Question

Let $\Psi_{1}:[0, \infty) \rightarrow R, \Psi_{2}:[0, \infty) \rightarrow R, f:[0, \infty) \rightarrow R$ and $g:[0, \infty) \rightarrow R$ be functions such that $f (0)= g (0)=0$,
$\psi_{1}(x)=e^{-x}+x, x \geq 0, $
$\psi_{2}(x)=x^{2}-2 x-2 e^{-x}+2, x \geq 0,$
$f(x)=\int\limits_{-x}^{x}\left(|t|-t^{2}\right) e^{-t^{2}} d t, x > 0$
and $g(x)=\int\limits_{0}^{x^{2}} \sqrt{t} e^{-t} d t, x>0$.
Which of the following statements is TRUE?

• A. $f(\sqrt{\ln 3})+g(\sqrt{\ln 3})=\frac{1}{3}$
• B. For every $x > 1$, there exists and $\alpha \in(1, x)$ such that $\psi_{1}(x)=1+\alpha x$
• C. For every $x > 0$, there exists a $\beta \in(0, x)$ such that $\psi_{2}(x)=2 x\left(\psi_{1}(\beta)-1\right)$
• D. $f$ is an increasing function on the interval $\left[0, \frac{3}{2}\right]$

Answer 1

Answer: (C)

$\psi_{2}( x )= x ^{2}-2 x -2 e ^{- x }+2, x \geq 0$
$\psi_{2}^{\prime}( x )=2 x -2+2 e ^{- x }=2\left( x + e ^{- x }-1\right) $
$\psi_{2}^{\prime}(\beta)=2\left(\psi_{1}(\beta)-1\right)$
Since $\psi_{2}(x)$ is a continuous and differentiable function $\forall x \in[0, x]$
$\psi_{2}(0)=0, \psi_{2}(x)=x^{2}-2 x-2 e^{-x}+2$
Hence according to LMVT there exist at least one $\beta \in(0, x)$ such that
$\left(\frac{\psi_{2}( x )-\psi_{2}(0)}{ x }\right)=\psi_{2}^{\prime}(\beta)$
$=\frac{\psi_{2}( x )}{ x }=2\left(\psi_{1}(\beta)-1\right)$
$=\psi_{2}( x )=2 x \left(\psi_{1}(\beta)-1\right)$