1}(x) = {e^{ - x}} + x$. Note that ${\psi _1}(0) = {e^0} + 0 = 1$.
2. The given equation can be rewritten as ${\psi _1}(x) - 1 = \alpha x$, or $\alpha = \frac{{{\psi _1}(x) - 1}}{x} = \frac{{{\psi _1}(x) - {\psi _1}(0)}}{{x - 0}}$.
3. According to the Mean Value Theorem (MVT) applied to the function ${\psi _1}(t)$ on the interval $[0, x]$, there exists a $c \in (0, x)$ such that ${\psi _1}'(c) = \frac{{{\psi _1}(x) - {\psi _1}(0)}}{{x - 0}}$.
4. So, we must have $\alpha = {\psi _1}'(c)$. Let's find the derivative: ${\psi _1}'(t) = -e^{-t} + 1 = 1 - e^{-t}$.
5. Thus, $\alpha = 1 - e^{-c}$ for some $c \in (0, x)$.
6. Since $c > 0$, we have $e^{-c} < 1$, which means $1 - e^{-c} < 1$. So, $\alpha < 1$.
7. The statement requires $\alpha \in (1, x)$, which implies $\alpha > 1$. This is a contradiction. Therefore, statement B is FALSE.

C: For every x > 0, there exists a $\beta\in$(0, x) such that ${\psi _2}(x) = 2x({\psi _1}(\beta ) - 1)$

1. Given ${\psi _2}(x) = {x^2} - 2x - 2{e^{ - x}} + 2$. Let's check its value at $x=0$: ${\psi _2}(0) = 0 - 0 - 2e^0 + 2 = -2 + 2 = 0$.
2. The given equation can be rewritten as $\frac{{{\psi _2}(x)}}{x} = 2({\psi _1}(\beta ) - 1)$ for $x>0$.
3. Since ${\psi _2}(0) = 0$, we can write the left side as $\frac{{{\psi _2}(x) - {\psi _2}(0)}}{{x - 0}}$.
4. By the MVT applied to ${\psi _2}(t)$ on the interval $[0, x]$ (note that ${\psi _2}(t)$ is continuous and differentiable everywhere), there exists a $\beta \in (0, x)$ such that: ${\psi _2}'(\beta ) = \frac{{{\psi _2}(x) - {\psi _2}(0)}}{{x - 0}} = \frac{{{\psi _2}(x)}}{x}$.
5. Let's compute the derivative of ${\psi _2}(t)$: ${\psi _2}'(t) = \frac{d}{{dt}}({t^2} - 2t - 2{e^{ - t}} + 2) = 2t - 2 + 2{e^{ - t}} = 2(t - 1 + {e^{ - t}})$.
6. Now let's look at the expression ${\psi _1}(t) - 1$. We have ${\psi _1}(t) = {e^{ - t}} + t$, so ${\psi _1}(t) - 1 = {e^{ - t}} + t - 1$.
7. Comparing this with ${\psi _2}'(t)$, we see that ${\psi _2}'(t) = 2({\psi _1}(t) - 1)$.
8. Substituting this into the MVT result from step 4: $2({\psi _1}(\beta ) - 1) = \frac{{{\psi _2}(x)}}{x}$.
9. Multiplying by x, we get ${\psi _2}(x) = 2x({\psi _1}(\beta ) - 1)$ for some $\beta \in (0, x)$. This is exactly the given statement. Therefore, statement C is TRUE.

D: f is an increasing function on the interval $\left[ {0,{3 \over 2}} \right]$

1. To determine if f is increasing, we need to analyze the sign of its derivative, $f'(x)$. $f(x) = \int_{ - x}^x {(|t| - {t^2}){e^{ - {t^2}}}dt}$
2. Using the Leibniz rule for differentiation of an integral: $f'(x) = \left( {(|x| - {x^2}){e^{ - {x^2}}}} \right) \cdot \frac{d}{{dx}}(x) - \left( {(| - x| - {{( - x)}^2}){e^{ - {{( - x)}^2}}}} \right) \cdot \frac{d}{{dx}}( - x)$ $f'(x) = (|x| - {x^2}){e^{ - {x^2}}} - (|x| - {x^2}){e^{ - {x^2}}}( - 1)$ $f'(x) = 2(|x| - {x^2}){e^{ - {x^2}}}$
3. For the interval $[0, 3/2]$, we have $x \ge 0$, so $|x|=x$. $f'(x) = 2(x - {x^2}){e^{ - {x^2}}} = 2x(1 - x){e^{ - {x^2}}}$
4. The function f is increasing when $f'(x) \ge 0$. Since $2{e^{ - {x^2}}} > 0$ and $x \ge 0$ in the interval, the sign of $f'(x)$ is determined by the sign of $(1-x)$. $1 - x \ge 0 \implies x \le 1$.
5. So, $f'(x) \ge 0$ for $x \in [0, 1]$ and $f'(x) < 0$ for $x > 1$.
6. This means f is increasing on $[0, 1]$ and decreasing on $[1, \infty)$. The interval $[0, 3/2]$ includes the subinterval $(1, 3/2]$ where f is decreasing. Therefore, f is not an increasing function on the entire interval $[0, 3/2]$. Statement D is FALSE.

Conclusion: Only statement C is true.